# Weak closure of $\{\sqrt n e_n|n\in \mathbb N\}$ and metrizability of weak topology

Let $\{e_n|n\in \mathbb N\}$ be an orthonormal basis of Hilbert space $\mathcal H$ and put $I = \left\{\sqrt n e_n|n\in \mathbb N\right \}$. Show that $0$ belongs to the weak closure of I but no sequence from $I$ is weakly convergent to $0$. Conclude from this that weak topology on $\mathcal H$ does not satisfy the first axiom of countability and hence is not metrizable.

• What does "n1/2en" mean? Is it supposed to be $\sqrt n e_n$? Is your question the same as this one: math.stackexchange.com/q/42337? Commented Nov 6, 2012 at 4:13
• that power of n is half and en is e sub script n
– math
Commented Nov 6, 2012 at 4:16
• Do you see how the conclusion follows if you can show the first part? The first part is a duplicate of the linked question. Commented Nov 6, 2012 at 4:17
• You should really pick a better title.
– Ted
Commented Nov 6, 2012 at 4:27

• $$0$$ is an element of the weak closure of $$I$$: fix a basic neighborhood of $$0$$ for the weak topology, say $$O:=\bigcap_{j=1}^N\{x,|f_j(x)|<\delta\}$$ where $$\delta>0$$ and $$f_j$$ are linear continuous functionals. We have to show that $$O$$ contains an element $$\sqrt n\cdot e_n$$ for some $$n$$. If not, representing the linear functional $$f_j$$ by the vector $$y_j$$, we would have $$\sum_{j=1}^N\|y_j\|^2=\sum_{j=1}^N\sum_{n=1}^\infty|\langle e_n,y_{j}\rangle|^2=\sum_{n=1}^\infty\sum_{j=1}^N|\langle e_n,y_{j}\rangle|^2\geqslant\sum_{n=1}^\infty\frac{\delta^2}{n}=\infty.$$

• Let $$\{x_n\}$$ a sequence of $$I$$. If there are infinitely many different terms, say $$\{\sqrt k\,e_k,\ k\in A\}$$ where $$A\subset\Bbb N$$ is infinite, write the sequence $$x_k:=\sqrt{n_k}e_{n_k}$$ where $$n_k$$ is an increasing sequence of integers. The sequence $$\{x_k\}$$ is not bounded an so cannot be weakly convergent. If there are only finitely many different terms, we extract a subsequence constant equal to one of them, proving we can't have convergence to $$0$$.

• If there were a decreasing countable basis of neighborhood at $$0$$ , say $$\{V_n,n\in\Bbb N\}$$ (for the weak topology), we would be able for each $$n$$, $$x_{k_n}\in I\cap V_n$$ by the first point (and the definition of the closure). And this sequence would converge weakly to $$0$$, a contradiction by the second item of the list.

• A metric space $$(S,d)$$ satisfies the first axiom of countability, as if $$x\in S$$, the collection $$\{B_d(x,n^{-1}),n\in\Bbb N^*\}$$ would be a countable basis of neighborhoods.

• i am confused with this proof could u make it clear for second and third . what about last point it is totally not fit in the proof.
– math
Commented Nov 12, 2012 at 17:42
• could you help me to join this thing sensibally so it seems good solution
– math
Commented Nov 12, 2012 at 17:47
• I've added details. If you have a problem, don't hesitate. Commented Nov 12, 2012 at 20:26
• Is it really possible to get $n_k\leq Nk$? There seems to be a problem, because maybe the first million entries are taken by a $j_j$ that only appears finitely many times. A fairly straightforward way of dealing with this is to notice that $$\sum_{j=1}^N\|y_j\|^2=\sum_{j=1}^N\sum_{n=1}^\infty|\langle \sqrt n\,e_n,y_{j}\rangle|^2=\sum_{n=1}^\infty\sum_{j=1}^N|\langle \sqrt n\,e_n,y_{j}\rangle|^2\geq\sum_{n=1}^\infty\frac{\delta^2}{n}=\infty,$$ a contradiction. Commented May 12, 2021 at 14:29
• @MartinArgerami Thank you very much for having pointed this out. This is indeed a good way to solve the issue and I am grateful to you for having proposed it. Commented May 13, 2021 at 14:03