# Prove the inequality?

Let $f$ be an analytic function in the unit disc without zeros satisfying $|f|\leqq 1$. Prove that $$\sup_{|z\leqq{1/5}|}|f(z)|^2\leqq \inf_{|z|\leqq{1/7}}|f(z)|$$ Help me please. These questions are in the book Banach spaces of analytic functions in page 40, exercises 4 (author Kenneth Hoffman). Thanks for help.

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have you tried something with Harnack's Inequality? – MBM Mar 31 '13 at 10:50

MBM's comment is right on: this is an exercise on Harnack's inequality. It requires a positive harmonic function, which you should be able to obtain from $\log |f|$.
A positive harmonic function $h$ in the unit disk is represented by the integral $$h(z)=\frac{1}{2\pi} \int_{\mathbb T} \frac{1-|z|^2}{|z-\zeta|^2} d\mu(\zeta) \tag1$$ where $\mu$ is a positive measure on the unit circle $\mathbb T$. By the triangle inequality, $1-|z|\le |z-\zeta|\le 1+|z|$. Plug this into (1) to get $$\frac{1}{2\pi} \frac{1-|z|^2}{(1+|z|)^2} \mu(\mathbb T) \le h(z)\le \frac{1}{2\pi} \frac{1-|z|^2}{(1-|z|)^2} \mu(\mathbb T) \tag2$$ The fact that we don't know $\mu$ is not a problem for estimating the ratio $h(z_1)/h(z_2)$, because $\mu(\mathbb T)$ cancels out: $$\frac{h(z_1)}{h(z_2)}\le \frac{1-|z_1|^2}{(1-|z_1|)^2} \frac{(1+|z_2|)^2}{1-|z_2|^2} \tag3$$ I leave it for you to do the arithmetics with $1/5$ and $1/7$.