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Exercise :

Let $X$ be a Banach space, $u_n to u$ and $x^*_n xrightarroww^* x^*$. Show that $langle x^*_n, u_nrangle to langle x^*,urangle$.

Attempt-Discussion :

I know that a sequence $x_n^* in X^*$ converges to $x^*$ provided that $x_n^*(u) to x^*(u)$ for all $u in X$. Knowing that the $w^*-$topology coincides the topology of pointwise convergence of linear functionals, this implies that $x_n^* xrightarroww^* x^*$.

Now, I know that the duality brackets given are essentialy the expression of the linear functional, aka : $u mapsto langle x^*, u rangle := x^*(u)$. So essentially, since we have haev that the pointwise convergence is coinciding, we essentialy need to prove that $x^*_n(u_n) to x^*(u)$ and I guess since we have the functional convergence, it boils down to $x^*(u_n) to x^*(u)$ ?

How would one work further to prove the convergence asked ?

Alternatively, we would be interested in showing that $|langle x^*_n, u_nrangle - langle x^*,urangle| < varepsilon$ but I think I am missing something in re-writting the expression by breaking it up to form an arbitrarily small inequality.

Any hints will be greatly appreciated.

Note that by the Banach-Steinhaus theorem $sup_n|x_n^*|<infty$. We find

$$lim_nrightarrowinfty|langle x_n^*,u_nrangle-langle x^*,urangle|=lim_nrightarrowinfty|langle x_n^*,u_n-urangle+langle x_n^*,urangle-langle x^*,urangle|$$
$$leqlim_nrightarrowinfty|langle x_n^*,u_n-urangle|+|langle x_n^*-x^*,urangle|leq lim_nrightarrowinfty|x_n^*||u_n-u|+lim_nrightarrowinfty|langle x_n^*-x^*,urangle|=0$$
as $x_n^*stackrelw^*rightarrowx^*$, $u_nrightarrow u$ and $(|x_n^*|)$ is bounded.