# Prove or disprove existence of a sequence converging weakly to $0$ in an infinite dim Hilbert space

This is a problem on an old analysis qual, the prompt is:

"Prove or give a counter example: if $H$ is an infinite dimensional Hilbert space and $0$ is the zero vector in $H$, then there exists a sequence $\{x_n\}$ in $H$ so that $||x_n|| \ge 1$ and $\{x_n\}$ converges weakly to the zero vector $0$ in $H$."

I know that the unit ball is not necessarily weakly compact in an infinite dimensional space if it is not reflexive. But this is specifying the existence of a single sequence, which doesn't say anything about every sequence having a convergent subsequence etc.

Since it is a Hilbert space I know it is equivalent to $(x_n,y) \rightarrow 0$ for all $y \in H$ for such a space. I was tempted to assume a countable orthonormal use Parseval's Identity to show $||x_n||^2$ could be made less than 1, but this would seem to require $(x_n,e_k)$ to converge uniformly (ie independently of $k$ where the $e_k$'s are the orthonormal set).

Anyway, I am stuck. Any suggestions? Thanks!

• The statement is true. You may assume $H$ is separable. Take an orthornormal basis of $H$ and show, by Parseval e.g., that is weakly convergent to $0$. (Or just take an infinite orthonormal sequence in $H$ and use Bessel.) Aug 17, 2013 at 19:04
• Why may I assume it is separable? Aug 17, 2013 at 19:06
• Sorry, it's a bit inelegant to do so. Better to take the second, parenthetical, approach I suggested in my first comment. Aug 17, 2013 at 19:13
• Ahhhhh, I think I see Aug 17, 2013 at 19:29
• @Fractal20 then you can answer your own question! Aug 17, 2013 at 19:29

Okay, based on David Mitra's comment we can construct an orthonormal sequence. For ease of showing the required result let's call this $\{x_n\}$. Then Bessel's inequality gives for all $y \in H$ that:

$\sum_{n=1} ^\infty |(x_n,y)|^2 \le ||y||^2$. Assume $||y||^2$ is not infinite, which seems reasonable then this is a convergent sequence which implies the terms much approach zero, hence $\lim_{n \rightarrow \infty} (x_n,y) = 0$, thus by definition we have a sequence that converges to the 0 vector (and since it is orthonormal $||x_n|| = 1$ so it satisfies $||x_n|| \ge 1$.

I wonder why the question specified the norm being greater than or equal to 1. Just to throw to make it more confusing?

• Presumably the requirement $\|x_n\| \ge 1$ is to keep you from just picking a sequence that converges to 0 in norm. Aug 17, 2013 at 21:39
• That makes sense. I suppose I meant that I was wondering why it could be greater than one, ie the proof relied on it being an orthonormal sequence which then would all have norm of 1. Aug 18, 2013 at 2:05
• Well, you could use $\{2x_n\}$. Aug 18, 2013 at 4:52
• Can you explain why we can assume ||y||^2 < infinity?
– Bob
Nov 18, 2014 at 10:44
• @Bob follows by definition, since $y\in H$. Jul 21, 2023 at 13:11