3

Some Important Theorems

3.1 Closed Graph Theorem

We know that if A : x → Y is a continuous linear operator between
normed linear spaces X and Y , then for every sequence (xn ) in X,

xn → x =⇒ Axn → Ax.

In applications there are plenty of situations in which a linear op-

erator A may be defined only in a subspace X0 of a known space,
usually a Banach space. In such cases a sequence (xn ) in X0 may
be convergent in X, but the limit need not be in X0 , but the image
sequence (Axn ) can still converge. So, a natural question would be:

If A is a linear operator defined on a subspace X0 of a

normed linear space X with values in a normed linear
space Y , and if (xn ) is a sequence in X0 such that

xn → x in X and Axn → y,

then do we have x ∈ X0 and y = Ax?

In view of the question raised above, we have the following defi-

nition.
Definition 3.1.1 Let X and Y be normed linear spaces, X0 be a
subspace of X and A : X0 → Y be a linear operator. Then A is
called a closed operator if for every sequence (xn ) is in X0 ,

xn → x in X and Axn → y =⇒ x ∈ X0 and y = Ax.

♦
We now give a characterization of a closed operator in terms of
the closedness of the graph of the operator in a product space.

106
Closed Graph Theorem 107

Definition 3.1.2 Suppose X and Y are normed linear spaces. Then

k(x, y)k := kxkX + kykY , (x, y) ∈ X × Y,

defines a norm on the product space X × Y , called the product

norm on X × Y , and X × Y with this product norm is called a
product space. ♦
Observe:
• For each p with 1 ≤ p ≤ ∞,
(kxkpX + kykpY )1/p , 1 ≤ p < ∞,

k(x, y)k :=
max{kxkX , kykY }, p = ∞,

Now, the following characterization of a closed operator is imme-

diate.
Proposition 3.1.1 Let X and Y be normed linear spaces and X0 be
a subspace of X. A linear operator A : X0 → Y is a closed operator
if and only if its graph,

G(A) := {(x, Ax) : x ∈ X0 },

is a closed subset of the product space X × Y .

Example 3.1.1 Let X = Y = C[a, b] with k · k∞ and X0 = C 1 [a, b].
Then A : X0 → Y defined by

Ax = x0 , x ∈ X0

is a closed operator: To see this, let (xn ) in X0 be such that

kxn − xk∞ → 0 and kx0n − yk∞ → 0

for some x, y ∈ C[a, b], i.e., (xn ) converges to x uniformly and (x0n )
converges to y uniformly. Then, by a result in real analysis (see Rudin
[9]), we know that x is differentiable and x0 = y. Thus, x ∈ X0 and
Ax = y. 
Example 3.1.2 Let X be an infinite dimensional Hilbert space
and E0 = {un : n ∈ N} be an orthonormal set in X. Let (λn ) be a
sequence of scalars. Let
n ∞
X
X0 = x ∈ X : |λn |2 |hx, un i|2 < ∞}.
n=1
108 Some Important Theorems

For x ∈ X0 , let
∞
X
Ax = λj hx, uj iuj .
j=1

By Riesz-Fischer theorem (Theorem 1.5.4), we see that Ax ∈ X for

every x ∈ X0 , and A : X0 → X is a linear operator. We show that

A is a closed operator, and it is a bounded operator if

and only if (λn ) is a bounded sequence.

Let (xn ) in X0 be such that

xn → x and Axn → y

for some (x, y) ∈ X × X. Then, for each j ∈ N,

hxn , uj i → hx, uj i and λj hxn , uj i = hAxn , uj i → hy, uj i.

Thus,
λj hx, uj i = hy, uj i ∀ j ∈ N.
Hence, x ∈ X0 . Also, if E0 is an orthonormal basis of X, we obtain
Ax = y. Suppose E0 is not an orthonormal basis. Let E be an
orthonormal basis which contains E0 . Then for every u ∈ E \ E0 , we
have hAxn , ui = 0 so that hAxn , ui = 0 for all n ∈ N and hy, ui =
lim hAxn , ui = 0. Thus,
n→∞

hAx, ui = hy, ui ∀u ∈ E;

consequently, Ax = y. Thus, we have showed that A is a closed

operator.
Clearly, if (λn ) is bounded sequence with β := supn∈N |λn |, then
∞
X
|λn |2 |hx, uj i|2 ≤ β 2 kxk2 ∀ x ∈ X,
j=1

so that X0 = X, A ∈ B(X) and kAk ≤ β. Conversely, if A ∈ B(X),

then we have

|λn | = kλn un k = kAun k ≤ kAk ∀n ∈ N

so that (λn ) is bounded. 

Closed Graph Theorem 109

The proof of the following theorem is easy and hence it is left as

exercise.

Theorem 3.1.2 Let X and Y be normed linear spaces, X0 be a

subspace of X and A : X0 → Y be a closed linear operator. Then the
following hold.

(i) N (A) is a closed subspace of X.

(ii) If A is injective, then A−1 : R(A) → X is a closed operator.

Is every continuous linear operator a closed operator?

The answer, in general, is not affirmative. Indeed, if X0 is a
non-closed subspace of a normed linear space X then the inclusion
operator I0 : X0 → X defined by

I0 x = x ∀ x ∈ X0 ,

is a continuous linear operator, which is not a closed operator. One

may also look at the following example which the reader must have
seen in real analysis.
Example 3.1.3 Let X0 = R[a, b], the space (over R) of all Riemann
integrable functions on [a, b], X = L1 [a, b], Y = R, and A : X0 → Y
be defined by
Z b
(Ax)(s) = x(t)dt, x ∈ X0 .
a
Let {r1 , r2 , . . .} be an enumeration of rational numbers in [a, b] and
for n ∈ N, let xn : [a, b] → R be defined by

0, t ∈ {r1 , . . . , rn },
xn (t) =
1, t 6∈ {r1 , . . . , rn }.

Then, it can be seen easily that xn ∈ R[a, b] and xn → x in L1 [a, b],

where x : [a, b] → R is defined by

0, t ∈ Q,
x(t) =
1, t 6∈ Q.

