About the trace of Sobolev functions

Let $\Omega\subset\mathbb{R}^d$ be a Lipschitz domain with compact boundary and $1<p<\infty$.$$\gamma_0:W^{1,p}(\Omega)\mapsto W^{1-1/p,p}(\partial\Omega)$$ is the trace operator. $W^{1-1/p,p}(\partial\Omega)$ is defined as the image of $W^{1,p}(\Omega)$ and is equipped with the norm $$\Vert v\Vert_{W^{1-1/p,p}(\partial\Omega)}=\underset{u\in W^{1,p}\\\gamma_0(u)=v}{\rm{inf}}\Vert u\Vert_{W^{1,p}(\Omega)}$$

Assume now that $\Omega$ is a bounded domain of class $\mathcal{C}^{1,1}$ and that $\nu$ is the outward unit normal vector. We know that $\nu$ can be extended to a Lipschitz continuous function $\tilde{\nu}$ on $\bar{\Omega}$.

Then my question is:

For any $u\in W^{2,p}(\Omega)$, do we have that $\gamma_0(\nabla u)\cdot\nu\in W^{1-1/p,p}(\partial\Omega)$? If so,do we have $\gamma_0(\nabla u)\cdot\nu=\gamma_0(\nabla u\cdot\tilde{\nu})$ in $W^{1-1/p,p}(\partial\Omega)$? And why?

I think it's true that $$\Vert \gamma_0(\nabla u)\cdot\nu-\gamma_0(\nabla u\cdot\tilde{\nu})\Vert_{L^p(\partial\Omega)}=0,$$ but how about $$\Vert \gamma_0(\nabla u)\cdot\nu-\gamma_0(\nabla u\cdot\tilde{\nu})\Vert_{W^{1-1/p,p}(\partial\Omega)}?$$

Maybe I miss understand something, but this bothers me a lot!

Another thing make me mad is that I am not able to calculate $\Vert 1\Vert_{W^{1-1/p,p}(\partial\Omega)}$, even when $p=2$ !

Anyone could help me? I'd greatly appreciate it, thanks!

• Just to make things clear, $\gamma_0(\nabla u)=(\gamma_0 u_{x_1},...,\gamma_0 u_{x_d})$? – Tomás Oct 15 '13 at 16:29
• @Tomás:Yes, that's what I mean. – Y.Z Oct 15 '13 at 16:35
• I think that you have to ask that $\Omega$ is bounded, because if $\Omega$ is not bounded, then $\tilde{\nu}$ can be unbounded and hence the product $\nabla u\cdot\tilde{\nu}$ does not need to belong to $W^{1,p}$. – Tomás Oct 15 '13 at 16:48
• @Tomás:Corrected,thanks! – Y.Z Oct 16 '13 at 1:53
• An answer depends on your definition of $W^{1-1/p,p}(\partial \Omega)$. For example if you define this space as the image of $W^{1,p}(\Omega)$ under the trace operator, how would you define the product $gu$ where $g$ is a Lipschitz function on $\partial \Omega$ and $u\in W^{1-1/p,p}(\partial \Omega)$ (meaning how do you show that $gu\in W^{1-1/p,p}$)? – Jose27 Oct 16 '13 at 2:08

It's easier to deal with a more general problem (in any case take $v=\nabla u$ and $f=\nu_i$ for each $i=1,\ldots,n$).

So your problem is basically this: given $v\in W^{1,p}(\partial \Omega)$ and $f:\partial \Omega \to \mathbb{R}$ a Lipschitz function, with $F:\bar{\Omega}\to \mathbb{R}$ an extension of $f$, is the following identity true?

$$\gamma_0(v)f = \gamma_0(vF).\tag{1}$$

Clearly this is true if $v\in C^\infty(\bar{\Omega})$, since in this case $\gamma_0(vF)= v|_{\partial \Omega} f$ and similarly for $u$ and $F$ (remember that $F\in W^{1,\infty}(\Omega)$). Now we have the following estimate

$$\| vF \|_{W^{1,p}} \leq \| v\|_{W^{1,p}}\| F\|_{W^{1,\infty}}.$$

Now take $v$ arbitrary and a sequence $v_k\in C^\infty(\bar{\Omega})$ with $v_k\to v$ in $W^{1,p}$. It follows from the estimate that $v_k F\to vF$ and, by continuity of the trace operator, that $\gamma_0(v_kF) \to \gamma_0(vF)$ in $W^{1-1/p,p}$. Again by continuity of the trace operator we have that $\gamma_0(v_k) f \to \gamma_0(v)f$ in $L^p(\partial \Omega)$. Collecting all this we obtain

$$\gamma_0(v)f \overset{L^p(\partial \Omega)}{\leftarrow}\gamma_0(v_k)f=\gamma_0(v_kF) \overset{W^{1-1/p,p}}{\rightarrow} \gamma_0( vF)$$ and (1) follows from this.

Also notice that if two functions $v_1,v_2 \in W^{1-1/p,p}$ satisfy $\| v_1-v_2\|_{L^p(\partial \Omega)}$ then $\| v_1-v_2\|_{W^{1-1/p,p}}=0$. This is because, by definition, $0\in W^{1,p}(\Omega)$ has $0$ trace so it's a representative of $v_1-v_2$.