Operators on Hilbert Spaces - user web page - AIMS

<strong>Operators</strong> on Hilbert Spaces
Margaret Modupe Kajotoni (meg@aims.ac.za)
African Institute for Mathematical Sciences (AIMS)
Supervised by Sonja Mouton
University of Stellenbosch, South Africa
June 4, 2007

Abstract
The concepts of arbitrary vector spaces can be generalised to inner product spaces and complete
inner product spaces, called Hilbert spaces. The inner product is a generalisation of the dot
product in R n . The dot product and orthogonality are important in many applications. The
theory of inner product spaces and Hilbert spaces is richer than that of general normed and
Banach spaces.
We recall that a Banach space is a complete normed vector space. Hence, every Hilbert space is
a Banach space but the converse is not generally true. A necessary and sufficient condition for a
Banach space to be a Hilbert space is that the parallelogram equality, which will be discussed in
a subsequent section, should hold.
The distinguishing features between Hilbert spaces and Banach spaces are
1. Representation of a Hilbert space as a direct sum of a closed subspace and its orthogonal
complement.
2. Orthogonal sets and sequences and corresponding representations of the elements of the
Hilbert space.
3. The Riesz representation of bounded linear functionals by inner products.
4. The Hilbert-adjoint operator T ∗ of the bounded linear operator T.
Spectral theory is a very broad but important aspect of applied functional analysis which extends
to the eigenvector and eigenvalue theory of a single square matrix. The name was introduced
by David Hilbert in his original formulation of Hilbert space theory. It was later discovered that
spectral theory could explain features of atomic spectral in quantum mechanics. In this essay, we
will consider the spectrum of a Hermitian (or self-adjoint) operator.
i

Contents
Abstract i
1 Introduction 1
2 Basic Concepts 2
2.1 Inner product spaces and Hilbert spaces . . . . . . . . . . . . . . . . . . . . . . 2
2.2 Properties of inner product space . . . . . . . . . . . . . . . . . . . . . . . . . 3
2.3 Orthogonal complement and direct sum . . . . . . . . . . . . . . . . . . . . . . 5
3 <strong>Operators</strong> 10
3.1 Functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
3.2 Riesz Representation of functionals . . . . . . . . . . . . . . . . . . . . . . . . 11
3.3 Sesquilinear functionals and Riesz representation . . . . . . . . . . . . . . . . . 12
3.4 Hilbert-adjoint and self-adjoint operators . . . . . . . . . . . . . . . . . . . . . 15
4 Spectral Analysis of a Self-Adjoint Operator 21
5 Conclusion and further work 25
Bibliography 27
ii

1. Introduction
The equation Lf = g occurs frequently in fields like mathematics, physics and engineering. Here
g is a known function of space and time variables, L is a linear differential operator or mapping
and f is a function which is unknown but required to satisfy some initial or boundary conditions.
The major problem will be that of finding f. No matter the trick used, it is not certain to work
for all equations. Moreover, it is unlikely to obtain a general solution in a form explicit enough
to tell everything about the system. We will be faced with questions like, does every equation
has a solution? If so, is it unique? In dealing with these questions, we might seek inspiration
from linear algebra since Lf = g resembles Ax = b, where A is a matrix and x, b are vectors.
In finite dimensions, the solution exists and is unique if and only if the determinant of L is not
equal to zero (det L = 0). However, the differences between finite and infinite dimensions are
great. In infinite dimensions, determinants can only play a very limited role. Nineteenth-century
analysts made great progress with these questions. In trying to present the result with simplicity
and generality brought about the evolution of functional analysis.
In modern view, functional analysis is seen as the study of spaces of functions over the real or
complex numbers. Hilbert spaces are the most useful spaces in practical applications of functional
analysis. The notion of the Hilbert space was initiated by D. Hilbert, a German mathematician,
in 1912. Completeness is an extremely important property of Hilbert spaces as we shall see.
Complete spaces posses many useful properties that are absent in incomplete spaces. Hilbert
spaces are useful in partial differential equations, quantum mechanics and signal processing.
<strong>Operators</strong> or mappings such as L in the equation above can be defined on Hilbert spaces. In
particular, bounded linear operators can be defined on Hilbert spaces and this is what we seek
to study in this essay. We shall acquaint ourselves with properties of normed and inner product
spaces. We shall discuss mappings or operators from one space to another. We need at least
two sets in order to define an operator. If the two sets are vector spaces, we can introduce the
concept of a linear operator, if the sets are normed spaces, we can construct a theory of bounded
linear operators on such spaces. <strong>Operators</strong> that map members of a specified space into the real
or complex numbers are called functionals. We shall also discuss the Hilbert-adjoint operators as
well as the self-adjoint, unitary and normal operators. Finally, we shall look at the spectral analysis
of the Hermitian (self-adjoint) operators which is an important aspect of functional analysis.
1

2. Basic Concepts
2.1 Inner product spaces and Hilbert spaces
To make a definition that will be applicable to both real and complex vector spaces, we need to
examine some properties of the complex plane C n . Recall that if λ = a + bi, where a, b ∈ R,
then the absolute value of λ is defined by |λ| = √ a 2 + b 2 and the complex conjugate of λ is
defined by λ = a − bi. Also, |λ| 2 = λλ. For z = (z1, . . . , zn) ∈ C n , we define the norm of z by
z = |z1| 2 + · · · + |zn| 2 . We want z to be nonnegative hence we take the absolute values
of its elements. Note that z 2 = z1z1 + · · · + znzn. Also for any w = (w1, . . . , wn) ∈ C n , the
inner product of w with z is equal to the complex conjugate of the inner product of z with w.
Definition 2.1.1 ([Kre78], 3·1-1 (Inner product))
An inner product on a vector space X is a function that takes each ordered pair (x, y) of elements
of X to number 〈x, y〉 of the elements of K (where K is the scalar field of X). That
is, with every pair of vector (x, y), there is associated a scalar 〈x, y〉 with the following properties:
1. 〈x, x〉 ≥ 0 for all x ∈ X,
2. 〈x, x〉 = 0 if and only if x = 0,
3. 〈x + y, z〉 = 〈x, z〉 + 〈y, z〉 for all x, y, z ∈ X,
4. 〈ay, z〉 = a〈y, z〉 for all a ∈ K, where K is the scalar field of X and y, z ∈ X,
5. 〈y, z〉 = 〈z, y〉 for all y, z ∈ X.
An inner product on X defines a norm on X given by x = 〈x, x〉 = 〈x, x〉 1
2 and a metric on
X given by d(x, y) = x − y = 〈x − y, x − y〉.
An inner product space (or pre-Hilbert space) is a vector space with inner product 〈x, y〉 defined
on it.
Definition 2.1.2 ([Kre78], 3·1-2 (Orthogonality))
An element x of an inner product space X is said to be orthogonal to an element y ∈ X if
〈x, y〉 = 0. It is denoted by x ⊥ y and we say that x and y are orthogonal. Similarly, for subsets
A, B ⊂ X, we write x ⊥ A if x ⊥ a for all a ∈ A, and A ⊥ B if a ⊥ b for all a ∈ A and all
b ∈ B.
Definition 2.1.3 (Hilbert space)
A Hilbert space is a complete inner product space.
2

