A metric space $X$ is defined to be an *absolute Lipschitz extensor for compacta* if each Lipschitz map $f:K\to K$ defined on a compact subset $K\subset X$ extends to a Lipschitz map $\bar f: X\to X$.


**Question.** Is each Banach space an absolute Lipschitz extensors for compacta?


I admit that the answer to this question is negative. In this case the Question can be refined to


**Problem.** Which Banach spaces are absolute Lipschitz extensors for compacta?


**Remark.** The class of absolute Lipschitz extensors for compacta includes all absolute Lipschitz retracts, so it includes all Hilbert spaces and all Banach spaces $C\_u(M)$ of bounded continuous functions on a metric space $M$. What about uniformly rotund Banach spaces (are they absolute Lipschitz extensors for compacta)? Are the Banach spaces $L\_p(\mu)$ absolute Lipschitz extensors for compacta?



Which Banach spaces are absolute Lipschitz extensors for compacta?
Here is one way you can prove that a space $X$ is not an absolute extensor for compacta: Find sequences $(E\_n)$ and $(F\_n)$ of finite dimensional subspaces of $X$ and a constant $C$ so that for every $n$ there is a linear isomorphism $T\_n$ from $E\_n$ onto $F\_n$ s.t. $\|T\_n\|=1$, $\|T\_n^{-1}\|\le C$, each $F\_n$ is the range of a projection of norm at most $C$, but the $E\_n$ are not uniformly complemented. If the restriction of $T\_n $ to the unit ball $B\_{E\_n}$ of $E\_n$ has a $D$ Lipschitz extension from $B\_X$ into $X$, then it has a $CD$-Lipschitz extension into $F\_n$, and the positively homogenous extension of this mapping is a $3CD$-Lipschitz extension of $T\_n$ to a mapping from $X$ into $F\_n$. By Lindenstrauss' 1964 paper on non linear projections (see the book of Benyamini-Lindenstrauss), there is then a linear extension $S\_n:X \to F\_n$ with $\|S\_n\|\le 3DC$ so that $T\_n^{-1}S\_n$ is a projection from $X$ onto $E\_n$ having norm at most $3DC^2$. So if such an X is an absolute extensor for compacta, there is no uniform Lipschitz bound on extensions of non expansive linear mappings from finite dimensional subspaces. 


It remains to show that if a space $X$ is an absolute extensor for compacta, then there is a uniform bound on extensions of non expansive linear mappings defined on finite dimensional subspaces. Suppose there is no such uniformity for $X$. Notice that then if $Y$ is a finite codimensional subspace of $X$, then there also is no uniformity for extensions of non expansive linear mappings defined on unit balls of finite dimensional subspaces of $Y$ into $X$. Now use the Mazur technique for constructing basic sequences to build a finite dimensional Schauder decomposition $(E\_n)$ for some subspace of $X$ and non expansive linear mappings $f\_n: E\_n \to X$ s.t. any extension of $f\_n$ to $X$ has Lipschitz constant at least $n $. Let $K = \cup\_n n^{-1} B\_{E\_n}$ and define $F:K\to X$ by $f(x) = f\_n(x)$ if $x\in n^{-1} E\_n$. Then $K$ is compact and $f$ is Lipschitz and $f$ has no Lipschitz extension to $X$. 


There are many spaces that contain sequences $(E\_n)$ and $(F\_n)$ of finite dimensional subspaces for which there is a constant $C$ so that for every $n$ there is a linear isomorphism $T\_n$ from $E\_n$ onto $F\_n$ s.t. $\|T\_n\|=1$, $\|T\_n^{-1}\|\le C$, each $F\_n$ is the range of a projection of norm at most $C$, but the $E\_n$ are not uniformly complemented. 
If $X$ is $ \ell\_p$ or $L\_p$, $1\le p \not=2 <\infty$, then $E\_n$ can be taken to be uniformly isomorphic to $\ell\_p^n$. If $X$ is super reflexive (or even just has non trivial type) but does not have type $p$ for some $p<2$, then $E\_n$ can be taken to be uniformly isomorphic to $\ell\_2^n$. These results are fairly deep, BTW, but well known to researchers in Banach space theory. 


