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.