Which Banach spaces are absolute Lipschitz extensors for compacta? 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?
 A: 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. 
