The Trace Theorem for $W^{1,p}$ functions

Question

I'm trying to understand the proof of the trace theorem. This is from a course I am taking, so I will write out what we have done explicitly.

$\textbf{Trace Theorem}$ Suppose $\Omega \subset\mathbb{R}^n$ is bounded, open with $C^1$ boundary. Then there exists a bounded linear operator $Tr:W^{1,p}(\Omega )\rightarrow L^p(\partial\Omega )$. Further, if $u\in C(\overline\Omega )\cap W^{1,p}(\Omega )$ then $Tru=\overline{u}$ for all $x\in\partial\Omega$, where $\overline{u}$ denotes the uniformly continuous extension of $u$ to $\overline{\Omega}$.

To prove this, we need the extension theorem:

$\textbf{Extension Theorem}$ Suppose $\Omega \subset\mathbb{R}^n$ is bounded, open with $C^1$ boundary. Suppose further that $\overline{\Omega}\subset V$ where $V\subset\mathbb{R}^n$ is bounded. Then there exists a bounded linear operator $E:W^{1,p}(\Omega )\rightarrow W^{1,p}_0(V)$ such that $Eu = u$ a.e. for all $x\in \Omega$. Further, if $u\in C(\overline\Omega )\cap W^{1,p}(\Omega )$ then $Eu \in C(\overline{V})$.

Now we are in a position to prove the theorem. I won't go in to too much detail but the idea is standard. We prove the first part of the theorem for test functions. We then have the result for $Eu$ by density, so I define $Tr(u) = \lim _{n\rightarrow \infty}Tr(u_n)$ where $(u_n)$ is my approximating sequence of test functions for $Eu$ and $Tr(u_n):={u_n|}_{\partial \Omega}$

Now suppose $u$ is uniformly continuous. We want to show the two constructions coincide. Firstly, the extension theorem tells me that $Eu$ is a uniformly continuous extension of $u$ to $\overline{\Omega}$, and since this extension is unique we have that $Eu=\overline{u}$ on the boundary. So I need to show that $Eu=Tr(u)$ on the boundary. We know that $J_\epsilon\ast u\rightarrow u$ uniformly on $\partial\Omega$ as $\epsilon\rightarrow 0$, where $J_\epsilon$ is the standard mollifier. This is where I am stuck. In the notes it says that $J_\epsilon\ast u$ also converges in $L^p(\partial\Omega )$ and therefore this limit must be $Tr(u)$. Why? Why isn't it just $Eu$ again?

Any help much appreciated!

Answer 1

I don't know how your prove can work without prove the trace estimation...

Here is a more standard idea. (You can find it on Evans PDE book or Leoni's Sobolev space. I would say Evans, it is easier)

Suppose $\Omega$ is an bounded extension domain so that the extension theorem, which you quote in your post, will work. Next, for $u\in W^{1,p}(\Omega)$ you actually can obtain a sequence $(u_n)\subset C^\infty(\bar\Omega)\cap W^{1,p}(\Omega)$ such that $u_n\to u$ in $W^{1,p}$ norm. This is an standard result of global approximation and you could prove it easily by using extension theorem.

Next, you could prove that, and this is an very important result in trace theorem (I would say the most important), the Trace estimation $$ \|T [u_n]\|_{L^p(\partial \Omega)}\leq C \|u_n\|_{W^{1,p}(\Omega)}\tag 1$$ Remember here $u_n\in C^\infty(\bar\Omega)$, hence $T[u_n]=u\lfloor_{\partial\Omega}$ is well defined. (This estimation also )

Hence, you have $$ \|T [u_n]-T[u_m]\|_{L^p(\partial \Omega)}\leq C \|u_n-u_m\|_{W^{1,p}(\Omega)}\tag 2 \to 0$$ which implies that $T[u_n]$ is a cauchy sequence in $L^p(\partial\Omega)$ and hence you could define the limit and this limit is $T[u]$.

Finally, you have
$$\|T [u_n]-T[u]\|_{L^p(\partial \Omega)}\leq C \|u_n-u\|_{W^{1,p}(\Omega)}\to 0$$ which shows $u_n\to u$ a.e. on $\partial \Omega$, which addressed your question.