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Kirszbraun theorem 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.

Now let's take $H_2$ to be the Euclidean space $\mathbb{R}^n$. My question is: Is there way to explicitly construct this extension? 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.

Some remarks:
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.

2) For $n > 1$, performing the above extension for each component of $f$ results in blowing up the Lipschitz constant by a factor of $\sqrt{n}$.

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Offhand I don't know how constructive this is, but are you aware of the paper of Lang-Pavlovic-Schroeder's springerlink.com/content/5pd0u4yr5frrvbyk ? –  Theo Buehler May 10 '11 at 16:11
    
The key step in the proof of Kirszbraun's theorem involves extending the function to one more point. You write down the conditions on an extension which make the extension have the same Lipschitz constant and show that it is possible to satisfy the conditions. It is easy to make the extension explicit. TBC –  Bill Johnson May 10 '11 at 18:29
    
If the space is separable, you can choose a countable dense set of the complement of the domain of the function and recursively define the extension to include that countable dense set and then extend to the whole space by continuity. The entire proof is then explicit once you have the countable dense set and an ordering on it. –  Bill Johnson May 10 '11 at 18:31
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If the dimension is finite, there is no need to use Hausdorff's maximal principle. Choose a dense countable set in U and extend the map to set of all rational points. The obtained map can be extended to a Lipschitz one on whole space. –  Anton Petrunin May 10 '11 at 23:06
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1 Answer 1

up vote 4 down vote accepted

I like a recent proof by Akopyan and Tarasov:

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.

I could not find this paper in the open web, but there is a copy behind a paywall: http://dx.doi.org/10.1134/S000143460811014X

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.)

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I learned from Akopyan, that the same construction was done earlier (but independently) in Brehm, U., Extensions of distance reducing mappings to piecewise congruent mappings on $R^m$. J. Geom. 16 (1981), no. 2, 187--193 –  Anton Petrunin May 10 '11 at 21:54
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