Spectral Theory in Hilbert Space

Spectral Theory in Hilbert 

Space 

Lectures fall 2008 

Christer Bennewitz

Copyright c○ 1993–2008 by Christer Bennewitz

Preface 

The aim of these notes is to present a reasonably complete exposition 

of Hilbert space theory, up to and including the spectral theorem 

for the case of a (possibly unbounded) selfadjoint operator. As an application, 

eigenfunction expansions for regular and singular boundary 

value problems of ordinary differential equations are discussed. We 

first do this for the simplest Sturm-Liouville equation, and then, using 

very similar methods of proof, for a fairly general type of first order 

systems, which include so called Hamiltonian systems. 

Prerequisites are modest, but a good understanding of Lebesgue 

integration is assumed, including the concept of absolute continuity. 

Some previous exposure to linear algebra and basic functional analysis 

(uniform boundedness principle, closed graph theorem and maybe 

weak ∗ compactness of the unit ball in a (separable) Banach space) is 

expected from the reader, but in the two places where we could have 

used weak ∗ compactness, a direct proof has been given. The standard 

proofs of the Banach-Steinhaus and closed graph theorems are given 

in Appendix A. A brief exposition of the Riemann-Stieltjes integral, 

sufficient for our needs, is given in Appendix B. A few elementary facts 

about ordinary linear differential equations are used. These are proved 

in Appendix C. In addition, some facts from elementary analytic function 

theory are used. Apart from this the lectures are essentially selfcontained. 

Egevang, August 2008 

Christer Bennewitz 

i

Contents 

Preface i 

Chapter 0. Introduction 1 

Chapter 1. Linear spaces 5 

Exercises for Chapter 1 8 

Chapter 2. Spaces with scalar product 9 


Chapter 3. Hilbert space 15 


Chapter 4. Operators 23 


Chapter 5. Resolvents 31 


Chapter 6. Nevanlinna functions 35 

Chapter 7. The spectral theorem 39 


Chapter 8. Compactness 45 


Chapter 9. Extension theory 51 

1. Symmetric operators 51 

2. Symmetric relations 53 


Chapter 10. Boundary conditions 59 


Chapter 11. Sturm-Liouville equations 69 


Chapter 12. Inverse spectral theory 83 

1. Asymptotics of the m-function 84 

2. Uniqueness theorems 87 

Chapter 13. First order systems 91 

iii

iv CONTENTS 


Chapter 14. Eigenfunction expansions 101 


Chapter 15. Singular problems 105 


Appendix A. Functional analysis 117 

Appendix B. Stieltjes integrals 121 

Exercises for Appendix B 127 

Appendix C. Linear first order systems 129 

Appendix. Bibliography 133

CHAPTER 0 

Introduction 

Hilbert space is the most immediate generalization to the infinite 

dimensional case of finite dimensional Euclidean spaces (i.e., essentially 

R n for real, and C n for complex vector spaces). Probably its most important 

uses, and certainly its historical roots, are in spectral theory. 

Spectral theory for differential equations originates with the method of 

separation of variables, used to solve many of the equations of mathematical 

physics. This leads directly to the problem of expanding an 

‘arbitrary’ function in terms of eigenfunctions of the reduced equation, 

which is the central problem of spectral theory. A simple example 

is that of a vibrating string. The string is supposed to be stretched 

over an interval I ⊂ R, be fixed at the endpoints a, b and vibrate 

transversally (i.e., in a direction perpendicular to the interval I) in a 

plane containing I. The string can then be described by a real-valued 

function u(x, t) giving the location at time t of the point of the string 

which moves on the normal to I through the point x ∈ I. In appropriate 

units the function u will then (for sufficiently small vibrations, i.e., 

we are dealing with a linearization of a more accurate model) satisfy 

the following equation: 

(0.1) 

⎧ 

⎪⎨ 

∂ 

⎪⎩ 

2u ∂x2 = ∂2u ∂t2 (wave equation) 

u(a, t) = u(b, t) = 0 for t > 0 

u(x, 0) and ut(x, 0) given 

(boundary conditions) 

(initial conditions). 

The idea in separating variables is first to disregard the initial conditions 

and try to find solutions to the differential equation that satisfy 

the boundary condition and are standing waves, i.e., of the special form 

u(x, t) = f(x)g(t). The linearity of the equation implies that sums of 

solutions are also solutions (the superposition principle), so if we can 

find enough standing waves there is the possibility that any solution 

might be a superposition of standing waves. By substituting f(x)g(t) 

for u in (0.1) it follows that f ′′ (x)/f(x) = g ′′ (t)/g(t). Since the left 

hand side does not depend on t, and the right hand side not on x, both 

sides are in fact equal to a constant −λ. Since the general solution of 

the equation g ′′ (t) + λg(t) = 0 is a linear combination of sin( √ λ t) and 

1

2 0. INTRODUCTION 

cos( √ λ t), it follows that 

 

′′ 

−f = λf in I 

(0.2) 

f(a) = f(b) = 0, 

and that g(t) = A sin( √ λ t) + B cos( √ λ t) for some constants A and 

B. As is easily seen, (0.2) has non-trivial solutions only when λ is 

an element of the sequence {λj} ∞ 1 , where λj = ( π 

b−a j)2 . The numbers 

λ1, λ2, . . . are the eigenvalues of (0.2), and the corresponding solutions 

(non-trivial multiples of sin(j π (x − a))), are the eigenfunctions of 

b−a 

(0.2). The set of eigenvalues is called the spectrum of (0.2). In general, 

a superposition of standing waves is therefore of the form u(x, t) = 

 

(Aj sin( √ λj t) + Bj cos( √ λj t)) sin( λj (x − a)). If we assume that 

we may differentiate the sum term by term, the initial conditions of 

(0.1) therefore require that 

 

Bj sin( π 

b−aj(x − a)) and Aj π π j sin( j(x − a)) 

b−a b−a 

are given functions. The question of whether (0.1) has a solution which 

is a superposition of standing waves for arbitrary initial conditions, is 

then clearly seen to amount to the question whether an ‘arbitrary’ 

function may be written as a series uj, where each term is an eigenfunction 

of (0.2), i.e., a solution for λ equal to one of the eigenvalues. 

We shall eventually show this to be possible for much more general 

differential equations than (0.1). 

The technique above was used systematically by Fourier in his Theorie 

analytique de la Chaleur (1822) to solve problems of heat conduction, 

which in the simplest cases (like our example) lead to what are 

now called Fourier series expansions. Fourier was never able to give a 

satisfactory proof of the completeness of the eigenfunctions, i.e., the 

fact that essentially arbitrary functions can be expanded in Fourier 

series. This problem was solved by Dirichlet somewhat later, and at 

about the same time (1830) Sturm and Liouville independently but 

simultaneously showed weaker completeness results for more general 

ordinary differential equations of the form −(pu ′ ) ′ + qu = λu, with 

boundary conditions of the form Au + Bpu ′ = 0, to be satisfied at the 

endpoints of the given interval. Here p and q are given, sufficiently regular 

functions, and A, B given real constants, not both 0 and possibly 

different in the two interval endpoints. The Fourier cases correspond 

to p ≡ 1, q ≡ 0 and A or B equal to 0. 

For the Fourier equation, the distance between successive eigenvalues 

decreases as the length of the base interval increases, and as the 

base interval approaches the whole real line, the eigenvalues accumulate 

everywhere on the positive real line. The Fourier series is then 

replaced by a continuous superposition, i.e., an integral, and we get 

the classical Fourier transform. Thus a continuous spectrum appears,

0. INTRODUCTION 3 

and this is typical of problems where the basic domain is unbounded, 

or the coefficients of the equation have sufficiently bad singularities at 

the boundary. 

In 1910 Hermann Weyl [12] gave the first rigorous treatment, in the 

case of an equation of Sturm-Liouville type, of cases where continuous 

spectra can occur. Weyl’s treatment was based on the then recently 

proved spectral theorem by Hilbert. Hilbert’s theorem was a generalization 

of the usual diagonalization of a quadratic form, to the case of 

infinitely many variables. Hilbert applied it to certain integral operators, 

but it is not directly applicable to differential operators, since 

these are ‘unbounded’ in a sense we will discuss in Chapter 4. With 

the creation of quantum mechanics in the late 1920’s, these matters became 

of basic importance to physics, and mathematicians, who had not 

advanced much beyond the results of Weyl, took the matter up again. 

The outcome was the general spectral theorem, generally attributed to 

John von Neumann (1928), although essentially the same theorem had 

been proved by Torsten Carleman in 1923, in a less abstract setting. 

Von Neumann’s theorem is an abstract result, and detailed applications 

to differential operators of reasonable generality had to wait until 

the early 1950’s. In the meantime many independent results about 

expansions in eigenfunctions had been given, particularly for ordinary 

differential equations. 

In these lectures we will prove von Neumann’s theorem. We will 

then apply this theorem to differential equations, including those that 

give rise to the classical Fourier series and Fourier transform. Once one 

has a result about expansion in eigenfunctions a host of other questions 

appear, some of which we will discuss in these notes. Sample questions 

are: 

• How do eigenvalues and eigenfunctions depend on the domain 

I and on the form of the equation (its order, coefficients etc.)? 

A partial answer is given if one can calculate the asymptotic 

distribution of the eigenvalues, i.e., approximate the growth of 

λj as a function of j. For simple ordinary differential operators 

this can be done by fairly elementary means. The first such 

result for a partial differential equation was given by Weyl in 

1912, and his method was later improved and extended by 

Courant. 

• How well does the expansion converge when expanding different 

classes of functions? Again, for ordinary differential 

operators some questions of this type can be handled by elementary 

methods, but in general the answer lies in the explicit 

asymptotic behavior of the so called spectral projectors. The 

first such asymptotic result was given by Carleman in 1934, 

and his method has been the basis for most later results.

4 0. INTRODUCTION 

• Can the equation be reconstructed if the spectrum is known? If 

not, what else must one know? If different equations can have 

the same spectrum, how many different equations? What do 

they have in common? Questions like these are part of what is 

called inverse spectral theory. Really satisfactory answers have 

only been obtained for the equation −u ′′ + qu = λu, notably 

by Gelfand and Levitan in the early 1950:s. Pioneering work 

was done by Göran Borg in the 1940:s. 

• Another aspect of the first point is the following: Given a ‘base’ 

equation (corresponding to a ‘free particle’ in quantum mechanics) 

and another equation, which outside some bounded 

region is close to the base equation (an ‘obstacle’ has been introduced), 

how can one relate the eigenfunctions for the two 

equations? The main questions of so called scattering theory 

are of this type. 

• Related to the previous point is the problem of inverse scattering. 

Here one is given scattering data, i.e., the answer to 

the question in the previous point, and the question is whether 

the equation is determined by scattering data, whether there 

is a method for reconstructing the equation from the scattering 

data, and similar questions. Many questions of this kind 

are of great importance to applications.

CHAPTER 1 

Linear spaces 

This chapter is intended to be a quick review of the basic facts 

about linear spaces. In the definition below the set K can be any field, 

although usually only the fields R of real numbers and C of complex 

numbers are of interest. 

Definition 1.1. A linear space or vector space over K is a set L 

provided with an addition +, which to every pair of elements u, v ∈ L 

associates an element u + v ∈ L, and a multiplication, which to every 

λ ∈ K and u ∈ L associates an element λu ∈ L. The following rules 

for calculation hold: 

(1) (u + v) + w = u + (v + w) for all u, v and w in L.(associativity) 

(2) There is an element 0 ∈ L such that u + 0 = 0 + u = u for 

every u ∈ L. (existence of neutral element) 

(3) For every u ∈ L there exists v ∈ L such that u+v = v +u = 0. 

One denotes v by −u. (existence of additive inverse) 

(4) u + v = v + u for all u, v ∈ L. (commutativity) 

(5) λ(u + v) = λu + λv for all λ ∈ K and all u, v ∈ L. 

(6) (λ + µ)u = λu + µu for all λ, µ ∈ K and all u ∈ L. 

(7) λ(µu) = (λµ)u for all λ, µ ∈ K and all u ∈ L. 

(8) 1u = u for all u ∈ L. 

If K = R we have a real linear space, if K = C a complex linear 

space. Axioms 1–3 above say that L is a group under addition, axiom 

4 that the group is abelian (or commutative). Axioms 5 and 6 are 

distributive laws and axiom 7 an associative law related to the multiplication 

by scalars, whereas axiom 8 gives a kind of normalization for 

the multiplication by scalars. 

Note that by restricting oneself to multiplying only by real numbers, 

any complex space may also be viewed as a real linear space. 

Conversely, every real linear space can be ‘extended’ to a complex linear 

space (Exercise 1.1). We will therefore only consider complex linear 

spaces in the sequel. 

Let M be an arbitrary set and let C M be the set of complex-valued 

functions defined on M. Then C M , provided with the obvious definitions 

of the linear operations, is a complex linear space (Exercise 1.2). 

In the case when M = {1, 2, . . . , n} one writes C n instead of C {1,2,...,n} . 

An element u ∈ C n is of course given by the values u(1), u(2), . . . , u(n) 

5

6 1. LINEAR SPACES 

of u so one may also regard C n as the set of ordered n-tuples of complex 

numbers. The corresponding real space is the usual R n . 

If L is a linear space and V a subset of L which is itself a linear 

space, using the linear operations inherited from L, one says that V is 

a linear subspace of L. 

Proposition 1.2. A non-empty subset V of L is a linear subspace 

of L if and only if u + v ∈ V and λu ∈ V for all u, v ∈ V and λ ∈ C. 

The proof is left as an exercise (Exercise 1.3). If u1, u2, . . . , uk are 

elements of a linear space L we denote by [u1, u2, . . . , uk] the linear 

hull of u1, u2, . . . , uk, i.e., the set of all linear combinations λ1u1 +· · ·+ 

λkuk, where λ1, . . . , λk ∈ C. It is not hard to see that linear hulls are 

always subspaces (Exercise 1.5). One says that u1, . . . , uk generates 

L if L = [u1, . . . , uk], and any linear space which is the linear hull 

of a finite number of its elements is called finitely generated or finitedimensional. 

A linear space which is not finitely generated is called 

infinite-dimensional. It is clear that if, for example, u1 is a linear 

combination of u2, . . . , uk, then [u1, . . . , uk] = [u2, . . . , uk]. If none of 

u1, . . . , uk is a linear combination of the others one says that u1, . . . , uk 

are linearly independent. It is clear that any finitely generated space 

has a set of linearly independent generators; one simply starts with 

a set of generators and goes through them one by one, at each step 

discarding any generator which is a linear combination of those coming 

before it. A set of linearly independent generators for L is called a basis 

for L. A given finite-dimensional space L can of course be generated 

by many different bases. However, a fundamental fact is that all such 

bases of L have the same number of elements, called the dimension of 

L. This follows immediately from the following theorem. 

Theorem 1.3. Suppose u1, . . . , uk generate L, and that v1, . . . , vj 

are linearly independent elements of L. Then j ≤ k. 

Proof. Since u1, . . . , uk generate L we have v1 = k 

s=1 x1sus, 

for some coefficients x11, . . . , x1k which are not all 0 since v1 = 0. 

By renumbering u1, . . . , uk we may assume x11 = 0. Then u1 = 

1 

x11 v1 − k x1s 

s=2 x11 us, and therefore v1, u2, . . . , uk generate L. In particular, 

v2 = x21v1 + k s=2 x2sus for some coefficients x21, . . . , x2k. We 

can not have x22 = · · · = x2k = 0 since v1, v2 are linearly independent. 

By renumbering u2, . . . , uk, if necessary, we may assume x22 = 0. It 

follows as before that v1, v2, u3, . . . , uk generate L. We can continue in 

this way until we run out of either v:s (if j ≤ k) or u:s (if j > k). But 

if j > k we would get that v1, . . . , vk generate L, in particular that 

vj is a linear combination of v1, . . . , vk which contradicts the linear 

independence of the v:s. Hence j ≤ k. 

For a finite-dimensional space the existence and uniqueness of coordinates 

for any vector with respect to an arbitrary basis now follows

1. LINEAR SPACES 7 

easily (Exercise 1.6). More importantly for us, it is also clear that L 

is infinite dimensional if and only if every linearly independent subset 

of L can be extended to a linearly independent subset of L with arbitrarily 

many elements. This usually makes it quite easy to see that a 

given space is infinite dimensional (Exercise 1.7). 

If V and W are both linear subspaces of some larger linear space 

L, then the linear span [V, W ] of V and W is the set 

[V, W ] = {u | u = v + w where v ∈ V and w ∈ W }. 

This is obviously a linear subspace of L. If in addition V ∩ W = {0}, 

then for any u ∈ [V, W ] there are unique elements v ∈ V and w ∈ W 

such that u = v + w. In this case [V, W ] is called the direct sum of V 

and W and is denoted by V ˙+W . The proof of these facts is left as an 

exercise (Exercise 1.9). 

If V is a linear subspace of L we can create a new linear space 

L/V , the quotient space of L by V , in the following way. We say 

that two elements u and v of L are equivalent if u − v ∈ V . It is 

immediately seen that any u is equivalent to itself, that u is equivalent 

to v if v is equivalent to u, and that if u is equivalent to v, and v to 

w, then u is equivalent to w. It then easily follows that we may split 

L into equivalence classes such that every vector is equivalent to all 

vectors in the same equivalence class, but not to any other vectors. 

The equivalence class containing u is denoted by u + V , and then 

u + V = v + V precisely if u − v ∈ V . We now define L/V as the set of 

equivalence classes, where addition is defined by (u + V ) + (v + V ) = 

(u+v)+V and multiplication by scalar as λ(u+V ) = λu+V . It is easily 

seen that these operations are well defined and that L/V becomes a 

linear space with neutral element 0 + V (Exercise 1.10). One defines 

codim V = dim L/V . We end the chapter by a fundamental fact about 

quotient spaces. 

Theorem 1.4. dim V + codim V = dim L. 

We leave the proof for Exercise 1.11.

8 1. LINEAR SPACES 

Exercises for Chapter 1 

Exercise 1.1. Let L be a real linear space, and let V be the set 

of ordered pairs (u, v) of elements of L with addition defined componentwise. 

Show that V becomes a complex linear space if one defines 

(x + iy)(u, v) = (xu − yv, xv + yu) for real x, y. Also show that L can 

be ‘identified’ with the subset of elements of V of the form (u, 0), in 

the sense that there is a one-to-one correspondence between the two 

sets preserving the linear operations (for real scalars). 

Exercise 1.2. Let M be an arbitrary set and let C M be the set 

of complex-valued functions defined on M. Show that C M , provided 

with the obvious definitions of the linear operations, is a complex linear 

space. 

Exercise 1.3. Prove Proposition 1.2. 

Exercise 1.4. Let M be a non-empty subset of R n . Which of the 

following choices of L make it into a linear subspace of C M ? 

(1) L = {u ∈ C M | |u(x)| < 1 for all x ∈ M}. 

(2) L = C(M) = {u ∈ C M | u is continuous in M}. 

(3) L = {u ∈ C(M) | u is bounded on M}. 

(4) L = L(M) = {u ∈ C M | u is Lebesgue integrable over M}. 

Exercise 1.5. Let L be a linear space and uj ∈ L, j = 1, . . . , k. 

Show that [u1, u2, . . . , uk] is a linear subspace of L. 

Exercise 1.6. Show that if e1, . . . , en is a basis for L, then for 

each u ∈ L there are uniquely determined complex numbers x1, . . . , xn, 

called coordinates for u, such that u = x1e1 + · · · + xnen. 

Exercise 1.7. Verify that L is infinite dimensional if and only if 

every linearly independent subset of L can be extended to a linearly 

independent subset of L with arbitrarily many elements. Then show 

that u1, . . . , uk are linearly independent if and only if λ1u1+· · ·+λkuk = 

0 only for λ1 = · · · = λk = 0. Also show that C M is finite-dimensional 

if and only if the set M has finitely many elements. 

Exercise 1.8. Let M be an open subset of R n . Verify that L is 

infinite-dimensional for each of the choices of L in Exercise 1.4 which 

make L into a linear space. 

Exercise 1.9. Prove all statements in the penultimate paragraph 

of the chapter. 

Exercise 1.10. Prove that if L is a linear space and V a subspace, 

then L/V is a well defined linear space. 

Exercise 1.11. Prove Theorem 1.4.

CHAPTER 2 

Spaces with scalar product 

If one wants to do analysis in a linear space, some structure in addition 

to the linearity is needed. This is because one needs some way 

to define limits and continuity, and this requires an appropriate definition 

of what a neighborhood of a point is. Thus one must introduce a 

topology in the space. We will not deal with the general notion of topological 

vector space here, but only the following particularly convenient 

way to introduce a topology in a linear space. This also covers most 

cases of importance to analysis. A metric space is a set M provided 

with a metric, which is a function d : M × M → R such that for any 

x, y, z ∈ M the following holds. 

(1) d(x, y) ≥ 0 and = 0 if and only if x = y. (positive definite) 

(2) d(x, y) = d(y, x). (symmetric) 

(3) d(x, y) ≤ d(x, z) + d(z, y). (triangle inequality) 

A neighborhood of x ∈ M is then a subset O of M such that for some 

ε > 0 the set O contains all y ∈ M for which d(x, y) < ε. An open 

set is a set which is a neighborhood of all its points, and a closed set 

is one with an open complement. One says that a sequence x1, x2, . . . 

of elements in M converges to x ∈ M if d(xj, x) → 0 as j → ∞. 

The most convenient, but not the only important, way of introducing 

a metric in a linear space L is via a norm (Exercise 2.1). A norm 

on L is a function · : L → R such that for any u, v ∈ L and λ ∈ C 

(1) u ≥ 0 and = 0 if and only if u = 0. (positive definite) 

(2) λu = |λ|u. (positive homogeneous) 

(3) u + v ≤ u + v. (triangle inequality) 

The usual norm in the real space R 3 is of course obtained from the dot 

product (x1, x2, x3) · (y1, y2, y3) = x1y1 + x2y2 + x3y3 by setting x = 

√ x · x. For an infinite-dimensional linear space L, it is sometimes 

possible to define a norm similarly by setting u = 〈u, u〉, where 

〈·, ·〉 is a scalar product on L. A scalar product is a function L×L → C 

such that for all u, v and w in L and all λ, µ ∈ C holds 

(1) 〈λu + µv, w〉 = λ〈u, w〉 + µ〈v, w〉. (linearity in first argument) 

(2) 〈u, v〉 = 〈v, u〉. (Hermitian symmetry) 

(3) 〈u, u〉 ≥ 0 with equality only if u = 0. (positive definite) 

If instead of (3) holds only 

(3’) 〈u, u〉 ≥ 0, (positive semi-definite) 

9

10 2. SPACES WITH SCALAR PRODUCT 

one speaks about a semi-scalar product. Note that (2) implies that 

〈u, u〉 is real so that (3) makes sense. Also note that by combining (1) 

and (2) we have 〈w, λu + µv〉 = λ〈w, u〉 + µ〈w, v〉. One says that the 

scalar product is anti-linear in its second argument (Warning: In the 

so called Dirac formalism in quantum mechanics the scalar product is 

instead anti-linear in the first argument, linear in the second). Together 

with (1) this makes the scalar product into a sesqui-linear (=1 1 

2 -linear) 

form. In words: A scalar product is a Hermitian, sesqui-linear and 

positive definite form. We now assume that we have a scalar product 

on L and define u = 〈u, u〉 for any u ∈ L. To show that this 

definition makes · into a norm we need the following basic theorem. 

Theorem 2.1. (Cauchy-Schwarz) If 〈·, ·〉 is a semi-scalar product 

on L, then for all u, v ∈ L holds |〈u, v〉| 2 ≤ 〈u, u〉〈v, v〉. 

Proof. For arbitrary complex λ we have 0 ≤ 〈λu + v, λu + v〉 = 

|λ| 2 〈u, u〉 + λ〈u, v〉 + λ〈v, u〉 + 〈v, v〉. For λ = −r〈v, u〉 with real r 

we obtain 0 ≤ r 2 |〈u, v〉| 2 〈u, u〉 − 2r|〈u, v〉| 2 + 〈v, v〉. If 〈u, u〉 = 0 but 

〈u, v〉 = 0 this expression becomes negative for r > 1〈v, 

v〉|〈u, v〉|−2 

2 

which is a contradiction. Hence 〈u, u〉 = 0 implies that 〈u, v〉 = 0 so 

that the theorem is true in the case when 〈u, u〉 = 0. If 〈u, u〉 = 0 

we set r = 〈u, u〉 −1 and obtain, after multiplication by 〈u, u〉, that 

0 ≤ −|〈u, v〉| 2 + 〈u, u〉〈v, v〉 which proves the theorem. 

In the case of a scalar product, defining u = 〈u, u〉, we may 

write the Cauchy-Schwarz inequality as |〈u, v〉| ≤ uv. In this 

case it is also easy to see when there is equality in Cauchy-Schwarz’ 

inequality. To see that · is a norm on L the only non-trivial point 

is to verify that the triangle inequality holds; but this follows from 

Cauchy-Schwarz’ inequality (Exercise 2.4). 

Recall that in a finite dimensional space with scalar product it is 

particularly convenient to use an orthonormal basis since this makes 

it very easy to calculate the coordinates of any vector. In fact, if 

x1, . . . , xn are the coordinates of u in the orthonormal basis e1, . . . , en, 

then xj = 〈u, ej〉 (recall that e1, . . . , en is called orthonormal if all basis 

elements have norm 1 and 〈ej, ek〉 = 0 for j = k). Given an arbitrary 

basis it is easy to construct an orthonormal basis by use of the Gram- 

Schmidt method (see the proof of Lemma 2.2). 

In an infinite-dimensional space one can not find a (finite) basis. 

The best one can hope for are infinitely many vectors e1, e2, . . . such 

that each finite subset is linearly independent, and any vector is the 

limit in norm of a sequence of finite linear combinations of e1, e2, . . . . 

Again, it will turn out to be very convenient if e1, e2, . . . is an orthonormal 

sequence, i.e., ej = 1 for j = 1, 2, . . . and 〈ej, ek〉 = 0 for j = k. 

The following lemma is easily proved by use of the Gram-Schmidt procedure.

2. SPACES WITH SCALAR PRODUCT 11 

Lemma 2.2. Any infinite-dimensional linear space L with scalar 

product contains an orthonormal sequence. 

Proof. According to Chapter 1 we can find a linearly independent 

sequence in L, i.e., a sequence u1, u2, . . . such that u1, . . . , uk are linearly 

independent for any k. Put e1 = u1/u1 and v2 = u2 −〈u2, e1〉e1. 

Next put e2 = v2/v2. If we have already found e1, . . . , ek, put 

vk+1 = uk+1 − k 

j=1 〈uk+1, ej〉ej and ek+1 = vk+1/vk+1. I claim 

that this procedure will lead to a well defined orthonormal sequence 

e1, e2, . . . . This is left for the reader to verify (Exercise 2.6). 

Supposing we have an orthonormal sequence e1, e2, . . . in L a natural 

question is: How well can one approximate (in the norm of L) an 

arbitrary vector u ∈ L by finite linear combinations of e1, e2, . . . . Here 

is the answer: 

Lemma 2.3. Suppose e1, e2, . . . is an orthonormal sequence in L 

and put, for any u ∈ L, ûj = 〈u, ej〉. Then we have 

k 

(2.1) u − λjej 2 = u 2 k 

− |ûj| 2 k 

+ |λj − ûj| 2 

j=1 

for any complex numbers λ1, . . . , λk. 

The proof is by calculation (Exercise 2.7). The interpretation of 

Lemma 2.3 is very interesting. The identity (2.1) says that if we want 

to choose a linear combination k 

j=1 λjej of e1, . . . , ek which approximates 

u well in norm, the best choice of coefficients is to take λj = ûj, 

j = 1, . . . , k. Furthermore, with this choice, the error is given exactly 

by u − k 

j=1 ûjej 2 = u 2 − k 

j=1 |ûj| 2 . One calls the coefficients 

û1, û2, . . . the (generalized) Fourier coefficients of u with respect to the 

orthonormal sequence e1, e2, . . . . The following theorem is an immediate 

consequence of Lemma 2.3 (Exercise 2.8). 

Theorem 2.4 (Bessel’s inequality). For any u the series ∞ 2 

j=1 |ûj| 

converges and one has 

∞ 

j=1 

j=1 

|ûj| 2 ≤ u 2 . 

Another immediate consequence of Lemma 2.3 is the next theorem 

(cf. Exercise 2.9). 

Theorem 2.5 (Parseval’s formula). The series ∞ 

j=1 ûjej converges 

(in norm) to u if and only if ∞ 

j=1 |ûj| 2 = u 2 . 

There is also a slightly more general form of Parseval’s formula. 

Corollary 2.6. Suppose ∞ 

j=1 |ûj| 2 = u 2 for some u ∈ L. Then 

∞ 

j=1 ûjˆvj = 〈u, v〉 for any v ∈ L. 

j=1


Proof. Consider the following form on L. 

∞ 

[u, v] = 〈u, v〉 − ûjˆvj . 

Since |ûjˆvj| ≤ 1 

2 (|ûj| 2 + |ˆvj| 2 ) by the arithmetic-geometric inequality, 

Bessel’s inequality shows that the series is absolutely convergent. It 

follows that [·, ·] is a Hermitian, sesqui-linear form on L. Because 

of Bessel’s inequality it is also positive (but not positive definite). 

Thus [·, ·] is a semi-scalar product on L. Applying Cauchy-Schwarz’ 

inequality we obtain |[u, v]| 2 ≤ [u, u][v, v]. By assumption [u, u] = 

u 2 − ∞ 

j=1 |ûj| 2 = 0 so that the corollary follows. 

It is now obvious that the closest analogy to an orthonormal basis 

in an infinite-dimensional space with scalar product is an orthonormal 

sequence with the additional property of the following definition. 

Definition 2.7. An orthonormal sequence in L is called complete 

if the Parseval identity u 2 = ∞ 

1 |ûj| 2 holds for every u ∈ L. 

It is by no means clear that we can always find complete orthonormal 

sequences in a given space. This requires the space to be separable. 

Definition 2.8. A metric space M is called separable if it has a 

dense, countable subset. This means a sequence u1, u2, . . . of elements 

of M, such that for any u ∈ M, and any ε > 0, there is an element uj 

of the sequence for which d(u, uj) < ε. 

The vast majority of spaces used in analysis are separable (Exercise 

2.10), but there are exceptions (Exercise 2.12). 

Theorem 2.9. A infinite-dimensional linear space with scalar product 

is separable if and only if it contains a complete orthonormal sequence. 

The proof is left as an exercise (Exercise 2.11). Suppose e1, e2, . . . is 

a complete orthonormal sequence in L. We then know that any u ∈ L 

may be written as u = ∞ j=1 ûjej, where the series converges in norm. 

Furthermore the numerical series ∞ 

j=1 |ûj| 2 converges to u 2 . The 

following question now arises: Given a sequence λ1, λ2, . . . of complex 

numbers for which ∞ 

j=1 |λj| 2 converges, does there exist an element 

u ∈ L for which λ1, λ2, . . . are the Fourier coefficients? Equivalently, 

does ∞ 

j=1 λjej converge to an element u ∈ L in norm? As it turns 

out, this is not always the case. The property required of L is that it 

is complete. Warning: This is a totally different property from the 

completeness of orthonormal sequences we discussed earlier! To explain 

what it is, we need a few definitions. 

Definition 2.10. A Cauchy sequence in a metric space M is a 

sequence u1, u2, . . . of elements of M such that d(uj, uk) → 0 as j, k → 

j=1

EXERCISES FOR CHAPTER 2 13 

∞. More exactly: To every ε > 0 there exists a number ω such that 

d(uj, uk) < ε if j > ω and k > ω. 

It is clear by use of the triangle inequality that any convergent 

sequence is a Cauchy sequence. Far more interesting is the fact that 

this implication may sometimes be reversed. 

Definition 2.11. A metric space M is called complete if every 

Cauchy sequence converges to an element in M. 

A normed linear space which is complete is called a Banach space. 

If the norm derives from a scalar product, ∞ j=1 |λj| 2 converges and 

e1, e2, . . . is an orthonormal sequence we put uk = k j=1 λjej. If k < n 

we then have (the second equality is a special case of Lemma 2.3) 

un − uk 2 = 

n 

j=k+1 

λjej 2 = 

n 

|λj| 2 = 

j=k+1 

n 

|λj| 2 − 

j=1 

k 

|λj| 2 . 

Since ∞ 

j=1 |λj| 2 converges the right hand side → 0 as k, n → ∞. Hence 

u1, u2, . . . is a Cauchy sequence in L. It therefore follows that if L is 

complete, then ∞ 

j=1 λjej actually converges in norm to an element 

of L. On the other hand, if L is not complete and e1, e2, . . . is an 

orthonormal sequence, then λ1, λ2, . . . may be chosen so that the series 

∞ 

j=1 λjej does not converge in L although ∞ 

j=1 |λj| 2 is convergent 

(Exercise 2.14). 


Exercise 2.1. Show that if · is a norm on L, then d(x, y) = 

u − v is a metric on L. 

Exercise 2.2. Show that d(x, y) = arctan|x − y| is a metric on 

R which can be extended to a metric on the set of extended reals 

R = R ∪ {−∞} ∪ {∞}. 

Exercise 2.3. Consider the linear space C 1 [0, 1], consisting of 

complex-valued, differentiable functions with continuous derivative, defined 

in [0, 1]. Show that the following are all norms on C 1 [0, 1]. 

• u∞ = sup0≤x≤1|u(x)| , 

• u1 = 1 

|u| , 0 

• u1,∞ = u ′ ∞ + u∞. 

Invent some more norms in the same spirit! 

Exercise 2.4. Find all cases of equality in Cauchy-Schwarz’ inequality 

for a scalar product! Then show that ·, defined by u = 

〈u, u〉, where 〈·, ·〉 is a scalar product, is a norm. 

j=1


Exercise 2.5. Show that 〈u, v〉 = 1 

u(x)v(x) dx is a scalar prod- 

0 

uct on the space C[0, 1] of continuous, complex-valued functions defined 

on [0, 1]. 

Exercise 2.6. Finish the proof of Lemma 2.2. 

Exercise 2.7. Prove Lemma 2.3. 

Exercise 2.8. Prove Bessel’s inequality! 

Exercise 2.9. Prove Parseval’s formula! 

Exercise 2.10. It is well known that the set of step functions which 

are identically 0 outside a compact subinterval of an interval I are dense 

in L 2 (I). Use this to show that L 2 (I) is separable. 

Exercise 2.11. Prove Theorem 2.9. 

Hint: Use Gram-Schmidt! 

Exercise 2.12. Let L be the set of complex-valued functions u of 

the form u(x) = k j=1 λjeiαjx where α1, . . . , αk are (a finite number of) 

different real numbers and λ1, . . . , λk are complex numbers. Show that 

L is a linear subspace of C(R) (the functions continuous on the real 

uv serves as a scalar product. 

line) on which 〈u, v〉 = limT →∞ 1 

2T 

T 

−T 

Then show that the norm of e iαx is 1 for any α ∈ R and that e iαx is 

orthogonal to e iβx as soon as α = β. Conclude that L is not separable. 

Exercise 2.13. Show that as metric spaces the set Q of rational 

numbers is not complete but the set R of reals is. 

Exercise 2.14. Suppose L is a space with scalar product which is 

not complete, and that e1, e2, . . . is a complete orthonormal sequence 

in L. Show that there exists a sequence λ1, λ2, . . . of complex numbers, 

such that |λj| 2 < ∞ but λjej does not converge to any element 

of L.

CHAPTER 3 

Hilbert space 

A Hilbert space is a linear space H (we will as always assume that 

the scalars are complex numbers) provided with a scalar product such 

that the space is also complete, i.e., any Cauchy sequence (with respect 

to the norm induced by the scalar product) converges to an element 

of H. We denote the scalar product of u and v ∈ H by 〈u, v〉 and 

the norm of u by u = 〈u, u〉. It is usually required, and we will 

follow this convention, that the space be separable as well, i.e., there 

is a countable, dense subset. Recall that this means that any element 

can be arbitrarily well approximated in norm by elements of this dense 

subset. In the present case this means that H has a complete orthonormal 

sequence, and conversely, if the space has a complete orthonormal 

sequence it is separable (Theorem 2.9). As is usual we will also assume 

that H is infinite-dimensional. 

Example 3.1. The space ℓ 2 consists of all infinite sequences u = 

(u1, u2, . . . ) of complex numbers for which |uj| 2 < ∞, i.e., which 

are square summable. The scalar product of u with v = (v1, v2, . . . ) is 

defined as 〈u, v〉 = ujvj. This series is absolutely convergent since 

|ujvj| ≤ (|uj| 2 + |vj| 2 )/2 and u, v are square summable. Show that ℓ 2 

is a Hilbert space (Exercise 3.1)! 

The space Hilbert himself dealt with was ℓ 2 . Actually, any Hilbert 

space is isometrically isomorphic to ℓ 2 , i.e., there is a bijective (oneto-one 

and onto) linear map H ∋ u ↦→ û ∈ ℓ 2 such that 〈u, v〉 = 〈û, ˆv〉 

for any u and v in H (Exercise 3.2). This is the reason any complete, 

separable and infinite-dimensional space with scalar product is called 

a Hilbert space. However, there are infinitely many isomorphisms that 

will serve, and none of them is ‘natural’, i.e., in general to be preferred 

to any other, so the fact that all Hilbert spaces are isomorphic is not 

particularly useful in practice. 

Example 3.2. The most important example of a Hilbert space is 

L 2 (Ω, µ) where Ω is some domain in R n and µ is a (Radon) measure 

defined there; often µ is simply Lebesgue measure. The space consists 

of (equivalence classes of) complex-valued functions on Ω, measurable 

with respect to µ and with integrable square over Ω with respect to µ. 

That this space is separable and complete is proved in courses on the 

theory of integration. 

15

16 3. HILBERT SPACE 

Given a normed space one may of course ask whether there is a 

scalar product on the space which gives rise to the given norm in the 

usual way. Here is a simple criterion. 

Lemma 3.3. (parallelogram identity) If u and v are elements of H, 

then 

u + v 2 + u − v 2 = 2u 2 + 2v 2 . 

Proof. A simple calculation gives u±v 2 = 〈u±v, u±v〉 = u 2 ± 

(〈u, v〉 + 〈v, u〉) + v 2 . Adding this for the two signs the parallelogram 

identity follows. 

The name parallelogram identity comes from the fact that the 

lemma can be interpreted geometrically, as saying that the sum of the 

squares of the lengths of the sides in a parallelogram equals the sum 

of the squares of the lengths of the diagonals. This is a theorem that 

can be found in Euclid’s Elements. Given a normed space, Lemma 3.3 

shows that a necessary condition for the norm to be associated with 

a scalar product is that the parallelogram identity holds for all vectors 

in the space. It was proved by von Neumann in 1929 that this is 

also sufficient (Exercise 3.3). We shall soon have another use for the 

parallelogram identity. 

