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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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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.) –  Steve Huntsman Jul 31 '13 at 15:40
    
You ought to have a look at the book Asymptotic expansions of integrals by Bleistein and Handelsman, especially Chapter 4. –  Liviu Nicolaescu Jul 31 '13 at 17:15

1 Answer 1

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.

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Thanks! It would still be interesting to get an optimal answer (or an example that shows that this one is optimal). –  H A Helfgott Aug 3 '13 at 13:16

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