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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!

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    $\begingroup$ 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.) $\endgroup$ Aug 17, 2013 at 19:04
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    $\begingroup$ Why may I assume it is separable? $\endgroup$
    – Fractal20
    Aug 17, 2013 at 19:06
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    $\begingroup$ Sorry, it's a bit inelegant to do so. Better to take the second, parenthetical, approach I suggested in my first comment. $\endgroup$ Aug 17, 2013 at 19:13
  • $\begingroup$ Ahhhhh, I think I see $\endgroup$
    – Fractal20
    Aug 17, 2013 at 19:29
  • $\begingroup$ @Fractal20 then you can answer your own question! $\endgroup$
    – Norbert
    Aug 17, 2013 at 19:29

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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?

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    $\begingroup$ Presumably the requirement $\|x_n\| \ge 1$ is to keep you from just picking a sequence that converges to 0 in norm. $\endgroup$ Aug 17, 2013 at 21:39
  • $\begingroup$ 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. $\endgroup$
    – Fractal20
    Aug 18, 2013 at 2:05
  • $\begingroup$ Well, you could use $\{2x_n\}$. $\endgroup$ Aug 18, 2013 at 4:52
  • $\begingroup$ Can you explain why we can assume ||y||^2 < infinity? $\endgroup$
    – Bob
    Nov 18, 2014 at 10:44
  • $\begingroup$ @Bob follows by definition, since $y\in H$. $\endgroup$ Jul 21, 2023 at 13:11

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