In practice it is quite common that one has a space with scalar product 

which is not complete (such a space is often called a pre-Hilbert 

space). In order to use Hilbert space theory, one must then embed 

the space in a larger space which is complete. The process is called 

completion and is fully analogous to the extension of the rational numbers 

to the reals, which is also done to make the Cauchy convergence 

principle valid. In very brief outline the process is as follows. Starting 

with a (not complete) normed linear space L let Lc be the set of all 

Cauchy sequences in L. The set Lc is made into a linear space in the 

obvious way. We may embed L in Lc by identifying u ∈ L with the 

sequence (u, u, u, . . . ). In Lc we may introduce a semi-norm · (i.e., 

a norm except that there may be non-zero elements u in the space for 

which u = 0) by setting (u1, u2, . . . ) = limuj. Now let Nc be the 

subspace of Lc consisting of all elements with semi-norm 0, and put 

H = Lc/Nc, i.e., elements in Lc are identified whenever the distance 

between them is 0. One may now prove that · induces a norm on H 

under which H is complete, and that through the identification above 

we may consider the original space L as a dense subset of H. If the 

original norm came from a scalar product, then so will the norm of H. 

We leave to the reader to verify the details, using the hints provided 

(Exercise 3.4). 

The process above is satisfactory in that it shows that any normed 

space may be ‘completed’ (in fact, the same process works in any metric 

space). Equivalence classes of Cauchy sequences are of course rather abstract 

objects, but in concrete cases one can often identify the elements

3. HILBERT SPACE 17 

of the completion of a given space with more concrete objects. So, for 

example, one may view L 2 (Ω, µ) as the completion, in the appropriate 

norm, of the linear space C0(Ω) of functions which are continuous in Ω 

and 0 outside a compact subset of Ω. 

In the sequel H is always assumed to be a Hilbert space. There are 

two properties which make Hilbert spaces far more convenient to deal 

with than more general spaces. The first is that any closed, linear subspace 

has a topological complement which can be chosen in a canonical 

way (Theorem 3.7). The second, a Hilbert space can be identified with 

its topological dual (Theorem 3.8). Both these properties are actually 

true even if the space is not assumed separable (and of course if the 

space is finite-dimensional), as our proofs will show. To prove them we 

start with the following definition. 

Definition 3.4. A set M is called convex if it contains all linesegments 

connecting two elements of the set, i.e., if u and v ∈ M, then 

tu + (1 − t)v ∈ M for all t ∈ [0, 1]. 

A subset of a metric space is of course called closed if all limits 

of convergent sequences contained in the subset are themselves in the 

subset. It is easily seen that this is equivalent to the complement of 

the subset being open, in the sense that it is a neighborhood of all its 

points (check this!). 

Lemma 3.5. Any closed, convex subset K of H has a unique element 

of smallest norm. 

Proof. Put d = inf{u | u ∈ K}. Let u1, u2, . . . be a minimizing 

sequence, i.e., uj ∈ K and uj → d. By the parallelogram identity 

we then have 

uj − uk 2 = 2uj 2 + 2uk 2 − 4(uj + uk)/2 2 . 

On the right hand side the two first terms both tend to 2d 2 as j, k → ∞. 

By convexity (uj + uk)/2 ∈ K so the last term is ≥ 4d 2 . Therefore 

u1, u2, . . . is a Cauchy sequence, and has a limit u which obviously has 

norm d and is in K, since K is closed. If u and v are both minimizing 

elements, replacing uj by u and uk by v in the calculation above immediately 

shows that u = v, so the minimizing element is unique. 

Lemma 3.6. Suppose M is a proper ( i.e., M = H) closed, linear 

subspace of H. Then there is a non-trivial normal to M, i.e., an 

element u = 0 in H such that 〈u, v〉 = 0 for all v ∈ M. 

Proof. Let w /∈ M and put K = w + M. Then K is obviously 

closed and convex so it has a smallest element u which is non-zero 

since 0 /∈ K. Let v = 0 be in M so that u + av ∈ K for any scalar 

a. Hence u 2 ≤ u + av 2 = u 2 + 2 Re(a〈v, u〉) + |a| 2 v 2 . Setting 

a = −〈u, v〉/v 2 we obtain −(|〈u, v〉|/v) 2 ≥ 0 so that 〈u, v〉 = 0.


Two subspaces M and N are said to be orthogonal if every element 

in M is orthogonal to every element in N. Then clearly M ∩ N = {0} 

so the direct sum of M and N is defined. In the case at hand this is 

called the orthogonal sum of M and N and denoted by M ⊕ N. Thus 

M ⊕ N is the set of all sums u + v with u ∈ M and v ∈ N. If M and 

N are closed, orthogonal subspaces of H, then their orthogonal sum is 

also a closed subspace of H (Exercise 3.5). If A is an arbitrary subset 

of H we define 

A ⊥ = {u ∈ H | 〈u, v〉 = 0 for all v ∈ A}. 

This is called the orthogonal complement of A. It is easy to see that 

A ⊥ is a closed linear subspace of H, that A ⊂ B implies B ⊥ ⊂ A ⊥ and 

that A ⊂ (A ⊥ ) ⊥ (Exercise 3.6). 

When M is a linear subspace of H an alternative way of writing 

M ⊥ is H ⊖ M. This makes sense because of the following theorem of 

central importance. 

Theorem 3.7. Suppose M is a closed linear subspace of H. Then 

M ⊕ M ⊥ = H. 

Proof. M ⊕ M ⊥ is a closed linear subspace of H so if it is not 

all of H, then it has a non-trivial normal u by Lemma 3.6. But if u is 

orthogonal to both M and M ⊥ , then u ∈ M ⊥ ∩ (M ⊥ ) ⊥ which shows 

that u cannot be = 0. The theorem follows. 

A nearly obvious consequence of Theorem 3.7 is that M ⊥⊥ = M 

for any closed linear subspace M of H (Exercise 3.7). 

A linear form ℓ on H is complex-valued linear function on H. Naturally 

ℓ is said to be continuous if ℓ(uj) → ℓ(u) whenever uj → u. The 

set of continuous linear forms on a Banach space B (or a more general 

topological vector space) is made into a linear space in an obvious way. 

This space is called the dual of B, and is denoted by B ′ . A continuous 

linear form on a Banach space B has to be bounded in the sense that 

there is a constant C such that |ℓ(u)| ≤ Cu for any u ∈ B. For 

suppose not. Then there exists a sequence of elements u1, u2, . . . of 

B for which |ℓ(uj)|/uj → ∞. Setting vj = uj/ℓ(uj) we then have 

vj → 0 but |ℓ(vj)| = 1 → 0, so ℓ can not be continuous. Conversely, if 

ℓ is bounded by C then |ℓ(uj) − ℓ(u)| = |ℓ(uj − u)| ≤ Cuj − u → 0 if 

uj → u, so a bounded linear form is continuous. The smallest possible 

bound of a linear form ℓ is called the norm of ℓ, denoted ℓ. 

It is easy to see that provided with this norm B ′ is complete, so the 

dual of a Banach space is a Banach space (Exercise 3.8). A familiar 

example is given by the space L p (Ω, µ) for 1 ≤ p < ∞, where Ω is 

a domain in R n and µ a Radon measure defined in Ω. The dual of 

this space is Lq (Ω, µ), where q is the conjugate exponent to p, in the 

sense that 1 1 + = 1. A simple example of a bounded linear form on 

p q 

a Hilbert space H is ℓ(u) = 〈u, v〉, where v is some fixed element of

3. HILBERT SPACE 19 

H. By Cauchy-Schwarz’ inequality |ℓ(u)| ≤ vu so ℓ ≤ v. But 

ℓ(v) = v 2 so actually ℓ = v. The following theorem, which has 

far-reaching consequences for many applications to analysis, says that 

this is the only kind of bounded linear form there is on a Hilbert space. 

In other words, the theorem allows us to identify the dual of a Hilbert 

space with the space itself. 

Theorem 3.8 (Riesz’ representation theorem). For any bounded 

linear form ℓ on H there is a unique element v ∈ H such that ℓ(u) = 

〈u, v〉 for all u ∈ H. The norm of ℓ is then ℓ = v. 

Proof. The uniqueness of v is clear, since the difference of two 

possible choices of v must be orthogonal to all of H (for example to 

itself). If ℓ(u) = 0 for all u then we may take v = 0. Otherwise we set 

M = {u ∈ H | ℓ(u) = 0} which is obviously linear because ℓ is, and 

closed since ℓ is continuous. Since M is not all of H it has a normal w = 

0 by Lemma 3.6, and we may assume w = 1. If now u is arbitrary in 

H we put u1 = u − (ℓ(u)/ℓ(w))w so that ℓ(u1) = ℓ(u) − ℓ(u) = 0, i.e., 

u1 ∈ M so 〈u1, w〉 = 0. Hence 〈u, w〉 = (ℓ(u)/ℓ(w))〈w, w〉 = ℓ(u)/ℓ(w) 

so ℓ(u) = 〈u, v〉 where v = ℓ(w)w. We have already proved that ℓ = 

v. 

So far we have tacitly assumed that convergence in a Hilbert space 

means convergence in norm, i.e., uj → u means uj − u → 0. This 

is called strong convergence; one writes s-lim uj = u or uj → u. There 

is also another notion of convergence which is very important. By definition 

uj tends to u weakly, in symbols w-lim uj = u or uj ⇀ u, if 

〈uj, v〉 → 〈u, v〉 for every v ∈ H. It is obvious that strong convergence 

implies weak convergence to the same limit (the scalar product is continuous 

in its arguments by Cauchy-Schwarz), but the converse is not 

true (Exercise 3.9). We have the following important theorem. 

Theorem 3.9. Every bounded sequence in H has a weakly convergent 

subsequence. Conversely, every weakly convergent sequence is 

bounded. 

Proof. The first claim is a consequence of the weak ∗ compactness 

of the unit ball of the dual of a Banach space. Since we do not 

want to assume knowledge of this, we will give a direct proof. To this 

end, suppose v1, v2, . . . is the given sequence, bounded by C, and let 

e1, e2, . . . be a complete orthonormal sequence in H. The numerical 

sequence {〈vj, e1〉} ∞ j=1 is then bounded and so has a convergent subsequence, 

corresponding to a subsequence {v1j} ∞ j=1 of the v:s, by the 

Bolzano-Weierstrass theorem. The numerical sequence {〈v1j, e2〉} ∞ j=1 

is again bounded, so it has a convergent subsequence, corresponding 

to a subsequence {v2j} ∞ j=1 of {v1j} ∞ j=1. Proceeding in this manner we 

get a sequence of sequences {vkj} ∞ j=1, k = 1, 2, . . . , each element of 

which is a subsequence of those preceding it, and with the property that


ˆvn = limj→∞〈vnj, en〉 exists. I claim that {vjj} ∞ j=1 converges weakly to 

v = ˆvnen. Note that {〈vjj, en〉} ∞ j=1 converges to ˆvn since it is a subsequence 

of {〈vnj, en〉} ∞ j=1 from j = n on. Furthermore N 

n=1 |ˆvn| 2 ≤ C 2 

for all N since it is the limit as j → ∞ of N 

n=1 |〈vNj, en〉| 2 which 

by Bessel’s inequality is bounded by vNj 2 ≤ C 2 . It follows that 

∞ 

n=1 |ˆvn| 2 ≤ C 2 so that v is actually an element of H. 

To show the weak convergence, let u = ûnen be arbitrary in 

H. Suppose ε > 0 given arbitrarily. Writing u = u ′ + u ′′ where 

u ′ = N 

n=1 ûnen we may now choose N so large that u ′′ < ε so that 

|〈vjj, u ′′ 〉| < Cε. Furthermore |〈v, u ′′ 〉| < Cε and 〈vjj, u ′ 〉 → 〈v, u ′ 〉 

so limj→∞|〈vjj, u〉 − 〈v, u〉| ≤ 2Cε. Since ε > 0 is arbitrary the weak 

convergence follows. 

The converse is an immediate consequence of the Banach-Steinhaus 

principle of uniform boundedness. 

Theorem 3.10 (Banach-Steinhaus). Let ℓ1, ℓ2, . . . be a sequence of 

bounded linear forms on a Banach space B which is pointwise bounded, 

i.e., such that for each u ∈ B the sequence ℓ1(u), ℓ2(u), . . . is bounded. 

Then ℓ1, ℓ2, . . . is uniformly bounded, i.e., there is a constant C such 

that |ℓj(u)| ≤ Cu for every u ∈ B and j = 1, 2, . . . . 

Assuming Theorem 3.10 (for a proof, see Appendix A), we can 

complete the proof of Theorem 3.9, since a weakly convergent sequence 

v1, v2, . . . can be identified with a sequence of linear forms ℓ1, ℓ2, . . . 

by setting ℓj(u) = 〈u, vj〉. Since a convergent sequence of numbers 

is bounded it follows that we have a pointwise bounded sequence of 

linear functionals. By Theorem 3.10 there is a constant C such that 

|〈u, vj〉| ≤ Cu for every u ∈ H and j = 1, 2, . . . . In particular, 

setting u = vj gives vj ≤ C for every j.



Exercise 3.1. Prove the completeness of ℓ 2 ! 

Hint: Given a Cauchy sequence show first that each coordinate converges. 

Exercise 3.2. Prove that any Hilbert space is isometrically isomorphic 

to ℓ 2 , i.e., there is a bijective (one-to-one and onto) linear 

map H ∋ u ↦→ û ∈ ℓ 2 such that 〈u, v〉 = 〈û, ˆv〉 for any u and v in H. 

Exercise 3.3. Suppose L is a linear space with norm · which 

satisfies the parallelogram identity for all u, v ∈ L. Show that 〈u, v〉 = 

1 

4 

3 

k=0 ik u + i k v 2 is a scalar product on L. 

Hint: Show first that 〈u, u〉 = u 2 , that 〈v, u〉 = 〈u, v〉 and that 

〈iu, v〉 = i〈u, v〉. Then show that 〈u + v, w〉 − 〈u, w〉 − 〈v, w〉 = 0 and 

from that 〈λu, v〉 = λ〈u, v〉 for any rational number λ. Finally use 

continuity. 

Exercise 3.4. Show that the semi-norm on the space Lc defined 

in the text is well-defined, i.e., that the limit limuj exists for any 

element (u1, u2, . . . ) ∈ Lc. Then verify that H = Lc/Nc can be given a 

norm under which it is complete, that L may be viewed as isometrically 

and densely embedded in H, and that H is a Euclidean space (a space 

with scalar product) if L is. 

Exercise 3.5. Show that if M and N are closed, orthogonal subspaces 

of H, then also M ⊕ N is closed. 

Exercise 3.6. Show that is A ⊂ H, then A ⊥ is a closed linear 

subspace of H, that A ⊂ B implies B ⊥ ⊂ A ⊥ and that A ⊂ (A ⊥ ) ⊥ . 

Exercise 3.7. Verify that M ⊥⊥ = M for any closed linear subspace 

M of H, and also that for an arbitrary set A ⊂ H the smallest closed 

linear subspace containing A is A ⊥⊥ . 

Exercise 3.8. Show that a bounded linear form on a Banach space 

B has a least bound, which is a norm on B ′ , and that B ′ is complete 

under this norm. 

Exercise 3.9. Show that an orthonormal sequence does not converge 

strongly to anything but tends weakly to 0. Conclude that if in a 

Euclidean space every weakly convergent sequence is convergent, then 

the space is finite-dimensional. 

Hint: Show that the distance between two arbitrary elements in the 

sequence is √ 2 and use Bessel’s inequality to show weak convergence 

to 0.

CHAPTER 4 

Operators 

A bounded linear operator from a Banach space B1 to another Banach 

space B2 is a linear mapping T : B1 → B2 such that for some 

constant C we have T u2 ≤ Cu1 for every u ∈ B1. The smallest 

such constant C is called the norm of the operator T and denoted by 

T . Like in the discussion of linear forms in the last chapter it follows 

that the boundedness of T is equivalent to continuity, in the sense that 

T uj − T u2 → 0 if uj − u1 → 0 (Exercise 4.1). If B1 = B2 = B one 

says that T is an operator on B. The operator-norm defined above has 

the following properties (Here T : B1 → B2 and S are bounded linear 

operators, and B1, B2 and B3 Banach spaces). 

(1) T ≥ 0, equality only if T = 0, 

(2) λT = |λ|T for any λ ∈ C, 

(3) S + T ≤ S + T if S : B1 → B2, 

(4) ST ≤ ST if S : B2 → B3. 

We leave the proof to the reader (Exercise 4.1). Thus we have made the 

set of bounded operators from B1 to B2 into a normed space B(B1, B2). 

In fact, B(B1, B2) is a Banach space (Exercise 4.2). We write B(B) 

for the bounded operators on B. Because of the property (4) B(B) is 

called a Banach algebra. 

Now let H1 and H2 be Hilbert spaces. Then every bounded operator 

T : H1 → H2 has an adjoint 1 T ∗ : H2 → H1 defined as follows. Consider 

a fixed element v ∈ H2 and the linear form H1 ∋ u ↦→ 〈T u, v〉2 

which is obviously bounded by T v2. By the Riesz’ representation 

theorem there is therefore a unique element v ∗ ∈ H1, such that 

〈T u, v〉2 = 〈u, v ∗ 〉1. By the uniqueness, and since 〈T u, v〉2 depends 

anti-linearly on v, it follows that T ∗ : v ↦→ v ∗ is a linear operator from 

H2 to H1. It is also bounded, since v ∗ 2 1 = 〈T v ∗ , v〉2 ≤ T v ∗ 1v2, 

so that T ∗ ≤ T . The adjoint has the following properties. 

Proposition 4.1. The adjoint operation B(H1, H2) ∋ T ↦→ T ∗ ∈ 

B(H2, H1) has the properties: 

(1) (T1 + T2) ∗ = T ∗ 1 + T ∗ 2 , 

(2) (λT ) ∗ = λT ∗ for any complex number λ, 

(3) (T2T1) ∗ = T ∗ 1 T ∗ 2 if T2 : H2 → H3, 

(4) T ∗∗ = T , 

1 Also operators between general Banach spaces, or even more general topological 

vector spaces, have adjoints, but they will not concern us here. 

23

24 4. OPERATORS 

(5) T ∗ = T , 

(6) T ∗ T = T 2 . 

Proof. The first four properties are very easy to show and are 

left as exercises for the reader. To prove (5), note that we already 

have shown that T ∗ ≤ T and combining this with (4) gives the 

opposite inequality. Use of (5) shows that T ∗ T ≤ T ∗ T = 

T 2 and the opposite inequality follows from T u 2 2 = 〈T ∗ T u, u〉1 ≤ 

T ∗ T u1u1 ≤ T ∗ T u 2 1 so (6) follows. The reader is asked to fill 

in the details missing in the proof (Exercise 4.3). 

If H1 = H2 = H3 = H, then the properties (1)–(4) above are the 

properties required for the star operation to be called an involution 

on the algebra B(H), and a Banach algebra with an involution, also 

satisfying (5) and (6), is called a B ∗ algebra. There are no less than 

three different useful notions of convergence for operators in B(H1, H2). 

We say that Tj tends to T 

• uniformly if Tj − T → 0, denoted by Tj ⇒ T , 

• strongly if Tju−T u2 → 0 for every u ∈ H1, denoted Tj → T , 

and 

• weakly if 〈Tju, v〉2 → 〈T u, v〉2 for all u ∈ H1 and v ∈ H2, 

denoted Tj ⇀ T . 

It is clear that uniform convergence implies strong convergence and 

strong convergence implies weak convergence, and it is also easy to see 

that neither of these implications can be reversed. 

Of particular interest are so called projection operators. A projection 

P on H is an operator in B(H) for which P 2 = P . If P is 

a projection then so is I − P , where I is the identity on H, since 

(I − P )(I − P ) = I − P − P + P 2 = I − P . Setting M = P H and 

N = (I − P )H it follows that M is the null-space of I − P since M 

clearly consist of those elements u ∈ H for which P u = u. Similarly 

N is the null-space of P . Since P and I − P are bounded (i.e., continuous) 

it therefore follows that M and N are closed. It also follows 

that M ∩ N = {0} and the direct sum M ˙+N of M and N is H (this 

means that any element of H can be written uniquely as u + v with 

u ∈ M and v ∈ N). Conversely, if M and N are linear subspaces of H, 

M ∩ N = {0} and M ˙+N = H, then we may define a linear map P satisfying 

P 2 = P by setting P w = u if w = u +v with u ∈ M and v ∈ N. 

As we have seen P can not be bounded unless M and N are closed. 

There is a converse to this: If M and N are closed, then P is bounded. 

This follows immediately from the closed graph theorem (Exercise 4.4). 

In the case when the projection P , and thus also I − P , is bounded, 

the direct sum M ∔ N is called topological. If M and N happen to be 

orthogonal subspaces P is called an orthogonal projection. Obviously 

N = M ⊥ then, since the direct sum of M and N is all of H. We have 

the following characterization of orthogonal projections.

4. OPERATORS 25 

Proposition 4.2. A projection P is orthogonal if and only if it 

satisfies P ∗ = P . 

Proof. If P ∗ = P and u ∈ M, v ∈ N, then 〈u, v〉 = 〈P u, v〉 = 

〈u, P ∗ v〉 = 〈u, P v〉 = 〈u, 0〉 = 0 so M and N are orthogonal. Conversely, 

suppose M and N orthogonal. For arbitrary u, v ∈ H we 

then have 〈P u, v〉 = 〈P u, P v〉 + 〈P u, (I − P )v〉 = 〈P u, P v〉 so that 

also 〈u, P v〉 = 〈P u, P v〉. Hence 〈P u, v〉 = 〈u, P v〉 holds generally, i.e., 

P ∗ = P . 

An operator T for which T ∗ = T is called selfadjoint. Hence an 

orthogonal projection is the same as a selfadjoint projection. We will 

have much more to say about selfadjoint operators in a more general 

context later. Another class of operators of great interest are the unitary 

operators. This is an operator U : H1 → H2 for which U ∗ = U −1 . 

Since 〈Uu, Uv〉2 = 〈U ∗ Uu, v〉1 = 〈u, v〉1 the operator U preserves the 

scalar product; such an operator is called isometric. If U is isometric 

we have 〈u, v〉1 = 〈Uu, Uv〉2 = 〈U ∗ Uu, v〉1, so that U ∗ is a left 

inverse of U for any isometric operator. If dim H1 = dim H2 < ∞, 

then a left inverse of a linear operator is also a right inverse, so in 

this case isometric and unitary (orthogonal in the case of a real space) 

are the same thing. If dim H1 = dim H2 or both spaces are infinitedimensional, 

however, this is not the case. For example, in the space 

ℓ 2 we may define U(x1, x2, . . . ) = (0, x1, x2, . . . ), which is obviously 

isometric (this is a so called shift operator), but the vector (1, 0, 0, . . . ) 

is not the image of anything, so the operator is not unitary. Its adjoint 

is U ∗ (x1, x2, . . . ) = (x2, x3, . . . ), which is only a partial isometry, 

namely an isometry on the vectors orthogonal to (1, 0, 0, . . . ). See also 

Exercise 4.8. 

It is never possible to interpret a differential operator as a bounded 

operator on some Hilbert space of functions. We therefore need to 

discuss unbounded operators as well. Similarly, we will need to discuss 

operators that are not defined on all of H. Thus we now consider a 

linear operator T : D(T ) → H2, where the domain D(T ) of T is some 

linear subset of H1. T is not supposed bounded. Another such operator 

S is said to be an extension of T if D(T ) ⊂ D(S) and Su = T u for 

every u ∈ D(T ). We then write T ⊂ S. We must discuss the concept 

of adjoint. The form u ↦→ 〈T u, v〉2 is, for fixed v ∈ H2, only defined for 

u ∈ D(T ), and though linear not necessarily bounded, so there may not 

be any v ∗ ∈ H1 such that 〈T u, v〉2 = 〈u, v ∗ 〉1 for all u ∈ D(T ). Even 

if there is, it may not be uniquely determined, since if w ∈ D(T ) ⊥ we 

could replace v ∗ by v ∗ + w with no change in 〈u, v ∗ 〉. We therefore 

make the basic assumption that D(T ) ⊥ = {0}, i.e., D(T ) is dense in 

H1. T is then said to be densely defined 2 . In this case v ∗ ∈ H1 is 

ter 9. 

2 We will discuss the case of an operator which is not densely defined in Chap


clearly uniquely determined by v ∈ H2, if it exists. It is also obvious 

that v ∗ depends linearly on v, so we define D(T ∗ ) to be those v ∈ H2 

for which we can find a v ∗ ∈ H1, and set T ∗ v = v ∗ . There is no reason 

to expect the adjoint T ∗ to be densely defined. In fact, we may have 

D(T ∗ ) = {0}, so T ∗ may not itself have an adjoint. To understand this 

rather confusing situation it turns out to be useful to consider graphs 

of operators. 

The graph of T is the set GT = {(u, T u) | u ∈ D(T )}. This set is 

clearly linear and may be considered a linear subset of the orthogonal 

direct sum H1 ⊕ H2, consisting of all pairs (u1, u2) with u1 ∈ H1 and 

u2 ∈ H2 with the natural linear operations and provided with the scalar 

product 〈(u1, u2), (v1, v2)〉 = 〈u1, v1〉1 + 〈u2, v2〉2. This makes H1 ⊕ H2 

into a Hilbert space (Exercise 4.6). 

We now define the boundary operator U : H1 ⊕ H2 → H2 ⊕ H1 by 

U(u1, u2) = (−iu2, iu1) (the terminology is explained in Chapter 9). It 

is clear that U is isometric and surjective (onto H2 ⊕ H1). It follows 

that U is unitary. If H1 = H2 = H it is clear that U is selfadjoint and 

involutary (i.e., U 2 is the identity). Now put 

(4.1) (GT ) ∗ := U((H1 ⊕ H2) ⊖ GT ) = (H2 ⊕ H1) ⊖ UGT . 

The second equality is left to the reader to verify who should also 

verify that (GT ) ∗ is a graph of an operator (i.e., the second component 

of each element in (GT ) ∗ is uniquely determined by the first) if and 

only if T is densely defined. If T is densely defined we now define T ∗ 

to be the operator whose graph is (GT ) ∗ . This means that T ∗ is the 

operator whose graph consists of all pairs (v, v ∗ ) ∈ H2 ⊕ H1 such that 

〈T u, v〉2 = 〈u, v ∗ 〉1 for all u ∈ D(T ), i.e., our original definition. An 

immediate consequence of (4.1) is that T ⊂ S implies S ∗ ⊂ T ∗ . 

We say that an operator is closed if its graph is closed as a subspace 

of H1 ⊕ H2. This is an important property; in many ways the 

property of being closed is almost as good as being bounded. An everywhere 

defined operator is actually closed if and only if it is bounded 

(Exercise 4.7). It is clear that all adjoints, having graphs that are orthogonal 

complements, are closed. Not all operators are closeable, i.e., 

have closed extensions; for this is required that the closure GT of GT 

is a graph. But it is clear from (4.1) that the closure of the graph is 

(GT ∗)∗ . So, we have proved the following proposition. 

Proposition 4.3. Suppose T is a densely defined operator in a 

Hilbert space H. Then T is closeable if and only if the adjoint T ∗ is 

densely defined. The smallest closed extension (the closure) T of T is 

then T ∗∗ . 

The proof is left to Exercise 4.9. Note that if T is closed, its domain 

D(T ) becomes a Hilbert space if provided by the scalar product 

〈u, v〉T = 〈u, v〉1 + 〈T u, T v〉2.


In the rest of this chapter we assume that H1 = H2 = H. A densely 

defined operator T is then said to be symmetric if T ⊂ T ∗ . In other 

words, if 〈T u, v〉 = 〈u, T v〉 for all u, v ∈ D(T ). Thus 〈T u, u〉 is always 

real for a symmetric operator. It therefore makes sense to say that 

a symmetric operator is positive if 〈T u, u〉 ≥ 0 for all u ∈ D(T ). A 

densely defined symmetric operator is always closeable since T ∗ is automatically 

densely defined, being an extension of T . If actually T = T ∗ 

the operator is said to be selfadjoint. This is an important property 

because these are the operators for which we will prove the spectral 

theorem. In practice it is usually quite easy to see if an operator is 

symmetric, but much more difficult to decide whether a symmetric operator 

is selfadjoint. When one wants to interpret a differential operator 

as a Hilbert space operator one has to choose a domain of definition; 

in many cases it is clear how one may choose a dense domain so that 

the operator becomes symmetric. With luck this operator may have 

a selfadjoint closure 3 , in which case the operator is said to be essentially 

selfadjoint. Otherwise, given a symmetric T , one will look for 

selfadjoint extensions of T . If S is a symmetric extension of T , we get 

T ⊂ S ⊂ S ∗ ⊂ T ∗ so that any selfadjoint extension of T is a restriction 

of the adjoint T ∗ . There is now obviously a need for a theory of 

symmetric extensions of a symmetric operator. We will postpone the 

discussion of this until Chapter 9. Right now we will instead study 

some very simple, but typical, examples. 

Example 4.4. Consider the differential operator d 

on some open 

dx 

interval I. We want to interpret it as a densely defined operator in 

the Hilbert space L2 (I) and so must choose a suitable domain. A convenient 

choice, which would work for any differential operator with 

smooth coefficients, is the set C∞ 0 (I) of infinitely differentiable functions 

on I with compact support, i.e., each function is 0 outside some 

compact subset of I. It is well known that C∞ 0 (I) is dense in L2 (I). 

Let us denote the corresponding operator T0; it is usually called the 

minimal operator for d . Sometimes it is the closure of this operator 

dx 

which is called the minimal operator, but this will make no difference 

to the calculations in the sequel. We now need to calculate the adjoint 

of the minimal operator. 

Let v ∈ D(T ∗ 0 ). This means that there is an element v∗ ∈ L2 (I) 

such that 

I ϕ′ v = 

I ϕv∗ for all ϕ ∈ C∞ 0 (I) and that T ∗ 0 v = v∗ . In- 

tegrating by parts we have 

I ϕv∗ = − 

I (ϕ′ v∗ ) since the boundary 

terms vanish. Here v∗ denotes any integral function of v∗ . Thus we 

have 

I ϕ′ (v + v∗ ) = 0 for all ϕ ∈ C∞ 0 (I). We need the following 

lemma. 

3 This is the same as T ∗ being selfadjoint. Show this!


Lemma 4.5 (du Bois Reymond). Suppose u is locally square integrable 

on R, i.e., u ∈ L 2 (I) for every bounded real interval I. Also 

suppose that uϕ ′ = 0 for every ϕ ∈ C ∞ 0 (R). Then u is (almost everywhere) 

equal to a constant. 

Assuming the truth of the lemma for the moment it follows that, 

choosing the appropriate representative in the equivalence class of v, 

v + v ∗ is constant. Hence v is locally absolutely continuous with 

derivative −v ∗ . It follows that D(T ∗ 0 ) consists of functions in L 2 (I) 

which are locally absolutely continuous in I with derivative in L 2 (I), 

and that T ∗ 0 v = −v ′ . Conversely, all such functions are in D(T ∗ 0 ), 

as follows immediately by partial integration in 

I ϕ′ v = 

I ϕv∗ . The 

operator T ∗ 0 is therefore also a differential operator, generated by − d 

dx . 

The differential operator − d 

dx is called the formal adjoint of d 

dx and 

the operator T ∗ 0 is called the maximal operator belonging to − d 

dx 

. In 

the same way any linear differential operator (with sufficiently smooth 

coefficients) has a formal adjoint, obtained by integration by parts. 

For ordinary differential operators with smooth coefficients one can 

always calculate adjoints in essentially the way we just did; for partial 

differential operators matters are more subtle and one needs to use the 

language of distribution theory. 

Proof of Lemma 4.5. Let ψ ∈ C∞ 0 (R) and assume that ψ = 1. 

Given ϕ ∈ C∞ 0 (R) we put ϕ0(x) = ψ(x) ∞ 

−∞ϕ and Φ(x) = x 

−∞ (ϕ−ϕ0). 

It is clear that Φ is infinitely differentiable. It also has compact support 

(why?), so ∞ 

−∞uΦ′ = 0 by assumption. But ∞ 

−∞uΦ′ = ∞ 

uϕ − 

−∞ ∞ 

−∞uψ ∞ 

−∞ϕ so that ∞ 

−∞ (u − K)ϕ = 0 where K = ∞ 

uψ does not 

−∞ 

depend on ϕ. Since C∞ 0 (R) is dense in L2 (R) this proves that u = K 

a.e., so that u is constant. 

For the minimal operator of a differential operator to be symmetric 

it is clear that the differential operator has to be formally symmetric, 

i.e., the formal adjoint has to coincide with the original operator. 

In Example 4.4 D(T0) ⊂ D(T ∗ 0 ) but there is a minus sign preventing 

T0 from being symmetric. However, it is clear that had we started 

with the differential operator −i d 

dx 

instead, then the minimal opera- 

tor would have been symmetric, but the domains of the minimal and 

maximal operators unchanged. One may then ask for possible selfadjoint 

extensions of the minimal operator, or equivalently for selfadjoint 

restrictions of the maximal operator. 

Example 4.6. Let T1 be the maximal operator of −i d 

dx 

on the 

interval I. Let u, v ∈ D(T1) and a, b ∈ I. Then b 

a T1uv − b 

a uT1v = 

−i b 

a (u′ v +uv ′ ) = iu(a)v(a)−iu(b)v(b). Since u, v, T1u and T1v are all 

in L 2 (I) the limit of uv exists in both endpoints of I. Consider the case 

I = R. Since |u(x)| 2 has limits as x → ±∞ and is integrable, the limits


must both be 0. Hence 〈T1u, v〉 − 〈u, T1v〉 = 0 for any u, v ∈ D(T1), so 

the maximal operator is symmetric and therefore selfadjoint (how does 

this follow?). It also follows that the maximal operator is the closure of 

the minimal operator so the minimal operator is essentially selfadjoint. 

Example 4.7. Consider the same operator as in Example 4.6 but 

for the interval (0, ∞). If u ∈ D(T1) we obtain 〈T1u, u〉 − 〈u, T1u〉 = 

i|u(0)| 2 . To have a symmetric restriction of T1 we must therefore require 

u(0) = 0, and with this restriction on the domain of T1 we obtain a 

maximal symmetric operator T . If now u ∈ D(T ) and v ∈ D(T1) we 

obtain 〈T u, v〉−〈u, T1v〉 = iu(0)v(0) = 0 so that T ∗ = T1. T is therefore 

not selfadjoint so no matter how we choose the domain the differential 

, though formally symmetric, will not be selfadjoint in 

operator −i d 

dx 

L2 (0, ∞). One says that −i d 

dx has no selfadjoint realization in L2 (0, ∞). 

Example 4.8. We finally consider the operator of Example 4.6 for 

the interval (−π, π). We now have 

(4.2) 〈T1u, v〉 − 〈u, T1v〉 = −i(u(π)v(π) − u(−π)v(−π)). 

In particular, for u = v it follows that for u to be in the domain of a 

symmetric restriction of T1 we must require |u(π)| = |u(−π)| so that u 

satisfies the boundary condition u(π) = e iθ u(−π) for some real θ. From 

(4.2) then follows that if v is in the domain of the adjoint, then v will 

have to satisfy the same boundary condition. On the other hand, if we 

impose this condition, then the resulting operator will be selfadjoint 

(because its adjoint will be symmetric). It follows that restricting the 

domain of T1 by such a boundary condition is exactly what is required 

to obtain a selfadjoint restriction. Each θ in [0, 2π) gives a different 

selfadjoint realization, but there are no others. 

The examples show that there may be a unique selfadjoint realization 

of our formally symmetric differential operator, none at all, or 

infinitely many depending on circumstances. It can be a very difficult 

problem to decide which of these possibilities occur in a given case. 

In particular, much effort has been devoted to decide whether a given 

differential operator on a given domain has a unique selfadjoint realization.



Exercise 4.1. Prove that boundedness is equivalent to continuity 

for a linear operator between normed spaces. Then prove the properties 

of the operator norm listed at the beginning of the chapter. 

Exercise 4.2. Suppose B1 and B2 are Banach spaces. Show that 

so is B(B1, B2). 

Exercise 4.3. Fill in the details of the proof of Proposition 4.1. 

Exercise 4.4. Show that if M and N are closed subspaces of H 

with M ∩N = {0} and M ˙+N = H, then the corresponding projections 

onto M and N are bounded operators. 

Hint: The closed graph theorem! 

Exercise 4.5. Show that a non-trivial (i.e., the range is not {0}) 

projection is orthogonal if and only if its operator norm is 1. 

Exercise 4.6. Suppose H1 and H2 are Hilbert spaces. Show that 

the orthogonal direct sum H1 ⊕ H2 is also a Hilbert space. 

Exercise 4.7. Show that a bounded, everywhere defined operator 

is automatically closed. Conversely, that an everywhere defined, closed 

operator is bounded. 


Exercise 4.8. Show that if U is unitary, then all eigen-values λ of 

U have absolute value |λ| = 1. Also show that if e1 and e2 are eigenvectors 

corresponding to eigen-values λ1 and λ2 respectively, then e1 

and e2 are orthogonal if λ1 = λ2. 

Exercise 4.9. Show that if T is densely defined and closeable, then 

the closure of T is T ∗∗ .

CHAPTER 5 

Resolvents 

We now consider a closed, densely defined operator T in the Hilbert 

space H. We define the solvability and deficiency spaces of T at λ by 

Sλ = {u ∈ H | (T − λ)v = u for some v ∈ D(T )} 

Dλ = {u ∈ D(T ∗ ) | T ∗ u = λu}. 

The following basic lemma is valid. 

Lemma 5.1. Supposed T is closed and densely defined. Then 

(1) D λ = H ⊖ Sλ. 

(2) If T is symmetric and Im λ = 0, then Sλ is closed and H = 

Sλ ⊕ D λ 

(3) If T is selfadjoint and Im λ = 0, then (T − λ)v = u is uniquely 

solvable for any u ∈ H ( i.e., Sλ = H), T has no non-real 

eigen-values ( i.e., Dλ = {0}), and v ≤ 1 

|Im λ| u. 

Proof. Any element of the graph of T is of the form (v, λv + u), 

where u ∈ Sλ. To see this, simply put u = T v − λv for any v ∈ D(T ). 

Now 〈T v, w〉 − 〈v, λw〉 = 〈u + λv, w〉 − 〈v, λw〉 = 〈u, w〉, so it follows 

that (w, λw) ∈ GT ∗, i.e., w ∈ Dλ , if and only if w is orthogonal to Sλ. 

This proves (1). 

If T is symmetric and (v, λv+u) ∈ GT , then 〈λv+u, v〉 = 〈v, λv+u〉, 

i.e., Im λv2 = Im〈v, u〉, which is ≤ vu by Cauchy-Schwarz’ inequality. 

