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How can I prove that the function:

$$u:\Omega\to\mathbb{R},\ u(x)=\begin{cases} 0, x\in\omega \\[3mm] v(x), x\in\Omega\setminus\omega\end{cases}$$ is not in $H^1(\Omega)$, knowing that $v\geq 1$ is any function from $L^{\infty}(\Omega)$, where $\Omega$ is an open, nonempty and bounded subset of $\mathbb{R}^N,N\geq 2$ and $\omega\subset\Omega$ is any measurable set with $\omega$ and $\Omega\setminus\omega$ both having strict positive measure).

It is a kind of characteristic function, which I know that doesn' belong to $H^1(\Omega)$. But how to prove it rigurously?

Bogdan
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  • Take $\omega = \emptyset$, then clearly $u\not\in H^1(\Omega)$. On the other hand, if $\omega = \Omega$, then $u\in H^1(\Omega)$ as $u\equiv 0$. – stange Jan 04 '24 at 12:48
  • I edited my post. – Bogdan Jan 04 '24 at 12:51
  • Your question is again flawed, since you can just choose $v(x)$ in a way, such that $u \in H^1$ again. Tink of $\Omega \setminus \omega$ to be the support of a smooth bump function $v$. – F. Conrad Jan 04 '24 at 13:57
  • It is essential that $v\geq 1$. I don't think you are right...There is a gap between them and somehow a $H^1$ function is absolutel continuous on lines and somehow this will prove it, but I don't know how exactly. – Bogdan Jan 04 '24 at 14:07
  • I don't think your assumptions are enough to guarantee $u\not\in H^1(\Omega)$. After all there exist discontinuous Sobolev functions in dimensions $N\ge 2$. – Stratos supports the strike Jan 04 '24 at 19:44

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