Let $H$ be an infinite dimensional Hilbert space. Given $x\in H$ with $\| x\| \leq1$, show there exists an orthonormal sequence $(x_n)$ such that $(x_n)$ converges weakly to $x$.
Below are my ideas and thoughts so far:
I thought about using the orthonormal basis to construct such sequence. But since we don't know if $H$ is countable, we can't assume there exists an orthonormal basis.
Also note that using Bessel's inequality, if we have an orthonormal sequence we have
$\sum_{n} |\langle x,x_n\rangle|^2 \leq \| x\|^2=1$.
So $\lim _{n \rightarrow\infty} \langle x,x_n\rangle^2 =0$.
Hence $\lim _{n \rightarrow\infty} \langle x,x_n\rangle =0$, which tells us $x_n$ converges weakly to zero.
But I'm not sure if this helps us with the question…
Any hints or ideas will be appreciated!
Thank you
Best Answer
This is only possible if $x=0$.
Suppose that $(x_n)$ converges weakly to $x$ and is an orthonormal sequence. Then you have already shown that $$ \lim_{n\to\infty} \langle x,x_n\rangle =0 $$ holds. From the weak convergence $x_n \rightharpoonup x$ we conclude $$ \|x\|^2 = \langle x,x\rangle = \lim_{n\to\infty} \langle x,x_n\rangle =0. $$ Therefore, $x=0$.
Constructing an orthonormal sequence $(x_n)$ which converges weakly to $0$ can be done in the usual way.