# Separable Hilbert space convergence problem

Suppose that $\{x_j \}_{1}^{\infty}$ is a sequence of separable Hilbert space $X$ and that $\|x_j\| \leq 1$ for all $j$. Show that there is a subsequence $\{x_{j_k} \}_{k=1}^{\infty}$ such that for every $y\in X$, $\{\langle y,x_{j_k}\rangle \}_{k=1}^{\infty}$ is convergent.

Attempted proof - Since $X$ is separable, it has a countable dense subset $\{y_n \}_{1}^{\infty}$ Then by Cauchy-Scwarz inequality and the fact that $\|x_j\| \leq 1$ $$\{\langle y_1,x_{j}\rangle \}_{j=1}^{\infty} \subset \{z\in\mathbb{C}: |z| \leq \|y\|_{1} \}$$ Since the set on the right is compact there exists a convergent subsequence $\{\langle y_1,x_{j_{1,k}} \}_{k=1}^{\infty}$. Now choose a subsequence $\{x_{j_{l,k}} \}_{k=1}^{\infty}$ for $l = 1,\ldots,s$ then choose $\{x_{j_{s+1,k}} \}_{k=1}^{\infty}$ to be a subsequence of $\{x_{j_{s,k}} \}_{k=1}^{\infty}$ such that $\{\langle y_{s+1},x_{j_{s+1,k}}\rangle \}_{k=1}^{\infty}$ is convergent . We then have that the subsequence $\{\langle y_n,x_{j_k,k}\rangle \}_{k=1}^{\infty}$ is convergent for every $n$.

Not sure if I am right any suggestions is greatly appreciated.

• How does this prove that for every $y\in X$ this subsequence is indeed convergent? Commented Apr 27, 2016 at 20:59
• yea I don't think my proof is complete yet Commented Apr 27, 2016 at 21:01
• Ok so basically you can select a subsequence $z_k$ such that for all $n$ the sequence $\{\langle y_n, z_k \rangle\}_{k\in \mathbb{N}}$ is convergent. Take an $y\in X$ then for all $\epsilon$ there's an $y_n$ such that $|y_n -y| < \epsilon$. Then use that $\langle y, z_k \rangle = \langle y-y_n +y_n, z_k \rangle$. Commented Apr 27, 2016 at 21:08
• @Andrew your a boss, thanks man Commented Apr 27, 2016 at 21:09
• When in doubt, add and subtract the same thing! Commented Apr 27, 2016 at 21:10

## 1 Answer

@Wolfy , Your proof is correct, but it is not complete. Here is a complete detailed proof based on your proof as a first part.

Suppose that $\{x_j \}_{1}^{\infty}$ is a sequence of separable Hilbert space $X$ and that $\|x_j\| \leq 1$ for all $j$. Show that there is a subsequence $\{x_{j_k} \}_{k=1}^{\infty}$ such that for every $y\in X$, $\{\langle y,x_{j_k}\rangle \}_{k=1}^{\infty}$ is convergent.

Proof - Since $X$ is separable, it has a countable dense subset $\{y_n \}_{1}^{\infty}$ Then by Cauchy-Scwarz inequality and the fact that $\|x_j\| \leq 1$ $$\{\langle y_1,x_{j}\rangle \}_{j=1}^{\infty} \subset \{z\in\mathbb{C}: |z| \leq \|y\|_{1} \}$$ Since the set on the right is compact there exists a convergent subsequence $\{\langle y_1,x_{j_{1,k}} \}_{k=1}^{\infty}$. Now choose a subsequence $\{x_{j_{l,k}} \}_{k=1}^{\infty}$ for $l = 1,\ldots,s$ then choose $\{x_{j_{s+1,k}} \}_{k=1}^{\infty}$ to be a subsequence of $\{x_{j_{s,k}} \}_{k=1}^{\infty}$ such that $\{\langle y_{s+1},x_{j_{s+1,k}}\rangle \}_{k=1}^{\infty}$ is convergent . We then have that the subsequence $\{\langle y_n,x_{j_k,k}\rangle \}_{k=1}^{\infty}$ is convergent for every $n$.

Let $l_n=\lim_{k \to \infty}\langle y_n,x_{j_k,k}\rangle$.

Given $y \in X$, we have let $\{y_{n_m}\}$ be a subsequnce of $\{y_n\}$ converging to $y$. Then
$$|l_{n_p} -l_{n_q}|=|\lim_{k \to \infty}\langle y_{n_p},x_{j_k,k}\rangle- \lim_{k \to \infty}\langle y_{n_q},x_{j_k,k}\rangle|=|\lim_{k \to \infty}\langle y_{n_p}- y_{n_q},x_{j_k,k}\rangle |\leq \|y_{n_p}- y_{n_q}\|$$ So $\{l_{n_m}\}_m$ is a Cauchy sequence in $\mathbb{C}$ and so it is convergent. Let $l=\lim_{m \to \infty}l_{n_m}$

Note that \begin{align*} |l-\langle y,x_{j_k,k}\rangle | & \leq |l -l_{n_m}|+|l_{n_m} - \langle y_{n_m},x_{j_k,k}\rangle| + |\langle y_{n_m},x_{j_k,k}\rangle - \langle y,x_{j_k,k}\rangle | =\\ & = |l -l_{n_m}|+|l_{n_m} - \langle y_{n_m},x_{j_k,k}\rangle| + |\langle y_{n_m}-y,x_{j_k,k}\rangle | \leq \tag{1}\\ & \leq |l -l_{n_m}|+|l_{n_m} - \langle y_{n_m},x_{j_k,k}\rangle| + \| y_{n_m}-y \| \end{align*} Given $\epsilon >0$, there is $M\in \mathbb{N}$ such that for all $m\geq M$, $$|l -l_{n_m}|<\epsilon /3 \quad \textrm{ and } \quad \| y_{n_m}-y \| <\epsilon /3$$ and there is $K\in \mathbb{N}$ such that for all $k\geq K$, $$|l_{n_M} - \langle y_{n_M},x_{j_k,k}\rangle|<\epsilon/3$$ So, using $(1)$, we have that, given $\epsilon >0$, there is $K\in \mathbb{N}$ such that for all $k\geq K$, $$|l-\langle y,x_{j_k,k}\rangle | <\epsilon$$ So $\{\langle y,x_{j_k,k}\rangle\}_{k=1}^\infty$ converges to $l$.

Remark: In other word we have proved that $$\lim_{k \to \infty}\lim_{m \to \infty}\langle y_{n_m},x_{j_k,k}\rangle=\lim_{k \to \infty}\langle y,x_{j_k,k}\rangle=l=\lim_{m \to \infty}l_{n_m}=\lim_{m \to \infty}\lim_{k \to \infty}\langle y_{n_m},x_{j_k,k}\rangle$$