Timeline for Is $f(x,y)=\sum_{n\in\mathbb{Z}\backslash\{0\}}\frac{1}{n}e^{2\pi i(xn+yn^2)}$ essentially bounded?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 7, 2015 at 17:03 | vote | accept | Tony B | ||
| Jul 7, 2015 at 17:02 | answer | added | Tony B | timeline score: 5 | |
| Jul 7, 2015 at 1:58 | comment | added | Tony B | @RobertIsrael You are right. The function is indeed well-defined a.e. Thanks. | |
| Jul 7, 2015 at 0:50 | comment | added | Robert Israel | Why not? The set of $(x,y)$ such that the series fails to converge is measurable and its intersection with every line $y = constant$ has measure $0$, therefore that set has $2$-dimensional measure $0$. | |
| Jul 7, 2015 at 0:09 | answer | added | Terry Tao | timeline score: 16 | |
| Jul 6, 2015 at 23:24 | comment | added | Tony B | @RobertIsrael Good point. But it does not follow that $f(x,y)$ is well-defined a.e. | |
| Jun 19, 2015 at 2:10 | comment | added | Tony B | @PaataIvanisvili:there are two steps: first show the function is well-defined a.e. and then show it is essentially bounded. | |
| Jun 18, 2015 at 14:51 | comment | added | Robert Israel | Also, it may be worth pointing out that for any $y$, $f(\cdot,y)$ is the Fourier series of an $L^2$ function, and so by Carleson's theorem it converges pointwise almost everywhere. | |
| Jun 18, 2015 at 4:20 | comment | added | Robert Israel | Actually, for the case $y=0$ the sum is essentially bounded: $f(x,0) = (1-2x)\pi i$ for $0 < x < 1$ (a "sawtooth wave") | |
| Jun 17, 2015 at 18:15 | comment | added | Paata Ivanishvili | @Michael Renardy: but why is this enough? OP wants to be essentially bounded but you give counterexample for zero set measure.\\ Also for me question does not make sense: Dony, I assume you are asking whether the function $f(x,y)=\limsup_{N \to \infty}|\sum_{|j|\leq N, j\neq 0} \frac{e^{i(xn+yn^{2})}}{n}|$ is essentially bounded. | |
| Jun 17, 2015 at 17:26 | review | Close votes | |||
| Jun 17, 2015 at 19:53 | |||||
| Jun 17, 2015 at 17:11 | comment | added | Michael Renardy | No. Set y=0. You can do the sum explicitly in that case. | |
| Jun 17, 2015 at 15:10 | comment | added | Tony B | Yes. I edited the question. | |
| Jun 17, 2015 at 15:10 | history | edited | Tony B | CC BY-SA 3.0 |
added 15 characters in body; edited title
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| Jun 17, 2015 at 6:43 | comment | added | Robert Israel | I presume you want to leave out $n=0$. | |
| Jun 17, 2015 at 3:27 | history | asked | Tony B | CC BY-SA 3.0 |