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Let $\Omega\subset\mathbb{R}^n$ be a bounded domain, with Lipschitz smooth boundary. Then a well known result by Stein gives that there exists an extension operator $E: H^k(\Omega)\rightarrow H^k(\mathbb{R}^n)$ such that

  1. $Eu(x)=u(x)$ for all $x\in\Omega$

  2. $\left\|Eu\right\|_{H^k(\mathbb{R}^n)}\leq C\left\|u\right\| _ {H^k(\Omega)} $

where the constant $C$ depends on $k$, $n$ and Lipschitz constant of $\Omega$. See the Thesis: A Degree-Independent Sobolev Extension Operator by Luke Rogers for a really nice summary of extension operators in the integer case.

DeVore and Sharpley in Besov spaces on bound domains of $\mathbb{R}^n$ extended this to the fractional case by real interpolation. There result is:

There exists an extension operator $E: H^\tau(\Omega)\rightarrow H^\tau(\mathbb{R}^n)$ such that:

  1. $Eu(x)=u(x)$ for all $x\in\Omega$

  2. $\left\|Eu\right\|_{H^\tau(\mathbb{R}^n)}\leq C\left\|u\right\| _ {H^\tau(\Omega)}$

but this time the constant depends on $\tau$, $n$ and $\Omega$. So the dependence is on $\Omega$ and not on the Lipschitz condition it satisfies.

This to me seems a highly unsatisfactory situation. Does there exist an extension operator so that the bound in the fractional Sobolev setting depends on $\Omega$ only through its Lipschitz condition? Does anyone know a reference which has this result?

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(I took the liberty of fixing the LaTeX; it seems that inserting a space works: \|u\| _{H^\tau(\Omega)} ) –  Pietro Majer Jun 15 '11 at 18:47
    
I think the answer is yes and stated in Satz 6.40 in "Angewandte Funtionalanalysis" by M. Dobrowolski (but unfortunately in German). As far as I have seen, the proof imitates the one for the integer case: First provide estimates for the extension from a half space to the whole space and then use localization and Lipschitz-continuous deformation. –  Dirk Jun 15 '11 at 18:48
    
@Pietro Majer: Thanks for fixing the latex! :D @Dirk: I am a little confused by Satz 6.40 is it really stating only dependence on the Lipschitz condition. It seems to be imposing that the boundary it $C^1$ smooth. –  alext87 Jun 15 '11 at 19:17
    
The boundary regularity needed is $C^{m,1}$ for an extension of an $H^{s,p}$-function with $0\leq s\leq m+1$. Hence Lipschitz works for $0\leq s\leq 1$. This is similar to what Dobrowolski proves for the integrer case in Satz 6.11. However, I just had a breif view on Stein's result and he seems to use a very different technique and only needs Lipschitz for arbitrary high $m$, right? –  Dirk Jun 16 '11 at 6:59
    
Yes, Stein's result only requires Lipschitz for arbitrary high $m$. At least Dobrowolski's result proves it is true for balls and annuli. Actually, I'm not completely convinced because I couldn't find where he explicitly stated what the constant depends on. My high school German can't cope with all the maths terminology so I may have missed it. –  alext87 Jun 16 '11 at 7:20
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