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Let $X$ be a subspace of $\ell_1^4$ (i.e. $\mathbb{R}^4$ equipped with the $\ell^1$ norm). Can one always extend a linear operator $l:X\rightarrow \ell_1^4$ to $L:\ell_1^4\rightarrow \ell_1^4$ such that $L$ has the same operator norm as $l$?

Thanks!

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    @Arturo, what Bill said is for two spaces, $X$ and $Y$. Here $X$ and $Y$ are the same space. The question really is: what if $X$ is not \ell{\infty}^n$?2010-12-14
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    Dear Arturo, what did you say a couple of minutes ago?2010-12-14
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    @user4730: Something only half right. I said we can assume $||l||=1$, which is true enough (just scale), and something false.2010-12-14
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    Dear Arturo, that is generally a good strategy. I like it.2010-12-14
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    To be simple, concrete, and computable, the space is set to be just $\ell_1^4$.2010-12-14
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    If $l$ is bounded, then I think you can use Hahn-Banach if I'm not mistaken.2011-01-13
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    @user1736: No. The Hahn-Banach theorem applies only to linear forms (functionals), not to operators.2011-03-14
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    I cleaned up the comment thread a bit to remove some orphaned and no-longer useful ones.2011-05-13
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    @Willie: Can't we just close this question? It is both completely uninteresting and unmotivated and pops up every now and again on MO and here e.g. in [this version](http://mathoverflow.net/questions/50600/an-existence-question-on-linear-map). The OP got an answer by Fedja in the comments on MO in one version of this question, which he didn't understand then deleted the question etc. I'm tired of this.2011-05-13
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    @user4730: If you deface the question again, I will lock it.2011-10-31

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