Uncertainty principle (really for Mellin, but never mind that!) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T19:23:30Z http://mathoverflow.net/feeds/question/111826 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111826/uncertainty-principle-really-for-mellin-but-never-mind-that Uncertainty principle (really for Mellin, but never mind that!) H A Helfgott 2012-11-08T15:51:48Z 2012-11-09T18:51:24Z <p>Is there a smooth funtion $f:\mathbb{R}\to \mathbb{C}$ such that</p> <p>(a) $f(x)$ decreases faster than $e^{-e^x}$ when $x\to \infty$,</p> <p>(b) $\widehat{f}(t)$ decreases faster than $e^{-|t|}$ when $t\to \pm\infty$?</p> <p>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.</p> <p>PS. Yes, this does come from a Mellin transform, as the double exponential probably gives away.</p> http://mathoverflow.net/questions/111826/uncertainty-principle-really-for-mellin-but-never-mind-that/111858#111858 Answer by Ralph Furmaniak for Uncertainty principle (really for Mellin, but never mind that!) Ralph Furmaniak 2012-11-09T00:57:01Z 2012-11-09T18:51:24Z <p>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</p> <p>Edit.</p> <p>Or you can look at a shift of your original example: $f(x+\log A)$ to get something on the lines of $$e^{-A e^x} e^{-B x^2} e^{C x}$$</p>