Regarding Poisson kernel being in $h^p(\mathbb{D})$ Let $\mathbb{D}$ denote the open unit disc in the complex plane. Let $h^p(\mathbb{D})$ be the space of all harmonic functions $f$ on $\mathbb{D}$ such that $$\sup_{0\leq r<1}\left(\frac{1}{2\pi}\int_{0}^{2\pi}|f(r e^{i\theta})|^p d\theta\right)^{1/p}$$ is finite. It is said that the Poisson kernel $$P(r e^{i\theta})=\frac{1-r^2}{1+r^2-2r \cos\theta}$$does not belong to $h^p(\mathbb{D})$ for $p>1$. Can you tell why?
 A: This is not an answer, just a heuristic way to think about this problem.  I'm setting $2\pi:=1$ for brevity in what follows.
We know either from Cauchy's integral formula or from the Dirichlet problem on the disk that $P_r(\theta)=P(re^{i\theta})$ is an approximate delta function, in the sense that $(P_r\ast f)(\theta)\to f(\theta)$ as $r\nearrow 1$, where $(f\ast g)(\theta)=\int_0^{2\pi}f(\theta-\phi)g(\phi)d\phi$ is convolution over the circle.  Moreover, $P_r$ is positive, so $P_r=|P_r|$, so in the $p=1$ case, the Hardy norm is just
$$
\sup_{r\in(0,1)}\int P_r(\theta)d\theta=\sup_r (P_r\ast 1)(0) = 1(0)=1.
$$
Now the $p>1$ case is clear, since we can write
$$
\int |P_r(\theta)|^pd\theta=(P_r\ast |P_r|^{p-1})(0).
$$
Formally, as $r\nearrow 1$, or $r=1-\delta$ this becomes $P_{1-\delta}^{p-1}(0)=\frac{1}{\delta^{p-1}}$, which blows up.
A: The denominator of $P(re^{i\theta})$ equals $(1-r)^2 + 2r(1-\cos \theta)$ and $0\le 1-\cos \theta \le \theta^2/2.$ Thus
$$P(re^{i\theta}) \ge \frac{1-r^2}{(1-r)^2+r\theta^2} \ge \frac{1-r}{(1-r)^2+\theta^2}.$$
Therefore
$$\tag 1\int_0^{2\pi}P(r e^{i\theta})^p\,d\theta \ge (1-r)^p\int_0^{2\pi}\frac{d\theta}{[(1-r)^2+\theta^2]^p}.$$
To finish, make the change of variables $\theta = (1-r)t.$ This will show the left side of $(1)$ is greater than a constant times $(1-r)^{1-p},$ hence tends to $\infty.$
