Let $1\leqslant p EDIT: I made a correction. EDIT 2: You are right. Please delete my question.
Operators from $\ell_p$ to $\ell_q$
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$\begingroup$
functional-analysis
banach-spaces
operator-theory
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0I don't understand how your inclusions are supposed to work, maybe you want to expand on that? We have $\ell^p \subset \ell^q$. Now take $x \in \ell^{q} \smallsetminus \ell^p$ and take a rank one operator on $\ell^q$ of norm one projecting on the span of $x$. How does that give an operator on $\ell^p$ with values in $\ell^p$? – 2012-05-20
1 Answers
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$B_{L(\ell^p)}$ is not even contained in $L(\ell^q)$.
Take $p=1$, $q=2$, and define $T : \ell^1 \to \ell^1$ by $$Tx = \left( \sum_{i=1}^\infty x(i), 0, 0, \dots\right).$$ $T$ is certainly a bounded operator on $\ell^1$, with operator norm 1, but it does not have a continuous extension to $\ell^2$. Let $x_n = (1, \frac{1}{2}, \dots, \frac{1}{n}, 0, 0, \dots)$. Then $x_n$ converges in $\ell^2$, but $\|T x_n\|_{\ell^2} \to \infty$.
Edit: The question was edited to reverse the desired inclusion, but this doesn't hold either, for even more trivial reasons. A bounded operator on $\ell^q$ may map some elements of $\ell^p$ to elements of $\ell^q \setminus \ell^p$.
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0Apologies, I meant the opposite inequality. – 2012-05-20
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0@MarkNeuer: I think that makes it worse. See my edit. – 2012-05-20
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0Ah, we're in agreement here :) – 2012-05-20
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0Sure, can you please delete it? :) – 2012-05-20
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0@MarkNeuer Why are you so keen on deleting your question? Others might benefit from it if it stays. – 2012-05-20