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Given a continuous function $f:\mathbb{R}\to\mathbb{R}$, with either $f(t)$ or $f(-t)$ diverging to $\pm\infty$ as $t\to\infty$, is there a way to define its Fourier transform in a way that conserves $$\mathcal{F}^{-1}\left\{\mathcal{F}_{\mathfrak{R}}\{f(t)\}_s\right\}_{t}=f(t)$$ For some non-degenerate interval of $t$?

As the usual integral definition lacks the capacity to control a function's growth like Laplace, Mellin or even Hillbert transforms.

The closest I have found is for $f\left(e^{-t}\right)$ with an inverse $f^{-1}(t)$ defined for $t\in f\left(\mathbb{R}^+\right)$, a semi-regularization: $$\mathcal{F}\left\{f\left(e^{-t}\right)\right\}_s=\frac{i}{s}\int_{f\left(\mathbb{R}^+\right)}f^{-1}(t)^{is}dt$$ Which doesnt converge by itself, needs either weight functions or a Wick rotation.

Its a re-formulation of the Mellin-Fourier duality: $$\mathcal{F}\left\{f\left(e^{-t}\right)\right\}_{-is}=\mathcal{M}\{f\}_s$$ But this one only works if $f$ grows as a power of $x$, and doesnt allow exponential or higher growth as the semi-regularization.

Cad mio
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    You can get a Fourier transform of any integrable function as a distribution. The most natural objects on which to talk about the Fourier Transform is on distributions. – Thomas Andrews Jul 03 '25 at 23:39
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    As far as I have understood, the usual distributions can extend for some functions that dont grow faster than any polynomial, if there was a sense on which one could talk about F{te^e^t} then that would be an end to my question, thanks for commenting :> – Cad mio Jul 04 '25 at 01:04
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    The distributions I am thinking of include any integrable function on a finite interval. It also includes things that are not functions, like the (misnamed) Dirac delta function. – Thomas Andrews Jul 04 '25 at 01:15
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    This is the functional analysis view of the Fourier transform. – Thomas Andrews Jul 04 '25 at 01:19
  • Any material you could recommend for me?, if the framework is indeed capable of integrating any C^{\infty} function, even if the distribution lives outside S', would answer my question – Cad mio Jul 04 '25 at 01:43
  • It technically isn't integrating, but it is equivalent to integrating when the Fourier transform exists as a function. It was a long time ago that I learned about it, so I'm not sure what to recommend. And, just as wit integration, the fact that it is known to give a result doesn't actually help you find the resulting distribution. – Thomas Andrews Jul 04 '25 at 02:19
  • One can define Fourier Transform pair $-|\omega|, \frac{1}{t^2}$ (with some scaling constant factors omitted). – Srini Jul 05 '25 at 00:10

1 Answers1

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First, as a summary, a relatively simple class of generalized functions on which to define Fourier transform (with image in the same class of things), is tempered distributions... not arbitrary distributions. Though, yes, things like (integrate against) $e^x e^{ie^x}$ are not of moderate growth, but (being classical-derivative(s) of moderate-growth fcns,) are tempered distributions.

Second, if we insist on taking Fourier transforms of (integrate-against...) arbitrary smooth functions: compactly supported smooth functions (=test functions) have Fourier transforms in Paley-Wiener spaces of entire functions. So, certainly, Fourier transforms of arbitrary distributions (and, certainly, smooth functions...) are in the dual of the Paley-Wiener spaces.

This is perhaps not very helpful for some purposes, but it does show that "things are (kinda?) under control/understandable" :)

EDIT:

To be clear(er), yes, a convenience of compactly-supported distributions is that they are in the dual space of all smooth functions... which includes the exponentials relevant to Fourier transforms... so that the Fourier transforms can be evaluated in the "obvious" way. E.g., the Fourier transform coefs of Dirac delta are $\delta(e^{2\pi i \xi x})$. For not-compactly-supported (but still tempered) distributions, the Fourier transform is subtler, since it cannot generally be evaluated that way.

But, as above, even without temperedness, the Fourier transform integral on arbitrary distributions can be "regularized" (naturally) to map to the dual of the Paley-Wiener space.

Yes, this may be unsatisfying, but is inescapably true, whatever other interpretations you/we may find. :)

paul garrett
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  • Thanks for the answer c:, but is there a way for non compactly supported functions or distributions?, more than for practical purposes, I'm interested in seeing how far one can push convergence into divergent Fourier transforms – Cad mio Jul 04 '25 at 19:35