# Weak convergence in the Hilbert space

Is it true that, if $$\{u_n\}$$ is a sequence in a Hilbert space $$H$$ that converges weakly to its limit $$u \in H$$ and the sequence satisfies $$\limsup_{n \rightarrow \infty} \|u_n \| \leq \|u\|$$ then $$\lim_{n \rightarrow \infty} \|u_n \| = \|u\|.$$

My attempt:

I wanted to show $$\|u\| \le \liminf_{n \to \infty} \|u_n\|,$$ if this is true, then combined with $$\liminf \|u\| \le \limsup \|u\|$$ the desired result follows.

If the forward implication is true, then to my understanding this would in fact be an "if and only if" relation because the converse implication is trivial.

It is in fact true that $$\|u\| \le \liminf_{n \to \infty} \|u_n\|$$, and so the claim at the top of the question does indeed follow.
Hint to get you started on the proof: $$\|u\|^2 = \langle u,u \rangle = \lim_{n \to \infty} \langle u_n, u \rangle$$
Incidentally, when $$u_n \to u$$ weakly and $$\limsup_{n \to \infty} \|u_n\| \le \|u\|$$, then you can prove even more: in fact $$u_n \to u$$ in norm.
• I did a slightly different approach that is sort of based on what you said: If $e_k$ is an orthonormal basis of $H$, then $$\|u\|^2=\sum_k |\langle u,e_k \rangle|^2=\lim_{n \to \infty} \sum_k |\langle u_n,e_k\rangle|^2 \le \liminf_{n \to \infty} \|u_n\|^2=(\liminf_{n \to \infty}\|u_n\|)^2,$$ where I used Bessel's inequality on $u_n$ and took infimum over $n$ on the right-hand side, then send $n \to \infty$ on both sides. Is this what you intended me to think about? – Cookie Sep 23 '19 at 3:26
• @Cookie: I had in mind something a little simpler. By Cauchy-Schwarz, we have $\langle u_n, u \rangle \le \|u_n\| \|u\|$. Now take the liminf of both sides. – Nate Eldredge Sep 23 '19 at 12:10
• Then I would eventually have to divide both sides by $\|u\|$, which only makes sense if $u$ is nonzero. But I think this is fine because there is a separate trivial argument for the case $u=0$. – Cookie Sep 23 '19 at 17:49
• Also, to prove $u_n \to u$ in norm, I think that means to prove $\|u_n-u\| \to 0$ as $n \to \infty$. So I did $\|u_n-u\|^2=\langle u_n-u,u_n-u \rangle = \|u_n\|^2-2\operatorname{Re}\langle u_n,u\rangle +\|u\|^2 \to \|u\|^2-2\langle u,u\rangle+\|u\|^2=0$ as $n \to \infty$. – Cookie Sep 23 '19 at 17:51