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Definition of the problem

Let $\mathcal{H}$ be a Hilbert space, and let $\left(x_{n}\right)_{n\in\mathbb{N}}\subset\mathcal{H}$ be a sequence. Prove the following:

If $\left(f\left(x_{n}\right)\right)_{n\in\mathbb{N}}\subset\mathbb{K}$ converges for each $f\in\mathcal{H}^{\star}$, then there exists $x\in\mathcal{H}$ such that $\left(x_{n}\right)_{n\in\mathbb{N}}$ converges weakly to $x$.

My idea

Define $\varphi:\mathcal{H}^{\star}\rightarrow\mathbb{K},\quad f\mapsto\lim\limits _{n\rightarrow\infty}f\left(x_{n}\right).$ We know that Hilbert spaces are reflexive, then the canonical mapping is surjective (from $\mathcal{H}$ to $\mathcal{H}^{\star\star}$).

Claim: $\varphi\in\mathcal{H}^{\star\star}$.

If we know that $\varphi\in\mathcal{H}^{\star\star}$, then $\varphi\left(f\right)=\lim\limits _{n\rightarrow\infty}f\left(x_{n}\right)=f\left(x\right)$, since the canonical mapping is surjective.

My questions

How could I prove now that my $\varphi$ is an element of $\mathcal{H}^{\star\star}$?

How do you find my proof so far? Is it complete? How would you improve it?

Thanks a lot, Franck.

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5  
Hint: Banach-Steinhaus theorem. –  Siminore Jun 18 '12 at 14:19
1  
(Also known as Uniform Boundedness Principle) –  user31373 Jun 18 '12 at 14:24
2  
Identify $x_n$ with $\phi_n$ in $H^{**}$ and apply UBP. Then use that normbounded sets are weakly compact. Now go back to $H$. –  user20266 Jun 18 '12 at 14:50
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We will make you solve this today ;-) –  user20266 Jun 18 '12 at 14:51
1  
The only missing step is to show that the $\phi$ you have defined is continuous. By assumption, the (single element) set $\{\phi(f)\}$ is bounded, for any $f$, so apply BS to conclude that $\phi$ is bounded. Then you are finished (using reflexivity). –  copper.hat Jun 18 '12 at 14:54

2 Answers 2

By corollary 2.4 of Banach-Steinhaus theorem in $\textbf{Brezis}$, we have that $x_n$ is bounded. Then as $H$ is Hilbert, hence reflexive, there exists a $x \in H $ such that $x_n \rightarrow x$ converges weakly to $x$. Then there exists a subsequence $x_{n_j}$ such that $f(x_{n_j}) \rightarrow f(x)$ for all $f \in H^{\star}$. But as $((f(x_n))_{n}$ converges we have that $f(x_n) \rightarrow f(x)$ for all $f \in H^{\star}$.

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up vote 1 down vote accepted

Define $\varphi:\mathcal{H}^{\star}\rightarrow\mathbb{K},\quad f\mapsto\lim\limits _{n\rightarrow\infty}f\left(x_{n}\right).$ We know that Hilbert spaces are reflexive, then the canonical mapping is surjective (from $\mathcal{H}$ to $\mathcal{H}^{\star\star}$).

Claim: $\varphi\in\mathcal{H}^{\star\star}$.

Proof of the claim:

We have to show that $\varphi$ is a continuous bounded operator. Define $\left\{ \varphi\left(f\right)\right\} $ to be a single element set, containing $\varphi$. By assumption, this set is bounded, for any $f\in\mathcal{H}^{\star}$. We can now apply Banach-Steinhaus theorem, and we obtain easily that $\varphi$ is bounded, since $\left\{ \varphi\left(f\right)\right\} $ is bounded aswell. Moreover, we know that norm-bounded sets are weakly compact, then $\varphi$ is also continuous. So $\varphi$ is element of $\mathcal{H}^{\star\star}$.

It follows directly that if $\varphi\in\mathcal{H}^{\star\star}$, then $\varphi\left(f\right)=\lim\limits _{n\rightarrow\infty}f\left(x_{n}\right)=f\left(x\right)$, since the canonical mapping is surjective.

We have then that for all $f\in\mathcal{H}^{\star}$, $f\left(x_{n}\right)\rightarrow f\left(x\right)$, so by definition of weak convergence, $x_{n}$ converges weakly to $x\in\mathcal{H}$.

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