Integral inequality involving an analytic function

Question

I have been trying to prove for any $\delta>0,$ $$ \int_0^{2\pi}\left|1+ e^{i\theta}f(e^{i\theta})\right|^{\delta}d\theta\leq \int_0^{2\pi}\left|1+e^{i\theta}\right|^{\delta}d\theta $$ for any analytic function $f(z)$ in $|z|\leq 1$ with $|f(e^{i\theta})|\leq 1.$ Can anyone help me in this?

Answer 1

Let $F(z)=1+z$ and $\phi(z)=zf(z)$. Then $\phi$ is an analytic function mapping the unit disk into itself, and $\phi(0)=0$, and your inequality becomes the subordination inequality $$\int_0^{2\pi}|F\circ\phi(e^{i\theta})|^\delta d\theta\leq\int_0^{2\pi}|F(e^{i\theta})|^\delta d\theta,$$ see, for example, J. E. Littlewood, Lectures on the Theory of Functions. Oxford University Press, 1944, Theorem 210 on p. 164.

Here is a short proof of the subordination inequality. Let $u$ be subharmonic in the disk, $v$ the least harmonic majorant (=harmonic function matching $u$ on the unit circle), and let $\phi$ be analytic, mapping the unit disk into itself, $\phi(0)=0$. Denote by $I$ the average over the unit circle. Then: $$I(u\circ \phi)\leq I(v\circ\phi)=v\circ\phi(0)=v(0)=I(v)=I(u),$$ where we used the average property of harmonic functions twice: once for $v$ and another for $v\circ\phi$.

In our application $u=|F|^\delta$.