# Given $\|x\| \leq 1$ in an infinite dimensional Hilbert space, show there exists a orthonormal sequence that converges weakly to $x$

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

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

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.
• 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! Oct 22, 2020 at 22:40
• 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. Oct 22, 2020 at 22:59
• May I ask where did you use that $H$ is infinite dimensional? Oct 22, 2020 at 23:24
• 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). Oct 23, 2020 at 0:47
• @John Just pick one. Any orthonormal sequence converges to $0$. See also en.wikipedia.org/wiki/… Oct 26, 2020 at 18:59