Conjugate Poisson kernel on the circle ($Q_r(t)=2r\sin t/(1-2r\cos t+r^2)$)

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.