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!
Best Answer
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.
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.