Unif. continuity of Laplace transform

Question

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.

Just edited my question. It's the usual laplace transform, where $f$ is the density function of a random variable — andereBen, Jan 21 '20 at 23:25
You seem to have started your proof in a way that's different to the answer for the Fourier transform. — Oliver Jones, Jan 21 '20 at 23:27
Yes, the fact is that the complex exponential has modulus 1 and somehow simplifies the proof there — andereBen, Jan 21 '20 at 23:27
Do you follow the proof that was given there? Do you know what the DCT is? — Oliver Jones, Jan 21 '20 at 23:32
It uses the fact that the modulus of the exponential is one, but here I can't use it, that's why I get stuck — andereBen, Jan 21 '20 at 23:44
Yes, he takes $h$ to $0$, IF I could do that, then the claim follows. To be able to do that, I need to use DCT, — andereBen, Jan 21 '20 at 23:55

score 1 · Accepted Answer · answered Jan 21 '20 at 23:30

1

If $t_n \to t$ then $Ee^{-t_nX} \to Ee^{-tX}$ by DCT, so $L_X$ is continuous. Also $L_X(t) \to 0$ as $ t\to \infty$ by DCT and any continuous function on $[0,\infty)$ which vanishes at $\infty$ is uniformly continuous.

answered Jan 21 '20 at 23:30

Kavi Rama Murthy

311,013

Thanks @Kavi Rama Murthy, but I am not allowed to use the tool you had in your answer. I need some explicit computation as the one I tried to do above – andereBen Jan 21 '20 at 23:32
By the way, the DCT applies because for $t>0$ I have $e^{-t X(\omega)}| \leq 1 $ since $X(\omega) \leq 0$ for every $\omega \in \Omega$, right? – andereBen Jan 22 '20 at 08:11
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@andereBen Yes. DCT applies because $0 <e^{-tX} \leq 1$. – Kavi Rama Murthy Jan 22 '20 at 08:18

Unif. continuity of Laplace transform

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