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Chapter 3<br />
<strong>Bounded</strong> <strong>Linear</strong> <strong>Operators</strong><br />
3.1 The Space B(X,Y)<br />
In this chapter we study linear operators (also called linear maps or linear<br />
transformations) between normed spaces, especially operators with the<br />
following additional property.<br />
Definition 3.1.1. Let X,Y be normed vector spaces and T : X → Y a<br />
linear operator. We say that T is bounded if there exists a number c > 0<br />
such that<br />
for all x ∈ X.<br />
TxY ≤ cxX<br />
Theorem 3.1.1. Let X,Y be normed vector spaces and T : X → Y linear.<br />
Then the following are equivalent.<br />
(i) T is bounded.<br />
(ii) T is uniformly continuous.<br />
(iii) T is continuous at 0.<br />
Proof. (i) ⇒ (ii) Suppose that there is a number c > 0 satisfying TxY ≤<br />
cxX for all x ∈ X. Let ǫ > 0 and choose δ = ǫ<br />
c . If we have x,y ∈ X with<br />
x−yX ≤ δ, then<br />
Tx−TyY = T(x−y)Y ≤ cx−yX ≤ ǫ.<br />
So T is uniformly continuous.<br />
(ii) ⇒ (iii) This is trivial.<br />
(iii) ⇒ (i) Choose δ > 0 such that<br />
xX ≤ δ ⇒ TxY ≤ 1.<br />
23
24 CHAPTER 3. BOUNDED LINEAR OPERATORS<br />
Then for any x ∈ X\{0},<br />
Hence T is bounded.<br />
TxX = xX<br />
δ<br />
<br />
<br />
<br />
T <br />
δx Y<br />
xX<br />
≤ xX<br />
.<br />
δ<br />
Definition 3.1.2. Let X,Y be normed vector spaces. Then B(X,Y) is the<br />
spacecomprisingallboundedlinearoperatorsT : X → Y. ForT ∈ B(X,Y),<br />
we define<br />
T = sup TxY.<br />
x∈X<br />
xX≤1<br />
This is called the operator norm of T. We also write B(X) = B(X,X).<br />
Sometimes we write T B(X,Y) or T B(X) for the operator norm in<br />
order to distinguish it from other norms. Note that the linearity implies the<br />
identities<br />
unless X = {0}.<br />
TxY<br />
T = sup TxY = sup ,<br />
x∈X x∈X\{0} xX<br />
xX=1<br />
Theorem 3.1.2. (i) If X,Y are normed vector spaces, then B(X,Y),<br />
equipped with the operator norm, is a normed vector space.<br />
(ii) If X is a normed vector space and Y is a Banach space, then B(X,Y)<br />
is a Banach space.<br />
(iii) Let X,Y,Z be normed vector spaces and suppose that S ∈ B(X,Y)<br />
and T ∈ B(Y,Z). Then TS ∈ B(X,Z) with<br />
TS ≤ TS.<br />
Proof. (i) is routine.<br />
(ii) Suppose that Y is complete. Let (Tn)n∈N be a Cauchy sequence in<br />
B(X,Y). Then for any fixed x ∈ X, we have<br />
Tmx−TnxY ≤ Tm −TnxX → 0 as m,n → ∞.<br />
Hence (Tnx)n∈N is a Cauchy sequence in Y and the limit<br />
Tx := lim<br />
n→∞ Tnx<br />
exists. This gives rise to a map T : X → Y. It is linear, because for x,y ∈ X<br />
and α ∈ F, we have<br />
T(αx+y) = lim<br />
n→∞ Tn(αx+y) = α lim<br />
n→∞ Tnx+ lim<br />
n→∞ Tny = αTx+Ty.
