A sort of converse of Banach-Steinhaus theorem.

Question

$(X, \|•\|) $ and $(Y, \|•\|') $ be two normed space.

$\begin{align} {\scr{B}}{(X, Y) }&=\{T\in {\scr{L}}{(X,Y)}: T \text{ is bounded } \}\end{align}$

$\|T\|_{op}=\sup\{\|Tx\|':\|x\|\le 1 \}$

Question: $\forall (T_n) \subset {\scr{B}}{(X, Y)}$ be such that $T_n\to T $ pointwise $[$ i.e $\forall x\in x,$$ T_nx\to Tx $ in the space $(Y, \|•\|') ]$ implies $T\in {\scr{B}}{(X, Y)}$.

Does this implies $(X, \|•\|) $ is a Banach space?

The converse is well known ( Banach- Steinhaus theorem) . But i think the above question can be answered negatively , I mean there is some counter examples but neither I can prove it not cite a counter example.

I sincerely need help. Thanks.

Answer 1

The answer is no. The Banach-Steinhauss Theorem is a consequence of the uniform boundedness principle which in turn holds whenever $X$ is a Baire space. Non-complete normed spaces which are Baire are therefore counter-examples to your question.