I was reading some papers, and I don't know how to prove following statement: Let $f$ be conformal map of $\mathbb{D}$ onto an unbounded domain,such that, $f(0)=0$ and $dA$ denote Lebesgue measure on $\mathbb{D}$. Can you help me to prove: if $\iint_{\mathbb{D}}|f(z)|^{p-2}|f'(z)|^2\log\frac{1}{|z|}dA(z)<\infty$ for some p>0, then $f\in H^p(\mathbb{D})$?
Author of papers (where I found this statement) said that this follows from Shinji Yamashita's criterion for functions to be of Hardy class H^p (https://pdfs.semanticscholar.org/6857/938a86acd170061b7a22e9a07559e24ff08d.pdf?_ga=2.254660301.438877210.1589970143-1780474168.1589970143 ), but I don't understand why.
Please, can anyone help me?
The paper referred shows that $\iint_{\mathbb{D}}|f(z)|^{p-2}|f'(z)|^2(1-|z|)dA(z)<\infty$ is equivalent with $f \in H^p$; now notice that for a fixed $r<1$, $\iint_{\mathbb{D_r}}|f(z)|^{p-2}|f'(z)|^2(1-|z|)dA(z) < \infty$ for any analytic $f$ in the unit disc and the same is true for $\iint_{\mathbb{D_r}}|f(z)|^{p-2}|f'(z)|^2\log\frac{1}{|z|}dA(z)$ (where $\mathbb{D_r}$ is the disc of radius $r$) because the function being continuos on the closed disc now, the only trouble would be at a zero of $f$ (when $0<p<2$) or at $0$ for the second integral because of the infinities there.
But if $f(w)=0, w \ne 0$ and $w$ has order $k$ the integral near $w$ is like the integral of $|z-w|^{k(p-2)}|z-w|^{2k-2}=|z-w|^{kp-2} = |z-w|^{-2+\delta}$ and that is integrable in two dimensions (using local polar coordinates the integral becomes a one variable integral of $r^{-1+\delta}$ near $0$)
For $f(0)=0$ we get an extra $-\log r$ near $0$ but that doesn't matter since as above $r^{-1+\delta}\log r$ is still integrable near zero
So the gist of the matter is the behavior as $r \to 1$ but then if $t=1-r, -\log r= -\log(1-t)=t+O(t^2), t \to 0$ which means that the two integrals are equivalent as behaaviour, so in other words $\iint_{\mathbb{D_r}}|f(z)|^{p-2}|f'(z)|^2\log\frac{1}{|z|}dA(z) < \infty$ is equivalent to $\iint_{\mathbb{D}}|f(z)|^{p-2}|f'(z)|^2(1-|z|)dA(z)<\infty$ and hence by the result in the paper, equivalent with $f \in H^p$
As an aside it is instructive to do the case $p=2$ by hand as then the integral is explicit using polar coordinates, namely $-2\pi \int_0^1\sum {n^2|a_n|^2r^{2n+1}\log r}dr$, where $f(z)=\sum {a_nz^n}$, which by parts becomes
$-2\pi\sum {n^2|a_n|^2r^{2n+2}/(2n+1)\log r}|_0^1+2\pi \int_0^1\sum {n^2|a_n|^2r^{2n+1}/(2n+1)}dr$
But the evaluation is $0$ since at $1$ we have the $\log r$ term that vanishes (same as $1-r$ in the paper), so we remain with:
$2\pi\sum {n^2|a_n|^2r^{2n+2}/(2n+1)^2}|_0^1=2\pi \sum \frac{n^2}{(2n+1)^2}|a_n|^2$ and it is now clear that the series we get is comparable with $\sum |a_n|^2$ since $1/5 \le \frac{n^2}{(2n+1)^2} \le 1, n \ge n_0$
For general $p$ the paper shows how to do it directly using what are called the Hardy-Stein identities (the relation before $2.1$)