Bounded Linear Operators

Chapter 3 

Bounded Linear Operators 

3.1 The Space B(X,Y) 

In this chapter we study linear operators (also called linear maps or linear 

transformations) between normed spaces, especially operators with the 

following additional property. 

Definition 3.1.1. Let X,Y be normed vector spaces and T : X → Y a 

linear operator. We say that T is bounded if there exists a number c > 0 

such that 

for all x ∈ X. 

TxY ≤ cxX 

Theorem 3.1.1. Let X,Y be normed vector spaces and T : X → Y linear. 

Then the following are equivalent. 

(i) T is bounded. 

(ii) T is uniformly continuous. 

(iii) T is continuous at 0. 

Proof. (i) ⇒ (ii) Suppose that there is a number c > 0 satisfying TxY ≤ 

cxX for all x ∈ X. Let ǫ > 0 and choose δ = ǫ 

c . If we have x,y ∈ X with 

x−yX ≤ δ, then 

Tx−TyY = T(x−y)Y ≤ cx−yX ≤ ǫ. 

So T is uniformly continuous. 

(ii) ⇒ (iii) This is trivial. 

(iii) ⇒ (i) Choose δ > 0 such that 

xX ≤ δ ⇒ TxY ≤ 1. 

23

24 CHAPTER 3. BOUNDED LINEAR OPERATORS 

Then for any x ∈ X\{0}, 

Hence T is bounded. 

TxX = xX 

δ 

 

 

 

T 

δx Y 

xX 

≤ xX 

. 

δ 

Definition 3.1.2. Let X,Y be normed vector spaces. Then B(X,Y) is the 

spacecomprisingallboundedlinearoperatorsT : X → Y. ForT ∈ B(X,Y), 

we define 

T = sup TxY. 

x∈X 

xX≤1 

This is called the operator norm of T. We also write B(X) = B(X,X). 

Sometimes we write T B(X,Y) or T B(X) for the operator norm in 

order to distinguish it from other norms. Note that the linearity implies the 

identities 

unless X = {0}. 

TxY 

T = sup TxY = sup , 

x∈X x∈X\{0} xX 

xX=1 

Theorem 3.1.2. (i) If X,Y are normed vector spaces, then B(X,Y), 

equipped with the operator norm, is a normed vector space. 

(ii) If X is a normed vector space and Y is a Banach space, then B(X,Y) 

is a Banach space. 

(iii) Let X,Y,Z be normed vector spaces and suppose that S ∈ B(X,Y) 

and T ∈ B(Y,Z). Then TS ∈ B(X,Z) with 

TS ≤ TS. 

Proof. (i) is routine. 

(ii) Suppose that Y is complete. Let (Tn)n∈N be a Cauchy sequence in 

B(X,Y). Then for any fixed x ∈ X, we have 

Tmx−TnxY ≤ Tm −TnxX → 0 as m,n → ∞. 

Hence (Tnx)n∈N is a Cauchy sequence in Y and the limit 

Tx := lim 

n→∞ Tnx 

exists. This gives rise to a map T : X → Y. It is linear, because for x,y ∈ X 

and α ∈ F, we have 

T(αx+y) = lim 

n→∞ Tn(αx+y) = α lim 

n→∞ Tnx+ lim 

n→∞ Tny = αTx+Ty.

3.1. THE SPACE B(X,Y) 25 

It is bounded, because 

TxY = lim 

n→∞ TnxY ≤ 

 

lim 

n→∞ Tn 

 

xX. 

It remains to prove that Tn → T in B(X,Y) (which is stronger than pointwise 

convergence). 

Let ǫ > 0 and choose N ∈ N such that Tm − Tn ≤ ǫ for m,n ≥ N. 

Then for every x ∈ X, 

Letting m → ∞, we obtain 

Tmx−TnxY ≤ ǫxX. 

Tx−TnxY ≤ ǫxX. 

That is, we have T −Tn ≤ ǫ, as required. 

