If I have a positive random variable $X$, how can I show that its Laplace transform $$L_X(t)=E[e^{-tX}] = \int_{\mathbb{R}} e^{-tx}f(x)dx, \quad t>0$$ where $f(x)$ is the density of the r.v. $X$, is uniformly continuous ?
I just know the $\epsilon$ - $\delta$ definition of uniform continuity.
I would start with $$|L(t+h) - L(t)|=||\int_{\mathbb{R}} f_X(x) e^{-tx} \bigl( e^{-hx} - 1 \bigr)dx|| \leq \int_\mathbb{R}|| f_X(x)|| e^{-tx} |e^{-hx} - 1 |dx$$ but then I don't know how to go on.
Actually, I wanted to use the same trick used for the fact that the Fourier transform is uniformly continuous as explained in this answer, but I get stuck.