Section 2.2. Properties of inner product space Page 3
Example 2.1.4 ([Kre78], 3·1-3 (Examples of Hilbert spaces))
1. The Euclidean space Rn is a Hilbert space with inner product defined by
〈x, y〉 = x1y1 + · · · + xnyn where x = (x1, ..., xn) and y = (y1, · · · , yn). We obtain that
x = 〈x, x〉 1
2 = (x 2 1 + · · · + x 2 n) 1
2 and the metric defined on it is given by
d(x, y) = x − y = 〈x − y, x − y〉 1
2 = (x1 − y1) 2 1
+ · · · + (xn − yn)
2 2 .
2. The unitary space Cn : 〈x, y〉 = x1y1 + · · · + xnyn. 3. The Hilbert space l2
=
, with inner product defined by 〈x, y〉 =
∞ j=1 |xj| 2 < ∞
∞ j=1 xjyj and the norm x = 〈x, x〉 1
2 = ( ∞ j=1 |xj| 2 ) 1
2 .
2.2 Properties of inner product space
The norm on an inner product space X satisfies the following properties.
Theorem 2.2.1 ([Erd80], Lemma 1.4 (Parallelogram law))
Proof.
If x, y ∈ X, then
x + y 2 + x − y 2 = 2(x 2 + y 2 ). (2.1)
x + y 2 + x − y 2 = 〈x + y, x + y〉 + 〈x − y, x − y〉
= x 2 + y 2 + 〈x, y〉 + 〈y, x〉 + x 2 + y 2 − 〈x, y〉 − 〈y, x〉
= 2(x 2 + y 2 ).
Remark 2.2.2 If a norm does not satisfy the parallelogram equality, then it cannot be obtained
from an inner product space.
Theorem 2.2.3 ([Erd80], Lemma 1.5 (Pythagorean theorem)) If x, y are orthogonal in an inner
product space X, then
Proof.
Suppose x, y are orthogonal vectors in X. Then
x + y 2 = x 2 + y 2 . (2.2)
x + y 2 = 〈x + y, x + y〉
= x 2 + y 2 + 〈x, y〉 + 〈y, x〉
= x 2 + y 2 . (2.3)

Section 2.2. Properties of inner product space Page 4
The Cauchy-Schwarz inequality gives one of the most important inequalities in Mathematics.
Before we state and prove the theorem, let us examine how we can decompose two orthogonal
vectors. If x, y ∈ X, then we would like to write x as a scalar multiple of y plus a vector w
orthogonal to y. This is called the orthogonal decomposition. Now let a ∈ K (where K is the
scalar field of X). Then x = ay + (x − ay). We choose a so that y is orthogonal to (x − ay).
That is, we want
(where y = 0). Hence we write
0 = 〈x − ay, y〉 = 〈x, y〉 − ay 2 thus, a =
x =
〈x, y〉
y + x −
y2 We will use this to prove the Cauchy-Schwarz inequality.
〈x, y〉
y .
y2 〈x, y〉
y 2
Theorem 2.2.4 ([Erd80], Theorem 1.1 (Cauchy-Schwarz inequality)) The Cauchy Schwarz inequality
states that if x, y ∈ X, then
|〈x, y〉| ≤ xy. (2.4)
Proof.
Let x, y ∈ X. If y = 0, then both sides of (2.4) will be equal to 0 and the inequality will hold.
Suppose y = 0. Consider the orthogonal decomposition x = 〈x,y〉
y2 y + w, where w is orthogonal
to y. By the Pythagorean theorem,
x 2
=
〈x, y〉
y
y2
2
+ w 2 =
|〈x, y〉|2
y 2
+ w2 ≥
|〈x, y〉|2
y 2 . (2.5)
Multiplying both sides of equation (2.5) by y 2 and then taking square roots yield the Cauchy-
Schwarz inequality.
Remark 2.2.5 Equation (2.4) is an equality if and only if (2.5) is an equality. This will happen
if and only if w = 0. But w = 0 if and only if x is a multiple of y. Hence equation (2.4) is an
equality if and only if one of x or y is a scalar multiple of the other.
Theorem 2.2.6 ([Axl97], 6.9 (Triangular inequality)) The triangular inequality states that the
length of any side of a triangle is less than the sum of the lengths of the other two sides. If x,y
∈ X, then
x + y ≤ x + y. (2.6)
This theorem can be used to show that the shortest path between two points is a straight line
segment.

Section 2.3. Orthogonal complement and direct sum Page 5
Proof.
Let x, y ∈ X. Then
and by Cauchy-Schwarz inequality,
x + y 2 = 〈x + y, x + y〉
= 〈x, x〉 + 〈y, y〉 + 〈x, y〉 + 〈y, x〉
= 〈x, x〉 + 〈y, y〉 + 〈x, y〉 + 〈x, y〉
= x 2 + y 2 + 2Re〈x, y〉
≤ x 2 + y 2 + 2|〈x, y〉| (2.7)
≤ x 2 + y 2 + 2xy
= (x + y) 2
Therefore x + y ≤ x + y.
Theorem 2.2.7 ([Kre78], 3·2-2 Lemma (Continuity of inner product))
In an inner product space, if xn → x and yn → y, then
〈xn, yn〉 → 〈x, y〉. (2.8)
Proof.
The condition 〈xn, yn〉 → 〈x, y〉 is equivalent to |〈xn, yn〉 − 〈x, y〉| → 0 as n → ∞.
|〈xn, yn〉 − 〈x, y〉| = |〈xn, yn〉 − 〈xn, y〉 + 〈xn, y〉 − 〈x, y〉|
≤ |〈xn, yn − y〉| + |〈xn − x, y〉| (By triangular inequality)
≤ xnyn − y + xn − xy (By Schwarz inequality)
Now (xn) is convergent and hence bounded, so that xn ≤ K for all n ∈ N. Therefore
|〈xn, yn〉 − 〈x, y〉| ≤ Kyn − y + xn − xy.
Since yn − y → 0 and xn − x → 0, the last expression tends to zero as n tends to infinity.
2.3 Orthogonal complement and direct sum
In a metric space X, the distance δ from an element x ∈ X to a nonempty subset M ⊂ H is
given by
δ = inf d(x, y). (2.9)
y∈M
In a normed space it becomes
δ = inf x − y. (2.10)
y∈M

