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Let $\Omega \subset \mathbb{R}^N$ be a bounded domain with Lipschitz boundary. Let $u,v$ be function defined on $H^1(\Omega)$ such that $u=v$ on boundary of $\Omega(\partial \Omega)$. Then can we say this: $$|u-v|_{H^1(\Omega)}=0$$ My idea is like this: $$|u-v|_{H^1(\Omega)}^2=\int_{\Omega}(\triangledown (u-v))^2dx=\int_{\partial \Omega}(u-v)^2ds=0$$ where the 2nd last equality holds from Stoke's Theorem.Any suggestions are welcome.

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  • $\begingroup$ why is $|w|_{H^1(\Omega)} = \int_\Omega \nabla w$? So $|-x|_{H^1(0,1)} < 0$? $\endgroup$ – Calvin Khor Feb 19 '18 at 10:42
  • $\begingroup$ Whats a square of a vector, and how does Stokes apply? $\endgroup$ – Calvin Khor Feb 19 '18 at 10:47
  • $\begingroup$ Got my mistake! $\endgroup$ – Abhinav Jha Feb 19 '18 at 10:48
  • $\begingroup$ (2nd comment is about your edit) $\endgroup$ – Calvin Khor Feb 19 '18 at 10:52
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No, Let $u=0$ and $v$ be any smooth nonzero function compactly supported in $\Omega$. $|u-v|_{H^1} ≥ |v|_{L^2} > 0 $.

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