# 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?

• $\ell_2 \oplus C[0,1]$ gives a negative answer to your question. Map almost isometric $\ell_2^n$ subspaces of $C[0,1]$ linearly isometrically onto subspaces of $\ell_2 \oplus\{0\}$ and restrict to the unit ball to get compact domains. If there is a Lipschitz extension with norm $A_n$, there is by Lindenstrauss non linear paper a linear extension $B_n$ with norm $3\|A_n\|$ (or maybe $\|A_n\|$). The norm of $B_n$ must go to infinity at rate at least $n^{1/2}$. Argue that if extensions of contractions always exist, there must be a bound on the Lipschitz constants of the best extensions. – Bill Johnson Feb 2 '18 at 14:35
• No time now for a detailed answer, but I imagine what I wrote is enough for you. – Bill Johnson Feb 2 '18 at 14:37
• @BillJohnson Thank you for the answer of the Question. You can write it as an answer and then I will accept it. By the way, the existence of the bound for Lipschitz extension (implying from the existence of Lipschitz extensions). Is it written anywhere? I want to cite this result and not prove it myself. – Taras Banakh Feb 2 '18 at 16:07

## 1 Answer

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.

• Hi Professor Johnson. I would like to know more about the construction of $E_n$ and $F_n$ in your answer. May i know any reference of it? Thanks. – Idonknow Apr 15 '18 at 11:42