Laplace transform on a continuous function on [0,1]

Question

Let g be a continuous function. I would like to show that $\lim_{h\to \infty} h \int_{t=0}^1 e^{-ht}g(t)dt=g(0)$. I try to solve it by integration by part, but that does not work. Is there any way to show it except using DCT.

Answer 1

Make the susbstitution $s=ht$. We have to show that $\int_0^{h}e^{-s} g(\frac s h)ds \to g(0)$. Use the following steps:

1. Let $r=\frac 1 {\sqrt h}$. By continuity of $g$ at $0$ we see that $|\int_0^{rh} e^{-s} [g(\frac s h)-g(0)]ds|<\epsilon \int_0^{\infty} e^{-s} ds/2 =\epsilon/2$ if $h$ is large enough.

2. $|\int_{rh}^{h} e^{-s} [g(\frac s h)-g(0)]ds|\leq C \int_{rh}^{h} e^{-s}ds$ where $C$ is bound for $g(\frac s h)-g(0)$. Note that $\int_{rh}^{h} e^{-s}ds=e^{-rh}-e^{-h} <\epsilon/{2C}$ if $e^{-rh} <\epsilon/2C$.

3. Finally note that $\int_0^{h} e^{-s} g(0)ds \to g(0)$ as $ h \to \infty$.