# Integrability of the Poisson kernel

Let $$P_t(x)=\frac{2t}{\omega_{n+1} (|x|^2+t^2)^\frac{n+1}{2}}$$ denote to the Poisson kernel on the upper half space $\mathbb{R}^n\times \mathbb{R}_+$. How do I see $P_t \in L^1 \cap L^\infty$ and why can I conclude that $P_t \in L^p\ \forall\ 1\leq p \leq \infty$?

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In spherical  coordinates the problem reduces to a one-dimensional integral. – Andrew Jul 11 '12 at 16:40

## 2 Answers

Being $P_t$ continuous, in order to prove $P_t\in L^1\cap L^{\infty}$ it is sufficient the relation $$P_t(x)=\mathcal{O}\left(\frac{1}{|x|^{n+1}}\right),\text{ as }|x|\to\infty.$$

About your latest question, it is a special instance of the interpolation theorem.

Let be $E$ a measurable subset of $\mathbb{R}^N$ and $p,q,r\in[1,\infty]$ with $p\leq r\leq q.$ If $f\in L^p(E)\cap > L^q(E)$ then $f\in L^r(E),$ with $$||f||_r\leq||f||_p^{\alpha}||f||_q^{1-\alpha}$$ where $\alpha\in > [0,1]$ is determined through $\frac{\alpha}{p}+\frac{1-\alpha}{q}=\frac{1}{r}.$

As you can figure out it is an easy application of Hoelder inequality.

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As for boundedness, can you see which value of x yields the maximum of the kernel? x is in the denominator... And once a function f is bounded, say, by M, you get $|f|^p\le M^{...}|f|$. (Trying to leave something for you to work out. )

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