Also, we have
Z b
xn (t)dt = b − a ∀ n ∈ N.
a
110 Some Important Theorems

Thus,
xn → x ∈ X and Axn → y := b − a;
but x 6∈ X0 . 
Theorem 3.1.3 Let X and Y be normed linear spaces, X0 be a
subspace of X and A : X0 → Y be a continuous linear operator.

(i) If X0 is a closed subspace, then A is a closed operator.

(i) If Y is a Banach space and A is a closed operator, then X0 is

closed in X.
Proof. (i) Let (xn ) be a sequence in X0 such that xn → x and
Axn → y for some x ∈ X and y ∈ Y . If X0 is closed in X, then
x ∈ X, so that, by continuity of A, we obtain Axn → Ax.
(ii) Suppose Y is a Banach space and A is a closed operator. Let
(xn ) be a sequence in X0 such that xn → x. By continuity of A,
we see that (Axn ) is a Cauchy sequence in Y . Since Y is complete,
there exists y ∈ Y such that Axn → y. Now, by the closedness of A,
x ∈ X0 .

The following example illustrates how Theorem 3.1.2(ii) and The-

orem 3.1.3 (ii) can also be used to show certain operator is a closed
operator.
Example 3.1.4 Let X be an infinite dimensional separable Hilbert
space and {un : n ∈ N} be an orthonormal basis of X. Let (λn ) be
a sequence of nonzero scalars such that
d := inf |λn | > 0.
n∈N

Let A be as in Example 3.1.2. Since {un : n ∈ N} is an orthonormal

basis, it follows that, for x ∈ X0 ,
Ax = 0 =⇒ x = 0
so that A is one-one. Further, for every y ∈ X,
∞
X |hy, un i|2 kyk2
≤
|λn |2 d2
n=1

so the the series

∞
X hy, un i
un
λn
n=1
Closed Graph Theorem 111

hy,un i
converges. Hence, x := ∞
P
n=1 λn un satisfies the equation Ax = y.
In other words, A is onto as well. Note that
∞
X |hy, un i|2 kyk2
kA−1 yk2 = ≤ ∀ y ∈ X.
|λn |2 d2
n=1

Thus, A−1 is continuous with closed domain, the whole of X. By

Theorem 3.1.3 (ii), A−1 is a closed operator, and hence, by Theorem
3.1.2(ii), A is a closed operator. 
In view of Theorem 3.1.3, a question naturally arises is the follow-
ing: When is a closed operator continuous? Theorem 2.1.6, together
with Theorem 3.1.2(i), shows that every closed linear functional is
continuous.
What about if dim (Y ) > 1? Closed graph theorem gives an an-
swer.
Theorem 3.1.4 (Closed graph theorem) Let X and Y be Ba-
nach spaces and A : X → Y be a closed operator. Then A is contin-
uous.
Proof. In order to show that A is continuous, it is enough to show
(Why?) that there exists c > 0 such that

B0 ⊆ {x ∈ X : kAxk ≤ c},

where B0 = {x ∈ X : kxk < 1}. For α > 0, let

Vα := {x ∈ X : kAxk ≤ α}.

Then we have X = ∪∞ j=1 Vj . Since X is complete, by the Baire

category theorem (Theorem 1.3.8), there is some k ∈ N such that
the interior of cl (Vk ) is nonempty. Thus, there is some x0 ∈ X and
r > 0 such that B(x0 , r) ⊆ cl (Vk ). Then it can be seen (Verify!)
that B0 ⊆ cl (V2k/r ). We show that

B0 ⊆ V2k/r . (∗)

Let us denote V2k/r by W . Let x ∈ B0 and 0 < ε < 1. Since

B0 ⊆ cl W , there exists x1 ∈ W such that kx − x1 k < ε. Hence,
ε−1 (x − x1 ) ∈ B0 . By the same argument, there exists x2 ∈ W such
that kε−1 (x − x1 ) − x2 k < ε, i.e.,

kx − (x1 + εx2 )k < ε2 .

112 Some Important Theorems

Continuing this argument, we obtain a sequence (xn ) in W with

x − (x1 + εx2 + ε2 x3 + · · · + εn−1 xn ) < εn

for every positive integer n. In particular, taking sn := nj=1 εj−1 xj ,

P
n ∈ N, the sequence (sn ) converges to x. Recall that xj ∈ W implies
kAxj k ≤ 2k/r. Hence, for n > m, we have
n n
X 2k X j−1
kAsn − Asm k ≤ εj−1 kAxj k ≤ ε .
r
j=m+1 j=m+1

Thus, (Asn ) is a Cauchy sequence in Y . Since Y is also a Banach

space, the sequence (Asn ) converges to some y ∈ Y . Since A is a
closed operator, we have y = Ax = limn→∞ Asn . Note that
n n
X 2k X j−1 2k
kAsn k ≤ εj−1 kAxj k ≤ ε ≤ .
r r(1 − ε)
j=1 j=1

Hence,
2k
kAxk = lim kAsn k ≤ .
n→∞ r(1 − ε)
This is true for all ε > 0. Hence, kAxk ≤ 2k/r. Thus, (∗) is proved,
which completes the proof.

The following corollary, which is also called closed graph theorem,

can be deduced from Theorem 3.1.3 and Theorem 3.1.4.
Corollary 3.1.5 (Closed graph theorem) Let X and Y be Ba-
nach spaces and X0 be a subspace of X. Then a closed operator
A : X0 → Y is continuous if and only if X0 is closed in X.
Here is an application of Theorem 3.1.4.
Theorem 3.1.6 Let X be a Hilbert space and A : X → X be a self
adjoint operator. Then A ∈ B(X).
Proof. By closed graph theorem, it is enough to prove that A is
a closed operator. So, let (xn ) in X be such

xn → x and Axn → y

for some x, y ∈ X. Using the property of A, we have

hAxn , ui = hxn , Aui ∀ u ∈ X.

Closed Graph Theorem 113

Hence, taking limit as n → ∞, and again using the property of A,

we obtain
hy, ui = hx, Aui = hAx, ui ∀ u ∈ X.
Hence, y = Ax. This proves that A is a closed operator.

Now we give examples to show that completeness assumption in

Closed Graph Theorem cannot be dropped.