Section 2.3. Orthogonal complement and direct sum Page 6
Theorem 2.3.1 ([Kre78], [Red53], 3·3-1 Theorem (Minimising vector)) Let X be an inner product
space and let M = ∅ be a convex subset which is complete (in the metric induced by the
inner product). Then for every given x ∈ X, there exists a unique y ∈ M such that
δ = inf
ˆy∈M
x − ˆy = x − y. (2.11)
Proof.
Existence
We choose a sequence (yn) in M such that δn → δ, where δn = x − yn (by definition of
infimum). We will show that (yn) is a Cauchy sequence and then we can make some deductions.
Using the parallelogram law,
But 1
yn − ym 2 = (yn − x) − (ym − x) 2
= 2yn − x 2 + 2ym − x 2 − (yn − x) + (ym − x) 2
= 2δn 2 + 2δm 2 − 2 2 1
2 (yn + ym) − x 2 .
2 (yn + ym) ∈ M so that 1
2 (yn + ym) − x2 ≥ δ2 . As n and m tends to infinity, we have
yn − ym ≥ 0, which implies that (yn) is a Cauchy sequence. Now since M is complete, the
sequence (yn) will converge to a limit say y0 ∈ M so that x − y0 ≥ δ. Also
x − y0 = x − yn + yn − y0 ≤ x − yn + yn − y0
= δn + yn − y0.
As n tends to infinity, yn − y0 tends to zero and δn tends to δ. Hence x − y0 ≤ δ and we
conclude that x − y0 = δ.
Uniqueness
We assume that y1 ∈ M and y2 ∈ M both satisfy x − y1 = δ and x − y2 = δ. We show
that y1 = y2. By the parallelogram equality,
y1 − y2 2 = (y1 − x) − (y2 − x) 2
= 2y1 − x 2 + 2y2 − x 2 − (y1 − x) + (y2 − x) 2
= 2δ 2 + 2δ 2 − 2 2 1
2 (y1 + y2) − x 2
But 1
2 (y1 + y2) ∈ M so that 1
2 (y1 + y2) − x 2 ≥ δ 2 , which implies that
Hence
2δ 2 + 2δ 2 − 2 2 1
2 (y1 + y2) − x 2 ≤ 4δ 2 − 4δ 2 .
y1 − y2 ≤ 0.
Clearly y1 − y2 ≥ 0, which means that y1 = y2.

Section 2.3. Orthogonal complement and direct sum Page 7
Theorem 2.3.2 ([Kre78], 3·3-2 Lemma (Orthogonality)) In Theorem 2.3.1, let M be a complete
subspace Y and x ∈ X fixed. Then z = x − y is orthogonal to Y.
Proof.
Suppose by contradiction that z is not orthogonal to Y. Then there would be a y1 ∈ Y such that
y1 = 0 since otherwise 〈z, y1〉 = 0.
Now for any scalar α,
z − αy1 2 = 〈z − αy1, z − αy1〉
〈z, y1〉 = β = 0. (2.12)
= 〈z, z〉 − ¯α〈z, y1〉 − α[〈y1, z〉 − ¯α〈y1, y1〉]
= 〈z, z〉 − ¯αβ − α[ ¯ β − ¯α〈y1, y1〉]. (2.13)
Let us choose α = β
〈y1,y1〉 , so that ¯α = ¯ β
〈y1,y1〉 . Then the expression ¯ β − ¯α〈y1, y1〉 becomes zero.
From Theorem 2.3.1, we have that z = x − y = δ for some y ∈ Y, so that equation (2.13)
becomes
z − αy1 2 = z 2 − |β|2
〈y1, y1〉 < δ2 .
But this is impossible since z − αy1 = x − y2, where y2 = y + αy1 ∈ Y so that z − αy1 ≥ δ,
by definition of δ. Hence equation (2.12) cannot hold, and the lemma is proved.
Definition 2.3.3 ([Erd80], Orthogonal complement)
Given a subspace M of H, the orthogonal complement M ⊥ is defined by
M ⊥ = {x ∈ X : 〈x, m〉 = 0 for all m ∈ M } . (2.14)
Lemma 2.3.4 ([Erd80], 1·9 Lemma) For subsets A and B of a Hilbert space H, if B ⊃ A, then
(i) A ⊂ A ⊥⊥ ,
(ii) B ⊥ ⊂ A ⊥ ,
(iii) A ⊥⊥⊥ = A ⊥ .
Proof.
(i) Let a ∈ A. Then a ⊥ A ⊥ and by Definition 2.3.3 a ∈ A ⊥⊥ . Hence A ⊂ A ⊥⊥ .
(ii) Let y ∈ B ⊥ and a ∈ A. Then A ⊂ B implies that a ∈ B. Thus y ⊥ a. Hence y ∈ A ⊥ .
(iii) Applying (i) to A ⊥ gives A ⊥ ⊂ A ⊥⊥⊥ . Also from (i), A ⊂ A ⊥⊥ and applying (ii) gives
A ⊥ ⊃ A ⊥⊥⊥ . Hence A ⊥⊥⊥ = A ⊥ .

Section 2.3. Orthogonal complement and direct sum Page 8
Definition 2.3.5 (([Kre78], 3·3-3 Definition (Direct Sum)) A vector space X is said to be the
direct sum of two subspaces Y and Z of X if each x ∈ X has a unique representation
We denote the direct sum of Y and Z by
x = y + z, with y ∈ Y and z ∈ Z. (2.15)
X = Y ⊕ Z. (2.16)
Our main interest is to represent a Hilbert space H as a direct sum of a closed subspace Y and
its orthogonal complement Y ⊥ = {z ∈ H |z ⊥ Y}. This leads us to the next theorem which is
sometimes called the projection theorem. Before we proceed, we will state and prove two useful
lemmas.
Lemma 2.3.6 If A is a subspace of a Hilbert space H, then A ⊥ is closed.
Proof. Clearly, A ⊥ is a vector subspace. To show that it is closed, let tn ∈ A ⊥ be a sequence
converging to t. Then by the continuity of inner product, Theorem 2.2.7, for all m ∈ A,
〈t, m〉 = limn→∞〈tn, m〉 = 0 so that t ∈ A ⊥ .
Lemma 2.3.7 ([Erd80], Corollary 1·8) If N is a closed subspace of a Hilbert space H, then
N ⊥ = {0} ⇔ N = H. (2.17)
Proof.
Clearly if N = H then N ⊥ = {0}. Now suppose that N ⊥ = {0} and N = H. Take h /∈ N.
Then there exists n0 ∈ N such that d(h, N) = h − n0 = 0. Moreover by Theorem 2.3.2,
0 = h − n0 ⊥ N, so that N ⊥ = {0} which contradicts our assumption that N ⊥ = {0}. Hence
N = H.
Theorem 2.3.8 ([Erd80], Theorem 1·11) Let N be any closed subspace of a Hilbert space H.
Then
H = N ⊕ N ⊥ . (2.18)
Proof.
If N is a closed subspace of H, then by Theorem 2.3.8, N ⊕ N ⊥ is also a closed subspace of
H. Also if x ∈ (N ⊕ N ⊥ ) ⊥ then x ∈ N ⊥ ∩ N ⊥⊥ so that x = {0}. Hence from Theorem 2.3.7,
N ⊕ N ⊥ = H.
Corollary 2.3.9 A subspace A of a Hilbert space is closed if and only if A = A ⊥⊥ .

Section 2.3. Orthogonal complement and direct sum Page 9
Proof.
Let A be a closed subspace of a Hilbert space H. Then by Lemma 2.3.4 (i), A ⊂ A ⊥⊥ .
Let x ∈ A ⊥⊥ . Since A is a closed subspace of H, then by Theorem 2.3.8, H = A ⊕ A ⊥ and for
some y ∈ A and z ∈ A ⊥ we have that x = y + z. Now since A ⊂ A ⊥⊥ , we have that y ∈ A ⊥⊥
and z = x − y ∈ A ⊥⊥ . But z ∈ A ⊥ , so that z = 0. Hence x = y ∈ A and A ⊥⊥ = A.
Conversely let A = A ⊥⊥ . Since A ⊥ ⊂ H, by Lemma 2.3.6 we have that (A ⊥ ) ⊥ = A is closed.