If Im λ = 0 we obtain v ≤ 1 u, so that v is uniquely 

|Im λ| 

determined by u; in particular T has no non-real eigen-values. Furthermore, 

suppose that u1, u2, . . . is a sequence in Sλ converging to u, and 

that (vj, λvj + uj) ∈ GT . Then v1, v2, . . . is also a Cauchy sequence, 

since vj − vk ≤ 1 

|Im λ| uj − uk. Thus vj tends to some limit v, and 

since T is closed we have (v, λv + u) ∈ GT . Hence u ∈ Sλ, so that Sλ 

is closed and (2) follows. 

Finally, if T is self-adjoint, then T ∗ = T is symmetric so it has no 

non-real eigen-values. If Im λ = 0 it follows that Dλ = {0} so that (3) 

follows and the proof is complete. 

In the rest of this chapter we assume that T is a selfadjoint operator. 

We define the resolvent set of T as 

ρ(T ) = {λ ∈ C | T − λ has a bounded, everywhere defined inverse} , 

31

32 5. RESOLVENTS 

and the spectrum σ(T ) of T as the complement of ρ(T ). By Lemma 5.1.3 

the spectrum is a subset of the real line. For every λ ∈ ρ(T ) we now 

define the resolvent of T at λ as the operator Rλ = (T − λ) −1 . The 

resolvent has the following properties. 

Theorem 5.2. The resolvent of a selfadjoint operator T has the 

properties: 

(1) Rλ ≤ 1/|Im λ| if Im λ = 0. 

(2) (Rλ) ∗ = R λ for λ ∈ ρ(T ). 

(3) Rλ − Rµ = (λ − µ)RλRµ for λ and µ ∈ ρ(T ). 

The last statement is called the (first) resolvent relation. 

Proof. The first claim is simply a re-statement of Lemma 5.1.3. 

Note that all elements of GT are of the form (Rλu, λRλu + u). Now 

w ∗ = (Rλ) ∗ w precisely if 〈Rλu, w〉 = 〈u, w ∗ 〉 for all u ∈ H. Adding 

λ〈Rλu, w ∗ 〉 to both sides we obtain 〈Rλu, λw ∗ + w〉 = 〈λRλu + u, w ∗ 〉, 

so that (w ∗ , λw ∗ + w) ∈ GT , i.e., w ∗ = R λ w. This proves (2). Finally, 

suppose (w, µw+u) ∈ GT . Since GT is linear it follows that (v, λv+u) ∈ 

GT if and only if (v, λv+u)−(w, µw+u) = (v−w, λ(v−w)+(λ−µ)w) ∈ 

GT . But this means exactly that (3) holds. 

Theorem 5.3. The resolvent set ρ(T ) is open, and the function 

λ ↦→ Rλ is analytic in the uniform operator topology as a B(H)-valued 

function. This means (by definition) that Rλ can be expanded in a 

power series with respect to λ around any point in ρ(T ) and that the 

series converges in operator norm in a neighborhood of the point. In 

fact, if µ ∈ ρ(T ), then λ ∈ ρ(T ) for |λ − µ| < 1/Rµ and 

Rλ = 

∞ 

k=0 

(λ − µ) k R k+1 

µ 

for |λ − µ| < 1/Rµ . 

Finally, the function ρ(T ) ∋ λ ↦→ 〈Rλu, v〉 is analytic for all u, v ∈ H, 

and for u = v it maps the upper and lower half-planes into themselves. 

Proof. The series is norm convergent if |λ − µ| < 1/Rµ since 

(λ − µ) k R k+1 

µ ≤ Rµ(|λ − µ|Rµ) k , which is a term in a convergent 

geometric series. Writing T − λ = T − µ − (λ − µ) and applying this 

to the series from the left and right, one immediately sees that the 

series represents the inverse of T − λ. We have verified the formula 

for Rλ and it also follows that ρ(T ) is open. Now by Theorem 5.2 

we have 2i Im〈Rλu, u〉 = 〈Rλu, u〉 − 〈u, Rλu〉 = 〈(Rλ − R λ )u, u〉 = 

2i Im λ〈R λ Rλu, u〉 = 2i Im λRλu 2 . It follows that Im〈Rλu, u〉 has 

the same sign as Im λ. The analyticity of 〈Rλu, v〉 follows since we 

have a power series expansion of it around any point in ρ(T ), by the 

series for Rλ. Alternatively, from Theorem 5.2.3 it easily follows that 

u, v〉 (Exercise 5.1). 

d 

dλ 〈Rλu, v〉 = 〈R 2 λ


Analytic functions that map the upper and lower halfplanes into 

themselves have particularly nice properties. Our proof of the general 

spectral theorem will be based on the fact that 〈Rλu, u〉 is such a 

function, so we will make a detailed study of them in the next chapter. 

That ρ(T ) is open means of course that the spectrum is always 

a closed subset of R. It is customary to divide the spectrum into (at 

least) two disjoint subsets, the point spectrum σp(T ) and the continuous 

spectrum σc(T ), defined as follows. 

σp(T ) = {λ ∈ C | T − λ is not one-to-one} 

σc(T ) = σ(T ) \ σp(T ). 

This means that the point spectrum consists of the eigen-values of T , 

and the continuous spectrum of those λ for which Sλ is dense in H 

but not closed. This follows since (T − λ) −1 is automatically bounded 

if Sλ = H, by the closed graph theorem (Exercise 5.2). For nonselfadjoint 

operators there is a further possibility; one may have Sλ 

non-dense even if λ is not an eigenvalue. Such values of λ constitute 

the residual spectrum which by Lemma 5.1 is empty for selfadjoint 

operators. 

An eigenvalue for a selfadjoint operator is said to have finite multiplicity 

if the eigenspace is finite-dimensional. Removing from the 

spectrum all isolated points which are eigenvalues of finite multiplicity 

leaves one with the essential spectrum. The name comes from the 

fact that the essential spectrum is quite stable under perturbations 

(changes) of the operator T , but we will not discuss such matters here. 


Exercise 5.1. Suppose that Rλ is the resolvent of a self-adjoint 

operator T in a Hilbert space H. Show directly from Theorem 5.2.3 that 

if u, v ∈ H, then λ ↦→ 〈Rλu, v〉 is analytic (has a complex derivative) 

for λ ∈ ρ(T ), and find an expression for the derivative. Also show that 

if u ∈ H, then λ ↦→ 〈Rλu, u〉 is increasing in every point of ρ ∩ R. 

Exercise 5.2. Show that if T is a closed operator with Sλ = H 

and λ /∈ σp(T ), then λ ∈ ρ(T ). 


Exercise 5.3. Show that if T is a self-adjoint operator, then U = 

(T +i)(T −i) −1 = I +2iR−i is unitary. Conversely, if U is unitary and 1 

is not an eigen-value, then T = i(U + I)(U − I) −1 is selfadjoint. What 

can one do if 1 is an eigen-value? This transform, reminiscent of a 

Möbius transform, is called the Cayley transform and was the basis for 

von Neumann’s proof of the spectral theorem for unbounded operators.

CHAPTER 6 

Nevanlinna functions 

Our proof of the spectral theorem is based on the following representation 

theorem. 

Theorem 6.1. Suppose F is analytic in C \ R, F (λ) = F (λ), 

and F maps each of the upper and lower half-planes into themselves. 

Then there exists a unique, left-continuous, increasing function ρ with 

ρ(0) = 0 and ∞ dρ(t) 

1+t2 < ∞, and unique real constants α and β ≥ 0, 

such that 

−∞ 

(6.1) F (λ) = α + βλ + 

∞ 

−∞ 

where the integral is absolutely convergent. 

 

1 t 

− 

t − λ 1 + t2 

dρ(t), 

For the meaning of such an integral, see Appendix B. Functions 

F with the properties in the theorem are usually called Nevanlinna, 

Herglotz or Pick functions. I am not sure who first proved the theorem, 

but results of this type play an important role in the classical book 

Eindeutige analytische Funktionen by Rolf Nevanlinna (1930). We will 

tackle the proof through a sequence of lemmas. 

Lemma 6.2 (H. A. Schwarz). Let G be analytic in the unit disk, 

and put u(R, θ) = Re G(Reiθ ). For |z| < R < 1 we then have: 

(6.2) G(z) = i Im G(0) + 1 

π 

Re 

2π 

iθ + z 

Reiθ u(R, θ) dθ. 

− z 

Proof. According to Poisson’s integral formula (see e.g. Chapter 6 

of Ahlfors: Complex Analysis (McGraw-Hill 1966)), we have 

Re G(z) = 1 

π 

R 

2π 

2 − |z| 2 

|Reiθ u(R, θ) dθ . 

− z| 2 

−π 

The integral here is easily seen to be the real part of the integral in 

(6.2). The latter is obviously analytic in z for |z| < R < 1, so the two 

sides of (6.2) can only differ by an imaginary constant. However, for 

z = 0 the integral is real, so (6.2) follows. 

The formula (6.2) is not applicable for R = 1, since we do not 

know whether Re G has reasonable boundary values on the unit circle. 

−π 

35

36 6. NEVANLINNA FUNCTIONS 

However, if one assumes that Re G ≥ 0 the boundary values exist at 

least in the sense of measure, and one has the following theorem. 

Theorem 6.3 (Riesz-Herglotz). Let G be analytic in the unit circle 

with positive real part. Then there exists an increasing function σ on 

[0, 2π] such that 

G(z) = i Im G(0) + 1 

2π 

π 

−π 

eiθ + z 

eiθ dσ(θ) . 

− z 

With a suitable normalization the function σ will also be unique, 

but we will not use this. To prove Theorem 6.3 we need some kind 

of compactness result, so that we can obtain the theorem as a limiting 

case of Lemma 6.2. What is needed is weak ∗ compactness in the 

dual of the continuous functions on a compact interval, provided with 

the maximum norm. This is the classical Helly theorem. Since we assume 

minimal knowledge of functional analysis we will give the classical 

proof. 

Lemma 6.4 (Helly). 

(1) Suppose {ρj} ∞ 1 is a uniformly bounded1 sequence of increasing 

functions on an interval I. Then there is a subsequence 

converging pointwise to an increasing function. 

(2) Suppose {ρj} ∞ 1 is a uniformly bounded sequence of increasing 

functions on a compact interval I, converging pointwise to ρ. 

Then 

 

(6.3) 

f dρj → f dρ as j → ∞, 

I 

for any function f continuous on I. 

I 

Proof. Let r1, r2, . . . be a dense sequence in I, for example an enumeration 

of the rational numbers in I. By Bolzano-Weierstrass’ theorem 

we may choose a subsequence {ρ1j} ∞ 1 of {ρj} ∞ 1 so that ρ1j(r1) converges. 

Similarly, we may choose a subsequence {ρ2j} ∞ 1 of {ρ1j} ∞ 1 such 

that ρ2j(r2) converges; as a subsequence of ρ1j(r1) the sequence ρ2j(r1) 

still converges. Continuing in this fashion, we obtain a sequence of sequences 

{ρkj} ∞ j=1, k = 1, 2, . . . such that each sequence is a subsequence 

of those coming before it, and such that ρ(rn) = limj→∞ ρkj(rn) exists 

for n ≤ k. Thus ρjj(rn) → ρ(rn) as j → ∞ for every n, since ρjj(rn) is 

a subsequence of ρnj(rn) from j = n on. Clearly ρ is increasing, so if 

x ∈ I but = rn for all n, we may choose an increasing subsequence rjk , 

k = 1, 2, . . . , converging to x, and define ρ(x) = limk→∞ ρ(rjk ). 

Suppose x is a point of continuity of ρ. If rk < x < rn we get 

ρjj(rk) − ρ(rn) ≤ ρjj(x) − ρ(x) ≤ ρjj(rn) − ρ(rk). Given ε > 0 we may 

1 i.e., all the functions are bounded by a fixed constant

6. NEVANLINNA FUNCTIONS 37 

choose k and n such that ρ(rn) − ρ(rk) < ε. We then obtain 

−ε ≤ lim (ρjj(x) − ρ(x)) ≤ lim (ρjj(x) − ρ(x)) ≤ ε . 

j→∞ 

j→∞ 

Hence {ρjj} ∞ 1 converges pointwise to ρ, except possibly in points of 

discontinuity of ρ. But there are at most countably many such discontinuities, 

ρ being increasing. Hence repeating the trick of extracting 

subsequences, and then using the ‘diagonal’ sequence, we get a subsequence 

of the original sequence which converges everywhere in I. We 

now obtain (1). 

If f is the characteristic function of a compact interval whose endpoints 

are points of continuity for ρ and all ρj it is obvious that (6.3) 

holds. It follows that (6.3) holds if f is a stepfunction with all discontinuities 

at points where ρ and all ρj are continuous. If f is continuous 

and ε > 0 we may, by uniform continuity, choose such a stepfunction 

g so that supI|f − g| < ε. If C is a common bound for all ρj we then 

obtain | 

I (f − g) dρ| < 2Cε and similarly with ρ replaced by ρj. It 

follows that limj→∞| 

I f dρj − 

f dρ| ≤ 4Cε and since ε is arbitrary 

I 

positive (2) follows. 

Proof of Theorem 6.3. According to Lemma 6.2 we have, for 

|z| < 1, 

G(Rz) = i Im G(0) + 1 

2π 

π 

−π 

e iθ + z 

e iθ − z dσR(θ) , 

where σR(θ) = θ 

−π Re G(Reiϕ ) dϕ. Hence σR is increasing, ≥ 0 and 

bounded from above by σR(π). Now Re G is a harmonic function so it 

has the mean value property, which means that σR(π) = 2π Re G(0). 

This is independent of R, so by Helly’s theorem we may choose a sequence 

Rj ↑ 1 such that σR converges to an increasing function σ. Use 

of the second part of Helly’s theorem completes the proof. 

To prove the uniqueness of the function ρ of Theorem 6.1 we need 

the following simple, but important, lemma. 

Lemma 6.5 (Stieltjes’ inversion formula). Let ρ be complex-valued 

of locally bounded variation, and such that ∞ dρ(t) 

−∞ t2 is absolutely con- 

+1 

vergent. Suppose F (λ) is given by (6.1). Then if y < x are points of 

continuity of ρ we have 

ρ(x) − ρ(y) = lim 

ε↓0 

1 

2πi 

x 

y 

(F (s + iε) − F (s − iε) ds 

= lim 

ε↓0 

1 

π 

x 

y 

∞ 

−∞ 

ε dρ(t) 

(t − s) 2 ds . 

+ ε2

38 6. NEVANLINNA FUNCTIONS 

Proof. By absolute convergence we may change the order of integration 

in the last integral. The inner integral is then easily calculated 

to be 

1 

(arctan((x − t)/ε) − arctan((y − t)/ε)). 

π 

This is bounded by 1, and also by a constant multiple of 1/t2 if ε is 

bounded (verify this!). Furthermore it converges pointwise to 0 outside 

[y, x], and to 1 in (y, x) (and to 1 for t = x and t = y). The theorem 

2 

follows by dominated convergence. 

Proof of Theorem 6.1. The uniqueness of ρ follows immediately 

on applying the Stieltjes inversion formula to the imaginary part 

of (6.1) for λ = s + iε. 

We obtain (6.1) from the Riesz-Herglotz theorem by a change of 

variable. The mapping z = 1+iλ maps the upper half plane bijectively 

1−iλ 

to the unit disk, so G(z) = −iF (λ) is defined for z in the unit disk 

and has positive real part. Applying Theorem 6.3 we obtain, after 

simplification, 

F (λ) = Re F (i) + 1 

2π 

π 

−π 

1 + λ tan(θ/2) 

tan(θ/2) − λ 

dσ(θ) . 

Setting t = tan(θ/2) maps the open interval (−π, π) onto the real axis. 

For θ = ±π the integrand equals λ, so any mass of σ at ±π gives rise 

to a term βλ with β ≥ 0. After the change of variable we get 

 

F (λ) = α + βλ + 

∞ 

−∞ 

1 + tλ 

t − λ 

dτ(t) , 

where we have set α = Re F (i) and τ(t) = σ(θ)/(2π). Since 

1 + tλ 

t − λ = 

 

1 t 

− 

t − λ 1 + t2 

(1 + t 2 ) 

we now obtain (6.1) by setting ρ(t) = t 

0 (1 + s2 ) dτ(s). 

It remains to show the uniqueness of α and β. However, setting 

λ = i, it is clear that α = Re F (i), and since we already know that ρ 

is unique, so is β. 

Actually one can calculate β directly from F since by dominated 

convergence Im F (iν)/ν → β as ν → ∞. It is usual to refer to β as the 

‘mass at infinity’, an expression explained by our proof. Note, however, 

that it is the mass of τ at infinity and not that of ρ!

CHAPTER 7 

The spectral theorem 

Theorem 7.1. (Spectral theorem) Suppose T is selfadjoint. Then 

there exists a unique, increasing and left-continuous family {Et}t∈R of 

orthogonal projections with the following properties: 

• Et commutes with T , in the sense that T Et is the closure of 

EtT . 

• Et → 0 as t → −∞ and Et → I (= identity on H) as t → ∞ 

(strong convergence). 

• T = ∞ 

−∞ t dEt in the following sense: u ∈ D(T ) if and only if 

∞ 

∞ 

−∞ t2 d〈Etu, u〉. 

−∞ t2 d〈Etu, u〉 < ∞, 〈T u, v〉 = ∞ 

−∞ t d〈Etu, v〉 and T u 2 = 

The family {Et}t∈R of projections is called the resolution of the identity 

for T . The formula T = ∞ 

−∞t dEt can be made sense of directly by 

introducing Stieltjes integrals with respect to operator-valued increasing 

functions. This is a simple generalization of the scalar-valued case. 

Although we then, formally, get a slightly stronger statement, it does 

not appear to be any more useful than the statement above. We will 

therefore omit this. 

For the proof we need two lemmas, the first of which actually contains 

the main step of the proof. 

Lemma 7.2. For f, g ∈ H there is a unique left-continuous function 

σf,g of bounded variation, with σf,g(−∞) = 0, and the following 

properties: 

• σf,g is Hermitian in f, g ( i.e., σf,g = σg,f and is linear in f), 

and σf,f is increasing. 

• dσf,g is a bounded sesquilinear form on H. In fact, we even 

have ∞ 

−∞ |dσf,g| ≤ fg. 

• 〈Rλf, g〉 = ∞ dσf,g 

−∞ t−λ . 

Proof. The uniqueness of σf,g follows from the Stieltjes inversion 

formula, applied to F (λ) = 〈Rλf, g〉. Since 〈Rλf, g〉 is sesqui-linear in 

f, g and R ∗ λ = R λ , it then follows that σf,g is Hermitian if it exists. 

However, by Theorem 5.3 the function λ ↦→ 〈Rλf, f〉 is a Nevanlinna 

39

40 7. THE SPECTRAL THEOREM 

function of λ for any f, so we have 

∞ 

 

1 t 

(7.1) 〈Rλf, f〉 = α + βλ + − 

t − λ 1 + t2 

dσf,f(t), 

−∞ 

where σf,f is increasing and α, β may depend on f. Since Rλ ≤ 1 

|Im λ| , 

we find that f 2 is an upper bound for ν〈Riνf, f〉 for ν ∈ R, the 

imaginary part of which is βν 2 + ∞ 

−∞ 

ν2 dσf,f (t) 

t2 +ν2 . Hence β = 0, and by 

Fatou’s lemma we get, as ν → ∞, that ∞ 

−∞ dσf,f ≤ f 2 . A more 

elementary argument is the following: For ν, ε > 0 we have 

1 

1 + ε2 νε 

∞ 

ν 

dσf,f ≤ 

2 

t2 + ν2 dσf,f(t) ≤ f 2 , 

−νε 

−∞ 

1 since 1+ε2 ≤ ν2 

ν2 +t2 for |t| ≤ νε, so letting ν → ∞, and then ε → 0, we 

obtain the same bound. We may now assume σf,f to be normalized so 

as to be left-continuous and σf,f(−∞) = 0. Clearly ∞ t 

−∞ 1+t2 dσf,f(t) 

is absolutely convergent, so this part of the integral in (7.1) may be 

incorporated in the constant α. So, with absolute convergence, we have 

〈Rλf, f〉 = α ′ + ∞ dσf,f (t) 

. However, for λ → ∞ along the imaginary 

−∞ t−λ 

axis, both the left hand side and the integral → 0 (Exercise 7.1), so we 

must have α ′ = 0. The proof is now finished in the case f = g. 

By the polarization identity (Exercise 7.2) 

〈Rλf, g〉 = 1 

3 

i 

4 

k 〈Rλ(f + i k g), f + i k g〉 , 

k=0 

so we obtain 〈Rλf, g〉 = ∞ dσf,g(t) 

−∞ t−λ 

σf,g = 1 

4 

3 

k=0 

by setting 

i k σ f+i k g,f+i k g. 

The function σf,g has the correct normalization, so only the bound on 

the total variation remains to be proved. But if ∆ is an interval, then 

∆ dσf,g is a semi-scalar product on H, so Cauchy-Schwarz’ inequality 

 

 

∆ dσf,g 

 

2 ≤ 

∆ dσf,f 

 

∆ dσg,g 

is valid. For ∆ = R this shows that 

R dσf,g is bounded by fg. If {∆j} ∞ 1 is a partition of R into disjoint 

intervals we obtain 

 

 

dσf,g 

 

≤ 

∆j 

∆j 

dσf,f 

 

 

 

≤ 

∆j 

 

∆j 

dσg,g 

dσf,f 

1 

2 

1 

2 

∆j 

dσg,g 

1 

2 

≤ fg,

7. THE SPECTRAL THEOREM 41 

where the second inequality is Cauchy-Schwarz’ inequality in ℓ 2 . The 

proof is complete. 

Lemma 7.3. ∞ 

−∞ dσf,g = 〈f, g〉 for any f, g ∈ H. 

Proof. Assume first that f ∈ D(T ) so that f = Rλ(v−λf), where 

v = T f. Thus 〈f, g〉 = −λ〈Rλf, g〉 + 〈Rλv, g〉. Since −iν ∞ 

∞ 

−∞ 

dσf,g(t) 

t−iν → 

−∞ dσf,g as ν → ∞ by bounded convergence (Exercise 7.1), the lemma 

is true for f ∈ D(T ), which is dense in H. But ∞ 

−∞ dσf,g is a bounded 

Hermitian form on H since | ∞ 

−∞ dσf,g| ≤ ∞ 

−∞ |dσf,g| ≤ fg by 

Lemma 7.2, so the general case follows by continuity. 

Proof of the spectral theorem. We first show the uniqueness 

of the resolution of identity. So, assume a resolution of the identity 

with all the properties claimed exists. Then EtEs = Emin(s,t), so if 

w ∈ D(T ) and s fixed we obtain 

s 

−∞ 

d〈EtT w, v〉 = 〈EsT w, v〉 

= 〈T Esw, v〉 = 

∞ 

−∞ 

t d〈EtEsw, v〉 = 

s 

−∞ 

t d〈Etw, v〉. 

Thus d〈EtT w, v〉 = t d〈Etw, v〉 as measures. Now suppose w = Rλu. 

We then get 

d〈Etu, v〉 

t − λ = d〈Et(T − λ)Rλu, v〉 

t − λ 

= d〈EtRλu, v〉. 

It follows that 〈Rλu, v〉 = d〈Etu,v〉 

t−λ . The uniqueness of the spectral 

projectors therefore follows from the Stieltjes inversion formula. 

The linear form f ↦→ σf,g(t) is bounded for each g ∈ H (by g, 

according to Lemma 7.2). By Riesz’ representation theorem it is therefore 

of the form 〈f, gt〉, where gt ≤ g. It is obvious that gt depends 

linearly on g so gt = Etg where Et is a linear operator with norm ≤ 1, 

which is selfadjoint since σf,g is Hermitian. Furthermore Etf ⇀ 0 as 

t → −∞ by the normalization of σf,g and Etf ⇀ f as t → ∞ (weak 

convergence) by Lemma 7.3. 

Suppose we knew that Et is a projection. Since Et is selfadjoint it is 

then an orthogonal projection. It follows that Etf 2 = 〈Etf, f〉 → 0 

as t → −∞ and similarly f − Etf 2 = 〈f − Etf, f〉 → 0 as t → ∞. 

Hence we only need to show that Et is a projection increasing with t, 

and the statements about T .


as 

The resolvent relation Rλ − Rµ = (λ − µ)RλRµ may be expressed 

∞ 

−∞ 

1 

t − λ 

d〈Etf, g〉 

t − µ = 

∞ 

−∞ 

d〈EtRµf, g〉 

t − λ 

(check this!), so the uniqueness of the Stieltjes transform shows that 

〈EtRµf, g〉 = t d〈Esf,g〉 

. But 

−∞ s−µ 

〈EtRµf, g〉 = 〈Rµf, Etg〉 = 

∞ 

−∞ 

d〈Esf, Etg〉 

. 

s − µ 

So, again by uniqueness, 〈Esf, Etg〉 = 〈Euf, g〉 where u = min(s, t), 

i.e., EtEs = Emin(s,t). For s = t this shows that Et is a projection, and 

if t > s we get 0 ≤ (Et − Es) ∗ (Et − Es) = (Et − Es) 2 = Et − Es so that 

{Et}t∈R is an increasing family of orthogonal projections. 

Now suppose f ∈ D(T ) and v = T f. For any non-real λ we then 

have f = Rλ(v−λf) or Rλv = f +λRλf. Since 1+λ/(t−λ) = t/(t−λ) 

we therefore obtain 

∞ 

−∞ 

dσv,g(t) 

t − λ = 

∞ 

−∞ 

t dσf,g 

t − λ 

so that σv,g(t) = t 

−∞ s dσf,g(s). In particular, 〈T f, g〉 = ∞ 

−∞t d〈Etf, g〉. 

We also get σv,v(t) = t 

−∞ s dσf,v(s) = t 

−∞ s2 dσf,f(s), so that T f2 = 

∞ 

−∞s2d〈Esf, f〉. 

Next we prove that any u ∈ H for which ∞ 

−∞s2 d〈Esu, u〉 < ∞ is 

in D(T ). To see this, note that 

 

 

 

 

 

|d〈Esu, v〉| ≤ d〈Esu, u〉 d〈Esv, v〉 

∆ 

∆ 

if ∆ is a finite union of intervals. This follows just as in the proof of 

Lemma 7.2. Now let ∆k = {s | 2 k−1 < |s| ≤ 2 k }, k ∈ Z. Then 

 

 

 

 

 

 

 

s d〈Esu, v〉 

≤ 2k |d〈Esu, v〉| 

 

 

∆k 

∆k 

≤ 2 k 

 

 

 

 

 

 

 

d〈Esu, u〉 d〈Esv, v〉 ≤ 2 

∆k 

∆k 

∆k 

∆ 

s2 

d〈Esu, u〉 

∆k 

d〈Esv, v〉. 

If now ∞ 

−∞s2 d〈Esu, u〉 < ∞ we obtain from this by adding over all k 

 

 

and using Cauchy-Schwarz’ inequality for sums that ∞ 

−∞s d〈Esu, 

 

 

v〉 ≤ 

∞ 

2 −∞s2d〈Esu, u〉v so that the anti-linear form v ↦→ ∞ 

−∞s d〈Esu, v〉

7. THE SPECTRAL THEOREM 43 

is bounded on H. It is therefore, by Riesz’ representation theorem, 

a scalar product 〈u1, v〉. It is obvious that u1 depends linearly on u, 

i.e., there is a linear operator S so that u1 = Su. It is clear that S is 

symmetric and an extension of T , so we have T ⊂ S ⊂ S∗ ⊂ T ∗ = T . 

Hence S = T so the claims about D(T ) are verified. 

Finally, we must prove that T Et is the closure of EtT . From what 

we just proved it follows that if u ∈ D(T ) then Etu ∈ D(T ). For v ∈ H 

we then have 〈T Etu, v〉 = ∞ 

−∞s d〈EsEtu, v〉 = ∞ 

−∞s d〈Esu, Etv〉 = 

〈T u, Etv〉 = 〈EtT u, v〉 so T Et is an extension of EtT . Since Et is 

bounded and T closed it follows that T Et is closed (Exercise 7.3). Now 

suppose Etu ∈ D(T ). We must find uj ∈ D(T ) such that uj → u and 

EtT uj → T Etu. Since D(T ) is dense in H we can find vj ∈ D(T ) so 

that vj → u. Now set uj = vj −Etvj +Etu. Clearly uj ∈ D(T ), uj → u 

and EtT uj = T Etuj = T Etu and the proof is complete. 

The operator Et is called the spectral projector for the interval 

(−∞, t). The spectral projector for the interval (a, b) is E(a,b) = Eb − 

Ea+ where Ea+ is the right hand limit at a of Et. Similarly E[a,b] = 

Eb+ − Ea, etc. For a general Borel set M ⊂ R the spectral projector is 

defined to be EM = 

M dEt. Show that this is actually an orthogonal 

projection for any Borel set B! 

Obviously the various parts of the spectrum (point spectrum etc.) 

are determined by the behavior of the spectral projectors. We end this 

chapter with a theorem which makes explicit this connection. 

Theorem 7.4. 

(1) λ ∈ σp(T ) if and only if Et jumps at t, i.e., E{λ} = E[λ,λ] = 0. 

(2) λ ∈ ρ(T ) ∩ R if and only if Et is constant in a neighborhood 

of t = λ. 

It follows that the continuous spectrum consists of those points of 

increase of Et which are not jumps 1 . 

Proof. If Et jumps at λ we can find a unit vector e in the range 

of E{λ}, i.e., such that E{λ}e = e. It follows immediately from the 

spectral theorem that e ∈ D(T ) and (T −λ)e = 0. Conversely, suppose 

that e is a unit vector with T e = λe. Then 

0 = (T − λ)e 2 ∞ 

= (t − λ) 2 d〈Ete, e〉, 

−∞ 

so that the support of the non-zero, non-negative measure d〈Ete, e〉 

is contained in {λ}. Hence Et jumps at λ, and the proof of (1) is 

complete. 

Now assume Et is constant in (λ − ε, λ + ε). Then λ is not an 

eigenvalue of T so Sλ is dense in H. Thus the inverse of T − λ exists 

1 A point of increase for Et is a point λ such that E∆ = 0 for every open ∆ ∋ λ.


as a closed, densely defined operator. We need only show that this 

inverse is bounded, to see that its domain is all of H so that λ ∈ ρ(T ). 

But (T − λ)u2 = ∞ 

−∞ (t − λ)2 d〈Etu, u〉 ≥ ε2 ∞ 

−∞d〈Etu, u〉 = ε2u2 so the inverse of T − λ is bounded by 1/ε. Conversely, assume that 

Et is not constant near λ. Then there are arbitrarily short intervals ∆ 

containing λ such that E∆ = 0, i.e., there are non-zero vectors u such 

that E∆u = u. But then (T − λ)u ≤ |∆|u, where |∆| is the length 

of ∆. Hence we can find a sequence of unit vectors uj, j = 1, 2, . . . 

for which (T − λ)uj → 0 (a ‘singular sequence’). Consequently either 

T −λ is not injective, or else the inverse is unbounded so λ /∈ ρ(T ). 


Exercise 7.1. Suppose σ is increasing and ∞ 

−λ ∞ 

−∞ 

dσ(t) 

t−λ → ∞ 

−∞ 

the origin. In particular, ∞ 

−∞ 

−∞ 

dσ < ∞. Show that 

dσ as λ → ∞ along any non-real ray originating in 

dσ(t) 

t−λ 

→ 0. 

Exercise 7.2. Suppose B(·, ·) is a sesqui-linear form on a complex 

linear space. Show the polarization identity 

B(u, v) = 1 

3 

i 

4 

k B(u + i k v, u + i k v) . 

k=0 

Exercise 7.3. Show that if T is a closed operator on H and S is 

bounded and everywhere defined, then T S, but not necessarily ST , is 

closed. 

Exercise 7.4. Show that if T is selfadjoint and f is a continuous 

function defined on σ(T ), then f(T ) = ∞ 

−∞f(t) dEt defines a densely 

defined operator, which is bounded if f is and selfadjoint if f is realvalued. 

Also show that (f(T )) ∗ = f(T ), that (f(T )) ∗ has the same domain 

as f(T ) and commutes with it in a reasonable sense, and that fg(T ) = 

f(T )g(T ). This is the functional calculus for a selfadjoint operator, and 

also makes sense for arbitrary Borel functions. The integral is made 

sense of in the same way as in the statement of the spectral theorem. 

Exercise 7.5. Let T be selfadjoint and put H(t) = e −itT , t ∈ R, 

the exponential being defined as in the previous exercise. Show that 

H(t + s) = H(t)H(s) for real t and s (a group of operators), that 

H(t) is unitary and that if u0 ∈ D(T ), then u(t) = H(t)u0 solves the 

Schrödinger equation T u = iu ′ t with initial data u(0) = u0. 

Similarly, if T ≥ 0 and t ≥ 0, show that K(t) = e −tT is selfadjoint 

and bounded, that K(t + s) = K(t)K(s) for s ≥ 0 and t ≥ 0 (a semigroup 

of operators) and that if u0 ∈ H then u(t) = K(t)u0 solves the 

heat equation T u = u ′ t for t > 0 with initial data u(0) = u0.

CHAPTER 8 

Compactness 

If a selfadjoint operator T has a complete orthonormal sequence of 

eigen-vectors e1, e2, . . . , then for any f ∈ H we have f = ˆ fjej where 

ˆfj = 〈f, ej〉 are the generalized Fourier coefficients; we have a generalized 

Fourier series. However, σp(T ) can still be very complicated; it 

may for example be dense in R (so that σ(T ) = R), and each eigenvalue 

can have infinite multiplicity. We have a considerably simpler 

situation, more similar to the case of the classical Fourier series, if the 

resolvent is compact. 

Definition 8.1. 

• A subset of a Hilbert space is called precompact (or relatively 

compact) if every sequence of points in the set has a strongly 

convergent subsequence. 

• An operator A : H1 → H2 is called compact if it maps 

bounded sets into precompact ones. 

Note that in an infinite dimensional space it is not enough for a set 

to be bounded (or even closed and bounded) for it to be precompact. 

For example, the closed unit sphere is closed and bounded, and it 

contains an orthonormal sequence. But no orthonormal sequence has 

a strongly convergent subsequence! 

The second point means that if {uj} ∞ 1 is a bounded sequence in H1, 

then {Auj} ∞ 1 has a subsequence which converges strongly in H2. 

Theorem 8.2. 

(1) The operator A is compact if and only if every weakly convergent 

sequence is mapped onto a strongly convergent sequence. 

Equivalently, if uj ⇀ 0 implies that Auj → 0. 

(2) If A : H1 → H2 is compact and B : H3 → H1 bounded, then 

AB is compact. 

(3) If A : H1 → H2 is compact and B : H2 → H3 bounded, then 

BA is compact. 

(4) If A : H1 → H2 is compact, then so is A ∗ : H2 → H1. 

Proof. If uj ⇀ u then uj − u ⇀ 0, and if A(uj − u) → 0 then 

Auj → Au. Thus the last statement of (1) is obvious. By Theorem 3.9 

every bounded sequence has a weakly convergent subsequence, so if A 

maps weakly convergent sequences into strongly convergent ones, then 

A is compact. Conversely, suppose uj ⇀ u and A is compact. Since 

45

46 8. COMPACTNESS 

weakly convergent sequences are bounded (Theorem 3.9), any subsequence 

of {Auj} ∞ 1 has a convergent subsequence. Suppose Aujk → v. 

Then for any w ∈ H we have 〈v, w〉 = lim〈Aujk , w〉 = lim〈ujk , A∗w〉 = 

〈u, A∗w〉 = 〈Au, w〉, so that v = Au. Hence the only point of accumulation 

of {Auj} ∞ 1 is Au, so Auj → Au1 . This completes the proof 

of (1). We leave the rest of the proof as an exercise for the reader 

(Exercise 8.1). 

Theorem 8.3. Suppose T is selfadjoint and its resolvent Rµ is 

compact for some µ. Then Rλ is compact for all λ ∈ ρ(T ), and T has 

discrete spectrum, i.e., σ(T ) consists of isolated eigenvalues with finite 

multiplicity. 

Proof. By the resolvent relation Rλ = (I + (λ − µ)Rλ)Rµ where 

I is the identity so the first factor to the right is bounded. Hence Rλ 

is compact by Theorem 8.2.3. 

 

∆ 

Now let ∆ be a bounded interval. If u ∈ E∆H then Rλu 2 = 

d〈Etu,u〉 

|t−λ| 2 

≥ Ku 2 where K = inft∈∆|t − λ| −2 > 0 (verify this calcu- 

lation!). We have Rλuj → 0 if uj ⇀ 0, so the inequality shows that 

any weakly convergent sequence in E∆H is strongly convergent (the 

identity operator is compact). This implies that E∆H has finite dimension 

(for example since an orthonormal sequence converges weakly 

to 0 but is not strongly convergent). In particular eigenspaces are 

finite-dimensional. It also follows that any bounded interval can only 

contain a finite number of points of increase for Et, because projections 

belonging to disjoint intervals have orthogonal ranges (Exercise 8.2). 

This completes the proof. 

Resolvents for different selfadjoint extensions of a symmetric operator 

are closely related. In particular, we have the following theorem. 

Theorem 8.4. Suppose a densely defined symmetric operator T0 

has a selfadjoint extension with compact resolvent and that dim Dλ < 

∞ for some λ ∈ C \ R. Then every selfadjoint extension of T0 has 

compact resolvent. 

Proof. Let Im λ = 0 and Rλ, ˜ Rλ be resolvents of selfadjoint extensions 

of T0. Then A = Rλ − ˜ Rλ has its range in Dλ, since Rλu and 

˜Rλu both solve the equation T ∗ 0 v = λv + u. It follows that A is a compact 

operator, since if {uj} ∞ 1 is a bounded sequence in H, then {Auj} ∞ 1 

is a bounded sequence in a finite-dimensional space. By the Bolzano- 

Weierstrass theorem there is therefore a convergent subsequence. If ˜ Rλ 

is compact it therefore follows that Rλ = ˜ Rλ + A is compact. 

1 If not, there would be a neighborhood O of Au and a subsequence of {Auj} ∞ j=1 

that were outside O. But we could then find a convergent subsequence which does 

not converge to Au.

8. COMPACTNESS 47 

A natural question is now: How do I, in a concrete case, recognize 

that an operator is compact? One class of compact operators which 

are sometimes easy to recognize, are the Hilbert-Schmidt operators. 

Definition 8.5. A : H → H is called a Hilbert-Schmidt operator if 

for some complete orthonormal sequence e1, e2, . . . we have Aej 2 < 

∞. The number ||A|| = Aej 2 is called the Hilbert-Schmidt norm 

of A. 

Lemma 8.6. ||A|| is independent of the particular complete orthonormal 

sequence used in the definition, it is a norm, ||A|| = |||A ∗ ||, and 

any Hilbert-Schmidt operator is compact. The set of Hilbert-Schmidt 

operators on H is a Hilbert space in the Hilbert-Schmidt norm. 

