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Let E be a Banach space.

Let $(x^∗_n )$ be a sequence in $E^∗$ verifying $(<x^∗_n , x>)$ 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.

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

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.

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  • $\begingroup$ 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||$? $\endgroup$ – PerelMan Apr 14 at 15:05
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    $\begingroup$ No $||x^*|| = sup _{||x||\leq 1} |x^*(x)|$ $\endgroup$ – Ignorant Mathematician Apr 14 at 15:12
  • $\begingroup$ Thank you !that makes sense! $\endgroup$ – PerelMan Apr 14 at 15:30

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