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Let $\lbrace e_n \rbrace$ for the standard unit vectors in $l_2$.

I want to show that $0$ is in weak closure of $\lbrace\sqrt{n}e_n\rbrace$ but no subsequence of $\lbrace \sqrt{n}e_n\rbrace$ weakly converges to 0.

For the second assertion, I have the following answer.

If $\lbrace \sqrt{n_k}e_{n_k}\rbrace$ be weakly convergent subsequence, then it must be norm-bounded. However, $\| \sqrt{n_k}e_{n_k}\|=\sqrt{n_k}\to \infty$ as $k\to \infty$ which is a contradiction.

However, for the first assertion, I cannot figure it out that the difference between the condition that 0 lies in weak closure of sequence and the condition that there is a subsequence weakly converges to 0.

Thanks.

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The weak topology is not first countable, so you cannot expect to answer all topological questions about it by considering sequences. Instead, try going back to the definition: the weak topology is the smallest topology that makes all the bounded linear functionals continuous. This characterizes the weakly open sets, which should help. Showing that 0 is in the weak closure is equivalent to showing that every weakly open set that contains 0 must also contain a point of your sequence. –  Nate Eldredge May 31 '11 at 14:52
    
Ok, now I got it. Thanks. –  Forthepiece May 31 '11 at 16:06
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2 Answers

up vote 2 down vote accepted

The hint from Pedersen, Analysis Now: If $x = \sum \alpha_n e_n$ in $l_2$, then there is no $\epsilon > 0$ such that $|(y|x)| > \epsilon$ for every $y = \sqrt{n} e_n$.

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Now, I am done. Thanks. –  Forthepiece May 31 '11 at 16:05
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Problem 28 of Halmos's A Hilbert space problem book says:

The weak topology of an infinite-dimensional Hilbert space is not metrizable.

It comes with a warning, "The shortest proof of this is tricky." Then there is the hint, "Construct a sequence that has a weak cluster point but whose norms tend to $\infty$." Finally, there is the solution, which begins by stating and solving your problem. The example is attributed by Halmos to Shields.

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I agree with Halmos that Shields's sequence $\sqrt n e_n$ is a bit simpler than von Neumann's solution $x_{k,l} = e_k + k e_l$ with $l \gt k$. However, I think von Neumann's example is more straightforward to come up with and analyze: we know from Bessel that $e_k \to 0$ weakly, hence $x_{k,l} \xrightarrow{l\to\infty} e_k$ weakly, so each $e_k$ is in the weak closure and hence so is $0$. On the other hand, a bounded subsequence of $x_{k,l}$ must have $k$ bounded by $K$, say, hence it is separated from the origin by the functional represented by $e_{1} + \cdots + e_{K}$. –  Martin Jan 10 '13 at 11:36
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