Proof. It is clear that ||| · || is a norm. Now suppose {ej} ∞ 1 and 

{fj} ∞ 1 are arbitrary complete orthonormal sequences. Using Parseval’s 

formula twice it follows that 

jAej2 = 

j,k |〈Aej, fk〉| 2 = 

 

j,k |〈ej, A∗fk〉| 2 = 

kA∗fk2 . Thus the Hilbert-Schmidt norm has 

the claimed properties. To see that A is compact, suppose uj ⇀ 0 and 

let ε > 0. Choose N so large that ∞ 

N A∗ ej 2 < ε and let C be a 

bound for the sequence {uj} ∞ 1 . By Parseval’s formula we then have 

Auk 2 = |〈Auk, ej〉| 2 = |〈uk, A ∗ ej〉| 2 . We obtain 

Auk 2 ≤ 

N 

1 

|〈uk, A ∗ ej〉| 2 + C 2 ε → C 2 ε 

as k → ∞ since |〈uk, A ∗ ej〉| ≤ CA ∗ ej . It follows that Auk → 0 

so that A is compact. We leave the proof of the last statement as an 

exercise for the reader (Exercise 8.4). 

It is usual to consider a differential operator defined in some domain 

Ω ⊂ R n as an operator in the space L 2 (Ω, w) where w > 0 is 

measurable 

 

and the scalar product in the space is given by 〈u, v〉 = 

u(x)v(x)w(x) dx. In all reasonable cases the resolvent of such an 

Ω 

operator can be realized as an integral operator, i.e., an operator of 

the form 

 

(8.1) Au(x) = g(x, y)u(y)w(y) dy for x ∈ Ω. 

Ω 

The function g, defined in Ω ×Ω, is called the integral kernel of the operator 

A. The integral kernel of the resolvent of a differential operator 

is usually called Green’s function for the operator. 

Theorem 8.7. Assume g(x, y) is measurable as a function of both 

its variables and that y ↦→ g(x, y) is in L 2 (Ω, w) for a.a. x ∈ Ω. Then 

the operator A of (8.1) is a Hilbert-Schmidt operator in L 2 (Ω, w) if and

48 8. COMPACTNESS 

only if g ∈ L 2 (Ω, w) ⊗ L 2 (Ω, w), i.e., if and only if 

 

Ω×Ω 

|g(x, y)| 2 w(x)w(y) dx dy < ∞. 

Proof. Let {ej} ∞ 1 be a complete orthonormal sequence in the 

space L 2 (Ω, w). For fixed x ∈ Ω we may view Aej(x) as the j:th 

Fourier coefficient of g(x, ·) so Parseval’s formula gives |Aej(x)| 2 = 

 

Ω |g(x, y)|2 w(y) dy for a.a x ∈ Ω. By monotone convergence the product 

of this function by w is in L 1 (Ω) if and only if the Hilbert-Schmidt 

norm of A is finite. The theorem now follows by an application of 

Tonelli’s theorem (i.e., a positive, measurable function is integrable 

over Ω × Ω if and only if the iterated integral is finite). 

Example 8.8. Consider the operator T in L 2 (−π, π) with domain 

D(T ), consisting of those absolutely continuous functions u with deriva- 

tive in L 2 (−π, π) for which u(π) = u(−π), and given by T u = −i du 

dx (cf. 

Example 4.8). This operator is self-adjoint and its resolvent is given 

by Rλu(x) = π 

g(x, y, λ)u(y) dy where Green’s function g(x, y, λ) is 

−π 

given by 

⎧ 

⎪⎨ − 

g(x, y, λ) = 

⎪⎩ 

e−iλπ 

2 sin λπ eiλ(x−y) y < x, 

− eiλπ 

2 sin λπ eiλ(x−y) y > x. 

The reader should verify this! Since |g(x, y, λ)| 2 dx dy < ∞ for noninteger 

λ the resolvent is a Hilbert-Schmidt operator, so it is compact. 

Now consider the operator of Example 4.6. Green’s function is now 

only defined for non-real λ and given by 

 

Im λ i |Im λ| (8.2) g(x, y, λ) = 

eiλ(x−y) if (x − y) Im λ > 0, 

0 otherwise. 

The reader should verify this as well! In this case there is no value of λ 

for which g(·, ·, λ) ∈ L 2 (R 2 ) so the resolvent is not a Hilbert-Schmidt 

operator.



Exercise 8.1. Prove Theorem 8.2(2)–(4). 

Exercise 8.2. Show that if ∆1 and ∆2 are disjoint intervals and 

{Et}t∈R a resolution of the identity, then the ranges of E∆1 and E∆2 

are orthogonal. Generalize to the case when ∆1 and ∆2 are arbitrary 

Borel sets in R. 

Exercise 8.3. Show the converse of Theorem 8.3, i.e., if the spectrum 

consists of isolated eigen-values of finite multiplicity, then the 

resolvent is compact. 

Hint: Let λ1, λ2, . . . be the eigenvalues ordered by increasing absolute 

value and repeated according to multiplicity and let the corresponding 

normalized eigen-vectors be e1, e2, . . . . Show that Rλu2 = |〈u,ej〉| 2 

|λ−λj| 2 

and use this to see that Rλuk → 0 if uk ⇀ 0. 

Exercise 8.4. Prove the last statement of Lemma 8.6. 

Exercise 8.5. Verify all claims made in Example 8.8. 

Exercise 8.6. Let T be a selfadjoint operator. Show that if the 

resolvent Rλ of T is a Hilbert-Schmidt operator and λj, j = 1, 2, . . . 

are the non-zero eigenvalues of T , then ∞ j=1 λ−2 j < ∞.

CHAPTER 9 

Extension theory 

We will here complete the discussion on selfadjoint extensions of 

a symmetric operator begun in Chapter 4. This material is originally 

due to von Neumann although our proofs are different, and we will also 

discuss an extension of von Neumann’s theory needed in Chapter 13. 

1. Symmetric operators 

We shall find criteria for the existence of selfadjoint extensions of a 

densely defined symmetric operator, which according to the discussion 

just before Example 4.4 must be a restriction of the adjoint operator. 

We shall deal extensively with the graphs of various operators and it 

will be convenient to use the same notation for the graph of an operator 

T as for T itself. Note that if T is a closed operator on the Hilbert 

space H, then its graph is a closed subspace of H ⊕ H, so in this case 

T is itself a Hilbert space. 

Recall that with the present notation we have 

T ∗ = U(H ⊕ H) ⊖ T ) = (H ⊕ H) ⊖ UT 

according to (4.1), where U : H ⊕ H ∋ (u, v) ↦→ (−iv, iu) is the boundary 

operator introduced in Chapter 4. Also recall that U is selfadjoint, 

unitary and involutary on H ⊕ H. 

So, assume we have a densely defined symmetric operator T . We 

want to investigate what selfadjoint extensions, if any, T has. Since 

T ⊂ T ∗ the adjoint is densely defined and thus the closure T = T ∗∗ 

exists (Proposition 4.3) and is also symmetric. We may therefore as well 

assume that T is closed to begin with. Recall that if S is a symmetric 

extension of T , then it is a restriction of T ∗ since we then have T ⊂ 

S ⊂ S ∗ ⊂ T ∗ . Now put 

D±i = {U ∈ T ∗ | UU = ±U}. 

Note that U ∈ T ∗ means exactly that U = (u, T ∗ u) for some u ∈ D(T ∗ ). 

It is immediately seen that Di and D−i consist of the elements of T ∗ 

of the form (u, iu) and (u, −iu) respectively, so that u satisfies the 

equation T ∗ u = iu respectively T ∗ u = −iu. We may therefore identify 

these spaces with the deficiency spaces D±i introduced in Chapter 5. 

Also D±i are therefore called deficiency spaces. 

Theorem 9.1 (von Neumann). If T is closed and symmetric operator, 

then T ∗ = T ⊕ Di ⊕ D−i. 

51

52 9. EXTENSION THEORY 

Proof. The facts that Di and D−i are eigenspaces of the unitary 

operator U for different eigenvalues and 〈T, UT ∗ 〉 = 0 imply that T , 

Di and D−i are orthogonal subspaces of T ∗ (cf. Exercise 4.8). It 

remains to show that Di ⊕ D−i contains T ∗ ⊖ T . However, U ∈ T ∗ ⊖ T 

implies U ∈ H2 ⊖ T and thus UU ∈ T ∗ . Denoting the identity on 

H2 by I and using U 2 = I one obtains U+ = 1 

2 (I + U)U ∈ Di and 

U− = 1 

2 (I − U)U ∈ D−i. Clearly U = U+ + U− so this proves the 

theorem. 

We define the deficiency indices of T to be 

n+ = dim Di = dim Di and n− = dim D−i = dim D−i 

so these are natural numbers or ∞. We may now characterize the 

symmetric extensions of T . 

Theorem 9.2. If S is a closed, symmetric extension of the closed 

symmetric operator T , then S = T ⊕ D where D is a subspace of 

Di ⊕ D−i such that 

D = {u + Ju | u ∈ D(J) ⊂ Di} 

for some linear isometry J of a closed subspace D(J) of Di onto part of 

D−i. Conversely, every such space D gives rise to a closed symmetric 

extension S = T ⊕ D of T . 

The proof is obvious after noting that if u+, v+ ∈ Di and u−, v− ∈ 

D−i, then 〈u+, v+〉 = 〈u−, v−〉 precisely if (u+ + u−, U(v+ + v−)) = 0. 

Some immediate consequences of Theorem 9.2 are as follows. 

Corollary 9.3. The closed symmetric operator T is maximal symmetric 

precisely if one of n+ and n− equals zero and selfadjoint precisely 

if n+ = n− = 0. 

Corollary 9.4. If S is the symmetric extension of the closed symmetric 

operator T given as in Theorem 9.2 by the isometry J with domain 

D(J) ⊂ Di and range RJ ⊂ D−i, then the deficiency spaces for 

S are Di(S) = Di ⊖ D(J) and D−i(S) = D−i ⊖ RJ respectively. 

Proof. If D ⊂ Di ⊕ D−i and S = T ⊕ D is symmetric, then 

u ∈ Di(S) ⊂ Di precisely if 〈T ⊕ D, Uu〉 = 0. But 〈T, Uu〉 = 0 and 

if u+ + u− ∈ D with u+ ∈ Di, u− ∈ D−i then 〈u+ + u−, u〉 = 〈u+, u〉 

which shows that Di(S) = Di ⊖ D(J). Similarly the statement about 

D−i(S) follows. 

Corollary 9.5. Every symmetric operator has a maximal symmetric 

extension. If one of n+ and n− is finite, then all or none of the 

maximal symmetric extensions are selfadjoint depending on whether 

n+ = n− or not. If n+ = n− = ∞, however, some maximal symmetric 

extensions are selfadjoint and some are not.

2. SYMMETRIC RELATIONS 53 

We will now generalize Theorem 9.1. To do this, we use the notation 

of Lemma 5.1. Define 

Dλ = {(u, λu) ∈ T ∗ } = {(u, λu) | u ∈ Dλ} 

Eλ = {(u, λu + v) ∈ T ∗ | v ∈ D λ } . 

It is clear that Eλ for non-real λ is the direct sum of Dλ and D λ since 

if a = i 

we have (u, λu + v) = a(v, λv) + (u − av, λ(u − av)). This 

2 Im λ 

direct sum is topological (i.e., the projections from Eλ onto Dλ and 

Dλ are bounded) since all three spaces are obviously closed. Thus 

the assertion follows from the closed graph theorem. Carry out the 

argument as an exercise! We can now prove the following theorem. 

Theorem 9.6. For any non-real λ we have T ∗ = T ˙+Eλ as a topological 

direct sum. 

Proof. Since all involved spaces are closed it is enough to show 

the formula algebraically (the reason is as above). Let (u, v) ∈ T ∗ . By 

Lemma 5.1.1 H = Sλ ⊕ D λ so we may write v − λu = w0 + w λ with 

w λ ∈ D λ and w0 ∈ Sλ. We can find u0 ∈ H such that (u0, λuo+w0) ∈ T 

so (u, v) = (u0, λu0 + w0) + (u − u0, λ(u − u0) + w λ ). The last term is 

obviously in Eλ. 

If (u, v) ∈ T ∩ Eλ we have v − λu ∈ Sλ ∩ D λ = {0} so that λ is an 

eigenvalue of T if u = 0. 

Corollary 9.7. If Im λ > 0 then dim Dλ = n+, dim D λ = n−. 

Proof. Suppose U = (u, T ∗ u) and V = (v, T ∗ v) are in T ∗ . The 

boundary form 

〈U, UV 〉 = i(〈u, T ∗ v〉 − 〈T ∗ u, v〉) 

is a bounded Hermitian form on T ∗ . It is immediately verified that 

it is positive definite on Dλ, negative definite on D λ , non-positive on 

T ˙+Dλ and non-negative on T ˙+D λ . 

Let µ be a complex number with Im µ > 0. We get a linear map 

of Dµ into Dλ in the following way. Given u ∈ Dµ we may write 

u = u0 +uλ +u λ uniquely with u0 ∈ T , uλ ∈ Dλ and u λ ∈ D λ according 

to Theorem 9.6. Let the image of u in Dλ be uλ. Then uλ can not be 0 

unless u is since the boundary form is positive definite on Dµ but nonpositive 

on T ˙+D λ . It follows that dim Dµ ≤ dim Dλ. By symmetry the 

dimensions of Dλ and Dµ are then equal, i.e., dim Dλ = n+. Similarly 

one shows that dim D λ = n−. 

2. Symmetric relations 

This section is a simplified version of Section 1 of [2]. Most of it can 

also be found in [1]. The theory of symmetric and selfadjoint relations 

is an easy extension of the corresponding theory for operators, but will 

be essential for Chapters 13 and 14.


We call a (closed) linear subspace T of H 2 = H⊕H a (closed) linear 

relation on H. This is a generalization of the concept of (the graph 

of) a linear operator which will turn out to be useful in the following 

chapters. We still denote by U the boundary operator on H 2 and define 

the adjoint of the linear relation T on H by 

T ∗ = H 2 ⊖ UT = U(H 2 ⊖ T ) . 

Clearly T ∗ is a closed linear relation on H. Note that by not insisting 

that T and T ∗ are graphs we can, for example, now deal with adjoints 

of non-densely defined operators. Naturally T is called symmetric if 

T ⊂ T ∗ and selfadjoint if T = T ∗ . 

Proposition 9.8. Let T ⊂ S be linear relations on H. Then S ∗ ⊂ 

T ∗ . The closure of T is T = T ∗∗ and (T ) ∗ = T ∗ . 

The reader should prove this proposition as an exercise. It is very 

easy to obtain a spectral theorem for selfadjoint relations as a corollary 

to the spectral theorem of Chapter 7. Given a relation T we call the set 

D(T ) = {u ∈ H | (u, v) ∈ T for some v ∈ H} the domain of T . Now let 

HT be the closure of D(T ) in H and put H∞ = {u ∈ H | (0, u) ∈ T ∗ } 

One may view H∞ as the ‘eigen-space’ of T ∗ corresponding to the 

‘eigen-value’ ∞. 

Proposition 9.9. H = HT ⊕ H∞. 

Proof. We have 〈(u, v), U(0, w)〉 = i〈u, w〉 so that (0, w) ∈ T ∗ 

precisely when w ∈ H ⊖ D(T ). The proposition follows. 

Now assume T is selfadjoint and put T∞ = {0} × H∞ and ˜ T = 

T ∩ H 2 T . Then it is clear that T = ˜ T ⊕ T∞ so we have split T into 

its many-valued part T∞ and ˜ T which is called the operator part of T 

because of the following theorem. 

Theorem 9.10 (Spectral theorem for selfadjoint relations). If T is 

selfadjoint, then ˜ T is the graph of a densely defined selfadjoint operator 

in HT with domain D(T ). 

Proof. ˜ T is the graph of a densely defined operator on HT since 

(0, w) ∈ ˜ T implies w ∈ H∞ ∩ HT = {0}. ˜ T is selfadjoint since ˜ T = 

T ⊖ T∞ so its adjoint (in HT ) is ˜ T ∗ = H 2 T ∩ (T ∗ ⊕ UT∞) = H 2 T ∩ T = ˜ T 

(check this calculation carefully!). 

It is now clear that we get a resolution of the identity for T by 

adjoining the orthogonal projector onto H∞ to the resolution of the 

identity for ˜ T . 

Assume we have a symmetric relation T . We want to investigate 

what selfadjoint extensions, if any, T has. Since the closure T of T is 

also symmetric we may as well assume that T is closed to begin with. 

Just as is the case for operators, if S is a symmetric extension of T ,

2. SYMMETRIC RELATIONS 55 

then it is a restriction of T ∗ since we then have T ⊂ S ⊂ S ∗ ⊂ T ∗ . 

Now put 

D±i = {u ∈ T ∗ | Uu = ±u} . 

It is immediately seen that Di and D−i consist of the elements of T ∗ of 

the form (u, iu) and (u, −iu) respectively. We call them the deficiency 

spaces of T . The following generalizes von Neumann’s formula. 

Theorem 9.11. For any closed and symmetric relation T holds 

T ∗ = T ⊕ Di ⊕ D−i. 

The proof is the same as for Theorem 9.1 and is left to Exercise 9.6 

As before we define the deficiency indices of T to be 

n+ = dim Di = dim Di and n− = dim D−i = dim D−i 

so these are again natural numbers or ∞. The next theorem is completely 

analogous to Theorem 9.2 with essentially the same proof, so 

we leave this as Exercise 9.7 

Theorem 9.12. If S is a closed, symmetric extension of the closed 

symmetric relation T , then S = T ⊕ D where D is a subspace of Di ⊕ 

D−i such that 

D = {u + Ju | u ∈ D(J) ⊂ Di} 

for some linear isometry J of a closed subspace D(J) of Di onto part of 

D−i. Conversely, every such space D gives rise to a closed symmetric 

extension S = T ⊕ D of T . 

The following consequences of Theorem 9.12 are completely analogous 

to Corollaries 9.3–9.5, and their proofs are left as Exercise 9.8 

Some immediate consequences of 

Corollary 9.13. The closed symmetric relation T is maximal 

symmetric precisely if one of n+ and n− equals zero and selfadjoint 

precisely if n+ = n− = 0. 

Corollary 9.14. If S is the symmetric extension of the closed 

symmetric relation T given as in Theorem 9.12 by the isometry J with 

domain D(J) ⊂ Di and range RJ ⊂ D−i, then the deficiency spaces 

for S are Di(S) = Di ⊖ D(J) and D−i(S) = D−i ⊖ RJ respectively. 

Corollary 9.15. Every symmetric relation has a maximal symmetric 

extension. If one of n+ and n− is finite, then all or none of the 

maximal symmetric extensions are selfadjoint depending on whether 

n+ = n− or not. If n+ = n− = ∞, however, some maximal symmetric 

extensions are selfadjoint and some are not. 

We will now prove a theorem generalizing Theorem 9.6. To do this, 

first note that Lemma 5.1 remains valid for relations, with the obvious


definitions of Sλ and Dλ and identical proofs. We now define 

Dλ = {(u, λu) ∈ T ∗ } = {(u, λu) | u ∈ Dλ} 

Eλ = {(u, λu + v) ∈ T ∗ | v ∈ D λ } . 

As before it is clear that Eλ for non-real λ is the direct sum of Dλ and 

Dλ since if a = i we have (u, λu+v) = a(v, λv)+(u−av, λ(u−av)) 

2 Im λ 

and that this direct sum is topological (i.e., the projections from Eλ 

onto Dλ and Dλ are bounded). 

Theorem 9.16. For any non-real λ holds T ∗ = T ˙+Eλ as a topological 

direct sum. 

Corollary 9.17. If Im λ > 0 then dim Dλ = n+, dim D λ = n−. 

The proofs of Theorems 9.16 and 9.17 is the same as for Theorems 

9.6 and 9.7 respectively, and are left as exercises.



Exercise 9.1. Fill in all missing details in the proofs of Theorem 

9.2 and Corollaries 9.3–9.5. 

Exercise 9.2. Show that if n+ = n− < ∞, then if one selfadjoint 

extension of a symmetric operator has compact resolvent, then every 

other selfadjoint extension also has compact resolvent. 

Hint: The difference of the resolvents for two selfadjoint extensions of 

a symmetric operator has range contained in Dλ. 

Exercise 9.3. Suppose T is a closed and symmetric operator on 

H, that λ ∈ R and that Sλ is closed. Show that if λ is not an eigenvalue 

of T , then T ∗ is the topological direct sum of T and Eλ and that 

n+ = n−. 

You may also show that if Sλ is closed but λ is an eigen-value of T , 

then one still has n+ = n−. 

Exercise 9.4. Suppose T is a symmetric and positive operator, 

i.e., 〈T u, u〉 ≥ 0 for every u ∈ D(T ). Use the previous exercise to show 

that T has a selfadjoint extension (this is a theorem by von Neumann). 

Exercise 9.5. Suppose T is a symmetric and positive operator. By 

the previous exercise T has at least one selfadjoint extension. Prove 

that there exists a positive selfadjoint extension (the so called Friedrichs 

extension). This is a theorem by Friedrichs. 

Hint: First define [u, v] = 〈T u, v〉 + 〈u, v〉 for u, v ∈ D(T ), show that 

this is a scalar product, and let H1 be the completion of D(T ) in the 

corresponding norm. Next show that H1 may be identified with a 

subset of H and that for any u ∈ H the map H1 ∋ v ↦→ 〈v, u〉 is a 

bounded linear form on H1. Conclude that 〈u, v〉 = [u, Gv] for u ∈ H1 

and v ∈ H, where G is an operator on H with range in H1. Finally 

show that G −1 − I, where I is the identity, is a positive selfadjoint 

extension of T . 



Exercise 9.8. Prove Corollaries 9.13–9.15. 


Exercise 9.10. Prove Theorem 9.17.

CHAPTER 10 

Boundary conditions 

A simple example of a formally symmetric differential equation is 

given by the general Sturm-Liouville equation 

(10.1) −(pu ′ ) ′ + qu = wf. 

Here the coefficients p, q and w are given real-valued functions in a 

given interval I. Standard existence and uniqueness theorems for the 

initial value problem are valid if 1/p, q and w are all in Lloc(I). There 

are (at least) two Hermitian forms naturally associated with this equa- 

tion, namely 

I (pu′ v ′ + quv) and 

I 

uvw. Under appropriate positivity 

conditions either of these forms is a suitable choice of scalar product for 

a Hilbert space in which to study (10.1). The corresponding problems 

are then called left definite and right definite respectively. We will not 

discuss left definite problems in these lectures. 

If p is not differentiable it is most convenient to interpret (10.1) as 

a first order system 

 

0 1 

U 

−1 0 

′ 

q 0 

+ 

0 − 1 

 

w 0 

U = V . 

p 0 0 

 

u 

This equation becomes equivalent to (10.1) on setting U = 

−pu ′ 

 

and letting the first component of V be f. It is a special case of a 

fairly general first order system 

(10.2) Ju ′ + Qu = W v 

where J is a constant n × n matrix which is invertible and skew- 

Hermitian (i.e., J ∗ = −J) and the coefficients Q and W are n × n 

matrix-valued functions which are locally integrable on I. In addition 

Q is assumed Hermitian and W positive semi-definite, and u, v are 

n × 1 matrix-valued functions. We shall study such systems in Chapters 

13–15. 

Here we shall just deal with the case of the simple inhomogeneous 

scalar Sturm-Liouville equation 

(10.3) −u ′′ + qu = λu + f 

and the corresponding homogeneous eigenvalue problem −u ′′ +qu = λu. 

The latter is often called the one-dimensional Schrödinger equation. In 

later chapters we shall then see that with minor additional technical 

complications we may deal with the first order system (10.2) in much 

59

60 10. BOUNDARY CONDITIONS 

the same way. This will of course include the more general Sturm- 

Liouville equation (10.1). 

We shall study (10.3) in the Hilbert space L 2 (I) where I is an 

interval and the function q is real-valued and locally integrable in I, 

i.e., integrable on every compact subinterval of I. In L 2 (I) the scalar 

product is 〈u, v〉 = 

I uv. 

Before we begin we need to quote a few standard facts about Sturm- 

Liouville equations. Basic for what follows is the following existence 

and uniqueness theorem. 

Theorem 10.1. Suppose q is locally integrable in an interval I and 

that c ∈ I. Then, for any locally integrable function f and arbitrary 

complex constants A, B and λ the initial value problem 

 

−u ′′ + qu = λu + f in I, 

u(c) = A, u ′ (c) = B 

has a unique, continuously differentiable solution u with locally absolutely 

continuous derivative defined in I. If A, B are independent of λ 

the solution u(x, λ) and its x-derivative will be entire functions of λ, 

locally uniformly in x. 

We shall use this only if f is actually locally square integrable. The 

theorem has the following immediate consequence. 

Corollary 10.2. Let q, λ and I be as in Theorem 10.1. Then the 

set of solutions to −u ′′ + qu = λu in I is a 2-dimensional linear space. 

If one rewrites −u ′′ + qu = λu + f as a first order system according 

to the prescription before (10.1), then Theorem 10.1 and Corollary 10.2 

become special cases of the theorems for first order systems given in 

Appendix C. 

In order to get a spectral theory for (10.3) we need to define a minimal 

operator, show that it is densely defined and symmetric, calculate 

its adjoint and find the selfadjoint restrictions of the adjoint. 

We define Tc to be the operator u ↦→ −u ′′ +qu with domain consisting 

of those continuously differentiable functions u which have compact 

support, i.e., they are zero outside some compact subinterval of the interior 

of I, and which are such that u ′ is locally absolutely continuous 

with −u ′′ + qu ∈ L 2 (I). 

We will show that Tc is densely defined and symmetric and then 

calculate its adjoint, but first need some preparation. If u, v are differentiable 

functions we define [u, v] = u(x)v ′ (x) − u ′ (x)v(x). This is 

called the Wronskian of u and v. It is clear that [u, v] = −[v, u], in 

particular [u, u] = 0. The following elementary fact is very important. 

Proposition 10.3. If u and v are linearly independent solutions of 

−v ′′ + qv = λv on I, then the Wronskian [u, v] is a non-zero constant 

on I.

10. BOUNDARY CONDITIONS 61 

Proof. Differentiating we obtain [u, v] ′ = uv ′′ − u ′′ v = u(q − λ)v − 

(q−λ)uv = 0 so that the Wronskian is constant. If the constant is zero, 

then given any point c ∈ I the vectors (u(c), u ′ (c)) and (v(c), v ′ (c)) are 

proportional. Since the initial value problem has a unique solution this 

implies that u and v are proportional. 

Now let v1 and v2 be solutions of −v ′′ + qv = λv in I such that 

[v1, v2] = 1 in I. There are certainly such solutions. We may for 

example pick a point c ∈ I and specify initial values v1(c) = 1, v ′ 1(c) = 0 

respectively v2(c) = 0, v ′ 2(c) = 1. By Proposition 10.3 the Wronskian 

is constant equal to its value at c, which is 1. The following theorem 

states a version of the classical method known as variation of constants 

for solving the inhomogeneous equation in terms of the solutions of the 

homogeneous equation. 

Lemma 10.4. Let v1, v2 be solutions of −v ′′ +qv = λv with [v1, v2] = 

1, let c ∈ I and suppose f is locally integrable in I. Then the solution 

u of −u ′′ + qu = λu + f with initial data u(c) = u ′ (c) = 0 is given by 

x 

x 

(10.4) u(x) = v1(x) v2(y)f(y) dy − v2(x) v1(y)f(y) dy. 

c 

Proof. With u given by (10.4) clearly u(c) = 0. Differentiating 

we obtain 

u ′ (x) = v ′ x 

1(x) v2f − v ′ x 

2(x) v1f, 

c 

since the two other terms obtained cancel. Thus u ′ (c) = 0. Differentiating 

again we obtain 

u ′′ (x) = v ′′ 

x 

1(x) v2f−v ′′ 

x 

2(x) v1f−[v1, v2]f(x) = (q(x)−λ)u(x)−f(x), 

c 

c 

which was to be proved. 

Corollary 10.5. If f ∈ L 1 (I) with compact support in I then 

−u ′′ + qu = f has a solution u with compact support in I if and only if 

 

I vf = 0 for all solutions v of the homogeneous equation −v′′ +qv = 0. 

Proof. If we choose c to the left of the support of f, then by 

Lemma 10.4 the function u given by (10.4) is the only solution of 

−u ′′ + qu = f which vanishes to the left of c. Since v1, v2 are linearly 

independent the equation has a solution of compact support if and only 

if f is orthogonal to both v1 and v2, which are a basis for the solutions 

of the homogeneous equation. The corollary follows. 

Lemma 10.6. The operator Tc is densely defined and symmetric. 

Furthermore, if u ∈ D(T ∗ c ) and f = T ∗ c u, then u is differentiable with 

c 

c


locally absolutely continuous derivative and satisfies −u ′′ + qu = f. 

Conversely, if u, f ∈ L 2 (I) and this equation is satisfied, then u ∈ 

D(T ∗ c ) and T ∗ c u = f. 

Proof. Let u1 be a solution of −u ′′ 

1 + qu1 = f. Assume u0 is in 

the domain of Tc and put f0 = Tcu0. Integrating by parts twice we get 

 

 

 

(10.5) 

I 

u0f = 

I 

u0(−u ′′ 

1 + qu1) = 

I 

(−u ′′ 

0 + qu0)u1 = 

I 

f0u1. 

So, if f is orthogonal to the domain of Tc, then u1 is orthogonal to all 

compactly supported elements f0 ∈ L2 (I) for which there is a solution 

u0 of −u ′′ 

0 + qu0 = f0 with compact support. By Corollary 10.5 it 

follows that u1 solves −v ′′ + qv = 0 so that f = 0. Thus Tc is densely 

defined. 

The calculation (10.5) also proves the converse part of the lemma. 

Furthermore, if u is in the domain of T ∗ c with T ∗ c u = f we obtain 

0 = 〈u0, f〉 − 〈f0, u〉 = 〈f0, u1 − u〉. Just as before it follows that u1 − u 

solves the equation −v ′′ + qv = 0. It follows that u solves the equation 

−u ′′ + qu = f so that Tc ⊂ T ∗ c . The proof is complete. 

Being symmetric and densely defined Tc is closeable, and we define 

the minimal operator T0 as the closure of Tc and denote the domain 

of T0 (the minimal domain) by D0. Similarly, the maximal operator 

T1 is T1 := T ∗ c with domain D1 ⊃ D0. Thus the maximal domain D1 

consists of all differentiable functions u ∈ L 2 (I) such that u ′ is locally 

absolutely continuous function for which T1u = −u ′′ + qu ∈ L 2 (I). 

We can now apply the theory of Chapter 9. The deficiency indices 

of T0 are accordingly the number of solutions of −u ′′ + qu = iu and 

−u ′′ + qu = −iu respectively which are linearly independent and in 

L 2 (I). Since there are only 2 linearly independent solutions for each of 

these equations the deficiency indices can be no larger than 2. For the 

equation (10.3) the deficiency indices are always equal, since if u solves 

−u ′′ + qu = λu, then u solves the equation with λ replaced by λ, and 

linear independence is preserved when conjugating functions. Thus, for 

our equation there are only three possibilities: The deficiency indices 

may both be 2, both may be 1, or both may be 0. We shall see later 

that all three cases can occur, depending on the choice of q and I. 

We will now take a closer look at how selfadjoint realizations are 

determined as restrictions of the maximal operator. Suppose u1 and 

u2 ∈ D1. Then the boundary form (cf. Chapter 9) is 

 

(10.6) 〈(u1, T1u1), U(u2, T1u2)〉 = i (u1T1u2 − T1u1u2) 

 

= i 

I 

(−u1u ′′ 

2 + u ′′ 

 

1u2) = −i 

I 

I 

[u1, u2] ′ 

 

= −i lim [u1, u2] , 

K→I 

K


the limit being taken over compact subintervals K of I. We must 

restrict T1 so that this vanishes. In some sense this means that the 

restriction of T1 to a selfadjoint operator T is obtained by boundary 

conditions since the limit clearly only depends on the values of u1 and 

u2 in arbitrarily small neighborhoods of the endpoints of I. This is of 

course the motivation for the terms boundary operator and boundary 

form. 

The simplest case is when an endpoint is an element of I. This 

means that the endpoint is a finite number, and that q is integrable 

near the endpoint. Such an endpoint is called regular; otherwise the 

endpoint is singular. If both endpoints are regular, we say that we are 

dealing with a regular problem. We have a singular problem if at least 

one of the endpoints is infinite, or if q /∈ L 1 (I). 

Consider now a regular problem. It is clear that the deficiency indices 

are both 2 in the regular case, since all solutions of −u ′′ +qu = iu 

are continuous on the compact interval I and thus in L 2 (I). We 

shall investigate which boundary conditions yield selfadjoint restrictions 

of T1. The boundary form depends only on the boundary values 

(u(a), u ′ (a), u(b), u ′ (b)), and the possible boundary values constitute a 

linear subspace of C 4 . On the other hand, the boundary form is positive 

definite on Di and negative definite on D−i, both of which are 

2-dimensional spaces. The boundary values for the deficiency spaces 

therefore span two two-dimensional spaces which do not overlap. It follows 

that as u ranges through D1 the boundary values range through 

all of C 4 . 

The boundary conditions need to restrict the 4-dimensional space 

Di⊕D−i to the 2-dimensional space D of Theorem 9.2, so two independent 

linear conditions are needed. This means that there are 2 × 2 ma- 

trices A and B such that the boundary conditions are given by AU(a)+ 

u 

BU(b) = 0, where U = 

−u ′ 

 

. Linear independence of the conditions 

means that the 2 × 4 matrix (A, B) must have linearly independent 

rows. Consider first the case when A is invertible. Then the condition 

is of the form U(a) = SU(b), where S = −A−1 0 1 

B. If J = ( −1 0 ) the 

boundary form is −i{(U2(a)) ∗JU1(a) − (U2(b)) ∗JU1(b)}, so symmetry 

requires this to vanish. Inserting U(a) = SU(b) the condition becomes 

(U2(b)) ∗ (S∗JS − J)U1(b) = 0 where U1(b) and U2(b) are arbitrary 2 × 1 

matrices. Thus it follows that the condition U(a) = SU(b) gives a 

selfadjoint restriction of T1 precisely if S satisfies S∗JS = J. Such a 

matrix S is called symplectic. 

Important special cases are when S is plus or minus the unit matrix. 

These cases are called periodic and antiperiodic boundary conditions 

respectively. Another valid choice is S = J. Since det J = 1 = 0 it 

is clear that any symplectic matrix S satisfies | det S| = 1 (see also 

Exercise 10.1). In particular, it is invertible. It is clear that the inverse


of a symplectic matrix is also symplectic (show this!), so it follows 

that assuming the matrix B to be invertible again leads to boundary 

conditions of the form U(a) = SU(b) with a symplectic S. 

It remains to consider the case when neither A nor B is invertible. 

Neither A nor B can then be zero, since then the other matrix must 

be invertible. Thus A and B both have linearly dependent rows, one 

of which has to be non-zero. We may assume the first row in A to be 

non-zero, and then adding an appropriate multiple of the first row to 

the second in (A, B) we may assume the second row of A to be zero. 

The second row of B will then be non-zero since the rows of (A, B) 

are linearly independent, and then adding an appropriate multiple of 

the second row to the first we may cancel the first row of B. At this 

point the first row gives a condition on U(a) and the second a condition 

on U(b). Such boundary conditions are called separated. We end the 

discussion of the regular case by determining what separated boundary 

conditions give rise to selfadjoint restrictions of T1. 

Separated boundary conditions require u1u ′ 2 − u ′ 1u2 to vanish in 

each endpoint. One possibility is of course to require u1 (and u2) to 

vanish there. Such a boundary condition is called a Dirichlet condition. 

If there is an element u1 in the domain of the selfadjoint realization 

for which u1(a) does not vanish in a we obtain for u2 = u1 

that 0 = u ′ 1(a)u1(a) − u1(a)u ′ 1(a) = 2i Im u ′ 1(a)u1(a) so that u ′ 1(a)u1(a) 

is real. Equivalently u ′ 1(a)/u1(a) is real, say = −h ∈ R. If u is 

any other element of the domain the condition for symmetry becomes 

0 = u ′ (a)u1(a)−u(a)u ′ 1(a) = (u ′ (a)+hu(a))u1(a) so that we must have 

u ′ (a) + hu(a) = 0. On the other hand, imposing this condition on all 

elements of the maximal domain clearly makes the boundary form at a 

vanish. In particular, if h = 0 we have a Neumann boundary condition. 

We may of course find α ∈ (0, π) such that h = cot α, and multiplying 

through by sin α the boundary condition becomes 

(10.7) u(a) cos α + u ′ (a) sin α = 0, 

and then α = 0 gives a Dirichlet condition. For α = π/2 we obtain a 

Neumann condition, and any separated, selfadjoint boundary condition 

at a is given by (10.7) for some α ∈ [0, π). 

To summarize: Separated, symmetric boundary conditions for a 

Sturm-Liouville equation are of the form (10.7) at a with a similar 

condition at b (possibly for a different value of α, of course). Important 

special cases are α = 0, a Dirichlet condition, and α = π/2, a 

Neumann condition. Every other selfadjoint realization is given by 

coupled boundary conditions U(a) = SU(b) for a symplectic matrix S. 

Important special cases are periodic and antiperiodic boundary conditions.


Let us now consider the singular case. We will then first consider 

the case when one endpoint is regular and the other singular. So, 

assume that I = [a, b) with a regular and b possibly singular. 

Lemma 10.7. There are elements of D1 which vanish in a neighborhood 

of b and have arbitrarily prescribed initial values u(a) and u ′ (a). 

Proof. Let c ∈ (a, b) and f ∈ L2 (a, b) vanish in (c, b). Now solve 

−u ′′ + qu = f with initial data u(c) = u ′ (c) = 0 so that u vanishes in 

(c, b). It follows that u ∈ D1, and we need to show that u(a) and u ′ (a) 

can be chosen arbitrarily by selection of f. Note that if −v ′′ + qv = 0 

integrating by parts twice shows that 

c 

〈f, v〉 = (−u ′′ + qu)v = [−u ′ v + uv ′ ] c a = u ′ (a)v(a) − u(a)v ′ (a). 

a 

If v1 and v2 are solutions of −v ′′ +qv = 0 satisfying v1(a) = 1, v ′ 1(a) = 0 

and v2(a) = 0, v ′ 2(a) = −1 respectively, we obtain u(a) = 〈f, v2〉 and 

u ′ (a) = 〈f, v1〉. Since v1, v2 are linearly independent we can choose 

f to give arbitrary values to this, for example by choosing f as an 

appropriate linear combination of v1 and v2 in [a, c]. 

