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.

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    $\begingroup$ 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 $l_a$. $\endgroup$ – Velobos Nov 12 '20 at 8:58
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    $\begingroup$ 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. $\endgroup$ – Velobos Nov 12 '20 at 8:59

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.

  • $\begingroup$ 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. $\endgroup$ – WoolierThanThou Nov 12 '20 at 9:19
  • $\begingroup$ 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 $\lim_{k\to\infty}\sum_{n=1}^N|\langle x_{n_k},e_n\rangle|^2\leq\limsup_{k\to\infty}\|x_{n_k}\|^2$? $\endgroup$ – autumnriddle Nov 14 '20 at 23:28
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    $\begingroup$ We have that $\sum_{n=1}^N |\langle x_{n_k},e_n \rangle|^2\leq \|x_ {n_k}\|^2$ by Parseval. $\endgroup$ – WoolierThanThou Nov 15 '20 at 16:16

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