# Proving an inequality regarding a holomorphic function

Let $$f$$ be holomorphic in $$\mathbb{D}$$ and $$0. Prove that for every $$p\in(0,\infty)$$:$$\frac{1}{2\pi}\int_0^{2\pi}|f(re^{i\theta})|^pd\theta \geq|f(0)|^p$$

My attempt: First I assume that $$p=1$$. Then this result is a direct consequence of Cauchy's Integral formula. Now I let $$p>0$$ and $$f$$ that doesn't vanish on $$\mathbb{D}$$. In this case, $$g(z)=f^p(z)$$ is holomorphic in $$\mathbb{D}$$ and the result again follows from Cauchy's integral formula.

Now to the general case - let $$p>0$$ and $$f$$ holomorphic in $$\mathbb{D}$$. If $$f(0)=0$$ then the result is trivial, meaning we can assume $$f(0)\neq0$$. In this case, since $$f$$ is holomorphic we have some $$\delta>0$$ s.t $$f(z)\neq 0$$ for every $$z\in\mathbb{D}_\delta(0)$$, and we can use the previous result on $$\frac{1}{2\pi}\int_0^{2\pi}|f(\delta e^{i\theta})|^pd\theta$$. Now what I want to claim that if we take $$\delta, we have|: $$\frac{1}{2\pi}\int_0^{2\pi}|f(re^{i\theta})|^pd\theta \geq \frac{1}{2\pi}\int_0^{2\pi}|f(\delta e^{i\theta})|^pd\theta \geq|f(0)|^p$$ But I don't know how to show it (or even if it's true).

P.S I know that this has something to do with Hardy's theorem and Volume Integral Means, but I can't use such methods (I'm supposed to use a more fundamental complex analysis tools).

Any hint would be appreciated.

• Standard argument: First, replacing $f$ by a dilate and taking a limit as above shows that wlog $f$ has only finitely many zeroes. Now $f=Bg$, where $B$ is a Blaschke product and $g$ has no zero... Jul 2, 2022 at 11:44
• @DavidC.Ullrich I'm not familiar with Blaschke product
– GBA
Jul 2, 2022 at 12:00
• Use Holder inequality plus what you have proved for $p=1$. By Holder inequality I mean $\int |f| <= (\int |f|^p)^{1/p}$ which holds for $p>=1$. Jul 2, 2022 at 13:20
• Can you use that $\log |f(z)|$ is subharmonic? Or Jensen's formula? Jul 2, 2022 at 13:32
• @MartinR I can use the fact that this function is subharmonic. How would one go about it then?
– GBA
Jul 2, 2022 at 14:11

The inequality holds trivially if $$f(0) = 0$$, so we'll assume that $$f(0) \ne 0$$. Then $$\log|f(0)| \le \frac{1}{2 \pi}\int_0^{2 \pi} \log |f(r e^{i \theta}| \, d\theta \, .$$ since $$\log |f(z)|$$ is subharmonic, or because of Jensen's formula.
It follows that $$\log\left(|f(0)|^p\right) = p \log|f(0)| \le \frac{p}{2 \pi}\int_0^{2 \pi} \log |f(r e^{i \theta}| \, d\theta \\ = \frac{1}{2 \pi}\int_0^{2 \pi} \log \left(|f(r e^{i \theta}|^p\right) \, d\theta \le \log \left(\frac{1}{2 \pi}\int_0^{2 \pi} |f(r e^{i \theta}|^p \, d\theta \right) \, .$$ The last step is Jensen's inequality applied to the concave function $$t \mapsto \log(t)$$.