$\mathbb D$ is the unitary disk centered at $0$. Does the following function belong to $H^1(\mathbb D)$? .$$f_\epsilon(z) = \frac{1}{(1-z)\left(\frac{1}{z}\log\frac{1}{1-z}\right)^{1+\epsilon}}, z\in\mathbb D, \epsilon>0 $$
First of all it's easy to see that $f_\epsilon$ is holomorphic on the unitary disk, because it's singularity at $z=0$ is removable. We need to check that the integral $$\int_{0}^{2\pi}\left|\,f_\epsilon(re^{i\theta})\,\right|\,d\theta $$ remains bounded for all positive values of r, in the range of $[0,1)$. I try to solve this problem as follows $$\int_0^{2\pi} \frac{1}{|1-re^{i\theta}|}\frac{|re^{i\theta}|^{1+\epsilon}}{\left|\log\frac{1}{1-re^{i\theta}}\right|^{1+\epsilon}}\,d\theta \leq \int_0^{2\pi} \frac{1}{|1-re^{i\theta}|}\frac{2^{1+\epsilon}}{\left|\log\left|\frac{1}{1-re^{i\theta}}\right|+i Arg\left(\frac{1}{1-re^{i\theta}}\right)\right|^{1+\epsilon}}\,d\theta. \tag1$$ $\left|\log\left|\frac{1}{1-re^{i\theta}}\right|+i Arg\left(\frac{1}{1-re^{i\theta}}\right)\right| = \left(\log^2\left|\frac{1}{1-re^{i\theta}}\right|+Arg^2\left(\frac{1}{1-re^{i\theta}}\right)\right)^{\frac{1}{2}}$ and $-\frac{\pi}{2}\leq Arg\left(\frac{1}{1-re^{i\theta}}\right)\leq\frac{\pi}{2} \Rightarrow$
$\Rightarrow 0\leq Arg^2\left(\frac{1}{1-re^{i\theta}}\right)\leq \dfrac{\pi^2}{4}$. So integral (1) $$\leq \int_{0}^{2\pi}\frac{2^{1+\epsilon}}{|1-r\epsilon^{i\theta}|\left|\log\left|\frac{1}{1-re^{i\theta}}\right|\right|^{1+\epsilon}}\,d\theta=\int_{A_r}...+\int_{B_r}...\tag2 $$where $$A_r=\{\theta:|1-re^{i\theta}|\geq1\}\,\, \hbox{and}\,\,B_r=\{\theta:|1-re^{i\theta}|<1\}. $$ Now I shall use things like $$|1-re^{i\theta}|^2= (1-r)^2 + 4\sin^2\left(\frac{\theta}{2}\right)\geq 4\sin^2\left(\frac{\theta}{2}\right)\Rightarrow \frac{1}{|1-re^{i\theta}|}\leq\frac{1}{2\sin\frac{\theta}{2}}$$. But I don't end up to something that I can make a conclusion from it. Any help would be appreciated. The theory is very rich, so any other method would be really accepted.
Prove at first that $f_\epsilon$ belongs to Nevanlinna class; i.e $f_\epsilon$ can be written as a quotient of two holomorphic, bounded functions. Thus, the radial limits $f_\epsilon(e^{i\theta})$ will exist almost everywhere. Then, prove that $f_\epsilon(e^{i\theta})\in L^1(\mathbb{T})$.
For the first one, observe that $f_\epsilon(z)=\frac{\phi(z)}{\psi(z)}$, where $\phi(z)=\big(\frac{1}{z}\log\frac{1}{1-z}\big)^{-1-\epsilon}$ and $\psi(z)=1-z$ and $\phi, \psi\in H^\infty$, because they are holomorphic on $\mathbb{D}$ and $|\psi(z)|\leq 2$ και $|\phi(z)|\leq 1$, since \begin{eqnarray*} |\phi(re^{i\theta})| & = & \frac{1}{\bigg(\frac{1}{r}\big|\log\frac{1}{1-re^{i\theta}}\big|\bigg)^{1+\epsilon}}=\frac{1}{\bigg(\frac{1}{r}\big|\log\frac{1}{|1-re^{i\theta}|}+iArg(\frac{1}{1-re^{i\theta}})\big|\bigg)^{1+\epsilon}}\leq \frac{1}{\bigg(\frac{1}{r}\big|\log\frac{1}{|1-re^{i\theta}|}\big|\bigg)^{1+\epsilon}}\\ &=& \frac{1}{\bigg(\frac{1}{r}\big|\log|1-re^{i\theta}|\big|\bigg)^{1+\epsilon}}\leq \frac{1}{\bigg(\frac{1}{r}\big|\log|1-r|\big|\bigg)^{1+\epsilon}}=\frac{1}{\bigg(-\frac{\log(1-r)}{r}\bigg)^{1+\epsilon}} \end{eqnarray*} But $g(r)=-\frac{\log(1-r)}{r}$ is an increasing function of r, thus, it is easy to see that $|\phi(z)|\leq 1$.
