TITLE: Uncertainty principle for Mellin transform
QUESTION [22 upvotes]: Let $f:\mathbb{R}^+\to \mathbb{C}$. Let $Mf$ be its Mellin transform: $Mf(s) = \int_0^\infty f(x) x^{s-1} dx$.
(a) Some time ago, I convinced myself that
$f(t)$, $Mf(\sigma+it)$ and $Mf(\sigma-it)$ cannot all decrease faster than exponentially as $t\to +\infty$. This should be a variation on G. H. Hardy's version of the uncertainty principle. I presume this is well-known? Does anybody have a reference? Does anybody know an easy proof (one that can be sketched here)?
(b) The example $f(t) = e^{-t^r}$ shows that $f(t)$ can be made to decrease much faster than exponentially, while $Mf(\sigma+it)$ and $Mf(\sigma-it)$ still decrease exponentially (meaning $\sim e^{-(\pi/2r) |t|}$ - that is, the exponent does degrade as $r$ increases). Is this in any sense optimal? That is, can one show that, if $f(t)$ decreases faster than $e^{-t^r}$ for every $r$, then $Mf(\sigma+it)$ and $Mf(\sigma-it)$ cannot both decrease exponentially?

REPLY [6 votes]: There is something the above posts missed (perhaps because the wording didn't make it explicit): the condition on the function $f(x)$ is one sided (i.e., it assumes something on the decay as $x\to \infty$, not as $x\to -\infty$), and thus, once we take logarithms to make the Mellin transform into a Fourier transform, we end up with a one-sided condition as well.
Now, one-sided versions of the uncertainty principle do exist - notably that of Nazarov (1993); see also Bonami and Demange (2006). In the end, it turns out to be best to rework the proofs given there to get an uncertainty principle for the Mellin transform of the shape I was asking for. And yes, one does get something essentially optimal in that way, with no spurious factors of log in the exponent.