TITLE: How do you prove that a function $f \in L^1( \bf R)$ and its Fourier transform cannot simultaneously be very small at infinity?
QUESTION [1 upvotes]: In a Research Paper it is stated that(Without any proof):

It is well known that a function and its Fourier transform cannot simultaneously be very small at infinity?

In my First course of Harmonic Analysis I've never seen such result.Can someone explain me the precise meaning of above statement and give me some idea to prove this ?

REPLY [1 votes]: I think what they mean is that the function and its Fourier transform cannot both be decreasing faster than $e^{-x^2}$ when $x \to \infty$ :
the theorem is that $e^{-\pi x^2},e^{-\pi f^2}$ is the Fourier transform couple maximizing the concentration of the energy at the origin :
$$\min \int_{-\infty}^\infty |h(x)|^2 x^2 dx + \int_{-\infty}^\infty |\text{FT}\{h\}(f)|^2 f^2 df$$
$$\text{such that } \int_{-\infty}^\infty |h(x)|^2 dx = 1$$
then the solution is $\displaystyle h(x) = c e^{-\pi x^2}$ with $|c|=1$.
(note that I didn't talk about $\displaystyle \min \int_{-\infty}^\infty |h(x)|^2 x^{2k} dx + \int_{-\infty}^\infty |\text{FT}_h(f)|^2 f^{2k} df$ $\text{such that } \int_{-\infty}^\infty |h(x)|^2 dx = 1$ for $k = 2,3, \ldots$
which should be interesting too, if someone knows the solutions ? I guess it is enough to prove that the solution is an eigenfunction of the Fourier transform $h = FT\{h\}$ to get $h(x) =  e^{-\pi x^2}$ again... )
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