3. <strong>Operators</strong>
3.1 Functionals
Definition 3.1.1 ([Red53], 3·2 (Linear functional)) A linear functional on a Hilbert space H is
a linear map from H to C. That is
ϕ : H → C. (3.1)
Definition 3.1.2 ([Red53], 5·2 (Bounded linear functional)) A linear functional ϕ is bounded,
or continuous, if there exists a constant k such that
.
The norm of a bounded linear functional is
If y ∈ H, then
is a bounded linear functional on H, with ϕy = y.
Example 3.1.3 The function ϕ : L 2 (a, b) → R defined by
|ϕ(x)| ≤ kx for all x ∈ H. (3.2)
ϕ = sup |ϕ(x)|. (3.3)
x=1
ϕy(x) = 〈y, x〉 (3.4)
ϕ(u) =
b
a
ϕ u(x)dx. (3.5)
is a functional. Where L2
(a, b) = u integrable | b
a |ϕ|2
≤ +∞ and 〈ϕ, u〉 = b
ϕ u(x)dx
a
defines an inner product on L2 . Thus for every u ∈ L2
b
(a, b), ϕL2 = a ϕ2u2 (x)dx. Clearly,
for ϕ = 1, ϕ is a linear functional since 〈ϕ, αu+βv〉 = α〈ϕ, u〉+β〈ϕ, v〉. Further more, applying
the Cauchy-Schwarz inequality on L2 we have
b
|〈ϕ, u〉| =
1 · u(x)dx
≤ 1L2uL2 = |b − a|uL2. Hence ϕ is bounded.
a
Remark 3.1.4 One of the fundamental facts about Hilbert spaces is that all bounded linear
functionals are of the form given in equation (3.4). This is the basis for the next theorem.
10

Section 3.2. Riesz Representation of functionals Page 11
3.2 Riesz Representation of functionals
Theorem 3.2.1 ([Red53], 5·4 Theorem (Riesz representation theorem)) Every bounded linear
functional ϕ on a Hilbert space H can be represented in terms of the inner product
ϕ(x) = 〈x, u〉, (3.6)
where u depends on ϕ and is uniquely determined by ϕ and has norm ϕ = u.
We want to prove that
(a) ϕ has a representation ϕ(x) = 〈x, u〉,
(b) u is unique,
(c) ϕ = u holds.
Proof.
If ϕ = 0 then (a), (b) and (c) hold, if we take u = 0. So suppose that ϕ = 0.
(a) Now N(ϕ) is a closed subspace of H. Furthermore since ϕ = 0, we have that N(ϕ) = H,
which implies that N(ϕ) ⊥ = 0 by Theorem 2.3.7. Hence there must be at least one nonzero
element, say u0 in N(ϕ) ⊥ .
Now set z = ϕ(x)u0 − ϕ(u0)x, then apply ϕ to give ϕ(z) = ϕ(x)ϕ(u0) − ϕ(u0)ϕ(x) =
0, and so z ∈ N(ϕ). Also, since u0 ∈ N(ϕ) ⊥ we have,
where
which implies that
If we set
then ϕ(x) = 〈x, u〉. Hence (a) is proved.
0 = 〈z, u0〉 = 〈ϕ(x)u0 − ϕ(u0)x, u0〉
= ϕ(x)〈u0, u0〉 − ϕ(u0)〈x, u0〉
〈u0, u0〉 = u0 2 = 0,
ϕ(x) = ϕ(u0)〈x, u0〉
u02 .
u = ϕ(u0)u0
,
u02 (b) To prove that u is unique, suppose that for all x ∈ H we have,
Then
ϕ(x) = 〈x, u1〉 = 〈x, u2〉.
〈x, u1 − u2〉 = 0 for all x.

Section 3.3. Sesquilinear functionals and Riesz representation Page 12
If we choose x such that x = u1 − u2, then
〈x, u1 − u2〉 = 〈u1 − u2, u1 − u2〉 = u1 − u2 2 = 0.
Hence u1 − u2 = 0, which implies that u1 = u2. Hence u is unique.
(c) Let ϕ = 0. Then u = 0. From (a) with x = u,
which implies that
By Cauchy-Schwarz inequality and (a),
u 2 = 〈u, u〉 = ϕ(u) ≤ ϕu
u ≤ ϕ. (3.7)
|ϕ(x)| = |〈x, u〉| ≤ xu
which implies that ϕ = sup |〈x, u〉| ≤ u.
x=1
Hence
ϕ ≤ u. (3.8)
Combining equations (3.7) and (3.8) gives
which yields (c).
u = ϕ, (3.9)
Remark 3.2.2 The Riesz theorem is named after Frigyes Riesz a Hungarian mathematician.
This representation is important in the theory of operators on Hilbert spaces. In particular it is
important in the representation of the Hilbert-adjoint operator of a bounded linear operator. In
the mathematical treatment of quantum mechanics, the theorem can be seen as a justification
for the popular bra-ket notation. When the theorem holds, every ket |ψ〉 has a corresponding bra
〈ψ|.
3.3 Sesquilinear functionals and Riesz representation
Definition 3.3.1 ([Kre78], 3·8-3 Definition (Sesquilinear functional)) Let X and Y be vector
spaces over the same field K (R or C). A sesquilinear form (or sesquilinear functional) is defined
as a mapping
h : X × Y → K (3.10)
such that for all x, x1, x2 ∈ X and y, y1, y2 ∈ Y and all scalars α and β,
1. h(x1 + x2, y) = h(x1, y) + h(x2, y),

Section 3.3. Sesquilinear functionals and Riesz representation Page 13
2. h(x, y1 + y2) = h(x, y1) + h(x, y2),
3. h(αx, y) = αh(x, y),
4. h(x, βy) = ¯ βh(x, y).
Remark 3.3.2 Notice that h is linear in the first argument and conjugate linear in the second
one. If X and Y are real then (4) becomes h(x, βy) = βh(x, y). Clearly, h is bilinear in both
arguments.
Definition 3.3.3 ([Swa97], Definition 8 (Bounded sesquilinear functional)) A sesquilinear functional
h is bounded if there exists K > 0 such that
The norm of of h is defined to be
|h(x, y)| ≤ Kxy, for every x ∈ X, y ∈ Y. (3.11)
|h(x, y)|
h = sup
x=0 xy
y=0
= sup
x=1
y=1
|h(x, y)|. (3.12)
It follows that |h(x, y)| ≤ hxy for all x ∈ X, y ∈ Y. The inner product is sesquilinear
and it is bounded.
Lemma 3.3.4 ([Kre78], 3·8-2 Lemma (Equality))
If 〈x1, y〉 = 〈x2, y〉 for all y in an inner product space X, then x1 = x2. In particular if 〈x1, w〉 = 0
for all w ∈ X, then x1 = 0.
Proof.
For all y,
〈x1 − x2, y〉 = 〈x1, y〉 − 〈x2, y〉 = 0.
If we choose y = x1 − x2, then x1 − x2 2 = 0. Hence we have x1 − x2 = 0 implying that
x1 = x2. In particular, if 〈x1, y〉 = 0 with y = x1 we get x1 2 = 0, giving x1 = 0.
Theorem 3.3.5 ([Kre78], 3·8-4 Theorem (Riesz representation))
Let H1, H2 be Hilbert spaces and let h : H1 × H2 → K be a bounded sesquilinear form. Then h
has a representation
h(x, y) = 〈Sx, y〉 for x ∈ H1, y ∈ H2
(3.13)
where S : H1 → H2 is a bounded linear operator and S is uniquely determined and has norm
S = h.

