Poisson kernel

Remarks

The kernel is positive, and $P(z, \zeta) = \operatorname{Re} \dfrac{\zeta + z}{\zeta - z}$, so $z \mapsto P(z, \zeta)$ is harmonic 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 $t$.