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Let $\Omega$ be a domain with $C^1$-boundary and $1 \leq p < \infty$. Then there's exactly one bounded linear operator $$tr_{ \partial \Omega}: W^1_p(\Omega) \rightarrow L^p(\partial \Omega), ~ u \mapsto u|_{\partial \Omega}$$

for all $u \in C^1_b(\bar \Omega) \cap W^1_p(\Omega)$.

Our lecturer said that $\partial \Omega$ is a set of measure zero and as such, this map would not be welldefined if we only had $u \in W^1_p(\Omega)$, but with $u \in C^1_b(\bar \Omega)$ as well, it is well-defined.

Could someone please explain this statement to me?

SallyOwens
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I believe what your lecturer meant is that functions in $W^1_p$ are only defined up to sets of measure zero. That's because functions in $W^1_p$ are technically equivalence classes of functions. Thus, you can't simply define the trace operator to be the value of the function on the boundary, because it has measure zero.

This definition however works for functions that are also continuous. The trick then is to show that the trace operator is continuous, and to extend the definition of the operator to all of $W^1_p$ using the fact that $C^1$ is dense in $W^1_p$.

Here's how the extension of the operator to $W^1_p$ works. Suppose we have already proved the following inequality for functions $u\in C^1(\Omega)$: $$ \|\left. u\right|_{\partial\Omega}\|_{L^p(\partial\Omega} \leq \|u\|_{W^1_p(\Omega)}. $$ Now suppose we want to define the trace $\left.v\right|_{\partial\Omega}$ for some $v\in W^1_p$ which is not necessarily in $C^1(\Omega)$. What we can do is find a sequence of functions $u_n$ which converges to $v$ in $W^1_p$ (since continuous functions are dense in $W^1_p$). Then we have $$ \|\left. (u_n-u_m)\right|_{\partial\Omega}\|_{L^p(\partial\Omega} \leq \|u_n-u_m\|_{W^1_p(\Omega)}. $$ Thus the sequence of the traces on $\partial\Omega$ is a Cauchy sequence in $L^p(\partial\Omega)$. We can take its limit to be the definition of the trace for $v$. This definition now allows us to consider the trace operator as a continuous linear operator defined on all of $W^1_p$.

felipeh
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  • Thank you! The first part of your answer about the $W^1_p$ is clear but I wanted to check whether I understood the second part correctly: If I have a function that is continous, I can consider it on a set of measure zero and as I know that $C^1 \subseteq W^1_p$ and in particular dense, I can say that I restrict this operator onto functions in the intersection of those two sets? – SallyOwens Aug 28 '17 at 13:04
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    I've edited the answer to better explain how the extension to $W^1_p$ works. The idea is that by using continuity we can eventually define the trace on all of $W^1_p$, even though at first we only knew how to define it on $C^1$. Hopefully that helps. – felipeh Aug 28 '17 at 13:14
  • Now it is clear - thank you very much :) – SallyOwens Aug 28 '17 at 13:17
  • Glad I could help! – felipeh Aug 28 '17 at 13:20