[Kirszbraun theorem](https://en.wikipedia.org/wiki/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 $\sqrt{n}$.



Explicit extension of Lipschitz function (Kirszbraun theorem)
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.)


