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I worked in this problem too:

On the weak closure

But after that I could not think of anything that can help me about the $l_p$ case. I mean, $\{ n^{1/p} e_n \}$ has $0$ as a weak accumulation point but no subsequence of this set weakly converges to $0$ (with $p \in [1,\infty)$) ?

I tried to use something about my $l_2$ proof but it uses strongly facts about Hilbert spaces.

For example, to prove that $0$ is an accumulation point. Given a weak neighborhood $W$ of $0$ and a natural $n_0$ if I suppose that, for all $n \geq n_0$, is true that $\sqrt{n}e_n \notin W$ I could find an absurd using the definition of $W$. But I used the fact that $\{ e_n \}$ is a hilbert basis.

So, do you have any hint about the $l_p$ case?

obs: I'm still studying english, sorry about my errors =p


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I believe it's more natural to consider the sequence $\{ n^{1/q}e_n\}$, where ${1\over p}+{1\over q}=1$.

In spirit, showing that $0$ is in the weak closure of this set is exactly the same as in the case for a Hilbert space:

Recall that the dual of $\ell_p$ is $\ell_q$; and given $y=(y_1,y_2,\ldots)$ in $\ell_q$, its action on $x=(x_1,x_2,\ldots)$ in $\ell_p$ is $$ y(x)=\sum_{i=1}^\infty y_ix_i. $$

A weak nhood of $0$ in $\ell_p$ has the form $$ O=\{ x\in\ell_p\mid |x_i^*(x)|<\epsilon, 1\le i\le n\} $$ for some positive integer $n$, $\epsilon>0$ and elements $x_1^*$, $x_2^*$, $\ldots\,$, $x_n^*$ in $\ell_q$.

Now suppose such an $O$ is given and that no element $m^{1/q}e_m$ is in $O$. Then for each $m$, it would follow that $$ \sum_{i=1}^n m |x_i^*(e_m)|^q= \sum_{i=1}^n |x_i^*(m^{1/q}e_m)|^q\ge \max_{1\le i\le n} |x_i^*(m^{1/q}e_m)|^q\ge \epsilon^q. $$

From this, we would then have $$ \sum_{m=1}^\infty\sum_{i=1}^n|x_i^*(e_m)|^q \ge\sum_{m=1}^\infty {\epsilon^q\over m}=\infty. $$ But $$ \sum_{m=1}^\infty\sum_{i=1}^n |x_i^*(e_m)|^q = \sum_{i=1}^n \sum_{m=1}^\infty|x_i^*(e_m)|^q = \sum_{i=1}^n\Vert x_i^*\Vert_q^q<\infty; $$ and thus the supposition that no element $m^{1/q}e_m$ is in $O$ cannot hold.

To show that no subsequence of $\{ n^{1/q}e_n\}$ converges to $0$, use the fact that weakly convergent sequences are norm bounded.

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I should've thought about $\{ n^{1/q} e_n \}$ instead of $\{ n^{1/p} e_n \}$. Thanks!

I think I can see two ways to prove the second statement now. I'm at work right now, later i'm gonna write it and post it here!

I'd appreciate your suggestions about my solution.

Thanks again!

Best, Thiago.

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