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Hi,

i know that the following statement is used extensively, but i cannot find a proof anywhere:

For $\Omega$ a Lipschitz domain with boundary $\Gamma$, the space $H^{1/2}(\Gamma)$ is dense in $L_2(\Gamma)$.

Here, the space $H^{1/2}(\Gamma)$ is defined as the trace space, i.e. as $\gamma_0(H^1(\Omega))$, where $\gamma_0$ is the trace operator.

I read that one has to use density of $C^\infty(\overline\Omega)$ in $H^1(\Omega)$, but i dont see how - i can't extend an $L_2(\Gamma)$ function to a function in $H^1(\Omega)$ in a bounded way, right?

I tried Google and looked in all the books I have access to, but didn't find a proof. Does anyone have a hint?

Thanks, Mike

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1 Answer 1

up vote 1 down vote accepted

Lipschitz continuous functions on $\Gamma$ are dense in $L^2(\Gamma)$ and are contained in $H^{1/2}(\Gamma)$.

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Hi, so there exists an extension operator from the set of Lipschitz continuous functions on $\Gamma$ to $H^1(\Omega)$? –  Mike Apr 21 '12 at 8:28
    
If $f$ is Lipschitz on $\Gamma$, use a partition of unity to work locally, then use a Lipschitz continuous local change of variables to straighten the boundary, and you are reduced to the case $\Gamma$ is $x _n=0$. If you have a Lipschitz function on $x _n=0$ it is trivial to extend it to a Lipschitz function on the whole space (just take $F(x',x_n)=f(x')$). –  Piero D'Ancona Apr 21 '12 at 9:10
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Also, a $k$-Lipschitz function $f$ on a subset $Y$ of a metric space $(X,d)$ has a cheap $k$-Lipschitz extension $$\tilde f(x):=\inf_{y\in Y} f(y)+kd(x,y)$$ –  Pietro Majer Apr 21 '12 at 12:36
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