Basic Analysis – Gently Done Topological Vector Spaces

10: Fréchet Spaces 121
Since v n ∈ V n , we have that d(v n ,0) < 2 −n r and so v n → 0. By the continuity of
T (which has not been invoked so far), it follows that T(v n ) → 0, and therefore
we see that y n → 0. Furthermore, s n → x implies that T(s n ) → T(x) so that
y 1 = T(x) ∈ T(V) and we conclude that T(V 1 ) ⊆ T(V).
Combining this with the first part, we see that there is some neighbourhood
W of 0 in Y such that W ⊆ T(V 1 ) ⊆ T(V). Thus 0 is an interior point of T(V),
and the proof is complete.
Corollary 10.22 (Inverse mapping theorem) Let X and Y be Fréchet spaces
and suppose that T : X → Y is a one-one continuous linear mapping from X onto
Y. Then T −1 : Y → X is continuous. In particular, for any given continuous
seminorm p on X, there is a continuous seminorm q on Y such that
p(x) ≤ q(T(x)) for all x ∈ X.
Proof The inverse T −1 exists because T is a bijection, by hypothesis. Write S
for T −1 , so that S : Y → X. Let G be any set in X. Then
S −1 (G) = {y ∈ Y : S(y) ∈ G} = {y ∈ Y : T −1 (y) ∈ G}
= {y ∈ Y : y = T(x) for some x ∈ G} = T(G).
By the theorem, T is open, so that if G is open so is T(G). Therefore T −1 is
continuous.
Now suppose that p is a continuous seminorm on X. Then y ↦→ p(S(y)) is a
continuous seminorm on Y. It follows that there is some C > 0 and seminorms
q 1 ,...,q m on Y (belonging to any separating familyof seminorms which determine
the topology on Y) such that
p(S(y)) ≤ C(q 1 (y)+···+q m (y)), y ∈ Y.
Setting q(y) = C(q 1 (y) + ···+q m (y)), for y ∈ Y, we see that q is a continuous
seminorm on Y and, replacing y by T(x), we get
for all x ∈ X, as required.
p(x) ≤ q(T(x))