The fact that T1 and T0 are closed means that their domains are 

Hilbert spaces with norm-square u 2 1 = u 2 +T1u 2 . We shall always 

view D1 and D0 as spaces in this way. We also note that if u ∈ D1, then 

u is continuously differentiable. If K is a compact interval we define 

C 1 (K) to be the linear space of continuously differentiable functions 

provided with the norm uK = sup K |u|+sup K |u ′ |. Convergence for a 

sequence {uj} ∞ 1 in this space therefore means uniform convergence on 

K of uj and u ′ j as j → ∞. This space is easily seen to be complete, and 

thus a Banach space. As we noted above, if K is a compact subinterval 

of I, then the restriction to K of any element of D1 is in C 1 (K). We 

will need the following fact. 

Lemma 10.8. For every compact subinterval K ⊂ I there exists a 

constant CK such that uK ≤ CKu1 for every u ∈ D1. In particular, 

the linear forms D1 ∋ u ↦→ u(x) and D1 ∋ u ↦→ u ′ (x) are locally 

uniformly bounded in x. 

Proof. The restriction map D1 ∋ u ↦→ u ∈ C 1 (K) is linear and 

we will show that this map is closed. By the closed graph theorem (see 

Appendix A) it then follows that this map is bounded, which is the 

statement of the lemma. 

To show that the map is closed we must show that if uj → u in 

D1 and the restrictions to K of uj converge to ũ in C 1 (K), then the 

restriction to K of u equals ũ. But this is clear, since if uj converges 

in L 2 (I) to u, then their restrictions to K converge in L 2 (K) to the 

restriction of u to K. At the same time the restrictions to K converge


uniformly to ũ, so that 

K |uj−u| 2 converges both to 0 and to 

K |ũ−u|2 . 

It follows that u = ũ a.e. in K. 

A bounded Hermitian form on a Hilbert space H is a map H × H ∋ 

(u, v) ↦→ B(u, v) ∈ C such that |B(u, v)| ≤ Cuv for some constant 

C. It is clear that the boundedness of a Hermitian form is equivalent 

to it being continuous as a function of its arguments. The boundary 

form i(〈u, T1v〉 − 〈T1u, v〉) is a bounded Hermitian form on D1, i.e., 

it is a Hermitian form in u, v and is bounded by u1v1, and by 

Lemma 10.8 the boundary form at a, i.e., u ′ (a)v(a) − u(a)v ′ (a), is also 

a bounded Hermitian form (bounded by 2CKu1v1 if a ∈ K). Since 

i(〈u, T1v〉 − 〈T1u, v〉) = −i lim 

x→b [u, v](x) + i[u, v](a) 

we see that i limx→b(u ′ (x)v(x) − u(x)v ′ (x)), the boundary form at b, 

is also a bounded Hermitian form on D1. Since the forms at a and b 

vanish if u is in the domain of Tc, i.e., if u vanishes near a and b, it 

follows that they also vanish if u ∈ D0. In particular, if u ∈ D0, then 

u(a) = u ′ (a) = 0. Now T0 is the adjoint of T1 so it follows that this 

is the only condition at a for an element of D1 to be in D0, since this 

guarantees that the form at a vanishes. Of course, u ∈ D0 also requires 

that the form at b vanishes. 

Now let Ta be the closure of the restriction of T1 to those elements 

of D1 which vanish near b, and let Da be the domain of Ta. 

Then Lemma 10.7 and the boundedness of the forms at a and b show 

that the boundary form at b vanishes on Da and that dim D1/D0 ≥ 

dim Da/D0 ≥ 2. We obtain the following theorem. 

Theorem 10.9. If the interval I has one regular endpoint a, then 

n+ = n− ≥ 1. If n+ = n− = 1, then the boundary form at the singular 

endpoint vanishes on D1, and any selfadjoint restriction of T1 is given 

by a boundary condition of the form (10.7) at a and no condition at all 

at b. 

Proof. If n+ = n− = 0 then T1 = T0 so that T1 is selfadjoint, 

according to Theorem 9.1. But then we can not have dim D1/D0 ≥ 2. 

If n+ = n− = 1, then 2 = dim D1/D0 ≥ dim Da/D0 ≥ 2 so that we 

must have D1 = Da. Thus the boundary form at the singular endpoint 

vanishes on D1, and the boundary form at a vanishes precisely if we 

impose a boundary condition of the form (10.7). 

If n+ = n− = 2 we obtain a selfadjoint restriction of T1 by imposing 

two appropriate boundary conditions. One of them can be a condition 

of the form (10.7), and then a condition at the singular endpoint also 

has to be imposed. There are also selfadjoint restrictions obtained by 

imposing coupled boundary conditions. See Exercise 10.3. 

Whether one obtains deficiency indices 1 or 2 when one endpoint is 

regular clearly only depends on conditions near the singular endpoint.


It is customary to say that a singular endpoint is in the limit point condition 

if the deficiency indices are 1 and in the limit circle condition if 

the deficiency indices are 2. The terminology derives from the methods 

Weyl [12] used in 1910 to construct the resolvent of a Sturm-Liouville 

operator. 

If an interval has two singular endpoints in the limit point condition 

it is clear that T1 is selfadjoint, since the boundary form vanishes on 

D1. No boundary conditions are therefore required in this case, and 

one often says that the operator Tc is essentially selfadjoint, since its 

closure T0 = T1 is selfadjoint. If one or both of the endpoints are in the 

limit circle condition, we have a situation similar to when the endpoint 

is regular, and need to impose boundary conditions of a similar type. 

Note, however, that at a limit circle endpoint the limits of u(x) and 

u ′ (x) for an element u ∈ D1 do not necessarily exist. To formulate the 

boundary conditions in explicit terms one may instead use the idea of 

Exercise 10.3. 

It is clearly an important problem to find explicit conditions on the 

interval and the coefficient q which guarantee limit point or limit circle 

conditions. A large number of different criterions for this are available 

today. We end the chapter by proving a simple criterion, known already 

to Weyl (with a more complicated proof), for the limit point condition 

at an infinite interval endpoint. 

Theorem 10.10. Suppose q is bounded from below near ∞. Then 

(10.3) is in the limit point condition at ∞. 

Proof. Suppose q > C on [a, ∞). Let u be the solution of −u ′′ + 

qu = (C + i)u with initial data u(a) = 1, u ′ (a) = 0. We have 

(Re(u ′ u)) ′ = Re(|u ′ | 2 + u ′′ (x)u(x)) 

= |u ′ | 2 + Re((q − C − i)|u| 2 = |u ′ | 2 + (q − C)|u| 2 ≥ 0. 

Thus Re(u ′ u) is increasing and Re(u ′ (0)u(0)) = 0 so Re(u ′ u) ≥ 0. But 

(|u| 2 ) ′ = 2 Re(u ′ u) so |u| 2 is increasing. It follows that |u| 2 ≥ 1 so that 

one can not have u ∈ L 2 (0, ∞). Thus deficiency indices are < 2. 

For a more general result, see Exercise 10.4. 


Exercise 10.1. Show that a symplectic 2 × 2 matrix S is of the 

form e iθ P where θ ∈ R and P is a real 2 × 2 matrix with determinant 

1. Also show that the inverses and adjoints of symplectic matrices are 

symplectic. 

Exercise 10.2 (Hard!). Suppose that all solutions of −v ′′ +qv = λv 

are in L 2 (I) for some real or non-real λ. Show that this is then true 

for all complex λ.


Hint: If −u ′′ + qu = µu, write this as −u ′′ + (q − λ)u = (µ − λ)u 

and use the variation of constants formula, thinking of (µ − λ)u as an 

inhomogeneous term, to write down an integral equation for u in terms 

of solutions of −v ′′ + qv = λv. Using an initial point sufficiently close 

to an endpoint use estimates in this integral equation to show that u 

is square integrable near the endpoint. 

Exercise 10.3. Show that 

[u1, v1][u2, v2] − [u1, v2][u2, v1] = [u1, u2][v1, v2] 

for differentiable functions u1, u2, v1, v2. Next show that if [v1, v2] = 1, 

then the boundary form for u1, u2 ∈ D1 at b equals limb([u1, v1][u2, v2]− 

[u1, v2][u2, v1]). Furthermore, show that if −v ′′ + qv = λv and −u ′′ + 

qu = f then ([u, v]) ′ = (f − λu)v. Finally show that if all solutions of 

−v ′′ + qv = λv are in L 2 (a, b) and if u ∈ D1 then the limit at b of [u, v] 

exists. 

Conclude that in the case n+ = n− = 2 selfadjoint boundary conditions 

may be described by conditions on the values of [u, v1], [u, v2] at 

the endpoints of exactly the same form as we described them for the 

regular case on the values of u, u ′ . 

Exercise 10.4. Show that (10.3) is in the limit point condition at 

∞ if q = q0 + q1 where q0 is bounded from below and q1 ∈ L 1 (0, ∞). 

Hint: First show that |u(x)| 2 ≤ 2 x 

0 |u′ ||u| if u(0) = 0, then multiply 

by |q1| and integrate. Conclude that there is a constant A so that 

x 

0 |q1||u| 2 ≤ x 

0 |u′ | 2 + A x 

0 |u|2 for all x > 0. Now show, similar to the 

proof of Theorem 10.10, that |u| 2 is increasing if u(0) = 0, u ′ (0) = 1 

and u satisfies −u ′′ + qu = λu for an appropriate λ.

CHAPTER 11 

Sturm-Liouville equations 

The spectral theorem we proved in Chapter 7 is very powerful, but 

sometimes its abstract nature is a drawback, and one needs a more explicit 

expansion, analogous to Fourier series or Fourier transforms. A 

general theorem of this type was proved by von Neumann in 1949, but 

it is still of a fairly abstract nature. It can be applied to elliptic partial 

differential equations (G˚arding around 1952), but gives more satisfactory 

results when applied to ordinary differential equations. How to 

do this was described by G˚arding in an appendix to John, Bers and 

Schechter: Partial Differential Equations, (1964). A slightly more general 

situation was treated in [2]. For Sturm-Liouville equations one 

can, however, as easily obtain an expansion theorem directly. We will 

do that in this chapter. 

As in our proof of the spectral theorem, we will deduce our results 

from properties of the resolvent, but now need to have a more explicit 

description of the resolvent operator. The first step is to prove that the 

resolvent is actually an integral operator. First note that all elements 

of D1 are continuously differentiable with locally absolutely continuous 

derivative, and according to Lemma 10.8 point evaluations of elements 

of D1 (and their derivatives) are locally uniformly bounded linear forms 

on D1. 

If T is a selfadjoint realization of (10.3) in L 2 (I) its resolvent Rλ is a 

bounded operator on L 2 (I) for every λ in the resolvent set. If E denotes 

the identity on L 2 (I) we have (T − λ)Rλ = E so that T Rλ = E + λRλ. 

Thus T Rλ ≤ 1 + |λ|Rλ. Since Rλu is in the domain of T we 

may also view the resolvent as an operator Rλ : L 2 (I) → D1, where 

D1 is viewed as a Hilbert space provided with the graph norm, as 

on page 65. This operator is bounded since Rλu 2 1 = Rλu 2 + 

T Rλu 2 ≤ (Rλ 2 + (|λ|Rλ + 1) 2 )u 2 . It is also clear that the 

analyticity of Rλ implies the analyticity of T Rλ = E + λRλ, and therefore 

the analyticity of Rλ : L 2 (I) → D1. We obtain the following 

theorem. 

Theorem 11.1. Suppose I is an interval, and that T is a selfadjoint 

realization in L 2 (I) of the equation (10.3). Then the resolvent Rλ of T 

may be viewed as a bounded linear map from L 2 (I) to C 1 (K), for any 

compact subinterval K of I, which depends analytically on λ ∈ ρ(T ), 

in the uniform operator topology. Furthermore, there exists Green’s 

69

70 11. STURM-LIOUVILLE EQUATIONS 

function g(x, y, λ), which is in L 2 (I) as a function of y for every x ∈ I 

and such that Rλu(x) = 〈u, g(x, ·, λ)〉 for any u ∈ L 2 (I). There is also 

a kernel g1(x, y, λ) in L 2 (I) as a function of y for every x ∈ I such 

that (Rλu) ′ (x) = 〈u, g1(x, ·, λ)〉 for any u ∈ L 2 (I). 

Proof. We already noted that ρ(T ) ∋ λ ↦→ Rλ ∈ B(L 2 (I), D1) is 

analytic in the uniform operator topology. Furthermore, the restriction 

operator IK : D1 → C 1 (K) is bounded and independent of λ. Hence 

ρ(T ) ∋ λ → IKRλ is analytic in the uniform operator topology. In 

particular, for fixed λ ∈ ρ(T ) and any x ∈ I, the linear form L 2 (I) ∋ 

u ↦→ (IKRλu)(x) = Rλu(x) is (locally uniformly) bounded. By Riesz’ 

representation theorem we have Rλu(x) = 〈u, g(x, ·, λ)〉, where y ↦→ 

g(x, y, λ) is in L 2 (I). Similarly, since L 2 ∋ u ↦→ (Rλu) ′ (x) is a bounded 

linear form for each x ∈ I the kernel g1 exists. 

Among other things, Theorem 11.1 tells us that if uj → u in L 2 (I), 

then Rλuj → Rλu in C 1 (K), so that Rλuj and its derivative converge 

locally uniformly. This is actually true even if uj just converges weakly, 

but all we need is the following weaker result. 

Lemma 11.2. Suppose Rλ is the resolvent of a selfadjoint relation T 

as above. Then if uj ⇀ 0 weakly in L 2 (I), it follows that both Rλuj → 0 

and (Rλuj) ′ → 0 pointwise and locally boundedly. 

Proof. Rλuj(x) = 〈uj, g(x, ·, λ)〉 → 0 since y ↦→ g(x, y, λ) is in 

L 2 (I) for any x ∈ I. Now let K be a compact subinterval of I. A 

weakly convergent sequence in L 2 (I) is bounded, so since Rλ maps 

L 2 (I) boundedly into C 1 (K), it follows that Rλuj(x) is bounded independently 

of j and x for x ∈ K. Similarly for the sequence of derivatives. 

 

Corollary 11.3. If the interval I is compact, then any selfadjoint 

restriction T of T1 has compact resolvent. Hence T has a complete 

orthonormal sequence of eigenfunctions in L 2 (I). 

Proof. Suppose uj ⇀ 0 weakly in L 2 (I). If I is compact, then 

Lemma 11.2 implies that Rλuj → 0 pointwise and boundedly in I, 

and hence by dominated convergence Rλuj → 0 in L 2 (I). Thus Rλ is 

compact. The last statement follows from Theorem 8.3. 

For a different proof, see Corollary 11.7. 

If T has compact resolvent, then the generalized Fourier series of 

any u ∈ L 2 (I) converges to u in L 2 (I). For functions in the domain of 

T much stronger convergence is obtained. 

Corollary 11.4. Suppose T has a complete orthonormal sequence 

of eigenfunctions in L 2 (I). If u is in the domain of T , then the generalized 

Fourier series of u, as well as the differentiated series, converges 

locally uniformly in I. In particular, if I is compact, the convergence 

is uniform in I.

11. STURM-LIOUVILLE EQUATIONS 71 

Proof. Suppose u is in the domain of T , i.e., T u = v for some 

v ∈ L 2 (I), and let ˜v = v − iu, so that u = Ri˜v. If e is an eigenfunction 

of T with eigenvalue λ we have T e = λe or (T + i)e = (λ + i)e so that 

R−ie = e/(λ + i). It follows that 〈u, e〉 e = 〈Ri˜v, e〉 e = 〈˜v, R−ie〉 e = 

1 

λ−i 〈˜v, e〉 e = 〈˜v, e〉Rie. If sNu denotes the N:th partial sum of the 

Fourier series for u it follows that sNu = RisN ˜v, where sN ˜v is the N:th 

partial sum for ˜v. Since sN ˜v → ˜v in L 2 (I), it follows from Theorem 11.1 

and the remark after it that sNu → u in C 1 (K), for any compact 

subinterval K of I. 

The convergence is actually even better than the corollary shows, 

since it is absolute and uniform (see Exercise 11.2). 

Example 11.5. Consider the equation −u ′′ = λu, first in L2 (−π, π), 

with periodic boundary conditions u(−π) = u(π), u ′ (−π) = u ′ (π). The 

general solution is u(x) = A cos( √ λ x) + B sin( √ λ x), where A, B are 

constants. The boundary conditions may be viewed as linear equations 

for determining the constants A and B, and if there is going to be a 

non-trivial solution, the determinant must vanish. The determinant is 

 

 

0 2 sin( 

 

√ λ π) 

−2 sin( √ 

 

 

λ π) 0 = 4 sin2 ( √ λ π) 

so that λ = k 2 , where k ∈ N. For each eigenvalue k 2 > 0 we have 

two linearly independent eigenfunctions cos(kx) and sin(kx). For the 

eigenvalue 0 the eigenfunction is 1 √ . These functions are orthonormal 

2 

if we use the scalar product 〈u, v〉 = 1 

π 

uv (check!). We obtain the 

π −π 

classical (real) Fourier series f(x) = a0 

2 + ∞ k=1 (ak cos kx + bk sin kx), 

where a0 = 〈f, 1〉, ak = 〈f(x), cos kx〉 for k > 0, and bk = 〈f(x), sin kx〉. 

In this case Corollary 11.4 states that the series for u as well as that 

for u ′ converge uniformly if u is continuously differentiable with an 

absolutely continuous derivative such that u ′′ ∈ L 2 (−π, π). 

Now consider the same equation in L 2 (0, π), with separated boundary 

conditions u(0) = 0 and u(π) = 0. Applying this to the general 

solution we obtain first B = 0 and then A sin √ λ π) = 0, so 

a non-trivial solution exists only if λ = k 2 for a positive integer k. 

Thus the eigenfunctions are sin x, sin 2x, . . . . These are orthonormal 

if the scalar product used is 〈u, v〉 = 2 

π 

uv. We obtain a sine se- 

π 0 

ries f(x) = ∞ k=1 bk sin(kx), where bk = 〈f(x), sin kx〉. This is the 

series expansion relevant to the vibrating string problem discussed in 

Chapter 0 (if the length of the string is π). 

Finally, consider the same equation, still in L2 (0, π), but now with 

separated boundary conditions u ′ (0) = 0 and u ′ (π) = 0. Applying this 

to the general solution we obtain first A = 0 and then B sin( √ λ π) = 0, 

so a non-trivial solution requires λ = k2 for a non-negative integer k. 

Thus the eigenfunctions are 1 

√ 2 , cos x, cos 2x, . . . . These are orthonormal 

with the same scalar product as in the previous example. We


obtain a cosine series f(x) = a0 

2 + ∞ and ak = 〈f(x), cos kx〉. 

k=1 ak cos(kx), where a0 = 〈f, 1〉 

We have thus retrieved some of the classical versions of Fourier 

series, but is clear that many other variants are obtained by simply 

varying the boundary conditions, and that many more examples are 

obtained by choosing a non-zero q in (10.3). 

We now have a satisfactory eigenfunction expansion theory for regular 

boundary value problems, so we turn next to singular problems. 

We then need to take a much closer look at Green’s function. We shall 

here primarily look at the case of separated boundary conditions for 

I = [a, b) where a is a regular endpoint and b possibly singular, and 

refer the reader to the theory of Chapter 15 for the general case. With 

this assumption Green’s function has a particularly simple structure. 

Assume that ϕ, θ are solutions of −u ′′ + qu = λu with initial data 

ϕ(a, λ) = − sin α, ϕ ′ (a, λ) = cos α and θ(a, λ) = cos α, θ ′ (a, λ) = sin α. 

Theorem 11.6. Suppose I = [a, b) with a regular, and that T is 

given by the separated condition (10.7) at a, and another separated condition 

at b if needed, i.e., if b is regular or in the limit circle condition. 

If Im λ = 0, then g(x, y, λ) = ϕ(min(x, y), λ)ψ(max(x, y), λ) where ψ is 

called the Weyl solution and is given by ψ(x, λ) = θ(x, λ)+m(λ)ϕ(x, λ). 

Here m(λ) is called the Weyl-Titchmarsh m-coefficient and is a Nevanlinna 

function in the sense of Chapter 6. The kernel g1 is g1(x, y, λ) = 

ϕ ′ (x, λ)ψ(y, λ) if x < y and g1(x, y, λ) = ϕ(y, λ)ψ ′ (x, λ) if x > y. 

Proof. It is easily verified that [θ, ϕ] = 1. Now ϕ satisfies the 

boundary condition at a and can therefore only satisfy the boundary 

condition at b if λ is an eigenvalue and thus real. On the other hand, 

there will be a solution in L2 (a, b) satisfying the boundary condition 

at b, since if deficiency indices are 1 there is no condition at b, and if 

deficiency indices are 2, then the condition at b is a linear, homogeneous 

condition on a two-dimensional space, which leaves a space of dimension 

1. Thus we may find a unique m(λ) so that ψ = θ + mϕ satisfies the 

boundary condition at b. It follows that [ψ, ϕ] = [θ, ϕ] + m[ϕ, ϕ] = 1. 

Now setting v(x) = 〈u, g(x, ·, λ)〉 and assuming that u ∈ L2 (a, b) 

has compact support we obtain 

x 

b 

v(x) = ψ(x, λ) uϕ(·, λ) + ϕ(x, λ) uψ(·, λ), 

a 

so that v(a) = − sin α b 

uψ(·, λ). Differentiating we obtain 

a 

(11.1) v ′ (x) = ψ ′ x 

(x, λ) uϕ(·, λ) + ϕ ′ b 

(x, λ) uψ(·, λ), 

a 

x 

x


since the other two terms obtained cancel. Thus v ′ (a) = cos α b 

uψ(·, λ) 

a 

so v satisfies the boundary condition at a. If x is to the right of the support 

of u we obtain v(x) = ψ(x, λ) b 

uϕ(·, λ) so that v also satisfies the 

a 

boundary condition at b, being a multiple of ψ near b. Differentiating 

again we obtain 

−v ′′ (x) + (q(x) − λ)v(x) = [ψ, ϕ]u(x) = u(x). 

It follows that v = Rλu and, since compactly supported functions are 

dense in L2 (a, b), that g(x, y, λ) is Green’s function for our operator. 

From (11.1) now follows that the kernel g1 is as stated. 

It remains to show that m is a Nevanlinna function. If u and v both 

have compact supports in I we have 

 

〈Rλu, v〉 = g(x, y, λ)u(y)v(x) dxdy, 

the double integral being absolutely convergent. Similarly 

 

〈u, Rλv〉 = g(y, x, λ)u(y)v(x) dxdy, 

and since the integrals are equal for all u, v by Theorem 5.2.2 we obtain 

g(x, y, λ) = g(y, x, λ) or, if x < y, 

ϕ(x, λ)θ(y, λ) + ϕ(x, λ)ϕ(y, λ)m(λ) 

= ϕ(x, λ)θ(y, λ) + ϕ(x, λ)ϕ(y, λ)m(λ), 

since ϕ(·, λ) = ϕ(·, λ) and similarly for θ. Since ϕ(x, λ) = 0 for non-real 

λ (why?) it follows that m(λ) = m(λ). Now λ ↦→ Rλu(x) is analytic 

for non-real λ and for compactly supported u 

 

Rλu(x) = θ(x, λ) 

a 

x 

 

uϕ(·, λ) + ϕ(x, λ) 

x 

b 

uθ(·, λ) 

 

+ m(λ)ϕ(x, λ) 

a 

b 

uϕ(·, λ). 

The first two terms on the right are obviously entire functions according 

to Theorem 10.1, as is the coefficient of m(λ), and since by choice 

of u we may always assume that this coefficient is non-zero in a neighborhood 

of any given λ it follows that m(λ) is analytic for non-real 

λ. 

Finally, integration by parts shows that 

 

λ 

a 

x 

|ψ| 2 = −ψ ′ ψ x 

a + 

x 

a 

(|ψ ′ | 2 + q|ψ| 2 ).


Taking the imaginary part of this and using the fact that ψ satisfies 

the boundary condition at b so that Im(ψ ′ ψ) → 0 at b we obtain 

(11.2) 0 ≤ 

b 

a 

|ψ(·, λ)| 2 = 

Im m(λ) 

Im λ , 

since a simple calculation shows that Im(ψ ′ (a, λ)ψ(a, λ)) = Im m(λ). It 

follows that m has all the required properties of a Nevanlinna function. 

 

Before we proceed, we note the following corollary, which completes 

our results for the case of a discrete spectrum. 

Corollary 11.7. Suppose both endpoints of I are either regular or 

in the limit circle condition. Then for any selfadjoint realization T the 

resolvent is compact. Thus there is a complete orthonormal sequence 

of eigenfunctions. 

Proof. By Theorem 8.4 it is enough to prove the corollary when 

T is given by separated boundary conditions. But as in the proof 

of Theorem 11.6 we can then find non-trivial solutions ψ−(·, λ) and 

ψ+(·, λ) of −v ′′ + qv = λv satisfying the boundary conditions to the 

left and right respectively. If Im λ = 0 the solutions ψ±(·, λ) can not be 

linearly dependent, since this would give a non-real eigenvalue for T . 

We may therefore assume [ψ+, ψ−] = 1 by multiplying ψ−, if necessary, 

by a constant. But then it is seen that ψ−(min(x, y), λ)ψ+(max(x, y), λ) 

is Green’s function for T just as in the proof of Theorem 11.6. 

It is clear that the assumption implies that deficiency indices equal 

2, so that ψ± are in L 2 (I). However, an easy calculation now shows 

that 

I×I 

|g(x, y, λ)| 2 dxdy ≤ 2ψ− 2 ψ+ 2 < ∞. 

Thus, according to Theorem 8.7, the resolvent is a Hilbert-Schmidt 

operator, so that it is compact. 

If at least one of the interval endpoints is singular and in the limit 

point condition the resolvent may not be compact (but it can be!). In 

this case the only boundary condition will be a separated boundary 

condition at the other endpoint, unless this is also in the limit point 

condition, when no boundary conditions at all are required. 

We now return to the situation treated in Theorem 11.6 when I = 

[a, b) with a regular, and T is given by the separated condition (10.7) 

at a, and another separated condition at b if needed. Since the mcoefficient 

is a Nevanlinna function there is a unique increasing and 

left-continuous matrix-valued function ρ with ρ(0) = 0 and unique real

numbers A and B ≥ 0 such that 

 

m(λ) = A + Bλ + 


∞ 

−∞ 

( 1 t 

− 

t − λ t2 ) dρ(t). 

+ 1 

The spectral measure dρ gives rise to a Hilbert space L 2 ρ, which 

consists of those functions û which are measurable with respect to dρ 

and for which û 2 ρ = ∞ 

−∞ |û|2 is finite. Alternatively, we may think of 

L 2 ρ as the completion in this norm of compactly supported, continuous 

functions. These alternative definitions give the same space, but we 

will not prove this here. We denote the scalar product in L 2 ρ by 〈·, ·〉ρ. 

The main result of this chapter is the following. 

Theorem 11.8. 

(1) If u ∈ L 2 (a, b) the integral x 

0 uϕ(·, t) converges in L2 ρ as x → b. 

The limit is called the generalized Fourier transform of u and 

is denoted by F(u) or û. We write this as û(t) = 〈u, ϕ(·, t)〉, 

although the integral may not converge pointwise. 

(2) The mapping u ↦→ û is unitary between L 2 (a, b) and L 2 ρ so that 

the Parseval formula 〈u, v〉 = 〈û, ˆv〉ρ is valid if u, v ∈ L 2 (a, b). 

(3) The integral 

K û(t)ϕ(x, t) dρ(t) converges in L2 (a, b) as K → 

R through compact intervals. If û = F(u) the limit is u, so 

the integral is the inverse of the generalized Fourier transform. 

Again, we write u(x) = 〈û, ϕ(x, ·)〉ρ for u ∈ L2 (a, b), although 

the integral may not converge pointwise. 

(4) Let E∆ denote the spectral projector of T for the interval ∆. 

Then E∆u(x) = 

∆ 

ûϕ(x, ·) dρ. 

(5) If u ∈ D(T ) then F(T u)(t) = tû(t). Conversely, if û and tû(t) 

are in L 2 ρ, then F −1 (û) ∈ D(T ). 

Before we prove this theorem, let us interpret it in terms of the 

spectral theorem. If the interval ∆ shrinks to a point t, then E∆ tends 

to zero, unless t is an eigenvalue, in which case we obtain the projection 

on the eigenspace. By (4) this means that eigenvalues are precisely 

those points at which the function ρ has a (jump) discontinuity; continuous 

spectrum thus corresponds to points where ρ is continuous, but 

which are still points of increase for ρ, i.e., there is no neighborhood of 

the point where ρ is constant. In terms of measure theory, this means 

that the atomic part of the measure dρ determines the eigenvalues, and 

the diffuse part of dρ determines the continuous spectrum. 

We will prove Theorem 11.8 through a long (but finite!) sequence 

of lemmas. First note that for u ∈ L 2 (a, b) with compact support in 

[a, b) the function û(λ) = 〈u, ϕ(·, λ)〉 is an entire function of λ since 

ϕ(x, λ) is entire, locally uniformly in x, according to Theorem 10.1. 

Lemma 11.9. The function 〈Rλu, v〉 − m(λ)û(λ)ˆv(λ) is entire for 

all u, v ∈ L 2 (a, b) with compact supports in [a, b).


Proof. If the supports are inside [a, c], direct calculation shows 

that the function is 

c 

x 

c 

 

θ(x, λ) uϕ(·, λ) + ϕ(x, λ) uθ(·, λ) v(x) dx . 

a 

a 

This is obviously an entire function of λ. 

Lemma 11.10. Let σ be increasing and differentiable at 0. Then 

1 

−1 ds 1 dσ(t) 

√ 

−1 t2 +s2 converges. 

Proof. Integrating by parts we have, for s = 0, 

1 

−1 

dσ(t) 

√ t 2 + s 2 

= σ(1) − σ(−1) 

√ 1 + s 2 

− 

1 

−1 

x 

σ(t) − σ(0) 

(t 

t 

d 

dt 

1 

√ t 2 + s 2 

) dt . 

The first factor in the last integral is bounded since σ ′ (0) exists, and the 

second factor is negative since (t2 + s2 1 

− ) 2 decreases with |t|. Furthermore, 

the integral with respect to t of the second factor is integrable 

with respect to s, by calculation (check this!). Thus the double integral 

is absolutely convergent. 

As usual we denote the spectral projectors belonging to T by Et. 

Lemma 11.11. Let u ∈ L 2 (a, b) have compact support in [a, b) and 

assume c < d to be points of differentiability for both 〈Etu, u〉 and ρ(t). 

Then 

(11.3) 〈Edu, u〉 − 〈Ecu, u〉 = 

d 

c 

|û(t)| 2 dρ(t). 

Proof. Let Γ be the positively oriented rectangle with corners in 

c ± i, d ± i. According to Lemma 11.9 

 

 

〈Rλu, u〉 dλ = û(λ)û(λ)m(λ) dλ 

Γ 

if either of these integrals exist. However, by Lemma 11.9, 

 

 

û(λ)û(λ)m(λ) dλ = 

∞ 

û(λ)û(λ) ( 1 t 

− 

t − λ t2 ) dρ(t) dλ. 

+ 1 

Γ 

Γ 

Γ 

The double integral is absolutely convergent except perhaps where t = 

λ. The difficulty is thus caused by 

1 

−1 

ds 

µ+1 

µ−1 

−∞ 

û(µ + is)û(µ − is) dρ(t) 

t − µ − is


for µ = c, d. However, Lemma 11.10 ensures the absolute convergence 

of these integrals. Changing the order of integration gives 

 

Γ 

û(λ)û(λ)m(λ) dλ = 

∞ 

−∞ 

 

Γ 

û(λ)û(λ)( 1 t 

− 

t − λ t2 ) dλ dρ(t) 

+ 1 

 

= −2πi 

c 

d 

|û(t)| 2 dρ(t) 

since for c < t < d the residue of the inner integral is −|û(t)| 2 dρ(t) 

whereas t = c, d do not carry any mass and the inner integrand is 

regular for t < c and t > d. 

Similarly we have 

 

Γ 

〈Rλu, u〉 dλ = 

∞ 

−∞ 

 

d〈Etu, u〉 

Γ 

dλ 

t − λ 

 

= −2πi 

c 

d 

d〈Etu, u〉 

which completes the proof. 

Lemma 11.12. If u ∈ L2 (a, b) the generalized Fourier transform 

uϕ(·, t) as x → b. Furthermore, 

û ∈ L 2 ρ exists as the L 2 ρ-limit of x 

a 

〈Etu, v〉 = 

t 

−∞ 

ûˆv dρ . 

In particular, 〈u, v〉 = 〈û, ˆv〉ρ if u and v ∈ L 2 (a, b). 

Proof. If u has compact support Lemma 11.11 shows that (11.3) 

holds for a dense set of values c, d since functions of bounded variation 

are a.e. differentiable. Since both Et and ρ are left-continuous we 

obtain, by letting d ↑ t, c → −∞ through such values, 

〈Etu, v〉 = 

t 

−∞ 

ûˆv(t) dρ(t) 

when u, v have compact supports; first for u = v and then in general 

by polarization. As t → ∞ we also obtain that 〈u, v〉 = 〈û, ˆv〉ρ when u 

and v have compact supports. 

For arbitrary u ∈ L 2 (a, b) we set, for c ∈ (a, b), 

uc(x) = 

 

u(x) for x < c 

0 otherwise 

and obtain a transform ûc. If also d ∈ (a, b) it follows that ûc −ûdρ = 

uc − ud, and since uc → u in L 2 (a, b) as c → b, Cauchy’s convergence 

principle shows that ûc converges to an element û ∈ L 2 ρ as c → b. The 

lemma now follows in full generality by continuity.


Note that we have proved that F is an isometry from L 2 (a, b) to 

L 2 ρ. 

Lemma 11.13. The integral 

K ûϕ(x, ·) dρ is in L2 (a, b) if K is a 

compact interval and û ∈ L 2 ρ, and as K → R the integral converges 

in L 2 (a, b). The limit F −1 (û) is called the inverse transform of û. 

If u ∈ L 2 (a, b) then F −1 (F(u)) = u. F −1 (û) = 0 if and only if û is 

orthogonal in L 2 ρ to all generalized Fourier transforms. 

Proof. If û ∈ L 2 ρ has compact support, then u(x) = 〈û, ϕ(x, ·)〉ρ 

is continuous, so uc ∈ L 2 (a, b) for c ∈ (a, b), and has a transform ûc. 

We have 

uc 2 = 

c 

a 

∞ 

−∞ 

ûϕ(x, ·) dρ u(x) dx. 

Considered as a double integral this is absolutely convergent, so changing 

the order of integration we obtain 

uc 2 = 

∞ 

−∞ 

c a 

 

uϕ(·, t) û(t) dρ(t) 

= 〈û, ûc〉ρ ≤ ûρûcρ = ûρuc, 

according to Lemma 11.12. Hence uc ≤ ûρ, so u ∈ L 2 (a, b), and 

u ≤ ûρ. If now û ∈ L 2 ρ is arbitrary, this inequality shows (like in the 

proof of Lemma 11.12) that 

K û(t)ϕ(x, t) dρ(t) converges in L2 (a, b) as 

K → R through compact intervals; call the limit u1. If v ∈ L 2 (a, b), 

ˆv is its generalized Fourier transform, K is a compact interval, and 

c ∈ (a, b), we have 

 

K 

c a 

 

v(x)ϕ(x, t) dx û(t) dρ(t) = 

c 

a 

 

v(x) 

K 

û(t)ϕ(x, t) dρ(t) dx 

by absolute convergence. Letting c → b and K → R we obtain 〈û, ˆv〉ρ = 

〈u1, v〉. If û is the transform of u, then by Lemma 11.12 u1 − u is 

orthogonal to L 2 (a, b), so u1 = u. Similarly, u1 = 0 precisely if û is 

orthogonal to all transforms. 

We have shown the inverse transform to be the adjoint of the transform 

as an operator from L 2 (a, b) into L 2 ρ. The basic remaining difficulty 

is to prove that the transform is surjective, i.e., according to 

Lemma 11.13, that the inverse transform is injective. The following 

lemma will enable us to prove this. 

Lemma 11.14. The transform of Rλu is û(t)/(t − λ).


Proof. By Lemma 11.12, 〈Etu, v〉 = t 

−∞ 

〈Rλu, v〉 = 

∞ 

−∞ 

d〈Etu, v〉 

t − λ = 

∞ 

−∞ 

By properties of the resolvent 

Rλu 2 = 

1 

2i Im λ 〈Rλu − R λ u, u〉 = 

û(t)ˆv(t) dρ(t) 

t − λ 

∞ 

−∞ 

ûˆv dρ, so that 

= 〈û(t)/(t − λ), ˆv(t)〉ρ. 

d〈Etu, u〉 

|t − λ| 2 = û(t)/(t − λ)2 ρ. 

Setting v = Rλu and using Lemma 11.12, it therefore follows that 

û(t)/(t−λ) 2 ρ = 〈û(t)/(t−λ), F(Rλu)〉ρ = F(Rλu) 2 ρ. It follows that 

we have û(t)/(t − λ) − F(Rλu)ρ = 0, which was to be proved. 

Lemma 11.15. The generalized Fourier transform is unitary from 

L 2 (a, b) to L 2 ρ and the inverse transform is the inverse of this map. 

Proof. According to Lemma 11.13 we need only show that if 

û ∈ L2 ρ has inverse transform 0, then û = 0. Now, according to 

Lemma 11.14, F(v)(t)/(t − λ) is a transform for all v ∈ L2 (a, b) and 

non-real λ. Thus we have 〈û(t)/(t − λ), F(v)(t)〉ρ = 0 for all non-real λ 

if û is orthogonal to all transforms. But we can view this scalar product 

as the Stieltjes-transform of the measure t 

ûF(v) dρ, so applying the 

−∞ 

inversion formula Lemma 6.5 we have 

ûF(v) dρ = 0 for all compact 

K 

intervals K, and all v ∈ L2 (a, b). Thus the cutoff of û, which equals 

û in K and 0 outside, is also orthogonal to all transforms, i.e., has 

inverse transform 0 according to Lemma 11.13. It follows that 

 

v(x) = û(t)ϕ(x, t) dρ(t) 

K 

is the zero-element of L2 (a, b) for any compact interval K. Differentiating 

under the integral sign we also see that v ′ (x) = 

K ûϕ′ (x, ·) dρ 

is the zero element of L2 (a, b). But these functions are continuous, so 

they are pointwise 0. Now 0 = v ′ (a) cos α − v(a) sin α = 

û dρ. Thus 

K 

û dρ is the zero measure, so that û = 0 as an element of L2 ρ. 

Lemma 11.16. If u ∈ D(T ), then F(T u)(t) = tû(t). Conversely, if 

û and tû(t) are in L 2 ρ, then F −1 (û) ∈ D(T ). 

Proof. We have u ∈ D(T ) if and only if u = Rλ(T u − λu), which 

holds if and only if û(t) = (F(T u)(t) − λû(t))/(t − λ), i.e., F(T u)(t) = 

tû(t), according to Lemmas 11.14 and 11.15. 

