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Suppose we have a function $f: \mathbb{R} \rightarrow \mathbb{R}$ whose Fourier transform exists and is given by $\hat{f}$. Then the convolution theorem tells us that for any $k \in \mathbb{N}$, we have:

$$\displaystyle \widehat{f^k} = \underbrace{\hat{f} \ast ... \ast \hat{f}}_{k \text{ terms}},$$

where $f \ast g$ denotes the convolution of two functions on $\mathbb{R}.$ However, this does not tell us anything about non-integer values of $k$. For instance, suppose we could prove the existence of $\widehat{f^{1/2}}$. Is there any way in which we can compute the Fourier transform of $f^{1/2}$ explicitly, given only an explicit form for $\hat{f}$?

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    $\begingroup$ Possibly, by putting that $\widehat{f^{1/2}} \ast \widehat{f^{1/2}} = \widehat{f} $ and solving for the coefficients of $\widehat{f^{1/2}}$ $\endgroup$ – G Cab Sep 21 '16 at 20:03
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    $\begingroup$ It is the opposite : you can define $g^{\ast k} = \underbrace{g \ast \ldots \ast g}_k$ for $k\not \in \mathbb{N}$ thanks to the Fourier transform $g^{\ast k} = FT^{-1}[FT[g]^k]$ (the case $k=-1$ is important, it is the convolutive inverse). Note that taking $g = \delta'$, you get a fractional derivative/integral operator : $\frac{d^k}{dx^k} h = \underbrace{\delta' \ast \ldots \ast \delta'}_k \ast h$ $\endgroup$ – reuns Sep 21 '16 at 20:17
  • $\begingroup$ @user1952009 Interesting. What is $\delta'$? $\endgroup$ – user363087 Sep 21 '16 at 22:01
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    $\begingroup$ The (distributional) derivative of the Dirac delta $\delta$, so that $h'(x) = \delta' \ast h(x)$. For being convinced, show that $f' \ast g(x) = f \ast g'(x)$ by integrating by parts. Finally, write that $h'(x) = \delta \ast h'(x)$. $\endgroup$ – reuns Sep 21 '16 at 22:04

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