# Growth of the Fourier transform of a compact function

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 ..." (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.

• Could you provide more information on $f$? Is it also smooth or just continuous or an $L^p$ ($p\in[1,2]$) function? – hal4math Sep 2 '19 at 17:49
• @hal4math Sorry I meant compactly supported and smooth (as V in the screenshot) – Thomas Sep 2 '19 at 18:26
• Maybe this result is what you are looking for: Paley-Wiener Theorem – Daniel Sep 2 '19 at 18:49
• 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? – 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 $$[-c,c]$$.