# 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?

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 $Y$.

• For those who can read german, this argument can be found in Funktionalanalysis by Dirk Werner. – Vincent Boelens Mar 7 '15 at 12:07
• I believe this is also essentially the proof given in Real Analysis by Royden and Fitzpatrick in Chapter 16. I also agree that Riesz representation does not need AC, though I am also not an expert. – Ian Mar 7 '15 at 12:31
• The argument by Royden and Fitzpatrick seems to me to be the same as well. The diagonal argument is given in Chapter 8 (Helley's theorem). – Vincent Boelens Mar 7 '15 at 13:03

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.

• I upvoted, thank you for your answer. I will wait some time before I accept just to see if anyone has any more comments in this! – user167889 Mar 6 '15 at 8:31
• (I'm slightly skeptical about the equivalence you mention because Alaoglu is a statement about any normed vector spaces whereas the statement in my question is about Hilbert spaces so there is more machinery to use...) – user167889 Mar 6 '15 at 8:36
• True, there is more machinery. However, I have not seen any proof using any of it. Even books focusing on hilbert spaces, like Halmos' problem book for instance, use the normed space proof as the proof of the theorem. – BigMathTimes Mar 6 '15 at 8:57
• If the proof of the Riesz representation theorem does not require Choice (which I think it does not, but I am not an expert) it can be done. See my answer. – Vincent Boelens Mar 7 '15 at 12:10
• @BigMathTimes Any sequence in a compact space $X$ has a convergent subsequence, for instance, provided $X$ is first countable, but not in general. For instance, a Čech-Stone compactification $\beta N$ of a space $N$ of positive integers with the discrete topology, has no non-trivial convergent sequences (see, for instance, Corollary 3.6.15 of Engelking's “General Topology” (2nd ed., Heldermann, Berlin, 1989)). – Alex Ravsky Jun 2 '17 at 18:04