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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.

2 Answers 2

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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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    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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    Is$n$'t that a little bit over$k$ill?2011-10-24