Weak Convergence in Infinite Hilbert space Let $H$ be an infinite Hilbert space.
Show: For all $x \in H$ with $\|x\|\leq1$, there exists a sequence $(u_n)$ in $H$ with $\| u_n\|=1 $ such that $u_n \rightharpoonup x$.
My attempt:
Since $H$ is infinite, there exists a countable subspace $K$ with $x\in K$.
By Gram-Schmidt, we can find a orthonormal basis $(y_n)$ for $K$.
Hence, $x=\sum_{k=1}^\infty a_k y_k$ for some $a_k \in \mathbb{F}$
Let $u_n= \frac {\sum_{k=1}^n a_k y_k}{\|\sum_{k=1}^n a_k y_k \|}$.
Then $\| u_n\|=1$.
Hence, we are done.
Could someone please check my proof, and let me know if it makes sense?
If not, could you please let me know where it went wrong?
Thanks!
 A: Your problem actually holds generally in Banach spaces that are not Schur spaces—Schur spaces are spaces for which weakly converging sequences converges in norm. ((By well-known results due to James and Rosenthal, we know that infinite-dimensional Schur spaces contain a copy of $\ell^1$, hence they are not reflexive; in particular, Hilbert spaces are not Schur spaces)). I will therefore state your problem in its entirety and prove it in the affirmative.
First, trivially, if $x\in S_X$ (that is, $x$ lies in the unit sphere), then we only need set $x_n:=x$ and we are done; hence suppose  $x\in B_X$ (that is, $x$ lies in the open unit ball). We have the following theorem.

THEOREM: Let $X$ be a Banach space that is not a Schur space and let $x\in B_X$. Then there exists $\{x_n\}\subset S_X$ such that $x_n\rightharpoonup x$.

The Special Case $x=0$:
Since $X$ is not a Schur space, it is necessarily infinite dimensional and has a weakly converging sequence, say $u_n\rightharpoonup u$ but $u_n\not\to u$. Without loss of generality, assume $u_n\ne u$ for all $n$, then define $$ x_n:=\frac{u_n-u}{\|u_n-u\|}\,.$$
Clearly $x_n\in S_X$ and $x_n\rightharpoonup 0$, and we’re done.
The General Case:
Now suppose $x\in B_X$, and thanks to the Special Case, let $y_n\in S_X$ and $y_n\rightharpoonup 0$. Define $$\alpha_n:=\sup\{\alpha>0:\alpha y_n\in B_X-x\}\,.$$
Observe that since $x\notin S_X$, then $\alpha_n>0$ and because $y_n\in S_X$ then $\alpha_n\le 2$ for all $n$. Now define $$x_n:=\alpha_n y_n +x$$ and note that by definition of $\alpha_n $ we must necessarily have $x_n\in S_X$; however, $\alpha_n$ is bounded and $y_n\rightharpoonup 0$, thus we have $x_n\rightharpoonup x$, as desired.
