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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.

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    You should really pick a better title.2012-11-06

1 Answers 1

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  • $0$ 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 $f_j$ by $y_j$, we would have $\sqrt n|\langle y_{j_n},e_n\rangle|\geqslant\delta$ for $j_n\in\{1,\dots,N\}$. In particular, there is $A\subset \Bbb N$ infinite and $j\in \{1,\dots,N\}$ such that $\sqrt n\cdot |\langle y_j,e_n\rangle|\geqslant \delta$ for all $n\in A$. Writing it as a sequence $n_k$ of integers and noticing we can choose $n_k\leq Nk$, we have $|\langle y_j,e_{n_k}\rangle|\geqslant \frac{\delta}{\sqrt{kN}}$, a contradiction, as $\sum_k|\langle y_j,e_{n_k}\rangle|^2$ should be finite.

  • Let $\{x_n\}$ a sequence of $I$. If there are infinitely many different terms, say $\{k\sqrt e_k,k\in A\}$ 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.

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    I've added details. If you have a problem, don't hesitate.2012-11-12