# Show that there is a subsequence $\{ x_{n_{k}}\}$ such that the sequence $\{ \langle x_{n_{k}},y\rangle\}$ converges

Let $$\{x_n\}$$ be a bounded sequence in a Hilbert space $$\mathbb{H}$$. Prove that for each $$y \in \mathbb{H}$$, there is a subsequence $$\{ x_{n_{k}}\}$$ such that the sequence $$\{ \langle x_{n_{k}},y\rangle\}$$ converges.

Now, I know that the inner product in a Hilbert space is jointly continuous i.e. $$x_n \rightarrow x , y_n \rightarrow y$$ $$\Rightarrow (x_n, y_n) \rightarrow (x,y)$$

But how do I use this fact to solve the above problem. Is there some other way to solve this question. If yes then please share!

• Perhaps the simplest way to use joint continuity is to say the boundedness of $x_n$ implies the boundedness of $(x_n,y)$ in $\mathbb R$. – hardmath Oct 2 '16 at 11:42

Set $U := \sup_{n \in \{1, 2, \dots\}} \|x_n\|$. Since the $x_n$'s are bounded, $U \in [0, \infty)$. By the Cauchy-Schwartz inequality, for every $n \in \{1, 2, \dots\}$, $$\left|\langle x_n, y\rangle\right| \leq \|x_n\|\|y\| \leq U\|y\|.$$ So the sequence $\left(\langle x_n, y\rangle\right)_{n \in \{1, 2, \dots\}}$ is a bounded sequence of real numbers, hence, by the Bolzano-Weierstrass theorem, has a convergent subsequence.
Using the Cauchy-Schwarz inequality you can show that the sequence $(\langle x_n, y\rangle)_{n\in\mathbb{N}}$ is a bounded sequence of real numbers, so it has a convergent subsequence.
You can prove something even stronger: There exists a subsequence $(x_{n_k})_k$ such that $\langle x_{n_k}, y\rangle$ converges for every $y$. That is, the subsequence is the same for every choice of $y$. But you need heavier tools for this. (Reflexivity + Riesz Theorem + Eberlein–Šmulian Theorem).