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Are there examples of functions $f(x)$ for which the inverse Fourier transform integral

$$f(x)=\mathcal{F}_{\omega}^{-1}[F(\omega)](x)=\int\limits_{-\infty}^\infty F(\omega)\, e^{2 \pi i x \omega}\, d\omega\tag{1}$$

converges in a normal sense but the related Fourier transform integral

$$F(\omega)=\mathcal{F}_x[f(x)](\omega)=\int\limits_{-\infty}^\infty f(x)\, e^{-2 \pi i \omega x}\, dx\tag{2}$$

doesn't converge in a normal sense?

Obviously both can converge in a normal sense, and there are examples where both exist but neither converges in a normal sense, but I'm wondering if its possible for one to converge in a normal sense and the other to converge in a distributional sense or as a Cauchy principal value.

The Wikipedia article Fourier inversion theorem indicates the following:

The theorem holds if both $f$ and its Fourier transform are absolutely integrable (in the Lebesgue sense) and $f$ is continuous at the point $x$. However, even under more general conditions versions of the Fourier inversion theorem hold. In these cases the integrals above may not converge in an ordinary sense.

When I used the term "normal sense" I was thinking in terms of evaluations using the Mathematica Integrate function without using the PrincipalValue option which is perhaps equivalent to "ordinary sense" in the quote above. The Mathematica FourierTransform function and InverseFourierTranform function many times give results when computing the transform integrals via the Mathematica Integrate function indicates the transform integral doesn't converge, but I don't want to make this question about Mathematica.

The Wikipedia article quoted above is a bit vague as "ordinary sense", "general conditions", and "versions of the Fourier inversion theorem" aren't defined, and "may not converge in an ordinary sense" doesn't necessarily rule out convergence in an "ordinary sense" under "more general conditions".

I guess perhaps what I'm really trying to understand is what are all of the conditions under which a function $f(x)$ is actually recoverable from its inverse Fourier transform $F(\omega)$. $f(x)$ and $F(\omega)$ both being Schwartz functions seems sufficient but not necessary. $f(x)$ and $F(\omega)$ both being absolutely integrable also seems sufficient but not necessary. So I'm wondering what are the necessary and sufficient conditions for recovering $f(x)$ from $F(\omega)$.

Steven Clark
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  • The wikipedia might mean "ordinary sense" as opposed to "distributional sense" for which we can make sense out of things like the Fourier transform of $f(x)\equiv 1$ which is not integral in any sense (Riemann or Lebesgue) – podiki Jan 05 '24 at 20:16
  • @podiki That was kind of what I meant by "normal sense", but I was also excluding Cauchy Principal Value. I'm not sure whether or not "ordinary sense" in the Wikipedia article includes Cauchy Principal Value. – Steven Clark Jan 05 '24 at 20:23
  • So, you are looking for sufficient and necessary conditions, correct? If you are restricting things that exclude distributions, then the function must be absolutely integrable. But even so, its FT might not be. But all of this is quite a different line of questioning from the original one, which I did provide a reasonable answer/example. – Mark Viola Jan 06 '24 at 16:12
  • @StevenClark Steve, have you had a look at THIS Wikipedia article on the inverse FT theorem? It might be instructive. – Mark Viola Jan 06 '24 at 16:28
  • @MarkViola Yes, in particular the Conditions on the function section which states the two conditions I mentioned in my last paragraph. But these conditions seem sufficient but not necessary as for example the function $f(x)=2, \text{sinc}(2 \pi x)$ (which is not absolutely integrable) is recoverable from its Fourier transform $F(\omega)=\frac{1}{2} (\text{sgn}(1+\omega)+\text{sgn}(1-\omega))$, is it not? – Steven Clark Jan 06 '24 at 16:45
  • I don't believe that there are conditions that are both necesssary and sufficient (i.e, "If and only if" conditions). What is motivating your asking this question? – Mark Viola Jan 06 '24 at 16:48
  • @MarkViola Interesting you should say that as the context for this question is my investigation of a conjectured "if and only if" theorem which is the subject of another question that I've been working on for awhile and plan to post soon. My motivation for this question was to identify all of the conditions that potentially need to be investigated before posting my other question, but as this seems elusive perhaps I'll just go ahead and post my other question when I'm done editing it. – Steven Clark Jan 06 '24 at 17:31
  • @StevenClark Thank you for the reply. I'll look forward to reading your other question. – Mark Viola Jan 06 '24 at 17:36
  • @MarkViola In case you haven't seen it, here's my follow-on question. – Steven Clark Jan 08 '24 at 19:45
  • @StevenClark Hi Steven. Thank you. Yes, I did see it. Since my post here has not been upvotes, shall I delete it? It answered your original question. – Mark Viola Jan 08 '24 at 21:31
  • @MarkViola I upvoted and accepted since you addressed my original question. – Steven Clark Jan 08 '24 at 21:39
  • @StevenClark Much appreciated. Hopefully, you'll find better answers than you've found in the Wikipedia articles. I'm not sure that there are conditions that are both necessary and sufficient. There are necessary conditions, and there are sufficient conditions. But never the twain shall meet. – Mark Viola Jan 08 '24 at 21:54
  • @MarkViola Perhaps a definition of "recoverable from Fourier transform" would help, but it seems the approach is to define it by example conditions, and not being able to specify the necessary and sufficient conditions leaves the definition a bit vague in my mind. My conjectured if-and-only-if theorem defines it precisely in my mind. I'd be interested if you think you have a counterexample to my conjectured if-and-only-if theorem. – Steven Clark Jan 09 '24 at 00:51

