Below is a question that I'm attempting to do but so far have made no progress. Any suggestions would be helpful.
Show that whenever $0 < \alpha < \frac{1}{2}$, then $\left( \frac{1+z}{1-z} \right)^{\alpha}$ belongs to $H^2(U)$, where $U$ denotes the open unit disk.
I was thinking the writing the function in terms of its power series of as $$ e^{\log \left(\frac{1+z}{1-z} \right)^{\alpha} }, $$
but neitther has really helped. I wanted to try to use Littlewood's Subordination Principle.
We evaluate the integral $$ \int_{-\pi}^\pi |f(re^{i\theta })|^2 d\theta $$ for $f(z)=\left(\frac{1+z}{1-z}\right)^\alpha .$
Let $z=re^{i\theta }$, $-\pi\le\theta \le \pi,$ $\,0<r<1$, then \begin{align} \left|\frac{1+z}{1-z}\right|^2&=\frac{1+2r\cos \theta +r^2}{1-2r\cos\theta +r^2}\le \frac{1}{r}\cdot\frac{1+2r+r^2}{\frac{1}{r}+r-2\cos\theta }\\ &\le \frac{1}{r}\cdot\frac{4}{2-2\cos\theta }= \frac{2}{r}\cdot\frac{1}{1-\cos\theta } \end{align} Using the well-known inequality $$\sin\theta \ge \frac{2}{\pi}\,\theta \quad (0\le \theta \le \frac{\pi}{2}),$$ we have $$ 1-\cos\theta =2\sin^2 \left(\frac{\theta }{2}\right)\ge \frac{2}{\pi^2}\,\theta ^2\quad (-\pi\le\theta \le\pi).$$ Therefore (note that $0<2\alpha <1$) \begin{align} \int_{-\pi}^\pi |f(re^{i\theta })|^2 d\theta &\le \int_{-\pi}^\pi \left(\frac{\pi^2}{r}\cdot\frac{1}{\theta ^2}\right)^\alpha d\theta\\ &=2\left(\frac{\pi^2}{r}\right)^\alpha \int_{0}^\pi \frac{1}{\theta ^{2\alpha} }d\theta \\ &=2\left(\frac{\pi^2}{r}\right)^\alpha\frac{\pi^{1-2\alpha }}{1-2\alpha }, \end{align} which implies $$\lim_{r\to 1} \left(\frac{1}{2\pi}\int |f(re^{i\theta })|^2d\theta \right)^{1/2}<+\infty.$$