If a function $f$ is smooth and compactly supported in $[a, b]$, can we assume anything on the exponential growth of $\hat f$?

For the context, I'm trying to understand how the author of this paper infers "$\hat G(x) D(...)$ is of exponential type in x ..." enter image description here (page 14 in the paper; D is a Dirichlet polynomial)

I know that if $f$ is in the Schwartz space, so is its Fourier transform, but this results looks stronger.

Thanks in advance!

  • $\begingroup$ Could you provide more information on $f$? Is it also smooth or just continuous or an $L^p$ ($p\in[1,2]$) function? $\endgroup$ – hal4math Sep 2 '19 at 17:49
  • $\begingroup$ @hal4math Sorry I meant compactly supported and smooth (as V in the screenshot) $\endgroup$ – Thomas Sep 2 '19 at 18:26
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    $\begingroup$ Maybe this result is what you are looking for: Paley-Wiener Theorem $\endgroup$ – Daniel Sep 2 '19 at 18:49
  • $\begingroup$ I think all you can assume is that $\widehat{f}$ is in the Schwartz space. Sorry, I am not too familiar with the concept of exponential type, but I would think that any Schwartz function will fall under this category since they even grow less than any polynomial ? So this specific exponent might come from the Dirichlet polynomial? $\endgroup$ – hal4math Sep 2 '19 at 18:50

For $f \in C^\infty_c(\Bbb{R})$ with $[a,b]$ the convex closure of its support, $c = \max(|a|,|b|)$

Then $$\hat{f}(z) = \int_{-\infty}^\infty f(t) e^{-2i \pi z t}dt, \qquad z \in \Bbb{C}$$ is entire, it is Schwartz on every hozirontal line/strip,

Moreover $|\hat{f}(z)| \le \|f\|_\infty e^{2 \pi c |z|}$ and $\hat{f}(z) \ne O(e^{2 \pi (c-\epsilon) |z|})$

Thus it is an entire function of exponential type $2\pi c$.

Any entire function $G$ of exponential type $2\pi c$ which is Schwartz on every horizontal strip (ie. the decay at $\infty$ is locally uniform) is of this form : letting $y \to sign(t)\infty$ in $g(t) = \int_{-\infty}^\infty G(x+iy) e^{2i \pi (x+iy)t}dx$ shows that $g$ is supported on $[-c,c]$.

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