(iii) Let x ∈ X. Then 

This implies the desired inequality. 

TSxZ ≤ TSxY ≤ TSxX. 

Definition 3.1.3. If X is a normed vector space over F, then X ∗ = B(X,F) 

is called the dual space of X. Its elements are called bounded linear functionals 

on X. 

Example 3.1.1. Let a 

Tf = f(c). This is a linear operator and we have 

|Tf| = |f(c)| ≤ f C 0 ([a,b]). 

So T ∈ (C 0 ([a,b])) ∗ and T ≤ 1. Since T1 = 1, we have in fact T = 1. 

Example 3.1.2. Let g ∈ C 0 ([a,b]) and define T : C 0 ([a,b]) → C 0 ([a,b]) by 

Tf = gf. Then 

Tf C 0 ([a,b]) = gf C 0 ([a,b]) ≤ g C 0 ([a,b])f C 0 ([a,b]). 

Hence T ∈ B(C 0 ([a,b])) and T ≤ g C 0 ([a,b]). Since T1 = g, we have in 

fact T = g C 0 ([a,b]). 

Example 3.1.3. Let p ∈ [1,∞] and consider the operators 

R: ℓ p → ℓ p , (x1,x2,x3,...) ↦→ (0,x1,x2,...), 

L: ℓ p → ℓ p , (x1,x2,x3,...) ↦→ (x2,x3,x4,...). 

For x = (x1,x2,x3,...) ∈ ℓ p , we clearly have 

Rxℓp = xℓp and Lxℓp ≤ xℓp. So R,L ∈ B(ℓ p ). It also follows immediately that R = 1 and L ≤ 1. 

In fact we have L = 1, because L maps (0,1,0,0,...) (with norm 1) to 

(1,0,0,...) (still with norm 1).


Example 3.1.4. Let p,q be conjugate exponents. Fix y = (y1,y2,y3,...) ∈ 

ℓq . Define T : ℓp → F by 

∞ 

Tx = xnyn. 

Then by the Hölder inequality, 

n=1 

|Tx| ≤ xℓpyℓq for every x ∈ ℓp . So T ∈ (ℓp ) ∗ with T ≤ yℓq. We will see later that 

T = yℓq. 3.2 The Uniform Boundedness Principle 

It is not so easy to come up with an example of a linear operator between 

Banach spaces that is not bounded. Nevertheless, boundedness is a severe 

restriction and has some deep consequences. We now discuss some of them. 

Recall from MA40043: a set Y in a metric space X is called nowhere 

dense if Y ◦ = ∅. The following is a result from MA40043. 

Theorem 3.2.1 (Baire Category Theorem). A complete, non-empty metric 

space cannot be a countable union of nowhere dense subsets. 

As a consequence we obtain a result on bounded linear operators. 

Theorem 3.2.2 (Banach-Steinhaus/Uniform Boundedness Principle). Let 

X be a Banach space and Y a normed vector space. Let T ⊂ B(X,Y) be 

such that for all x ∈ X, the set {Tx: T ∈ T } is bounded in Y. Then T is 

bounded in B(X,Y). 

In other words, if we have pointwise bounds, then we automatically get 

a bound in the norm of B(X,Y) as well. The second property appears much 

stronger and implies in particular that we have uniform bounds on bounded 

subsets of X. That is, if A ⊂ X with 

then 

sup 

T∈T 

Proof. For n ∈ N, define 

supx 

< ∞, 

x∈A 

 

supTxY 

≤ sup T 

x∈A T∈T 

supxX 

x∈A 

 

< ∞. 

Xn = {x ∈ X: TxY ≤ n for all T ∈ T } = 

{x ∈ X: TxY ≤ n}. 

T∈T

3.2. THE UNIFORM BOUNDEDNESS PRINCIPLE 27 

Then Xn is closed in X. By the hypothesis, we have 

X = 

∞ 

Xn. 

n=1 

By Theorem 3.2.1, there exists an n ∈ N such that Xn is not nowhere dense. 

