---
tag: W19567AVYG
kind: definition
name: Poisson kernel
msc: [31A10, 31A05]
defines: TCYSWCTYYM
uses: [8G5ABXRK85, 0DEB0JMN0W]
sources:
  - url: https://en.wikipedia.org/wiki/Poisson_kernel
---

## Definition

Let $\mathbb{D} = \set{z \in \C : \abs{z} < 1}$ be the [[unit disc|8G5ABXRK85]] and $\mathbb{T} = \set{\zeta \in \C : \abs{\zeta} = 1}$ its boundary circle.
The *Poisson kernel* of $\mathbb{D}$ is the function $P \colon \mathbb{D} \times \mathbb{T} \to \R$ given by
$$P(z, \zeta) = \frac{1 - \abs{z}^2}{\abs{\zeta - z}^2}.$$
In polar form, for $0 \le r < 1$ and real $\theta$, one writes $P_r(\theta) = P(r e^{i\theta}, 1) = \dfrac{1 - r^2}{1 - 2r\cos\theta + r^2}$.

## Remarks

The kernel is positive, and $P(z, \zeta) = \operatorname{Re} \dfrac{\zeta + z}{\zeta - z}$, so $z \mapsto P(z, \zeta)$ is [[harmonic|0DEB0JMN0W]] on $\mathbb{D}$.
It has the expansion $P_r(\theta) = \sum_{n \in \Z} r^{\abs{n}} e^{i n\theta}$, from which $\frac{1}{2\pi}\int_0^{2\pi} P_r(\theta)\, d\theta = 1$.
Example: at $z = 0$ one gets $P(0, \zeta) = 1$ for every $\zeta$, so the Poisson integral at the centre is the plain mean over the circle.
For a continuous function $\varphi$ on $\mathbb{T}$, the Poisson integral $u(z) = \frac{1}{2\pi}\int_0^{2\pi} P(z, e^{it}) \varphi(e^{it})\, dt$ is harmonic on $\mathbb{D}$ and extends continuously to $\varphi$ on $\mathbb{T}$.
Non-example: the function $(z, \zeta) \mapsto 1/\abs{\zeta - z}^2$ is not the Poisson kernel; it lacks the factor $1 - \abs{z}^2$, it is not harmonic in $z$, and its integral over the circle is $1/(1 - \abs{z}^2)$, not $1$.
For the upper half-plane, the Poisson kernel is $\dfrac{1}{\pi}\dfrac{y}{(x - t)^2 + y^2}$ for $x + iy$ with $y > 0$ and real $t$.