3
$\begingroup$

Given a real number $c\ge 1$ let us say that a metric space $X$ is a $c$-Lipschitz comp-extensor if each Lipschitz self-map $f:K\to K$ of a compact subset $K\subset X$ extends to a Lipschitz self-map $\bar f:X\to X$ with Lipschitz constant $Lip(\bar f)\le c\cdot Lip(f)$.

It is well-known that for any set $\Gamma$ the Banach space $\ell_\infty(\Gamma)$ of bounded real-valued functions on $\Gamma$ is 1-Lipschitz comp-extensor. By the Kirszbraun Theorem the Hilbert space $\ell_2$ is 1-Lipschitz comp-extensor.

Using the fact that each compact subset of $c_0$ is contained in a cube $\prod_{n\in\omega}[-a_n,a_n]$ with $\lim_{n\to\infty}a_n=0$, it can be shown that the Banach space $c_0$ is 1-Lipschitz comp-extensor.

By a result of Kalton, for any compact metric space $K$ the Banach space $C(K)$ of real-valued continuous functions on $K$ is a $2$-Lipschitz retract of the Banach space $\ell_\infty(K)$. This implies that $C(K)$ is a 2-Lipschitz comp-extensor.

Problem. Let $K$ be a compact metrizable space. Is the Banach space $C(K)$ a $1$-Lipschitz comp-extensor (or at lest a $(1+\varepsilon)$-Lipschitz comp-extensor for some/any positive $\varepsilon<1$)?

Remark. By Johnson's answer to this MO-question, not all Banach spaces are Lipschitz comp-extensors.

$\endgroup$
  • 1
    $\begingroup$ It is very likely the answer is yes, as $C(K)$-spaces (more generally, $L_1$-preduals) are $(1+\varepsilon)$-injective for the class of compact operators. Probably Lindenstrauss' proof can be adjusted for this setting too. $\endgroup$ – Tomek Kania Apr 24 '18 at 7:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.