Let $ Lf(s) = \int_0^\infty e^{-sx}f(x)~dx$ be the Laplace transform of a measurable function $f$ on $[0,\infty).$
I would like to be able to show the following:
If $f\in L^1[0,\infty]$ , then $Lf(s)$ exists and is bounded for all $s\geq 0$.
Suppose $\{f_n\}~n\geq 1$ and $f\in L^1[0,\infty)$. If $f_n \to f$ in $L^1$-norm, then $Lf_n\to Lf$ uniformly on $[0,\infty)$.
For (1), these are my thoughts. Since $f\in L^1[0,\infty)$ and $f\geq 0$, $f$ is finite a.e. So there is an $M \gt 0$ such that $|f(x)|\leq M$. Infact, I can choose $M$ large enought so that $|f(x)|\leq M e^{at}$ for some $a\geq 0$. Then $\int_0^\infty |e^{-sx}f(x)|~dx \leq M\int_0^\infty e^{(a-s)x}~dx =\frac{M}{s-a}.$