Example 3.1.5 Let X = C 1 [0, 1] and Y = C[0, 1], both with k · k∞

and X0 = X. Let A : X0 → Y be defined by

Ax = x0 , x ∈ X0 .

As in Example 3.1.1, we see that A is a closed operator. We have

seen in Example 2.1.7 that A is not a continuous operator. Note that
X is not a Banach space. 
Example 3.1.6 Let X be an infinite dimensional Banach space
and E := {uλ : λ ∈ Λ} be a basis of X with kuλ k = 1 for all λ ∈ Λ.
Then E is an uncountable set (Why?).
P Since E is a basis of X, every
x ∈ X can be written as x = λ∈Λ x̂(λ)uλ , where x̂(λ) are scalars
such that x̂(λ) = 0 for all but a finite number of λ’s. Define
X
kxk∗ := |x̂(λ)|, x ∈ X.
λ∈Λ

Then it can be seen easily that k · k∗ is also a norm on X and

kxk ≤ kxk∗ ∀ x ∈ X.

We first show that k · k∗ is not complete.

Consider a sequence (λn ) of distinct elements from Λ. For each
n ∈ N, let
n
X uλj
xn = .
j2
j=1

Then, for every n, m ∈ N with n > m, we have

n
X 1
kxn − xm k∗ = .
j2
j=m+1

Hence, (xn ) is a Cauchy sequence with respect to the norm k · k∗ .

We claim that (xn ) does not converge with respect to k · k∗ . On the
114 Some Important Theorems

contrary, suppose there exists x ∈ X such that kx − xn k∗ → 0 as

n → ∞. Then we have
X 1
kx − xn k∗ := |x̂(λ) − xn (λ)| ≥x̂(λj ) − ∀j ∈ N.

j
λ∈Λ

Since kx − xn k∗ → 0, it follows that x̂(λj ) = 1/j for all j ∈ N, which

is not possible. Thus, (xn ) is not convergent with respect to k · k∗ .
Next, let X∗ be the linear space X with k · k∗ . Then the identity
operator I : X → X∗ is a closed operator. But, it is not continuous,
because, if it is is continuous, then there would exist c > 0 such
that kxk∗ ≤ ckxk for all x ∈ X, which would imply that the norms
k · k and k · k∗ are equivalent; a contradiction to the fact that k · k is
complete and k · k∗ is not complete. 
Now, let us derive some important consequences of closed graph
theorem.

3.1.1 Bounded inverse theorem

Theorem 3.1.7 (Bounded inverse theorem) Suppose X and Y
are Banach spaces, X0 is a subspace of X and A : X0 → Y is a
closed operator. Suppose A is injective. Then A−1 : R(A) → X is
continuous if and only if R(A) is closed.
Proof. Suppose A is injective. By Theorem 3.1.2, the operator
A−1 : R(A) → X is a closed operator. Hence, by Corollary 3.1.5,
A−1 : R(A) → X is continuous if and only if R(A) is closed.

The proof of the following corollary, which is also known as bounded

inverse theorem, is immediate from Theorem 3.1.7.
Corollary 3.1.8 (Bounded inverse theorem) Suppose X and Y
are Banach spaces and A ∈ B(X, Y ). If A is bijective, then A−1 :
Y → X is a bounded operator.
Here is another consequence of Theorem 3.1.7.
Corollary 3.1.9 Suppose k · k1 and k · k2 are complete norms on
a normed linear space X such that one of them is stronger than the
other. Then they are equivalent.
Proof. Suppose k · k1 is stronger than k · k2 , that is, there exists
c > 0 such that
kxk2 ≤ ckxk1 ∀ x ∈ X.
Closed Graph Theorem 115

Let X1 and X2 be the space X with norms k · k1 and k · k2 , respec-

tively. By the above inequality, the identity map from X1 to X2 is
continuous. Since this map is bijective, and since X1 and X2 are Ba-
nach spaces, by Corollary 3.1.8, its inverse is also continuous. Hence,
there exists c0 > 0 such that

kxk1 ≤ c0 kxk2 ∀ x ∈ X.

Thus, k · k1 and k · k2 are equivalent.

For the next theorem we shall make use of the following lemma.
Lemma 3.1.10 Let X, Y be normed linear spaces and A ∈ B(X, Y ).
Then à : X/N (A) → Y defined by

Ã[x] = Ax, [x] ∈ X/N (A),

is an injective bounded linear operator with kÃk = kAk. Further,

A ∈ K(X, Y ) =⇒ Ã ∈ K(X/N (A), Y ).

Proof. Note that for every x ∈ X,

kÃ[x]k = kA(x − u)k ≤ kAk kx − uk, u ∈ N (A).

Hence,

kÃ[x]k ≤ kAkdist(x, N (A)) = kAk k[x]k ∀ x ∈ X.

so that à is a bounded linear operator and kÃk ≤ kAk. Also, for

every x ∈ X,

kAxk = kÃ[x]k ≤ kÃk k[x]k ≤ kÃk kxk.

Hence, kAk ≤ kÃk. Thus, kÃk = kAk.

Next, assume that A ∈ K(X, Y ). Let (ξn ) be a bounded sequence
in X/N (A), so that there exists M > 0 such that kξn k ≤ M for every
n ∈ N. Thus,
dist (xn , N (A)) ≤ M ∀ n ∈ N,
where ξn = [xn ], n ∈ N. Hence, there exists a sequence (un ) in N (A)
such that kxn − un k ≤ 2M for all n ∈ N. In particular, (xn − un ) is a
bounded sequence in X. Since A(xn − un ) = Axn for all n ∈ N and
A is a compact operator, (Axn ) has a convergent subsequence. But,
Ãξn = Axn for all n ∈ N. Thus, we have proved that à is a compact
operator.
116 Some Important Theorems

Theorem 3.1.11 Let X and Y be Banach spaces and A ∈ K(X, Y ).

Then R(A) is closed if and only if rank (A) < ∞.
Proof. Suppose Y0 := R(A) is closed. Let à : X/N (A) → Y0 be
defined by
Ã[x] = Ax, x ∈ X.
By Lemma 3.1.10, Ã is a bijective compact operator. Hence, by
Theorem 3.1.8, inverse of à is a bounded operator. Therefore, the
identity operator on X/N (A) is a compact operator, as it is a com-
position of a bounded operator with a compact operator. Hence,
by Theorem 1.3.4, X/N (A) is finite dimensional; consequently, Y0 is
finite dimensional.

