Counter example for M. Riesz conjugate function

Question

M. Riesz claim than for $1 < p < \infty$, that $A_p$ is constant: $$|| \psi h||_p \leq A_p||h||_p$$ $$ \psi h(z) = \int_T \frac{e^{it}+z}{e^{it}-z} h(e^{it})d\rho $$ The counter example for p = $\infty$ is given by conformal mapping of U to a vertical strip, But I cannot figure out p = 1 one.

Answer 1

The traditional argument for $p=1$ is to use duality, showing that if the inequality held for $p=1$ then it would hold for $p=\infty$. One can give an explicit example: Let $$f(z)=\frac{1}{1-z}=\frac{1-\overline z}{|1-z|^2}=u+iv\quad(|z|<1).$$So $$u(re^{it})=\frac{1-r\cos(t)}{1-2r\cos(t)+r^2}, v(re^{it})=\frac{-r\sin(t)}{1-2r\cos(t)+r^2}\quad(0\le r<1).$$

Now since $u\ge0$ it's clear that $$\frac1{2\pi}\int_0^{2\pi}|u(re^{it})|\,dt=u(0).$$(Thanks to Conrad for pointing this out.) Otoh $$\frac1{2\pi}\int_0^{2\pi}|v(re^{it})|\,dt$$must be unbounded, because otherwise $f\in H^1(\Bbb D)$, which would imply that the boundary values of $f$ on the circle were integrable.