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 $L^2$.



Fourier transform inversion theorem for a function not in L1 or L2
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.