3.1.2 Open mapping theorem

Recall that a function from a metric space to another metric space
is said to be an open map if image of every open set is open. In the
case of bounded linear operators between Banach spaces we have a
nice characterization of open maps. First we prove the following.
Lemma 3.1.12 Let X be a normed linear space and X0 be a closed
subspace of X. Let η : X → X/X0 be the quotient map, i.e.,

η(x) = x + X0 ∀ x ∈ X.

Then η is linear, continuous, onto and open.

Proof. Clearly, η is linear and onto . Note that

kη(x)k = dist(x, X0 ) ≤ kxk ∀ x ∈ X.

Hence, η is continuous. To show that it is open, let G be an open

subset of X. We have to show that η(G) is open in X/X0 . For this,
it is enough to show that for every x ∈ G, there exists r > 0 such
that

y ∈ X, k(x + X0 ) − (y + X0 )k < r =⇒ y + X0 ∈ η(G).

So, let x ∈ G. Since G is open, there exists r > 0 such that

y ∈ X, kx − yk < r =⇒ y ∈ G.

Now, let y ∈ X be such that k(x + X0 ) − (y + X0 )k < r. Then there

exists u ∈ X0 such that kx − y + uk < r. Then y − u ∈ G, and hence
y + X0 = y − u + X0 ∈ η(G).
Closed Graph Theorem 117

Theorem 3.1.13 (Open mapping theorem) Suppose X and Y

are Banach spaces and A : X → Y is a bounded linear operator.
Then A is an open map if and only if it is onto.

Proof. Suppose A is an open map, i.e., A maps every open subset

of X onto an open subset of Y . In particular, R(A), the image of
the open set X is a nonempty open subset of Y . By Theorem 1.3.9,
this is possible only if R(A) = Y , i.e., A is onto.
Conversely, suppose that A is onto. Then the linear operator
Ae : X/N (A) → Y defined by

A[x]
e = Ax, [x] ∈ X/N (A),

is a bijective bounded linear operator between Banach spaces X/N (A)

and Y (cf. Lemma 3.1.10). Hence, by Bounded Inverse Theorem, in-
verse of Ae is also continuous. In particular, A
e is an open map. Since

A = η ◦ A,
e

where η : X → X/N (A) is the quotient map as in Lemma 3.1.12, we

obtain that A is also an open map.

3.1.3 Uniform boundedness principle

From analysis we know that if a sequence (fn ) of real valued contin-
uous functions defined on a metric space Ω converges uniformly to
a function f : Ω → R, then f is also continuous. However, if the
uniform convergence is replaced by pointwise convergence, i.e.,

fn (x) → f (x) as n → ∞ for each x ∈ Ω,

then the function f need not be continuous. This is the case even
for continuous linear functionals on a normed linear space. However,
if the domain space is a Banach space, the the limiting functional is
continuous. We shall derive this fact as a consequence of the following
general result.

Theorem 3.1.14 (Uniform boundedness principle) Let X be a

Banach space, Y be a normed linear space and A be subset of B(X, Y )
such that {Ax : A ∈ A} is bounded for each x ∈ X. Then A is a
bounded subset of B(X, Y ).
118 Some Important Theorems

Proof. Let us denote the norms on X and Y by k · kX and k · kY ,

respectively. Since {Ax : A ∈ A} is bounded in X, for each x ∈ X,
supA∈A kAxkY is a well defined non-negative real number. Define

kxk∗ := kxkX + sup kAxkY , x ∈ X.

A∈A

It is easily seen that k · k∗ is also a norm on X. Further, k · k∗ is

stronger than k · kX . Now, we show that k · k∗ is complete. For this,
let (xn ) be a Cauchy sequence in X with respect to k · k∗ . Since k · k∗
is stronger than k · kX , (xn ) is a Cauchy sequence in X with respect
to k · kX as well. Using the completeness of k · kX , there exists x ∈ X
such that kxn − xkX → 0 as n → ∞. Hence, for every A ∈ A, by
its continuity, kAxn − Axk → 0 as n → ∞. Now, let ε > 0 be given.
Since (xn ) is a Cauchy sequence in X with respect to k·k∗ , for ε > 0,
there exists N ∈ N such that

kxn − xm kX + sup kAxn − Axm kY < ε ∀ n ≥ N.

A∈A

Let A ∈ A. From the above inequality, we have

kxn − xm kX + kAxn − Axm kY < ε ∀ n ≥ N, (i)

and since, kAxn − AxkX → 0 as n → ∞, letting m → ∞ in (i), we

have
kxn − xkX + kAxn − AxkY ≤ ε ∀ n ≥ N.
This is true for every A ∈ A. Therefore,

kxn − xk∗ = kxn − xkX + sup kAxn − AxkY ≤ ε ∀ n ≥ N.

A∈A

Thus, we have shown that (xn ) converges with respect to k · k∗ , and

consequently, k · k∗ is complete. Hence, by Corollary 3.1.9, k · k∗ and
k · kX are equivalent, so that there exists c > 0 such that

kxk∗ ≤ ckxkX ∀ x ∈ X.

In particular,
sup kAxkY ≤ ckxkX ∀ x ∈ X.
A∈A

Hence, kAk ≤ c for all A ∈ A, and the proof is complete.

Hahn-Banach Extension Theorem 119

Remark 3.1.1 The proof of Uniform Boundedness Principle given

above is different from that usually appear in standard text books
on Functional Analysis. This proof was conveyed to the author by
Professor S. Ramaswamy [7]. ♦

Corollary 3.1.15 (Banach–Steinhaus theorem) Let X be a Ba-

nach space, Y be a normed linear space and (An ) be a sequence of
operators in B(X, Y ) which converges pointwise on X. Then (kAn k)
is bounded and the operator A : X → Y defined by

Ax := lim An x, x ∈ X,
n→∞

belongs to B(X, Y ).

