Does there exist a function which is holomorphic in $|z|<1,$ continuous in $|z|\leq1$ and such that the series $\sum |a_n|$ is divergent, where $a_n$'s coefficients in the Taylor series expansion of $f?$
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8 Yes there are lots of classical examples by Hardy and Littlewood like say $f(z)=\sum _{n \ge 1}e^{in \log n}\frac{z^n}{n^{3/4}}$.
Using the second derivative test for exponential sums with $f(u)=(u\log u+u\theta)/(2\pi)$, so $f''(u) \sim 1/M$ when $u \sim M$, the exponential sum $\sum_{n=M}^{2M}e^{in\log n}e^{in\theta} =O(\sqrt M)$ uniformly in $\theta$ and an easy dyadic argument shows that $\sum_{n=1}^Ne^{in\log n}e^{in\theta}=O(\sqrt N)$ also uniformly in $\theta$, so it follows that $f$ converges uniformly on the unit circle and obviously satisfies $\sum |a_n|=\infty$
- $\begingroup$ Can you please provide some references where such examples are given? $\endgroup$Nik– Nik2023-01-18 17:08:33 +00:00Commented Jan 18, 2023 at 17:08
- $\begingroup$ Zygmund classic book on Trigonometric Series 3rd edition Chapter 5 has this and many others with references to the original papers. See cambridge.org/core/books/trigonometric-series/… $\endgroup$Conrad– Conrad2023-01-18 17:10:11 +00:00Commented Jan 18, 2023 at 17:10
- $\begingroup$ I have few doubts. (1) why you are taking $f(u)=(u\log u+ u\theta)/2\pi?$ (2) how $\sum |a_n|=\infty?$ $\endgroup$Nik– Nik2023-01-18 17:56:59 +00:00Commented Jan 18, 2023 at 17:56
- $\begingroup$ the exponential sums are generally taken as $\sum_{n \sim M}e(f(n))$ where $e(x)=e^{2\pi ix}$ so here since we have $e^{in\log n +in\theta}$ the corresponding $f(u)=u\log u+\theta u$ requires the $2\pi$ normalization, but that is just a convention; since $|a_n|=n^{3/4}$ clearly $\sum |a_n|=\infty$ $\endgroup$Conrad– Conrad2023-01-18 18:04:30 +00:00Commented Jan 18, 2023 at 18:04
- $\begingroup$ Isn't $a_n=e^{in\log n}/n^{3/4}$ here? $\endgroup$Nik– Nik2023-01-18 18:14:52 +00:00Commented Jan 18, 2023 at 18:14