Decay of Fourier Transform of a Schwartz Function Suppose we have a function $f(x)\in \mathcal{S}(\mathbb R)$; that is, it is a function in Schwartz space. Further, suppose we know that $$|f(x)|\leq Ce^{-|x|}.$$If it is helpful, we can actually replace the exponent on $|x|$ by any $1<c<2$ (in other words, it doesn't seem to be "too far" from a Gaussian). With this information, is there anything we can say about the decay of the Fourier transform of $f(x)$ beyond the fact that it is in the Schwartz class? In particular, does it necessarily decay like $f(x)$, or could it decay much slower, say like $\exp(-(1+x^2)^c)$ for some arbitrarily small $c>0$?
I've tried looking online and haven't found much. What I have found is:


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*The Fourier transform is a linear isomorphism of the Schwartz space; in particular, we know that the Fourier transform is also in the Schwartz space

*The Gaussian, $g(x)=e^{-x^2}$, is essentially a fixed point of this isomorphism (we introduce some constants, but the decay of the function and the decay of the transform is identical - since I'm only worried about the decay, I'm using the term "fixed point" a bit loosely).
Some more information that might be helpful, though I couldn't find any way to use it specifically:


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*$f(x)$ is essentially the characteristic function of a given random variable, which means that the Fourier transform is the corresponding density function. Specifically, this means the Fourier transform takes a maximum value at $0$ (which is equal to $1$) and decreases to $0$ as $|x|\to\infty$.


Even without anything specific, references would be appreciated. I've tried looking in Folland's book as well as Stein/Shakarchi's books, but these have not offered any insight for this problem.
 A: That condition says nothing about the decay of $\hat f$. As a general rule conditions on the decay of $f$ give smoothness for $\hat f$ (here for example it follows that $\hat f$ is holomorphic in a horizontal strip).
Edit: Previous version had  a gap. On reflection I realized that an example filling the gap would also be a counterexample to the question, so that answer actually had very little content.
Actual counterexample: Let $(z+2i)^{1/4}$ denote the principal branch of the fourth root, so in particular if $y\ne-2$ is fixed then $$(x+iy+2i)^{1/4}\sim x^{1/4}\quad(x\to+\infty).$$Note that $$\Re(-x+iy+2i)^{1/4}\sim \frac1{\sqrt 2}|x|^{1/4}\quad(x\to+\infty).$$Let $$F(z)=e^{-(z+2i)^{1/4}}$$and define $$F_y(x)=F(x+iy).$$Then $$F_y(x)=O(|x|^{-N}),$$uniformly for $|y|\le1$; so a little bit of complex analysis shows that $$F_0\in\mathcal S.$$
So there exists $f\in\mathcal S$ with $F_0=\hat f$, and we're done if we show that $f(t)=O(e^{-|t|})$. But, assuming $2\pi=1$, Cauchy's Theorem shows that $$f(t)=\int F_0(x)e^{ixt}\,dx
=\int F_1(t)e^{i(x+i)t}\,dx=e^{-t}\int F_1(x)e^{ixt}\,dx,$$hence $f(t)=O(e^{-t})$. Similarly $f(t)=O(e^t)$.
