We have the result that : $\Omega$ is open set .If $u \in H^1(\Omega)$ then $|u|\in H^1(\Omega)$.My question is :
If $u \in H^2(\Omega) $ .Does $|u| \in H^2(\Omega)$?
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No. Consider for example $\Omega = (-1,1)$ and $u(x) =x$. Then $w(x) = \vert u(x) \vert = \vert x \vert$ satisfies $w'(x) = 1$ for $x>0$ and $w'(x) =-1$ for $x \le 0$. If $w'$ were in $H^1((-1,1))$, it would be continuous, which it obviously isn't.
Glitch
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I could be wrong, but if we're looking at Sobolev spaces, do we even need $w'$ to be continuous? My gut says we just need the weak derivative to exist and be square-integrable. – πr8 Dec 12 '16 at 19:57
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Sure, in general continuity is not required. My point here is simply that we have the embedding $H^1((-1,1)) \hookrightarrow C^0((-1,1))$, and so in this particular case it's impossible for $w'$ to belong to $H^1$. Certainly if $w \in H^2$ then $w' \in H^1$. – Glitch Dec 12 '16 at 20:49
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No. The function $u(x) =x$ is in $H^2$, but $w= \vert u \vert$ is not. – Glitch Dec 12 '16 at 21:02
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Yep sorry that's on me, had slightly misinterpreted which spaces we were talking about. – πr8 Dec 12 '16 at 21:05