Basic Analysis – Gently Done Topological Vector Spaces

10: Fréchet Spaces 117
If a family E of linear maps is equicontinuous, then certainly each member of
E is continuous at 0. However, by linearity of the mappings, this is equivalent to
each mapping being continuous. In particular, if E consists of just a single member
T, then equicontinuity of E = {T} is equivalent to the continuity of T. The point
of the definition is that the neighbourhood V should not depend on any particular
member of E.
In normed spaces, equicontinuous families are the uniformly bounded families
as the next result indicates.
Proposition 10.16 A family E of linear maps from a normed space X into a
normed space Y is equicontinuous if and only if there is C > 0 such that
for x ∈ X and all T ∈ E.
�Tx� ≤ C�x�
Proof Suppose that E is an equicontinuous family of linear maps from the normed
space X into the normed space Y. Then, by definition, there is a neighbourhood
of 0 in X such that T(U) ⊆ {y : �y� < 1} for all T ∈ E. But there is some r > 0
such that {x : �x� < r} ⊆ U, and so �Tx� < 1 whenever �x� < r, T ∈ E. Putting
C = 2/r, it follows that �Tx� ≤ C�x� for all x ∈ X and for all T ∈ E.
Conversely, suppose that there is C > 0 such that �Tx� ≤ C�x�, for all x ∈ X
and T ∈ E. Let W be any neighbourhood of 0 in Y. There is ε > 0 such that
{y : �y�ε} ⊆ W and so �Tx� < ε for all T ∈ E whenever �x� < ε/C. That is, if we
set V = {x : �x� < ε/C}, then T(V) ⊆ W for all T ∈ E. Thus, E is equicontinuous
as claimed.
Proposition 10.17 Suppose that Eisan equicontinuous familyoflinear maps from
a topological vector space X into a topological vector space Y. For any bounded
subset E in X there is a bounded set F in Y such that T(E) ⊆ F for every T ∈ E.
Proof Suppose that E is a bounded set in X. Then there is some s > 0 such that
E ⊂ tV for all t > s. Let F = �
T∈E T(E). For t > s, we have
T(E) ⊆ T(tV) = tT(V) ⊆ tW
so that F ⊆ tW and therefore F is bounded.