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Let $1\le p< \infty$. Show that a sequence $t_k = ({t_{kj}})_{j=1}^{\infty}\in l^p$ converges weakly to 0 iff $||t_k||_p$ is bounded and $\lim_k t_{kj}=0$. I proved that if $t_k$ converges weakly to 0 then we conclude that. I want to prove the reciprocal.

Let's assume that $1<p<\infty$ If I assume that $(t_k)$ it's weakly cauchy I can prove that it's weakly convergento to 0, but I don't know how to prove that. Under that assumption I used the reflexivity of the space $l^p$ ($l^1$ is not reflexive)

If $p=1$ since $l^1$ is not reflexive my arguments are not valid here. I don't really now if the result it's also true here. Please help me!

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You can find a proof of more general result here.

Appply that theorem to the case $p\in(1,+\infty)$ with $S=\{f_j\in (\ell_p)^*:j\in\mathbb{N}\}$, where $$ f_j:\ell_p\to\mathbb{K}: t\mapsto t_j $$

For $p=1$ see this answer, where it was proved that weak convergence is equivalent to strong convergence. It is remains to note that every strongly convergent sequence is bounded and pointwise convergent.

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  • $\begingroup$ Thanks Norbert for the complete answer! But I have one question! why the span of the projections are dense? $\endgroup$
    – Shanks
    Oct 17 '13 at 5:14
  • $\begingroup$ span of projections is the space of finitely supported vectors usually denoted by $c_{00}$. It is dense in $\ell_p$ by the following argument. For a given $\varepsilon$ and $t\in\ell_p$ there exists $N\in\mathbb{N}$ such that $\sum_{j=N+1}^\infty|t_j|^p<\varepsilon^p$. Then consider $y\in c_{00}$ such that $y_j=t_j$ for $j=1,\ldots,N$ and $y_j=0$ otherwise. Then $\Vert t- y\Vert_p<\varepsilon$. $\endgroup$
    – Norbert
    Oct 17 '13 at 6:19
  • $\begingroup$ You are right, thanks! $\endgroup$
    – Shanks
    Oct 17 '13 at 6:32
  • $\begingroup$ But note that for the case $p=1$ the dual of $l^1$ is $l^{\infty}$ and here $c_{00}$ is not dense on $l^{\infty}$ so the span of the projections are not dense on $(l^1)'$ $\endgroup$
    – Shanks
    Oct 17 '13 at 7:26
  • $\begingroup$ @Shanks, you are right. See edits to my answer. $\endgroup$
    – Norbert
    Oct 17 '13 at 9:06

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