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 $\sqrt{n}$.
Best Answer
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: https://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.)