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At the origin, the Poisson kernel is one, so the mean value property for harmonic functions givesFor $0<r<1$, the denominator in the Poisson kernel satisfies $1-2r\cos \vartheta +r^2\ge (1-r{)}^2$. ThusThe nonnegativity of $f$ allows this upper bound to be inserted under the Poisson integral on the unit disk:This is the upper bound in the Harnack inequality on the unit disk. The same argument with the minimum of the Poisson kernel gives the corresponding lower bound $(1-r)u(0)/(1+r)$.

The outward normal of the exterior region points into the excluded disc, so it is $-\mathbf{r}$ on the unit circle. Differentiating the Dirichlet Green function from (e), and writing $s=|{\mathbf{r}}^{\prime }|$ and $d=|\mathbf{r}-{\mathbf{r}}^{\prime }|$, givesHere $|\mathbf{r}-\mathbf{p}{|}^2=d^2/s^2$ was used. The concentrated boundary data have unit integral, so the boundary representation evaluates this exterior Poisson kernel at $\mathbf{r}=(-1,0)$. Thus the bounded solution on either part of the exterior $x$-axis isIt is positive on both intervals, diverges as $x^{\prime }\to -1^{-}$ at the boundary source, and tends to $1/(2\pi )$ as $|x^{\prime }|\to \infty$. Boundedness at infinity selects this solution; otherwise one could add an arbitrary multiple of $\log |{\mathbf{r}}^{\prime }|$ without changing the circle's boundary data.

The Poisson integral convolves boundary data with the Poisson kernel to produce a harmonic function in a half-space. In the upper half-plane,

For $F\in L^1(\partial \mathbb{D})$ and a Lebesgue point ${\xi }_0$, its Poisson integral tends to $F({\xi }_0)$ through every fixed Stolz region. If $\delta =1-|z|$, the Poisson kernel there is at most $C_A\delta /({\delta }^2+d(\xi ,{\xi }_0{)}^2)$, where $d$ is shorter angular distance. Dyadic annuli about ${\xi }_0$ bound the near contribution by small local averages of $|F-F({\xi }_0)|$ times a summable sequence $C_A{2}^{-k}$. The contribution outside a fixed small arc is $O(\delta )\Vert F-F({\xi }_0){\Vert }_1$ and vanishes. This is the local kernel argument underlying the Fatou theorem for bounded holomorphic functions.

For integrable boundary data $g$ on the unit circle, this follows from the Poisson kernel bound $C_A\delta /({\delta }^2+d^2)$ in a fixed Stolz region. The arc of radius $\delta$ and successive dyadic annuli contribute at most $C_A{2}^{-k}Mg(\xi )$, which sum to the displayed estimate. This links harmonic extension to the Hardy-Littlewood maximal function and its strong $L^p$ bounds for $p>1$.

For radial SLE at parameter two, the normalized slit-domain Poisson kernel at the growing tip is a local martingale. This observable provides the coupling to loop-erasure of planar Brownian motion, oriented from the boundary point to the interior target. Its discrete counterpart is a loop-erased random walk.

For a positive harmonic function on the unit disc, choose a global harmonic conjugate and apply the Carathéodory derivative bound for the right half-plane to $u+iv$. This gives $|\nabla u(0)|\le 2u(0)$ by the Cauchy-Riemann equations. A disk automorphism sending zero to $z$ has derivative modulus $1-|z{|}^2$, so the chain rule gives the displayed bound. Composing a boundary-pole Poisson kernel with a disk automorphism attains equality, proving the coefficient is optimal at every point.