7
$\begingroup$

By Theorem 1.6 in the book "Geometric Nonlinear Functional Analysis" by Benyamini and Lindenstrauss, the Banach space $C[0,1]$ is a Lipschitz retract of the Banach space $\ell_\infty[0,1]$. Unfortunately, the proof does not give any upper bounds on possible Lipschitz constants of the retraction. So, the

Problem. Give some upper bounds on the smallest Lipschitz constant $L$ of a retraction $r:\ell_\infty[0,1]\to C[0,1]$. Is $L\le 20$?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
11
$\begingroup$

Yes, as we have this theorem of Nigel Kalton:

Let $K$ be a compact metric space. Then $C(K)$ is an absolute 2-Lipschitz retract.

Please see [1] for details.

[1] Kalton, Nigel J. "Extending Lipschitz maps into C (K)-spaces." Israel Journal of Mathematics 162.1 (2007): 275-315.

Improve this answer
$\endgroup$
4
  • $\begingroup$ @TarasBanakh, you are most welcome. I remember that I had used this result some time ago. $\endgroup$
    – Tomasz Kania
    Apr 22, 2018 at 16:01
  • $\begingroup$ On the first page of this paper the constant 20 (by Lindenstrauss) is mentioned. It is the same as was asked by OP. May I wonder, is it something special to have Lipschitz constant 20? $\endgroup$
    – Fedor Petrov
    Apr 22, 2018 at 16:04
  • $\begingroup$ @FedorPetrov, I think it simply the estimate that follows from the Lindenstrauss' (non-optimal) proof. Kalton wanted to make a case that his result indeed strengthens the original result. $\endgroup$
    – Tomasz Kania
    Apr 22, 2018 at 16:05
  • $\begingroup$ @FedorPetrov The constant 20 appears in the book of Benyamini and Lindenstrauss, where they construct a uniformly continuous retraction of $\ell_\infty(K)$ onto $C(K)$ which is locally Lipschitz with constant 20 at a neighborhood of $C(K)$ in $\ell_\infty(K)$. But even this local constant 20 cannot be traced to give a reasonable global Lipschitz constant. So the result of Kalton is very nice and exact (2 cannot be further improved). $\endgroup$
    – Taras Banakh
    Apr 22, 2018 at 16:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.