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 $?