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.

  • $\begingroup$ 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? $\endgroup$ Commented Nov 6, 2012 at 4:13
  • $\begingroup$ that power of n is half and en is e sub script n $\endgroup$
    – math
    Commented Nov 6, 2012 at 4:16
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    $\begingroup$ Do you see how the conclusion follows if you can show the first part? The first part is a duplicate of the linked question. $\endgroup$ Commented Nov 6, 2012 at 4:17
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    $\begingroup$ You should really pick a better title. $\endgroup$
    – Ted
    Commented Nov 6, 2012 at 4:27

1 Answer 1

  • $0$ 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.

  • $\begingroup$ 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. $\endgroup$
    – math
    Commented Nov 12, 2012 at 17:42
  • $\begingroup$ could you help me to join this thing sensibally so it seems good solution $\endgroup$
    – math
    Commented Nov 12, 2012 at 17:47
  • $\begingroup$ I've added details. If you have a problem, don't hesitate. $\endgroup$ Commented Nov 12, 2012 at 20:26
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    $\begingroup$ 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. $\endgroup$ Commented May 12, 2021 at 14:29
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    $\begingroup$ @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. $\endgroup$ Commented May 13, 2021 at 14:03

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