# On weak closure again.

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

Thanks!

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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!