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:

  • 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:

  • $f(x)$ 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.

  • $\begingroup$ Came up with an actual counterexample... $\endgroup$ Jun 1, 2019 at 17:12
  • 1
    $\begingroup$ To rephrase David : if $f$ is complex analytic on $\Im(z) \in (-c,c)$ and Schwartz on horitonal lines then $\mathcal{F}[f(x)](\omega)$ is Schwartz and $O(e^{-(c-\epsilon)|\omega|})$. The analycity condition on $f$ says nothing of its decay $\endgroup$
    – reuns
    Jun 1, 2019 at 17:49
  • $\begingroup$ @reuns Indeed. I don't know exactly who first stated that version, possibly Bochner - the result with $L^2$ in place of $\mathcal S$ is Bochner. (Just doing the special case seemed simpler than stating a precisely correct general version and verifying the hypotheses...) $\endgroup$ Jun 2, 2019 at 1:04

1 Answer 1


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 $f(t)=O(e^t)$.

  • $\begingroup$ You've written $F_y(x)=O(|x|^{-N})$, but isn't more true? Specifically that $F_y(x)=O\big(\exp(-|x|^{1/4})\big)$? Which causes this to fail to be a counterexample? $\endgroup$ Jun 3, 2019 at 17:29
  • $\begingroup$ Yes, more is true (although I think there may be a constant missing in your "more"). So what? Nothing causes it to fail to be a counterexample - it is a counterexample. That's why I called it a counterexample. $\endgroup$ Jun 3, 2019 at 17:39
  • $\begingroup$ Yes, there would be a constant, but the exponent on $|x|$ cannot be made arbitrarily small so the function still appears to have exponential decay (slower exponential decay, granted, but it still decays exponentially). Just to be clear you're saying you have a function that decays like $e^{-|t|}$ whose Fourier transform decays like $\exp(-|x|^{1/4})$, correct? Not that it decays at a rate like $\exp(-|x|^{1/\log\log(10+x^2)})$. $\endgroup$ Jun 3, 2019 at 17:48
  • $\begingroup$ "Which causes this to fail to be a counterexample?" I misread that a minute ago. The $f$ such that $\hat f=F_0$ is a counterexample to the statement that $f=O(e^{-|t|})$ implies $\hat f(x)=O(e^{-|x|})$. You must be missing a minus sign or something; $e^{=|t|^{1/4}}$ is not $O(e^{-|t|})$. $\endgroup$ Jun 3, 2019 at 17:51
  • $\begingroup$ We never defined "exponential decay". I wouldn't say $e^{-|t|^{1.4}}$ has exponential decay. If you feel it does, what about $e^{-\ln|t|}$? $\endgroup$ Jun 3, 2019 at 17:54

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