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Let $\Omega \subset \mathbb{R}^n$, $n\geq 1$ be a bounded domain with $\partial \Omega \in C^1$ and $u \in C^2(\Omega)\cap C^1(\overline\Omega)$ a harmonic function.

Show that for all $v\in C^2(\Omega)\cap C^1(\overline\Omega)$ with $v=u$ on $\partial \Omega$ the following holds: $$ \int_\Omega |\nabla u|^2 dx\leq \int_\Omega |\nabla v|^2dx $$

I got the hint that $0\leq \int_\Omega |\nabla (u-v)|^2 dx$ but I simply couldn't come up with anything useful from that.

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i)

\begin{align*} 0& =\int_{\partial \Omega } \ (u-v)\nabla (u-v)\cdot n \\&=_{divergence\ theorem} \int \ {\rm div}\ (u-v)\nabla (u-v) \\&=\int\ |\nabla (u-v)|^2+(u-v)\Delta (-v) \end{align*}

Hence $$ A:=\int\ u\Delta v -v\Delta v \geq 0 $$

ii) $(u-v)\nabla v$ : $$ 0= \int_{\partial \Omega }\ n\cdot (u-v)\nabla v = \int\ {\rm div}\ (u-v)\nabla v = \int \ \nabla u\cdot \nabla v-|\nabla v|^2 +A$$ so that $$\int\ |\nabla v|^2\geq \int\ \nabla u\cdot \nabla v $$

iii) $v\nabla u$ : Here $$ \int_{\partial\Omega }\ n\cdot u\nabla u= \int\ {\rm div}\ v\nabla u = \int\ \nabla v\cdot \nabla u $$

And $$ \int_{\partial \Omega }\ n\cdot u \nabla u = \int\ |\nabla u|^2$$

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