Suppose the Hilbert space $H$ has a countable (I assume Hilbert?) basis. Let $x \in H$ be such that $\lvert x \rvert \leq 1$. Show that there exists a sequence $\{u_{n}\}$ in $H$ with $\lvert u_{n} \rvert = 1$ for all $n$ such that $\{u_{n}\}$ converges weakly to $x$.
By hypothesis, for the Hilbert basis $\{v_{n}\}$, $$x = \sum_{n = 1}^{\infty} a_{n}v_{n}.$$ I think my issue is with the requirement that $u_{n}$ have unit norm. The only unit sequence I could come up with was $\lvert s_{m}\rvert^{-1}s_{m}$, where $$s_{m} = \sum_{n = 1}^{m} a_{n}v_{n},$$ which apparently converges to $\lvert x \rvert^{-1} x$. This seems to work if $\lvert x \rvert$ is on the unit sphere, but I'm not sure what to do otherwise. Also, I've noticed that I haven't really used the fact that the sequence should converge weakly.