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If $\Omega$ is a cube in $\mathbb{R}^n$ and $f\in C^1(\overline\Omega)$. By reflection one can extend such a function to all of $\mathbb{R}^n$ and the extenstion is in $C^1(\mathbb{R}^n)$. If $\Omega$ is a polygon, has piecewise $C^1$ boundary (so edges and corneres are not to wild) or is a convex set this still seems to be possible. Can this be extended to arbitrary Lipschitz domains?

Are there examples and or references for these cases (starting from polygons)?

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    $\begingroup$ How exactly do you define $C^1(\overline \Omega)$? A standard tool in this business is the Whitney extension theorem, e.g., Theorem 2.1 here. $\endgroup$
    – user127096
    Commented Mar 30, 2014 at 22:07
  • $\begingroup$ $f$ and all its partial derivatives are continuous up to the boundary. So my problem when it comes to Whitney's theorem is: Can I guarantee (with my assumption) that the limit (in the theorem) has this uniformity property? If the domain is $C^1$, I will obtain this. But for mere Lipschitz boundary this is not so clear, isn't it? Otherwise, one could always extend any $C^1$-function from Lipschitz domains which sounds a bit odd. You also might check the comment of Giuseppe Negro here math.stackexchange.com/questions/723686/… $\endgroup$ Commented Apr 1, 2014 at 17:50
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    $\begingroup$ @Thibaut Dumont There are different kinds of reflection, besides the one you have in mind. Some preserve some smoothness. $\endgroup$
    – user127096
    Commented Apr 6, 2014 at 4:34
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    $\begingroup$ You need arbitrarily close points in the domain to be connected by arbitrarily short curves. If this is the case, then the requirements of Whitney's theorem will be satisfied. $\endgroup$ Commented Apr 6, 2014 at 13:51
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    $\begingroup$ @Quickbeam2k1: Sorry, I made a mistake there, and what you state is exactly what you need. The conditions for Whitney's theorem will be satisfied as long as there is a $C > 0$ such that any two points $x,y$ in the domain are joined (in the domain) by a curve of length no more than $C\lvert x-y\rvert$. Just requiring that the length is arbitrarily small for arbitrarily small $\lvert x-y\rvert$ is not enough. $\endgroup$ Commented Apr 7, 2014 at 0:12

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As George Lowther pointed out, such an extension is possible for any quasiconvex domains (in particular, for any Lipschitz domain). This is the main result in a short paper by Whitney from 1934:

Whitney, Hassler. Functions differentiable on the boundaries of regions. Ann. of Math. (2) 35 (1934), no. 3, 482–485.

whitney

Property P is what we now call quasiconvexity:

quasiconvexity

This result, and many later developments, are presented in section 2.5 of the book

Brudnyi, Alexander; Brudnyi, Yuri. Methods of geometric analysis in extension and trace problems. Volume 1. Monographs in Mathematics, 102. Birkhäuser/Springer Basel AG, Basel, 2012.

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  • $\begingroup$ thank you so much, is there a way to award George Lowther and you some extra reputation? $\endgroup$ Commented Apr 8, 2014 at 5:19
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    $\begingroup$ @Quick Don't worry about it. If you feel like doing something about this thread, please change the title into something less likely to mislead casual readers into thinking the result is false. $\endgroup$
    – user127096
    Commented Apr 8, 2014 at 5:33
  • $\begingroup$ corrected it :) $\endgroup$ Commented Apr 8, 2014 at 5:57
  • $\begingroup$ Just for clearification: In the article and also in the book the class $C^m$ is merely the class of functions such that the $m^{th}$-partial derivatives are considered (i.e. a "homogeneous" semi norm is taken into account). How, e.g. in the case $m=1$ can one return to the full case? Or are the extensions for $m=0$ and $m=1$ "compatible"? $\endgroup$ Commented Apr 8, 2014 at 7:40
  • $\begingroup$ @Quick. Continuity of partial derivatives implies the function itself is continuous. So, continuity of m-th derivatives implies continuity of lower orders too. If boundedness becomes an issue, multiply by a smooth cutoff function. $\endgroup$
    – user127096
    Commented Apr 8, 2014 at 12:17

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