weak star convergence [closed]

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 $\lim_n(\lim_m f_n(x_m)) = \lim_n f_n(x)$?!

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closed as off topic by Nate Eldredge, Emil Jeřábek, Andreas Blass, Willie Wong, Bill JohnsonJun 6 '13 at 16:41

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This site is for research-level questions, which this question is not. It would be fine at math.stackexchange.com however. –  Nate Eldredge Jun 6 '13 at 12:59

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.
So I could just use $x^n = (0,...,0,1,1,...)$ converging weakely to $0$, and $f_m(y) = sum_{i=1}^m y_i$ converging to $f(y) = sum_{i=1}^\infty y_i$ then $\lim_n f_n(x^n) = 1$. thx –  Bohem Jun 6 '13 at 12:31
It's enough if $X_n \to x$ in norm. Proof: by the uniform boundedness principle $\|f_n\|$ is bounded. Now write $|f_n(x_n) - f(x)| \le \|f_n\| \|x_n - x\| + |f_n(x) - f(x)|$. –  Nate Eldredge Jun 6 '13 at 13:00