Section 3.3. Sesquilinear functionals and Riesz representation Page 14
We will apply Riesz Theorem 3.2.1 in the proof of this theorem. We will show that
(a) h has a representation h(x, y) = 〈Sx, y〉,
(b) h = S, and
(c) S is unique.
Proof.
(a) Let us take a fixed x and then we will show that φ : y → h(x, y) is a linear map. Suppose
φ(y) = h(x, y), then by Definition 3.3.1,
Moreover,
φ(y1 + y2) = h(x, y1 + y2)
= h(x, y1) + h(x, y2)
= φ(y1) + φ(y2).
φ(αy) = h(x, αy)
= αh(x, y)
= αh(x, y)
= αφ(y).
Hence h(x, y) is linear.
Applying Theorem 3.2.1 yields a representation h(x, y) = 〈y, z〉. Hence
h(x, y) = 〈z, y〉. (3.14)
Our z ∈ H2 is unique but depends on our fixed x ∈ H1. Equation (3.14) with variable x defines
an operator S : H1 → H2 given by z = Sx. Hence equation (3.14) becomes h(x, y) = 〈Sx, y〉,
giving (3.13).
Now let us show that S is linear. From equation (3.13), and the sesquilinearity we obtain,
for all y in H2, so that by Lemma 3.3.4,
〈S(αx1 + βx2), y〉 = h(αx1 + βx2, y)
= αh(x1, y) + βh(x2, y)
= α〈Sx1, y〉 + β〈Sx2, y〉
= 〈αSx1 + βSx2, y〉
S(αx1 + βx2) = αSx1 + βSx2.
Hence S is linear.
(b) We show that S is bounded. Leaving the trivial case S = 0, we have
|〈Sx, y〉|
h = sup
x=0 xy
y=0
|〈Sx, Sx〉|
≥ sup
x=0 xSx
Sx=0
Sx
= sup
x=0 x
= S. (3.15)

Section 3.4. Hilbert-adjoint and self-adjoint operators Page 15
Now by Cauchy-Schwarz inequality,
so that
|〈Sx, y〉|
sup
x=0 xy
y=0
Sxy
≤ sup
x=0 xy
y=0
= S, (3.16)
h ≤ S. (3.17)
Combining equations (3.15) and (3.17) gives h = S.
(c) Let us assume that T : H1 → H2 is a linear operator such that for all x ∈ H1 and y ∈ H2,
we have
h(x, y) = 〈Sx, y〉 = 〈Tx, y〉.
Then, by Lemma 3.3.4, Sx = Tx for all x ∈ H1. Hence S = T by definition.
Remark 3.3.6 An important consequence of the Riesz representation theorem is the existence
of the adjoint of a bounded operator on a Hilbert space.
3.4 Hilbert-adjoint and self-adjoint operators
Definition 3.4.1 ([Kre78], 3·9-1 Definition (Hilbert-adjoint operator, T ∗ ))
Given two Hilbert spaces, H1 and H2, let T : H1 → H2 be a bounded linear operator. Then the
Hilbert-adjoint operator T ∗ of T is the operator T ∗ : H2 → H1 such that for all x ∈ H1 and
y ∈ H2,
〈Tx, y〉 = 〈x, T ∗ y〉. (3.18)
Let us show that there exists such an operator T ∗ .
Theorem 3.4.2 ([Kre78], 3·9-2 Theorem (Existence)) The Hilbert-adjoint operator T ∗ of T
exist, is unique and is a bounded linear operator with norm
Proof.
Consider
T ∗ = T. (3.19)
h(y, x) = 〈y, Tx〉. (3.20)

Section 3.4. Hilbert-adjoint and self-adjoint operators Page 16
We will show that h is sesquilinear. Now h is linear in the first argument and conjugate linear in
the second argument since,
h(y, αx1 + βx2) = 〈y, T(αx1 + βx2)〉
= 〈y, αTx1 + βTx2〉
= ¯α〈y, Tx1〉 + ¯ β〈y, Tx2〉
= ¯αh(y, x1) + ¯ βh(y, x2).
Hence since the inner product is sesquilinear, we conclude that h is sesquilinear. By the Cauchy-
Schwarz inequality,
|h(y, x)| = |〈y, Tx〉| ≤ yTx ≤ Txy
which implies that |h(y,x)|
xy
Also,
≤ T and by equation (3.12) we have that
|〈y, Tx〉|
h = sup
x=0 yx
y=0
h ≤ T. (3.21)
|〈Tx, Tx〉|
≥ sup
x=0 Txx
Tx=0
= T. (3.22)
Combining equations (3.21) and (3.22) gives h = T. From Theorem 3.3.5, substituting T ∗
for S, we have
h(y, x) = 〈T ∗ y, x〉, (3.23)
where T ∗ : H2 → H1 is a uniquely determined, bounded linear operator with norm
Combining (3.20) and (3.23), we get
Taking the conjugate gives equation (3.18).
T ∗ = h = T. (3.24)
〈y, Tx〉 = 〈T ∗ y, x〉.
Lemma 3.4.3 ([Kre78], 3·9-3 Lemma (Zero operator)) Let X and Y be inner product spaces
and T : X → Y a bounded linear operator. Then:
1. T = 0 if and only if 〈Tx, y〉 = 0 for all x ∈ X and y ∈ Y.
2. For X a complex vector space, if T : X → X and 〈Tx, x〉 = 0 for all x ∈ X, then T = 0.
Proof.
1. If T = 0, then for all x ∈ X, Tx = 0, and for any u ∈ X we have,
〈Tx, y〉 = 〈0, y〉 = 0〈u, y〉 = 0.

