1. ## Inequality

Lemma. Let $\displaystyle f(z)$ be analytic inside and on the boundary of the disk $\displaystyle |z-z_0| < \rho_0$ and satisfy $\displaystyle |f(z)| \leq M_0$ for $\displaystyle |z-z_0| \leq \rho_0$. If $\displaystyle |f(z_0)| = M_0$ then $\displaystyle |f(z)| = M_0$ for all $\displaystyle z$ in the disk.

Proof. Consider any circle $\displaystyle |z-z_0| = \rho \leq \rho_0$. By the Cauchy Integral Formula we know that $\displaystyle f(z_0) = \frac{1}{2 \pi i} \oint_{|\xi-z_0| = \rho} \frac{f(\xi) \ d \xi}{\xi - z_0}$. For each $\displaystyle \xi$ on this circle with $\displaystyle |f(\xi)| < M_0$ or $\displaystyle |f(\xi)| = M_0$. Assume that there is a $\displaystyle \xi_1$ on this circle for which $\displaystyle |f(\xi_1)| < M_0$. Then since $\displaystyle |f(\xi)|$ is continuous there must be a whole arc $\displaystyle C_1$ of the circle on which $\displaystyle |f(\xi)| < M_0$ for $\displaystyle \xi$ on $\displaystyle C_1$. Also $\displaystyle |f(\xi)| \leq M_0$ on the rest of the circle. Write $\displaystyle f(z_0) = \frac{1}{2 \pi i} \left[ \int_{C_1} \frac{f(\xi) \ d \xi}{\xi- z_0}+ \int_{\text{rest}} \frac{f(\xi)}{\xi-z_0} \ d \xi \right]$.

From here how do we establish that $\displaystyle |f(z_0)| < M_0$, a contradiction? From the M-L Inequality, we know that $\displaystyle |f(z_0)| < \frac{1}{2 \pi i} \left(2 \pi \rho M_0+ 2 \pi \rho M_0 \right) = \frac{2 \rho M_0}{i}$. How do we get this to $\displaystyle M_0$?

2. Originally Posted by manjohn12
Lemma. Let $\displaystyle f(z)$ be analytic inside and on the boundary of the disk $\displaystyle |z-z_0| < \rho_0$ and satisfy $\displaystyle |f(z)| \leq M_0$ for $\displaystyle |z-z_0| \leq \rho_0$. If $\displaystyle |f(z_0)| = M_0$ then $\displaystyle |f(z)| = M_0$ for all $\displaystyle z$ in the disk.
This is similar to the maximum-modulos theorem. By Cauchy's theorem we know that:
$\displaystyle f(z_0) = \frac{1}{2\pi i} \oint_{|z-z_0|=r} \frac{f(z)}{z-z_0}dz \text{ where }0<r\leq \rho$
Therefore, (by definition of contour integral)
$\displaystyle f(z_0) = \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta}) d\theta$
Thus,
$\displaystyle M_0 = |f(z_0)| \leq \frac{1}{2\pi } \int_0^{2\pi} |f(z_0 + re^{i\theta}) |d\theta \leq \max\{ |f(z_0+re^{i\theta})| : \theta\in [0,2\pi]\}$

We have shown that on maximum value of $\displaystyle |f|$ on the disk $\displaystyle |z-z_0| = r$ is at least $\displaystyle M_0$. However, $\displaystyle |f|\leq M_0$ and so the maximum value of $\displaystyle |f|$ must be equal to $\displaystyle M_0$. Therefore, it follows that,
$\displaystyle \frac{1}{2\pi} \int_0^{2\pi} |f(z_0 + re^{i\theta})| d\theta = M_0$
The only way that integral can equal to $\displaystyle M_0$ is it at each $\displaystyle z$ on $\displaystyle |z-z_0|=r$ we have $\displaystyle |f|=M_0$.
Thus, $\displaystyle |f| = M_0$ for $\displaystyle |z-z_0| = r$.
But $\displaystyle r$ was arbitrary and by shrinking and expanding $\displaystyle r$ we get the whole disk $\displaystyle |z-z_0|\leq \rho$ with $\displaystyle |f| = M_0$.