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Does there exist a sequence $(x_n)$ belonging to $\ell_1\cap\ell_2$ which converges in one but not the other? $(x_n)$ is of course a sequence in these spaces, so it's a sequence of sequences.

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Try $x_n(k)=\frac1n$ if $k\leqslant n$ and $0$ otherwise. Then $\|x_n\|_2=\frac1{\sqrt{n}}$ hence $x_n\to0$ in $\ell_2$ but $\|x_n\|_1=1$ hence $(x_n)$ does not converge in $\ell_1$. This proves that convergence in $\ell_2$ does not imply convergence in $\ell_1$.

On the other hand, if $x$ and $y$ are sequences such that $\|x-y\|_1\leqslant1$ then $|x(k)-y(k)|\leqslant1$ for every $k$ hence $|x(k)-y(k)|^2\leqslant|x(k)-y(k)|$ for every $k$ and $\|x-y\|_2^2\le\|x-y\|_1$. This proves that convergence in $\ell_1$ implies convergence in $\ell_2$ (to the same limit).

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    You're just too fast :)2011-10-23
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    The "identity" map is continuous from $\ell_1$ norm to $\ell_2$ norm, so there is no counterexample the other way.2011-10-23
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    @GEdgar, unless I am missing something, asserting this continuity to prove the result is a circular reasoning since it is what is asked, is it not? Rather, note that if $\|x-y\|_1\leqslant1$ then $\|x-y\|_2^2\le\|x-y\|_1$.2011-10-23
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    @t.b. Sorry about that... :-) But I note you proposed a different example so you might leave your answer, if you ask me.2011-10-23
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    No problem :) Thanks, it's not as if I'm especially proud of that example. If you want to incorporate it into your answer, feel free to do that. I removed my answer because it gave much more away than your initial response and I didn't want to undermine the pedagogical idea of leaving the possibility open (if there was one). *off-topic* I'm more than jealous of your celebrated W, however, I'd have written `\mathsf{W}` which gives $\mathsf{W}$ (had I had the idea and had I dared). Anyway, congratulations!2011-10-23
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    @t.b. Thanks. Using mathsf is a good idea, I will apply it.2011-10-23
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    Another question: Why does it follow from the fact that $||x_n||_1=1$ that $(x_n)$ diverges in $\ell_1$?2011-10-24
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    FPP, it does not follow right away, you are right that a (small) supplementary step is needed. Here it is. Assume that $(x_n)$ converges in $\ell_1$. Then $(x_n)$ converges in $\ell_2$ to the same limit hence $(x_n)$ converges to the sequence $0$ in $\ell_1$. But $\|x_n-0\|_1$ does not go to zero. This is absurd, hence the hypothesis was wrong, which proves that $(x_n)$ diverges in $\ell_1$.2011-10-24
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I would refer you to the interpolation theorems, such as that of Marcinkiewicz. Look in Cora Sadowsky's wonderful book on Harmonic Analysis. Also there are works by Stein and Weiss. There is a whole machinery on questions of this ilk.

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    Isn't that a little bit overkill?2011-10-24