M. Riesz's Theorem for $L^2(T)$

Question

Suppose that $f \in L^2(T)$ and define \begin{equation} F(z)=\frac{1}{2\pi}\int_T \frac{e^{it}+z}{e^{it}-z} f(e^{it}) dt \end{equation} for $z \in U$. M. Riesz's Theorem states that there exists a constant $C_2$ such that $\|F\|_2 \le C_2\|f\|_2$. I wonder whether we can apply the Parseval Theorem to prove this result and find the best value of $C_2$. Unfortunately, I don't know how to start. Any suggestion?

Answer 1

Let $$ P(z, e^{it}) = \frac{e^{it}+z}{e^{it}-z} $$

$$\frac{1}{2\pi} \int_T |F(re^{i\theta})|^2 d\theta = \frac{1}{2\pi} \int_T |\frac{1}{2\pi}\int_T P(re^{i\theta}, e^{it}) f(e^{it}) dt|^2 d\theta$$

$$ <\frac{1}{2\pi} \int_T |\frac{1}{2\pi}\int_T |P(re^{i\theta}, e^{it}) f(e^{it})| dt|^2 d\theta $$ Schwarz inequality $$ < \frac{1}{2\pi} \int_T \frac{1}{2\pi}\int_T |P(re^{i\theta}, e^{it})|^2 dt] ||f||_2d\theta $$ $$ < ||f||_2\frac{1}{2\pi} \int_T\frac{1}{2\pi}\int_T |P(re^{i\theta}, e^{it})|^2 dt d\theta $$ $$ C_2 = inf_{0\leq r < 1} \int_T\frac{1}{2\pi}\int_T |P(re^{i\theta}, e^{it})|^2 dt d\theta $$