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A subset $F$ of a metric space $M$ is called a Lipschitz retract of $M$ if there is a Lipschitz map from $M$ onto $F$ which coincides with the identity on $F$. A metric space which is a Lipschitz retract of any metric space containing it is called an absolute Lipschitz retract.

The typical example of separable Banach space which is an absolute Lipschitz retract is $c_0$. More generally, $C_u(M)$ (the space of real-valued, uniformly continuous bounded functions on the metric space $M$ with the sup norm) is an absolute Lipschitz retract (see e.g. Benyamini and Lindenstrauss' book). If $M$ is compact, then $C_u(M)=C(M)$ is separable, but if it is not compact it is not separable since it contains $\ell_\infty$.

Question 1. What other separable Banach spaces which are absolute Lipschitz retracts do we have, except for $C(K)$ spaces with $K$ compact metric?

This question can be considered quite "wiki". My second related question is more specific.

Question 2. Suppose that $X$ is a Banach space which is an absolute Lipschiz retract. Does $X$ contain a copy of $c_0$?

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  • $\begingroup$ Thank you @BillJohnson for the edit, and for the remarks that unfortunately vanished with the deleted answer. I never thought about whether or not all Banach spaces which are a.l.r. should be $\mathcal{L}_\infty$-spaces, it is an interesting question! Do you remember the name of the user who deleted his own answer? $\endgroup$ – Pedro Kaufmann Mar 17 '14 at 10:21
  • $\begingroup$ Nevermind my last question in the comment, I've found him. $\endgroup$ – Pedro Kaufmann Mar 17 '14 at 10:51
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This does not answer the original questions, just something from the comments: if a Banach space is an absolute Lipschitz retract then it is an $\mathcal{L}_\infty$-space.

Let $X$ be a Banach space which is an absolute Lipschitz retract. Let $Y \subset Z$ be Banach spaces and $t : Y \to X$ a bounded linear map. Since $X$ is an absolute Lipschitz retract, there is a Lipschitz extension $\tau : Z \to X$ of $t$. By Theorem 7.2 in the Benyamini-Lindenstrauss book, there is a bounded linear map $T : Z \to X^{**}$ that coincides with $\tau$ on $Y$, so then $T$ is in fact a linear extension of $t$.

Now, by a result of Lindenstrauss (Theorem 2.1 in Extension of compact operators. Mem. Amer. Math. Soc. No. 48 1964), this implies that $X^{**}$ is an injective space. But then both $X^{**}$ and $X$ are $\mathcal{L}_\infty$-spaces (Theorem F.2(v) in Benyamini-Lindenstrauss).

Note: this is an adaptation of the arguments at the beginning of the proof of Proposition 3.5 in Avilés, Antonio; Cabello Sánchez, Félix; Castillo, Jesús M. F.; González, Manuel; Moreno, Yolanda On separably injective Banach spaces. Adv. Math. 234 (2013), 192–216.

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    $\begingroup$ AFAIK, it is open whether every absolute Lipschitz retract is isomorphic to a $C(K)$ space, and also open whether every (say, separable) $\mathcal{L}_\infty$ space is an absolute Lipschitz retract. $\endgroup$ – Bill Johnson Mar 17 '14 at 14:07
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    $\begingroup$ Thanks @Alejandro. Another way to see what you just proved for the separable case is the following. Let $X$ be a separable Banach space which is an ALR. $X$ is a subspace of $C[0,1]$, and there is a Lipschitz retract from $C[0,1]$ onto $X$. By Corollary 7.3 in Benyamini-Lindenstrauss, there is a linear projection from $C[0,1]^{**}$ onto $X^{**}$. $C[0,1]^{**}$ is an $\mathcal{L}_\infty$-space since this property is preserved by taking biduals, and since it is preserved by taking complemented subspaces, so is $X^{**}$. By the principle of local reflexivity, $X$ is a $\mathcal{L}_\infty$ space. $\endgroup$ – Pedro Kaufmann Mar 18 '14 at 10:55

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