Poisson integral on the unit disk ($P_r[u](t)$)

The disk Poisson integral is the harmonic function with boundary data $u$, given byIt is the real part of the Schwarz integral on the unit disk for real data. The nonnegative kernel has integral one; smooth boundary data extend smoothly to the circle. This disk formula should be distinguished from the half-plane formula.

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 $0<r<1$, the geometric series gives $P_r(t)=1+2{\sum }_{n\ge 1}{r}^n\cos nt$. Hence this positive even kernel has normalized integral one, and normalized convolution with it multiplies the $n$th Fourier coefficient by $r^{|n|}$. Its mass outside any fixed neighbourhood of zero tends to zero as $r\uparrow 1$. If an integrable function has finite one-sided limits $A,B$ at zero, evenness gives half the mass to each side and the convolution tends to $(A+B)/2$.

The Poisson kernel on the circle is nonnegative, has normalized integral one, and its mass outside any fixed arc about zero tends to zero as $r\uparrow 1$. Split the convolution error into a short arc, controlled by the modulus of continuity, and its complement, controlled by the vanishing kernel mass. This proves uniform convergence for every continuous periodic function. Its Fourier multiplier is $r^{|n|}$, so the convolution is the Abel sum of the Fourier series.

The geometric series gives $Q_r(t)=2{\sum }_{n\ge 1}{r}^n\sin nt$. Its convolution multiplier is $-i\mathrm{sgn}(n)r^{|n|}$ for nonzero frequencies and zero at frequency zero. Since the series is uniformly absolutely convergent for $r<1$, these identities hold for every Lebesgue integrable function by termwise integration. If all negative Fourier coefficients vanish, $Q_r\ast f=-i(P_r\ast f-\hat{f}(0))$.

Let an integrable function have finite one-sided limits $A,B$ at zero. Oddness gives $(Q_r\ast f)(0)=(2\pi {)}^{-1}{\int }_0^{\pi }{Q}_r(t)[f(-t)-f(t)]dt$. Also ${\int }_0^{\delta }{Q}_r(t)dt=\log [(1-2r\cos \delta +r^2)/(1-r{)}^2]=2\log (1/(1-r))+O_{\delta }(1)$. Thus $(Q_r\ast f)(0)/\log (1/(1-r))\to (A-B)/\pi$: make the near-zero error in $f(-t)-f(t)$ arbitrarily small and bound the remaining integral uniformly. When negative Fourier coefficients vanish, the conjugate-kernel identity makes this convolution bounded because $P_r\ast f$ tends to $(A+B)/2$. Therefore $A=B$. A removable discontinuity caused solely by altering a point value is not excluded.