weak convergence in Hilbert space

Question

I have the following theorem from the book "Applied Analaysis" by Hunters.

Suppose that $x_n$ is a sequence in a Hilbert space $H$ and $D$ is a dense subset of $H$. Then, $$ converges to $$ for every $y\in H$ iff $\|x_n\|\leq M$ for some constant $M$ and $$ converges to $$ for $z\in D$.

Then, as an illustration of this theorem, the author says that:

If $\{u_{\alpha}\}$ is an orthonormal basis of a Hilbert space, then a sequence $x_n$ converges weakly to $x$ if and only if it is bounded and its coordinate converges, that is, $ \to $

But, I do not understand this example because an orthonormal basis is not dense in a Hilbert space. What am I missing?

Answer 1

All right. So, let $g$ be an element in the span of $u_{\alpha}$. Then, we can write $g = \sum c_\alpha u_\alpha$, where $c_\alpha$ are elements from the field over which the Hilbert space is constructed (mostly the real or complex numbers).

Note then, that $$ \langle y,g \rangle = \sum c_\alpha \langle y,u_\alpha \rangle $$ holds true for all elements $y$ of the Hilbert space.

Suppose that for all $\alpha$, $\langle x_n,u_\alpha \rangle \to \langle x,u_\alpha \rangle$ is true. Then: $$ \langle x_n,g \rangle=\sum c_\alpha\langle x_n,u_\alpha \rangle \to \sum c_\alpha\langle x,u_\alpha \rangle = \langle x,g \rangle $$

For all $g$ "in the span of $u_\alpha$", which is dense in the Hilbert space. By the given proposition, since $x_n$ are given bounded, we are done. I hope this explanation is sufficient.