Relationships between growth rates of a distribution and smoothness of its Fourier transform

Question

Let $f\in \mathcal{S}^\prime(\mathbb{R})$ be a tempered distribution, and $\hat{f}$ be its Fourier transform. It is known that when both $f$ and $\hat{f}$ are $L^2$ functions, there are relationships between the decay rate of $f$ at infinity and differentiability properties of $\hat{f}$. Are there any similar results in the tempered distribution case? More specifically, are these two propositions true?

(1) if $f$ is a function, and $f(x)<C(1+|x|)^N$ for some $C,N\geq 0$, then $\hat{f}=D^{N+c_0}g$ for some constant $c_0$ independent of $N$, and $g$ is a continuous function.\

(2) if $f\in C^m(\mathbb{R})$ , then $\hat{f}=D^{N}g$ where $g$ is a continuous function satisfying $g(\omega)<(1+|\omega|)^{-m+c_1}$ for some constant $c_1$ independent of $m$.\

I have not been able to find related results on Hormander and Strichartz.

Answer 1

Not a full answer, but it allows us to charasterise the behaviour of Fourier transform, and partially answers the point (a).

There's a characterisation of tempered distributions as $$T\in S'\iff T=\partial_x^{k}((1+x^2)^nf) \text{ for some } k,n\in \Bbb N \text{ and } f\in C_b.$$ So this gives you the behaviour of Fourier transform:

$$F[T](\xi) = i^k\xi^k(1-\Delta)^nF[f].$$

Answer 2

Necromancy! The answer to your specific question is no, because the integral of a decaying function need not decay. For example, $e^{-x^2}$ is not the derivative of any rapidly decaying function.

However, you can define decay rates properly using weighted Sobolev Spaces, and then a result of the kind you want can be proved. The relationship is proved for a pretty general situation in Theorem 2.1 of this paper. I found this while trying to answer my own question, and I provide a modest (and trivial) extension of the paper's result in my answer.