# Exercise: If $\{f(x_n)\}$ is Cauchy $\forall f \in X^\ast$ then $\exists x \in X : x_n \rightarrow x$ weakly

I'm working on this exercise (not homework) and I would gladly welcome some hints for how to solve it!

Excercise: $X$ is a reflexive Banach space and $\{x_n \} \in X$.

Prove that if $\{f(x_n)\}$ is Cauchy $\forall f \in X^\ast$ then $\exists x \in X : x_n \rightarrow x$ weakly (i.e. $f(x_n) \rightarrow f(x), \forall f \in X^\ast$)

My Idea: We have this mapping $T:X \rightarrow X^{\ast \ast}$ defined by $x \mapsto \hat{x}$ where $\hat{x}(f) = f(x)$ for $f \in X^\ast$. We know that it is isometric and $X$ is reflexive means nothing other than $T$ is surjective. So we can identify $X$ with $X^{\ast \ast}$.

Since $f(x_n) = \widehat{x_n}(f)$ we have that $\{\widehat{x_n}(f)\}$ is a Cauchy sequence for every $f \in X^\ast$. I realize we need to somehow move over to $X$ and utilize the completeness, but I don't know how to do this formally.

Any help is appreciated!

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Define $L(f):=\lim_{n\to +\infty}f(x_n)$: it's a linear functional on $X^*$. Show that this one is bounded using Baire categories theorem with the closed sets $$F_n:=\bigcap_{k\geqslant 1}\{f\in X^*,|f(x_k)|\leqslant n\}.$$
Since $\{\hat{x}_n(f)\}$ is Cauchy for fixed $f\in X^*$ you can show that there exist a limit $\hat{x}(f)$. You need to work a little to show that $\hat{x}\in X^{**}$. Then $T^{-1}(\hat{x})$ is the desired limit of $\{x_n\}$.