Solution [pdf]

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� < ∞. �