On an inequality for harmonic functions

Question

I'm trying to understand the following;

$| u(z)|^{p}= |\int P_{z}u \ d\lambda |^{p} \le (\int P_{z} |u |^{p} \ d\lambda) (\int P_{z})^{p-1} $

where $P_z$ is the Poisson kernel and $u$ is harmonic in unit discs of bounded hardy $p$-norm, it's supposed to follow by Hölder but I cant see it. I can't get the $P_z$ into the the integral $\int P_{z} |u |^{p} \ d\lambda$ or $q$ modulus off $|P_{z} |^{q}$

Answer 1

Hölder says that $$\bigg|\int fg\bigg| \le \bigg(\int |f|^p\bigg)^{1/p} \bigg(\int |g|^q\bigg)^{1/q}$$ when $\frac1p+\frac1q=1$. Apply this with $f = (P_z)^{1/p}u$ and $g=(P_z)^{1/q}$, and then raise both sides to the $p$th power.

Answer 2

Hint: Write $$P_qu=(P_q)^{1/p}u\cdot(P_q)^{1-1/p}$$ and apply Hölder. You will find useful rewriting the Hölder pair relation as $$1/q=1-1/p$$