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It is well known that the Fourier transform $\mathcal{F}$ maps $L^1(\mathbb{R}^d)$ into, but not onto, $\overline{C_0^0}(\mathbb{R}^d)$, where the closure is taken in the $L^\infty$ norm. This is a consequence of the open mapping theorem, for instance.

My question is: what's an explicit example of a function in $\overline{C_0^0}(\mathbb{R}^d)$ which is not in the image of $L^1(\mathbb{R}^d)$ under the Fourier transform?

I would also like to know whether there is a useful characterization of $\mathcal{F}(L^1(\mathbb{R}^d))$.

Remark: it is easy to see that the Banach space $\overline{C_0^0}(\mathbb{R}^d)$ consists of all continuous functions $f$ on $\mathbb{R}^d$ such that $f(\xi)\rightarrow 0$ as $|\xi|\rightarrow\infty$.

Ben McKay
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user17240
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  • This one is pretty close to your first question: http://mathoverflow.net/questions/3764/does-there-exist-a-continuous-function-of-compact-support-with-fourier-transform 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. – Darsh Ranjan Dec 07 '09 at 08:02
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    I've been told that the answer to your second question is that none is known, but I don't know a good reference. – Jonas Meyer Dec 07 '09 at 08:16
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    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) – Gian Maria Dall'Ara Dec 07 '09 at 08:24
  • @fpqc, it isn't. See http://www.jstor.org/pss/2036333 for example. – Jonas Meyer Dec 07 '09 at 08:48
  • @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. – Jonas Meyer Dec 07 '09 at 09:34
  • 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. – Harry Gindi Dec 07 '09 at 10:03
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    @fpqc- Really? Do not recognize how confusing that is for people who didn't read your original comment? – Ben Webster Dec 07 '09 at 14:49
  • I guess, but people reading my misunderstanding of abstract don't really gain anything. – Harry Gindi Dec 07 '09 at 14:58
  • 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. – Jonas Meyer Dec 09 '09 at 20:11
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    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.

     – Yemon Choi Dec 09 '09 at 23:51
  • You wound me, sir. – Harry Gindi Dec 09 '09 at 23:55
  • @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. – Jonas Meyer Dec 10 '09 at 03:39
  • @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 – Yemon Choi Dec 10 '09 at 04:16
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    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 $G = \mathbb{R}^d$, $\widehat{G} \cong \mathbb{R}^d$. – Thomas Winckelman Apr 16 '21 at 23:19

4 Answers4

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For the first question I commented above that the function $\sqrt(1-|x|^2)$ extended to $0$ outside the unit ball is not the fourier transform of any integrable function in dimension 2 or higher. In dimension $1$ there's "Further results" of Chapter I in Introduction to Fourier Analysis of Stein. In case you don't have access to the book, this is the construction: Observe that $|\int_a^b \sin(x)/x\ dx| \leq C<\infty$ for any strictly positive $a$ and $b$. Now, if $f\in L^1$ and $F(f)$ is odd you have $F(f)(x) = \int f(t) \sin(xt)\ dt$ up to a multiplicative constant. Than it's easy to see from the previous estimate that $|\int_1^b F(f)(x)/xdx|\leq C'<\infty$ uniformly in $b$. So a function which is continuous, odd and which decays too slowly ($1/\log(x)$ will do) is not the Fourier transform of an integrable function.

Yemon Choi
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Gian Maria Dall'Ara
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    I looked at the pertinent part of Stein and Weiss (Intro to Fourier Analysis on Euclidean Space—not Stein and Shakarchi), and the argument given there does not seem to be correct. Specifically, they say that "using Fubini's theorem one easily deduces that..." but I was not able to verify that Fubini applies, unless one assumes the original function is slightly better than L^1. (Would need that f(t) times a logarithmic factor is still in L^1.) Does anyone know if the argument can be fixed? – Brian Hall Aug 30 '17 at 17:41
  • I suspect that if the odd function—say g(x)—with logarithmic decay is smooth, then one can argue that the inverse Fourier transform of g—let's call it f— is a tempered distribution that away from the origin is given by an ordinary function that decays at least quadratically at infinity. Thus, if f were an L^1 function, it would have the integrability needed for the Stein–Weiss argument to hold. – Brian Hall Aug 30 '17 at 17:52
  • @BrianHall I suspect that besides Fubini theorem, the Lebesgue dominated convergence theorem is also required, and that's why the boundedness of the inner integral is important. The Fubini theorem is used whenever the domain is finite, and the Lebesgue dominated convergence theorem is used to take the limit inside. A guided proof can also be found in Frank Jones, Lebesgue Integration on Euclidean Space, Revised edition, p.305. It also bears similar spirit in Elijah Liflyand, Functions of Bounded Variation and Their Fourier Transforms, p.18. – davyjones Feb 21 '21 at 04:44
  • @BrianHall did you ever get this issue completely cleared up? – D.R. Mar 13 '24 at 05:01
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It's not germane to your question, but I can't resist pointing out that it is very hard to exhibit any continuous linear bijection from $L^1$(sensible measure space) onto $C_0$(sensible topological space), and in fact if either space is infinite then I suspect this is never possible, just for reasons of Banach space geometry. Thus, although it doesn't help with what you want to look at, I thought it might be worth mentioning that one can know the answer to "is the FT onto?" must be "no", before looking for an example or using properties of the Fourier transform.

(My caveats are because I don't want to categorically state it can't be done, but in all cases I can think of no such bijection will exist. However, both my general measure theory and my general topology are not what they should be, so I can't remember how to do things precisely in the most general settings.)

Anyway. I claim that there is no continuous linear bijection between $L^1({\mathbb R}^d)$ and $C_0(X)$, where $X$ is locally compact Hausdorff (e.g. a metric space). The reason is that we have big powerful results telling us that

(i) every bounded linear operator from $C_0(X)$ to $L^1({\mathbb R}^d)$ is weakly compact;

(ii) if the identity map on a Banach space $E$ is weakly compact, then $E$ is reflexive;

(iii) $L^1({\mathbb R}^d)$ is not reflexive (ibid).

Unfortunately I can't locate a self-contained proof of the key fact (i). (It can be deduced as a corollary of a rather powerful, fundamental and beautiful result - due to some promising former student of Dieudonné and Schwartz, not sure if he ever went on to do anything important...)

Bill Johnson
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Yemon Choi
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See Page 68 (Theorem 3.2.2) in my notes on Fourier Analysis: https://home.iitm.ac.in/mtnair/FS-2018.pdf

Thamban Nair.M
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Any characterization of $\mathcal{F}(L^1(\mathbb{R}^d))$, as one of my harmonic analysis professors put it, will get you immediate tenure in a math department that cares about harmonic analysis. It is a well known open question in the field.

mathisfun
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  • For any $g \in L^\infty$ : $\widehat{fg}= \hat{f} \ast \hat{g}$ is uniformly continuous and vanishes at $\infty$ – reuns May 06 '17 at 07:12
  • @reruns The open problem is a characterization of the form $$f\in \mathcal{F}(L^1(\mathbb{R}^d))$$ if and only if ___________________. Yes, $\widehat{f}\in C_0$ for $f\in L^1$ but not all $f\in C_0$ are in the image of the FT when the domain is $L^1$ funtions. – mathisfun May 07 '17 at 21:07
  • I meant that $\mathcal{F}(L^1)$ is the subspace of $C_0$ generated by the norm$\displaystyle\sup_{|g|\infty \le 1} | \widehat{fg}|\infty = |f|_{L^1}$ (this is tautologic). – reuns May 07 '17 at 22:19