# Compactly supported function whose Fourier transform decays exponentially?

It's well known now that a function can not be compactly supported both on the space side and the frequency side (so-called uncertainty principle). On the other hand a function can have exponential decay on both sides, e.g. guassian function $e^{-x^2}$. My question is whether the intermediate case exists. More precisely, is there a function which is compactly supported on the space side and has exponential decay on the frequency side. Thanks!

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Not exactly what you're looking for, but this note gives a decay rate $\hat{f}(k) \sim k^{-3/4} e^{-\sqrt{k}}$ for the standard bump function $f(x) = \exp{(1/(1-x^2))}$. Note also that exponential decay of the Fourier transform should yield real analyticity of $f$ which doesn't go together well with the compact support requirement. I can't make this precise right now, though. –  t.b. Apr 29 '12 at 11:55

We claim that the zero function is the only function with this property. Suppose to the contrary that $f$ is compactly supported and $\hat f$ has exponential decay. Then $\hat f \in L^1$, so $f = (\hat f)^\vee$, the inverse fourier transform of $\hat f$. But, the exponential decay condition allows us to differentiate under the integral sign to obtain that

$$(\hat f)^\vee(z) = \int_{\mathbb R} \hat f(x) e^{ixz} dx$$

is defined and holomorphic for $z$ in a neighborhood of the real axis. But since this function has compact support on $\mathbb R$, it vanishes on a set with an accumulation point, hence is identically zero.

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If $\left|\hat{f}(\xi)\right|\leqslant Ce^{-k|\xi|}$ , then by Fourier Inversion, we have \begin{align} \left|f^{(n)}(x)\right| &=\left|\int_{\mathbb{R}}\hat{f}(\xi)\,(i2\pi\xi)^n\,e^{i2\pi\,x\xi}\,\mathrm{d}\xi\right|\\ &\leqslant\int_{\mathbb{R}}Ce^{-k|\xi|}(2\pi|\xi|)^n\,\mathrm{d}\xi\\ &=(2C/k)(2\pi/k)^nn!\tag{1} \end{align} Estimate $(1)$ implies the power series for $f$ has a radius of convergence of at least $\frac{k}{2\pi}$ everywhere. Thus, it is real analytic over all $\mathbb{R}$, and cannot have compact support unless it is identically $0$.

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Nice! Thank you. –  Syang Chen Apr 30 '12 at 3:00
Since I can only accept one, I think I will accept the earliest one.. hard to choose.. –  Syang Chen Apr 30 '12 at 3:03
@DavideGiraudo: I prefer $\le$, but I have nothing against $\leqslant$. In any case, it is a pretty small change; however, it moved the question to the front page and got another vote. :-) –  robjohn Jan 7 '13 at 22:51
Actually, in the "align", there was a $dx$ instead of $d\xi$, which I corrected. –  Davide Giraudo Jan 8 '13 at 9:21
@DavideGiraudo: ah I missed that. Thanks. –  robjohn Jan 8 '13 at 9:51