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Let $1\leqslant p<q<\infty$. Denote $L(\ell_p)$ the space of bounded operators on $\ell_p$. Let $B_{L(\ell_q)}$ [was $B_{L(\ell_p)}$ ] be the closed unit ball of $L(\ell_q)$ [was $L(\ell_p)$] considered as a subset of $L(\ell_p)$ [was $L(\ell_p)$]. Is it closed in $L(\ell_p)$ [was $L(\ell_p)$]?

EDIT: I made a correction.

EDIT 2: You are right. Please delete my question.

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I 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 $\ell^p$? –  t.b. May 20 '12 at 13:28
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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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Apologies, I meant the opposite inequality. –  MarkNeuer May 20 '12 at 13:21
    
@MarkNeuer: I think that makes it worse. See my edit. –  Nate Eldredge May 20 '12 at 13:31
    
Ah, we're in agreement here :) –  t.b. May 20 '12 at 13:32
    
Sure, can you please delete it? :) –  MarkNeuer May 20 '12 at 13:38
    
@MarkNeuer Why are you so keen on deleting your question? Others might benefit from it if it stays. –  Matt N. May 20 '12 at 16:07
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