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I found the following statement:

if $\Omega$ has $C^{\infty}$ boundary, then for all $u\in W^{k,p}(\Omega)$, we can find $u_k\in C^\infty_c(\bar\Omega)$ such that $\|u_k-u\|_{W^{k,p}(\Omega)}\to0$.

this statement is applied to $\Omega=\mathbb R^n_+$ to prove the trace theorem.

but this statement implies $W^{k,p}(\Omega)=W^{k,p}_0(\Omega)$ which is wrong. What's the problem?

by the way, I can only prove the statement for $u_k\in C^\infty(\bar\Omega)$

Lookout
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  • Like you said, the statement is incorrect, unless it's a matter of notation. It seems strange to start with a statement about general smooth domains to "prove" the trace theorem on the half-space, too. – mathematician Dec 23 '15 at 04:35

1 Answers1

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If by $C_c^\infty(\overline{\Omega})$, it is understood the space of functions which has all derivatives continuous up to the boundary and $u(x)=0$ on the boundary, then the statement is not true.

Indeed, let $I=(0,1)$, $p\in [1,\infty)$ and $u(x)=1$ for $x\in I$. Note that $u\in W^{1,p}(I)$.

Assume that there is a sequence $u_n\in C_c^\infty(\overline{I})$ such that $u_n\to u$ in $W^{1,p}(I)$. Once $W^{1,p}(I)$ is continuously embedded in $C(\overline{I})$, we have that $u_n\to u$ in $C(\overline{I})$, which implies that $$0=u_n(0)\to u(0)=1.$$

An absurd!

If you want to study the trace of a function $u\in W^{k,p}(\Omega)$, you should compare it with a function in $C^\infty(\overline{\Omega})$ and it is a know fact that $C^\infty(\overline{\Omega})$ is dense in $u\in W^{k,p}(\Omega)$.

Tomás
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