if $F$ is a Banach space and $f\_n \subset F^\* $ weak star convergent to $f\in F^\*$. If further $x\in F$ is the weak limit of $(x\_n)\_n \subset F$ does then $f\_n(x\_n) \longrightarrow f(x)$ hold?


We know that for all $n$: $\lim\_m f\_n(x\_m) = f\_n(x)$ and for all $m$: $\lim\_n f\_n(x\_m) = f(x\_m)$ so what can I infer about $\lim\_n\lim\_m f\_n(x\_m)$? I thought as the limit of $f\_n$ is again in $F^\*$ i could just put $\lim\_n(\lim\_m f\_n(x\_m)) = \lim\_n f\_n(x)$?!



weak star convergence 
The answer is no. Consider the simple case of an infinite dimensional Hilbert space with a sequence $(x\_n)\_n$ of the unit sphere weakly converging to zero.


