Let $H$ be an infinite dimensional Hilbert space.

It is well known that the weak-closure of the unit ball is the unit sphere. But I want to prove it as basicaly as possible, using the weakly-sequential deffinition of closure, by actually finding a convering sequence of members of B.

It is clear to me that $\overline S \subset B$ since $B$ is strong-closed and convex.

For the hard and interesting part, I figured out that I could pick an orthonormal basis for $H$,$\{e_a\}_{a\in A}$, and given a $x\in B$ write it as $x=\sum_{n=1}^\infty \langle x,e_n\rangle e_n$ for $(e_n)_{n=1}^{\infty} \subset \{e_a\}_{a\in A}$.

Couldn't go any further.

Please guide me a bit, and let me know if you think that what I'm trying to do is possible. A nice hint would be welcomed :)


  • $\begingroup$ Do you have H.Brezis book? This is Example 1 on page 59, in which you could find a Basic prove. $\endgroup$ – spatially Nov 21 '14 at 0:54
  • $\begingroup$ @wisher: Thanks. I did check this book, and it doesn't seem as basic nor as explicit as I'm looking for. To be more precise, I didn't like the use of linear functionals, and rather like a proof using the sequencial definition of weak-convergence. $\endgroup$ – user188400 Nov 21 '14 at 1:01
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    $\begingroup$ The weak closure of the (open or closed) unit ball is the closed unit ball. Not the unit sphere. $\endgroup$ – daw Nov 21 '14 at 10:23

Fix $x\in H$ with $\|x\|\leq1$. Now define $$ x_n=\sum_{k=1}^n\langle x,e_n\rangle\,e_n\ +(1-\sum_{k=1}^n|\langle x,e_n\rangle|^2)^{1/2}e_{n+1}. $$ Then $\|x_n\|=1$ for all $n$.

For any $y\in H$, $$ \langle x-x_n,y\rangle=[\langle x,e_n\rangle-(1-\sum_{k=1}^n|\langle x,e_n\rangle|^2)^{1/2}]\,\bar y_n+\sum_{k=n+1}^\infty \langle x,e_n\rangle\bar y_n\leq2\left(\sum_{k=n}^\infty|y_n|^2\right)^{1/2}\to0 $$

  • $\begingroup$ Thanks @Martin. I understood the building of these $x_n$, but had trouble with proving that it weakly-converges to x. First of all, did you assume here that $(e_n)$ acts like the standart basis for $\mathbb R^\infty$? It seems that way but I can't understand why would it. Also, the last inequallity wasn't clear, but I guess it is a use of "cauchy-schwarz"? Thanks! $\endgroup$ – user188400 Nov 21 '14 at 14:41
  • $\begingroup$ I'm using $e_n$ to denote an orthonormal basis, exactly as you did in your question. The last inequality is indeed Cauchy-Schwartz, coupled with the fact that $\|x\|\leq1$. $\endgroup$ – Martin Argerami Nov 21 '14 at 14:56
  • $\begingroup$ I wasn't very careful with the estimate, but the point is that we are doing the scalar product of a tail of $y$ with another vector, so it has to go to zero as $n\to\infty$. $\endgroup$ – Martin Argerami Nov 21 '14 at 15:06
  • $\begingroup$ I made a small edit, I was using $x_n$ for both the vectors in the sequence and for the entries of $x$. Corrected now. $\endgroup$ – Martin Argerami Nov 21 '14 at 16:12
  1. Let $\{e_n\}$ be an orthonormal sequence. Show that $e_n \to 0$ weakly. (Hint: Bessel's inequality says $\sum_{n=1}^\infty |\langle y, e_n \rangle|^2 \le \|y\|^2$.) This is a very useful example to keep in mind when thinking about the weak topology.

  2. Show that if $\{a_n\}$ is any bounded sequence of scalars, then $a_n e_n \to 0$ weakly.

  3. Fix $x \in B$. For each $n$, find a scalar $a_n$ such that $x + a_n e_n \in S$. Verify that $\{a_n\}$ is bounded.

  4. Conclude.


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