Can we extend a a.e. Lipschitz map defined on a closed subset of \bR^N to the whole space such that it is still a.e. Lipschitz - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T10:44:16Z http://mathoverflow.net/feeds/question/108315 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108315/can-we-extend-a-a-e-lipschitz-map-defined-on-a-closed-subset-of-brn-to-the-who Can we extend a a.e. Lipschitz map defined on a closed subset of \bR^N to the whole space such that it is still a.e. Lipschitz Changyu Guo 2012-09-28T06:03:06Z 2012-09-28T22:03:02Z <p>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 $\mathrm{Lip} f &lt; +\infty$ $L^N$-almost everywhere, can we then continuously extends f to the whole $\mathbb{R}^N$ such that $\mathrm{Lip} f &lt; +\infty$ $L^N$-almost everywhere? Here $\mathrm{Lip}$ is the local lipschitz constant of $f$.</p> http://mathoverflow.net/questions/108315/can-we-extend-a-a-e-lipschitz-map-defined-on-a-closed-subset-of-brn-to-the-who/108320#108320 Answer by Misha for Can we extend a a.e. Lipschitz map defined on a closed subset of \bR^N to the whole space such that it is still a.e. Lipschitz Misha 2012-09-28T06:53:03Z 2012-09-28T06:53:03Z <p>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: </p> <p>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.</p> <p>M.D. Kirszbraun, Uber die zusammenziehende und Lipschitzsche Transformationen, Fundamenta Math. 22 (1934), p. 77-108.</p> <p>If you do not read German, you can find <a href="http://www.math.psu.edu/petrunin/papers/akp-papers/lang-schroeder_kirszbraun.pdf" rel="nofollow">here</a> a generalization of Kirszbraun's theorem. </p> http://mathoverflow.net/questions/108315/can-we-extend-a-a-e-lipschitz-map-defined-on-a-closed-subset-of-brn-to-the-who/108323#108323 Answer by Tapio Rajala for Can we extend a a.e. Lipschitz map defined on a closed subset of \bR^N to the whole space such that it is still a.e. Lipschitz Tapio Rajala 2012-09-28T07:36:33Z 2012-09-28T07:42:06Z <p>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 <a href="http://mathoverflow.net/questions/100693" rel="nofollow">http://mathoverflow.net/questions/100693</a> 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).</p> <p>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. :)</p>