# $f$ is analytic, prove that for all $z$ with $|z|=1$, $\sum_{n=0}^\infty|a_nz^n|\leq 2\max\{|f(z)|:|z|=2\}$

Suppose that $$f(z)=\sum_{n=0}^\infty a_nz^n$$ for all $$z\in\mathbb{C}$$. Prove that for all $$z$$ with $$|z|=1$$, $$\sum_{n=0}^\infty|a_nz^n|\leq 2 \max\{|f(z)|:|z|=2\}$$.

My try: I have no idea about how to prove it. I guess it can proved by Cauchy estimates or use the relation $$f(z)=\sum_{n=0}^\infty\frac{f^n(0)}{n!}z^n.$$

• Are you sure you transcribed the homework correctly? – copper.hat Mar 17 at 17:58
• @copper.hat I am sure the HW is written like this. I totally have no idea how to bound the $\sum_{n=0}^\infty|a_nz^n|$. – whereamI Mar 17 at 18:00
• Try Cauchy's integral formula and note that if $|z| \le 1$ and $|w| = 2$ then $|z-w| \ge 1$. – copper.hat Mar 17 at 18:18
• Sorry, my comment was misleading and would only show that $|\sum_n a_n|$ is bounded by the quantity in the question. – copper.hat Mar 17 at 18:48

You were on the right track with a Cauchy estimate approach. For $$n=0,1,\dots,$$
$$a_n = \frac{f^{(n)}(0)}{n!} = \frac{1}{2\pi i}\int_{|z|=2}\frac{f(z)}{z^{n+1}}\,dz.$$
Let $$M= \max_{|z|=2}|f(z)|.$$ From the above, $$|a_n| \le \dfrac{M}{2^n}.$$ Thus
$$\sum_{n=0}^{\infty}|a_n| \le M\cdot \sum_{n=0}^{\infty}2^{-n} = 2M.$$
• You're welcome. Notice that $f$ entire is not needed; we only need $f$ analytic on $D(0,2+\epsilon).$ – zhw. Mar 17 at 18:50