0
$\begingroup$

Assume that for some smooth bounded open subset $\Omega$ in $\mathbb{R}^n$ and some $u\in H^{1,2}(\mathbb{R}^n)$ we know that $\chi_{\Omega} u\in H^{1,2}(\mathbb{R}^n)$. Is it then true that $u\in H^{1,2}_{0}(\Omega)$? Here $\chi_{\Omega}$ denotes the characteristic function which is one on $\Omega$ and zero elsewhere.

$\endgroup$

1 Answer 1

2
$\begingroup$

It is true that $$ H_0^{1,2}(\Omega)= \bigl\{u\bigr|_\Omega\bigm| u\in H^{1,2}(\mathbb R^n),\, \operatorname{supp}u\subseteq\overline\Omega\bigr\}. $$ Indeed, the inclusion $\subseteq$ is obvious. To see the inclusion $\supseteq$, notice that, for $u\in H^{1,2}(\mathbb R^n)$, taking the trace $u\bigr|_{\partial\Omega}$ from either side of $\partial\Omega$ yields the same result, and if $\operatorname{supp}u\subseteq \overline\Omega$, then $\bigl(u\bigr|_\Omega\bigr)\bigr|_{\partial\Omega} = \bigl(u\bigr|_{\mathbb R^n\setminus\overline\Omega}\bigr)\bigr|_{\partial\Omega}=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.