Let $f(z) = \displaystyle \sum_{n=0}^\infty a_n z^n$ be an entire function. Show that $$ \sum_{n=0}^\infty \frac{|a_n|}{n+1} \leq \pi \int_{0}^{2 \pi}|f(e^{i \theta})| \,d\theta. $$

The Cauchy integral formula gives $$ a_n = \frac{1}{2 \pi i} \int_0^{2 \pi} \frac{f(e^{i \theta})}{e^{i (n+1) \theta}} i e^{i \theta} \, d \theta = \frac{1}{2 \pi} \int_0^{2 \pi} f(e^{i \theta}) e^{-i n \theta} \, d \theta. $$ Then $$ \begin{aligned} \sum_{n=0}^\infty \frac{|a_n|}{n+1} &= \frac{1}{2 \pi} \sum_{n=0}^\infty \frac{1}{n+1} \left|\int_0^{2 \pi} f(e^{i \theta}) e^{-i n \theta} \, d \theta \right| \\ &\leq \frac{1}{2 \pi} \sum_{n=0}^\infty \frac{1}{n+1} \int_0^{2 \pi} \left| f(e^{i \theta}) \right| \left| e^{-i n \theta} \right| \, d \theta \\ &= \frac{1}{2 \pi} \sum_{n=0}^\infty \frac{1}{n+1} \int_0^{2 \pi} \left| f(e^{i \theta}) \right| \, d \theta. \end{aligned} $$ It seems that this method doesn't work. Is there any idea to estimate it?

  • $\begingroup$ Are you sure that's a $\pi$ out in front of the integral and not a $\frac{1}{\pi}$ or even $\frac{1}{2\pi}$? $\endgroup$ – Carl Schildkraut Aug 3 '18 at 3:29

It is a theorem due to Hardy. See for example:

  • Kenneth Hoffman, Banach space of analytic functions, the theorem at page 70
  • Peter Duren, Theory of $H^p$ spaces, corollary to theorem 3.15 at page 48

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