4
$\begingroup$

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

$\endgroup$
2
  • 1
    $\begingroup$ Since $x$ is not arbitrary it does not follow that $x_n$ converges weakly to $0$ in your argument. $\endgroup$
    – supinf
    Oct 22, 2020 at 22:32
  • 1
    $\begingroup$ The claim is false. Any infinite orthonormal sequence converges weakly to $0$. Hence if it happens to converge to some $x$ then $x=0$. $\endgroup$
    – Conifold
    Oct 22, 2020 at 23:01

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Are you saying that the statement is true if and only if $x=0$? Also, doesn't that mean $H$ is not necessarily infinite in this case? Thanks! $\endgroup$ Oct 22, 2020 at 22:40
  • 1
    $\begingroup$ Yes, I am saying the statement is true if and only if $x=0$. I was operating under the assumption that $H$ is infinite dimensional. $\endgroup$
    – supinf
    Oct 22, 2020 at 22:59
  • $\begingroup$ May I ask where did you use that $H$ is infinite dimensional? $\endgroup$ Oct 22, 2020 at 23:24
  • 1
    $\begingroup$ For constructing an orthonormal sequence in the case of $x=0$. If $H$ is finite dimensional, then there cannot be an orthonormal sequence (because by definition a sequence is infinite and due to orthogonality their elements are linear independent). $\endgroup$
    – supinf
    Oct 23, 2020 at 0:47
  • 1
    $\begingroup$ @John Just pick one. Any orthonormal sequence converges to $0$. See also en.wikipedia.org/wiki/… $\endgroup$
    – supinf
    Oct 26, 2020 at 18:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .