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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 $|\xi|\rightarrow\infty$.

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    $\begingroup$ I've been told that the answer to your second question is that none is known, but I don't know a good reference. $\endgroup$ Dec 7, 2009 at 8:16
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    $\begingroup$ 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) $\endgroup$ Dec 7, 2009 at 8:24
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    $\begingroup$ @fpqc- Really? Do not recognize how confusing that is for people who didn't read your original comment? $\endgroup$
    – Ben Webster
    Dec 7, 2009 at 14:49
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    $\begingroup$ 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. $\endgroup$
    – Yemon Choi
    Dec 9, 2009 at 23:51
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    $\begingroup$ 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 $G = \mathbb{R}^d$, $\widehat{G} \cong \mathbb{R}^d$. $\endgroup$ Apr 16, 2021 at 23:19

4 Answers 4

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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.

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    $\begingroup$ 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? $\endgroup$
    – Brian Hall
    Aug 30, 2017 at 17:41
  • $\begingroup$ 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. $\endgroup$
    – Brian Hall
    Aug 30, 2017 at 17:52
  • $\begingroup$ @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. $\endgroup$
    – davyjones
    Feb 21, 2021 at 4:44
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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...)

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    $\begingroup$ Easier, Yemon: $C_0$ contains a subspace isomorphic to $c_0$ while no $L_1$ space does (e.g. by cotype or weak sequential completeness or...). $\endgroup$ Jun 13, 2013 at 17:41
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    $\begingroup$ Ah, yes that is simpler. Thanks, Bill $\endgroup$
    – Yemon Choi
    Jun 13, 2013 at 17:58
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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.

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  • $\begingroup$ For any $g \in L^\infty$ : $\widehat{fg}= \hat{f} \ast \hat{g}$ is uniformly continuous and vanishes at $\infty$ $\endgroup$
    – reuns
    May 6, 2017 at 7:12
  • $\begingroup$ @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. $\endgroup$
    – mathisfun
    May 7, 2017 at 21:07
  • $\begingroup$ 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). $\endgroup$
    – reuns
    May 7, 2017 at 22:19
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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

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  • $\begingroup$ This is a nice example, but perhaps you could at least write out the description of this function in your answer, in case of future "linkrot"? $\endgroup$
    – Yemon Choi
    Dec 21, 2018 at 21:47
  • $\begingroup$ This is essentially the same as the answer by Gian Maria. $\endgroup$ Mar 31, 2020 at 11:30

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