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.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Yes it is another name of Banach Steinhaus theorem, right? $\endgroup$ – PerelMan Apr 14 at 14:41

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.