Section 3.4. Hilbert-adjoint and self-adjoint operators Page 17
Now let
〈Tx, y〉 = 0 for all x ∈ X, y ∈ Y.
Then by Lemma 3.3.4, Tx = 0 for all x ∈ X and T = 0.
2. If 〈Tx, x〉 = 0 for all x ∈ X, then for w = αx + y ∈ X we have
〈Tw, w〉 = 〈T(αx + y), αx + y〉
Now if we choose α = 1, then (3.25) becomes
= |α| 2 〈Tx, x〉 + 〈Ty, y〉 + α〈Tx, y〉 + ¯α〈Ty, x〉. (3.25)
〈Tw, w〉 = 〈Tx, x〉 + 〈Ty, y〉 + 〈Tx, y〉 + 〈Ty, x〉. (3.26)
Now 〈Tx, x〉 and 〈Ty, y〉 are equal to zero by our assumption. Hence equation (3.26) becomes
If we choose α = i, then ¯α = −i and equation (3.25) becomes
〈Tx, y〉 + 〈Ty, x〉 = 0. (3.27)
〈Tx, y〉 − 〈Ty, x〉 = 0. (3.28)
Adding equations (3.27) and (3.28) gives 〈Tx, y〉 = 0 and T = 0 follows from 1.
Theorem 3.4.4 ([Kre78], 3·9-4 Theorem (Properties of the Hilbert adjoint-operator)) Let H1
and H2 be Hilbert spaces, S : H1 → H2 and T : H1 → H2 be bounded linear operators and β
any scalar. Then
1. 〈T ∗ y, x〉 = 〈y, Tx〉 for any x ∈ H1, y ∈ H2,
2. (S + T) ∗ = S ∗ + T ∗ ,
3. (βT) ∗ = ¯ βT ∗ ,
4. (T ∗ ) ∗ = T,
5. T ∗ T = TT ∗ = T 2 ,
6. T ∗ T = 0 if and only if T = 0,
7. (ST) ∗ = T ∗ S ∗ (if H2 = H1).
Proof.
1. By Definition 3.4.1, we have
〈T ∗ y, x〉 = 〈x, T ∗ y〉 = 〈Tx, y〉 = 〈y, Tx〉. (3.29)

Section 3.4. Hilbert-adjoint and self-adjoint operators Page 18
2. By the definition of the Hilbert-adjoint operator in Definition (3.4.1), for all x and y,
〈x, (S + T) ∗ y〉 = 〈(S + T)x, y〉
= 〈Sx, y〉 + 〈Tx, y〉
= 〈x, S ∗ y〉 + 〈x, T ∗ y〉
= 〈x, (S ∗ + T ∗ )y〉.
Thus it follows from Lemma 3.3.4 that, (S + T) ∗ y = (S ∗ + T ∗ )y for all y ∈ H2, so that
(S + T) ∗ = S ∗ + T ∗ .
3. By Definition (3.4.1),
〈(βT) ∗ y, x〉 = 〈y, (βT)x〉
= 〈y, β(Tx)〉
= ¯ β〈y, Tx〉
= ¯ β〈T ∗ y, x〉
= 〈 ¯ βT ∗ y, x〉.
Hence by Lemma 3.4.3, (βT) ∗ y = ¯ βT ∗ y for all y ∈ H2, which implies that (βT) ∗ = ¯ βT ∗ .
4. By Definition 3.4.1 and from 1 we have, 〈(T ∗ ) ∗ x, y〉 = 〈x, T ∗ y〉 = 〈Tx, y〉 so that 〈((T ∗ ) ∗ −
T)x, y〉 and by Lemma 3.4.3, we have (T ∗ ) ∗ = T.
5. We know that T ∗ T : H1 → H1 and TT ∗ : H2 → H2. By the Cauchy-Schwarz inequality
and by the definition of the Hilbert-adjoint operator in Definition (3.4.1) we have
Tx 2 = 〈Tx, Tx〉 = 〈T ∗ Tx, x〉 ≤ T ∗ Txx ≤ T ∗ Tx 2 .
Taking the supremum over all x of norm 1 we obtain T 2 ≤ T ∗ T. Now by Theorem 3.4.2,
we have T ∗ T ≤ T ∗ T = T 2 . Hence T ∗ T = T 2 . We substitute T ∗ for T to get
T ∗∗ T ∗ = T ∗ 2 = T 2 , But by 4 we have (T ∗ ) ∗ = T, so that TT ∗ = T 2 .
6. From 5, if T ∗ T = 0, then T = 0 and conversely if T = 0, then T ∗ T = 0.
7. By Definition 3.4.1, 〈x, (ST) ∗ y〉 = 〈(ST)x, y〉 = 〈Tx, S ∗ y〉 = 〈x, T ∗ S ∗ y〉. Hence by Lemma
3.3.4 we obtain (ST) ∗ y = T ∗ S ∗ y for all y ∈ H2, so that (ST) ∗ = T ∗ S ∗ .
Definition 3.4.5 ([Kre78], 3·10-1 (Self-adjoint operator)) A bounded linear operator T : H → H
on a Hilbert space H is said to be self-adjoint or Hermitian if
Equivalently, a bounded linear operator T is said to be self-adjoint if
T ∗ = T. (3.30)
〈x, Ty〉 = 〈Tx, y〉 for all x, y ∈ H. (3.31)

Section 3.4. Hilbert-adjoint and self-adjoint operators Page 19
Example 3.4.6 A linear map on R n with matrix A is a self-adjoint if and only if A is symmetric
(A = A T ). A linear map on C n with matrix A is self-adjoint if and only if A is Hermitian
(A = A ∗ ).
Remark 3.4.7 Self-adjoint operators on Hilbert spaces are used in quantum mechanics to represent
physical observables such as position, momentum, angular momentum and spin. An
important example is the Hamiltonian operator given by
Hψ = − 2
2m ∇2 ψ + V ψ.
Definition 3.4.8 ([Kre78], 3·10-1 (Unitary operator)) A bounded linear operator T : H → H
on a Hilbert space H is said to be unitary if T is bijective and
Hence
TT ∗ = T ∗ T. (3.32)
T ∗ = T −1 . (3.33)
Definition 3.4.9 ([Kre78], 3·10-1 (Normal operators)) A bounded linear operator T : H → H
on a Hilbert space H is said to be normal if
TT ∗ = T ∗ T. (3.34)
Remark 3.4.10 If T is self-adjoint or unitary, then T is normal, but the converse is not generally
true. For example if I : H → H is the identity operator, then T = 2iI is normal since T ∗ = −2iI,
so that TT ∗ = T ∗ T = 4I but T ∗ = T and T ∗ = T −1 = − 1
2 iI.
Theorem 3.4.11 ([Kre78], 3·10-3 Theorem (Self-Adjointness)) Let T : H → H be a bounded
linear operator on a Hilbert space H. Then
1. If T is self-adjoint, then 〈Tx, x〉 ∈ R for all x ∈ H.
2. If H is complex and 〈Tx, x〉 ∈ R for all x ∈ H, then the operator T is self-adjoint.
Proof.
1. If T is self-adjoint, then for all x,
〈Tx, x〉 = 〈x, Tx〉. (3.35)
By definition, 〈Tx, y〉 = 〈x, T ∗ y〉 and since T is self-adjoint, we have
Combining equations (3.35) and (3.36) gives
〈Tx, x〉 = 〈x, Tx〉. (3.36)
〈Tx, x〉 = 〈Tx, x〉.
Hence 〈Tx, x〉 is equal to its complex conjugate which implies that it is real.

