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 $$ f(z)=\sum_{n=0}^\infty\frac{f^n(0)}{n!}z^n. $$

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

$$\sum_{n=0}^{\infty}|a_n| \le M\cdot \sum_{n=0}^{\infty}2^{-n} = 2M.$$

  • $\begingroup$ Thank you very much! $\endgroup$ – whereamI Mar 17 at 18:46
  • $\begingroup$ You're welcome. Notice that $f$ entire is not needed; we only need $f$ analytic on $D(0,2+\epsilon).$ $\endgroup$ – zhw. Mar 17 at 18:50

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