Non-negative functions on $[0,+\infty[$ whose Laplace transform is well-defined on $]0,1]$

Question

I am trying to determine the set $E$ of non-negative Lebesgue-measurable functions $f$ on $[0,+\infty[$ whose Laplace transform is defined on $]0,1]$, that is, such that the Lebesgue integral $\int_0^{+\infty} f(t) e^{-st} dt$ (an element of $[0,+\infty]$ as $f$ is non-negative) is finite. Obviously $E$ contains $L^1\left([0,+\infty[\right)$ and $L^{\infty}\left([0,+\infty[\right)$ but can we be more precise ?

Answer 1

Let $F(x) = \int_0^x f(t)\; dt$. Thus $F(x)$ is nondecreasing and $$\int_0^\infty f(t) e^{-st}\; dt \ge e^{-sx} F(x)$$

I claim that a necessary and sufficient condition is that $$\limsup_{x \to \infty} \frac{\ln(F(x))}{x} = 0 $$

If $\limsup_{x \to \infty} \ln(F(x))/x = r > 0$, then for $0 < s < s' < r$ there are arbitrarily large $x$ with $\ln(F(x))/x > s'$, hence $F(x) > e^{s'x}$, and $\int_0^\infty f(t) e^{-st}\; dt \ge e^{(s'-s) x}$, so $\int_0^\infty f(t) e^{-st}\; dt = \infty$.

On the other hand if $\limsup_{x \to \infty} \ln(F(x))/x = 0$, for every $s > 0$ we can take $s > s' >0$ and $R>0$ so $\ln(F(x))/x < s'$, i.e. $F(x) \le e^{s' x}$, for $x \ge R$, and then $$ \left. \int_0^x f(t) e^{-st}\; dt = F(t) e^{-st}\right|_{t=0}^x + \int_0^x F(t) e^{-st} \; dt \le 1 + \int_0^R F(t) e^{-st}\; dt + \int_{R}^\infty e^{-(s-s')t}\; dt < \infty $$