Section 3.4. Hilbert-adjoint and self-adjoint operators Page 20
2. If 〈Tx, x〉 ∈ R for all x ∈ H, then
Hence
〈Tx, x〉 = 〈Tx, x〉 = 〈x, T ∗ x〉 = 〈T ∗ x, x〉.
0 = 〈Tx, x〉 − 〈T ∗ x, x〉 = 〈(T − T ∗ )x, x〉
and by Lemma 3.4.3, T − T ∗ = 0. Therefore T = T ∗ .
Theorem 3.4.12 ([Kre78], 3·10-5 Theorem (Sequence of self-adjoint operators)) Let Tn be a
sequence of bounded self-adjoint linear operators Tn : H → H on a Hilbert space H. If Tn
converges to T, then T is a bounded self-adjoint linear operator.
Proof.
If Tn → T, then Tn − T → 0. Now by Theorem 3.4.4 and Theorem 3.4.2 we have that
Tn ∗ − T ∗ = (Tn − T) ∗ = Tn − T, so that
T − T ∗ ≤ T − Tn + Tn − T ∗
n + T ∗
n − T ∗
= T − Tn + Tn − T = 2Tn − T.
As n tends to infinity, Tn − T tends to zero. Hence T − T ∗ = 0 which implies that T ∗ = T.
Hence T is self-adjoint.

4. Spectral Analysis of a Self-Adjoint
Operator
Definition 4.0.13 ([Aup91], 2·2·4 Definition (Spectrum of T)) Let L(H) denote the set of all
bounded linear operators on H and let T ∈ L(H). We define the spectrum of T as the set of
λ ∈ C such that T − λI is not invertible in L(H). It is denoted by Sp T. So λ ∈ SpT if and
only if at least one of the following conditions hold.
1. the range of T − λI is not all of H, that is T − λI is not surjective.
2. T − λI is not injective.
Hence
SpT = {λ ∈ C : T − λI is not invertible in L(H)} .
Remark 4.0.14 The kernel and range of T are denoted by N(T) and R(T) respectively. If 2
holds, then λ is called the eigenvalue of T and N(T − λI) is the corresponding eigenspace which
is the set of all x ∈ X such that Tx = λx, where x = 0 is called an eigenvector corresponding
to the the eigenvalue λ.
Definition 4.0.15 ([Aup91]) The spectral radius of T ∈ L(H) is given by
ρ(T) = max {|λ| : λ ∈ SpT} . (4.1)
Theorem 4.0.16 ([Aup91], Theorem 2·3·1) Let H be a Hilbert space and let T ∈ L(H). Then
1. N(T) = R(T ∗ ) ⊥ and N(T ∗ ) = R(T) ⊥ ,
2. I + T ∗ T is invertible in L(H).
Proof.
1. If we choose x ∈ N(T), then for all y ∈ H,
〈0, y〉 = 〈Tx, y〉 = 〈x, T ∗ y〉 = 0,
so that x ∈ R(T ∗ ) ⊥ . Now applying the adjoint ∗ gives
N(T ∗ ) = R(T ∗∗ ) ⊥ ,
21

ut by Theorem 3.4.4 (4), T ∗∗ = T which implies that N(T ∗ ) = R(T) ⊥ .
2.To show that I + T ∗ T is invertible in L(H), we show that I + T ∗ T is
(i) injective and
(ii) show that it is surjective by showing that the range of I + T ∗ T is the whole of H.
(i) Consider T ∗ T and I + T ∗ T which are self-adjoint.
(I + T ∗ T)x 2 = 〈(I + T ∗ T)x, (I + T ∗ T)x〉
= 〈(I + T ∗ T)(I + T ∗ T)x, x〉
= 〈(I + T ∗ T) 2 x, x〉
= 〈I + 2T ∗ T + (T ∗ T) 2 x, x〉
= 〈(x + (2T ∗ T)x + (T ∗ T) 2 )x, x〉
= 〈x, x〉 + 2〈T ∗ (Tx), x〉 + 〈(T ∗ T) 2 x, x〉
= x 2 + 2〈Tx, Tx〉 + 〈T ∗ Tx, T ∗ Tx〉
Page 22
(4.2)
= x 2 + 2Tx 2 + T ∗ Tx 2 ≥ x 2 . (4.3)
It follows that if (I + T ∗ T)x = 0, then x = 0. Hence we conclude that N(I + T ∗ T) = {0}, so
that I + T ∗ T is injective.
(ii) We will first show that R(I + T ∗ T) is closed. Let ((I + T ∗ T)xn) be a Cauchy sequence in
R(I + T ∗ T). Then given ɛ ≥ 0, there exist N such that for n, m ≥ N we have (I + T ∗ T)(xn) −
(I + T ∗ T)(xm) ≤ ɛ. Now from inequality (4.3),
ɛ ≥ (I + T ∗ T)(xn) − (I + T ∗ T)(xm) = (I + T ∗ T)(xn − xm) ≥ xn − xm,
which implies that xn−xm ≤ ɛ. Hence xn is also a Cauchy sequence. Now since H is complete,
xn → x ∈ H. Since I + T ∗ T is bounded, then it is continuous. Hence lim (I + T
n→∞ ∗ T)(xn) =
(I + T ∗ T)( lim (xn)) = (I + T
n→∞ ∗ T)x. Hence (I + T ∗ T)x ∈ R(I + T ∗ T) and we conclude that
R(I + T ∗ T) is closed.
Now by direct sum, Theorem 2.3.8, H = R(I + T ∗ T) ⊕ R(I + T ∗ T) ⊥ . But by 1 we have
that R(I + T ∗ T) ⊥ = N(I + T ∗ T) = {0} . Therefore H = R(I + T ∗ T). Hence I + T ∗ T is
surjective.
We conclude from (i) and (ii) that I + T ∗ T is bijective and hence invertible.
Lemma 4.0.17 If S ∈ L(H) where H a Hilbert space, S = S ∗ and λ is an eigenvalue of S, then
λ ∈ R.
Proof.
If λ is an eigenvalue of S, then by Remark 4.0.14, S−λI is not injective. Hence N(S−λI) = {0}.
Hence for x ∈ N(S − λI), Sx = λx. Now since S is self-adjoint, then by Theorem 3.4.11 (1),
〈Sx, x〉 ∈ R for all x ∈ H and 〈x, Sx〉 = 〈x, λx〉 = λx 2 ∈ R. Hence λ = λ and we conclude
that λ ∈ R.

