Basic Analysis – Gently Done Topological Vector Spaces

118 Basic Analysis
Theorem 10.18 (Banach-Steinhaus theorem) Suppose that E is a collection
of continuous linear mappings from a Fréchet space X into a topological vector
space Y such that for each x ∈ X the set
B(x) = {Tx : T ∈ E}
is bounded in Y. Then E is equicontinuous.
Proof Let W be a neighbourhood of 0 in Y, which, without loss of generality,
we may assume to be balanced, and let U be a balanced neighbourhood of 0 such
that U +U +U +U ⊆ W. Since U ⊆ U +U, we have that U +U ⊆ W. Let
A = �
T −1 (U).
T∈E
For any x ∈ X, B(x) is bounded, by hypothesis, and so there is some n ∈ N such
that B(x) ⊆ nU. It follows that T((1/n)x) ∈ U for every T ∈ E which implies
that (1/n)x ∈ A, i.e., x ∈ nA. Hence
X = �
nA.
n∈N
Now, every T in E is continuous and so A is closed and so, therefore, is each nA,
n ∈ N. By the Baire Category theorem, Theorem 10.11, some nA has an interior
point; there is some m ∈ N, a ∈ A and some open set G with ma ∈ G ⊆ mA. We
see that a ∈ 1
1
mG ⊆ A so that writing V for mG−a, V is a neighbourhood of 0
such that V ⊆ A−a. It follows that
T(V) ⊆ T(A)−Ta ⊆ U ⊆ W
for every T ∈ E, which shows that E is equicontinuous.
Theorem 10.19 Let (T n ) n∈N be a sequence of continuous linear mappings from
a Fréchet space X into a topological vector space Y such that
Tx = lim
n→∞ T n x
exists for each x ∈ X. Then T is a continuous linear map from X into Y.
Proof It is clear that T : x ↦→ Tx = lim n T n x defines a linear map from X into
Y. We must show that T is continuous. Let W be any neighbourhood of 0 in Y
and let V be a neighbourhood of 0 such that V +V ⊆ W. For each x ∈ X, the
sequence (T n x) converges, so it is bounded. By the Banach Steinhaus theorem,
Theorem 10.18, there is a neighbourhood U of 0 in X such that T n (U) ⊆ V for all
n. Now T n u → Tu for each u ∈ U, so it follows that Tu ∈ V for u ∈ U. Hence
and we conclude that T is continuous.
T(U) ⊆ V ⊆ V +V ⊆ W
Department of Mathematics King’s College, London