A function $F: \mathbb R\_+ \rightarrow \mathbb R\_+$ is said to be long tailed if $F(\infty)=0$ and for all $y \geq 0$ $$\frac{F(x+y)}{F(x)} \rightarrow 1, \quad x\rightarrow \infty.$$


Let $\mu$ be a measure on $[0, \infty)$ with full support and denote $f(x)=\mathcal L(\mu,x)$ its Laplace transform. I am trying to figure out whether or not it is true that $f$ is long tailed however given $\mu$.


* If $\mu (x) \sim x^\gamma g(x) dx$, $\gamma>-1$ and $g$ analytic in 0, $g(0)\neq 0$ then there should be no problem since we can essentially invoke Watson's lemma, and then $f(x) \sim g(0)\Gamma(\gamma+1)x^{-\gamma-1}$.
* There are also things like $\mu(x)=e^{-1/x}dx$. Then $f(x) =2 K\_1(2 \sqrt{x})/\sqrt{x}$ with $K\_\nu$ the modified Bessel function, and $f(x) \sim \sqrt{\pi}e^{-2 \sqrt{x}}x^{-3/4}$. Even though the decay is faster than every polynomial order the long tail property is verified.


Purely discrete measures are ruled out by definition, I would initially be happy with the absolutely continuous case only (or a counterexample).



Long tail property of Laplace transforms
Yes, this is true, assuming that
\begin{equation\*}
 f(s):=\int\_{[0,\infty)} e^{-sx}\mu(dx)<\infty
\end{equation\*}
for some real $s\_0$ and all real $s\ge s\_0$. As in the OP, we are also assuming that $\mu$ is a measure on $[0, \infty)$ with full support. We want to show that then for each real $y\ge0$ we have
\begin{equation\*}
 r\_y(s):=\frac{f(s-y)}{f(s)}\to1 \tag{10}\label{10}
\end{equation\*}
(as $s\to\infty$).


To do that, take any real $h>0$ and write
\begin{equation\*}
 f(s)=I\_1(s)+I\_2(s), 
\end{equation\*}
where
\begin{equation\*}
 I\_1(s):=I\_{h,1}(s):=\int\_{[0,h)} e^{-sx}\mu(dx),\quad I\_2(s):=I\_{h,2}(s):=\int\_{[h,\infty)} e^{-sx}\mu(dx). 
\end{equation\*}
So, for real $s\ge s\_0$,
\begin{equation\*}
 r\_y(s)\le\frac{I\_1(s-y)}{I\_1(s)}\Big(1+\frac{I\_2(s-y)}{I\_1(s-y)}\Big). \tag{20}\label{20}
\end{equation\*}
Next,
\begin{equation\*}
 I\_1(s-y)=\int\_{[0,h)} e^{yx}e^{-sx}\mu(dx)\le e^{yh}I\_1(s) \tag{30}\label{30}
\end{equation\*}
and, for real $s\ge s\_0$,
\begin{equation\*}
 I\_2(s)\le e^{-(s-s\_0)h}f(s\_0)=Ce^{-sh},
\end{equation\*}
where $C:=e^{s\_0h}f(s\_0)\in[0,\infty)$. On the other hand,
\begin{equation\*}
 I\_1(s)\ge \int\_{[0,h/2)} e^{-sx}\mu(dx)\ge ce^{-sh/2}, 
\end{equation\*}
where $c:=\int\_{[0,h/2)} \mu(dx)>0$, since $\mu$ is a measure on $[0,\infty)$ with full support. So, for real $s\ge s\_0$ we have $I\_2(s)\le \frac Cc\, I\_1(s)\,e^{-sh/2}$.
So, $\dfrac{I\_2(s-y)}{I\_1(s-y)}\to0$ (as $s\to\infty$).


It now follows from \eqref{20} and \eqref{30} that $L\_y:=\limsup\_{s\to\infty}r\_y(s)\le e^{yh}$ for any real $h>0$, and hence $L\_y\le1$ for any real $y\ge0$. On the other hand, the function $f$ is nonincreasing and hence $r\_y(s)\ge1$ for any real $y\ge0$ and $s\ge s\_0$. Thus, \eqref{10} follows. $\quad\Box$




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It is easily seen from the above proof that the condition that $\mu$ is a measure on $[0, \infty)$ with full support can be relaxed to the condition that $0$ is in the support of $\mu$.


