# $X$ Banach, $u_n \to u$ and $x^*_n \xrightarrow{w^*} x^*$ implie that $\langle x^*_n, u_n\rangle \to \langle x^*,u\rangle$.

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.