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I know the following result is true in the case of strong convergence. But I don't know whether it is true in the case of weak convergence also. Let $p>1$. Suppose that each $x_n$ is a non negative sequence such that $\|x_n\|_p=1$ and $\stackrel{w}{x_{n}\rightarrow x}$ in $\ell^p$. Is it true then that $\stackrel{w}{x_n^p\rightarrow x^p}$ in $\ell^1$.

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If $x$ is an element of $\ell^p$, then what is $x^p$? Take the $p$th power of each entry? That's why you said "non-negative"? –  Gerald Edgar Nov 1 '10 at 16:40
    
Didn't I answer to some close question of yours on this matter? I can't find it though... It would help me to recall your problem. –  Pietro Majer Nov 1 '10 at 17:49
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May I ask you why you deleted your previous question with my answer? –  Pietro Majer Nov 1 '10 at 18:05
    
OK, no problem; it's your choice to delete the answers you got. In my opinion, in this case, had you leaved it, you could have provided this further question with more insight and motivation, making it easier for people to answer. (Also, I would have been glad to see other people's comments: I myself would like to profit of other people's knowledge). –  Pietro Majer Nov 2 '10 at 10:33

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up vote 5 down vote accepted

No. Consider the unit vector basis of $\ell_2$.

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So $x_n$ converges weakly to zero in $\ell_2$, but $x_n^2=x_n$ does not converge at all in $\ell_1$. –  Gerald Edgar Nov 1 '10 at 16:42

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