Exercise :
Let $X$ be a Banach space, $u_n \to u$ and $x^*_n \xrightarrow{w^*} x^*$. Show that $\langle x^*_n, u_n\rangle \to \langle x^*,u\rangle$.
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 $x_n^* \xrightarrow{w^*} x^*$.
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 $|\langle x^*_n, u_n\rangle - \langle x^*,u\rangle| < \varepsilon$ 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
$$\lim_{n\rightarrow\infty}|\langle x_{n}^{*},u_{n}\rangle-\langle x^{*},u\rangle|=\lim_{n\rightarrow\infty}|\langle x_{n}^{*},u_{n}-u\rangle+\langle x_{n}^{*},u\rangle-\langle x^{*},u\rangle|$$ $$\leq\lim_{n\rightarrow\infty}|\langle x_{n}^{*},u_{n}-u\rangle|+|\langle x_{n}^{*}-x^{*},u\rangle|\leq \lim_{n\rightarrow\infty}\|x_{n}^{*}\|\|u_{n}-u\|+\lim_{n\rightarrow\infty}|\langle x_{n}^{*}-x^{*},u\rangle|=0$$ as $x_{n}^{*}\stackrel{w^{*}}{\rightarrow}x^{*}$, $u_{n}\rightarrow u$ and $(\|x_{n}^{*}\|)$ is bounded.