3.1. THE SPACE B(X,Y) 25<br />
It is bounded, because<br />
TxY = lim<br />
n→∞ TnxY ≤<br />
<br />
lim<br />
n→∞ Tn<br />
<br />
xX.<br />
It remains to prove that Tn → T in B(X,Y) (which is stronger than pointwise<br />
convergence).<br />
Let ǫ > 0 and choose N ∈ N such that Tm − Tn ≤ ǫ for m,n ≥ N.<br />
Then for every x ∈ X,<br />
Letting m → ∞, we obtain<br />
Tmx−TnxY ≤ ǫxX.<br />
Tx−TnxY ≤ ǫxX.<br />
That is, we have T −Tn ≤ ǫ, as required.<br />
(iii) Let x ∈ X. Then<br />
This implies the desired inequality.<br />
TSxZ ≤ TSxY ≤ TSxX.<br />
Definition 3.1.3. If X is a normed vector space over F, then X ∗ = B(X,F)<br />
is called the dual space of X. Its elements are called bounded linear functionals<br />
on X.<br />
Example 3.1.1. Let a < b and c ∈ [a,b]. Define T : C 0 ([a,b]) → F by<br />
Tf = f(c). This is a linear operator and we have<br />
|Tf| = |f(c)| ≤ f C 0 ([a,b]).<br />
So T ∈ (C 0 ([a,b])) ∗ and T ≤ 1. Since T1 = 1, we have in fact T = 1.<br />
Example 3.1.2. Let g ∈ C 0 ([a,b]) and define T : C 0 ([a,b]) → C 0 ([a,b]) by<br />
Tf = gf. Then<br />
Tf C 0 ([a,b]) = gf C 0 ([a,b]) ≤ g C 0 ([a,b])f C 0 ([a,b]).<br />
Hence T ∈ B(C 0 ([a,b])) and T ≤ g C 0 ([a,b]). Since T1 = g, we have in<br />
fact T = g C 0 ([a,b]).<br />
Example 3.1.3. Let p ∈ [1,∞] and consider the operators<br />
R: ℓ p → ℓ p , (x1,x2,x3,...) ↦→ (0,x1,x2,...),<br />
L: ℓ p → ℓ p , (x1,x2,x3,...) ↦→ (x2,x3,x4,...).<br />
For x = (x1,x2,x3,...) ∈ ℓ p , we clearly have<br />
Rxℓp = xℓp and Lxℓp ≤ xℓp. So R,L ∈ B(ℓ p ). It also follows immediately that R = 1 and L ≤ 1.<br />
In fact we have L = 1, because L maps (0,1,0,0,...) (with norm 1) to<br />
(1,0,0,...) (still with norm 1).
26 CHAPTER 3. BOUNDED LINEAR OPERATORS<br />
Example 3.1.4. Let p,q be conjugate exponents. Fix y = (y1,y2,y3,...) ∈<br />
ℓq . Define T : ℓp → F by<br />
∞<br />
Tx = xnyn.<br />
Then by the Hölder inequality,<br />
n=1<br />
|Tx| ≤ xℓpyℓq for every x ∈ ℓp . So T ∈ (ℓp ) ∗ with T ≤ yℓq. We will see later that<br />
T = yℓq. 3.2 The Uniform <strong>Bounded</strong>ness Principle<br />
It is not so easy to come up with an example of a linear operator between<br />
Banach spaces that is not bounded. Nevertheless, boundedness is a severe<br />
restriction and has some deep consequences. We now discuss some of them.<br />
Recall from MA40043: a set Y in a metric space X is called nowhere<br />
dense if Y ◦ = ∅. The following is a result from MA40043.<br />
Theorem 3.2.1 (Baire Category Theorem). A complete, non-empty metric<br />
space cannot be a countable union of nowhere dense subsets.<br />
As a consequence we obtain a result on bounded linear operators.<br />
Theorem 3.2.2 (Banach-Steinhaus/Uniform <strong>Bounded</strong>ness Principle). Let<br />
X be a Banach space and Y a normed vector space. Let T ⊂ B(X,Y) be<br />
such that for all x ∈ X, the set {Tx: T ∈ T } is bounded in Y. Then T is<br />
bounded in B(X,Y).<br />
In other words, if we have pointwise bounds, then we automatically get<br />
a bound in the norm of B(X,Y) as well. The second property appears much<br />
stronger and implies in particular that we have uniform bounds on bounded<br />
subsets of X. That is, if A ⊂ X with<br />
then<br />
sup<br />
T∈T<br />
Proof. For n ∈ N, define<br />
supx<br />
< ∞,<br />
x∈A<br />
<br />
supTxY<br />
≤ sup T<br />
x∈A T∈T<br />
supxX<br />
x∈A<br />
<br />
< ∞.<br />
Xn = {x ∈ X: TxY ≤ n for all T ∈ T } = <br />
