Convolution and the square root of the Fourier Transform

Question

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:

1. What do the columns of the matrix in question look like? (Basis functions)
2. Is there an equivalent of the convolution theorem for this transform? What do pointwise multiplication and convolution become in the transformed domain?
3. Does this have an extension to continuous function spaces, or to distributions, as does the Fourier transform?

Answer 1

To take it one step further than DisintegratingByParts's answer:

The Fourier transform $\mathcal F$ is unitary on $L^2({\mathbb R})$ with spectrum $\{1,i,-1,-i\}$. This means that $\mathcal F$ effectively behaves like a finite-dimensional diagonal operator. In particular, (basically) any square root of $\mathcal F$ can be written as a polynomial of degree $3$ in $\mathcal F$ itself. These polynomials can be deduced by Lagrange interpolation and equal

$$p(x) = \pm \frac{1}{4} (x+1)(x^2+1) \pm \frac{1+i}{4 i \sqrt{2}} (x^2-1)(x+i) \pm \frac{i}{4} (x-1)(x^2+1) \pm \frac{-1+i}{4 i \sqrt{2}} (x^2-1)(x-i).$$

So the answers are:

1. On $\mathbb R$, the basis functions are Hermite functions like DisintegratingByParts mentioned. In higher dimensions, $L^2(\mathbb R^d)$ can be decomposed into invariant subspaces whose angular parts are given by spherical harmonics. The book of Stein and Weiss has a very nice exposition.

2. Generally convolution, pointwise multiplication, differentiaion, and multiplication by monomials will not transform into anything nice simply because the different terms in $p({\mathcal F})$ will transform differently. (It's easy to check that in none of the cases do coefficients magically vanish to produce unexpected symmetry).

3. Because it's a polynomial in $\mathcal F$, it rather directly extends to distributions.

Answer 2

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.