"Analytic Continuation" of the Convolution Operator?

Question

The convolution of two functions $f, g: \mathbb{R} \rightarrow \mathbb{R}$ is defined as:

$$(f \ast g)(t) := \displaystyle \int_{\mathbb{R}}f(\tau)g(t - \tau)d\tau,$$

and can be generalised to higher dimensions. A result of significant importance is the well-known convolution theorem, a version of which states that:

$$\displaystyle \hat{f} \ast \hat{g} = \widehat{f \cdot g},$$

where $\hat{f}$ denotes the Fourier transform of $f$. Since it can also be shown that $\ast$ is associative, then we also have:

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

for any positive integer $k$. Now, as silly as this sounds, I would like to investigate this for $k = 1/2$. That is, given that we know the Fourier coefficients of $f$, can we derive, in terms of convolutions, the Fourier coefficients of $\sqrt{f}$ ?

There is such an extension of the convolution operator, known as fractional convolution, for which several papers have been published, the majority of which fall in the category of signal processing. However, the "fractional convolution theorems" seem to depend on being able to write $f$ as a product of functions whose Fourier coefficients are known (see, for example, David Mustard's 1995 paper entitled Fractional Convolution). But this is not possible here.

Does anyone know of an analytic continuation of $\otimes_{k}$ to rational values of $k$ that would allow this to be possible? Otherwise, are there any other ways to find the Fourier coefficients of $\sqrt{f}$ given that the Fourier coefficients of $f$ are known?

Answer 1

One can write:

$$\displaystyle f^{\otimes_{k}} = \mathcal{F}^{-1}\{\mathcal{F}(f)^{k+1}\},$$

as an analytic continuation for $k \not \in \mathbb{N},$ where $\mathcal{F}(f)$ denotes the Fourier transform of $f$.

Answer 2

Existence of such a continuation is not a problem. You can simply define $\hat{f}^{\otimes_{k}} = \widehat{f^k}$ for arbitrary positive reals $k$. Of course you'll want $f^k$ to have a Fourier transform, so you'll want, say, $f \in L^{1/k}$. And you'll want to specify which branch of the $k$'th power to take when $f$ is not always positive, perhaps the principal branch.

But none of this addresses the problem of actually computing $\widehat{f^k}$ from $\widehat{f}$.