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I try to prove the following :

"If u $\in \mathcal{C}^2(\Omega)$ and for each ball is entirely in $\Omega$ $$\int_{\partial B} \frac{\partial u}{\partial \nu} =0$$ then u is harmonic". Where the $\frac{\partial u}{\partial \nu} $ is the normal derivative.

Any idea how to solve it?

I think there is a type error in exercise.

I also think the reverse is easy to check , so that if u is harmonic from divergence theorem $$\int_B \Delta u = \int_{\partial B}\frac{\partial u}{\partial \nu} $$ because of $\Delta u = 0 $ everywhere in $\Omega$.

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You have, by hypothesis, that $$\int_{\partial B} \frac{\partial u}{\partial \nu} =0.$$ From divergence theorem it is equivalent to say $$\int_B\Delta u=0.$$ Now, since $\Delta u$ is continous, if there exists $p\in \Omega$ such that $(\Delta u)(p)>0$ then there exists $B(p,r)$ such that $\Delta u>0$ on $B(p,r).$ But this contradicts $$\int_{B(p,r)}\Delta u=0.$$ So, it must be $\Delta u\equiv 0.$ That is, $u$ is harmonic.

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  • $\begingroup$ I am skeptical about the $\mathcal{C}^2(\Omega) $ can we prove it for less strength hypothesis ?? $\endgroup$ – chaviaras michalis Oct 17 '16 at 12:06

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