TITLE: Growth of (integral of) Laplace transform of a function of compact support as $Re \to -\infty$
QUESTION [6 upvotes]: Let $f:[0,\infty)\to \mathbb{R}$ be supported on $[0,1]$, with $\int_0^1 f(x) dx = 1$. Let $\mathcal{L} f$ be its Laplace transform. How slowly may
$$\int_{-\infty}^\infty |\mathcal{L} f(\sigma+i t)| dt$$
grow as $\sigma\to -\infty$? It is clear it could be $\ll e^{\epsilon |\sigma|}$ (just let $f$ be supported on $[0,\epsilon]$). Could it grow polynomially on $|\sigma|$? Linearly on $|\sigma|$?

REPLY [6 votes]: It has exponential growth: if your integral is $O(e^{-\sigma\epsilon}), \sigma\to -\infty,$ then the
support of your function $f$ is contained in $[0,\epsilon]$. This follows
from the inversion formula for the Laplace transform:
$$f(x)=\lim_{r\to\infty}\frac{1}{2\pi i}\int_{b-ir}^{b+ir}Lf(s)e^{sx}ds,$$
where the integration is on any vertical line $\Re s=b$.
Under your assumptions,
$$|f(x)\leq\frac{1}{2\pi} e^{-b\epsilon}e^{bx},$$
for every $b<0$, therefore, by letting $b\to-\infty$ we obtain $f(x)=0$ when $x>\epsilon$.

REPLY [6 votes]: The Fourier inversion formula says that $$f(x)e^{-\sigma x} = (2\pi)^{-1} \int \mathcal{L}f(\sigma+it) e^{itx} dt$$
so if your integral is $I(\sigma)$, one sees that
$$|f(x)| \le (2\pi)^{-1} e^{\sigma x} I(\sigma)$$
It follows that if $I(\sigma)$ is $O(e^{\varepsilon\sigma})$, then f is supported on $[0,\varepsilon]$.  So that can't hold for ALL $\varepsilon$. Thus the behavior that you mention is best possible.