Convolution and the square root of the Fourier Transform The Fourier transform is a linear transform. At least in the discrete, finite dimensional case, it can be represented as a unitary matrix, meaning that it has a nontrivial square root (although not necessarily a unique one). These square roots have the property that if you apply the transformation twice, you get the Fourier transform.
What can be said about these linear transformations? In particular I'd be interested in learning the following two things:


*

*What do the columns of the matrix in question look like? (Basis functions)

*Is there an equivalent of the convolution theorem for this transform? What do pointwise multiplication and convolution become in the transformed domain?

*Does this have an extension to continuous function spaces, or to distributions, as does the Fourier transform?


The Fourier transform $\mathcal{F} : L^2(\mathbb{R})\rightarrow L^2(\mathbb{R})$ is a unitary operator that is diagonalized by the orthonormal basis of Hermite functions:
$$
             h_n(x) = (-1)^n(2^n n!\sqrt{\pi})^{-1/2}e^{x^2/2}\frac{d^n}{dx^n} e^{-x^2}.
$$
Specifically,
$$
              (\mathcal{F}h_n)(k) = (-i)^n h_n(k).
$$
From this you can define a square root:
$$
               \sqrt{\mathcal{F}}f = \sum_{n=0}^{\infty}(-i)^{n/2}\langle f,h_n\rangle h_n,
$$
where $(-i)^{n/2}$ is either square root of $(-i)^n$, and $\langle f,h_n\rangle = \int_{-\infty}^{\infty}f(t)h_n(t)dt$ is the $L^2$ inner product of $f$ and $h_n$. I don't know the properties of such square roots.
