Analytic function and absolute convergence (H. Priestley , Introduction to Complex analysis, exercise 5.5)
Suppose $f(z)= \sum_{n=0}^\infty c_n z^n$ for $z \in \Bbb C$.
Prove that for all $R$: $$\sum_{n=0}^\infty |c_n| R^n \le 2M(2R)$$ where $M(r):=\sup\{|f(z)| : |z|=r\}$
I know that holomorphic functions attain their maximum on a boundary , so I tried $f(-R) + f(R)$ vs $f(2R)$ but I am still stuck
Please can someone help me ?
 A: many thanks for the hint zhw!
$$c_n=\frac{1}{2i\pi}\int\frac{f(z)}{z^{n+1}}dz$$
so estimating on the circle $\gamma(0,2R)$
$$|c_n|=|\frac{1}{2i\pi}\int\frac{f(z)}{z^{n+1}}dz| \le |\frac{1}{2i\pi}|\int|\frac{f(z)}{z^{n+1}}dz| \le \frac{1}{2\pi} \frac{M(2R)}{(2R)^{n+1}}2\pi(2R) $$
so
$$\sum_{n=0}^\infty |c_n| R^n \le  M(2R) \sum_{n=0}^\infty \frac{1}{(2R)^n}R^n=2M(2R)$$
A: I'm assuming you cannot use Laurents result here because the result would be trivial in that case. 
So my suggestion is to show this result for the real numbers, does the result hold? This is a major step in seeing how this is generalized. Now the interesting step, the assumption is our function is analytic? Well doesn't that entail something about what equations our function must satisfy? I think Cauchy-Riemann did something with that. Well if thats the case, it says something about this thing named after a guy called Lipschitz. I think that looks a great deal like the result you are trying to go for..strange.
Good luck.
A: Hint: Cauchy's estimates for $f^{(n)}(0)$.
