Proposition 3.1.19DAC835A39525

Proof. (1) Let $r<R'<R$. By theorem 3.1.18 on $B(0,R')$, both $h(x)$ and $h(0)$ are integrals of the nonnegative boundary values of $h$ against the Poisson kernel. For $|\zeta |=R'$ we have $R'-r\leq |x-\zeta |\leq R'+r$, so the ratio

lies between the two coefficients in (3.1.20) with $R'$ in place of $R$. Integrating and letting $R'\uparrow R$ gives (3.1.20).

(2) If a nonnegative harmonic function on a domain vanishes at a point, it attains its minimum there and is identically zero by proposition 3.1.10.

(3) We first apply (1), after a translation, to a ball $B(a,2t)\subset \Omega $: for $y\in B(a,t)$ we have $r=|y-a|<t$, and the right inequality in (3.1.20) on $B(a,2t)$ gives

Now let $y\in \Omega $. Since $\Omega $ is connected, $y$ can be joined to $x_0$ by a polygonal path in $\Omega $: the set of points that can be so joined is open, and so is its complement in $\Omega $, because balls are convex. The path is compact, so there is $t>0$ with $B(z,2t)\subset \Omega $ for every point $z$ of the path. Choose points $x_0=p_0,p_1,\ldots ,p_ N=y$ along the path with $|p_{i+1}-p_ i|<t$. Applying the ball estimate on $B(p_ i,2t)$ for $i=0,\ldots ,N-1$ gives $h(y)\leq C_0^ Nh(x_0)$, and once more on $B(y,2t)$ gives $h\leq C_0^{N+1}h(x_0)$ on $B(y,t)$. Here $t$ and $N$ depend on $y$, $x_0$ and $\Omega $, but not on $h$. Finitely many of the balls $B(y,t)$, $y\in K$, cover $K$, and the largest of the corresponding constants $C_0^{N+1}$ is the required $C$. $\square$