# Can we extract information about how fast a function decay from its Laplace transform?

My question is whether we can extract information about how fast an integrable function converges to zero by looking at the asymptotics of its Laplace transform.

More concrete case, let $f:\mathbb{R} \to \mathbb{R}_+$ be a smooth function in $L^1(\mathbb{R})$. If we know that its Laplace transform exists on the positive real axis and:

$\int_{\mathbb{R}} f(x) e^{sx} {\rm d}x \geq e^{\frac{s^2}{2}}, \quad \forall s > 0$,

can we conclude that the speed that $f$ converge to zero cannot be faster than $e^{-\frac{x^2}{2}}$, say,

$\liminf_{|x| \to \infty} \frac{f(x)}{e^{\frac{-x^2}{2} (1 - \epsilon)}} > 0$

for some small $\epsilon \in (0, 1)$? In a more probabilistic setup, if we know the moment generating function is lower bounded by that of Gaussian, can we conclude that it is "super-Gaussian"? I know that the other direction seems to be true and is called sub-Gaussian.

If the information on the right-half real axis is not enough, do we need to know more? Will Fourier transform be more helpful? How about the other direction, i.e., lower bound on the Laplace transform and upper bound on the decay of $f$? Thanks.

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You can extract some information but not as accurate as you suggested. For instance, there is no chance to say anything about your $\liminf$ but if you are happy with $\limsup$ instead, you are fine. To describe everything that is possible and everything that isn't will just take too much time. What exactly are you aiming at?

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Yes. I did got limsup myself: $\limsup \frac{\log\frac{1}{f(x)}}{x^2} \geq 1/2$, something like that. I am aiming at lower bound $f$ pointwise, or maybe lower bound it except for a set of small Lebesgue measure. –  mr.gondolier Feb 19 '10 at 4:07
That is just impossible to do looking at crude growth rates: take $f(s)$ equal to $e^{-s^2/2}$ on $[2n-1,2n]$ and $0$ on $[2n,2n+1]$. You'll see that the gaps don't really matter much for the growth rate but they occupy half the line. –  fedja Feb 19 '10 at 4:37

Can you control the oscillation of f(x) as x increases? If you can show that the ratio of f(x) to your 'simplified' form is 'slowly varying' then your asymptotics will probably work out.

A simple example of what you cannot afford is a log-periodic oscillation; this is because the limits of oscillation of the function and Laplace transform need not agree. The simplest example of a log-periodic oscillation is a complex exponential:

$\int_{0}^{\infty }e^{-st}t^{\alpha +i\beta }dt=\left[ \frac{\Gamma \left( \alpha +i\beta +1\right) }{\Gamma \left( \alpha +1\right) }e^{i\beta \log t}\right] \frac{\Gamma \left( \alpha +1\right) }{s}s^{-\alpha }$

In a sense you can view the imaginary part of the exponential as a 'wobbly constant' which changes more and more slowly. The point is the amplitude of the wobble in the transform depends on β but not for the function.

If you do have this sort of problem (it happens all the time in analysis of algorithms and chaotic dynamics) then you can for example resort to the 'gamma function method' of DeBruijn.

The same thing holds true for the moment question. If you look up a counterexample for the moment problem, (e.g. Feller volume II p. 227) you see the ubiquitous log-periodic oscillation.

Not surprisingly the log-periodic oscillation also shows up in convergence questions of Fourier series, but there it is not oscillating more and more slowly, but faster and faster.

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