Uncertainty principle (really for Mellin, but never mind that!)

2012-11-08T15:51:48Z

Is there a smooth funtion $f:\mathbb{R}\to \mathbb{C}$ such that

(a) $f(x)$ decreases faster than $e^{-e^x}$ when $x\to \infty$,

(b) $\widehat{f}(t)$ decreases faster than $e^{-|t|}$ when $t\to \pm\infty$?

Note there is no restriction on $f(x)$ for $x\to -\infty$ except that it decay fast enough for the Fourier transform $\widehat{f}$ to be well defined. The function $f_\epsilon(x) = e^{\epsilon x - e^x}$, $\epsilon>0$, decays almost a fast as $e^{-e^x}$ for $x\to \infty$ and $\widehat{f_\epsilon}(t)$ decays roughly as $e^{-|t|}$ for $t\to \pm \infty$ -- so I am really asking whether one can defeat $f_\epsilon$ on both the physical and Fourier aspect.

PS. Yes, this does come from a Mellin transform, as the double exponential probably gives away.

Answer by Ralph Furmaniak for Uncertainty principle (really for Mellin, but never mind that!)

2012-11-09T00:57:01Z

How about $$\frac{e^{-e^x}}{1+\epsilon x^2}$$ If you compute the Fourier transform you can shift the contour to height $\pm\pi/2$ to get an $e^{-|t|}$ times something decaying to 1, by Riemann-Lebesgue lemma

Edit.

Or you can look at a shift of your original example: $f(x+\log A)$ to get something on the lines of $$e^{-A e^x} e^{-B x^2} e^{C x}$$

Uncertainty principle (really for Mellin, but never mind that!) - MathOverflow

Uncertainty principle (really for Mellin, but never mind that!)

Answer by Ralph Furmaniak for Uncertainty principle (really for Mellin, but never mind that!)