$e^{\lambda \frac{1+z}{1-z}}\in H^2(\mathbb{D})$?

Question

for $\lambda>0$ i want to prove that

$$f_\lambda(z)=e^{\lambda \frac{1+z}{1-z}}\in H^2(\mathbb{D})$$

It is very complicated to calculate the integral: $$\int_{0}^{2\pi}|f_\lambda(re^{it})|^2dt$$

Honestly, i don't know how to start. Maybe we can show that $f_\lambda$ is bounded on $\mathbb{D}$? but how? Can someone give me a hint please ?

Answer 1

Suppose $f(z) \in H^2$. Then $$ f(z) = \sum_{n=0}^{\infty}a_n z^n,\;\;\; \|f\|_{H^2}^2=\sum_{n}|a_n|^2 < \infty, \\ |f(z)| \le \left(\sum_{n=0}^{\infty}|a_n|^2\right)^{1/2}\left(\sum_{n=0}^{\infty}|z|^{2n}\right)^{1/2} = \|f\|_{H^2}\frac{1}{\sqrt{1-|z|^2}}. $$ On the other hand, for $0 < t < 1$, and $\Re\lambda > 0$, $$ |f_{\lambda}(t)|=\exp\left(\Re\lambda\frac{1+t}{1-t}\right).$$ So $f_{\lambda}$ is not in $H^2$ for $\Re\lambda > 0$.