Every bounded sequence has a weakly convergent subsequence in a Hilbert space

I tried to prove the following theorem and was wondering if someone could please tell me if my proof can be fixed somehow...

Theorem: Let $H$ be a Hilbert space and $x_n\in H$ a bounded sequence. Then $x_n$ has a weakly convergent subsequence.

My idea for a proof:

The map $\phi: H \to H^\ast$ in the Riesz representation theorem is an isometry therefore $\varphi_n := \phi(x_n)$ is also bounded and therefore $\varphi_1(x_n)$ is a bounded sequencein $\mathbb R$. By Bolzano Weierstras it ha a convergent subsequence $\varphi_1(x_{n_{k_1}})$. (Say, $\varphi_1(x_{n_{k_1}})\to \varphi_1(x)$ for some $x$) Let $x_{n_1}$ be the argument of the first element in this sequence (apologies for the notation; the first element is also called $x_{n_1}$...).

The sequence $\varphi_2(x_{n_1})$ has a convergent subsequence $x_{n_{k_2}}$. Let $x_{n_2}$ be the first element in that sequence.

And so on. Then the resulting sequence $x_{n_k}$ has the property that for all $j$:

$$ \varphi_j(x_{n_k}) \to \varphi_j(x)$$

My only problem is that I only showed this limit for $\varphi_n$ that is, not for all $\varphi \in H^\ast$.

Can this argument be fixed somehow?

Solution 1:

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 $\lim_x'(x_{n_k})=\ell(x')=x'(x)$, 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 alle $y'\in Y'$. 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 $Y$.

Solution 2:

Unfortunately, I do not see an easy way to salvage this proof. Your question is essentially equivalent to the Banach-Alaoglu Theorem, which states that the unit ball is weakly compact in $H$. Unfortunately, the only proof I have ever seen of the Banach-Alaoglu Theorem uses Tychonoff's theorem on compact topological spaces. While certainly manageable, this is some rather heavy machinery. I cannot think of any similarly powerful techniques to fix the above proof, without using Tychonoff's theorem. My suggestion is to use or prove the Banach-Alaoglu theorem and apply it to your problem. The link included above includes a proof of the theorem.

As for why this theorem is equivalent to your question, this is some straightforward topology. Any sequence in a compact set in a topological space must have a convergent subsequence. As the sequence is bounded, it must lie in some ball about the origin. This ball will be compact because of Banach-Alaoglu.