Consider $X$ a Banach space. For some sequence $x_n\in X$, assume that for every $f\in X^*$, $f(x_n)\to c_f$. Does this imply that $\exists x\in X$ where $c_f = f(x)$? How might I go about finding such an $x$? (It's clearly not the limit of $x_n$ since no guarantees on strong convergence.)
Thank you!
In general, the limit $f \mapsto c_f$ will only be an element of the bidual $X^{\ast\ast}$.
If you additionally assume that $X$ is reflexive (i.e. $X=X^{\ast\ast}$, which is to be understood as the map $X \to X^{\ast\ast}, f \mapsto (g \mapsto g(f))$ being surjective), then you can conclude that the limit lies in $X$.
EDIT: As an example that in general the limit is not given by an element of $X$, consider the case $X = c_0 (\Bbb{N})$ (space of null sequences) and the sequence $x_n = (1,\dots,1,0,\dots)$, where the first $n$ entries are $1$.
Then $X^\ast = \ell^1 (\Bbb{N})$ (with the usual identifications) and it is easy to see that $f(x_n) = \sum_{m=1}^n f_m \to \sum_{m=1}^\infty f_m$ for all $f = (f_n)_n \in \ell^1$.
The "limit" is here given by the sequence $(1,\dots)$, which is not in $c_0$ (but in $X^{\ast\ast} = \ell^\infty$). You can also try to show directly that $\ell^1 \to \Bbb{C}, f \mapsto \sum_{m=1}^\infty f_m$ is not given by $f \mapsto f(x)$ for some $x \in c_0$.