Banach-Alaoglu and Riesz theorem are an overkilling, even if I really like it as a proof. Moreover, your statement is true in general in reflexive spaces and it is usually known as the easy part of Eberlein-Smulian theorem. Recalling the following definitions,
Definition. Let $X$ be a Banach space. $X$ is reflexive iff the canonical injection $J: X \to X''$, $\langle J, x \rangle_{X'',X} = \langle \cdot, x \rangle_{X',X}$, is surjective.
A fundamental result is
Theorem 1 (Kakutani). Let $X$ be a Banach space. They are equivalent
- $X$ is reflexive;
- $B_X$ is weakyl compact.
From Kakutani's theorem follow
Theorem 2. Let $X$ be reflexive, $M$ a closed linear subspace of $X$. Then $M$ is reflexive.
Theorem 3. $X$ reflexive $\iff$ $X'$ reflexive.
Corollary 1. Let $X$ be reflexive and $M$ a closed linear subspace of $X$. Then $X/M$ (quotient space) is reflexive.
Corollary 2. $X$ reflexive and separable $\iff$ $X'$ reflexive and separable.
Now we can state
Theorem 4. Let $X$ be reflexive and $\{ x_n \} \subseteq X$ a bounded sequence. Then there exist $x \in X$ and $\{ x_{n_k}\}$ s.t. $x_{n_k} \rightharpoonup x$ as $k\to\infty$.
Proof. Set $M = \overline{\mbox{span}_{\mathbb K}}\{x_n\}$. By Theorem 2, $M$ is reflexive and obviously is separable. Hence, by Corollary 2, $M'$ is reflexive and separable. Then $J(B_M) = B_{M''}$ is compact and metrizable with respect to the $\sigma(M'', M')$ topology, so there exists $\{x_{n_k}\}$ subsequence and $F = J(x)$ s.t. $J(x_{n_k}) \rightharpoonup J(x)$ when $k\to \infty$ with respect to the $\sigma(M'', M')$, i.e., by definition, $\langle J(x_{n_k}), f \langle_{M'',M'} \to \langle J(x_n), f \rangle_{M'',M'}$. Since $M$ is closed, $x \in M$ and $\langle f, x_{n_k} \rangle_{M',M} \to \langle f, x_n \rangle_{M',M}$ when $k\to\infty$. Thanks to the Hahn-Banach theorem, there exists $L \in X'$ s.t. $f := L_{|M}$ and hence we have $\langle L, x_{n_k} \rangle_{X',X} \to \langle L, x_{n_k} \rangle_{X',X}.$ $\square$
Now consider a Hilbert space $H$. The scalar product on $H$ induces a norm which satisfies the parallelogramm identity (trivial):
$$\| x + y \|^2 + \| x - y \|^2 = 2( \|x\|^2 + \|y\|^2)$$
for all $x,y\in H$. Such a norm is uniformly convex and hence, by
Theorem 5 (Milman). Let $X$ be a Banach space. If $X$ is uniformly convex, then $X$ is reflexive.
we have $H$ is reflexive. So the previous Theorem 4 holds and your claim follows as a corollary of it.
I know your method was quicker, but I listed a few of results you would probably know, provided you can apply Banach-Alaoglou theorem. By the way, in the previous work there is no needing of knowledge of Hilbert spaces theory, except for the definition. In this sense, this is the easiest proof, even is longer.
References.
- Brezis, Functional analysis, Sobolev spaces and partial differential equations.
- Yosida, Functional analysis.
Best Answer
I don't even think you need to apply Riesz to get this result.
So, note that $j\mapsto \langle x_j, e_n\rangle$ is a bounded, complex sequence for every $n$. Hence, there is some subsequence $x_{j,1}$ of the $x_j$ such that $\langle x_{j,1},e_1\rangle$ is convergent. Inductively, let $x_{j,l+1}$ be a subsequence of the $x_{j,l}$ such that $\langle x_{j,l+1},e_{l+1}\rangle$ is convergent. Defining $x_{n_k}=x_{k,k},$ we get a subsequence such that $\langle x_{n_k},e_n\rangle$ is convergent for every $n$. Denote the limit $\alpha_n$.
We wish to argue that $\sum_{n=1}^{\infty} |\alpha_n|^2<\infty$. Indeed, note that for every $N>0$ $$ \sum_{n=1}^N |\alpha_n|^2=\lim_{k\to\infty} \sum_{n=1}^N |\langle x_{n_k},e_n\rangle|^2\leq \limsup_{k\to\infty} \|x_{n_k}\|^2<\infty, $$ and the right-hand side is independent of $N$.
Now, define $x=\sum_{n=1}^{\infty} \alpha_n e_n$. This an absolutely convergent $\ell^2$-series and hence, $x$ defines an element of $H$.
Now, given $x'=\sum_{n=1}^{\infty} \beta_n e_n$ and $\varepsilon>0$, fix $N$ so large that $\sum_{n=N+1}^{\infty} |\beta_n|^2,\sum_{n=N+1}^{\infty} |\alpha_n|^2<\varepsilon^2$ and we get, defining $\tilde{x}=\sum_{n=1}^N \alpha_n e_n$ and $\tilde{x}'=\sum_{n=1}^N \beta_n e_n$,
\begin{align} |\langle x_{n_k},x'\rangle-\langle x,x'\rangle|&\leq |\langle x_{n_k},x'-\tilde{x}'\rangle|+|\langle x_{n_k}-\tilde{x},\tilde{x}'\rangle|+|\langle \tilde{x}-x,x'\rangle|\\ &\leq \sup_{k}\|x_{n_k}\| \varepsilon+|\langle x_{n_k}-\tilde{x},\tilde{x}'\rangle|+\varepsilon\|x'\| \end{align} Now, $|\langle x_{n_k}-\tilde{x},\tilde{x}'\rangle|$ goes to $0$ by our construction of the $\alpha_n$, and so, since $\varepsilon>0$ was arbitrary and the $x_{n_k}$ are bounded, we're done.