I know that the space $W^{1,p}_0(\Omega)$ is defined as the closure of $C^\infty_0(\Omega)$ in the $\|\cdot\|_{W^{1,p}}$ norm. Do functions in $W^{1,p}_0(\Omega)$ themselves have compact support in $\Omega$ (as the functions in $C^\infty_0(\Omega)$)?
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1No, they are interpreted as the functions such that "$u=0$ on $\partial\Omega$" which has to be understood in the trace sense – Peter Melech Dec 22 '22 at 11:58
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1If $T$ denotes the trace operator then $W_0^{1,p}(\Omega)={u\in W^{1,p}:Tu=0}=\ker(T)$ – Peter Melech Dec 22 '22 at 12:53
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A simple counterexample. Take $\Omega = B_1 \subseteq \mathbb{R}^n$ (unit ball) and a function $u \colon \mathbb{R}^n \to \mathbb{R}$ whose support is exactly $\overline{B_1}$. The $u$ doesn't lie in $C_0^\infty(\Omega)$ but each of the functions $u_\lambda(x) := u(\lambda x)$ for $\lambda > 1$ does, as $\operatorname{supp} u_\lambda = \overline{B_{1/\lambda}}$. And it's easy to convince yourself that $u_\lambda \to u$ in $W^{1,p}(\Omega)$ as $\lambda \to 1$.
Michał Miśkiewicz
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