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.
I think this can be done without invoking Banach-Alaoglu or the Axiom of Choice. I will sketch the proof. By the Riesz representation theorem (which as far as I can tell can be proven without Choice), a Hilbert space is reflexive. Furthermore, it is separable iff its dual is.
To show the weak convergence of the bounded sequence $(x_n)$ assume first that $H$ is separable and let $\{x'_1,x'_2,\ldots\}$ be a dense set in the dual space. Use a diagonal argument to extract a subsequence $(x_{n_k})$ such that $x'_m(x_{n_k})$ converges for all $m$. If $x'$ is any functional and for $\epsilon>0$, there is $x'_m$ such that $\|x'-x'_m\|<\epsilon$. Then,
\begin{align}\|x'(x_{n_k})-x'(x_{n_l})\|&\le \|x'(x_{n_k})-x'_m(x_{n_k})\|+\|x_m'(x_{n_k})-x'_m(x_{n_l})\|\\&+\|x'_m(x_{n_l})-x'(x_{n_l})\|<(2M+1)\epsilon,\end{align}
if $k$ and $l$ are large enough (define $M=\sup_n \|x_n\|$). Hence, $(x'(x_{n_k}))$
is a Cauchy sequence. It remains to be shown that the weak limit exists. Consider the linear map $\ell(x'):= \lim_k x'(x_{n_k})$. This is well-defined by the previous argument and bounded, since $\ell(x')\le \|x'\|M$. By reflexivity of $H$, there is $x\in H$ such that $\lim_kx'(x_{n_k})=\ell(x')=x'(x)$, which means exactly that $x$ is the weak limit of $(x_{n_k})$.
For the general case, let $Y$ be the closed linear span of $\{x_1,x_2,\ldots\}$. This is then a separable Hilbert space and by the previous argument, there is a subsequence $(x_{n_k})$ and $y\in Y$ such that $(y'(x_{n_k}))$ converges to $y'(y)$ for all $y'\in Y'$.
It remains to be shown that $(x'(x_{n_k}))$ converges to $x'(y)$ for all $x'\in H'$. But this is obvious, since the restriction of $x'$ to $Y$ is a functional on $Y$.
Best Answer
Okay, based on David Mitra's comment we can construct an orthonormal sequence. For ease of showing the required result let's call this $\{x_n\}$. Then Bessel's inequality gives for all $y \in H$ that:
$\sum_{n=1} ^\infty |(x_n,y)|^2 \le ||y||^2$. Assume $||y||^2$ is not infinite, which seems reasonable then this is a convergent sequence which implies the terms much approach zero, hence $\lim_{n \rightarrow \infty} (x_n,y) = 0$, thus by definition we have a sequence that converges to the 0 vector (and since it is orthonormal $||x_n|| = 1$ so it satisfies $||x_n|| \ge 1$.
I wonder why the question specified the norm being greater than or equal to 1. Just to throw to make it more confusing?