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For $\frac{1}{4}<a<1$ consider the following function:

$$f(x)=\frac{|x|^{\frac{1}{2}}}{(x^2+1)^{a+ib}}$$

If $1>a>\frac{1}{2}$ then $f(x) \in L^2$ and the Fourier inversion theorem can be applied ($\mathcal{F^{-1}}$ is the inverse Fourier transform):

$$\mathcal{F^{-1}} \circ \mathcal{F} (f)=f$$

But if $\frac{1}{4}<a<\frac{1}{2}$ ? The function is no longer in $L^2$ (and of course not in $L^1$).

I did not find a clear reference in literature to justify that the same equality holds for this type of functions.

Any reference for such a case ? I am also looking for a justification of Fourier inversion theorem for:

$$g(x)= \frac{1}{\sqrt{|x|}} K_{a+ib} (\frac{1}{|x|})$$

for $0<a<\frac{1}{2}$, where $K_{a+ib} $ is the K-Bessel function. Asymptotic at infinity of $g(x)$ is:

$$g(x)= k_1 x^{-\frac{1}{2}+a+ib} + k_1 x^{-\frac{1}{2}-a-ib} +o(\frac{1}{x})$$

So this function is not in $L^1$ or $L^2$.

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    $\begingroup$ $\mathcal{F^{-1}} \circ \mathcal{F} (f)=f$ holds for tempered distributions, and any continuous function increasing slower than a polynomial can be interpreted as a tempered distribution. $\endgroup$ – Nick S Nov 26 '17 at 17:14
  • $\begingroup$ Is this argument enough to conclude that the double integral defining $\mathcal{F}^{-1} \circ \mathcal{F} (f)$ is equal to $f(x)$ for all $x$ ? $\endgroup$ – Bertrand Nov 26 '17 at 17:23
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    $\begingroup$ @Bertrand: you got the wrong point of view. As Nick said you need to interpret this in the setting of temperate distributions. So the bad news is you have to learn that stuff, .e.g. here terrytao.wordpress.com/2009/04/19/245c-notes-3-distributions/… in order to get a satisfactory answer to your question. $\endgroup$ – Abdelmalek Abdesselam Nov 26 '17 at 19:51
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    $\begingroup$ To make your question well-defined, you need to say what the domain and codomain of your $\mathcal F$ and ${\mathcal F}^{-1}$ are! For a start, what are you defining ${\mathcal F}(f)$ to be? A function? a distribution? etc $\endgroup$ – Yemon Choi Nov 26 '17 at 20:28
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    $\begingroup$ @Bertrand If your function is not $L^1$, then the double integral defining $\mathcal{F}^{-1} \circ \mathcal{F} (f)$ is NOT defined. For non $L^1$ functions $\mathcal{F}^{-1} \circ \mathcal{F} (f)$ needs to be interpreted differently. For example, for $L^2$ functions the FT transform is defined as the unique norm-preserving extension from $L^1 \cap L^2$ to $L^2$, and then, this extension satisfies this formula. But the equality is not true as the double integral.... Same holds for distributions. $\endgroup$ – Nick S Nov 26 '17 at 23:32
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You can define the distributionial Fourier transform of a tempered distribution using all the abstract machinery established by Schwartz, and the thing you want to check is that it agrees with the integral Fourier transform defined by the usual integral.

For the first Fourier transform (from $x$ to $\xi$), you can first take the Fourier transform of $f'$, which is a good old $L^1$ function, so its integral Fourier transform is well defined and agrees with its distributional Fourier transform. The distributional Fourier transform of $f$ is just that of $f'$ divided by $\xi$ (at least away from 0), and you can check that the integral Fourier transform of $f$ (define away from 0) is the same thing using integration by parts (the bulk integral is in $L^1$, and the boundary term goes to 0 because $f(0)=0$ and $f(x)\to0$ as $x\to\infty$).

This leaves the question at zero, where the only possible way the two definitions differ is that the distributional Fourier transform could have an extra multiple of the delta function at zero. We will rule that out later.

Now we return to the integral Fourier transform of $f$. By repeated integration by parts, we know that it decreases faster than any polynomial at infinity. Next we turn to zero frequency. Since $f'\lesssim (1+x)^{-1-a}$, we can show that the integral Fourier transform of $f'$ is Holder continuous at 0 by splitting $\int e^{ix\xi}f(x)=\int (e^{ix\xi}-1)f(x)$ into two parts, one on $(-|\xi|^{-1/2},|\xi|^{-1/2})$ and the other on the rest of $\mathbb R$. This shows that the integral Fourier transform of $f$, although not defined at 0, is nevertheless in $L^1$. Then its distributional Fourier transform agrees with the integral Fourier transform. That means you can do the two Fourier transforms by iterated integration and it gives the same $f$, provided that the distributional Fourier transform of $f$ does not have an extra delta function at 0.

Now we can rule out this ghost delta function. Assume the contrary, then the distributional Fourier transform of $f$ would be its integral Fourier transform (which is an $L^1$ function) plus some delta function at 0. Then when we take the second distributional Fourier transform, we would get a $C_0$ function plus a constant function. But taking two distributional function is supposed to give the same $f$, so the extra constant function is 0, which means no ghost delta function in the distributional Fourier transform of $f$, which then agrees with the integral Fourier transform of $f$.

Example of distributional Fourier transform of $f$ different from what is obtained from (either) Fourier transform of $f'$: Let $f=1$. Then $f'=0$ and $\mathcal F(f')=0$. Then we know that the distributional Fourier transform of $f$ is 0 when $\xi\neq0$, but has a delta function at 0 (times a normalization factor). When we take the second (distributional) Fourier transform, we get 1 (the correct answer) and 0 (the wrong answer) respectively, and we can tell that we missed out a delta function in the distributional Fourier transform of $f$ from the fact that $f(x)\to1$ as $x\to\infty$.

Hope this clarifies the matter.

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  • $\begingroup$ Your answer is clear (except, I think, that we cannot integrate by parts more than one times so the Fourier transform $f$ decreases at infinity but not faster than any polynomial: it does not change the conclusion). Your explanation works also for the second function $g$. Do you know where I can find a reference for this result ? Or there no theorem / characterisation in literature giving the condition to have a couple $f$ and $\mathcal{F}(f)$ of classical functions (not distributions) ? $\endgroup$ – Bertrand Nov 27 '17 at 17:51
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Fourier transform has been generalized far beyond $L^1$ and $L^2$. The most useful generalization (Schwartz tempered distributions) is explained in

L. Schwartz, Mathematics for the physical sciences. Hermann, Paris; Addison-Wesley Publishing Co., Reading, Mass.-London-Don Mills, Ont. 1966.

This is enough for your purpose. For further generalizations the excellent source is a series of books I. M. Gelfand (with various co-authors) Generalized functions, volumes 1-5 (mutiple editions).

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