Proof. It can seen easily that A is a linear operator. Now, since

(An ) converges pointwise on X, by Theorem 3.1.14, (kAn k) is bounded,
say kAn k ≤ M for all n ∈ N for some M > 0.
Now, let x ∈ S be such that kxk ≤ 1. Let N ∈ N be such that
kAn x − Axk < 1 for all n ≥ N . Then we have

kAxk ≤ kAx − AN xk + kAN xk ≤ 1 + M.

This is true for all x ∈ S. Hence, A ∈ B(X, Y ).

3.2 Hahn-Banach Extension Theorem

We know that if X is a Hilbert space, then its dual can be identified
with X by a conjugate linear isometry.
What can we say about the dual of a general normed linear space?
Of course, if X is finite dimensional, then we know that X 0 is
of the same dimension as that of X. Also, in certain specific cases,
we can identify the dual space. In this context, we may recall from
Section 2.3 the following:

• For 1 ≤ p < ∞, the dual of `p is linearly isometric with `q .

• For 1 ≤ p < ∞, the dual of Lp [a, b] is linearly isometric with

Lq [a, b].

• The dual of C[a, b] with k·k∞ is linearly isometric with N BV [a, b].

Here q is the conjugate exponent of p, i.e., 1/p + 1/q = 1. However,

using the theory discussed so far, we are not in a position to say
120 Some Important Theorems

even that X 0 is nonzero whenever X is nonzero! Our attempt is

to prove a general theorem, called Hahn-Banach extension theorem,
using which we shall, in fact, show that

dim (X 0 ) ≥ dim (X).

3.2.1 The theorem and its consequences

Theorem 3.2.1 (Hahn-Banach extension theorem) (HBET)
Let X0 be a subspace of a normed linear space X. If f0 ∈ X00 , then
there exists f ∈ X 0 such that

f|X0 = f0 and kf k = kf0 k.

Before proving HBET, let us deduce some of its consequences.

Corollary 3.2.2 Let X be a nonzero normed linear space and x0 be

a nonzero element in X. Then there exists f ∈ X 0 such that

f (x0 ) = kx0 k and kf k = 1.

Proof. Let X0 = span {x0 }, and define f0 : X0 → K by

f0 (αx0 ) = αkx0 k, α ∈ K.

Clearly, f0 is a linear functional on X0 . Further, f0 ∈ X00 and kf0 k =

1 (Exercise). Hence, by HBET, there exists f ∈ X 0 such that f (x0 ) =
f0 (x0 ) = kx0 k and kf k = kf0 k = 1.

More generally we have the following.

Corollary 3.2.3 Let X0 be a closed proper subspace of a normed

linear space X and x0 ∈ X \ X0 . Then there exists f ∈ X 0 such that

f (x0 ) = dist (x0 , X0 ) kf k = 1 and f|X0 = 0.

Proof. Let X1 = span {x0 , X0 }, and define f0 : X1 → K by

f0 (αx0 + u) = αdist (x0 , X0 ), α ∈ K, u ∈ X0 .

Clearly, f0 is a linear functional on X1 . Further,

|f0 (αx0 + u)| = dist (αx0 , X0 ) = dist (αx0 + u, X0 ) ≤ kαx0 + uk

Hahn-Banach Extension Theorem 121

for all α ∈ K, u ∈ X0 . Thus, f0 ∈ X10 and kf0 k ≤ 1. Also, we have

dist (x0 , X0 ) = |f0 (x0 )| = |f0 (x0 − u)| ≤ kf0 kkx0 − uk ∀ u ∈ X0 .

Hence,
dist (x0 , X0 ) ≤ kf0 kdist (x0 , X0 ).
Since dist (x0 , X0 ) > 0, we have kf0 k ≥ 1. Thus, kf0 k = 1. Hence,
by HBET, there exists f ∈ X 0 satisfying

f|X1 = f0 and kf k = kf0 k.

In particular,

f|X0 = f0 |X = 0, kf k = kf0 k = 1
0

and
f (x0 ) = f0 (x0 ) = dist (x0 , X0 ).
This completes the proof.

An immediate consequence of the above corollary, which is often

used in applications, is the following.

Corollary 3.2.4 Let X0 be a subspace of a normed linear space X.

If there exists a nonzero f ∈ X 0 such that f (x) = 0 for every x ∈ X0 ,
then X0 is not dense in X.

Corollary 3.2.5 Let X be a normed linear space and {u1 , . . . , uk }

be a linearly independent subset of X. Then there exists a linearly
independent set {f1 , . . . , fk } ⊂ X 0 such that

fi (uj ) = δij , i, j = 1, . . . , k.

Proof. Let X0 = span {u1 , . . . , uk }. Then, from linear algebra,

we know that there exist linear functionals gi on X0 such that

gi (uj ) = δij ∀ i, j = 1, . . . , k.

In fact, gi is defined by
k
X 
gi αj uj = αi , αi ∈ K.
j=1
122 Some Important Theorems

Since X0 is finite dimensional, by Theorem 2.1.4, gi ∈ X00 . Hence,

by Theorem 3.2.1, each gi has a norm preserving extension fi to all
of X, so that fi ∈ X 0 and

fi (uj ) = gi (uj ) = δij , i, j = 1, . . . , k.

This completes the proof.

Corollary 3.2.6 Let X be a normed linear space and X0 be a finite

dimensional subspace of X. Then there exists a closed subspace Z of
X such that X = X0 + Z and X0 ∩ Z = {0}.

Proof. Let dim (X0 ) = n and let {u1 , . . . , un } be a basis of X0 .

Let f1 , . . . , fk be as in Corollary 3.2.5. Then every x ∈ X can be
expressed as x = y + z where
n
X n
\
y= fj (x)uj ∈ X0 and z = x − y ∈ Z := N (fj ).
j=1 j=1

Note that Z is a closed subspace of X and X0 ∩ Z = {0}.

