Proposition 3.1.10B304785CC98D

Proof. For the first assertion, suppose that $h$ attains its maximum $m$ at a point of $D$, and let $E=\{ x\in D:h(x)=m\} $. This set is nonempty and closed in $D$. If $x\in E$ and $\overline{B(x,r)}\subset D$, the ball mean-value property gives $\int _{B(x,r)}(m-h)\, \mathrm dy=0$. The integrand is continuous and nonnegative, so it vanishes on $B(x,r)$. Thus $E$ is open, and $E=D$ by connectedness. For the minimum, apply this to $-h$.

For the second assertion, let $\varepsilon >0$, and suppose that $h>\varepsilon $ somewhere in $D$; we derive a contradiction. Each $\zeta \in \partial D$ has a ball $B(\zeta ,r_\zeta )$ such that $h<\varepsilon $ on $D\cap B(\zeta ,r_\zeta )$. Since $D$ is bounded, $\partial D$ is compact, and finitely many of these balls cover it; let $W$ be their union. Then $K=\overline D\setminus W=D\setminus W$ is compact. Since $h<\varepsilon $ on $D\setminus K=D\cap W$, the set $K$ is nonempty. The maximum of $h$ over $K$ is attained at some point $x_0$, and $h(x_0)>\varepsilon $; hence $h(x_0)$ is also the maximum of $h$ over $D$. Let $D_0$ be the component of $D$ containing $x_0$; it is open, because every point of $D$ lies in a ball contained in $D$, and balls are connected. By the first assertion, $h=h(x_0)$ on $D_0$.

Since $\mathbb R^ d$ is connected and $D_0$ is a nonempty, bounded, open subset, $D_0$ is not closed, so it has a boundary point $\zeta $. This point does not lie in $D$: otherwise the component of $D$ containing $\zeta $ would be an open neighborhood of $\zeta $, so it would meet $D_0$ and hence equal $D_0$, which does not contain its boundary point $\zeta $. Thus $\zeta \in \partial D$ and $\varlimsup _{D\ni x\to \zeta }h(x)\geq h(x_0)>0$, contradicting the hypothesis. Hence $h\leq \varepsilon $ on $D$, and letting $\varepsilon \downarrow 0$ gives $h\leq 0$.

Finally, if two harmonic functions on a bounded open set are continuous on its closure and agree on its boundary, apply the second assertion to the real and imaginary parts of their difference and of its negative. $\square$