Let $X$ be Banach and $Y$ be a reflexive space with $T_n \in \mathscr{L}(X,Y)$ so that $\Lambda (T_nx) $ converges for all $x \in X$ and $\Lambda \in Y^{*}$.
Then there exists a $T \in \mathscr{L}(X,Y)$ so that $\Lambda (Tx) = \lim_{n \rightarrow \infty} \Lambda(T_nx)$ for all $x \in X$ and $\Lambda \in Y^{*} $.
I have already proven that for a Banach space $X$ and normed $Y$ the pointwise limit of a linear continous operator is again linear and continous but I am just not getting forward here. Thanks in advance.
For each $x$ in $X$, consider the linear map $$ \varphi _x:\Lambda \in Y^*\mapsto \lim_{n\to \infty }\Lambda \big (T_n(x)\big ) $$ By the Banach-Steinhaus theorem we have that $\varphi _x$ is continuous, so $\varphi _x\in Y^{**}$. Since $Y$ is reflexive there must be some vector $T(x)$ in $Y$ such that $$ \varphi _x(\Lambda ) = \Lambda \big (T(x)\big ),\quad \forall \Lambda \in Y^*. $$ It then follows that $$ \Lambda \big (T(x)\big ) = \varphi _x(\Lambda ) = \lim_{n\to \infty }\Lambda \big (T_n(x)\big ). $$ Can you pick up from here and prove that $T$ is bounded?