Solution [pdf]

HOMEWORK 9 SOLUTION
MATH 5051, FALL 2012
Exercise 1 (Folland, 5.31). Let X, Y be Banach spaces and let S : X → Y be
an unbounded linear map. Let Γ(S) be the graph of S, a subspace of X × Y.
a. Γ(S) is not complete.
Proof. Since S is unbounded, the Closed Graph Theorem implies
that Γ(S) is not closed. Hence, there exists a Cauchy sequence
(xn, Sxn) → (x, y) /∈ Γ(S), so Γ(S) is not complete. �
b. Define T : X → Γ(S) by T x = (x, Sx). Then T is closed but not
bounded.
Proof. Suppose xn → x and T xn = (xn, Sxn) → (x ′ , Sx ′ ) ∈ Γ(S).
Since �xn − x ′ � ≤ � � (xn, Sxn) − (x ′ , Sx ′ ) � � → 0, it follows that xn →
x ′ = x. Thus, (x ′ , Sx ′ ) = (x, Sx) = T x, so T is closed.
On the other hand, notice that S = π2 ◦ T . Since π2 is bounded
and S is unbounded, T must also be unbounded. �
c. T −1 : Γ(S) → X is bounded and surjective but not open.
Proof. Clearly T −1 = π1 is bounded, since �x� ≤ � � (x, Sx) � �, and
surjective since x = T −1 (T x) for all x ∈ X. However, if T −1 were
open, that would imply that T was continuous and hence bounded,
which is not the case. �
Exercise 2 (Folland, 5.32). Let �·�1 and �·�2 be norms on the vector space
X such that �·�1 ≤ �·�2. If X is complete with respect to both norms, then
the norms are equivalent.
Proof. Since �x�1 ≤ �x�2 for all x ∈ X, the identity mapping x ↦→ x is a
bounded bijection � � � �
X, �·�2 → X, �·�1 . Hence, by the Bounded Inverse
Theorem (Corollary 5.11), its inverse is also bounded, i.e., there exists C > 0
such that �x�2 ≤ C�x�1 for all x ∈ X. �
Exercise 3 (Folland, Exercise 5.37). Let X and Y be Banach spaces. If
T : X → Y is a linear map such that f ◦ T ∈ X ∗ for every f ∈ Y ∗ , then T is
bounded.
Proof. By the Closed Graph Theorem, it suffices to show that T is closed.
Suppose that xn → x ∈ X and T xn → y ∈ Y. Then for all f ∈ Y ∗ , it follows
that f(T xn) → f(y), and since f ◦ T ∈ X ∗ , we also have f(T xn) → f(T x).
Hence, f(y) = f(T x) for all f ∈ Y ∗ , but since bounded linear functionals
separate points, it follows that y = T x, and therefore T is closed.
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2 MATH 5051, FALL 2012
Alternatively, we could have applied the Uniform Boundedness Principle.
Since f ◦ T ∈ X ∗ for all f ∈ Y ∗ , we have
and therefore
sup �
�x�=1
� T x(f)� = sup �f(T x)� < ∞,
�x�=1
sup �T x� = sup �
�x�=1 �x�=1
� T x� < ∞,
so T is bounded. �
Exercise 4 (Folland, Exercise 5.44). If X is a first countable topological
vector space and every Cauchy sequence in X converges, then every Cauchy
net in X converges.
Proof. Let 〈xi〉i∈I be a Cauchy net, i.e., 〈xi − xj〉 (i,j)∈I×I → 0. Since X is
first countable, 0 has a countable neighborhood base {Un} ∞ n=1 , and we can
assume that U1 ⊃ U2 ⊃ · · · . Now, since 〈xi − xj〉 (i,j)∈I×I → 0, it follows
that there exists some (i1, j1) ∈ I × I such that (xi − xj) ∈ U1 whenever
i � i1 and j � j1. In particular, we can take k1 ∈ I such that k1 � i1, j1,
so (xi − xj) ∈ U1 whenever i, j � k1. We can repeat this for any n ∈ N, so
there exists kn ∈ I such that (xi − xj) ∈ Un whenever i, j � kn. Hence, the
sequence {xkn }∞n=1 is Cauchy, so xkn → x ∈ X, and it follows immediately
that xi → x as well. �
Exercise 5 (Folland, Exercise 5.46). If X is a vector space, Y a normed
linear space, T the weak topology on X generated by a family of linear maps
{Tα : X → Y}, and T ′ the topology defined by the seminorms � x ↦→ �Tαx� � ,
then T = T ′ .
Proof. The topology T has base elements � x : �Tαx − y0� < ɛ � , while T ′ has
base � x : �Tαx − Tαx0� < ɛ � . Clearly T ′ ⊂ T, so it suffices to show that the
reverse inclusion holds. If �Tαx0 − y0� < ɛ, then let δ = ɛ − �Tαx0 − y0�,
and it follows that �Tαx − Tαx0� < δ implies
�Tαx − y0� ≤ �Tαx − Tαx0� + �Tαx0 − y0� < ɛ.
Hence, � x ∈ X : �Tαx − Tαx0� < δ � ⊂ � x ∈ X : �Tαx − y0� < ɛ � . �
Exercise 6 (Folland, Exercise 5.47). Suppose that X and Y are Banach
spaces.
a. If {Tn} ∞ n=1 ⊂ L(X, Y) and Tn → T weakly (or strongly), then
sup n�Tn� < ∞.
Proof. By definition, Tn → T weakly means that f(Tnx) → f(T x)
for all x ∈ X, f ∈ Y∗ . Therefore, the sequence f(Tnx) = � �
Tnx(f) is
bounded, so sup ��
n Tnx(f) � � < ∞, and by the uniform boundedness
principle, supn� � Tnx� = supn�Tnx� < ∞. Finally, applying the
uniform boundedness principle again, we get supn�Tn� < ∞. �

HOMEWORK 9 SOLUTION 3
b. Every weakly convergent sequence in X, and every weak*-convergent
sequence in X∗ , is bounded (with respect to the norm).
Proof. If xn → x weakly in X, then f(xn) → f(x) for all f ∈ X∗ � , so
the sequence f(xn) = �xn(f) is bounded. That is, sup ��xn(f) n
� � < ∞
for all f ∈ X ∗ , so by the uniform boundedness principle, sup n��xn� =
sup n�xn� < ∞.
Similarly, if fn → f in X ∗ with the weak*-topology, then fn(x) →
f(x) �for
all x ∈ X, so the sequence fn(x) is bounded. Therefore,
sup �fn(x) n
� � < ∞ for all x ∈ X, and the uniform boundedness principle
implies supn�fn� < ∞. �