That is, we have X ◦ n = X ◦ 

n = ∅. In particular, the set Xn contains a ball 

B0 = {x ∈ X: x−x0 ≤ r} 

for some x0 ∈ X and r > 0. That is, for x ∈ B0 and T ∈ T , we have 

TxY ≤ n. 

Fix T ∈ T and suppose that x ∈ X with xX ≤ 1. Define y = rx+x0. 

Then y ∈ B0. So 

 

 

TxY = 

T 

y −x0 Y 

r 

 

 

= 

1 

(Ty −Tx0) 

r 

Y 

≤ 1 

r (TyY +Tx0Y) ≤ 2n 

r . 

Hence T ≤ 2n 

r . The right-hand side is independent of T, so T is bounded 

in B(X,Y). 

Corollary 3.2.1. Let X be a Banach space and Y a normed vector space. 

Suppose that (Tn)n∈N is a sequence in B(X,Y) such that (Tnx)n∈N is convergent 

for every x ∈ X. Then there exists a unique operator T ∈ B(X,Y) 

such that 

Tx = lim 

n→∞ Tnx 

for all x ∈ X. 

Proof. It is clear (by the assumption and the uniqueness of limits) that the 

formula defines a unique map T : X → Y. It is also clear that T is linear. 

We need to show that T is bounded. 

Let T = {Tn: n ∈ N}. For every x ∈ X, the sequence (Tnx)n∈N is 

convergent by the hypothesis; in particular it is bounded. Therefore, we can 

apply Theorem 3.2.2 to T . It follows that there exists a number M ≥ 0 

with the property that Tn ≤ M for every n ∈ N. Now for any x ∈ X, 

So T ≤ M and T ∈ B(X,Y). 

TxY = lim 

n→∞ TnxY ≤ MxX. 

Example 3.2.1. Let (an)n∈N be a sequence in R such that for every x ∈ 

ℓ 2 , the sequence (a1x1,a2x2,a3x3,...) also belongs to ℓ 2 . Then the linear 

operator 

T : ℓ 2 → ℓ 2 , (x1,x2,x3,...) ↦→ (a1x1,a2x2,a3x3,...),


is bounded. 

To prove this statement, define Tn : ℓ 2 → ℓ 2 with 

Tn(x1,x2,x3,...) = (a1x1,...,anxn,0,0,...). 

Then Tn ∈ B(ℓ 2 ) with Tn ≤ max{|a1|,...,|an|}. Moreover, for every 

x ∈ ℓ 2 , 

Tx = lim 

n→∞ Tnx. 

Thus the claim follows from Corollary 3.2.1. 

3.3 The Open Mapping Theorem 

Recall that a continuous map can be characterised by the property that the 

preimage of any open set is open. We now consider the reverse property, 

whichisuseful, e.g., whenwestudythecontinuityoftheinverseofabijective 

map. 

Definition 3.3.1. Let X,Y be metric spaces and f : X → Y a map. We say 

that f is open if for every open set U ⊂ X, the image f(U) = {f(x): x ∈ U} 

is open in Y. 

Theorem 3.3.1 (Open Mapping Theorem). Let X,Y be Banach spaces and 

T ∈ B(X,Y). If T is surjective, then it is open. 

Proof. Consider the closed unit balls 

BX = {x ∈ X: xX ≤ 1} and BY = {y ∈ Y : yY ≤ 1}. 

ForR > 0, alsoconsidertheballRBX ofradiusRinX. SinceT issurjective, 

we have 

∞ 

Y = T(nBX). 

n=1 

By the Baire category theorem (Theorem 3.2.1), there exists an n ∈ N such 

that T(nBX) is not nowhere dense. Hence there exist y0 ∈ Y and r > 0 

such that 

y0 +rBY ⊂ T(nBX). 

Let x0 ∈ X such that T(x0) = y0 and define 

Then 

R = 1 

r (n+x0X). 

rBY ⊂ T(nBX)−T(x0) = T(nBX −x0) ⊂ T(rRBX). 

