# Show this function is a good kernel on unit disk

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

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

2) $$\sup_{0

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!

## 1 Answer

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

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

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