This completes the proof of Theorem 11.8. We also have the following 

analogue of Corollary 11.4.


Theorem 11.17. Suppose u ∈ D(T ). Then the inverse transform 

〈û, ϕ(x, ·)〉ρ converges locally uniformly to u(x). 

Proof. The proof is very similar to that of Corollary 11.4. Put 

v = (T − i)u so that u = Riv. Let K be a compact interval, and put 

uK(x) = 

K û(t)ϕ(x, t) dP (t) = F −1 (χû)(x), where χ is the characteristic 

function for K. Define vK similarly. Then by Lemma 11.14 

RivK = F −1 ( χ(t)ˆv(t) 

) = F 

t − i 

−1 (χû) = uK. 

Since vK → v in L2 (a, b) as K → R, it follows from Theorem 11.1 

that uK → u in C1 (L) as K → R, for any compact subinterval L of 

[a, b). 

Example 11.18 (Sine and cosine transforms). Let us interpret Theorem 

11.8 for the case of the equation −u ′′ = λu on the interval [0, ∞). 

We shall look at the cases when the boundary condition at 0 is either 

a Dirichlet condition (α = 0 in (10.7)) or a Neumann condition (α = 

π/2). The general solution of the equation is u(x) = Ae √ −λx +Be − √ −λx . 

Let the root be the principal branch, i.e., the branch where the real 

part is ≥ 0. Then the only solutions in L 2 (0, ∞) are, unless λ ≥ 0, 

the multiples of e −√ −λx = cos(i √ −λx) + i sin(i √ −λx). It follows that 

the equation is in the limit point condition at infinity (this is also a 

consequence of Theorem 10.10). 

With a Dirichlet condition at 0 we have θ(x, λ) = cos(i √ −λx) 

and ϕ(x, λ) = −i sin(i √ −λx)/ √ −λ. It follows that the m-function is 

mD(λ) = − √ −λ. Similarly, the m-function in the case of a Neumann 

condition at 0 is mN(λ) = 1/ √ −λ, using again the principal branch of 

the root. 

Using the Stieltjes inversion formula Lemma 6.5 we √see that the 

t dt for t ≥ 

corresponding spectral measures are given by dρD(t) = 1 

π 

0, dρD = 0 in (−∞, 0), respectively dρN(t) = dt 

π √ t for t ≥ 0, dρN = 0 

in (−∞, 0). If u ∈ L2 (0, ∞) and we define û(t) = ∞ 

0 u(x) sin(√tx) √ dx, 

t 

as a generalized integral converging in L2 ρD , then the inversion formula 

reads u(x) = 1 

∞ 

π 0 û(t) sin(√tx) dt. 

In this case one usually changes variable in the transform and 

defines the sine transform S(u)(ξ) = ∞ 

0 u(x) sin(ξx) dx = ξû(ξ2 ). 

Changing variable to ξ = √ t in the inversion formula above then shows 

that u(x) = 2 

∞ 

S(u)(ξ) sin(ξx) dξ. 

π 0 

Similarly, if we set û(t) = ∞ 

0 u(x) cos(√tx) dx the inversion for- 

dt. In this case it is again 

mula obtained is u(x) = 1 

π 

∞ 

0 û(t) cos(√ tx) 

√ t 

common to use ξ = √ t as the transform variable, so one defines 

the cosine transform C(u)(ξ) = ∞ 

u(x) cos(ξx) dx. Changing vari- 

0 

ables in the inversion formula above then gives the inversion formula 

C(u)(ξ) cos(ξx) dξ for the cosine transform. 

u(x) = 2 

π 

∞ 

0


Note that there are no eigenvalues in either of these cases; the 

spectrum is purely continuous. 


Exercise 11.1. Show that if K is a compact interval, then C 1 (K) 

is a Banach space with the norm sup x∈K|u(x)| + sup x∈K|u ′ (x)|. 

If you know some topology, also show that if I is an arbitrary interval, 

then C(I) is a Fréchet space (a linear Hausdorff space with 

the topology given by a countable family of seminorms, which is also 

complete), under the topology of locally uniform convergence. 

Exercise 11.2. With the assumptions of Corollary 11.4 the Fourier 

series for u in the domain of T actually converges absolutely and locally 

uniformly to u. If λ1, λ2, . . . are the eigenvalues and e1, e2, . . . the corresponding 

orthonormal eigenfunctions, use Parseval’s formula to show 

that, pointwise in x, g(x, ·, λ)2 = | ej(x) 

λj−λ |2 , with natural notation. 

Then show that as an L2 (I)-valued function x ↦→ g(x, ·, λ) is locally 

bounded, i.e., x ↦→ g(x, ·, λ) is bounded on any compact subinterval 

of I. 

If v = Rλu and ûj is the j:th Fourier coefficient of u, then ˆvj = 

〈Rλu, ej〉 = 〈u, Rλej〉 = ûj/(λj 

− λ). Show that this implies that 

j>n |ˆvjej(x)| tends locally uniformly to 0.

CHAPTER 12 

Inverse spectral theory 

In this chapter we continue to study the simple Sturm-Liouville 

equation −u ′′ + qu = λu, on an interval with at least one regular 

endpoint. Our aim is to give some results on inverse spectral theory, 

i.e., questions related to the determination of the equation, in this 

case the potential q from spectral data, such as eigenvalues, spectral 

measures or similar things. Our object of study is the eigen-value 

problem 

(12.1) 

(12.2) 

− u ′′ + qu = λu on [0, b), 

u(0) cos α + u ′ (0) sin α = 0. 

Here α is an arbitrary, fixed number in [0, π), so that the boundary 

condition is an arbitrary separated boundary condition. We assume q ∈ 

L1 loc [0, b), i.e., q integrable on any interval [0, c] with c ∈ (0, b), so that 

0 is a regular endpoint for the equation. The other endpoint b may be 

infinite or finite, in the latter case singular or regular. If the deficiency 

indices for the equation in L2 (0, b) are (1, 1) the operator corresponding 

to (12.1), (12.2) is selfadjoint; if they are (2, 2) a boundary condition 

at b is required to obtain a selfadjoint operator. We assume that, if 

necessary, a choice of boundary condition at b is made, so that we are 

dealing with a self-adjoint operator which we will call T . 

If the deficiency indices are (2, 2) we know the spectrum is discrete 

(Theorem 11.7), but when the deficiency indices are (1, 1) the spectrum 

can be of any type. As in Chapter 11, let ϕ and θ be solutions of (12.1) 

satisfying initial conditions 

 

ϕ(0, λ) = − sin α 

(12.3) 

ϕ ′ (0, λ) = cos α , 

 

θ(0, λ) = cos α 

θ ′ (0, λ) = sin α . 

Then Green’s function for T is given by 

g(x, ·, λ) = ϕ(min(x, y), λ)ψ(max(x, y), λ) 

where ψ(x, λ) = θ(x, λ) + m(λ)ϕ(x, λ) and the Titchmarsh-Weyl mfunction 

m(λ) is determined so that ψ satisfies the boundary condition 

at b. In particular ψ ∈ L 2 (0, b). Let the Nevanlinna representation of 

m be 

m(λ) = A + Bλ + 

∞ 

−∞ 

1 t 

− 

t − λ t2 

dρ(t), 

+ 1 

83

84 12. INVERSE SPECTRAL THEORY 

where A ∈ R, B ≥ 0 and ρ increases (dρ is a positive measure) and 

∞ 

−∞ 

dρ(t) 

t 2 +1 < ∞. The transform space L2 ρ consists of those functions û, 

measurable with respect to dρ, for which û 2 ρ = ∞ 

−∞ |û|2 dρ is finite. 

The generalized Fourier transform of u ∈ L 2 (0, b) is 

û(t) = 

b 

0 

u(x)ϕ(x, t) dx, 

converging in L 2 ρ, and with inverse given by 

u(x) = 

∞ 

−∞ 

û(t)ϕ(x, t) dρ(t), 

which converges in L 2 (0, b). Furthermore, u = ûρ (Parseval) and 

u ∈ D(T ) if and only if û and tû(t) ∈ L 2 (0, b), and then T u(t) = tû(t). 

In the case when one has a discrete spectrum, which means that the 

spectrum consists of isolated eigenvalues (of finite multiplicity), the 

function ρ is a step function, with a step at each eigenvalue. Suppose 

the eigenvalues are λ1, λ2, . . . and that the size of the step is cj = 

limε↓0(ρ(λj + ε) − ρ(λj − ε)). Then the inverse transform takes the 

form 

u(x) = 

∞ 

û(λj)ϕ(x, λj)cj, 

j=1 

where û(λj) = 〈u, ϕ(·, λj〉. For u = ϕ(·, λj) the expansion becomes 

ϕ(x, λj) = ϕ(·, λj) 2 ϕ(x, λj)cj. It follows that cj = ϕ(·, λj) −2 . Note 

that ϕ(·, λj) is an eigenfunction associated with λj, so the jump cj of ρ 

at λj is the so called normalization constant for the eigenfunction. The 

name comes from the fact that a normalized eigenfunction is given by 

ej = √ cj ϕ(·, λj). We have shown the following proposition. 

Proposition 12.1. In the case of a discrete spectrum knowledge 

of the spectral function ρ is equivalent to knowing the eigenvalues and 

the corresponding normalization constants. 

1. Asymptotics of the m-function 

In order to discuss some results in inverse spectral theory we need 

a few results on the asymptotic behavior of the m-function for large λ. 

We denote by mα(λ) the m-function for the boundary condition (12.2) 

and some fixed boundary condition at b. The following theorem is a 

simplified version of a result from [3]. 

Theorem 12.2. We have 

m0(λ) = − √ −λ + o(|λ| 1/2 )

1. ASYMPTOTICS OF THE m-FUNCTION 85 

as λ → ∞ along any non-real ray 1 . Similarly, for 0 < α < π, 

mα(λ) = cot α + ( √ −λ sin 2 α) −1 + o(|λ| −1/2 ) 

as λ → ∞ along any non-real ray. 

By a non-real ray we always mean a half-line starting at the origin 

which is not part of the real line. Here and later the square root is 

always the principal branch, i.e., the branch with a positive real part 

Now note that, up to constant multiples, the Weyl solution ψ is 

determined by the boundary condition at b. For α = 0 we have 

ψ ′ (0, λ)/ψ(0, λ) = m0(λ), so keeping a fixed boundary condition at b 

we obtain m0(λ) = (sin α+mα(λ) cos α)/(cos α−mα(λ) sin α). Solving 

for mα gives 

mα(λ) = cos α m0(λ) − sin α 

sin α m0(λ) + cos α 

= cot α − (m0(λ) sin 2 α) −1 + 

cos α 

m0(λ) sin 2 α(m0(λ) sin α + cos α) . 

Thus, the formula for m0 immediately implies that for mα, 0 < α < π, 

so that we only have to prove the formula for m0. This will require 

good asymptotic estimates of the solutions ϕ and θ. 

Lemma 12.3. If u solves −u ′′ + qu = λu with fixed initial data in 0 

one has 

(12.4) u(x) = u(0)(cosh(x √ −λ) + O(1)(e x 

0 |q|/√ |λ| − 1)e x √ −λ ) 

uniformly in x, λ. 

+ u′ (0) 

√ −λ (sinh(x √ −λ) + O(1)(e x 

0 |q|/√ |λ| − 1)e x √ −λ ), 

Proof. Solving the equation u ′′ + λu = f and then replacing f by 

qu gives 

(12.5) u(x) = cosh(kx)u(0) + sinh(kx) 

k 

+ 

u ′ (0) 

x 

where we have written k for √ −λ. Setting 

0 

sinh(k(x − t)) 

q(t)u(t) dt, 

k 

g(x) = |u(x) − cosh(kx)u(0) − sinh(kx) 

u 

k 

′ −x Re k 

(0)|e 

1 If g is a positive function the notation f(λ) = o(g(λ)) as λ → ∞ means 

f(λ)/g(λ) → 0 as λ → ∞.


easy estimates give 

g(x) ≤ c(λ) 

|k| 

x 

0 

|q| + 1 

|k| 

x 

0 

|q|g, 

where c(λ) = |u(0)| + |u ′ (0)|/|k|. Integrating after multiplying by the 

integrating factor |q(x)| exp(− x 

|q|/|k|) we obtain 

0 

g(x) ≤ c(λ)(e x 

0 |q|/√ |λ| − 1). 

The estimate for u follows immediately from this. 

Proof of Theorem 12.2. As noted, we only need to prove the 

theorem for α = 0, so assume this. Now let λ = rµ, where µ is in 

some fixed, compact subset of C \ R, and r > 0 is large. We define 

ϕr(x, µ) = √ rϕ(x/ √ r, rµ) and θr(x, µ) = θ(x/ √ r, rµ). Then ϕr and 

θr satisfy the initial conditions (12.3) for α = 0 and satisfy the equation 

−u ′′ + qru = µu on (0, b √ r), where qr(x) = q(x/ √ r)/r (check!!). From 

Lemma 12.3 it immediately follows that, locally uniformly in x, µ, we 

have ϕr(x, µ) → sinh(x√−µ) √ and θr(x, µ) → cosh(x −µ 

√ −µ) as r → ∞. 

Now let mr(µ) = m(rµ) 

√ r and make a change of variable x = y/ √ r in 

(11.2). This gives b √ r 

|θr + mrϕr| 0 

2 = Im(mr(µ))/ Im µ, so if c > 0 we 

have 

(12.6) 

c 

0 

|θr(·, µ) + mr(µ)ϕr(·, µ)| 2 ≤ 

Im mr(µ) 

Im µ 

as soon as b √ r ≥ c. The inequality may be rewritten as 

|mr(µ) − Cr| ≤ Rr, 

where Cr and Rr are easily expressed in terms of c 

0 θrϕr, c 

0 |θr| 2 

and 

c 

0 |ϕr| 2 . The inequality therefore confines mr to a disk Kr(c), and is 

clear from Lemma 12.3 that as r → ∞ the coefficients converge, locally 

uniformly for µ ∈ C \ R, to those in the corresponding disk K(c) for 

the case q = 0. Therefore, given any neighborhood Ω of K(c), we must 

have mr(µ) ∈ Ω for all sufficiently large r. 

This is true for any c > 0, and it is obvious from (12.6) that K(c) 

decreases as a function of c. We shall show presently that only the 

point − √ −µ is common to all K(c), and then it follows that mr(µ) → 

− √ −µ, locally uniformly for µ ∈ C \ R. But this means that m(λ) = 

− √ −λ(1 + o(1)) as λ → ∞ in a closed, non-real sector with vertex at 

the origin, and thus proves the theorem. 

It remains to show that ∩c>0K(c) = − √ −µ. But any point ℓ 

in the intersection corresponds to a solution u(x) = cosh(x √ −µ) + 

ℓ sinh(x √ −µ)/ √ −µ with u ∈ L 2 (0, ∞), since we have c 

0 |u|2 ≤ 

Im ℓ 

Im µ for 

all c > 0. Thus the only possible value is ℓ = − √ −µ. On the other

2. UNIQUENESS THEOREMS 87 

hand, the equation with q = 0 has a Weyl solution on [0, ∞), so that 

in fact this value of ℓ gives a point which is in all K(c). This may of 

course also be verified directly (do it!). The proof is now complete. 

2. Uniqueness theorems 

Given q, b, and the boundary conditions, one may in principle determine 

m and thus dρ. We will take as our basic inverse problem to 

determine q (and possibly b and the boundary conditions) when dρ is 

given. Around 1950 Gelfand and Levitan [9] gave a rather complete 

solution to this problem. Their solution includes uniqueness, i.e., a 

proof that different boundary value problems can not yield the same 

spectral measure, reconstruction, i.e., a method (an integral equation) 

whereby one, at least in principle, can determine q from the spectral 

measure, and characterization, i.e., a description of those measures 

that are spectral measures for some equation. 

To discuss the full Gelfand-Levitan theory here would take us too 

far afield. Instead we will confine ourselves to the problem of uniqueness, 

i.e., to show that two different operators can not have the same 

spectral measure. This problem was solved independently by Borg [8] 

and Marčenko [10] just before the Gelfand-Levitan theory appeared. 

To state the theorem we introduce, in addition to the operator T , another 

similar operator ˜ T , corresponding to a boundary condition of the 

form (12.2), but with an angle ˜α ∈ [0, π), an interval [0, ˜ b), a potential 

˜q and, if needed, a boundary condition at ˜ b. Let the corresponding 

spectral measure be d˜ρ. 

Theorem 12.4 (Borg-Marčenko). If dρ = d˜ρ, then ˜ T = T , i.e., 

˜α = α, ˜ b = b and ˜q = q. 

A few years ago Barry Simon [11] proved a ‘local’ version of this 

uniqueness theorem. This was a product of a new strategy developed by 

Simon for obtaining the results of Gelfand and Levitan. I will give my 

own proof [6], which is quite elementary and does not use the machinery 

of Simon. We will use the same idea to prove Theorem 12.4. 

In order to state Simon’s theorem, one should first note that knowing 

m is essentially equivalent to knowing dρ, at least if the boundary 

condition (12.2) is known. Knowing m one can in fact find dρ 

via the Stieltjes inversion formula, and knowing dρ one may calculate 

the integral in the representation of m. By Theorem 12.2 we always 

have B = 0, and A may be determined (if α = 0) since we also have 

m(iν) → cot α as ν → ±∞. We denote the m-functions associated 

with T and ˜ T by m and ˜m respectively. Then Simon’s theorem is the 

following. 

Theorem 12.5 (Simon). Suppose that 0 < a ≤ min(b, ˜ b). Then 

α = ˜α and q = ˜q a.e. on (0, a) if (m(λ) − ˜m(λ))e 2(a−ε) Re √ −λ → 0 for


every ε > 0 as λ → ∞ along some non-real ray. Conversely, if α = ˜α 

and q = ˜q on (0, a), then (m(λ) − ˜m(λ))e 2(a−ε) Re √ −λ → 0 for every 

ε > 0 as λ → ∞ along any non-real ray. 

We will prove both theorems by the same method, the crucial point 

of which is the following lemma. 

Lemma 12.6. For any fixed x ∈ (0, b) holds ϕ(x, λ)ψ(x, λ) → 0 as 

λ → ∞ along a non-real ray. 

Note that ϕ(x, λ)ψ(x, λ) is Green’s function on the diagonal x = y. 

We shall postpone the proof a moment and see how the theorem follows 

from it. We first have a corollary. 

Corollary 12.7. Suppose α = ˜α = 0 or α = 0 = ˜α. Then both 

˜ϕ(x, λ)ψ(x, λ) and ϕ(x, λ) ˜ ψ(x, λ) tend to 0 as λ → ∞ along a non-real 

ray, locally uniformly in x. 

Proof. Clearly (12.4) implies that for fixed x and ˜α = 0 we have 

ϕ(x, λ)/ ˜ϕ(x, λ) → sin α/ sin ˜α as λ → ∞ 

along a non-real ray. If α = ˜α = 0 we instead obtain the limit 1, so the 

corollary follows from Lemma 12.6. 

We shall also need a standard theorem from complex analysis, which 

is a slight elaboration of the maximum principle. 

Theorem 12.8 (Phragmén-Lindelöf). Suppose f is analytic in a 

closed sector bounded by two rays from the origin, that it is bounded on 

the rays, and that |f(z)| ≤ AeB|z|1/2 in the sector, for some constants 

A and B. Then f is bounded in the sector. 

This is just one of the simplest versions of a general class of theorems, 

which are all known under the names of Phragmén and Lindelöf. 

Proofs are given in many textbooks on complex analysis, but for the 

reader’s convenience we also give a proof here. 

Proof. We may without loss of generality assume that the rays 

are given by the angles ±β. Let ε > 0 and F (z) = e−εzγ f(z), where 

1/2 < γ < π/(2β) and the branch of z γ is chosen to be positive real for 

positive real z. Now, for z = re ±iβ we have |F (z)| = e −εrγ cos(βγ) |f(z)|, 

where cos(βγ) > 0. Let M be a bound for f on the rays. Then we 

have |F (z)| ≤ M on the rays. 

For z = Re iδ with |δ| ≤ β we have 

|F (z)| ≤ A exp(BR 1/2 − εR γ cos(βγ)) 

which tends to 0 as R → ∞. Thus, on all circular sectors bounded 

by the rays we have |F (z)| ≤ M on the boundary if the radius R is 

sufficiently large. By the maximum principle this also holds in the 

interior of the circular sector. Since R can be chosen arbitrarily large,

2. UNIQUENESS THEOREMS 89 

the bound is valid in the entire domain bounded by the rays. It follows 

that if z is in this domain, then |f(z)| ≤ Meε|z|γ , and letting ε → 0 we 

obtain the desired result. 

Proof of Theorem 12.4. According to the Nevanlinna representation 

formula for m and ˜m their difference is constant = C, since 

the linear term Bλ is always absent by the asymptotic formulas of 

Theorem 12.2. In particular, since Dirichlet m-functions are always 

unbounded near ∞ on a non-real ray and all others are bounded, we 

must have either α = ˜α or α = 0 = ˜α if dρ = d˜ρ. Thus, according to 

Corollary 12.7, the difference ˜ϕ(x, λ)ψ(x, λ) − ϕ(x, λ) ˜ ψ(x, λ) tends to 

0 as λ → ∞ along a non-real ray. This difference is 

˜ϕ(x, λ)θ(x, λ) − ϕ(x, λ) ˜ θ(x, λ) + Cϕ(x, λ) ˜ϕ(x, λ), 

which is an entire function of λ tending to 0 along non-real rays, and it 

may be bounded by a multiple of eB|λ|1/2 for some constant B according 

to (12.4). By Theorem 12.8 such a function is bounded in the entire 

plane, and therefore constant by Liouville’s theorem, hence identically 

0 since the limit is zero along the rays. It follows that 

θ(x, λ)/ϕ(x, λ) = ˜ θ(x, λ)/ ˜ϕ(x, λ) + C 

for all x, λ. Differentiating with respect to x, using the fact that 

θ ′ ϕ − θϕ ′ = 1, we obtain ϕ 2 (x, λ) = ˜ϕ 2 (x, λ). Taking the logarith- 

mic derivative of this we obtain ϕ′ (x,λ) 

ϕ(x,λ) = ˜ϕ′ (x,λ) 

˜ϕ(x,λ) . 

For x = 0 this gives α = ˜α, and thus that m and ˜m are asymptotically 

the same. Thus C = 0, so that m = ˜m. Differentiating once more 

we obtain ϕ ′′ /ϕ = ˜ϕ ′′ / ˜ϕ which means that q = ˜q on min(b, ˜ b). From 

this follows that ϕ = ˜ϕ and θ = ˜ θ, and thus also ψ = ˜ ψ, on min(b, ˜ b). 

This implies that b = ˜ b, since otherwise ψ (or ˜ ψ) would satisfy selfadjoint 

boundary conditions both at b and ˜ b, so that ψ would be an 

eigenfunction to a non-real eigen-value for a selfadjoint operator. Since 

ψ = ˜ ψ also the boundary conditions at b = ˜ b (if any) are the same. It 

follows that T = ˜ T . 

Proof of Theorem 12.5. Our starting point is that if α = ˜α 

the functions ˜ϕ(x, λ)ψ(x, λ) and ϕ(x, λ) ˜ ψ(x, λ) tend to 0 as λ → ∞ 

along a non-real ray. Their difference is 

(12.7) ˜ϕ(x, λ)θ(x, λ) − ϕ(x, λ) ˜ θ(x, λ) + (m(λ) − ˜m(λ))ϕ(x, λ) ˜ϕ(x, λ). 

Suppose first that α = ˜α and q = ˜q on (0, a). Then the first two 

terms cancel on (0, a), so that (m(λ) − ˜m(λ))ϕ(x, λ) ˜ϕ(x, λ) → 0 as 

λ → ∞ along non-real rays if x ∈ (0, a). By (12.4) this implies that 

(m(λ) − ˜m(λ))e 2(a−ε) Re √ −λ ) → 0 as λ → ∞ along any non-real ray. 

Conversely, the estimate for m − ˜m implies first that α = ˜α and 

then that for 0 < x < a the last term of (12.7) tends to 0 according 

to assumption and (12.4), so that the entire function ˜ϕ(x, λ)θ(x, λ) −


ϕ(x, λ) ˜ θ(x, λ) of λ also tends to 0 along a non-real ray, and by symmetry 

also along its conjugate. However, as in the proof of Theorem 12.4 this 

entire function is bounded by eB|λ|1/2 for some constant B, so by the 

Phragmén-Lindelöf theorem it vanishes for all x ∈ (0, a). It follows 

that q = ˜q in (0, a) exactly as in the proof of Theorem 12.4. 

It only remains to prove Lemma 12.6. 

Proof of Lemma 12.6. Note that for α = 0 (Dirichlet’s boundary 

condition) we have ψ(0, λ) = 1 and ψ ′ (0, λ) = m(λ). Since only ψ 

and its multiples satisfy the boundary condition at b, we have m(λ) = 

u ′ (0, λ)/u(0, λ) for any solution of −u ′′ +qu = λu satisfying the boundary 

condition at b. But consider now the interval [a, b) for 0 < a < b 

and the corresponding operator generated by our differential equation 

in L 2 (a, b) with the Dirichlet boundary condition at a and the same 

boundary condition as before at b. It follows that its m-function is 

given by ψ ′ (a, λ)/ψ(a, λ). Similarly, −ϕ ′ (a, λ)/ϕ(a, λ) is the m-function 

corresponding to the interval (0, a], considering a as the initial point, 

provided with the Dirichlet boundary condition, and using the boundary 

condition (12.2) at 0. The change in sign is due to the fact that the 

initial point of the interval is now to the right of the other end point. 

Now, since ϕψ ′ − ϕ ′ ψ ≡ 1 we have 

1/(ϕψ) = (ϕψ ′ − ϕ ′ ψ)/(ϕψ) = ψ ′ /ψ − ϕ ′ /ϕ, 

so this is a sum of two Dirichlet m-functions. According to Theorem 

12.2 all such m-functions are asymptotic to − √ −λ as λ → ∞ along 

a non-real ray, which immediately implies that ϕ(a, λ)ψ(a, λ) → 0. 

We make some final remarks. One may generalize Simon’s theorem 

to the more general Sturm-Liouville equation −(pu ′ ) ′ +qu = λu, where 

1/p and q are realvalued and locally integrable, provided one can show 

appropriate growth estimates for the solutions and that ˜ϕψ → 0 as 

before. I showed in [4, 5] that ˜ϕψ → 0 in the appropriate manner, 

provided 1/p is in Lr loc for some r > 1 and q − ˜q is in Lr′ loc , where r′ is 

the conjugate exponent to r. For example, if 1/p is locally bounded it 

is enough with local integrability of q and ˜q. Simon’s theorem therefore 

generalizes to this situation. The condition on the m-functions then 

has to be replaced by m(λ) − ˜m(λ) = O(exp(−2 a−ε 

0 Re −λ/p)). 

As far as the original Borg-Marčenko theorem is concerned, it is 

now well known exactly to what extent the coefficients p, q and w in 

the equation (10.1), as well as the interval and boundary conditions, 

are determined by the spectral measure, see [7].

CHAPTER 13 

First order systems 

We shall here study the spectral theory of general first order system 

(13.1) Ju ′ + Qu = W v 

where J is a constant n × n matrix which is invertible and skew- 

Hermitian (i.e., J ∗ = −J) and the coefficients Q and W are n × n 

matrix-valued functions which are locally integrable on I. In addition 

Q is assumed Hermitian and W positive semi-definite. As we shall see, 

these properties ensure the proper symmetry of the differential expression. 

The functions u and v are n×1 matrix-valued on I. In the special 

case when n is even and J = 

0 I 

−I 0 , I being the unit matrix of order 

n/2, systems of the form (13.1) are usually called Hamiltonian systems. 

The following existence and uniqueness theorem is fundamental 

Theorem 13.1. Suppose A is an n×n matrix-valued function with 

locally integrable entries in an interval I, and that B is an n×1 matrixvalued 

function, also locally integrable in I. Assume further that c ∈ I 

and C is an n × 1 matrix. Then the initial value problem 

 

u ′ + Au = B in I, 

u(c) = C, 

has a unique n × 1 matrix-valued solution u with locally absolutely continuous 

entries defined in I. 

The theorem has the following immediate consequence. 

Corollary 13.2. The set of solutions to u ′ + Au = 0 in I is an 

n-dimensional linear space. 

Proofs for Theorem 13.1 and Corollary 13.2 are given in Appendix 

C. We will apply them for A = J −1 (Q − λW ), where λ ∈ C, and 

B = J −1W v. 

We shall study (13.1) in the Hilbert space L 2 W 

of equivalence classes 

of n × 1 matrix-valued Lebesgue measurable functions u for which 

u∗ W u is integrable over I. In this space the scalar product is 〈u, u〉 = 

I v∗W u. Two functions u and ũ are considered equivalent if the integral 

 

I (u − ũ)∗W (u − ũ) = 0. Note that this means that they can be 

very different pointwise. For example, in the case of the system equivalent 

to (10.1) the second component of an element of L2 W is completely 

undetermined. 

91

92 13. FIRST ORDER SYSTEMS 

Since W is assumed locally integrable it is clear that constant n × 

1 matrices are locally in L2 W so (each components of) W u is locally 

integrable if u ∈ L2 W . It is also clear that u and ũ are two different 

representatives of the same equivalence class in L2 W precisely if W u = 

W ũ almost everywhere (Exercise 13.1). 

Example 13.3. Any standard scalar differential equation may be 

written on the form (13.1) with a constant, skew-Hermitian J. If it 

is possible to do this so that Q and W are Hermitian, the differential 

equation is called formally symmetric. We have already seen this in 

the case of the Sturm-Liouville equation (10.1), which will be formally 

symmetric if p, q and w are real-valued. The first order scalar equation 

iu ′ + qu = wv is already of the proper form and formally symmetric if 

q and w are real-valued. The fourth order equation (p2u ′′ ) ′′ − (p1u ′ ) ′ + 

qu = wv may be written on the form (13.1) by setting 

 

 

U = 

, J = 

and Q = 

u 

hu ′ 

(p2u ′′ ) ′ −p1u ′ 

−p2u ′′ 

0 0 1 0 

0 0 0 1 

−1 0 0 0 

0 −1 0 0 

p0 0 0 0 

0 p1 1 0 

0 1 0 0 

0 0 0 −1/p2 

as is readily seen, and it will be formally symmetric if the coefficients 

w, p0, p1 and p2 are real-valued. 

In order to get a spectral theory for (13.1) it is convenient to use 

the theory of symmetric relations, since it is sometimes not possible 

to find a densely defined symmetric operator realizing the equation. 

Consequently, we must begin by defining a minimal relation, show that 

it is symmetric, calculate its adjoint and find the selfadjoint restrictions 

of the adjoint. We define the minimal relation T0 to be the closure in 

L 2 W ⊕ L2 W of the set of pairs (u, v) of elements in L2 W 

 

, 

with compact 

support in the interior of I (i.e., which are 0 outside some compact 

subinterval of the interior of I which may be different for different pairs 

(u, v)) and such that u is locally absolutely continuous and satisfies the 

equation Ju ′ + Qu = W v. This relation between u and v may or may 

not be an operator (Exercise 13.2). 

The next step is to calculate the adjoint of T0. In order to do this, 

we shall again use the classical variation of constants formula, now in 

a more general form than in Lemma 10.4. Below we always assume 

that c is a fixed (but arbitrary) point in I. Let F (x, λ) be a n × n 

matrix-valued solution of JF ′ + QF = λW F with F (c, λ) invertible. 

This means precisely that the columns of F are a basis for the solutions 

of (13.1) for v = λu. Such a solution is called a fundamental matrix for 

this equation. We will always in addition suppose that S = F (c, λ) is 

independent of λ and symplectic, i.e., S ∗ JS = J. We may for example 

take S equal to the n × n unit matrix or, if J is unitary, S = J. 

Lemma 13.4. We have F ∗ (x, λ)JF (x, λ) = J for any complex λ 

and x ∈ I. The solution u of Ju ′ + Qu = λW u + W v with initial data

u(c) = 0 is given by 

(13.2) u(x) = F (x, λ)J −1 

 

13. FIRST ORDER SYSTEMS 93 

c 

x 

F ∗ (y, λ)W (y)v(y) dy . 

Proof. We have (F ∗ (x, λ)JF (x, λ)) ′ = −(JF ′ (x, λ)) ∗ F (x, λ) + 

F ∗ (x, λ)JF ′ (x, λ) = 0 using the differential equation. It follows that 

F ∗ (x, λ)JF (x, λ) is constant. Since it equals J for x = c this is its 

value for all x ∈ I. It follows that J −1 F ∗ (x, λ) is the inverse matrix of 

JF (x, λ). Straightforward differentiation now shows that (13.2) solves 

the equation. 

Corollary 13.5. If v ∈ L2 W with compact support in I then (13.1) 

has a solution u with compact support in I if and only if 

I v∗W u0 = 0 

for all solutions u0 of the homogeneous equation (13.1) with v = 0. 

Proof. If we choose c to the left of the support of v, then by 

Lemma 13.4 the function u(x) = F (x)J −1 x 

c F ∗W v is the only solution 

of (13.1) which vanishes to the left of c. Since F (x)J −1 is invertible 

(13.1) has a solution of compact support if and only if 

I F ∗W v = 0. 

But the columns of F are linearly independent so they are a basis for 

the solutions of the homogeneous equation. The corollary follows. 

Lemma 13.6. Suppose (u, v) ∈ T ∗ 0 . Then there is a representative 

of the equivalence class u, also denoted by u, which is absolutely continuous 

and satisfies Ju ′ + Qu = W v. Conversely, if this holds, then 

(u, v) ∈ T ∗ 0 . 

Proof. Let u1 be a solution of Ju ′ 1 + Qu1 = W v and assume 

(u0, v0) ∈ T0 has compact support. Integrating by parts we get 

 

v ∗ 

W u0 = (Ju ′ 1 + Qu1) ∗ 

u0 = u ∗ 1(Ju ′ 

0 + Qu0) = u ∗ 1W v0. 

I 

I 

This proves the converse part of the lemma. We also have 0 = 〈u0, v〉− 

〈v0, u〉 = 〈v0, u1 − u〉. Here v0 is an arbitrary compactly supported 

element of L2 W for which there exists a compactly supported element 

u0 ∈ L2 W satisfying Ju′ 0 + Qu0 = W v0. By Corollary13.5 it follows that 

u1 − u solves the homogeneous equation, i.e., u solves (13.1). 

It now follows that T0 is symmetric and that its adjoint is given 

by the maximal relation T1 consisting of all pairs (u, v) in L2 W × L2W such that u is (the equivalence class of) a locally absolutely continuous 

function for which Ju ′ + Qu = W v. We can now apply the theory of 

Chapter 9.2. The deficiency indices of T0 are accordingly the number 

of solutions of Ju ′ + Qu = iW u and Ju ′ + Qu = −iW u respectively 

which are linearly independent in L2 W . Since there are altogether only 

I 

I


n (pointwise) linearly independent solutions of these equations the deficiency 

indices can be no larger than n; in particular they are both 

finite. We now make the following basic assumption. 

Assumption 13.7. If K is a sufficiently large, compact subinterval 

of I there is no non-trivial solution of Ju ′ +Qu = 0 with 

K u∗ W u = 0. 

Note that if there is a solution with u ∗ W u = 0, then W u = 0 so u 

actually also solves Ju ′ + Qu = λW u for any complex λ. The assumption 

automatically holds if (13.1) is equivalent to a Sturm-Liouville 

equation, or more generally an equation of the types discussed in Example 

13.3 and Exercise 13.3. One reason for making the assumption is 

that it ensures that the deficiency indices of T0 are precisely equal to the 

dimensions of the spaces of those solutions of Ju ′ + Qu = ±iW u which 

have finite norm, but the assumption will be even more important in 

the next chapter. 

According to Corollary 9.15 there will be selfadjoint realizations of 

(13.1) precisely if the deficiency indices are equal. We will in the rest 

of this chapter assume that a selfadjoint extension of T0 exists. Some 

simple criterions that ensure this are given in the following proposition, 

but if these do not apply it can, in a concrete case, be very difficult to 

determine whether there are selfadjoint realizations or not. 

Proposition 13.8. The minimal relation T0 has equal deficiency 

indices if either of the following conditions is satisfied: 

(1) J, Q and W are real-valued. 

(2) The interval I is compact. 

Proof. If u ∈ L 2 W satisfies Ju′ + Qu = λW u and the coefficients 

are real-valued, then conjugation shows that u is still in L 2 W and Ju′ + 

Qu = λW u. There is therefore a one-to-one correspondence between 

Dλ and D λ which obviously preserves linear independence. It follows 

that n+ = n−. 

If I is compact, then solutions of Ju ′ + Qu = λW u are absolutely 

continuous in I, and W is integrable in I, so that all solutions are in 

L 2 W . Thus n+ = n− = n. 

Example 13.9. Note that J ∗ = −J, so J can be real-valued only if 

n is even (show this!). Suppose u solves the equation m 

k=0 (pku (k) ) (k) = 

iwu where the coefficients p0, . . . pm are realvalued and w > 0. Then u 

satisfies m 

k=0 (pku (k) ) (k) = −iwu. It follows that if (13.1) is equivalent 

to an equation of this form, then its deficiency indices are always equal 

so that selfadjoint realizations exist. This is in particular the case for 

the Sturm-Liouville equation (10.1). 

We will now take a closer look at how selfadjoint realizations are 

determined as restrictions of the maximal relation. Suppose (u1, v1)


and (u2, v2) ∈ T1. Then the boundary form (cf. Chapter 9) is 

 

(13.3) 〈(u1, v1), U(u2, v2)〉 = i 

 

= i 

I 

I 

(v ∗ 2W u1 − u ∗ 2W v1) 

((Ju ′ 2 + Qu2) ∗ u1 − u ∗ 2(Ju ′ 1 + Qu1)) 

 

= −i 

I 

(u ∗ 2Ju1) ′ = −i lim 

K→I [u ∗ 2Ju1]K, 

the limit being taken over compact subintervals K of I. We must 

restrict T1 so that this vanishes. Like in Chapter 10 this means that 

the restriction of T1 to a selfadjoint relation T is obtained by boundary 

conditions since the limit clearly only depends on the values of u1 and 

u2 in arbitrarily small neighborhoods of the endpoints of I. 

An endpoint is called regular if it is a finite number and Q and W 

are integrable near the endpoint. Otherwise the endpoint is singular. 

If both endpoints are regular, we again say that we are dealing with 

a regular problem. We have a singular problem if at least one of the 

endpoints is infinite, or if at least one of Q and W is not integrable on 

I. 

