explicit extention of Lipschitz function (Kirszbraun theorem) - MathOverflow most recent 30 from http://mathoverflow.net2013-06-19T04:10:50Zhttp://mathoverflow.net/feeds/question/64511http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/64511/explicit-extention-of-lipschitz-function-kirszbraun-theoremexplicit extention of Lipschitz function (Kirszbraun theorem)mr.gondolier2011-05-10T15:49:42Z2011-05-10T17:30:06Z
<p><a href="http://en.wikipedia.org/wiki/Kirszbraun_theorem" rel="nofollow">Kirszbraun theorem</a> states that if $U$ is a subset of some Hilbert space $H_1$, and $H_2$ is another Hilbert space, and $f : U \to H_2$ is a Lipschitz-continuous map, then $f$ can be extended to a Lipschitz function on the whole space $H_1$ with the same Lipschitz constant. </p>
<p>Now let's take $H_2$ to be the Euclidean space $\mathbb{R}^n$. <strong>My question is: Is there way to explicitly construct this extension?</strong> Note that the standard proof (e.g. see Federer's geometric measure theory book or Schwartz's nonlinear functional analysis book) is an existence proof, which uses Hausdorff's maximal principle. </p>
<p>Some remarks:<br>
1) For $n = 1$, the extension can be constructed explicitly, which works even if $H_1$ is only a metric space (with metric $d$): $\tilde{f}(x) = \inf_{y \in U} \{ f(y) + {\rm Lip}(f) d(x,y) \}$. See for example Mattila's book p. 100.</p>
<p>2) For $n > 1$, performing the above extension for each component of $f$ results in blowing up the Lipschitz constant by a factor of $\sqrt{n}$.</p>
http://mathoverflow.net/questions/64511/explicit-extention-of-lipschitz-function-kirszbraun-theorem/64515#64515Answer by Sergei Ivanov for explicit extention of Lipschitz function (Kirszbraun theorem)Sergei Ivanov2011-05-10T17:30:06Z2011-05-10T17:30:06Z<p>I like a recent proof by Akopyan and Tarasov:</p>
<p>A. V. Akopyan, A. S. Tarasov, "A constructive proof of Kirszbraun's
theorem"(Russian), Mat. Zametki 84 (2008), no. 5, 781--784;
translation in Math. Notes 84 (2008), no. 5-6, 725–728; MR2500644.</p>
<p>I could not find this paper in the open web, but there is a copy behind a
paywall: <a href="http://dx.doi.org/10.1134/S000143460811014X" rel="nofollow">http://dx.doi.org/10.1134/S000143460811014X</a></p>
<p>What they do: if $U\subset\mathbb R^n$ is a finite set and $f:U\to\mathbb R^n$ is 1-Lipschitz, then they construct a piecewise-linear piecewise-isometric (and hence 1-Lipshitz) extension of $f$ to the whole space. The construction is explicit, but some combinatorics is involved, so I'm not sure how it works for an infinite $U$. (I haven't read the paper but learned the proof from a seminar talk by one of the authors.)</p>