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Well, I'm studying now Sobolev Spaces and I have a problem. Suppose an arbitrary open set $\Omega$, we will work there all the time. The space $W^{1,p}_{0}$ is the completeness of $C^{\infty}_{0}$ in $W^{1,p}$ and friend told me that this space is the functions in $W^{1,p}$ but also vanish at the boundary of $\Omega$. My questions are:

i) is this true?

ii) if $\Omega$ is an open set, the function is not define at the boundary, does it make sense that it is defined at the boundary?

iii) What happens if $\Omega$ is not bounded?

Thank you very much.

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I will give you answers for questions (i) and (ii).

i) is this true?

We have the following result: Let $\Omega$ be bounded with smooth boundary and $u\in W^{1,p}(\Omega)\cap C(\overline{\Omega})$. Then $u\in W^{1,p}_0(\Omega)$ if and only if $u=0$ on $\partial\Omega$. In other words, under the said assumptions, $$u\in W^{1,p}_0(\Omega)\quad\Longleftrightarrow \quad u|_{\partial\Omega}=0.$$

Note that $u|_{\partial\Omega}$ does not make sense in general for a function $u\in W^{1,p}(\Omega)$ (because $\partial\Omega$ has zero measure). This is why we assume $u\in C(\overline{\Omega})$. However, is it possible to prove that there exists a bounded linear operator $T: W^{1,p}(\Omega)\to L^p(\partial\Omega)$ such that $Tu=u|_{\partial\Omega}$ for all $u\in W^{1,p}(\Omega)\cap C(\overline{\Omega})$. Also, we have the following result for a function $u\in W^{1,p}(\Omega)$: $$u\in W^{1,p}_0(\Omega)\quad\Longleftrightarrow \quad Tu=0.$$ The operator $T$ is called "trace operator" and $Tu$ is called "trace of $u$". So, what we can say is:

For a bounded open set $\Omega$ with smooth boundary, $W^{1,p}_0(\Omega)$ is the space of the functions in $W^{1,p}(\Omega)$ which vanish at the boundary of $\Omega$ in the sense of the trace.

(Note that for a function in $C(\overline{\Omega})$, "vanishing at the boundary in the sense of the trace" is the same as "vanishing at the boundary in the usual sense".)

ii) if $\Omega$ is an open set, the function is not define at the boundary, does it make sense that it is defined at the boundary?

No, it doesn't. But the function has a trace. So, we can talk about the function on the boundary in the sense of the trace. (Note that for a function in $C(\overline{\Omega})$, "the function on the boundary in the sense of the trace" is the same as "the function on the boundary in the usual sense".)

Further reading: Section 5.5 of Evans book.

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