Weak convergence in Hilbert spaces 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.
 A: 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 $f \in H^{\star}$.
A: 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
$x\in\mathcal{H}$.