Corollary 3.2.7 Let A : X → Y be a finite rank linear operator

between normed linear spaces X and Y . Then A ∈ B(X, Y ) if and
only if there exist y1 , . . . , yn in Y and continuous linear functionals
f1 , . . . , fn on X such that
n
X
Ax = fi (x)yi , x ∈ X.
i=1

Proof. Let A ∈ B(X, Y ) be of finite rank, say rank (A) = k, and

let {y1 , . . . , yk } be a basis of R(A). Then, for every x ∈ X, there
exist scalars α1 (x), . . . , αk (x) such that
k
X
Ax = αj (x)yj . (∗)
j=1

Now, by Corollary 3.2.5, there exist continuous linear functionals

g1 , . . . , gk on R(A) such that gi (yj ) = δij . Thus, from (∗), we have
k
X
gi (Ax) = αj (x)gi (yj ) = αi (x) ∀ x ∈ X.
j=1
Hahn-Banach Extension Theorem 123

Thus, taking fi = gi ◦ A we see that fi ∈ X 0 and

k
X
Ax = fj (x)(x)yj . ∀ x ∈ X.
j=1

The converse part is obvious.

3.2.2 Proof of the theorem

We shall prove a theorem in a slightly general context and derive
Theorem 3.2.1 as a corollary to that. First a definition.
Definition 3.2.1 Let X be a linear space over C.
(i) A linear functional f : X → C is called a complex-linear
functional.
(ii) A function f : X → R is called a real-linear functional if
f is a linear functional considering X as a linear space over R, i.e., if

f (x + y) = f (x) + f (y), f (αx) = αf (x)

for all x, y in X and α ∈ R. ♦

We shall also make use of the following two lemmas.
Lemma 3.2.8 Let X be a linear space over C.
(i) Let f : X → C be a complex-linear functional. Then the
function ϕ : X → R defined by

ϕ(x) = Ref (x), x ∈ X,

is a real-linear functional and f (x) = ϕ(x) − iϕ(ix) for all x ∈ X.

(ii) Let ϕ : X → R be a real-linear functional. Then the function
f : X → C defined by

f (x) = ϕ(x) − iϕ(ix), x ∈ X,

is a complex-linear functional.
Proof. (i) It can be easily seen that

ϕ(x + y) = ϕ(x) + ϕ(y), ϕ(αx) = αϕ(x)

for all x, y in X and α ∈ R. Thus, ϕ is a real-linear functional. Next,

let
ψ(x) = Im f (x), x ∈ X.
124 Some Important Theorems

Then we have f (x) = ϕ(x) + iψ(x) for all x ∈ X. Also, for all x ∈ X,
since f (ix) = if (x), we have
ϕ(ix) + iψ(ix) = −ψ(x) + iϕ(x).
Therefore, ψ(x) = −ϕ(ix) so that
f (x) = ϕ(x) + iψ(x) = ϕ(x) − iϕ(ix) ∀ x ∈ X.
(ii) It can be easily seen that
f (x + y) = f (x) + f (y), f (αx) = αf (x)
for all x, y in X and α ∈ R. Also, for x ∈ X, we have
f (ix) = ϕ(ix) − iϕ(−x)
= ϕ(ix) + iϕ(x)
= i[ϕ(x) − iϕ(ix)]
= if (x).
Hence, for x ∈ X and α, β in R,
f (αx + iβx) = f (αx) + f (iβx) = αf (x) + βf (ix) = αf (x) + iβf (x).
Thus, for x ∈ X and λ ∈ C, we have
f (λx) = λf (x).
This completes the proof.

Lemma 3.2.9 Let X be a linear space over C, p : X → R be a

seminorm and f : X → C be a linear functional. Then
|f (x)| ≤ p(x) ∀ x ∈ X ⇐⇒ |Ref (x)| ≤ p(x) ∀ x ∈ X.
Proof. Clearly,
|f (x)| ≤ p(x) ∀ x ∈ X =⇒ |Ref (x)| ≤ p(x) ∀ x ∈ X.
Conversely, suppose |Ref (x)| ≤ p(x) for all x ∈ X. Now, if x ∈ X,
then |f (x)| = λf (x) for some λ ∈ C with |λ| = 1 and . Thus,
|f (x)| = λf (x) = f (λx)
so that |f (x)| = Ref (λx) and hence
|f (x)| = |Ref (λx)| ≤ p(λx) = |λ|p(x) = p(x).
This completes the proof.
Hahn-Banach Extension Theorem 125

We shall derive Theorem 3.2.1 from the following general version.

Theorem 3.2.10 (Hahn-Banach extension theorem) (HBET)

Let X0 be a subspace of a linear space X and p : X → R be a
seminorm. If g : X0 → K is a linear functional on X0 such that

|g(x)| ≤ p(x) ∀ x ∈ X0

then there exists a linear functional f : X → K on X such that

|f (x)| ≤ p(x) ∀ x ∈ X.

Proof. If X0 = X or g = 0, then we can take f = g. So assume

that X0 6= X and g 6= 0.
First we consider the case of K = R.
Let x0 ∈ X \ X0 . The idea of the proof is that first we extend
g to a linear functional on span {X0 , x0 } satisfying the requirements
and then use that result to extend to all of X. So, let

e0 := span {x0 ; X0 } = {u + αx0 : u ∈ X0 , α ∈ R}.

X

Note that for every u, v ∈ X0 ,

g(u) − g(v) = g(u − v) ≤ p(u − v) ≤ p(u − x0 ) + p(v − x0 )

so that

g(u) − p(u − x0 ) ≤ g(v) + p(v − x0 ) ∀ u, v ∈ X0 .

Hence,

sup {g(u) − p(u − x0 ) : u ∈ X0 } ≤ inf {g(u) + p(u − x0 ) : u ∈ X0 }.

Now, let r ∈ R be such that

sup {g(u) − p(u − x0 ) : u ∈ X0 } ≤ r ≤ inf {g(u) + p(u − x0 ) : u ∈ X0 }.

Then we have

g(u) − p(u − x0 ) ≤ r ∀ u ∈ X0 , r ≤ g(u) + p(u − x0 ) ∀ u ∈ X0 ,

so that
|g(u) − r| ≤ p(u − x0 ) ∀ u ∈ X0 .
126 Some Important Theorems

e0 → R be defined by
Let g̃ : X

g̃(u + αx0 ) = g(u) + αr, u ∈ X0 , α ∈ R.

Then, it can be easily verified that g̃ is a linear functional on X

e0 .
Further, for u ∈ X0 and α 6= 0, we obtain

|g(u) + αr| = |(−α)[g(−u/α) − r]|

≤ |α|p(−u/α − x0 )
= p(u + αx0 ).

