Problem: Suppose $f$ is analytic inside and on the unit circle $|z|=1$. Prove that if $|f(z)| \leq M$ for $|z|=1$, then $|f(0)| \leq M$ and $|f'(0)| \leq M$. What estimate can you give for $|f^n(0)|$?
Attempt:
In Cauchy's integral formula $f^{n}(z_0)=\frac{n!}{2 \pi i} \int_{|z|=1} \frac{f(z)}{(z-z_0)^{n+1}}dz$
To get $f$ as a function of $z$ replace $z$ with $\zeta$ and $z_0$ with $z$
So $f^{n}(z)=\frac{n!}{2 \pi i} \int_{|\zeta|=1} \frac{f(\zeta)}{(\zeta-z)^{n+1}}d\zeta$
Then $f^{n}(0)=\frac{n!}{2 \pi i} \int_{|\zeta|=1} \frac{f(\zeta)}{\zeta^{n+1}}d\zeta$
Then by the ML inequality
$|f^{n}(0)|=|\frac{n!}{2 \pi i} \int_{|\zeta|=1} \frac{f(\zeta)}{\zeta^{n+1}}d\zeta| \leq \frac{n!}{2 \pi }\frac{M 2\pi}{\zeta^{n+1}}$
The answer given in the back of the book says
$|f^{n}(0)| \leq \frac{n!}{2 \pi }M 2\pi=n!M$ thus $|f(0)| \leq M$ and $|f'(0)| \leq M$
Here I am confused why 1 is an upper bound for $\frac{1}{|\zeta^{n+1}|}$ I thought $\zeta$ would be a point on the circle $|\zeta|=1$ and since $|\zeta|$ would be a real number between $0$ and $1$, this would make the numerator larger so that $\frac{n!}{2 \pi }\frac{M 2\pi}{\zeta^{n+1}} \geq \frac{n!}{2 \pi }M 2\pi=n!M$
What's going on here? Also is it right to change the variables in Cauchy's formula from $z$ to $\zeta$ and from $z_0$ to $z$, and since the integral was along the contour $|z|=1$ would this change of variables imply the new contour would be $|\zeta|=1$?
Best Answer
You need the magnitude in the inequality. The correct inequality is $$\left|f^{n}(0)\right|=\left|\frac{n!}{2 \pi i} \int_{|\zeta|=1} \frac{f(\zeta)}{\zeta^{n+1}}d\zeta\right| \leq \left|\frac{n!}{2 \pi i}\right| \int_{|\zeta|=1} \frac{\left|f(\zeta)\right|}{\left|\zeta^{n+1}\right|}|d\zeta| \le \frac{n!}{2 \pi }\frac{M 2\pi}{1^{n+1}} = n!M$$
Note that the integrand is $\frac {f (\zeta)}{\zeta^{n + 1}}$, so while using the ML inequality you have to bound the magnitude of this expression as $\zeta$ ranges over the contour $|\zeta| = 1$. That turns out to be $\left|\frac{f (\zeta)}{\zeta^{n + 1}} \right| \le \frac M {1^{n + 1}}$.