When is the Laplace transform of a function continuous?

Question

Are there any general conditions on a function so that its Laplace transform is continuous? e.g. what conditions on $f:\mathbb {R} \rightarrow \mathbb {R}$ guarantee that its Laplace transform $$F\left(s\right)=\int_{0}^{\infty}f\left(t\right)e^{-st}dt$$ is continuous? It seems to me that if $f$ has compact support and is bounded, then it ought to be. The idea I have is that we get $$\left|F\left(s'\right)-F\left(s\right)\right|\leq\int_{0}^{y}\left|f\left(t\right)\right|\left|e^{-st}\right|dt$$ by making $s'$ and $s$ close enough, it seems we can make the RHS arbitrarily small. I haven't been able to complete a proof. Additionally, can we change the assumptions? Are there more natural assumptions to get continuity?

Answer 1

Assuming you want it continuous on all of $\mathbb R$ (or $\mathbb C$), the necessary and sufficient condition is that $f$ is measurable and $f(t) e^{-st}$ is integrable for all $s \in \mathbb R$. Then the Lebesgue Dominated Convergence Theorem gives you the desired continuity.

EDIT: More explicitly, suppose $f$ satisfies the above condition, and take any $s \in \mathbb C$. For $|s' - s| \le 1$ and $t \ge 0$, $$\left|f(t) e^{-st} - f(t) e^{-s' t}\right| \le 2 \left|f(t) e^{(1+|s|)t}\right|$$ the right side being integrable on $[0,\infty)$ by assumption. As $s \to s'$, $f(t) e^{-s't} - f(t) e^{-st} \to 0$ for all $t$, and the Dominated Convergence Theorem says $$\int_0^\infty f(t) e^{-s' t} \; dt \to \int_0^\infty f(t) e^{-st}\; dt $$