Timeline for Sets of divergence of Fourier series
Current License: CC BY-SA 3.0
18 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Aug 7, 2011 at 21:31 | vote | accept | Andrés E. Caicedo | ||
| Jun 7, 2011 at 10:27 | answer | added | Andrey Rekalo | timeline score: 5 | |
| Jun 7, 2011 at 2:16 | comment | added | Andrés E. Caicedo | (The horrible integral 3 comments above should read $\displaystyle \int_{x-r}^{x+r}$, of course.) | |
| Jun 7, 2011 at 2:11 | comment | added | Zen Harper | @Andres: OK, I see. That's interesting, but different to your question as I understood it; your question says "What is known about those sets...", without explicitly saying that you are assuming the sets to have measure zero. So it seems (to me) like you want a refinement of the Kahane/Katznelson result, and the Carleson result is a separate side-issue. | |
| Jun 7, 2011 at 1:59 | comment | added | Andrés E. Caicedo | @Zen: What I meant is that I would expect we should be able to prove by a direct construction something like the following: Given $E$ of measure zero and $F_{\sigma\delta}$, there is a continuous (or $L^2$ or whatever) $f$ such that the Fourier series of $f$ converges to $f$ precisely in $E$. | |
| Jun 7, 2011 at 1:57 | comment | added | Andrés E. Caicedo | @Zen: If we assume $f\in L^2\cap L^1$, we can choose a specific representative of $f$, by replacing $f$ with the function $\displaystyle x\mapsto \lim_{r\to 0}\frac1{2r}\int{x-r}^{x+r}f(y)dy$. I expect then we get that every $F_{\sigma\delta}$ set works, and perhaps even for $f$ continuous, with an argument that ought to be easier than what would be needed for the problem I linked to. | |
| Jun 7, 2011 at 1:51 | comment | added | Zen Harper | @Andres: sorry to keep leaving all these comments, but you say: "I imagine the question here can be answered by direct constructions that do not require the use or knowledge of the theorem". But, suppose you had the desired THEOREM: $E$ is a set of divergence for some $f$ if and only if...(DIV) The hypothetical criteria (DIV) must somehow imply zero measure (maybe non-trivially). However, for (DIV) to be a useful, usable criterion, we would expect it to be possible to prove directly that (DIV) for $E$ implies $m(E) = 0$. This would then give the Carleson theorem, wouldn't it? | |
| Jun 7, 2011 at 1:37 | comment | added | Zen Harper | ...but of course, the Fourier series question here is genuinely more general than for power series, I think. You can split any Fourier series into two power series in $z$ and $\bar{z}$ and apply the results for power series, but it's possible (maybe) for each to diverge, but the sum to converge. | |
| Jun 7, 2011 at 1:33 | comment | added | Zen Harper | @Andres: I've just noticed the other question "Behaviour of power series on their circle of convergence" you referred to above [together with your wonderful answer!] That question, of course, is just the special case of this question where the Fourier coefficients for negative $n$ are all zero. It seems like you've almost answered your own question! | |
| Jun 7, 2011 at 1:20 | history | edited | Andrés E. Caicedo | CC BY-SA 3.0 |
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| Jun 7, 2011 at 1:10 | answer | added | Robert Israel | timeline score: 1 | |
| Jun 7, 2011 at 1:09 | comment | added | Andrés E. Caicedo | @Zen: Yes, as far as I can see, the proofs I know of the C-H theorem do not give us an insight, but I imagine the question here can be answered by direct constructions that do not require the use or knowledge of the theorem. | |
| Jun 7, 2011 at 1:05 | history | edited | Andrés E. Caicedo | CC BY-SA 3.0 |
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| Jun 7, 2011 at 1:02 | comment | added | Zen Harper | I imagine this question will forever remain impossibly difficult for finite mathematicians; and even if it is someday solved, I personally am sure I will never understand the proof [or the criteria on $E$]. Having said that, it would be nice to be proved wrong! I mean, I don't understand it, but I thought that the Carleson-Hunt theorem tells us nothing whatsoever about the detailed structure of $E$, apart from having zero measure. If even this horribly difficult theorem gives no information, I am not optimistic. The $L^p$ spaces cannot distinguish between different $E$ with $m(E) = 0$. | |
| Jun 7, 2011 at 0:56 | comment | added | Andrés E. Caicedo | Hi Juris. Sure; in fact, for continuous functions, the set $E$ is Borel (of rather low complexity). But I do not even have a working conjecture on what the general answer should be (does it change as $p$ changes, where we require $f\in L^p$, for example?), or whether any kind of structure can be expected at all. | |
| Jun 6, 2011 at 23:08 | comment | added | Juris Steprans | Doesn't a cardinality argument yield a negative answer to your last question for continuous functions? (Also for L2 functions actually.) | |
| Jun 6, 2011 at 22:10 | history | asked | Andrés E. Caicedo | CC BY-SA 3.0 |