If $\sup\| h_n \| < \infty$ and $\langle h_n, e\rangle \to 0$ then $\langle h_n, h \rangle \to 0$ for all $h$

Question

Suppose $\mathscr H$ is a Hilbert space with orthonormal basis $\mathscr E$, and $\{ h_n \}$ is a sequence such that $\sup_{ n \in \mathbb N} \{ \| h_n \| \} < \infty$ and $\langle h_n, e\rangle \to 0$ for each $e \in \mathscr E$ then $\langle h_n, h \rangle \to 0$ for all $h \in \mathscr H$.

This is what I have come up so far $$ h = \sum_{e\in\mathscr E}\langle h, e \rangle e \implies \langle h_n, h \rangle =\sum_{e\in\mathscr E}\langle h, e \rangle \langle h_n, e \rangle \le \sqrt{\sum_{e\in \mathscr E} |\langle h, e \rangle |^2}\sqrt{\sum_{e\in \mathscr E} |\langle h_n, e \rangle |^2} $$ If each $\langle h_n, e \rangle \to 0$ with $\| h_n\| < \infty$ then how do I show the last part $\sum_{e\in \mathscr E} |\langle h_n, e \rangle |^2 \to 0$. I need to prove the converse as well but converse easily follows from uniform boundedness principle.

Answer 1

You propose to show that $\sum_{e\in\mathcal{E}}|\langle h_n,e\rangle|^2 = ||h_n||^2 \rightarrow 0.$ But this may not be true. Indeed if we take a countable basis $\{e_n:n\in\mathbb{N}\},$ and let $h_n=e_n,$ its true that $\langle h_n, e_j\rangle\rightarrow 0$ for all $e_j,$ but $\|h_n\| = 1$ for all $n.$ So we need to take a different strategy.

Here's my solution

Consider that for every $\varepsilon>0$ I can find a finite subset $E\subset \mathcal{E}$ such that $\sum_{e\in\mathcal{E}\setminus E}|\langle h,e\rangle|<\varepsilon.$ Let $m$ be the number of elements of $E$ and let $N$ be a large enough natural number so that $|\langle h_n, e\rangle| < \varepsilon/(m||h||)$ for every $e\in E$ and $n>N.$ Now consider that for every $n>N$ we have

$$\begin{aligned} |\langle h_n,h\rangle| &= |\sum_{e\in\mathcal{E}}\langle h,e\rangle\langle h_n,e\rangle| \\&\leq \sum_{e\in\mathcal{E}\setminus E}|\langle h,e\rangle||\langle h_n,e\rangle| + \sum_{e\in E}|\langle h,e\rangle||\langle h_n,e\rangle|\\&\leq ||h_n||\sum_{e\in\mathcal{E}\setminus E}|\langle h,e\rangle| + \sum_{e\in E}|\langle h,e\rangle||\langle h_n,e\rangle|\\&\leq \varepsilon||h_n|| + \sum_{e\in E}|\langle h,e\rangle||\langle h_n,e\rangle|\\&\leq \varepsilon||h_n|| + \max\{|\langle h_n,e\rangle|e\in E\}\sum_{e\in E}|\langle h,e\rangle| \\ &\leq \varepsilon||h_n|| + \dfrac{\varepsilon}{m||h||}m||h|| \\ &= \varepsilon||h_n|| + \varepsilon\end{aligned}$$

Since $\sup||h_n||<\infty$ this completes the proof

Answer 2

In your argument, those $\sum_{e \in \mathcal E}$'s are infinite sums, so any manipulations you do will require proof.

Can I suggest an alternative approach?

Step 1: Consider the special case where $h$ can be expressed as a linear combination of finitely many elements in $\mathcal E$, i.e. $h = \sum_{i=1}^N c_i e_i$, with $c_i \in \mathbb C$ and $e_i \in \mathcal E$. Prove the result for this special case.

Step 2: Since $\mathcal E$ is an orthonormal basis for $H$, the set of finite linear combinations of elements of $\mathcal E$ is dense in $H$. Said another way, given any $h \in H$ and given any $\varepsilon > 0$, we can find some $h' = \sum_{i=1}^N c_i e_i$ such that $\| h - h' \| < \varepsilon$. Go from here.