I have a quick question regarding power series. Let $\psi_1, \psi_2, \ldots$ denote the real-valued coefficients of a power series. I would like to see a proof (or a counterexample) to the following result:

$$\left| \sum_{j=0}^{\infty} \psi_j z^j \right |< \infty \quad \forall \: z \in \mathbb{C} \textrm{ such that } |z| \leq 1 \implies \sum_{j=0}^{\infty} |\psi_j| < \infty.$$

If the result is true: could one replace the complex plane $\mathbb{C}$ by the real line $\mathbb{R}$?

Thanks very much for your help.

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    $\begingroup$ If $\psi_j=\frac{e^{ij^2}}{j}$, does $\sum_{j=1}^\infty \psi_j e^{ij\theta}=\sum_{j=1}^\infty \frac{e^{ij^2}}{j} e^{ij\theta}$ converge for all $\theta$? Does $\sum_{j=1}^\infty \frac{1}{j}$ converge? $\endgroup$ – Mark Viola Feb 16 '17 at 4:02
  • $\begingroup$ Dr. MV, thanks for your reply. I am wondering if there is a counterexample with real coefficients $\psi_j$ (sorry i did not mention this in the original post. See the edited version above). $\endgroup$ – JLp Feb 16 '17 at 4:17
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    $\begingroup$ Replace $e^{ij^2}$ with $\sin(j^2)$. $\endgroup$ – Mark Viola Feb 16 '17 at 4:50

There is a power series converging uniformly but not absolutely on the unit circle (Hardy's example; see the book by Landau-Gaier MR 88d:01046, p. 68, or Theorem 2.28 in a book by A. Sasane Algebras of holomorphic functions and control theory. Dover Publications (ISBN 978-0-486-47465-6). ix, 140 p. (2009). If you want real coefficients,just take the real parts of the coefficients.


Take $\psi_n=1$ for all $n\in\mathbb N$. Then $\sum_{n=0}^{\infty} z^n$ converges for all $z\in\mathbb C$ such that $\left|z\right|<1$.

But the power series diverges at all point of the unit circle (for $z^n$ doesn't have limit $0$).

You can compute other examples, where the power series converges only at certain points, or everywhere but a few points...

  • $\begingroup$ Hi Nicolas, thanks for your reply; but I do not think your counterexample is correct. My assumption is that the power series converges for every $|z| \leq 1$. Setting $\psi_n=1$ for all $n$ does not satisfy this assumption. Am i right? $\endgroup$ – JLp Feb 16 '17 at 4:11
  • $\begingroup$ Oh, I see, I didn't note the $\le$ :-) I'll give it a little more thoughts :-) $\endgroup$ – Nicolas FRANCOIS Feb 16 '17 at 4:33

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