Strong maximum principle for harmonic functions

Suppose $u$ attains its global maximum $M$, and put $E=\{x\in \mathrm{\Omega }:u(x)=M\}$. This set is nonempty and relatively closed. Fix $z\in E$ and choose $r>0$ with ${\bar{B}}_{3r}(z)\subset \mathrm{\Omega }$. The set $E\cap {\bar{B}}_{2r}(z)$ is nonempty compact. If there were $y\in B_r(z)\setminus E$, its distance to this compact set would attain a positive minimum $R$ at a point $p$, with $0<R\le \mathrm{\mid }y-z\mathrm{\mid }<r$. The closed ball ${\bar{B}}_R(y)$ lies inside $B_{2r}(z)$, and $u<M$ on its interior while $u(p)=M$ on its boundary.

On the inner sphere $\mathrm{\mid }x-y\mathrm{\mid }=R\mathrm{/}2$, compactness and strict inequality give $a:=M-\max u>0$. Use the barrier $v$ from [F2] with $\alpha =2n\mathrm{/}{R}^2$. The function $u+av$ is subharmonic on the annulus: $\mathrm{\Delta }u=0$ and $\mathrm{\Delta }v\ge 0$. On its inner boundary $u+av=u+a\le M$, and on its outer boundary $u+av=u\le M$. By [F3], $u+av\le M$ throughout the annulus.

Let $\nu =(p-y)\mathrm{/}R$. For $0<t<R\mathrm{/}2$, step 2.1 implies $\frac{u(p)-u(p-t\nu )}{t}\ge a\frac{v(p-t\nu )}{t}\mathrm{.}$ As $t\downarrow 0$, the left side tends to $\mathrm{\nabla }u(p)\cdot \nu =0$ by [F4], because $p$ is an interior global maximum in $\mathrm{\Omega }$. Since $v(p)=0$, the right side tends to $-a{\mathrm{\partial }}_{\nu }v(p)>0$ by [F2], which is impossible. Thus there is no such $y$, so $B_r(z)\subseteq E$. Every $z\in E$ therefore has a neighborhood in $E$.

The nonempty set $E$ is both open and closed in the connected domain $\mathrm{\Omega }$, so [F4] gives $E=\mathrm{\Omega }$. Thus $u$ is constant. For a global minimum apply the same argument to $-u$, which is again harmonic. All choices above concern finitely many points and real parameters at a time; no mean-value or countable-choice hypothesis is used.