Weak convergence in Banach space

Question

If $X$ is a Banach space, $X'$ its dual and $x,x_n\in X$ and $x',x_n'\in X'$, then the following implication holds:

$x_n\to_w x$ (weak convergence) and $x_n'\to x'$ in $X'$ $\implies$ $x_n'(x_n)\to x'(x)$

I tried the following:

$$\left|x_n'(x_n)-x'(x)\right|\leq \left|x_n'(x)-x'(x)+x_n'(x_n-x) \right|\leq\|x_n'-x\|\|x\|+\left|x_n'(x_n-x)\right|$$

The first term tends to zero since $x_n'\to x'$. I'm not so sure about the second term. Does the weak convergence $x_n\to_w x$ imply that this term tends to zero as well?

How to proceed?

Answer 1

Note that $$ |x_n'(x_n)-x'(x)|\le|x_n'(x_n)-x'(x_n)+x'(x_n-x)|\le\|x_n'-x'\|\cdot\|x_n\|+|x'(x_n-x)| $$ and the sequence $\|x_n\|$ is bounded because it converges.

Answer 2

Of course this is quite simple, as Jonas showed. Here's a fun way to look at it.

Let $K=\{x_n'\}\cup\{x'\}$. Then $K$ is a compact metric space. Regard $x_n$ and $x$ as functions from $K$ to $\Bbb C$. Uniform Boundedness shows that $||x_n||$ is bounded, and this shows that our family of functions from $K$ to $\Bbb C$ is equicontinuous. And $x_n\to x$ pointwise on $K$; so Arzela-Ascoli shows that $x_n\to x$ uniformly on $K$.

(Don't think that's fun? Sorry, seems like fun to me...)