# Bounded sequence in a Hilbert space has weakly convergent subsequence

Suppose $$H$$ is a Hilbert space with an orthonormal basis $$(e_n)$$. Let $$(x_n)$$ be a bounded sequence in $$H$$.

1. Prove that $$(x_n)$$ contains a subsequence $$(x_{n_k})$$ and $$H$$ contains an element $$x$$ so that $$\langle x_{n_k},e_n\rangle\to\langle x,e_n\rangle$$ for each $$n$$.
2. Prove that $$\langle x_{n_k},x'\rangle\to\langle x,x'\rangle$$ for any $$x'\in H$$.

I am not sure how to proceed with this question. I am familiar wiht the Riesz representation theorem, which states that for every $$f\in H^*$$ there exists $$y\in H$$ so that $$f(x)=\langle x,y\rangle$$. But I don't know how to apply that here.

I tried defining a linear functional $$f(h)=\langle h,e_n\rangle$$ and seeing what I can do from there, but I am truly stuck. I have seen some proofs using the Banach-Alaoglu theorem, but I am wondering if there is a proof without using this result. Any help in this question will be highly appreciated - thanks in advance.

• Hint : let $a \in H$ fixed. The real sequences $(<x_n|a>)_n$ is bounded by cauchy schwarz and therefore you can extract a subsequence that converges toward a real $l_a$. – Velobos Nov 12 '20 at 8:58
• Now set $F : a \mapsto l_a$ and try to prove that $F$ is indeed a linear continuous map from $H$ to $\mathbb{R}$. Then you can apply Riesz theorem and you should have your result. – Velobos Nov 12 '20 at 8:59

## 1 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.

• If one wants to compare to Velobos' hint, Riesz could have helped us argue that $x$, indeed, defines an element of $H$ (you don't have to check the $\ell^2$-convergence). I don't think you get around checking the other things (making the diagonal argument, checking that you get the appropriate limit general elements $x'\in H$) somewhat manually. – WoolierThanThou Nov 12 '20 at 9:19
• Thank you for the very clear answer! I have one question: In showing that $\sum_n|\alpha_n|^2<\infty$, how do you know that $\lim_{k\to\infty}\sum_{n=1}^N|\langle x_{n_k},e_n\rangle|^2\leq\limsup_{k\to\infty}\|x_{n_k}\|^2$? – autumnriddle Nov 14 '20 at 23:28
• We have that $\sum_{n=1}^N |\langle x_{n_k},e_n \rangle|^2\leq \|x_ {n_k}\|^2$ by Parseval. – WoolierThanThou Nov 15 '20 at 16:16