holomorphic on unit disk, then $\frac{1}{2\pi}\int_{0}^{2\pi}\left|f\left(re^{i\theta}\right)\right|^{p}d\theta\geq\left|f\left(0\right)\right|^{p}$

Asked Feb 08 '23 at 11:46

Active Feb 08 '23 at 12:04

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Let $f$ be holomorphic on $\left\{ z\in\mathbb{C}:\left|z\right|<1\right\} $ and $0<r<1$. Then for every $p>0$: $$\frac{1}{2\pi}\int_{0}^{2\pi}\left|f\left(re^{i\theta}\right)\right|^{p}d\theta\geq\left|f\left(0\right)\right|^{p}$$

I solved for two cases, the first is when $p=1$ and we get the inequality by Cauchy integral formula. the second, is where $f$ not vanishing on the open disk. then $f$ can be written as $f\left(z\right)=e^{g\left(z\right)}$, and again applying Cauchy formula for $f^{p}$. I was told that the general case can be proven using those cases, yet I'm not sure how to proceed.

edited Feb 08 '23 at 12:04

asked Feb 08 '23 at 11:46

Simple Conjugate

I think you are missing a factor of $\frac1 {2\pi}$ on the left. – Kavi Rama Murthy Feb 08 '23 at 11:58
Indeed, fixed. thanks for noticing. – Simple Conjugate Feb 08 '23 at 12:04
For $p \geq 1$ this follows by Mean Value Property of analytic functions and Holder's inequality. I doubt its validity for $0<p<1$. – Kavi Rama Murthy Feb 08 '23 at 12:18
It follows because $|f|^p$ is subharmonic for every $p>0$ when $f$ is holomoprhic, so the submean property in discs holds; more generally the integrals above increase with $r$ for a fixed $p>0$ – Conrad Feb 08 '23 at 13:48
Related: https://math.stackexchange.com/q/3406949/42969 – Martin R Feb 08 '23 at 18:34

holomorphic on unit disk, then $\frac{1}{2\pi}\int_{0}^{2\pi}\left|f\left(re^{i\theta}\right)\right|^{p}d\theta\geq\left|f\left(0\right)\right|^{p}$

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