Consider now the regular case. Since it is clear that both deficiency 

indices equal n in the regular case there are always selfadjoint 

realizations. To see what they look like, let ũ be the boundary value 

 

of (u, v) ∈ T1, i.e., ũ = 

so that the 

u(a) 

u(b) 

 

. Also put B = iJ 0 

0 −iJ 

boundary form is ũ ∗ 2Bũ1. Now if u ∈ Di then 〈u, Uu〉 = 〈u, u〉 so that 

the boundary form is positive definite on Di. Similarly it is negative 

definite on D−i (cf., Corollary 9.17). Since dim Di ⊕D−i = 2n the rank 

of the boundary form is 2n on this space so that the boundary values 

of this space, and a fortiori those of T1, range through all of C 2n . Since 

〈T1, UT0〉 = 0 it follows that the boundary value of any element of T0 

is 0. 

Conversely, to guarantee that 〈T1, Uu〉 = 0 for some u ∈ T1 it is 

obviously enough that the boundary value of u vanishes. Hence the 

minimal relation consists exactly of those elements of the maximal relation 

which have boundary value 0. It is now clear that any maximal 

symmetric restriction of T1 is obtained by restricting the boundary 

values to a maximal subspace of C 2n for which the boundary form vanishes, 

a so called maximal isotropic space for B. We know, since the 

deficiency indices are finite and equal, that all such maximal symmetric 

restrictions are actually selfadjoint (Corollary 9.15). Since the problem 

of finding maximal isotropic spaces of B is a purely algebraic one 

we consider the problem of identifying all selfadjoint restrictions of T1 

solved in the regular case. See also Exercise 13.4.


Clearly all these restrictions are obtained by restricting the boundary 

values of elements in T1 to certain n-dimensional subspaces of C 2n , 

i.e., by imposing n linear, homogeneous boundary conditions on T1. 

We consider a few special cases. One selfadjoint realization is obtained 

by imposing periodic boundary conditions u(b) = u(a) or more generally 

u(b) = Su(a) where S is a fixed matrix satisfying S ∗ JS = J. As 

already mentioned, such a matrix S is often called symplectic, at least 

, so that n is even. 

in the case when S is real, and J = 0 I 

−I 0 

Another possibility occurs if the invertible Hermitian matrix iJ has 

an equal number of positive and negative eigen-values (this obviously 

requires n to be even). In that case we may impose separated boundary 

conditions, i.e., conditions that make both u∗ (a)Ju(a) and u∗ (b)Ju(b) 

vanish. Boundary conditions which are not separated are called coupled. 

It must be emphasized that for n > 2 there are selfadjoint realizations 

which are determined by some conditions imposed only on the 

value at one of the endpoints, and some conditions involving the values 

at both endpoints. 

Let us now turn to the general, not necessarily regular case. We 

first need to briefly discuss Hermitian forms of finite rank. If B is a 

Hermitian form on a linear space L we set LB = {u ∈ L | B(u, L) = 0} 

which is a subspace of L. The rank of B is codim LB (= dim L/LB ). 

In the sequel we assume that B has finite rank. If M is a subspace on 

which the form B is non-degenerate, i.e., there is no non-zero element 

u ∈ M such that B(u, v) = 0 for all v ∈ M, then we must have 

LB ∩ M = {0} so that M has to be finitedimensional. This means, of 

course, that after introducing a basis in M the form B is given on M 

by an invertible matrix. If B is non-degenerate on M, then for every 

u ∈ L there is a unique element v ∈ M (the B-projection of u on M) 

such that B(u − v, M) = 0 (Exercise 13.5). If B is non-degenerate on 

M, but not on any proper superspace of M, we say that M is maximal 

non-degenerate for B. Of course this means exactly that LB ∩M = {0} 

and dim M = rank B so that L = M ˙+L B as a direct sum. 

We call a subspace P of L on which B is positive definite a maximal 

positive definite space for B if P has no proper superspaces on 

which B is positive definite. If B is positive definite on P, then clearly 

dim P ≤ rank B. It follows that forms of finite rank always have maximal 

positive definite spaces. Similarly for negative definite spaces. 

Proposition 13.10 (Sylvester’s law of inertia). Suppose B is a 

Hermitian form of finite rank on a linear space L. Then all maximal 

positive definite subspaces for B have the same dimension. Similarly 

for maximal negative definite subspaces. 

Proof. Suppose P is maximal positive definite for B and that ˜ P 

is another positive definite space for B. Then the B-projection on P 

is injective as a linear map BP : ˜ P → P. For if not, there exists


a non-zero u ∈ ˜ P such that B(u, P) = 0. But then B is positive 

definite on the linear hull of u and P, since B(αu + βv, αu + βv) = 

|α| 2 B(u, u)+|β| 2 B(v, v) for any v ∈ P. This contradicts the maximality 

of P as a positive definite space. From the standard fact dim ˜ P = 

dim BP ( ˜ P) + dim{u ∈ ˜ P | BP u = 0} now follows that dim ˜ P ≤ dim P. 

By symmetry all maximal positive definite subspaces for B have the 

same dimension. Similarly, all maximal negative definite spaces for B 

have the same dimension. 

If P is any maximal positive definite subspace, and N any maximal 

negative definite subspace, for B, we set r+ = dim P and r− = dim N . 

The pair (r+, r−) is called the signature of the form B. 

Proposition 13.11. Suppose P and N are maximal as positive 

and negative definite subspaces for a Hermitian form B of finite rank. 

Then P ∩ N = {0}, the direct sum P ˙+N is a maximal non-degenerate 

space for B, and rank B = r+ + r−. 

Proof. Clearly B can not be both positive and negative on the 

same vector u, so P ∩M = {0}. B is obviously (check!) non-degenerate 

on P ˙+N , and if P ˙+N is not maximal there exists u /∈ P ˙+N such that 

B is non-degenerate on the linear hull M of u and P ˙+N . We may 

assume B(u, P ˙+N ) = 0, since otherwise we can subtract from u its Bprojection 

on P ˙+N . We cannot have B(u, u) = 0 since B would then 

be degenerate on M. But if B(u, u) > 0, then B would be positive 

definite on the linear hull of u and P, contradicting the maximality 

of P. Similarly, if B(u, u) < 0 we would get a contradiction to the 

maximality of N . Therefore P ˙+N is maximal non-degenerate so that 

r+ + r− = rank B. 

Two Hermitian forms Ba and Bb of finite rank are said to be independent 

if each has a maximal non-degenerate space Ma respectively 

Mb such that Ba(Mb, L) = Bb(Ma, L) = 0. It is then clear that 

Ma ∩ Mb = {0} and that Ma ˙+Mb is maximal non-degenerate for 

Bb − Ba. If (r a +, r a −) and (r b +, r b −) are the signatures of Ba and Bb respectively 

it follows that (r b + + r a −, r b − + r a +) is the signature of Bb − Ba. 

Now consider (13.3) and suppose I = (a, b). If u1 = (u1, v1) and 

u2 = (u2, v2) ∈ T1 then −iu ∗ 2Ju1 has a limit both in a and b by (13.3). 

We denote these limits Ba(u1, u2) and Bb(u1, u2) respectively and call 

them the boundary forms at a and b respectively. Clearly Ba and Bb are 

Hermitian forms on T1. Being limits of forms of rank n they both have 

ranks ≤ n (Exercise 13.6). They are also independent. This follows 

from the next lemma. 

Lemma 13.12. Suppose (u, v) ∈ T1. Then there exists (u1, v1) in 

T1 such that (u1, v1) = (u, v) in a right neighborhood of a and (u1, v1) 

vanishes in a left neighborhood of b.


Proof. Let [c, d] be a compact subinterval of I = (a, b) such that 

d 

c F ∗ (·, λ)W F (·, λ) is invertible and put v1 = v in (a, c] and v1 ≡ 0 in 

[d, b). Now let 

u1(x) = F (x, λ)(u(c) + J −1 

x 

F ∗ (y, λ)W (y)v1(y) dy) . 

It is clear that u1 = u in (a, c] and if we choose v1 appropriately in 

[c, d] we can achieve that u1 ≡ 0 in [d, b). In fact, setting v(x) = 

−F (x, λ)( d 

c F ∗ (·, λ)W F (·, λ)) −1 Ju(c) in this interval will do. 

It follows that (u − u1, v − v1) ∈ T1 is 0 near a and equals (u, v) 

near b. We can therefore find a maximal non-degenerate space for Ba 

consisting of elements of T1 vanishing near b. Similarly, a maximal nondegenerate 

space for Bb consisting of elements of T1 vanishing near a. 

Thus Ba and Bb are independent, as claimed. Since the signature of 

the complete boundary form Bb − Ba is (n+, n−) the independence of 

Ba and Bb implies that n+ = r a − + r b + and n− = r a + + r b −, using the 

notation introduced above for the signatures of Ba and Bb. According 

to Corollary 9.15 T1 has selfadjoint restrictions precisely if n+ = n−. 

Reasoning like in the regular case it follows that there are selfadjoint 

restrictions defined by separated boundary conditions precisely if r a + = 

r a − and r b + = r b −, from which n+ = n− follows. In fact, from any two of 

these relations the third clearly follows. 

Consider finally the case when a is a regular endpoint but b possibly 

is singular. In this case Ba is given by Ba(u1, u2) = iu2(a) ∗ Ju1(a), with 

notation as above. Clearly r a + is the number of positive eigenvalues of 

iJ and r a − the number of negative eigenvalues. It follows that selfadjoint 

restrictions of T1 defined by separated boundary conditions exist if and 

only if the deficiency indices are equal and iJ has an equal number 

of positive and negative eigenvalues; in particular n must be even. In 

the Sturm-Liouville case all these conditions are fulfilled, as we already 

know. 


Exercise 13.1. Show that u and ũ are elements of the same equivalence 

class in L2 W if and only if W u = W ũ a.e. 

Exercise 13.2. Verify that T0 is the graph of an operator if (13.1) 

is equivalent to an equation of the type (10.1) (or more generally an 

equation of the type discussed in Exercise 13.3) and w > 0 a.e. in 

I. Also show that in this case Assumption 13.7 holds. Try to show 

this assuming only that w ≥ 0 but w > 0 on a subset of I of positive 

measure (this is considerably harder). 

c


Exercise 13.3. Show that the differential equation iu ′′′ = wv (here 

i = √ −1) can be written on the form (13.1). 

Also show that the equation m 

k=0 (pku (k) ) (k) = wv can be written 

on this form if the coefficients w and p0, p1, . . . , pm satisfy appropriate 

conditions (state these conditions!). 

 

Hint: Put U = in the first case. In the second case, let U be 

u 

hu ′ 

hu ′′ 

the matrix with 2m rows u0, . . . , u2m−1 where uj = u (j) and um+j = 

(−1) j m k=j+1 (pku (k) ) (k−j−1) for j = 0, . . . , m − 1. 

Exercise 13.4. Find all selfadjoint realizations of a regular Sturm- 

Liouville equation. More generally, assume J −1 = J ∗ = −J and show 

that the eigen-values of B are ±1, both with multiplicity n. Then 

describe all maximal isotropic spaces for B. 

Exercise 13.5. Suppose B is a Hermitian form of finite rank on 

a Hilbert space L, and that B is non-degenerate on a subspace M. 

Show that for any u ∈ L there is a unique v ∈ M, the B-projection on 

M, such that B(u − v, M) = 0. Also show that if, and only if, M is 

maximal non-degenerate, then B(u − v, L) = 0. 

Exercise 13.6. Suppose B1, B2, . . . is a sequence of Hermitian 

forms on L with finite rank, all of signature (r+, r−), and suppose 

Bj(u, v) → B(u, v) as j → ∞, for any u, v ∈ L. Show that B is 

a Hermitian form on L of finite rank (s+, s−), where s+ ≤ r+ and 

s− ≤ r−.

CHAPTER 14 

Eigenfunction expansions 

Just as in Chapter 11, we will deduce our results for the system 

(13.1) from a detailed description of the resolvent. As before we will 

prove that the resolvent is actually an integral operator. To see this, 

first note that according to Lemma 13.6 all elements of D1 are locally 

absolutely continuous, in particular they are in C(I). The set C(I) becomes 

a Fréchet space if provided with the topology of locally uniform 

convergence; with a little loss of elegance we may restrict ourselves to 

consider C(K) for an arbitrary compact subinterval K ⊂ I. This is a 

Banach space with norm uK = supx∈K|u(x)|, |·| denoting the norm 

of an n × 1 matrix (Exercise 14.1). The set T1 is a closed subspace of 

H ⊕ H, since T1 is a closed relation. It follows from Assumption 13.7 

that the map T1 ∋ (u, v) ↦→ u ∈ C(I) is well defined, i.e., there can not 

be two different locally absolutely continuous functions u in the same 

L 2 W 

-equivalence class satisfying (13.1) for the same v. The restriction 

map IK : T1 ∋ (u, v) ↦→ u ∈ C(K) is therefore a linear map between 

Banach spaces. 

Proposition 14.1. For every compact subinterval K ⊂ I there 

exists a constant CK such that 

for any (u, v) ∈ T1. 

uK ≤ CK(u, v)W 

Proof. We shall show that the restriction map IK is a closed operator 

if K is sufficiently large. Since IK is everywhere defined in the 

Hilbert space T1 it follows by the closed graph theorem (Appendix A) 

that IK is a bounded operator, which is the statement of the proposition. 

Now suppose (uj, vj) → (u, v) in T1 and uj → ũ in C(K). We 

must show that IK(u, v) = ũ, i.e., u = ũ pointwise in K. We have 

0 ≤ 

K (u − uj) ∗W (u − uj) ≤ u − uj2 and by Lemma 13.4 

uj(x) = F (x, λ)(uj(c) + J −1 

x 

F ∗ (y, λ)W (y)vj(y) dy), 

so letting j → ∞ it is clear that 

K (u − ũ)∗W (u − ũ) = 0 and that ũ 

satisfies Jũ ′ + Qũ = W v, so Assumption 13.7 shows that u − ũ = 0 

pointwise in K if K is sufficiently large. Hence IK is closed, and we 

are done. 

101 

c

102 14. EIGENFUNCTION EXPANSIONS 

We can now show that the resolvent is an integral operator. First 

note that if T is a selfadjoint realization of (13.1), i.e., a selfadjoint 

restriction of T1, then setting HT = D(T ) the resolvent Rλ of the operator 

part ˜ T of T is an operator on HT , defined for λ ∈ ρ( ˜ T ). We define 

the resolvent set ρ(T ) = ρ( ˜ T ) and extend Rλ to all of L2 W by setting 

RλH∞ = 0, and it is then clear that the resolvent has all the properties 

of Theorems 5.2 and 5.3; the only difference is that the resolvent 

is perhaps no longer injective. Given u ∈ L 2 W 

we obtain the element1 

(Rλu, λRλu+u) ∈ T1, so we may also view the resolvent as an operator 

˜Rλ : L 2 W → T1. This operator is bounded since (Rλu, λRλu+u)W ≤ 

((1 + |λ|)Rλ + 1)uW . Hence ˜ Rλ ≤ (1 + |λ|)Rλ + 1, where Rλ 

is the norm of Rλ as an operator on HT . It is also clear that the analyticity 

of Rλ implies the analyticity of ˜ Rλ. We obtain the following 

theorem. 

Theorem 14.2. Suppose I is an arbitrary interval, and that T is a 

selfadjoint realization in L2 W of the system (13.1), satisfying Assumption 

13.7. Then the resolvent Rλ of T may be viewed as a bounded 

to C(K), for any compact subinterval K of I, 

linear map from L2 W 

which depends analytically on λ ∈ ρ(T ), in the uniform operator topology. 

Furthermore, there exists Green’s function G(x, y, λ), an n × n 

matrix-valued function, such that Rλu(x) = 〈u, G∗ (x, ·, λ)〉W for any 

u ∈ L2 W . The columns of y ↦→ G∗ (x, y, λ) are in HT = D(T ) for any 

x ∈ I. 

Proof. We already noted that ρ(T ) ∋ λ ↦→ ˜ Rλ ∈ B(L 2 W , T1) is 

analytic in the uniform operator topology. Furthermore, the restriction 

operator IK : T1 → C(K) is bounded and independent of λ. Hence 

ρ(T ) ∋ λ → IK ˜ Rλ is analytic in the uniform operator topology. In 

particular, for fixed λ ∈ ρ(T ) and any x ∈ I, the components of the 

linear map L 2 W ∋ u ↦→ (IK ˜ Rλu)(x) = Rλu(x) are bounded linear forms. 

By Riesz’ representation theorem we have Rλu(x) = 〈u, G ∗ (x, ·, λ)〉W , 

where the columns of y ↦→ G ∗ (x, y, λ) are in L 2 W . Since Rλu = 0 for 

u ∈ H∞ it follows that the columns of G ∗ (x, ·, λ) are actually in HT 

for each x ∈ I. 

Among other things, Theorem 14.2 tells us that if uj → u in L 2 W , 

then Rλuj → Rλu in C(K), so that Rλuj converges locally uniformly. 

This is actually true even if uj just converges weakly, but we only need 

is the following weaker result. 

Lemma 14.3. Suppose Rλ is the resolvent of a selfadjoint relation 

T as above. Then if uj ⇀ 0 weakly in L 2 W , it follows that Rλuj → 0 

pointwise and locally boundedly. 

1 u = uT + u∞ with uT ∈ HT and (0, u∞) ∈ T and ˜ T RλuT = ( ˜ T − λ)RλuT + 

λRλuT = uT + λRλu. Thus (Rλu, λRλu + u) = (Rλu, λRλu + uT ) + (0, u∞) ∈ T ⊂ 

T1.

14. EIGENFUNCTION EXPANSIONS 103 

Proof. Rλuj(x) = 〈uj, G∗ (x, ·, λ)〉W → 0 since the columns of 

y ↦→ G∗ (x, y, λ) are in L2 W for any x ∈ I. Now let K be a compact 

subinterval of I. A weakly convergent sequence in L2 W is bounded, so 

since Rλ maps L2 W boundedly into C(K), it follows that Rλuj(x) is 

bounded independently of j and x for x ∈ K. 

Corollary 14.4. If the interval I is compact, then any selfadjoint 

restriction T of T1 has compact resolvent. Hence T has a complete 

orthonormal sequence of eigenfunctions in HT . 

Proof. Suppose uj ⇀ 0 weakly in L2 W . If I is compact, then 

Lemma 14.3 implies that Rλuj → 0 pointwise and boundedly in I, 

and hence by dominated convergence Rλuj → 0 in L2 W . Thus Rλ is 

compact. The last statement follows from Theorem 8.3. 

If T has compact resolvent, then the generalized Fourier series of 

any u ∈ HT converges to u in L2 W ; if we just have u ∈ L2W the series 

converges to the projection of u onto HT . For functions in the domain 

of T much stronger convergence is obtained. 

Corollary 14.5. Suppose T has a complete orthonormal sequence 

of eigenfunctions in HT . If u ∈ D(T ), then the generalized Fourier series 

of u converges locally uniformly in I. In particular, if I is compact, 

the convergence is uniform in I. 

Proof. Suppose u ∈ D(T ) = D( ˜ T ), i.e., ˜ T u = v for some v ∈ HT , 

and let ˜v = v − iu, so that u = Ri˜v. If e is an eigenfunction of 

T with eigenvalue λ we have ˜ T e = λe or ( ˜ T + i)e = (λ + i)e so 

that R−ie = e/(λ + i). It follows that 〈u, e〉W e = 〈Ri˜v, e〉W e = 

〈˜v, R−ie〉W e = 1 

λ−i 〈˜v, e〉W e = 〈˜v, e〉Rie. If sNu denotes the N:th partial 

sum of the Fourier series for u it follows that sNu = RisN ˜v, where sN ˜v 

is the N:th partial sum for ˜v. Since sN ˜v → ˜v in HT , it follows from 

Theorem 14.2 and the remark after it that sNu → u in C(K), for any 

compact subinterval K of I. 

The convergence is actually even better than the corollary shows, 

since it is absolute and uniform (see Exercise 14.2). 

Example 14.6. Consider the operator of Example 4.8, which is 

−i d 

dx considered in L2 (−π, π), with the boundary condition u(−π) = 

u(π). This is a regular, selfadjoint realization of (13.1) for n = 1, 

J = −i, Q = 0 and W = 1, and it is clear that H∞ = {0}. Hence there 

is a complete orthonormal sequence of eigenfunctions in L 2 (−π, π). 

The solutions of −iu ′ = λu are the multiples of e iλx , and the bound- 

ary condition implies that λ is an integer. We obtain the classical 

(complex) Fourier series expansion u(x) = ∞ k=−∞ ûkeikx , where ûk 

= 

1 π 

2π −π u(x)e−ikx dx. According to our results, the series converges in 

L 2 (−π, π) for any u ∈ L 2 (−π, π), and uniformly if u is absolutely continuous 

with derivative in L 2 (−π, π).

104 14. EIGENFUNCTION EXPANSIONS 


Exercise 14.1. Show that if K is a compact interval, then C(K) is 

a Banach space with the norm sup x∈K|u(x)|. Also show that if I is an 

arbitrary interval, then C(I) is a Fréchet space (a linear Hausdorff space 

with the topology given by a countable family of seminorms, which is 

also complete), under the topology of locally uniform convergence. 

Exercise 14.2. With the assumptions of Corollary 14.5 the Fourier 

series for u ∈ D(T ) actually converges absolutely and uniformly to u. 

This may be proved just as for the case of a Sturm-Liouville equation, 

which was considered in Exercise 11.2. Do it!

CHAPTER 15 

Singular problems 

We now have a satisfactory eigenfunction expansion theory for regular 

boundary value problems, so we turn next to singular problems. 

We then need to take a much closer look at Green’s function. To do 

this, we fix an arbitrary point c ∈ I; if I contains one of its endpoints, 

this is the preferred choice for c. Next, let F (x, λ) be a fundamental 

matrix for JF ′ + QF = λW F with λ-independent, symplectic initial 

data in c. We will need the following theorem. 

Theorem 15.1. A solution u(x, λ) of Ju ′ + Qu = λW u with initial 

data independent of λ is an entire function of λ, locally uniformly with 

respect to x. 

This means that u(x, λ) is analytic as a function of λ in the whole 

complex plane, and that the difference quotients 1 (u(x, λ+h)−u(x, λ)) 

h 

converge locally uniformly in x as h → 0. The proof is given in Appendix 

C. We can now give the following detailed description of Green’s 

function. 

Theorem 15.2. Green’s function has the following properties: 

(1) For λ ∈ ρ(T ) we have Rλu(x) = 〈u, G ∗ (x, ·, λ)〉W . 

(2) As functions of y the columns of G ∗ (x, y, λ) satisfy the equation 

Ju ′ + Qu = λW u for y = x. 

(3) As functions of y, the columns of G ∗ (x, y, λ) satisfy the boundary 

conditions that determine T as a restriction of T1, for any 

x interior to I. 

(4) G ∗ (x, y, λ) = G(y, x, λ), for all x, y ∈ I and λ ∈ ρ(T ). 

(5) G(x, y, λ) − G(x, y, µ)) = (λ − µ)〈G ∗ (y, ·, µ), G ∗ (x, ·, λ)〉W = 

(λ − µ)RλG ∗ (y, ·, µ)(x), for all x, y ∈ I and λ, µ ∈ ρ(T ). 

Furthermore, there exists an n×n matrix-valued function M(λ), defined 

in ρ(T ) and satisfying M ∗ (λ) = M(λ), such that 

(15.1) G(x, y, λ) = F (x, λ)(M(λ) ± 1 

2 J −1 )F ∗ (y, λ), 

where the sign of 1J 

should be positive for x > y, negative for x < y. 

2 

Proof. We already know (1). Now let K be a compact subinterval 

of I, (u, v) ∈ T0 with support in K, and suppose x /∈ K. We have u ∈ 

D(T0) ⊂ D(T ) and (u, v) = (u, λu + (v − λu)) so that u = Rλ(v − λu). 

105

106 15. SINGULAR PROBLEMS 

We obtain 

0 = u(x) = Rλ(v − λu)(x) = 〈v − λu, G ∗ (x, ·, λ)〉W 

= 〈v, G ∗ (x, ·, λ)〉W − 〈u, λG ∗ (x, ·, λ)〉W . 

But according to Lemma 13.6 this means that each column of y ↦→ 

G ∗ (x, y, λ) is in the domain of the maximal relation for (13.1) on the 

intervals I ∩ (−∞, x) and I ∩ (x, ∞) and satisfies the equation Ju ′ + 

Qu = λW u on these intervals, so (2) follows. It also follows that we 

have 

G ∗ (x, y, λ) = 

 

F (y, λ)P ∗ +(x, λ), y < x 

F (y, λ)P ∗ −(x, λ), y > x, 

for some n × n matrix-valued functions P+ and P−. 

If u is compactly supported and in L2 W we have, for x outside the 

convex hull of the support of u, 

(15.2) Rλu(x) = P±(x, λ)〈u, F (·, λ)〉W . 

The function v = Rλu satisfies the equation Jv ′ + Qv = λW v + W u, 

so we may write P±(x, λ) = F (x, λ)H±(λ), and since Rλu ∈ D(T ) 

it certainly satisfies the boundary conditions determining T . If the 

support of u is large enough the scalar product in (15.2) can be any 

column vector, in view of Assumption 13.7, so for every y each column 

of x ↦→ G(x, y, λ) also satisfies the boundary conditions determining T . 

This proves (3). If the endpoints of I are a and b respectively we now 

have 

x 

 

Rλu(x) = F (x, λ) 

Differentiating this we obtain 

JRλu ′ + (Q − λW )Rλu 

a 

H+(λ)F ∗ (·, λ)W u + 

b 

x 

H−(λ)F ∗ (·, λ)W u . 

= JF (x, λ)(H−(λ) − H+(λ))F ∗ (x, λ)W (x)u(x), 

so JF (x, λ)(H−(λ) − H+(λ))F ∗ (x, λ) should be1 the unit matrix. In 

view of the fact that JF (x, λ) is the inverse of J −1F ∗ (x, λ) this means 

that H−(λ) − H+(λ) = J −1 . If we define M(λ) = (H−(λ) + H+(λ))/2 

we now obtain (15.1). 

If now u and v both have compact supports we have 

 

〈Rλu, v〉W = v ∗ (x)W (x)G(x, y, λ)W (y)u(y) dxdy, 

1 Actually, one must again argue using Assumption 13.7. We leave the details 

to the reader.

15. SINGULAR PROBLEMS 107 

the double integral being absolutely convergent. Similarly 

 

〈u, Rλv〉W = v ∗ (x)W (x)G ∗ (y, x, λ)W (y)u(y) dxdy, 

and since the integrals are equal by Theorem 5.2 (2) and G(x, y, λ) − 

G ∗ (y, x, λ) = F (x, λ)(M(λ) − M ∗ (λ))F ∗ (y, λ) we obtain 

〈F (·, λ), v〉W (M(λ) − M ∗ (λ))〈u, F ∗ (·, λ)〉W = 0. 

By Assumption 13.7 this implies that M(λ) = M ∗ (λ) and thus (4). 

Finally, to prove (5) we use the resolvent relation Theorem 5.2(3). 

For u ∈ L2 W this gives 

〈u, G ∗ (x, ·, λ) − G ∗ (x, ·, µ)〉W = Rλu(x) − Rµu(x) 

Now 

Thus 

= (λ − µ)RλRµu(x) = (λ − µ)〈Rµu, G ∗ (x, ·, λ)〉W 

= 〈u, (λ − µ)RµG ∗ (x, ·, λ)〉W . 

RµG ∗ (x, ·, λ)(y) = 〈G ∗ (x, ·, λ), G ∗ (y, ·, µ)〉W . 

G(x, y, λ) − G(x, y, µ) = (λ − µ)〈G ∗ (y, ·, µ), G ∗ (x, ·, λ)〉W 

= (λ − µ)RλG ∗ (y, ·, µ)(x), 

since both sides are clearly in D(T ). This proves (5). 

Before we proceed, we note the following corollary, which completes 

our results for the case of a discrete spectrum. 

Corollary 15.3. Suppose for some non-real λ that all solutions 

of Ju ′ + Qu = λW u and Ju ′ + Qu = λW u are in L2 W . Then for any 

selfadjoint realization T the resolvent of T is compact. 

In other words, if the deficiency indices are maximal, then the resolvent 

is compact. Actually, the assumptions are here a bit stronger 

than needed. In fact, it is not difficult to show (Exercise 15.1) that if 

all solutions are in L2 W for some λ, real or not, then the same is true 

for all λ. 

Proof. One could use a version of Theorem 8.7 valid for L2 W 

and show that Rλ is a Hilbert-Schmidt operator. Here is an alternative 

proof. Suppose uj ⇀ 0 weakly in L2 W and let I = (a, b). 

Then x 

a F ∗ (y, λ)W (y)uj(y) dy and b 

x F ∗ (y, λ)W (y)uj(y) dy are both 

bounded uniformly with respect to x by Cauchy-Schwarz and since the 

columns of F (·, λ) are in L2 W . The latter fact also shows that the integrals 

tend pointwise to 0 as j → ∞. Since also the columns of F (·, λ) 

are in L2 W it follows that Rλuj → 0 strongly in L2 W by dominated 

convergence.


We will give an expansion theorem generalizing the Fourier series 

expansion obtained for a discrete spectrum. The first step is the following 

lemma. 

Lemma 15.4. Let M(λ) be as in Theorem 15.2. Then there is a 

unique increasing and left-continuous matrix-valued function P with 

P (0) = 0 and unique Hermitian matrices A and B ≥ 0 such that 

(15.3) M(λ) = A + Bλ + 

∞ 

−∞ 

( 1 t 

− 

t − λ t2 ) dP (t). 

+ 1 

Proof. If S = F (c, λ) Theorem 15.2.(5) gives 

S(M(λ) − M(µ))S ∗ = (λ − µ)RλG ∗ (c, ·, µ)(c), 

where the constant matrix S is invertible. Thus M(λ) is analytic in 

ρ(T ), since the resolvent Rλ : L2 W → C(K) is. Furthermore, for µ = λ 

non-real we obtain 

1 

2i Im λ (M(λ) − M ∗ 1 

(λ)) = (M(λ) − M(λ)) 

2i Im λ 

= S −1 〈G ∗ (c, ·, λ), G ∗ (c, ·, λ)〉W (S −1 ) ∗ ≥ 0. 

Thus M is a ‘matrix-valued Nevanlinna function’. We now obtain 

the representation (15.3) by applying Theorem 6.1 to the Nevanlinna 

function m(λ, u) = u ∗ M(λ)u where u is an n × 1-matrix. Clearly the 

quantities α, β and ρ in the representation (6.1) are Hermitian forms 

in u, so (15.3) follows. 

The function P is called the spectral matrix for T . We now define 

the Hilbert space L2 P in the following way. We consider n × 1 matrixvalued 

Borel functions û, so that they are measurable with respect to 

all elements of dP , and for which the integral ∞ 

−∞û∗ (t) dP (t) û(t) < 

∞. The elements of L2 P are equivalence classes of such functions, two 

functions u, v being equivalent if they are equal a.e. with respect to 

dP , i.e., if dP (u − v) has all elements equal to the zero measure. We 

denote the scalar product in this space by 〈·, ·〉P and the norm by 

·P . Note that one may write the scalar product in a somewhat more 

familiar way by using the Radon-Nikodym theorem to find a measure 

dµ with respect to which all the entries in dP are absolutely continuous; 

one may for example let dµ be the sum of all diagonal elements in 

dP . One then has dP = Ω dµ, where Ω is a non-negative matrix of 

functions locally integrable with respect to dµ, and the scalar product 

is 〈û, ˆv〉P = ∞ 

−∞ˆv∗ Ωû dµ. Alternatively, we define L2 P as the completion 

of compactly supported, continuous n×1 matrix-valued functions with 

respect to the norm ·P . These alternative definitions give the same 

space (Exercise 15.2). The main result of this chapter is the following.


Theorem 15.5. 


K F ∗ (y, t)W (y)u(y) dy converges in L2 P for u ∈ 

as K → I through compact subintervals of I. The limit 

L2 W 

is called the generalized Fourier transform of u and is 

denoted by F(u) or û. We write this as û(t) = 〈u, F (·, t)〉W , 

although the integral may not converge pointwise. 

(2) The mapping u ↦→ û has kernel H∞ and is unitary between HT 

and L2 P so that the Parseval formula 〈u, v〉W = 〈û, ˆv〉P holds 

if u, v ∈ L2 W and at least one of them is in HT . 


K F (x, t)dP (t)û(t) converges in HT as K → R 

through compact intervals. If û = F(u) the limit is PT u, where 

PT is the orthogonal projection onto HT . In particular the integral 

is the inverse of the generalized Fourier transform on 

HT . Again, we write u(x) = 〈û, F ∗ (x, ·)〉P for u ∈ HT , although 

the integral may not converge pointwise. 

(4) Let E∆ denote the spectral projector of ˜ T for the interval ∆. 

Then E∆u(x) = 

F (x, t) dP (t) û(t). 

∆ 

(5) If (u, v) ∈ T then F(v)(t) = tû(t). Conversely, if û and tû(t) 

are in L2 P , then F −1 (û) ∈ D(T ). 

We will prove Theorem 15.5 through a sequence of lemmas. First 

note that for u ∈ L2 W with compact support, the function û(λ) = 

〈u, F (·, λ)〉W is an entire, matrix-valued function of λ since F (x, λ), 

and thus also F ∗ (x, λ), is entire, locally uniformly in x, according to 

Theorem 15.1. 

Lemma 15.6. The function 〈Rλu, v〉W − ˆv ∗ (λ)M(λ)û(λ) is entire 

for all u, v ∈ L2 W with compact supports. 

Proof. If the supports are inside [a, b], direct calculation shows 

that the function is 

1 

2 

b 

a 

x a 

 

− 

x 

b 

 

v ∗ (x)W (x)F (x, λ)J −1 F ∗ (y, λ)W (y)u(y) dy dx . 

This is obviously an entire function of λ. 

As usual we denote the spectral projectors belonging to T (i.e., 

those belonging to ˜ T ) by Et. 

Lemma 15.7. Let u ∈ L2 W have compact support and assume a < b 

to be points of differentiability for both 〈Etu, u〉 and P (t). Then 

(15.4) 〈Ebu, u〉 − 〈Eau, u〉 = 

b 

a 

û ∗ (t) dP (t) û(t).


Proof. Let Γ be the positively oriented rectangle with corners in 

a ± i, b ± i. According to Lemma 15.6 

 

 

〈Rλu, u〉 dλ = û ∗ (λ)M(λ)û(λ) dλ 

Γ 

Γ 

if either of these integrals exist. However, by Lemma 15.4, 

 

û ∗ 

(λ)M(λ)û(λ) dλ = û ∗ ∞ 

(λ) ( 1 t 

− 

t − λ t2 ) dP (t) û(λ) dλ . 

+ 1 

Γ 

Γ 

−∞ 

The double integral is absolutely convergent except perhaps where t = 

λ. The difficulty is thus caused by 

1 

−1 

ds 

µ+1 

µ−1 

û ∗ (µ − is)dP (t)û(µ + is) 

t − µ − is 

for µ = a, b. However, Lemma 11.10 ensures the absolute convergence 

of these integrals. Changing the order of integration gives 

 

Γ 

û ∗ (λ)M(λ)û(λ) dλ = 

∞ 

−∞ 

 

Γ 

û ∗ (λ)dP (t)û(λ)( 1 t 

− 

t − λ t2 ) dλ 

+ 1 

 

= −2πi 

a 

b 

û ∗ (t)dP (t)û(t) 

since for a < t inner integral is −û ∗ (t)dP (t)û(t) 

whereas t = a, b do not carry any mass and the inner integrand is 

regular for t < a and t > b. 

Similarly we have 

 

Γ 

〈Rλu, u〉 dλ = 

∞ 

−∞ 

 

d〈Etu, u〉 

Γ 

dλ 

t − λ 

 

= −2πi 

a 

b 

d〈Etu, u〉 

which completes the proof. 

Lemma 15.8. If u ∈ L2 W the generalized Fourier transform û ∈ L2P exists as the L2 

P -limit of K F ∗ (y, t)W (y)u(y) dy as K → I through 

compact subintervals of I. Furthermore, 

〈Etu, v〉W = 

t 

−∞ 

ˆv ∗ (t) dP (t) û(t). 

In particular, 〈PT u, v〉W = 〈û, ˆv〉P if u and v ∈ L 2 W .


Proof. If u has compact support Lemma 15.7 shows that (15.4) 

holds for a dense set of values a, b since functions of bounded variation 

are a.e. differentiable. Since both Et and P are left-continuous we 

obtain, by letting b ↑ t, a → −∞ through such values, 

t 

〈Etu, v〉 = ˆv ∗ (t) dP (t) û(t) 

−∞ 

when u, v have compact supports; first for u = v and then in general 

by polarization. If PT is the projection of L 2 W onto HT we obtain as 

t → ∞ also that 〈PT u, v〉W = 〈û, ˆv〉P when u and v have compact 

supports. 

For arbitrary u ∈ L 2 W 

we set, for a compact subinterval K of I, 

 

uK(x) = 

u(x) 

0 

for x ∈ K 

otherwise 

and obtain a transform ûK. If L is another compact subinterval of I it 

follows that ûK − ûLP = PT (uK − uL)W ≤ uK − uLW , and since 

uK → u in L2 W as K → I, Cauchy’s convergence principle shows that 

ûK converges to an element û ∈ L2 P as K → I. The lemma now follows 

in full generality by continuity. 

Note that we have proved that F is an isometry on HT , and a 

partial isometry on L 2 W . 

Lemma 15.9. The integral 

compact interval and û ∈ L 2 P 

K F (x, t) dP (t) û(t) is in HT if K is a 

, and as K → R the integral converges in 

HT . The limit F −1 (û) is called the inverse transform of û. If u ∈ L2 W 

then F −1 (F(u)) = PT u. F −1 (û) = 0 if and only if û is orthogonal in 

L2 P to all generalized Fourier transforms. 

Proof. If û ∈ L2 P has compact support, then u(x) = 〈û, F ∗ (x, ·)〉P 

is continuous, so uK ∈ L2 W for compact subintervals K of I, and has a 

transform ûK. We have 

uK 2 

W = u ∗ ∞ 

(x)W (x) F (x, t)dP (t)û(t) dx . 

K 

−∞ 

Considered as a double integral this is absolutely convergent, so changing 

the order of integration we obtain 

uK 2 W = 

∞ 

−∞ 

 

K 

F ∗ ∗ (x, t)W (x)u(x) dx dP (t) û(t) 

= 〈û, ûK〉P ≤ ûP ûKP ≤ ûP uKW , 

according to Lemma 15.8. Hence uKW ≤ ûP , so u ∈ L 2 W 

uW ≤ ûP . If now û ∈ L 2 P 

, and 

is arbitrary, this inequality shows (like


in the proof of Lemma 15.8) that 

K F (x, t)dP (t) û(t) converges in L2W as K → R through compact intervals; call the limit u1. If v ∈ L2 W , ˆv 

is its generalized Fourier transform, K is a compact interval, and L a 

compact subinterval of I, we have 

 

( F ∗ (x, t)W (x)v(x) dx) ∗ dP (t) û(t) 

K 

L 

 

= 

L 

v ∗ 

(x)W (x) 

K 

F (x, t)dP (t) û(t) dx 

by absolute convergence. Letting L → I and K → R we obtain 

〈û, ˆv〉P = 〈u1, v〉W . If û is the transform of u, then by Lemma 15.8 

u1 − u is orthogonal to HT , so u1 = PT u. Similarly, u1 = 0 precisely if 

û is orthogonal to all transforms. 

