Littlewood subordination theorem

Remarks

The function $\abs{F}^p$ is subharmonic. Let $u$ be the harmonic function on the disc $\abs{z} < r$ with boundary values $\abs{F}^p$ on $\abs{z} = r$, given by the Poisson kernel; then $\abs{F}^p \le u$ on $\abs{z} \le r$. By the Schwarz lemma, $\abs{\omega(z)} \le \abs{z}$, so $\omega$ maps the closed disc of radius $r$ into itself and $\abs{F \circ \omega}^p \le u \circ \omega$ there. Since $u \circ \omega$ is harmonic, its mean over $\abs{z} = r$ is $u(\omega(0)) = u(0)$, which is the mean of $\abs{F}^p$ over $\abs{z} = r$. The hypothesis $\omega(0) = 0$ is needed. For the constant map $\omega \equiv c$ with $0 < c < 1$ and $F(z) = z$, the left side is $2\pi c^p$ and the right side is $2\pi r^p$. So the inequality fails for $r < c$.

Going deeper

For a student

Taking the supremum over $r$ shows that the composition operator $F \mapsto F \circ \omega$ is a contraction on each Hardy space $H^p$ when $\omega(0) = 0$. For a general holomorphic self-map $\omega$ one composes with a disc automorphism moving $\omega(0)$ to $0$, which shows that every composition operator on $H^p$ is bounded. The proof only uses that $\abs{F}^p$ is subharmonic, so the same inequality holds for any subharmonic function in place of $\abs{F}^p$. This is the starting point of the theory of composition operators on spaces of analytic functions.