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 $\sqrt{n}$.

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 $\sqrt{n}$.