{x ∈ X: TxY ≤ n}.<br />
T∈T
3.2. THE UNIFORM BOUNDEDNESS PRINCIPLE 27<br />
Then Xn is closed in X. By the hypothesis, we have<br />
X =<br />
∞<br />
Xn.<br />
n=1<br />
By Theorem 3.2.1, there exists an n ∈ N such that Xn is not nowhere dense.<br />
That is, we have X ◦ n = X ◦<br />
n = ∅. In particular, the set Xn contains a ball<br />
B0 = {x ∈ X: x−x0 ≤ r}<br />
for some x0 ∈ X and r > 0. That is, for x ∈ B0 and T ∈ T , we have<br />
TxY ≤ n.<br />
Fix T ∈ T and suppose that x ∈ X with xX ≤ 1. Define y = rx+x0.<br />
Then y ∈ B0. So<br />
<br />
<br />
TxY = <br />
T <br />
y −x0 Y<br />
r<br />
<br />
<br />
= <br />
1 <br />
(Ty −Tx0) <br />
r<br />
Y<br />
≤ 1<br />
r (TyY +Tx0Y) ≤ 2n<br />
r .<br />
Hence T ≤ 2n<br />
r . The right-hand side is independent of T, so T is bounded<br />
in B(X,Y).<br />
Corollary 3.2.1. Let X be a Banach space and Y a normed vector space.<br />
Suppose that (Tn)n∈N is a sequence in B(X,Y) such that (Tnx)n∈N is convergent<br />
for every x ∈ X. Then there exists a unique operator T ∈ B(X,Y)<br />
such that<br />
Tx = lim<br />
n→∞ Tnx<br />
for all x ∈ X.<br />
Proof. It is clear (by the assumption and the uniqueness of limits) that the<br />
formula defines a unique map T : X → Y. It is also clear that T is linear.<br />
We need to show that T is bounded.<br />
Let T = {Tn: n ∈ N}. For every x ∈ X, the sequence (Tnx)n∈N is<br />
convergent by the hypothesis; in particular it is bounded. Therefore, we can<br />
apply Theorem 3.2.2 to T . It follows that there exists a number M ≥ 0<br />
with the property that Tn ≤ M for every n ∈ N. Now for any x ∈ X,<br />
So T ≤ M and T ∈ B(X,Y).<br />
TxY = lim<br />
n→∞ TnxY ≤ MxX.<br />
Example 3.2.1. Let (an)n∈N be a sequence in R such that for every x ∈<br />
ℓ 2 , the sequence (a1x1,a2x2,a3x3,...) also belongs to ℓ 2 . Then the linear<br />
operator<br />
T : ℓ 2 → ℓ 2 , (x1,x2,x3,...) ↦→ (a1x1,a2x2,a3x3,...),
28 CHAPTER 3. BOUNDED LINEAR OPERATORS<br />
is bounded.<br />
To prove this statement, define Tn : ℓ 2 → ℓ 2 with<br />
Tn(x1,x2,x3,...) = (a1x1,...,anxn,0,0,...).<br />
Then Tn ∈ B(ℓ 2 ) with Tn ≤ max{|a1|,...,|an|}. Moreover, for every<br />
x ∈ ℓ 2 ,<br />
Tx = lim<br />
n→∞ Tnx.<br />
Thus the claim follows from Corollary 3.2.1.<br />
3.3 The Open Mapping Theorem<br />
Recall that a continuous map can be characterised by the property that the<br />
preimage of any open set is open. We now consider the reverse property,<br />
whichisuseful, e.g., whenwestudythecontinuityoftheinverseofabijective<br />
map.<br />
Definition 3.3.1. Let X,Y be metric spaces and f : X → Y a map. We say<br />
that f is open if for every open set U ⊂ X, the image f(U) = {f(x): x ∈ U}<br />
is open in Y.<br />
Theorem 3.3.1 (Open Mapping Theorem). Let X,Y be Banach spaces and<br />
T ∈ B(X,Y). If T is surjective, then it is open.<br />
Proof. Consider the closed unit balls<br />
BX = {x ∈ X: xX ≤ 1} and BY = {y ∈ Y : yY ≤ 1}.<br />
ForR > 0, alsoconsidertheballRBX ofradiusRinX. SinceT issurjective,<br />
we have<br />
∞<br />
Y = T(nBX).<br />
n=1<br />
By the Baire category theorem (Theorem 3.2.1), there exists an n ∈ N such<br />
that T(nBX) is not nowhere dense. Hence there exist y0 ∈ Y and r > 0<br />
such that<br />
y0 +rBY ⊂ T(nBX).<br />
Let x0 ∈ X such that T(x0) = y0 and define<br />
Then<br />
R = 1<br />
r (n+x0X).<br />
rBY ⊂ T(nBX)−T(x0) = T(nBX −x0) ⊂ T(rRBX).<br />
Thus BY ⊂ T(RBX).
