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I have an exercise in P.D.E that I couldn't solve.

Let $\Omega \subset \mathbb{R}^n$ be a connected open set and $u:\Omega \to \mathbb{R}$ a continuous function that satisfies the following property:

(1) $\forall x \in \Omega \ \exists a(x) > 0$; $\forall 0 < r < a(x): u(x) =\displaystyle\frac{1}{\omega_n r^{n-1}} \int_{S_r(x)} u(y)d\sigma(y)$

Show that, if $u$ has a local maximum or a local minimum, then $u$ is constant in $\Omega$.

I know how to solve that exercise if $u$ has a global maximum, but I don't know how to change the solution for the case of a local maximum.

Of course I know that functions that satisfies the property (1) are harmonic (real-analytic!) but I can't use that because this exercise is just a step to show that functions that satisfies (1) are harmonic!

PS: I'm sorry for my really bad English. I'm from Brazil and almost never have to write in other language than Portuguese. :)

PS2: It would be good if I show how to solve this exercise for the case of global maximum so you guys can help me to adapt the solution?

EDIT: As @D.Thomine noticed, it is necessary to supose that f is continuous.

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    Argue *by contradiction*: assume $x$ is a strict local maximum for $u$; then for all $y\in B(x;a(x))$ (here and in what follows $B(x;r):=\{ y\in \mathbb{R}^N:\ |y-x| is the *open ball with radius $r>0$ centered in $x$*) you get: $$\tag{1} u(x)>u(y)\; ;$$ now, taking the integral mean in both sides of (**1**) with respect to the $y$ variable, you obtain a contradiction...2011-12-07
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    P.S.: I'm assuming that the set you call $S_r(x)$ is the one I named $B(x;r)$, i.e. the open ball. In such a case, in writing the integral mean you have to integrate with respect to $y$ and not w.r.t. $\sigma (y)$ (which I assume to be a *surface measure*). On the other hand, if $S_r(x) =\partial B(x;r)$ (i.e. if $S_r(x)$ is a sphere), then in writing the integral mean you have to divide by $N\omega_N\ r^{N-1}=\sigma (S_r(x))$.2011-12-07
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    Thanks. It was usefull, but and for x that is a local maximum, but not a _strict_ local maximum?2011-12-08
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    As stated, what you want to prove is false. For instance, take the function $f$ on $\mathbb{R}$ such that $f(x) = 0$ if $x<0$, then $f(x) = 1$ if $x>0$, and $f(x) = 1/2$ is $x=0$. I think that something like the continuity of $a$ or $u$ is required.2012-01-07

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