Prove that $$\sqrt{\frac{1+z}{1-z}}$$ is in the Hardy space $H^1(\mathbb{D})$, where we take the principal branch of the square root.
Answer- let $f(z) = \frac{(1+z)^{1/2}}{(1-z)^{1/2}}$ analytic on $|z|<1$ and continuous on $|z|\le 1,z\ne 1$
as $t\to 0$, $f(e^{it}) \sim C t^{-1/2}$
whence $f\in L^1(|z|=1)$.
The Hardy space $H^1(\mathbb{D})$ is the vector space of holomorphic functions $f$ on the open unit disk that satisfy: $$ \sup_{0\leq r<1}\left(\frac{1}{2\pi} \int_0^{2\pi}\left|f \left (re^{i\theta}\right )\right| \; \mathrm{d}\theta\right)<\infty.$$ Question Why to prove that $f\in H^1\mathbb{(D)}$, it is enough to prove that $f\in L^1(|z|=1)$?
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