Suppose $f \in L^\infty(\mathbb{R})$ and define the laplace transform $F:(0,\infty)\rightarrow \mathbb{R}$ by $$F(s) = \int_0^\infty f(t)e^{-st}dt.$$
Prove that $F$ is absolutely continuous on $[a,b]$ for any $b>a>0$.
So, I'm thinking that I might be able to do this by proving that $F$ is Lipschitz. I think I got most of the way to proving that it's Lipschitz, but I'm a little stuck on some algebra getting the final Lipschitz inequality in place.
What I have so far: Let $M$ be the essential upper bound of $f$ (which exists because $f\in L^\infty(\mathbb{R})$), and let $x,y \in [a,b]$. Then $$|F(x)-F(y)| = \left\vert \int_0^\infty f(t)e^{-xt}dt - \int_0^\infty f(t)e^{-yt}dt\right\vert \leq M \left\vert\int_0^{\infty}e^{-x}-e^{-y}\right\vert = $$ $$M\left\vert \frac{1}{x}-\frac{1}{y}\right\vert \leq M\left\vert \frac{1}{a}-\frac{1}{b}\right\vert.$$
And here's where it's not totally coming to me. I'm pretty sure I'm going to want to do something with $M\left\vert \frac{1}{a}-\frac{1}{b}\right\vert$ as my constant for the lipschitz condition. I'm just not totally sure how to incorporate it. I suppose I could just be completely wrong, and the laplace transform isn't lipschitz continuous. Given that $|F(x)-F(y)| \leq M\left\vert \frac{1}{x}-\frac{1}{y}\right\vert$ I think it could be proved directly from the definition of absolute continuity.
Any thoughts either way would be greatly appreciated. Thanks in advance.
You were almost there. For $x, y \in [a,b]$ $$\left| \frac{1}{x} - \frac{1}{y} \right| = \frac{|x-y|}{|xy|} \le \frac{1}{a^2} |x-y|$$
So that $F$ is $(M/a^2)$-Lipschitz.