Thus BY ⊂ T(RBX).

3.3. THE OPEN MAPPING THEOREM 29 

Next we claim that 

BY ⊂ T(2RBX). 

In order to verify this, fix y ∈ BY. Then there exists an x1 ∈ RBX with 

y −Tx1Y ≤ 1 

2 . 

Hence 2(y − Tx1) ⊂ BY ⊂ T(RBX). Therefore, there exists a point ˜x2 ∈ 

RBX such that 

2(y −Tx1)−T˜x2Y ≤ 1 

2 . 

Let x2 = 1 

2 ˜x2. Then 

y −T(x1 +x2)Y ≤ 1 

4 . 

Now we recursively define x3,x4,... such that xk ∈ 21−kRBX and 

k 1 

y −T(x1 +···+xk)Y ≤ . 

2 

Then 

Therefore, 

∞ ∞ 

xkX ≤ R 2 1−k = 2R. 

k=1 

x = 

k=1 

∞ 

k=1 

exists (cf. Exercise 2.4) and belongs to 2RBX. Moreover, 

y −TxY = lim 

k→∞ y −T(x1 +···+xk)Y = 0. 

So y = Tx. That is, we have proved BY ⊂ T(2RBX). 

Finally, let U ⊂ X be an open set and let V = T(U). Fix y0 ∈ V. There 

exists an x0 ∈ U with Tx0 = y0. As U is open, there exists a number ρ > 0 

such that x0 +ρBX ⊂ U. Then 

Hence V is open. 

V ⊃ T(x0 +ρBX) = y0 + ρ 

2R T(2RBX) ⊃ y0 + ρ 

2R BY. 

Corollary 3.3.1. Let X,Y be Banach spaces and T ∈ B(X,Y). If T is 

bijective, then T −1 ∈ B(Y,X). 

Proof. We have T −1 ∈ B(X,Y) if, and only if, the inverse T −1 is continuous 

by Theorem 3.1.1. A map is continuous if, and only if, its preimages of open 

sets are open. For an inverse map T −1 , this is equivalent to T being open. 

So the claim follows directly from Theorem 3.3.1. 

xk


If X,Y are normed spaces, then can equip the Cartesian product X×Y 

with the norm 

(x,y)X×Y = xX +yY. 

If X and Y are Banach spaces, then this makes X ×Y a Banach space as 

well. 

We now have the following criterion for boundedness of an operator 

between Banach spaces. 

Theorem 3.3.2 (Closed Graph Theorem). Let X,Y be Banach spaces and 

T : X → Y a linear operator. Then T ∈ B(X,Y) if, and only if, the graph 

G = {(x,Tx): x ∈ X} is closed in X ×Y. 

Proof. Suppose that T ∈ B(X,Y). Define L : X × Y → Y by L(x,y) = 

y −Tx. Then 

L(x,y)Y ≤ yY +TxX ≤ (1+T)(x,y)X×Y. 

Thus L ∈ B(X×Y,Y), which means that L is continuous. So G = L −1 ({0}) 

is closed. 

Now suppose that G is closed. Consider the projections PX : X×Y → X 

and PY : X ×Y → Y with PX(x,y) = x and PY(x,y) = y. We have 

PX(x,y)X = xX ≤ (x,y)X×Y. 

So PX ∈ B(X ×Y,X) with PX ≤ 1. Similarly, we see that PY ∈ B(X × 

Y,Y) with PY ≤ 1. 

Note that G is a linear subspace of a Banach space. As it is closed, 

this makes G a Banach space. The restriction PX|G of PX to G belongs to 

B(G,X). Furthermore, it is bijective. By Corollary 3.3.1, we have 

Finally, we note that 

Hence T ∈ B(X,Y). 

(PX|G) −1 ∈ B(X,G). 

T = PY ◦(PX|G) −1 .

Bounded Linear Operators

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