1 Answers1

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I preseume by "normal" sense, you mean "lebesgue Integrable." If so, then there are a voluminous set of examples that qualify.

Here is a classical expample. Let $f$ be the function defined as

$$f(x)=\begin{cases} 1&,|x|\le 1\\\\0&,|x|>1 \end{cases}$$

Then, we have

$$\begin{align} F(\omega)&=\int_{-\infty}^\infty f(x)e^{-i2\pi \omega x}\,dx\\\\ &=\int_{-1}^1 e^{-i2\pi \omega x}\,dx\\\\ &=\frac{\sin(2\pi \omega)}{\pi \omega} \end{align}$$

But, $F$ is not Lebesgue Integrable. However, the Principal Value integral exists with

$$\frac1{2\pi}\text{PV}\int_{-\infty}^\infty \frac{\sin(2\pi \omega)}{\pi \omega}e^{i2\pi \omega x}\,d\omega=f(x)$$

Mark Viola
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  • I updated my question above in an attempt to clarify my confusion. The function $F(\omega)=2 \text{sinc}(2 \pi \omega) $ has a removable singularity at $\omega=0$ and Mathematica gives the result $$\int\limits_{-\infty}^{\infty} 2 \text{sinc}(2 \pi \omega) , d\omega =1$$ without using the PrincipalValue option, so do you really need to specify the integral as a Cauchy principal value? – Steven Clark Jan 05 '24 at 20:07
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    The issue is that the positive and negative parts of sinc are not integrable, so hence the Lebesgue integral of sinc does not exist (it’s also not Lebesgue integrable in the sense that the integral of its absolute value also doesn’t exist). You need some other notion of improper integration to make sense of the integral of sinc, like the Cauchy principal value. – kieransquared Jan 05 '24 at 20:13
  • @StevenClark To reiterate kierandsquared's comment. the the sinc function is neither Riemann integrable not Lebesgue integrable. The isssue is not at the origin; it is at infinity. The integral does exist as a Cauchy Principal Value. – Mark Viola Jan 05 '24 at 20:19
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    @MarkViola It should exist as an improper Riemann integral, not just a Cauchy Principal Value. – podiki Jan 05 '24 at 20:26
  • I'm getting more-and-more lost in all of these types of integrals. :) – Steven Clark Jan 05 '24 at 20:28
  • I guess perhaps what I'm really trying to understand is what are all of the conditions under which a function is actually recoverable from its inverse Fourier transform, but perhaps there is no simple answer. – Steven Clark Jan 05 '24 at 20:29
  • The Schwartz class of funcitons is what you might be seeking. The Fourier Transform in a linear isomorphism of $\mathbb{S}\to \mathbb{S}$ – Mark Viola Jan 05 '24 at 20:39
  • @podiki In the example posted herein, the PV is not necessary. But in general, the PV is needed. Take for example, the FT of the sign function. It is the distribution $\text{PV}\left(\frac1k\right)$ (times a constant that depends on the convention used to define the FT). The funciton $\frac1x$ is NOT Riemann integrable. – Mark Viola Jan 06 '24 at 16:38