Thus,

|g̃(u + αx0 )| ≤ p(u + αx0 ) ∀ u ∈ X0 , α ∈ R.

e0 → R is a linear functional satisfying
Thus, we have proved that g̃ : X

|g̃(x)| ≤ p(x) ∀x ∈ X
e0 .

We shall use the above result, along with Zorn’s lemma, to obtain
a linear extension f : X → R of g such that |f (x)| ≤ p(x) for every
x ∈ X. For this purpose, consider the family S of all pairs (Y, h),
where Y is a subspace of X such that X0 ⊆ Y and h : Y → R is
a linear extension of g such that |h(x)| ≤ p(x) for all x ∈ Y . This
family S is non-empty, since (X e0 , g̃) obtained in the last paragraph
belongs to S. For (Y1 , h1 ), (Y2 , h2 ) in S, define (Y1 , h1 ) 4 (Y2 , h2 )
whenever Y1 ⊆ Y2 and h2 is an extension of h1 . It can be seen that
4 is a partial order on S. Suppose T is a totally ordered subset of
S. Then consider
Z = ∪{Y : (Y, h) ∈ T },
and define φ : Z → R such that φ(x) = h(x) whenever x ∈ Y ,
(Y, h) ∈ T . Then, we see that (Z, φ) ∈ S, and (Z, φ) is an upper
bound of T . Therefore, by Zorn’s lemma, S has a maximal element,
say (Y0 , f ). Now, we show that Y0 = X.
Suppose Y0 6= X, and let y0 ∈ X \ Y0 . Then, by the first part
of the proof, f has a linear extension, say f˜ to Ye0 := span {y0 ; Y0 }
satisfying |f˜(x)| ≤ p(x) for all x ∈ Ye0 . Thus, we have

(Y0 , f ) 4 (Ye0 , f˜) ∈ S, (Y0 , f ) 6= (Ye0 , f˜)

contradicting the maximality of (Y0 , f ). Therefore, Y0 = X, and f is

a linear extension of g satisfying |f (x)| ≤ p(x) for all x ∈ X.
Hahn-Banach Extension Theorem 127

Now we take up the case of K = C. Let g0 : X0 → R be defined

by
g0 (x) = Re g(x), x ∈ X0 .
Then by Lemma 3.2.8 and Lemma 3.2.9, g0 is a real-linear functional
on X0 satisfying |g0 (x)| ≤ ν(x) for all x ∈ X0 . Therefore, by the
earlier part, g0 has a real-linear extension f0 : X → R satisfying
|f0 (x)| ≤ ν(x) for all x ∈ X. Again, by Lemma 3.2.8 and Lemma
3.2.9, the function f : X → C defined by

f (x) = f0 (x) − if0 (ix), x ∈ X,

is a complex-linear functional satisfying |f (x)| ≤ ν(x) for all x ∈ X.

This f is an extension of g since, for every x ∈ X0 ,

f (x) = f0 (x) − if0 (ix) = g0 (x) − ig0 (ix) = g(x).

This completes the proof of the theorem.

Remark 3.2.1 A natural question that may come into mind is

whether there is an analogue for Hahn-Banach extension theorem for
general operators. In general, the answer is not in affirmative. To
see this suppose X is an infinite dimensional normed linear space and
X0 is a non-closed subspace of X. Let A0 : X0 → X0 be the identity
operator on X0 , i.e.,

A0 x = x ∀ x ∈ X0 .

Clearly, A0 ∈ B(X0 ) with kA0 k = 1. However, there is no norm

preserving extension for A0 to all of X. To see this, suppose there is
a norm preserving extension A : X → X0 . Then taking I0 : X0 → X
as the inclusion operator, we see that

P := I0 A

is a projection operator in B(X) with kP k = 1 and R(P ) = X0 .

This forces X0 to be a closed subspace, which is a contradiction to
the assumption on X0 . ♦

3.2.3 Further consequences

We have seen that every normed linear space is linearly isometric
with a dense subspace of a Banach space (cf. Theorem 1.3.16). This
fact is also a consequence of Hahn-Banach extension theorem, as the
following theorem shows.
128 Some Important Theorems

Theorem 3.2.11 Let X be a normed linear space. For each x ∈ X,

let x̂ : X 0 → K be defined by

x̂(f ) = f (x), f ∈ X 0.

Then x̂ ∈ X 00 for every x ∈ X, and the function J : X → X 00 defined

by
J(x) = x̂, x ∈ X,
is a linear isometry. In particular, closure of {x̂ : x ∈ X} is a
completion of X.
Proof. Let x ∈ X. Then we observe that

x̂(f + αg) = (f + αg)(x) = f (x) + αg(x) = x̂(f ) + αx̂(g)

for all f, g ∈ X and for all α ∈ K. Further,

|x̂(f )| = |f (x)| ≤ kf k kxk ∀ f ∈ X 0.

Hence, x̂ ∈ X 00 for each x ∈ X and kx̂k ≤ kxk. Clearly, if x = 0,

then x̂ = 0. If x 6= 0, then by Corollary 3.2.2, there exists fx ∈ X 0
such that
fx (x) = kxk and kfx k = 1.
Thus,
kxk = |fx (x)| = |x̂(fx )| ≤ kx̂k kfx k = kx̂k.
Thus, we have proved that

kx̂k = kxk ∀ x ∈ X.

It remains to show that J : x 7→ x̂ is a linear operator. For this, let

x, y ∈ X and α ∈ K. Then, for every f ∈ X 0 , we have

ϕx+αt (f ) = f (x + αy) = f (x) + αf (y)

= x̂(f ) + αϕy (f ) = (x̂ + αϕy )(f ).

The last part follows, because, X 00 is a Banach space. This competes

the proof.

Definition 3.2.2 Let X be a normed linear space.

1. The linear isometry J : X → X 00 obtained in Theorem 3.2.11
is called the canonical isometry from X to X 00 .
Hahn-Banach Extension Theorem 129

2. The space X is said to be a reflexive space if the canonical

isometry J : X → X 00 is surjective.

