How to express and prove a smoothness/decay rate relationship between tempered distributions and their Fourier transforms?

Question

In general, there is a relationship between a function and its Fourier transform. A smooth function has a quickly decaying Fourier transform, and a quickly decaying function has a smooth Fourier transform.

For example, see the old question: Smoothness and decay property of Fourier transformation

I want to make a more general inquiry. Say that you have $u\in H^s_{\textrm{loc}}(\mathbb{R}^n)$ where $s\in\mathbb{R}$ and $H^s_{\textrm{loc}}(\mathbb{R}^n)$ is the space of tempered distributions s.t. their restriction to any bounded open subset $\Omega\in\mathbb{R}^n$ is in $H^s(\Omega).$ What can we say about the decay rate of its Fourier transform? How, in fact, do we define the decay rate of the Fourier transform, given that we have no reason to expect the Fourier transform to be smooth or even $\mathbb{L}^2_{\textrm{loc}}$? For this purpose, let's say we know that $\hat{u}\in H^{-t}_{\textrm{loc}}(\mathbb{R}^n)$ where $t>0$ is a real number that could be aribtrarily large, to keep things simple.

I'm also interested in the inverse relationship. If we have a tempered distribution $w\notin H^s_{\textrm{loc}}(\mathbb{R}^n)$, I expect that we should be able to show that in some sense the Fourier Transform of $w$ decays slowly in some sense.

Answer 1

The solution for decay rate is to use a weighted Sobolev space. Define the weighted Sobolev space $H^r_s(\mathbb{R}^n)$ by $u\in H_s^r(\mathbb{R}^n)$ if $u(1+|x|^2)^{s/2}\in H^r(\mathbb{R}^n)$. For $r\ge0$ and $s\ge 0$, this paper proves in Theorem 2.1 that the Fourier transform imbeds $H_s^r$ in $H^s_r$ and vice versa.

We can easily extend the result to $H^{-r}_{-s}$ for $r\ge0$ and $s\ge 0$. $H^{-r}_{-s}$ is the dual of $H^r_s$, so for any $u\in H^{-r}_{-s}$, we have that $\left<u,\mathcal{F}w\right>$ exists for all $w\in H^s_r$. Letting $v$ be the linear functional taking $w\in H^s_r$ to $\left<u,\mathcal{F}w\right>$, it is immediate both that $v=\mathcal{F}u$ by definition and that $v\in H^{-s}_{-r}$ (because $v$ is in the dual space of $H^s_r$).

The Fourier transform is an embedding because by the theorem in the paper, (i) $\exists\,C>c>0$ s.t. $\forall\,w\in H^s_r$, $C\left\|w\right\|_{H^s_r}\ge\left\|\mathcal{F}w\right\|_{H^r_s}\ge c\left\|w\right\|_{H^s_r}$ (equivalent to saying that FT is an embedding for nonnegative $r,s$), so $\left|\left<u,\mathcal{F}w\right>\right|\le C\left\|u\right\|_{H^{-r}_{-s}}\left\|w\right\|_{H^s_r}$, while by definition of the norm in the dual space $\exists \hat{w}_0\in H_r^s$ with $\left\|\hat{w}_0\right\|_{H^r_s}=1$ and $|\left<u,\hat{w}_0\right>|\ge \frac{1}{2}$, and for $w_0:=\mathcal{F}^{-1}\hat{w}_0$, we then have $\left|\left<\mathcal{F}u,w_0\right>\right|=\left|\left<u,\hat{w}_0\right>\right|\ge \frac{c}{2}\left\|u\right\|$.

I do not know how to extend this result to $H^r_s$ when the signs of $r$ and $s$ differ. It can be shown that for $u\in H^r_s$ where $r>0$ and $s<0$, $\mathcal{F}u\in H^s_r$. To do this, simply apply $1-\Delta$ to $u$ enough times to get a function in a Sobolev space with two nonpositive indices, take the Fourier transform, and divide by $1+|\xi|^2$ to the power of the number of times we applied $1-\Delta$. Unfortunately, going the other way is much harder unless we can prove certain results about the inverse of $1-\Delta$.