Consider the following the theorem in the classical PDE book of Evans (chapter 2):

(Part of the strong maximum principle) Let $U$ a open set in $R^n$ and $u \in C^2 (U) \cap C(\overline{U})$, with $\Delta u = 0$ in $U$.

If $U$ is connected and there exists a point $x_0 \in U$ such that $$ u(x_0) = \displaystyle\max_{\overline{U}} u$$ then $u$ is constant within $U$.

Part of the proof:

Suppose there exists a point $x_0 \in U$ with $u(x_0) = M = \displaystyle\max_{\overline{U}} u . $ Then for $0 < r < \mbox{dist} (x_0 , \partial U)$, the mean value property asserts $$ M = u(x_0) = \displaystyle\frac{\displaystyle1}{|B(x_0, r)|}\int_{B(x_0,r) } u \ dy \leq M.$$

Then $$u = M\quad\text{in}\quad B(x_0 , r)\tag{$*$}.$$

I dont understand the equality in $(*)$. If I be non rigorous, for me is clear to see the equality in $(*)$. But I dont know how to prove the equality... Someone can give me a hint about how to prove the equality in $(*)$?

Thanks in advance!

  • $\begingroup$ Actually Im also wondering about the equation before, why we have $ (\ldots) \leq M$? $\endgroup$ – MJimitater Jul 9 '20 at 12:54

We can split the ball $B=B(x_0,r)=E_1\cup E_2\cup E_3$, where $$E_1=\{x:u(x)<M\},\qquad E_1=\{x:u(x)=M\},\qquad E_3=\{x:u(x)>M\}$$ these sets are disjoint so $$\int_Budx=\int_{E_1}udx+\int_{E_2}udx+\int_{E_3}udx$$ By the assumption on $x_0$ we must have $E_3=\emptyset$, but then $E_1$ must be empty too, and hence $B=E_2$.

  • $\begingroup$ why E1 must be empty $\endgroup$ – Raja Sekar Jun 11 '14 at 15:10
  • $\begingroup$ @RajaSekar That is a consequence from the mean value property of harmonic functions and the assumption $u(x_0)=M$ (this shows $E_1$ has measure $0$, and then it must be empty due to continuity of $u$). $\endgroup$ – AD. Jun 12 '14 at 21:44

Suppose that there exist a set $\Omega$ in $B(x_0,r)$, such that $u<M$ in $\Omega$. Note that $$\frac{1}{|B(x_0,r)|}\int_{B(x_0,r)}u dy=\frac{1}{|B(x_0,r)|}\left(\int_{\Omega}udy+\int_{B(x_0,r)\setminus\Omega}udy\right)<M$$

Can you conclude?

Hint for the last inequality: Note that in $\Omega$ we have that $u<M$ and in $B(x_0,r)\setminus \Omega$ we have that $u\leq M$, hence $$\frac{1}{|B(x_0,r)|}\left(\int_{\Omega}udy+\int_{B(x_0,r)\setminus\Omega}udy\right)<\frac{1}{|B(x_0,r)|}\left(M|\Omega|+M|B(x_0,r)\setminus\Omega|\right)$$

  • $\begingroup$ i dont understand your last inequality , can you explain to me ? please.. $\endgroup$ – math student Aug 20 '13 at 13:18
  • $\begingroup$ volume integral of a f over V < volume_of_V*maximum_of_f_in_V $\endgroup$ – nonlinearism Aug 20 '13 at 14:15

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