P-norm of the Poisson Kernel

Question

If we have $P_r(\theta)=\sum_{n=-\infty}^{\infty}r^{|n|}e^{in\theta}=\frac{1-r^2}{1-2r\cos(\theta)+r^2},$

Is there a quick way to compute the norm $\|P_r\|_p$, or an upper estimate?

Answer 1

"Compute" $||P_r||_p$? II doubt it; in fact I doubt that a closed form exists.

An upper estimate? Certainly. It's clear that $$||P_r||_\infty=\frac{1-r^2}{1-2r+r^2}=\frac{1+r}{1-r},$$and assuming we're talking about the normalized Lebesgue measure $dt/2\pi$ as usual it's clear that $$||P_r||_p\le||P_r||_\infty.$$

Of course that's a hugely bad upper estimate. You could get something marginally better by noting that $||P_r||_1=1$ (since $P_r>0$ and the Fourier series shows that $\int P_r=1$) and $$||P_r||_p^p\le||P_r||_1||P_r||_\infty^{p-1}.$$

Edit: I find it surprising that in fact it's easy to show that such a crude estimate is best possible, within a constant multiple. We have $||P_r||_p\le c/(1-r)^{1-1/p}$ and we want to show that $||P_r||_p\ge c/(1-r)^{1-1/p}$ of course with a different constant. Parseval shows that the second inequality holds for $p=2$, and now $$||P_r||_2^2\le||P_r||_p||P_r||_{p'}$$gives the lower estimate on $||P_r||_p$ from the upper estimate on $||P_r||_{p'}$.