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I have the following question:

Suppose that $f(z) = \sum_n^\infty a_nz^n$ is analytic for $|z| < R$ and is continous on $|z| = R$. Let $M = \sup_{|z| \leq R} |f(z)|$. Show that $|a_n|R^n \leq M$ for all $n$ and more generally that $\sum_n^\infty |a_n|^2R^{2n} \leq M^2$.

The first assertion follows from the Cauchy Integral Formula. However, I do not see how to incorporate the series in the second claim. There is a hint that says to consider $\int_{|z| = R} |f(z)|^2/z dz$.

Mike
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  • Just take $I(r)=\frac{1}{2\pi}\int_0^{2\pi}|f(re^{i\theta})|^2d\theta, r < R$ and compute it explicitly integrating term by term (allowed by absolute convergence and then only diagonal terms have non-zero integral by periodicity) and let $r \to R$ noting that trivially $I(r) \le M^2$ – Conrad Jun 09 '20 at 02:53
  • How can I integrate term by term? Wouldn't I need to push the absolute value side inside the series (which messes with the series)? – Mike Jun 09 '20 at 02:58
  • $|f(re^{i\theta})|^2=f(re^{i\theta}) \bar f(re^{i\theta})=\sum a_n \bar a_m r^{n+m}e^{i(n-m)\theta}$ and all the terms with $n-m \ne 0$ integrate to zero – Conrad Jun 09 '20 at 03:15
  • @Conrad. Are you allowed to let $r \to R$? We don't even know if the the series actually converges at $|z| = R$. – Mike Jul 12 '20 at 00:33
  • monotone convergence theorem applies – Conrad Jul 12 '20 at 11:56

1 Answers1

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For every $r \in (0, R)$ you can apply Parseval's identity (see for example Proof of Parseval's identity): $$ \sum_n |a_n|^2 r^{2n} = \frac{1}{2 \pi}\int_0^{2\pi}|f(re^{it})|^2dt \le \frac{1}{2 \pi}\int_0^{2\pi} M^2dt = M^2 \, . $$ The conclusion $ \sum_n |a_n|^2 R^{2n} \le M^2$ then follows by taking the limit for $r \to R$.

Martin R
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