Page 23
Lemma 4.0.18 Let S ∈ L(H) where H is a Hilbert space. Suppose that Sλ = λI − S with
λ = α + iβ and S = S ∗ . If Sλx ≥ |β|x, with β = 0, then Sλ is invertible.
Proof.
(i) To show that it is injective, let Sλx = 0, then |β|x ≤ 0 which implies that x = 0 for
β = 0. Hence N(Sλ) = {0} and we conclude that Sλ is injective.
(ii) Let us first show that R(Sλ) is closed. If (Sλ(xn)) is a Cauchy sequence so is (xn).
Now since H is complete, xn → x ∈ H. Since Sλ is bounded, then it is continuous. Hence
lim
n→∞ (Sλ(xn)) = (Sλ) lim (xn) = (Sλ)x. Hence (Sλ)x ∈ R(Sλ) and we conclude that R(Sλ) is
n→∞
closed.
Now by direct sum, Theorem 2.3.8, H = R(Sλ) ⊕ R(Sλ) ⊥ . By Theorem 4.0.16 (1), we have that
R(Sλ) ⊥ = N(Sλ ∗ ). We will show that Sλ ∗ is injective and deduce that R(Sλ) ⊥ = N(Sλ ∗ ) = {0}.
Now by Theorem 4.0.16 (1), R(Sλ) ⊥ = N(Sλ ∗ ). For β = 0, N((αI + iβI − S) ∗ ) = N(αI −
iβI − S) = {0} otherwise by Lemma 4.0.17, α − iβ would be an eigenvalue. But α − iβ /∈ R,
so that α − iβ cannot be an eigenvalue. This implies that N(Sλ ∗ ) = R(Sλ) ⊥ = {0}. Hence
H = R(Sλ), showing that Sλ is surjective. We conclude that Sλ is bijective hence invertible.
Theorem 4.0.19 ([Kre78], 7·5-5 (Theorem)) If T is a bounded linear operator on a complex
Banach space, then
ρ(T) = lim T
n→∞ n 1
n . (4.4)
Remark 4.0.20 The proof of this theorem is not trivial and falls outside the scope of this essay.
Corollary 4.0.21 ([Aup91], Corollary 2·3·2) Let H be a Hilbert space and let S be a self-adjoint
operator on H. Then
1. S 2 = S 2 and consequently S = ρ(S).
2. Sp S ⊂ R.
Proof.
1. From Theorem 3.4.4 (5), S ∗ S = S 2 . But S is self-adjoint. Hence S 2 = S 2 and by
induction we have S 2n
= S2n . Now by Theorem 4.0.19, we have that ρ(S) = lim S
n→∞ n 1
n =
S.
2. Let λ = a + iβ with α, β ∈ R and let Sλ = λI − S for all λ ∈ C. Then
Sα + iβI = αI − S + iβI
= (α + iβ)I − S
= λI − S.

Therefore Sα + iβI = Sλ. Thus for all x ∈ H and S self-adjoint we have,
Sλx 2 = 〈Sλx, Sλx〉
= 〈(Sα + iβI)x, (Sα + iβI)x〉
= 〈Sαx, Sαx〉 + 〈iβx, iβx〉 + 〈Sαx, iβx〉 + 〈iβx, Sαx〉
= 〈Sαx, Sαx〉 + 〈iβx, iβx〉 − iβ〈Sαx, x〉 + iβ〈Sαx, x〉
= 〈Sαx, Sαx〉 + |β| 2 x 2
= Sαx 2 + |β| 2 x 2 ≥ |β| 2 x 2 .
Page 24
Hence Sλx ≥ |β|x, and by Lemma 4.0.18, we conclude that for β = 0, Sλ = λI − S is
invertible and hence Sp S ⊂ R.

5. Conclusion and further work
The parallelogram equality gives Hilbert spaces an edge over Banach spaces. Moreover orthogonality,
continuity of the inner product and completeness are also important features of Hilbert
spaces.
<strong>Operators</strong> on Hilbert spaces is a basis for a comprehensive study of the spectral theory. In
particular, the Hermitian or self-adjoint operators are very important in many applications. In
this essay, we have seen that the minimising vector theorem, Riesz representation of functionals
and direct sum play important roles in this study. Given an inner product space X and M is
a nonempty complete convex subspace of X , the Minimising Vector Theorem (Theorem 2.3.1)
enables us to find a unique point y ∈ M that is closer to x ∈ X than any other point in M. The
Riesz theorem, Theorem 3.2.1 is important in the representation of Hilbert-adjoint operators and
also in quantum physics. The direct sum theorem, Theorem 2.3.8, makes it possible to represent a
Hilbert space H as the sum of any closed subspace and its orthogonal complement (Lemma 2.3.7).
The algebra of all n × n complex matrices is a special case of an algebra of bounded linear
operators on a Hilbert space and so ”Hermitian matrices” correspond to ”self-adjoint operators”.
Moreover the spectrum of an n × n matrix (considered as an operator) consists of exactly all
the eigenvalues of the matrix whereas the spectrum SpT of the self-adjoint operator T contains
other elements which are not eigenvalues of T. An element, λ ∈ SpT of the spectrum of the
self-adjoint operator is an eigenvalue of T if and only if T − λI is not injective. It is also well
known that the eigenvalues of a Hermitian matrix are real numbers. In this essay, we have generalised
this idea by showing that the spectrum of a self-adjoint operator on a Hilbert space also
consists entirely of real values (Theorem 4.0.21).
So far, we have studied a small aspect of the operators on Hilbert spaces. However, there
are other interesting facts about the self-adjoint operators like the bounded resolution of the
identity. Moreover, other aspects of the spectral theory includes
1. Integral equations, Fredholm theory, compact operators,
2. Sturm-Liouville theory, hydrogen atom,
3. Isospectral theory, Lax pairs,
4. Atiyah-Singer index theorem.
These are interesting facts about the spectral theory that can be studied in future.
25

Acknowledgements
Above all, I thank the God Almighty for His faithfulness. It is the Lord’s doing and it is marvellous
in my eyes.
I would like to appreciate the founder of AIMS, professor Neil Turok and its director, professor
Fritz Hahne for giving me this great opportunity to study at AIMS.
I wish to express my profound gratitude to my supervisor, Dr S. Mouton for her guidance and
interest in my work. I also thank my tutor, Dr Laure Gouba, my essay tutor, Mr Christian
Rivasseau, and all the tutors at AIMS. I really appreciate your zeal, which has spurred me to
the peak. Many thanks also goes to Mr Igsaan Kamalie, Mr Jan Groenewald and all staff and
lecturers at AIMS.
To my parents, Mr and Mrs R.O. Kajotoni, I pray that the good Lord will grant you long life
to reap the fruits of your labour. I also appreciate the contributions of my brothers Kayode
and Babatunde, my sister and her husband, Mr and Mrs Ojerinde and my little niece Eniola.
Furthermore, I appreciate the contributions of Mr and Mrs Abiodun Kajotoni, Mr and Mrs A.
Eniolorunda and Iyabo Bello for her relentless effort.
I own my fiance, Raphael olushola Folorunsho many thanks for his love, prayers, patience and
concern.
Finally, I want to appreciate the Nigerian students, juststudents2006, all AIMS alumni and others
too numerous to mention. God bless you all.
26

Bibliography
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[Axl97] Sheldon Axler, Linear algebra done right, Undergraduate text in Mathematics, Springer,
San Francisco State University, 1997.
[Erd80] John Erdos, <strong>Operators</strong> on hilbert spaces, King’s College London, 1980.
[Kre78] Erwin Kreyszig, Introductory functional analysis with applications, John Wiley and sons,
New York, 1978.
[Red53] B. Daya Reddy, Introductory functional analysis with applications to boundary value
problems and finite elements, Text in Applied Mathematics, no. 27, Springer, University
of Cape Town, 1953.
[Swa97] Charles Swartz, An introduction to functional analysis, Monographs and Textbooks, no.
157, Marcel Dekker, Inc. New York, 1997.
27