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I have the following question: If $A$ is a metrically oriented $n$-dimensional subset of $\mathbb{R}^N$ and $f$ is a continuous map from $A$ to $\mathbb{R}^M$ . We know that $\mathrm{Lip} f < +\infty$ $L^N$-almost everywhere, can we then continuously extends f to the whole $\mathbb{R}^N$ such that $\mathrm{Lip} f < +\infty$ $L^N$-almost everywhere? Here $\mathrm{Lip}$ is the local lipschitz constant of $f$.

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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. –  Changyu Guo Sep 28 '12 at 6:10
    
What does "metrically oriented" mean? $\:$ What do you mean by "L^n-almost everywhere" $\hspace{0.6 in}$ and "L^N-almost everywhere"? $\;\;$ –  Ricky Demer Sep 28 '12 at 6:18
    
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. –  Changyu Guo Sep 28 '12 at 6:27
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@Changyu: Please, use proper TeX syntax; this would make your question more readable. –  Misha Sep 28 '12 at 6:55
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2 Answers

up vote 4 down vote accepted

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. :)

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+1 just for the last paragraph (the edit) –  Yemon Choi Sep 29 '12 at 7:25
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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.

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Thanks. I will take a look at the paper. –  Changyu Guo Oct 1 '12 at 8:15
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