Timeline for Range of the Fourier transform on $L^1$
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
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| Apr 16, 2021 at 23:19 | comment | added | Thomas Winckelman | I believe this is answered by Theorem 1.6.3 in Fourier Analysis on Groups by Walter Rudin (page 27). Deciphering Rudin's notation, the theorem states that the set $\{\hat{f} : f \in L^1(G)\}$ "consists precisely of the convolutions $F_1*F_2$ with $F_1$ and $F_2$ in" $L^2(\widehat{G})$, where $\widehat{G}$ is the dual group of $G$. For instance, for $G = \mathbb{R}^d$, $\widehat{G} \cong \mathbb{R}^d$. | |
| Dec 21, 2018 at 16:44 | answer | added | Thamban Nair.M | timeline score: 5 | |
| May 5, 2017 at 19:05 | history | edited | Ben McKay | CC BY-SA 3.0 |
spelling, grammar, formatting
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| May 5, 2017 at 19:02 | answer | added | mathisfun | timeline score: 4 | |
| Dec 10, 2009 at 4:16 | comment | added | Yemon Choi | @Jonas It wasn't so much that: it's that I think of the existence of Helson sets as something difficult and fiddly, although people with different tastes in analysis may disagree. As such, a proof relying on facts about Helson sets is IMHO sweeping issues under the carpet. Mais chacun a son gout, etc | |
| Dec 10, 2009 at 3:39 | comment | added | Jonas Meyer | @Yemon Choi, I made no attempt to optimize merit, and will not attempt to mutate your view; it was basically the first appropriate reference I found in a google search. I agree that while the abstract is "correct," it could be ambiguous depending on what the domain and codomain are taken to be. | |
| Dec 9, 2009 at 23:55 | comment | added | Harry Gindi | You wound me, sir. | |
| Dec 9, 2009 at 23:51 | comment | added | Yemon Choi | Have deleted an old comment claiming that the abstract of the paper which Jonas links to was "fine" - as it happens, it was guilty of using shorthand that makes sense to some of us, but only because of our training not because of our perspicacity. Am not quite convinced about the merit of said paper, btw, but that's just my subjective and mutable view. Also: not knowing that the FT fails to surject onto $C_0(R)$ is fine, but from someone so au fait with higher stuff and prone to hasty & vehement judgment of others? Vaguely disappointing. | |
| Dec 9, 2009 at 23:47 | answer | added | Yemon Choi | timeline score: 13 | |
| Dec 9, 2009 at 20:11 | comment | added | Jonas Meyer | To those who are confused: The user formerly know as fpqc, now Harry Gindi, had initially commented with speculation that F is onto in case d=1, hence my second comment above. In between my second and third comments fpqc had asked for clarification of what jstor.org/pss/2036333 is supposed to show. | |
| Dec 7, 2009 at 14:58 | comment | added | Harry Gindi | I guess, but people reading my misunderstanding of abstract don't really gain anything. | |
| Dec 7, 2009 at 14:49 | comment | added | Ben Webster♦ | @fpqc- Really? Do not recognize how confusing that is for people who didn't read your original comment? | |
| Dec 7, 2009 at 13:15 | answer | added | Gian Maria Dall'Ara | timeline score: 25 | |
| Dec 7, 2009 at 10:03 | comment | added | Harry Gindi | The abstract was misleading. It claims something completely different from the actual paper. That will explain my above comment. I deleted my earlier remarks to clear up space in the comments section. | |
| Dec 7, 2009 at 9:34 | comment | added | Jonas Meyer | @fpqc, I'm not sure exactly what you're asking. The Fourier transform generalizes to each locally compact Hausdorff abelian group G, mapping the L^1 space of G with Haar measure to the C_0 space of the Pontryagin dual of G. In this case, G is R^d, and the dual is also R^d. The paper I linked to gives a proof of the fact that the Fourier transform on $L^1(G)$ is onto $C_0(\hat G)$ only if G is finite. | |
| Dec 7, 2009 at 8:48 | comment | added | Jonas Meyer | @fpqc, it isn't. See jstor.org/pss/2036333 for example. | |
| Dec 7, 2009 at 8:24 | comment | added | Gian Maria Dall'Ara | If you look for an explicit example look at the convolution kernel for Bochner-Riesz means. K(x) = sqrt(1-|x|^2) (and 0 outside the unit disc) in dimension 2 or higher, and F(K) is not integrable. (this was my answer to the question cited by Darsh Ranjan) | |
| Dec 7, 2009 at 8:16 | comment | added | Jonas Meyer | I've been told that the answer to your second question is that none is known, but I don't know a good reference. | |
| Dec 7, 2009 at 8:02 | comment | added | Darsh Ranjan | This one is pretty close to your first question: mathoverflow.net/questions/3764/… Yemon Choi has a nice construction there that isn't worked out completely but seems quite plausible. I have no idea about your second question, though. | |
| Dec 7, 2009 at 7:28 | history | edited | user17240 | CC BY-SA 2.5 |
added 14 characters in body
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| Dec 7, 2009 at 7:22 | history | asked | user17240 | CC BY-SA 2.5 |