Timeline for Estimating the growth of the Taylor coefficients given the growth of the function at the boundary
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 25, 2021 at 13:07 | history | edited | André Henriques | CC BY-SA 4.0 |
added 4 characters in body
|
| Nov 24, 2021 at 21:52 | history | became hot network question | |||
| Nov 24, 2021 at 21:40 | vote | accept | André Henriques | ||
| Nov 24, 2021 at 17:39 | answer | added | Christian Remling | timeline score: 12 | |
| Nov 24, 2021 at 17:00 | comment | added | Conrad | @Christian By partial summation using that $s_N(t)=\sum_{n=2}^N e^{in \log n}e^{int}$ are $O(\sqrt N)$ uniformly in $t$, the estimate coming from the standard Van der Corput second derivative test applied to $f(x)=(x \log x+xt)/(2\pi)$ | |
| Nov 24, 2021 at 16:56 | comment | added | Conrad | actually I may be wrong as there is a result of Pommerenke (?) proved on page 71 (Thm 3.3) in Hayman Multivalent Functions (2nd edition) that gives the required estimate $|a_n| \le C_pn^{-k-1}$ for $k >1/2$ and mean $p$-multivalent functions satisfying the estimate in the OP, while for $0 \le k<1/2$ one can get only $o(n^{-1/2})$ but not better; of course this doesn't prove the result in general for $k >1/2$ as one has to deal with functions that may be $\infty$ valent, but it provides counterxamples for $0 \le k <1/2$ | |
| Nov 24, 2021 at 16:51 | comment | added | Christian Remling | @Conrad: This is interesting. Can you please explain why $f$ is continuous? | |
| Nov 24, 2021 at 16:39 | comment | added | Conrad | For example $f(z)=\sum_{n \ge 2}\frac{e^{in \log n}}{\sqrt n \log^2n}z^n$ is continuous hence bounded in the closed unit disc but its coefficients are only $o(1/\sqrt n)$ and not $o(n^{-1/2-\epsilon})$ and similar examples should be manufactured for $k >0$ | |
| Nov 24, 2021 at 16:24 | comment | added | André Henriques | @ChristianRemling. Yes, I'm also interested in various variants of this question, such as $|f(z)|\le 1$, and $|f(z)|\le -\log(1-|z|)$. | |
| Nov 24, 2021 at 13:52 | history | asked | André Henriques | CC BY-SA 4.0 |