Actually someone asked this before (Click here). However, the answer given is too simple and I don't understand it. I tried to ask question in that post but received no reply. So I decided to ask again to seek for a more detailed and inspiring answer.
Suppose $f$ is an analytic inside and on a circle $C$ with $|f(z)|≤M$
on $C$ and suppose $z_0$ is a point inside $C$. Use Cauchy’s integral
formula to show that $|f(z_0)|^n≤KM^n$ where $K$ is independent of $n$.
Cauchy' Integral Formula:
Let $g:D(a;R) \to \mathbb{C}$ be an analytic
function. Choose $r>0$ such that $0<r<R$. Then, for all $z \in
D(a;r)$, we have $$g(z) = \frac{1}{2 \pi i}\int_{\partial D(a;r)}\frac{g(k)}{k-z}dk$$
I have trouble identifying the corresponding $r$ here. I know that I should let $D(a;R)$ be $C$ and $g=f^n$, but I have no idea how to proceed. Any help is appreciated.
Best Answer
You can integrate along $C$ itself. "Analytic inside and on" usually means analytic on a neighbourhood of $C$, so there is no problem justifying the use of Cauchy's integral formula. (Even if it just means analytic on the inside and continuous up to $C$, everything works fine.)
So let $z_0$ be inside $C$. Then by Cauchy's integral formula $$ f(z_0)^n = \frac{1}{2\pi i} \int_C \frac{f(z)^n}{z-z_0}\,dz $$ Hence by the "ML-inequality" or "standard estimation lemma" or whatever name you know it as:. $$ \lvert f(z_0)^n \rvert = \left\lvert \frac{1}{2\pi i} \int_C \frac{f(z)^n}{z-z_0}\,dz \right\rvert \le \frac{1}{2\pi} \cdot 2\pi r \cdot \frac{M^n}{\operatorname{dist}(z_0, C)}. $$
Note that the $K$ depends on $z_0$ but not on $n$. That's ok, because we want to take the $n$:th root of both sides and let $n \to \infty$ to recover the maximum modulus principle. More precisely, from the above $$ \lvert f(z_0) \rvert \le \left( \frac{r}{\operatorname{dist}(z_0, C)} \right)^{1/n} M $$ for all $n$. In partcicular (let $n \to \infty$), $$ \lvert f(z_0) \rvert \le M $$ and this estimate holds for all $z_0$ inside $C$.