♦
Clearly, a reflexive space has to be a Banach space. It is known
(cf. Nair [5]) that the spaces

• `p and Lp [a, b] for 1 < p < ∞, and Hilbert spaces are reflexive
spaces,

whereas the spaces

• `1 , `∞ , L1 [a, b], L∞ [a, b] and C[a, b] (with k · k∞ ) are not re-

flexive spaces.

Theorem 3.2.12 Let X be a non-zero normed linear space and let

Ω := {f ∈ X 0 : kf k = 1}.

For each x ∈ X, let ϕx : Ω → K be defined by

ϕx (f ) = f (x), f ∈ Ω.

Then ϕx ∈ Cb (Ω) for every x ∈ X, and the function T : X → Cb (Ω)

defined by
T (x) = ϕx , x ∈ X,

is a linear isometry. In particular,

b := cl {T (x) : x ∈ X}
X

is a closed subspace of Cb (Ω) and it is a completion of X.

Proof. Use the arguments as in Theorem 3.2.11.

Note that, for proving the last part of the Theorem 3.2.11, we
used the fact that dual of a normed linear space is a Banach space
(cf. Theorem 2.1.3). Now, we prove the converse of this statement.

Theorem 3.2.13 Let X and Y be normed linear spaces with X 6=

{0} and let B(X, Y ) be a Banach space. Then Y is a Banach space.
130 Some Important Theorems

Proof. Let (yn ) be a Cauchy sequence in Y . We have to show

that (yn ) converges to some element in Y . Since X 6= {0}, there
exists x0 ∈ X such that kx0 k = 1. By Corollary 3.2.2, there exists
f0 ∈ X 0 be such that

f0 (x0 ) = kx0 k = 1 and kf0 k = 1.

For each n ∈ N, let An : X → Y be defined by

An x = f0 (x)yn , x ∈ X.

Clearly, An is a linear operator for every n ∈ N. Further, for every

x ∈ X and n ∈ N, we have

kAn xk = kf0 (x)yn k = |f0 (x)| kyn k ≤ kyn kkxk

so that An ∈ B(X, Y ) and kAn k ≤ kyn k for all n ∈ N. Also, we have,

for every x ∈ X and n ∈ N,

k(An − Am )xk = kAn x − Am xk = kf0 (x)(yn − ym )k ≤ kyn − ym kkxk

so that kAn − Am k ≤ kyn − ym k for all n ∈ N. Consequently, (An )

is a Cauchy sequence in B(X, Y ). Since B(X, Y ) is Banach space,
there exists A ∈ B(X, Y ) such that kAn − Ak → 0 as n → ∞. Thus,
in particular, taking y0 := Ax0 , we have

kyn − y0 k = kAn x0 − Ax0 k → 0 as n → ∞.

This competes the proof.

3.2.4 Problems
1. Let X and Y be inner product spaces and A : X → Y be a
linear operator. Prove that, if A∗ exists then A∗ is a closed
operator.

2. Every self adjoint operator on an inner product space is a closed

operator. Why?

3. Let X be an inner product space, Y be Hilbert space and X0 be

a dense subspace of X. Let A : X0 → Y be a linear operator.
Prove that there exists a subspace Y0 of Y and a closed operator
B : Y0 → X such that

hAx, yi = hx, Byi ∀ x ∈ X0 , y ∈ Y0 .

Hahn-Banach Extension Theorem 131

4. Every linear functional on a normed linear space which is a

closed operator is continuous. Why?

5. Is every continuous linear functional on a normed linear space a

closed operator? Why?

6. Let X be a Banach space, Y be a normed linear space, X0 be

a subspace of X and A : X0 → Y be a closed operator. Prove
that if A is bounded below, then R(A) is a closed subspace.

7. Let X be a Banach space, X0 be a subspace of X and A : X0 →

Y be a linear operator which is bounded below. Prove that A
is a closed operator if and only if R(A) is closed.

8. Let X = L2 [0, 1] = Y and Ax = x0 , x ∈ X0 , where X0 is linear

space of all x ∈ L2 [0, 1] such that x is absolutely continuous
with x(0) = 0 and x0 ∈ L2 [0, 1]}. Prove that A : X0 → L2 [a, b]
is a bijective, closed operator.

9. Let X be a Banach space, Y be a normed linear space, X0 be

a subspace of X and A : X0 → Y be a closed operator. Prove
that the following are equivalent:

(a) A is bijective and A−1 is continuous,

(b) A is bounded below and R(A) is dense.

10. Let X be a Banach space and P ∈ B(X) be a projection oper-

ator. Prove that P ∈ K(X) if and only if rank (P ) < ∞.

11. Let X be a Banach space and P : X → X be a projection

operator. Prove that P ∈ B(X) if and only if R(P ) and N (P )
are closed subspaces of X.

12. Let X and Y be a Banach spaces, and (An ) be a sequence

of operators in B(X, Y ) which converges pointwise on a dense
subset of {x ∈ X : kxk ≤ 1}. Prove that, if S is a relatively
bounded subset of X, then

sup{kAn x − Axk : x ∈ S} → 0 as n → ∞,

that is, (An ) converges to A uniformly on S.

132 Some Important Theorems

13. Let X be a Banach space and (Pn ) in B(X) be a sequence

of finite rank projection operators which converges pointwise.
Prove that P : X → X defined by P x = lim Pn x is a projec-
n→∞
tion operator and P ∈ B(X).

14. Let X be a normed linear space and (xn ) be a sequence in X

such that, there is an x ∈ X satisfying f (xn ) → f (x) for every
f ∈ X. Prove that (xn ) is a bounded sequence. Is it necessary
that (xn ) converges to x? Why?

15. Let X be a normed linear space. Prove that if x, y ∈ X with

x 6= y, then there exists f ∈ X 0 such that f (x) 6= f (y).

16. Prove Corollary 3.2.4.

17. Let X be a normed linear space, and for a subset S of X, let

S a := {f ∈ X 0 : f (x) = 0 ∀ x ∈ S}.

Prove the following:

(a) S a is a closed subspace of X.

(b) If S is a subspace which is not dense in X, then S a 6= {0}.

18. Let X be a normed linear space and X0 be a finite dimensional

subspace of X. Prove that there exists a projection operator
P ∈ B(X) such that R(P ) = X0 .

19. Give details of the proof of Theorem 3.2.12.