We have shown the inverse transform to be the adjoint of the transform 

as an operator from L2 W into L2P . The basic remaining difficulty is 

to prove that the transform is surjective, i.e., according to Lemma 15.9, 

that the inverse transform is injective. The following lemma will enable 

us to prove this. 

Lemma 15.10. The transform of Rλu is û(t)/(t − λ). 

Proof. By Lemma 15.8, 〈Etu, v〉W = t 

−∞ ˆv∗ dP û, so that 

〈Rλu, v〉W = 

∞ 

−∞ 

d〈Etu, v〉 

t − λ = 

∞ 

−∞ 

By properties of the resolvent 

Rλu 2 = 

ˆv ∗ (t) dP (t) û(t) 

t − λ 

1 

2i Im λ 〈Rλu − R λ u, u〉W = 

∞ 

−∞ 

d〈Etu, u〉W 

|t − λ| 2 

= 〈û(t)/(t − λ), ˆv(t)〉P . 

= û(t)/(t − λ)2 P . 

Setting v = Rλu and using Lemma 15.8, it therefore follows that 

û(t)/(t − λ)2 P = 〈û(t)/(t − λ), F(Rλu)〉P = F(Rλu)2 P . It follows 

that we have û(t)/(t−λ)−F(Rλu)P = 0, which was to be proved. 

Lemma 15.11. The generalized Fourier transform is unitary from 

and the inverse transform is the inverse of this map. 

HT to L 2 P


Proof. According to Lemma 15.9 we need only show that if û ∈ 

L2 P has inverse transform 0, then û = 0. Now, according to Lemma 15.10, 

F(v)(t)/(t − λ) is a transform for all v ∈ L2 W and non-real λ. Thus 

we have 〈û(t)/(t − λ), F(v)(t)〉P = 0 for all non-real λ if û is orthogonal 

to all transforms. But we can view this scalar product as the 

Stieltjes-transform of the measure t 

−∞ F(v)∗ dP û, so applying the inversion 

formula Lemma 6.5 we have 

K F(v)∗dP û = 0 for all compact 

intervals K, and all v ∈ L2 W . Thus the cutoff of û, which equals û in 

K and 0 outside, is also orthogonal to all transforms, i.e., has inverse 

transform 0 according to Lemma 15.9. It follows that 

 

F (x, t)dP (t) û(t) 

K 

is the zero-element of L2 W for any compact interval K. Now multiply 

this from the left with F ∗ (x, s)W (x) and integrate with respect to x 

over a large compact subinterval L ⊂ I. We obtain 

 

B(s, t)dP (t) û(t) = 0 for every s, 

K 

where B(s, t) = 

L F ∗ (x, s)W (x)F (x, t) dx. Thus B(s, t)dP (t) û(t) is 

the zero measure for all s. By Assumption 13.7 the matrix B(s, t) 

is invertible for s = t, so by continuity it is, given s, invertible for t 

sufficiently close to s. Thus, varying s, it follows that dP (t) û(t) is the 

zero measure in a neighborhood of every point. But this means that 

û = 0 as an element of L2 P . 

Lemma 15.12. If (u, v) ∈ T , then ˆv(t) = tû(t). Conversely, if û 

and tû(t) are in L 2 P , then F −1 (û) ∈ D(T ). 

Proof. We have (u, v) ∈ T if and only if u = Rλ(v − λu), which 

holds if and only if û(t) = (ˆv(t) − λû(t))/(t − λ), i.e., ˆv(t) = tû(t), 

according to Lemmas 15.10 and 15.11. 

This completes the proof of Theorem 15.5. We also have the following 

analogue of Corollary 14.5. 

Theorem 15.13. Suppose u ∈ D(T ). Then the inverse transform 

〈û, F ∗ (x, ·)〉P converges locally uniformly to u(x). 

Proof. The proof is very similar to that of Corollary 14.5. Put 

v = ( ˜ T − i)u so that v ∈ HT and u = Riv. Let K be a compact 

interval, and put uK(x) = 

K F (x, t) dP (t) û(t) = F −1 (χû)(x), where 

χ is the characteristic function for K. Define vK similarly. Then by 

Lemma 15.10 

RivK = F −1 ( χ(t)ˆv(t) 

) = F 

t − i 

−1 (χû) = uK .


Since vK → v in L2 W as K → R, it follows from Theorem 14.2 that 

uK → u in C(L) as K → R, for any compact subinterval L of I. 

Example 15.14. Let us interpret Theorem 15.5 for the case of 

the operator of Example 4.6, Green’s function of which is given in 

Example 8.8. Comparing (8.2) with (15.1), we see that M(λ) = i/2 for 

λ in the upper half plane. By Lemma 6.5 the corresponding spectral 

measure is P (t) = limε→0 1 

t 

t 

Im M(µ + iε) dµ = . This means that 

π 0 2π 

if f ∈ L2 (R), then as a, b → ∞ the integral b 

−a f(x) e−ixt dt converges 

in the sense of L2 (R) to a function ˆ f ∈ L2 (R). Furthermore the integral 

 

1 b 

2π −a ˆ f(t) eixt dt converges in the same sense to f as a and b → ∞. 

We also conclude that ∞ 

−∞ |f|2 = 1 

∞ 

2π −∞ | ˆ f| 2 . Finally, if f is locally 

absolutely continuous and together with its derivative in L2 (R), then 

the transform of −if ′ is t ˆ f(t) and conversely, if ˆ f and t ˆ f(t) are both in 

L2 (R), then the inverse transform of ˆ f is locally absolutely continuous, 

and its derivative is in L2 (R) and is the inverse transform of it ˆ f(t). We 

also get from Theorem 15.13 that if f has these properties, then the 

inverse transform of ˆ f converges absolutely and locally uniformly to f. 

Actually, it is here easy to see that the convergence is uniform on the 

whole axis, but nevertheless it is clear that we have retrieved all the 

basic properties of the classical Fourier transform. 


Exercise 15.1. Use, e.g., estimates in the variation of constants 

formula Lemma 13.4 for v = (λ − µ)u to show that all columns of 

F (x, µ) are in L2 W , then so are those of F (x, λ). 

Exercise 15.2. Show that the two definitions of L2 P given in the 

text are equivalent. What needs to be proved is that any measurable 

n × 1 matrix-valued function with finite norm can be approximated in 

norm by a similar function which is C∞ 0 . 

Hint: Use a cut off and convolution with a C ∞ 0 -function of small support. 

Exercise 15.3. In Lemma 15.9 is claimed that for every compact 

interval K the integral 

K F (x, t) dP (t) û(t) ∈ HT , but this is never 

proved; or is it? Clarify this point! 

Exercise 15.4. Consider, as in the beginning of Chapter 10, the 

first order system corresponding to a general Sturm-Liouville equation 

−(pu ′ ) ′ + qu = λwu on [a, b), 

where 1/p, q and w are integrable on any interval [a, x], x ∈ (a, b). 

Also assume that p and q are real-valued functions and w ≥ 0 and not 

a.e. equal to 0. Consider a selfadjoint realization given by separated


boundary conditions (cf. Chapters 10 and 13). This will be a condition 

at a, and if the boundary form does not vanish at b, also a condition 

at b. Choose the point c = 0 and the fundamental matrix F such 

that its first column satisfies the boundary condition at a. Show that 

M(λ) = 

m(λ) 1 

2 

1 

2 

0 

 

, where the Titchmarsh-Weyl function m(λ) is a 

scalar-valued Nevanlinna function. 

ϕ θ 

Now write F = −pϕ ′ −pθ ′ 

 

. Show that there is a scalar Green’s 

function for the operator given by 

 

ϕ(x, λ)ψ(y, λ), x < y, 

g(x, y, λ) = 

ψ(x, λ)ϕ(y, λ), y < x, 

where ψ(x, λ) = θ(x, λ) + m(λ)ϕ(x, λ), with the property that the 

solution of −(pu ′ ) ′ + qu = λwu + wv which is in L 2 w and satisfies the 

boundary conditions is given by u(x) = Rλv(x) = ∞ 

g(x, y, λ)v(y) dy. 

0 

Show also that the spectral matrix P = 

ρ 0 

0 0 , where the spectral 

function ρ is the function in the representation (6.1) for the function 

m(λ), and that 

b 

Im m(λ) = Im λ |ψ(x, λ)| 2 dx. 

Finally show that the generalized Fourier transform of ψ is always given 

by ˆ ψ(t, λ) = 1/(t − λ). 

Thus the spectral theory for the general Sturm-Liouville equation 

has precisely the same basic features as for the simple case treated in 

Chapter 11. 

a

APPENDIX A 

Functional analysis 

In this appendix we will give the proofs of some standard theorems 

from functional analysis. They are all valid in more general situations 

than stated here. As is usual, our proofs will be based upon the following 

important theorem. We have stated it for a Banach space, but 

the proof would be the same in any complete, metric space. 

Theorem A.1 (Baire). Suppose B is a Banach space and F1, F2, . . . 

a sequence of closed subsets of B. If all Fn fail to have interior points, 

so does ∪ ∞ n=1Fn. In particular, the union is a proper subset of B. 

Proof. Let B0 = {x ∈ B | x − x0 ≤ R0} be an arbitrary closed 

ball. We must show that it can not be contained in ∪ ∞ n=1Fn. We do 

this by first selecting a decreasing sequence of closed balls B0 ⊃ B1 ⊃ 

B2 ⊃ · · · such that the radii Rn → 0 and Bn ∩ Fn = ∅ for each n. But 

if we already have chosen B0, . . . , Bn we can find a point xn+1 ∈ Bn (in 

the interior of Bn) which is not contained in Fn+1, since Fn+1 has no 

interior points. Since Fn+1 is closed we can choose a closed ball ⊂ Bn, 

centered at xn+1, and which does not intersect Fn+1. If we also make 

sure that the radius Rn+1 is at most half of the radius Rn of Bn, it 

follows by induction that we may find a sequence of balls as required. 

For k > n we have xk ∈ Bn so that xk − xn ≤ Rn → 0, so that 

x1, x2, . . . is a Cauchy sequence, and thus converges to a limit x. We 

have x ∈ Bn for every n since xk ∈ Bn for k > n and Bn is closed. 

Thus x is not contained in any Fn. B0 being arbitrary, it follows that 

no ball is contained in ∪ ∞ n=1Fn, which therefore has no interior points, 

and the proof is complete. 

A set which is a subset of the union of countably many closed 

sets without interior points, is said to be of the first category. More 

picturesquely such a set is said to be meager. Meager subsets of R n have 

many properties in common with, or analogous to, sets of Lebesgue 

measure zero. There is no direct connection, however, since a meager 

set may have positive measure, and a set of measure zero does not have 

to be meager. A set which is not meager is said to be of the second 

category, or to be non-meager (how about fat?). The basic properties 

of meager sets are the following. 

117

118 A. FUNCTIONAL ANALYSIS 

Proposition A.2. A subset of a meager set is meager, a countable 

union of meager sets is meager, and no meager set has an interior 

point. 

Proof. The first two claims are left as exercises for the reader to 

verify; the third claim is Baire’s theorem. 

The following theorem is one of the cornerstones of functional analysis. 

Theorem A.3 (Banach). Suppose B1 and B2 are Banach spaces 

and T : B1 → B2 a bounded, injective (one-to-one) linear map. If the 

range of T is not meager, in particular if it is all of B2, then T has a 

bounded inverse, and the range is all of B2. 

Proof. We denote the norm in Bj by ·j. Let 

An = {T x | x1 ≤ n} 

be the image of the closed ball with radius n, centered at 0 in B1. The 

balls expand to all of B1 as n → ∞, so the range of T is ∪ ∞ n=1An ⊂ 

∪ ∞ n=1An. The range not being meager, at least one An must have an 

interior point y0. Thus we can find r > 0 so that {y0 + y | y2 < r} ⊂ 

An. Since An is symmetric with respect to the origin, also −y0+y ∈ An 

if y2 < r. Furthermore, An is convex, as the closure of (the linear 

image of) a convex set. It follows that y = 1 

2 ((y0 +y)+(−y0 +y)) ∈ An. 

Thus 0 is an interior point of An. Since all An are similar (An = nA1), 

0 is also an interior point of A1. This means that there is a number 

C > 0, such that any y ∈ B2 for which y2 ≤ C is in A1. For 

such y we may therefore find x ∈ B1 with x1 ≤ 1, such that T x is 

arbitrarily close to y. For example, we may find x ∈ B1 with x1 ≤ 1 

such that y − T x2 ≤ 1 

2 C. For arbitrary non-zero y ∈ B2 we set 

˜y = C 

y2 y, and then have ˜y2 = C, so we can find ˜x with ˜x1 ≤ 1 

and ˜y − T ˜x2 ≤ 1 

2 

C. Setting x = y2 

C 

˜x we obtain 

(A.1) x1 ≤ 1 

C y2 and y − T x2 ≤ 1 

2y2. Thus, to any y ∈ B2 we may find x ∈ B1 so that (A.1) holds (for y = 0, 

take x = 0). 

We now construct two sequences {xj} ∞ j=0 and {yj} ∞ j=0, in B1 respectively 

B2, by first setting y0 = y. If yn is already defined, we define 

xn and yn+1 so that xn1 ≤ 1 

C yn2, yn+1 = yn − T xn, and yn+12 ≤ 

1 

2yn2. We obtain yn2 ≤ 2−ny2 and xn1 ≤ 1 

C 2−ny2 from this. 

Furthermore, T xn = yn+1 − yn, so adding we obtain T ( n 

j=0 xj) = 

y − yn+1 → y as n → ∞. But the series ∞ j=0xj1 converges, 

since it is dominated by 1 

C y2 

∞ 

j=0 2−j = 2 

C y2. Since B1 is com- 

plete, the series ∞ 

j=0 xj therefore converges to some x ∈ B1 satisfying 

x1 ≤ 2 

C y2, and since T is continuous we also obtain T x = y. In

A. FUNCTIONAL ANALYSIS 119 

other words, we can solve T x = y for any y ∈ B2, so the inverse of 

T is defined everywhere, and the inverse is bounded by 2 , so it is 

C 

continuous. The proof is complete. 

In these notes we do not actually use Banach’s theorem, but the 

following simple corollary (which is actually equivalent to Banach’s 

theorem). Recall that a linear map T : B1 → B2 is called closed if the 

graph {(u, T u) | u ∈ D(T )} is a closed subset of B1 ⊕B2. Equivalently, 

if uj → u in B1 and T uj → v in B2 implies that u ∈ D(T ) and T u = v. 

Corollary A.4 (Closed graph theorem). Suppose T is a closed 

linear operator T : B1 → B2, defined on all of B1. Then T is bounded. 

Proof. The graph {(u, T u) | u ∈ B1} is by assumption a Banach 

space with norm (u, T u) = u1 + T u2, where ·j denotes the 

norm of Bj. The map (u, T u) ↦→ u is linear, defined in this Banach 

space, with range equal to B1, and it has norm ≤ 1. It is obviously 

injective, so by Banach’s theorem the inverse is bounded, i.e., there is 

a constant so that (u, T u) ≤ Cu1. Hence also T u2 ≤ Cu1, so 

that T is bounded. 

In Chapter 3 we used the Banach-Steinhaus theorem, Theorem 3.10. 

Since no extra effort is involved, we prove the following slightly more 

general theorem. 

Theorem A.5 (Banach-Steinhaus; uniform boundedness principle). 

Suppose B is a Banach space, L a normed linear space, and M 

a subset of the set L(B, L) of all bounded, linear maps from B into L. 

Suppose M is pointwise bounded, i.e., for each x ∈ B there exists a 

constant Cx such that T xB ≤ Cx for every T ∈ M. Then M is uniformly 

bounded, i.e., there is a constant C such that T xB ≤ CxL 

for all x ∈ B and all T ∈ M. 

Proof. Put Fn = {x ∈ B | T xB ≤ n for all T ∈ M}. Then 

Fn is closed, as the intersection of the closed sets which are inverse 

images of the closed interval [0, n] under a continuous function B1 ∋ 

x ↦→ T xL ∈ R. The assumption means that ∪ ∞ n=1Fn = B. By 

Baire’s theorem at least one Fn must have an interior point. Since Fn 

is convex (if x, y ∈ Fn and 0 ≤ t ≤ 1, then tT x + (1 − t)T yL ≤ 

tT xL + (1 − t)T yL ≤ n) and symmetric with respect to the origin 

it follows, like in the proof of Banach’s theorem, that 0 is an interior 

point in Fn. Thus, for some r > 0 we have T xL ≤ n for all T ∈ M, if 

xB ≤ r. By homogeneity follows that T xL ≤ n 

r xB for all T ∈ M 

and x ∈ B.

APPENDIX B 

Stieltjes integrals 

The Riemann-Stieltjes integral is a simple generalization of the 

(one-dimensional) Riemann integral. To define it, let f and g be two 

functions defined on the compact interval [a, b]. For every partition 

∆ = {xj} n j=0 of [a, b], i.e., a = x0 < x1 < · · · < xn = b, we let the mesh 

of ∆ be |∆| = max(xk − xk−1). This is the length of the longest subinterval 

of [a, b] in the partition. We also choose from each subinterval 

[xk−1, xk] a point ξk and form the sum 

s = 

n 

f(ξk)(g(xk) − g(xk−1)) . 

k=1 

Now suppose that s tends to a limit as |∆| → 0 independently of the 

partition ∆ and choice of the points ξk. The exact meaning of this is 

the following: There exists a number I such that for every ε > 0 there 

is a δ > 0 such that |s − I| < ε as soon as |∆| < δ. In this case we say 

that the integrand f is Riemann-Stieltjes integrable with respect to the 

integrator g and that the corresponding integral equals I. We denote 

this integral by b 

a f(x) dg(x) or simply b 

f dg. The choice g(x) = x 

a 

gives us, of course, the ordinary Riemann integral. 

Proposition B.1. A function f is integrable with respect to a function 

g if and only if for every ε > 0 there exists a δ > 0 such that for 

any two partitions ∆ and ∆ ′ and the corresponding sums s and s ′ , we 

have |s − s ′ | < ε as soon as |∆| and |∆ ′ | are both < δ. 

This is of course a version of the Cauchy convergence principle. We 

leave the proof as an exercise (Exercise B.1). From the definition the 

following calculation rules follow immediately (Exercise B.2). 

(1) 

b 

a 

(2) C 

(3) 

b 

a 

f1 dg + 

b 

a 

b 

a 

f dg = 

 

f dg1 + 

a 

b 

a 

b 

f2 dg = 

Cf dg, 

f dg2 = 

b 

a 

b 

a 

(f1 + f2) dg, 

f d(g1 + g2), 

121

122 B. STIELTJES INTEGRALS 

(4) C 

(5) 

b 

a 

b 

a 

f dg = 

f dg = 

d 

a 

b 

a 

f d(Cg), 

f dg + 

b 

d 

f dg for a < d < b. 

where f, f1, f2, g, g1 and g2 are functions, C a constant and the 

formulas should be interpreted to mean that if the integrals to the left 

of the equality sign exist, then so do the integrals to the right, and 

equality holds. 

Proposition B.2 (Change of variables). Suppose that h is continuous 

and increasing and f is integrable with respect to g over [h(a), h(b)]. 

Then the composite function f ◦h is integrable with respect to g ◦h over 

[a, b] and 

 

h(b) 

h(a) 

f dg = 

b 

a 

f ◦ h d(g ◦ h). 

We leave the proof also of this proposition to the reader (Exercise 

B.3). The formula for integration by parts takes the following 

nicely symmetric form in the context of the Stieltjes integral. 

Theorem B.3 (Integration by parts). If f is integrable with respect 

to g, then g is also integrable with respect to f and 

b 

a 

g df = f(b)g(b) − f(a)g(a) − 

b 

a 

f dg. 

Proof. Let a = x0 < x1 < · · · < xn = b be a partition ∆ of [a, b] 

and suppose xk−1 ≤ ξk ≤ xk, k = 1, . . . , n. Set ξ0 = a, ξn+1 = b. Then 

a = ξ0 ≤ ξ1 ≤ · · · ≤ ξn+1 = b gives a partition ∆ ′ (one discards any 

ξk+1 which is equal to ξk) of [a, b] for which |∆ ′ | ≤ 2|∆| (check this!). 

We have ξk ≤ xk ≤ ξk+1 and 

s = 

n 

g(ξk)(f(xk) − f(xk−1)) = 

k=1 

n 

n−1 

g(ξk)f(xk) − g(ξk+1)f(xk) 

k=1 

= f(b)g(b) − f(a)g(a) − 

k=0 

n 

f(xk)(g(ξk+1) − g(ξk)). 

If |∆| → 0 we have |∆ ′ | → 0, so the last sum converges to b 

f dg 

a 

(note that if ξk+1 = ξk then the corresponding term in the sum is 0). 

It follows that s converges to f(b)g(b) − f(a)g(a) − b 

f dg and the 

a 

theorem follows. 

k=0

B. STIELTJES INTEGRALS 123 

Note that Theorem B.3 is a statement about the Riemann-Stieltjes 

integral; for more general (Lebesgue-Stieltjes) integrals it is not true 

without further assumptions about f and g. The reason is that the 

Riemann-Stieltjes integrals can not exist if f and g have discontinuities 

in common (Exercise B.4), whereas the Lebesgue-Stieltjes integrals exist 

as soon as f and g are, for example, both monotone. In such a case 

the integration by parts formula only holds under additional assumptions, 

for example if f is continuous to the right and g to the left in any 

common point of discontinuity, or if both f and g are normal, i.e., their 

values at points of discontinuity are the averages of the corresponding 

left and right hand limits. 

So far we don’t know that any function is integrable with respect 

to any other (except for g(x) = x which is the case of the Riemann 

integral). 

Theorem B.4. If g is non-decreasing on [a, b], then every continuous 

function f is integrable with respect to g and we have 

b 

 

f dg ≤ max|f|(g(b) 

− g(a)). 

[a,b] 

a 

Proof. Let ∆ ′ and ∆ ′′ be partitions a = x ′ 0 < x ′ 1 < · · · < x ′ m = b 

and a = x ′′ 

0 < x ′′ 

1 < · · · < x ′′ n = b of [a, b] and consider the corresponding 

Riemann-Stieltjes sums s ′ = m 

k=1 f(ξ′ k )(g(x′ k ) − g(x′ k−1 )) and s′′ = 

n 

k=1 f(ξ′′ 

k )(g(x′′ 

k )−g(x′′ 

k−1 )). If we introduce the partition ∆ = ∆′ ∪∆ ′′ , 

supposing it to be a = x0 < x1 < · · · < xp = b, we can write 

s ′ − s ′′ p 

= (f(ξ ′ kj ) − f(ξ′′ qj ))(g(xj) − g(xj−1)) 

j=1 

where kj = k for all j for which [xj−1, xj] ⊂ [x ′ k−1 , x′ k ] and qj = k for 

all j for which [xj−1, xj] ⊂ [x ′′ 

k−1 , x′′ k ] (check this carefully!). Thus, for 

all j, ξ ′ kj and xj are in the same subinterval of the partition ∆ ′ , and 

ξ ′′ 

qj and xj in the same subinterval of the partition ∆ ′′ . It follows that 

|ξ ′ − ξ′′ kj qj | ≤ |ξ′ kj − xj| + |ξ ′′ 

qj − xj| ≤ |∆ ′ | + |∆ ′′ | for all j. Since f 

is uniformly continuous on [a, b], this means that given ε > 0, then 

|f(ξ ′ ) − f(ξ′′ 

kj qj )| ≤ ε if |∆′ | and |∆ ′′ | are both small enough. It follows 

that |s ′ −s ′′ | ≤ ε p j=1 |g(xj)−g(xj−1| = ε(g(b)−g(a)) for small enough 

|∆ ′ | and |∆ ′′ |. Thus f is integrable with respect to g according to 

Proposition B.1. We also have |s ′ | ≤ n k=1 |f(ξ′ k )||g(x′ k ) − g(x′ k−1 )| ≤ 

max|f|(g(b) − g(a)) so the proof is complete. 

As a generalization of Theorem B.4 we may of course take g to be 

any function which is the difference of two non-decreasing functions. 

Such a function is called a function of bounded variation. We shall 

briefly discuss such functions; the main point is that they are characterized 

by having finite total variation.


Definition B.5. Let f be a real-valued function defined on [a, b]. 

Then the total variation of f over [a, b] is 

(B.1) V (f) = sup 

∆ 

n 

|f(xk) − f(xk−1)|, 

k=1 

the supremum taken over all partitions ∆ = {x0, x1, . . . , xn} of [a, b]. 

We have 0 ≤ V (f) ≤ +∞, and if V (f) is finite, we say that f has 

bounded variation on [a, b]. 

When the interval considered is not obvious from the context, one 

may write the total variation of f over [a, b] as V b 

a (f); another common 

notation is b 

|df|. As we mentioned above, a function of bounded 

a 

variation can also be characterized as a function which is the difference 

of two non-decreasing functions. 

Theorem B.6. 

(1) The total variation V b 

a (f) is an interval additive function, i.e., 

if a < x 

a (f) + V b 

x (f) = V b 

a (f). 

(2) A function of bounded variation on an interval [a, b] may be 

written as the difference of two non-decreasing functions. Conversely, 

any such difference is of bounded variation. 

(3) If f is of bounded variation on [a, b], then there are nondecreasing 

functions P and N, such that f(x) = f(a)+P (x)− 

N(x), called the positive and negative variation functions of f 

on [a, b], with the following property: For any pair of nondecreasing 

functions u, v for which f = u − v holds u(x) ≥ 

u(a) + P (x) and v(x) ≥ v(a) + N(x) for a ≤ x ≤ b. 

Proof. It is clear that if a < x 

[a, x] respectively [x, b], then ∆ ∪ ∆ ′ is a partition of [a, b]; the cor- 

responding sum is therefore ≤ V b 

a (f). Taking supremum over ∆ and 

then ∆ ′ it follows that V x 

a (f) + V b 

x (f) ≤ V b 

a (f). On the other hand, 

in calculating V b 

a (f), we may restrict ourselves to partitions ∆ con- 

taining x, since adding new points can only increase the sum (B.1). If 

∆ = {x0, . . . , xn} and x = xp we have p 

k=1 |f(xk) − f(xk−1)| ≤ V x 

a (f) 

respectively m 

k=p+1 |f(xk) − f(xk−1)| ≤ V b 

x (f). Taking supremum over 

all ∆ we obtain V b 

a (f) ≤ V x 

a (f) + V b 

x (f). The interval additivity of the 

total variation follows. 

Setting T (x) = V x 

a (f) the function T is finite in [a, b]; it is called 

the total variation function of f over [a, b]. Since by interval additivity 

T (y)−T (x) = V y 

x (f) ≥ |f(y)−f(x)| ≥ ±(f(y)−f(x)) if a ≤ x ≤ y ≤ b 

it also follows that T is non-decreasing, as are P = 1(T 

+ f − f(a)) 

2 

and N = 1(T 

− f + f(a)). But then f = (f(a) + P ) − N is a splitting 

2 

of f into a difference of non-decreasing functions. Note also that T = 

P + N. Conversely, if u and v are non-decreasing functions on [a, b]

B. STIELTJES INTEGRALS 125 

and {x0, . . . , xn} a partition of [a, x], a < x ≤ b, then 

n 

|(u(xk) − v(xk)) − (u(xk−1) − v(xk−1))| 

k=1 

≤ 

n 

|u(xk) − u(xk−1)| + 

k=1 

n 

|v(xk) − v(xk−1)| 

k=1 

= u(x) − u(a) + v(x) − v(a), 

so that V x 

a (u−v) ≤ u(x)+v(x)−(u(a)+v(a)). In particular, for x = b 

this shows that u − v is of bounded variation on [a, b]. The inequality 

also shows that if f = u − v, then 

P (x) = 1(T 

(x) + f(x) − f(a)) 

2 

≤ 1 

2 

(u(x) − u(a) + v(x) − v(a) + f(x) − f(a)) = u(x) − u(a) . 

Similarly one shows that N(x) ≤ v(x) − v(a) so that the proof is 

complete. 

We remark that a complex-valued function (of a real variable) is 

said to be of bounded variation if its real and imaginary parts are. If 

Tr and Ti are the total variation functions of the real and imaginary 

parts of f, then one defines the total variation function of f to be 

T = T 2 r + T 2 

i (sometimes the definition T = Tr + Ti is used). One 

may also use Definition B.5 for complex-valued functions, and then it 

is easily seen that T 2 r + T 2 

i ≤ T ≤ Tr + Ti. 

Since a monotone function can have only jump discontinuities, and 

at most countably many of them, also functions of bounded variation 

can have at most countably many discontinuities, all of them jump discontinuities. 

Moreover, it is easy to see that the positive and negative 

variation functions (and therefore the total variation function) are continuous 

wherever f is (Exercise B.7). 

Corollary B.7. If g is of bounded variation on [a, b], then every 

continuous function f is integrable with respect to g and we have 

b 

 

(B.2) 

f dg ≤ max|f|V 

[a,b] 

b 

a (g). 

a 

Proof. The integrability statement follows immediately from Theorem 

B.4 on writing g as the difference of non-decreasing functions. To 

obtain the inequality, consider a Riemann-Stieltjes sum 

s = 

n 

f(ξk)(g(xk) − g(xk−1)). 

k=1


We obtain 

n 

|s| ≤ |f(ξk)||g(xk) − g(xk−1)| 

k=1 

≤ max|f| 

[a,b] 

n 

k=1 

|g(xk) − g(xk−1)| ≤ max 

[a,b] 

|f|V b 

a (g) . 

Since this inequality holds for all Riemann-Stieltjes sums, it also holds 

for their limit, which is b 

f dg. 

a 

In some cases a Stieltjes integral reduces to an ordinary Lebesgue 

integral. 

Theorem B.8. Suppose f is continuous and g absolutely continuous 

on [a, b]. Then fg ′ ∈ L1 (a, b) and b 

a f dg = b 

a f(x)g′ (x) dx, where 

the second integral is a Lebesgue integral. 

The proof of Theorem B.8 is left as an exercise (Exercise B.8).

EXERCISES FOR APPENDIX B 127 

Exercises for Appendix B 

Exercise B.1. Prove Proposition B.1. 

Exercise B.2. Prove the calculation rules (1)–(5). 

Exercise B.3. Prove Proposition B.2. 

Exercise B.4. Show that if f and g has a common point of discontinuity 

in [a, b], then f is not Riemann-Stieltjes integrable with respect 

to g over [a, b]. 

Exercise B.5. Show that if f is absolutely continuous on [a, b], 

then f is of bounded variation on [a, b], and V b 

a (f) = b 

a |f ′ |. 

Hint: First show V b 

a (f) ≥ b 

a |f ′ |. To show the other direction, write 

(B.1) on the form b 

a ϕf ′ for a stepfunction ϕ and use Hölder’s inequality. 

Exercise B.6. Show that the set of all functions of bounded variation 

on an interval [a, b] is made into a normed linear space by setting 

f = |f(a)| + V b 

a (f). Convergence in this norm is called convergence 

in variation. Show that convergence in variation implies uniform convergence, 

and that the normed space just introduced is complete (any 

Cauchy sequence of functions in the space converges in variation to a 

function of bounded variation). 

Exercise B.7. Show that a monotone function can have at most 

countably many discontinuities, all of them jump discontinuities. Also 

show that if a function of bounded variation is continuous to the left 

(right) at a point, then so are its positive and negative variation functions, 

and that only if the function jumps up (down) will the positive 

(negative) variation function have a jump. 

Hint: How many jumps of size > 1/j can there be? 

Exercise B.8. Prove Theorem B.8. Also show that if g is absolutely 

continuous on [a, b], then any Riemann integrable f is integrable 

with respect to g and the same formula holds. 

Hint: f(ξk)(g(xk) − g(xk−1) = ϕg ′ where ϕ is a step function 

converging to f. 

Exercise B.9. Suppose f, g are continuous and ρ of bounded variation 

in (a, b). Put σ(t) = t 

f(s) dρ(s) for some c ∈ (a, b). Show that 

c 

b 

a 

g(t) dσ(t) = 

b 

a 

g(t)f(t) dρ(t) . 

Hint: Integrate both sides by parts, first replacing (a, b) by an arbitrary 

compact subinterval.

APPENDIX C 

Linear first order systems 

In this appendix we will prove some standard results about linear 

first order systems of differential equations which are used in the text. 

We will prove no more than we actually need, although the theorems 

have easy generalizations to non-linear equations, more complicated parameter 

dependence, etc. The first result is the standard existence and 

uniqueness theorem, Theorem 13.1, which also implies Theorem 10.1. 

Theorem. Suppose A is an n × n matrix-valued function with locally 

integrable entries in an interval I, and that B is an n × 1 matrixvalued 

function, locally integrable in I. Assume further that c ∈ I and 

C is an n × 1 matrix. Then the initial value problem 

(C.1) 

 

u ′ = Au + B in I, 

u(c) = C, 

has a unique n × 1 matrix-valued solution u with locally absolutely continuous 

entries defined in I. 

Corollaries 13.2 and 10.2 are immediate consequences of the theorem. 

Corollary. Let A and I be as in the previous theorem. Then the 

set of solutions to u ′ = Au in I is an n-dimensional linear space. 

Proof. It is clear that any linear combination of solutions is also 

a solution, so the set of solutions is a linear space. We must show 

that it has dimension n. Let uk solve the initial value problem with 

uk(c) equal to the k:th column of the n × n unit matrix. If u is any 

solution of the equation, and the components of u(c) are x1, . . . , xn, 

then the function x1u1+· · ·+xnun is also a solution with the same initial 

data. It therefore coincides with u, and it is clear that no other linear 

combination of u1, . . . , un has the same initial data as u. It follows 

that u1, . . . , un is a basis for the space of solutions, which therefore is 

n-dimensional. 

Finally we shall prove Theorem 15.1. 

Theorem. A solution u(x, λ) of Ju ′ + Qu = λW u with initial 

data independent of λ is an entire function of λ, locally uniformly with 

respect to x. 

129

130 C. LINEAR FIRST ORDER SYSTEMS 

If we integrate the differential equation in (C.1) from c to x, using 

the initial data, we get the integral equation 

(C.2) u(x) = H(x) + 

x 

c 

Au, 

where H(x) = C+ x 

B. Conversely, if u is continuous and solves (C.2), 

c 

then u has initial data H(c) = C and is locally absolutely continuous 

(being an integral function). Differentiation gives u ′ = Au + B, so that 

the initial value problem is equivalent to the integral equation (C.2). 

In the case of Theorem 13.1, we put A = J −1 (Q − λW ) and B = 0 

to get an equation of the form (C.1). We therefore need to show the 

following theorems. 

Theorem C.1. Suppose A has locally integrable, and H locally absolutely 

continuous, elements. Then the integral equation (C.2) has a 

unique, locally absolutely continuous solution. 

Theorem C.2. Suppose that A depends analytically on a parameter 

λ, in the sense that there is a matrix A ′ (x, λ) which is locally integrable 

with respect to x, and such that 

1 | J h (A(x, λ+h)−A(x, λ))−A′ (x, λ)| → 

0 as h → 0, for all compact subintervals J of I, and all λ in some open 

set Ω ⊂ C. Then the solution u(x, λ) of (C.2) is analytic for λ ∈ Ω, 

locally uniformly in x. 

Proof of Theorem C.1. We will find a series expansion for the 

solution. To do this, we set u0 = H, and if uk is already defined, we set 

uk+1(x) = x 

c Auk. It is then clear that uk is defined for k = 0, 1, . . . 

inductively, and all uk are (absolutely) continuous. I claim that 

sup|uk| 

≤ sup|H| 

[c,x] [c,x] 

1 

k! 

x c 

k |A| 

for k = 0, 1, . . . , 

for x > c, and a similar inequality with c and x interchanged for x < c. 

Here |·| denotes a norm on n-vectors, and also the corresponding subordinate 

matrix-norm (so that |Au| ≤ |A||u|). Indeed, the inequality 

is trivial for k = 0, and supposing it valid for k, we obtain 

|uk+1(x)| ≤ 

x 

c 

|A||uk| ≤ 1 

k! sup 

 

|H| 

[c,x] 

c 

x 

t k |A(t)| |A| dt 

= 

1 

c 

(k + 1)! sup 

[c,x] 

x |H| 

c 

k+1 |A| , 

for c < x, and a similar inequality for x < c. It follows that the series 

u = ∞ 

k=0 uk is absolutely and uniformly convergent on any compact

C. LINEAR FIRST ORDER SYSTEMS 131 

subinterval of I. Therefore u is continuous, and 

∞ 

∞ 

x 

u(x) = uk(x) = H(x) + Auk 

k=0 

k=0 

c 

= H(x) + 

x 

c 

A 

∞ 

uk = H(x) + 

k=0 

x 

c 

Au . 

Thus (C.2) has a solution. To prove the uniqueness, we need the following 

lemma. 

Lemma C.3 (Gronwall). Suppose f ∈ C(I) is real-valued, h is a 

non-negative constant, and g is a locally integrable and non-negative 

function. Suppose that 0 ≤ f(x) ≤ h + | x 

gf| for x ∈ I. Then 

c 

f(x) ≤ h exp(| x 

g|) for x ∈ I. 

c 

The uniqueness of the solution of (C.2) follows directly from this. 

For suppose v is the difference of two solutions. Then v(x) = x 

c Av, 

so setting f = |v| and g = |A| we obtain 0 ≤ f(x) ≤ | x 

gf|. Hence 

c 

f ≡ 0 by Lemma B.3, and thus v ≡ 0, so that (C.2) has at most one 

solution. 

It remains to prove the lemma. 

Proof of Lemma C.3. We will prove the lemma for c < x, leaving 

the other case as an exercise for the reader. Set F (x) = h + x 

c gf. 

Then f ≤ F and F ′ = gf so that F ′ ≤ gF . Multiplying by the integrating 

factor exp(− x 

d 

g) we get c dx (F (x) exp(− x 

g)) ≤ 0 so that 

c 

F (x) exp(− x 

c g) is non-increasing. Thus F (x) exp(− x 

g) ≤ F (c) = h 

c 

for x ≥ c. We obtain f(x) ≤ F (x) ≤ h exp( x 

g), which was to be 

c 

proved. 

Proof of Theorem C.2. It is clear by their definitions that the 

functions uk in the proof of Theorem C.1 are analytic in Ω as functions 

of λ, locally uniformly in x (this is a trivial induction). But the solution 

u is the locally uniform limit, in x, λ, of the partial sums j k=1 uk. Since 

uniform limits of analytic functions are analytic, we are done.

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