1
$\begingroup$

I have the following question. Let $A$ be a metrically oriented $n$-dimensional subset of $\mathbb{R}^N$ and $f$ a continuous map from $A$ to $\mathbb{R}^M$. We know that $\operatorname{Lip} f < +\infty$ holds $L^N$-almost everywhere, where $\operatorname{Lip} f$ is the local Lipschitz constant of $f$. Can we then continuously extend $f$ to the whole of $\mathbb{R}^N$ in a way that $\operatorname{Lip} f < +\infty$ holds $L^N$-almost everywhere?

$\endgroup$
4
  • $\begingroup$ Of course one can use the Tietze extension theorem to extend f continuously to the whole space. The problem then is that whether it keeps the a.e. finiteness of the Lipf. $\endgroup$ Sep 28, 2012 at 6:10
  • $\begingroup$ What does "metrically oriented" mean? $\:$ What do you mean by "L^n-almost everywhere" $\hspace{0.6 in}$ and "L^N-almost everywhere"? $\;\;$ $\endgroup$
    – user5810
    Sep 28, 2012 at 6:18
  • $\begingroup$ To "Ricky Demer": the term "metrically oriented" is from J.Heinonen and S.Rickman's paper "Geometric branched covers between generalized manifolds. Duke Math. J. 113 (2002)", it basically says that as A itself is a very good metric measurable space with n-dimensional Hausdorff measure (e.g. supports a (1,1)-Poincare inequality, n-Ahlfors regular, n-rectifiable), so Lipf<\infty L^n-a.e. means that as a metric measure space itself, Lipf is finite in Hausdorff n-measure a.e.. When you extend this map f, then f is defined on R^N which the corresponding measure should be L^N. $\endgroup$ Sep 28, 2012 at 6:27
  • 2
    $\begingroup$ @Changyu: Please, use proper TeX syntax; this would make your question more readable. $\endgroup$
    – Misha
    Sep 28, 2012 at 6:55

2 Answers 2

4
$\begingroup$

You only need to assume that $A$ is a closed subset of $\mathbb{R}^N$ and then construct an extension of $f$ so that it is locally Lipschitz outside $A$. Something like what I explained in Can we extend a continuous function with keeping Hausdorff dimension? should work (extending by hand using a Whitney decomposition). Now the extended mapping is locally Lipschitz exactly outside the same exceptional set as the original mapping. One has to be careful with the boundary points: if the original mapping was locally Lipschitz at the boundary, the extension is also (because of the way it is constructed).

Edit: Only now I noticed who was asking the question. You can drop by my office to discuss more, if there are any problems with the extension. :)

$\endgroup$
1
  • 1
    $\begingroup$ +1 just for the last paragraph (the edit) $\endgroup$
    – Yemon Choi
    Sep 29, 2012 at 7:25
2
$\begingroup$

This is not a solution to your problem as I do not know what "metrically oriented" sets are. However, you could try to use Kirszbraun's extension construction and see what it gives in the context of your question:

Kirszbraun's proves that every Lipschitz function $f: A \to {\mathbb R}$ defined on an arbitrary subset of ${\mathbb R}^m$ has a Lipschitz extension to ${\mathbb R}^m$ with the same Lipschitz constant.

M.D. Kirszbraun, Uber die zusammenziehende und Lipschitzsche Transformationen, Fundamenta Math. 22 (1934), p. 77-108.

If you do not read German, you can find here a generalization of Kirszbraun's theorem.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.