3.3. THE OPEN MAPPING THEOREM 29<br />
Next we claim that<br />
BY ⊂ T(2RBX).<br />
In order to verify this, fix y ∈ BY. Then there exists an x1 ∈ RBX with<br />
y −Tx1Y ≤ 1<br />
2 .<br />
Hence 2(y − Tx1) ⊂ BY ⊂ T(RBX). Therefore, there exists a point ˜x2 ∈<br />
RBX such that<br />
2(y −Tx1)−T˜x2Y ≤ 1<br />
2 .<br />
Let x2 = 1<br />
2 ˜x2. Then<br />
y −T(x1 +x2)Y ≤ 1<br />
4 .<br />
Now we recursively define x3,x4,... such that xk ∈ 21−kRBX and<br />
k 1<br />
y −T(x1 +···+xk)Y ≤ .<br />
2<br />
Then<br />
Therefore,<br />
∞ ∞<br />
xkX ≤ R 2 1−k = 2R.<br />
k=1<br />
x =<br />
k=1<br />
∞<br />
k=1<br />
exists (cf. Exercise 2.4) and belongs to 2RBX. Moreover,<br />
y −TxY = lim<br />
k→∞ y −T(x1 +···+xk)Y = 0.<br />
So y = Tx. That is, we have proved BY ⊂ T(2RBX).<br />
Finally, let U ⊂ X be an open set and let V = T(U). Fix y0 ∈ V. There<br />
exists an x0 ∈ U with Tx0 = y0. As U is open, there exists a number ρ > 0<br />
such that x0 +ρBX ⊂ U. Then<br />
Hence V is open.<br />
V ⊃ T(x0 +ρBX) = y0 + ρ<br />
2R T(2RBX) ⊃ y0 + ρ<br />
2R BY.<br />
Corollary 3.3.1. Let X,Y be Banach spaces and T ∈ B(X,Y). If T is<br />
bijective, then T −1 ∈ B(Y,X).<br />
Proof. We have T −1 ∈ B(X,Y) if, and only if, the inverse T −1 is continuous<br />
by Theorem 3.1.1. A map is continuous if, and only if, its preimages of open<br />
sets are open. For an inverse map T −1 , this is equivalent to T being open.<br />
So the claim follows directly from Theorem 3.3.1.<br />
xk
30 CHAPTER 3. BOUNDED LINEAR OPERATORS<br />
If X,Y are normed spaces, then can equip the Cartesian product X×Y<br />
with the norm<br />
(x,y)X×Y = xX +yY.<br />
If X and Y are Banach spaces, then this makes X ×Y a Banach space as<br />
well.<br />
We now have the following criterion for boundedness of an operator<br />
between Banach spaces.<br />
Theorem 3.3.2 (Closed Graph Theorem). Let X,Y be Banach spaces and<br />
T : X → Y a linear operator. Then T ∈ B(X,Y) if, and only if, the graph<br />
G = {(x,Tx): x ∈ X} is closed in X ×Y.<br />
Proof. Suppose that T ∈ B(X,Y). Define L : X × Y → Y by L(x,y) =<br />
y −Tx. Then<br />
L(x,y)Y ≤ yY +TxX ≤ (1+T)(x,y)X×Y.<br />
Thus L ∈ B(X×Y,Y), which means that L is continuous. So G = L −1 ({0})<br />
is closed.<br />
Now suppose that G is closed. Consider the projections PX : X×Y → X<br />
and PY : X ×Y → Y with PX(x,y) = x and PY(x,y) = y. We have<br />
PX(x,y)X = xX ≤ (x,y)X×Y.<br />
So PX ∈ B(X ×Y,X) with PX ≤ 1. Similarly, we see that PY ∈ B(X ×<br />
Y,Y) with PY ≤ 1.<br />
Note that G is a linear subspace of a Banach space. As it is closed,<br />
this makes G a Banach space. The restriction PX|G of PX to G belongs to<br />
B(G,X). Furthermore, it is bijective. By Corollary 3.3.1, we have<br />
Finally, we note that<br />
Hence T ∈ B(X,Y).<br />
(PX|G) −1 ∈ B(X,G).<br />
T = PY ◦(PX|G) −1 .