Suppose $X$ is a closed subspace of an $L^{1}$-space and $X$ is isometric to another $L^{1}$-space. Then we know that $X$ is in the range of a contractive projection on the $L^{1}$-space. Is there any way to extend this to other spaces?
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$\begingroup$ Are you asking when a closed linear subspace X of a Banach space admits a linear projector with norm 1? Or more generally, are you interested in non-linear maps? For instance in Hilbert, non-linear context, Kirszbraun theorem provides extensions of Lipschitz maps, keeping the same Lipschitz constants. $\endgroup$– Pietro MajerCommented Jun 10, 2010 at 6:30
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$\begingroup$ If I recall correctly, the result you state is false if we replace "isometric" with "isomorphic", due to a counterexample of Bourgain. Also, which other spaces did you have in mind? I´d prefer to have more evidence of prior thought and reading on the question, rather than (apparent) mere curiosity $\endgroup$– Yemon ChoiCommented Jun 10, 2010 at 10:50
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$\begingroup$ Right you are, Yemon. The isomorphic version is also false for $L_p$ with $1<p<\infty$, but the almost isometric versions are true ($p=1$ by Dor and $p>1$ by Schechtman). $\endgroup$– Bill JohnsonCommented Jun 10, 2010 at 10:55
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The same theorem is true for $L_p$ spaces, $1<p<\infty$. That is, if $X$ is a subspace of $L_p(\mu)$ and $X$ is isometrically isomorphic to $L_p(\nu)$, then $X$ is the range of a contractive projection. See e.g. volume 2 of Lindenstrauss-Tzafriri "Classical Banach spaces". It is also true that if a subspace $X$ of $L_p$ is contractively complemented, then $X$ is isometrically isomorphic to another $L_p$ space.
There is a huge literature on related things. What are you looking for?
answered Jun 10, 2010 at 7:20
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$\begingroup$ @ Professor Johnson. Just for the sake of curiosity. Is the same thing true for the family of their "cousins", namely the Sobolev spaces $W^{k,1}(\mathbb{R}^{n})$ with $k=1,2,...$ and $n\geq2$ , which are not $\mathcal{L}_{1}$- spaces ? Or, more generally, is it true that every Banach space having not GL-l.u.st. contains an isometric, non-complemented copy of itself ? $\endgroup$– AdyCommented Jun 11, 2010 at 11:36
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$\begingroup$ @ Ady: I don't know about the first question. The answer to the second question is "no"-- $\ell_2$ has an equivalent renorming so that the only isometries are plus and minus the identity. This was proved by W. J. Davis IIRC (and was discussed on another MO thread where I got the history wrong and now am not sure of the correct history). $\endgroup$ Commented Jun 12, 2010 at 2:53
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$\begingroup$ @ Professor Johnson. Thank you, but I do not understand the answer. Since I was speaking about into isometries. Also, I said "having not GL-l.u.st." $\endgroup$– AdyCommented Jun 12, 2010 at 5:12
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$\begingroup$ @ Ady, the theorem is that any separable space can be equivalently renormed so that plus and minus the identity are the only isometries, and I think that is for into isometries. Sorry I botched the comment above and that I don't recall the history or who proved the final theorem, but that is in some other MO thread... $\endgroup$ Commented Jun 12, 2010 at 15:35