# convergence for the weak-* topology

Let E be a Banach space.

Let $$(x^∗_n )$$ be a sequence in $$E^∗$$ verifying $$()$$ converges for any $$x ∈ E$$.

Prove that $$\exists x^∗ ∈ E^∗: (x^∗_n )$$ converges vers $$x^*$$ for the weak-∗ topology.

The solution I have states that it is a corollary of the Banach Steinhaus theorem but I don't see how it is related and I am not aware of such a corollary.

Many thanks for your help.

• What do u mean $<x_n^*,x>$ ? – Ignorant Mathematician Apr 14 at 14:36
• $E^*$ is the topological dual of $E$, so by $<x_n^*,x>$ I mean $x_n(x)$ as $x_n^*\in E^*$ – PerelMan Apr 14 at 14:40
• You know of the Uniform Boundedness Principle ? – Ignorant Mathematician Apr 14 at 14:41
• Yes it is another name of Banach Steinhaus theorem, right? – PerelMan Apr 14 at 14:41

Your condition says $$x_n^*(x)$$ is convergent and in particular bounded for any $$x \in E$$
So the uniform Boundedness Principle gives you $$sup_n ||x_n^*|| <\infty$$ .
Now if you define $$x^* :E \rightarrow \mathbb K$$ $$x\mapsto lim_n \ x_n^*(x)$$ then this map is bounded since $$||x^*(x) || \leq sup_n ||x_n^*||$$
Also $$x_n^* \rightarrow x^*$$ as $$n\rightarrow \infty$$ by definition of $$weak^*$$ topology.
• Thank you! I just don't see why the map is bounded? Could you please explain why $||x^*(x) || \leq sup_n ||x_n^*||$? I know that $||x^*(x) || \le sup_n ||x_n^*(x)||$ but I don't see how you get $sup_n ||x_n^*(x)|| \le sup_n ||x_n^*||$. By using Holder we get $||x_n^*(x)|| \le ||x_n^*||||x||\le sup_n ||x_n^*||||x||$ but still we need to get rid of ||x||$? – PerelMan Apr 14 at 15:05 • No$||x^*|| = sup _{||x||\leq 1} |x^*(x)|\$ – Ignorant Mathematician Apr 14 at 15:12