Prove that in a Hilbert space, $\lim_{k \rightarrow \infty} \langle x_{n_k} , y\rangle =\langle x,y \rangle$

Question

Let $\mathscr{H}$ be a Hilbert space and $(x_n)_n$ a bounded sequence in $\mathscr H$.

How can I show that there exists a subsequence $(x_{n_k})_k$ and an $x \in \mathscr{H}$ so that $$ \lim_{k \rightarrow \infty} \langle x_{n_k},y\rangle= \langle x,y\rangle $$ for all $ y \in \mathscr{H} $?

I know some of such proofs, but I don't know how to show that for a Hilbert space ! As a hint I got : using the diagonal consequences argument. Appreciate any help of you !

Answer 1

Since you mentioned diagonalization I'll give you a (sketch of) proof that uses the method. For this proof to work, we need to assume also that $\mathscr H$ is separable though.

Let $E=\{e_1,e_2,\dots\}$ be an orthonormal basis of $\mathscr H$ (which exists by our assumption that $\mathscr H$ is separable). For each fixed $i\in\Bbb N$, the sequence $(a_n)_n$ defined by $$ a^{(i)}_n:=\langle x_n,e_i\rangle $$ is a bounded sequence in $\Bbb R$ (or in $\Bbb C$, depending on what's your base field), hence we can extract a subsequence $a^{(i)}_{n_k}\to a^{(i)}$. This corresponds to a subsequence $(x_{n_k})_k$.

Can you see now which diagonal sequence you should take so that $\langle x_{n_k},e_i\rangle$ converges for all $i\in\Bbb N$?

We know that $E$ is an orthonormal basis, what do you need to do to conclude that $\langle x_{n_k},y\rangle$ converges for all $y\in \mathscr H$?

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Edit: Denote $(x^{(i)}_{n})_n$ to be the $i^{\text{th}}$ subsequence of $(x_n)_n$, i.e. we have $x^{(0)}_{n}=x_n$ and that $x^{(i+1)}_{k}=x^{(i)}_{n_k}$ be a subsequence of $x^{(i)}_{n}$.

Consider $$ \begin{matrix} x^{(1)}_1 & x^{(1)}_2 & x^{(1)}_3 & \dots & \text{ a subseq. such that $a^{(1)}_{n_k} \to a^{(1)}$} \\ x^{(2)}_1 & x^{(2)}_2 & x^{(2)}_3 & \dots & \text{ a subseq. such that $a^{(2)}_{n_k} \to a^{(2)}$}\\ x^{(3)}_1 & x^{(3)}_2 & x^{(3)}_3 & \dots & \text{ a subseq. such that $a^{(3)}_{n_k} \to a^{(3)}$} \\ \vdots \end{matrix} $$ we can define a diagonal sequence $x'_n:=x^{(n)}_n$. From construction, it is easy to see that $\langle x'_{n},e^{(i)}\rangle \to a^{(i)}$ for all $i\in\Bbb N$.

Thanks to the remark by Berci, actually my proof works for inseparable Hilbert space $\mathscr H$ as well. We just have to restrict our attention to the (separable) subspace $H:=\overline{\text{span}}\{x_n:n\in\Bbb N\}$ .