The Hardy space $H^p(\mathbf{D})$ is the vector space of holomorphic functions $f$ on the open unit disk that satisfy: $$ \sup_{0< r<1}\left(\frac{1}{2\pi} \int_0^{2\pi}\left|f \left (re^{i\theta}\right )\right|^p \; \mathrm{d}\theta\right)^\frac{1}{p}<\infty. $$Consider $$ f(z)=\frac{z}{1-z} \zeta(\frac {1}{1-z})$$ where $\zeta$ denotes the Riemann zeta function.
Balazard Saias and Yor in a paper used that $f\in H^{1/3}(\mathbf{D}) $
Question - Prove that $f\in H^{1/3}(\mathbf{D})$
My Try -
$$ |f(re^{i\theta})|=|\frac{re^{i\theta}}{1-re^{i\theta}} | |\zeta(\frac {1}{1-re^{i\theta}})|$$
$$ |f(re^{i\theta})|=\frac{r}{|1-re^{i\theta}|} |\zeta(\frac {1}{1-re^{i\theta}})| \ \ \ ...(1)$$
$$\zeta(s)= s\int_1^\infty \frac{1-x+[x]}{x^{s+1}}dx +\frac{1}{s-1}, \Re(s)>0$$
$$|\zeta(s)|\leq \frac{|s|}{\Re(s)}+\frac{1}{|s-1|} , \ \ \ \ \Re(s)>0\ \ \ \ \ ... (2)$$
$$s=\frac{1}{1-re^{i\theta}} \Rightarrow|s|=\frac{1}{|1-re^{i\theta}|}$$
$$\Re(s) = \Re(\frac{1}{1-re^{i\theta}}) = \frac{1-r \ cos\theta}{|1-re^{i\theta}|^2}$$
$$\frac{1}{|s-1|}= \frac{|1-re^{i\theta}|}{r}$$
By equation (2),
$$|\zeta(\frac {1}{1-re^{i\theta}})| \leq \frac{|1-re^{i\theta}|}{(1-rcos\theta)}+\frac{|1-re^{i\theta}|}{r}$$
Equation (1) gives,
$$|f(re^{i\theta})|\leq \frac{r}{|1-re^{i\theta}|} [\frac{|1-re^{i\theta}|}{(1-rcos\theta)}+\frac{|1-re^{i\theta}|}{r} ] $$
$$|f(re^{i\theta})|\leq \frac{1+r(1-cos\theta)}{1-rcos\theta}$$
Further $1+r(1-cos\theta) \leq 1+2r$
$$|f(re^{i\theta})|\leq \frac{1+2r}{1-rcos\theta}$$
After this I am stuck. Please solve
On $\Re(s)=1/2$ $$\zeta(s) = \frac{s}{s-1}+ s\int_1^\infty (\lfloor x\rfloor -x)x^{-s-1}dx= O(|s|)$$ therefore $$\int_{|z|=1} |f(z)|^{1/3}|dz|=\int_{|z|=1} O(\frac1{|1-z|^{2/3}})|dz| <\infty$$