Weak closure of orthonormal basis in Hilbert space

Question

Let $H$ be an infinite dimensional separable Hilbert space and let $E=\{e_{1},e_{2},\ldots\}$ be an orthonormal basis. I want to prove that $0_{H}\in\text{Clo}_{\text{w}}(E)$ (= weak closure of $E$). Using the Riesz representation theorem, I convinced myself that it suffices to prove that $\langle e_{n},h\rangle\to0$ for each $h\in H$. It is probably elementary Hilbert space theory, but I really don't see why the latter sequence converges to $0$. Any suggestions are greatly appreciated.

Answer 1

The other answers don't quite sit with me right since they implicitly assume convergence and it feels a bit circular.

Define $h_n = \displaystyle\sum_{i=1}^n \langle h, e_i\rangle e_i$, then

$$0\le\|h-h_n\|^2 = \langle h - h_n, h - h_n\rangle = \langle h, h\rangle - \langle h_n, h\rangle - \langle h, h_n\rangle + \langle h_n, h_n\rangle.$$

Let's work on the cross terms to see what these become.

$$\langle h, h_n\rangle = \sum_{i=1}^n \overline{\langle h, e_i\rangle} \langle h, e_i\rangle = \sum_{i=1}^n |\langle h,e_i\rangle|^2 $$

and similarly

$$\langle h_n, h\rangle = \sum_{i=1}^n \langle h, e_i\rangle \langle e_i, h\rangle = \sum_{i=1}^n |\langle h, e_i\rangle|^2. $$

The last term in the first equation simplifies nicely to

$$ \langle h_n, h_n \rangle = \sum_{i=1}^n |\langle h, e_i\rangle|^2 $$

since $e_n$ is orthonormal.

Assembling these pieces,

$$ 0 \le \langle h, h\rangle - 2 \sum_{i=1}^n |\langle h,e_i\rangle|^2 + \sum_{i=1}^n |\langle h, e_i\rangle|^2 = \langle h, h\rangle - \sum_{i=1}^n |\langle h, e_i\rangle|^2.$$

Moving to the other side,

$$ \sum_{i=1}^n |\langle h, e_i\rangle|^2 \le \langle h, h\rangle < \infty.$$

for all $n \ge 1$. Since these are uniformly bounded, what you must conclude is that $\langle h, e_n\rangle\to 0$ as $n\to\infty$.