For $\theta\in [-\pi,0)\cup(0,\pi]$ we have that $$ \frac{1}{|1-e^{i\theta}|}=\bigg(\frac{1}{|1-e^{i\theta}|^2}\bigg)^{1/2}=\bigg(\frac{1}{2-2\cos\theta}\bigg)^{1/2}=\bigg(\frac{1}{4\sin^2\frac{\theta}{2}}\bigg)^{1/2}=\frac{1}{2|\sin\frac{\theta}{2}|} $$ Since $\log\frac{1}{1-e^{i\theta}}=\log\frac{1}{\big|1-e^{i\theta}\big|}+i Arg\big(\frac{1}{1-e^{i\theta}}\big)$ and $-\frac{\pi}{2}<Arg\big(\frac{1}{1-e^{i\theta}}\big)<\frac{\pi}{2}$, \begin{eqnarray*} \bigg|\frac{1}{e^{i\theta}}\log\frac{1}{1-e^{i\theta}}\bigg|= \bigg|\log\frac{1}{|1-e^{i\theta}|} + iArg\big(\frac{1}{1-e^{i\theta}}\big) \bigg|&=&\sqrt{\log^2\frac{1}{|1-e^{i\theta}|}+Arg^2\big(\frac{1}{1-e^{i\theta}}\big)}\\ &=&\sqrt{\log^2\frac{1}{2|\sin\frac{\theta}{2}|}+Arg^2\big(\frac{1}{1-e^{i\theta}}\big)} \end{eqnarray*}
Thus for $\theta\in [-\pi,0)\cup(0,\pi]$ $$ |f_\epsilon(e^{i\theta})|=\frac{1}{|1-e^{i\theta}|}\frac{1}{\bigg|\frac{1}{e^{i\theta}}\log\frac{1}{1-e^{i\theta}}\bigg|^{1+\epsilon}}=\frac{1}{2|\sin\frac{\theta}{2}|}\frac{1}{\bigg(\log^2\frac{1}{2|\sin\frac{\theta}{2}|}+Arg^2\big(\frac{1}{1-e^{i\theta}}\big)\bigg)^{(1+\epsilon)/2}} $$ But, $$\frac{1}{1-e^{i\theta}} =\frac{1-e^{-i\theta}}{|1-e^{i\theta}|^2}=\frac{1-\cos\theta+i\sin\theta}{4\sin^2\frac{\theta}{2}}=\frac{2\sin^2\frac{\theta}{2}+i\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{4\sin^2\frac{\theta}{2}}=\frac{1}{2}+i\cot\frac{\theta}{2}$$ So, $Arg\big(\frac{1}{1-e^{i\theta}}\big)=\arctan(\cot\frac{\theta}{2})=\arctan\big(\tan(\frac{\pi}{2}-\frac{\theta}{2})\big)=\frac{\pi}{2}-\frac{\theta}{2}$. Then, \begin{eqnarray*} \int_{-\pi}^{\pi}|f_\epsilon(e^{i\theta})|d\theta & = & \int_{-\pi}^{\pi}\frac{1}{2|\sin\frac{\theta}{2}|}\frac{1}{\bigg(\log^2\frac{1}{2|\sin\frac{\theta}{2}|}+\big(\frac{\pi}{2}-\frac{\theta}{2}\big)^2\bigg)^{(1+\epsilon)/2}}d\theta\\ &=& \int_{-\pi/2}^{\pi/2}\frac{1}{|\sin u|}\frac{1}{\bigg(\log^2\frac{1}{2|\sin u|}+\big(\frac{\pi}{2}-u\big)^2\bigg)^{(1+\epsilon)/2}}du \end{eqnarray*} The function which is inside the integral is continuous on intervals of the form $[-\frac{\pi}{2},-\delta]\cup[\delta,\frac{\pi}{2}]$, where $\delta>0$, therefore it is bounded and the integral is finite on sets $[-\frac{\pi}{2},-\delta]\cup[\delta,\frac{\pi}{2}]$. It remains to prove that the integral is finite on $[-\delta,\delta]$. For $u\in [0,\delta]$, oserve that $$ \lim_{u\rightarrow 0}\frac{u(\log\frac{1}{u})^{1+\epsilon}}{\sin u \bigg(\log^2\frac{1}{2|\sin u|}+(\frac{\pi}{2}-u)^2\bigg)^{(1+\epsilon)/2}}=1 $$ Now take a $\delta$ such that $$ \frac{1}{2}\frac{1}{u\big(\log\frac{1}{u}\big)^{1+\epsilon}}<\frac{1}{\sin u \bigg(\log^2\frac{1}{2|\sin u|}+(\frac{\pi}{2}-u)^2\bigg)^{(1+\epsilon)/2}} <\frac{3}{2}\frac{1}{u\big(\log\frac{1}{u}\big)^{1+\epsilon}} $$ for $0<u<\delta$. Thus, $$ \int_0^\delta \frac{du}{\sin u \bigg(\log^2\frac{1}{2|\sin u|}+(\frac{\pi}{2}-u)^2\bigg)^{(1+\epsilon)/2}}<\infty \Leftrightarrow \int_0^\delta \frac{du}{u\big(\log\frac{1}{u}\big)^{1+\epsilon}}<\infty $$ which is valid since $$\int_0^\delta \frac{du}{u\big(\log\frac{1}{u}\big)^{1+\epsilon}}=\int_{\log\frac{1}{\delta}}^\infty \frac{1}{x^{1+\epsilon}}dx=\frac{1}{\epsilon \big(\log\frac{1}{\delta}\big)^\epsilon}<\infty$$ Similarly, $$ \int_{-\delta}^0 \frac{du}{\sin u \bigg(\log^2\frac{1}{2|\sin u|}+(\frac{\pi}{2}-u)^2\bigg)^{(1+\epsilon)/2}}<+\infty $$ Therefore, $f_\epsilon(e^{i\theta})\in L^1(\mathbb{T})$ and $f_\epsilon\in H^1$