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?

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    $\begingroup$ Have you tried to look at the Mellin transform as a modified Fourier transform? To wit, $\mathcal{M} = \sqrt{2\pi} \cdot \mathcal{F} \mathcal{E}$, where $\mathcal{F}$ is the unitary angular frequency Fourier transform and $\mathcal{E}f := f \circ \exp$. (Note that this requires writing $\mathcal{M}f(k)$ where $k = -is$, rather than $\mathcal{M}f(s)$ directly.) $\endgroup$ Jul 31 '13 at 15:40
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    $\begingroup$ You ought to have a look at the book Asymptotic expansions of integrals by Bleistein and Handelsman, especially Chapter 4. $\endgroup$ Jul 31 '13 at 17:15

Combining Steve's idea with the Beurling-Hormander uncertainty principle yields the kind of result you're looking for.

The BHUP is the following:

Beurling-Hormander Uncertainty Principle: If $f\in L^1(\mathbb{R})$ and $$ \int\int_{\mathbb{R}^2}|f(x)\hat f(y)|e^{|xy|}dxdy<\infty, $$ then $f=0$.

As a corollary, we have a Hardy uncertainty principle type result:

Corollary: If $\phi(x)$ and $\psi(y)$ are convex conjugates (so that $\phi(x)+\psi(y)\geq xy$), and if $$\int_{-\infty}^\infty|f(x)|e^{\phi(x)}dx<\infty\quad\mbox{and}\quad\int_{-\infty}^\infty |\hat f(y)|e^{\psi(y)}dy<\infty,$$ then $f=0$.

These theorems are stated for $\hat f(y)=\int_{-\infty}^\infty f(x)e^{-ixy}dx$. Note that $\psi(y)$ is essentially the Legendre transform of $\phi(x)$: $$\psi(y)=\sup_x xy-\phi(x).$$

Also, note that we can ignore a bounded part of the domain of $\hat f$, since $|\hat f(y)|\leq\Vert f\Vert_1$.

Now like Steve said, setting $f(x)=g(e^{-x})$, we have $\hat f(y)=Mg(iy)$. Note that $f\in L^1(\mathbb{R},dx)$ if and only if $g\in L^1(\mathbb{R}_{>0},\frac{dx}x)$. Thus the BHUP for Mellin Transforms is as follows:

BHUP for Mellin Transforms: Let $g$ be a function in $L^1(\mathbb{R}_{>0},\frac{dx}x)$.

If $$ \int_0^\infty\int_{-\infty}^\infty |g(x)||Mg(iy)|e^{|\ln(x)\cdot y|}\frac{dx}{x}dy<\infty,$$ then $g=0$.

If $\phi(x)$ and $\psi(y)$ are functions such that $\phi(x)+\psi(y)\geq xy$, then the Hardy uncertainty principle type result translates to:

Corollary: If $$\int_0^\infty |g(x)|e^{\phi(-\ln x)}\frac{dx}x<\infty\quad\mbox{and}\quad\int_{-\infty}^\infty |Mg(iy)|e^{\psi(y)}dy<\infty,$$ then $g=0$.

Now suppose that $g$ is a non-zero function in $L^1(\mathbb{R}_{>0},\frac{dx}x)$ such that $|g(t)|=O(e^{-t^r}/t^\delta)$ for some $\delta>0$. If we choose $\phi(x)=e^{-rx}$, then

$$\int_0^\infty |g(x)|e^{\phi(-\ln x)}\frac{dx}x <\infty.$$

Thus $Mg(it)$ cannot decay (significantly) faster than $|t|^{-1}e^{-\psi(t)}$, where $\psi(t)$ is (something like) the Legendre transform of $e^{-rx}$. What we need is $\psi(t)\geq |xt|-e^{-rx}$ for all $t$ outside some interval. I think that taking $$\psi(t)=\frac{|t|}r \ln\left(\frac{|t|}r\right)-\frac{|t|}r\qquad\mbox{for}\quad |t|>r$$ works.

Thus as $t\to\infty$, $|Mg(\pm it)|$ cannot decay faster than roughly $e^{-\frac tr\ln\frac tr}$.

This isn't quite as sharp as you were hoping for, but I probably haven't done the most careful analysis.

  • $\begingroup$ Thanks! It would still be interesting to get an optimal answer (or an example that shows that this one is optimal). $\endgroup$ Aug 3 '13 at 13:16

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.

  • $\begingroup$ To wit: if $f(x)$ decays like $exp(-x^r)$ as $x\to \infty$, then $M(\sigma+i\tau)$ cannot decay as fast as $exp(-c |\tau|)$ for any $c>\pi/2r$. (Well, unless $f$ is $0$ almost everywhere, obviously.) $\endgroup$ Jan 4 '16 at 20:10

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