3

I have got the following question. From Bochner's theorem we know that a continuous, positive-definite function is the Fourier transform of a positive, bounded measure on $\mathbb{R}^n$. It is also known that the Fourier transform of a function $f \in L^1(\mathbb{R}^n)$ is in $C_0(\mathbb{R}^n)$. Hence the Fourier transform of a positive $L^1$-function is a positive-definite function which is also in $C_0(\mathbb{R}^n)$. Is the converse true? I have the feeling that the answer should be yes but I am not able to prove it. I cannot come up with a counterexample either. Any hints how to get a proof/counterexample would be appreciated. Thanks in advance.

Jan M.
  • 125
  • Not true. In fact there is no simple characterizartion of the range of the FT on $L^{1}$. – Kavi Rama Murthy Aug 14 '21 at 12:46
  • Well, can you tell why it is not true? Moreover there may not be a simple characterisation of the image of $L^1$ but what I am asking for is whether the image of all positive $L^1$-function coincide with all positive-definite functions in $C_0$. I don't think that the two things necessarily relate to each other. The Bochner theorem also answers what happens to the positive measures without answering what happens to all measures. – Jan M. Aug 14 '21 at 13:02
  • @JanM. If $g$ is an integrable, continuous, bounded and positive definite, then there is a function $f\geq0$ and integrable such that $g=\widehat{f}$. This result is in Stein and Weiss' book on Fourier analysis on Euclidean spaces. – Mittens Aug 14 '21 at 13:26
  • @JanM. This posting may be of interest to you. I think it answers your question. – Mittens Aug 14 '21 at 14:28
  • @OliverDiaz As for the literature result you mentioned, given the Bochner theorem this is rather clear, isn't it? It tells you that the $g$ you're considering is the Fourier transform of some measure $\mu$. Moreover, since $g$ is integrable the inverse Fourier transform is in $C_0(\mathbb{R}^n)$. In particular, it is a function, so a measure absolutely continuous w.r.t. the Lebesgue measure. So it is in $L^1$. Concerning the link, thanks for it. INteresting. But the question there is slightly different. Here in our context we know that the object under consideration is a Fourier transform tbc – Jan M. Aug 14 '21 at 18:44
  • ... the question is just whether the measure we transformed is absolutely continuous w.r.t. the Lebesgue measure, i.e., whether it is in $L^1$. – Jan M. Aug 14 '21 at 18:46
  • @JanM. You are right, the posting I sent answers a slightly different question, namely that $\mathcal{F}:L_1(\mathbb{R},\lambda)\rightarrow\mathcal{C}_0(\mathbb{R})$ is not subjective. In any event, the answer to your question is no. There are in fact singular measures $\mu$ (with respect to Lebesgue measure) such that $\widehat{\mu}$ vanishes at infinity. This paper has several references to examples of that sort. – Mittens Aug 14 '21 at 18:56
  • @JanM. The result I stated earlier is used by Stein and Weiss to prove the Bochner-Herglotz theorem. That is why I out it in my comment. – Mittens Aug 14 '21 at 19:25

1 Answers1

1

The answer to you question is no in general. The paper Yakubovich, S., On the affirmative solution to Salem's problem gives several references to examples of singular Borel measures on $\mathbb{R}$ whose Fourier transform vanishes at infinity.

The paper Wiener, N. and Winter, A., Fourier Stieltjes transform and Singular Infinite Convolutions, American Journal of Mathematics, vol60, num 3, pp. 513-522, 1938 shows that for any $\varepsilon>0$, there exists a singular measure $\mu$ on $[-\pi,\pi]$ such that $|\widehat{\mu}(t)|=O(|t|^{-\tfrac12+\varepsilon})$

Mittens
  • 39,145
  • Awesome. Thanks for it. The paper you mentioned actually aims to be self-contained. Corollary 2 gives the answer. Very good. – Jan M. Aug 14 '21 at 19:39