The Poisson kernel realizes the sharp Harnack bounds on concentric discs

The function $H(z):=\frac{1+z}{1-z}$ is holomorphic on $\mathrm{\mid }z\mathrm{\mid }<1$ by [L2], and its real part is $\mathrm{Re}H(\rho e^{i\varphi })=\frac{1-{\rho }^2}{1-2\rho \cos \varphi +{\rho }^2}=P(\rho e^{i\varphi },1)$ by [L1]. Therefore $u(z):=P(z,1)$ is harmonic on the unit disc. Since $1-\mathrm{\mid }z{\mathrm{\mid }}^2>0$ and $\mathrm{\mid }1-z{\mathrm{\mid }}^2>0$ for $\mathrm{\mid }z\mathrm{\mid }<1$, it is positive there as well.

Substituting $\varphi =0$ and $\varphi =\pi$ into [L1] gives $u(r{e}^{i0})=\frac{1-r^2}{(1-r{)}^2}=\frac{1+r}{1-r},u(r{e}^{i\pi })=\frac{1-r^2}{(1+r{)}^2}=\frac{1-r}{1+r},$ and $u(0)=P(0,1)=1$.

Fix $R$ with $r<R<1$. The function $u$ is harmonic on a neighbourhood of $\overline{D(0,R)}$, so [L3] gives $\frac{R-r}{R+r}u(0)\le u(r{e}^{i\varphi })\le \frac{R+r}{R-r}u(0)\mathrm{.}$ Letting $R\to 1^{-}$ yields the unit-disc Harnack bounds $\frac{1-r}{1+r}\le u(r{e}^{i\varphi })\le \frac{1+r}{1-r}\mathrm{.}$ Step 1.2 shows equality at $\varphi =\pi$ and $\varphi =0$, respectively. Thus the Poisson kernel realizes both Harnack extremes on the circle $\mathrm{\mid }z\mathrm{\mid }=r$.