I am having a bit of trouble showing the following function is a good kernel on the unit disk:

$$ U_r(e^{i \theta}):=\frac{(1+r)^2(1-r)\theta\sin\theta}{(1-2r\cos\theta + r^2)^2}, \text{for}~~ 0<r<1 $$

I recall that $f_r \in L^1(T)$, where $T$ is the unit disk, is a good kernel if it has the following properties:

1)for all $0<r<1$, $$ \frac{1}{2\pi}\int_{-\pi}^{\pi}f_r(e^{it})dt=1 $$

2) $$\sup_{0<r<1}\left( \int_{-\pi}^{\pi}|f_r(e^{it})|dt \right) < \infty $$

3) for all $\delta \in (0,\pi)$,

$$ \lim_{r \rightarrow 1^{-}} \left( \int_{\delta<|t|<\pi}|f_r(e^{it})|dt \right) =0 $$

I already figured out 3) and since $U_r$ is positive, 1) implies 2).

I just can't find a way to prove $\frac{1}{2\pi}\int_{-\pi}^{\pi}U_r(e^{it})dt=1 $. Any hints ?

Thank you!


Let $P_r(\theta)=\frac{1-r^2}{1-2r\cos\theta + r^2}$, the usual Poisson Kernel. Then integrating by parts:

$\frac{1}{2\pi}\int_{-\pi}^{\pi}U_r(e^{it})dt=-\frac{1}{2\pi}\frac{1+r}{2r}(\pi P_r(\pi)-(-\pi) P_r(-\pi))+\frac{1+r}{2r}\frac{1}{2\pi}\int_{-\pi}^{\pi}P_r(e^{it})dt=-\frac{1-r}{2r}+\frac{1+r}{2r}=1$

  • $\begingroup$ I suspêcted the Poisson Kernel was the key to this problem, thanks a lot! $\endgroup$ – D666 Mar 20 at 21:06
  • $\begingroup$ you are welcome $\endgroup$ – Conrad Mar 20 at 21:08
  • $\begingroup$ To be sure, how did you write U_r = u dv ? $\endgroup$ – D666 Mar 20 at 21:49
  • $\begingroup$ $U_r(e^{it})=-\frac{1+r}{2r}tP_r(t)'$ $\endgroup$ – Conrad Mar 20 at 22:00
  • $\begingroup$ Marvelous! Thank you! $\endgroup$ – D666 Mar 20 at 22:09

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