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

-

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

Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

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