# Continuity of Laplace's transformation

I have a question related to Laplace's transformation. Given a function $$f:\mathbb{R}\rightarrow\mathbb{R}$$, we define the Laplace's transformation as the following function:

$$L(f)(s)=\int_{0}^{+\infty}{f(t)\cdot e^{-s\cdot t}\,dt},$$

for any $$s>0$$ when the integral exists. I've proved that if $$f\in L^1(0,+\infty)\cup L^2(0,+\infty)$$, then $$L(f)$$ is well defined for all $$s>0$$. Furthermore, when $$f\in L^1(0,+\infty)$$, then $$L(f)$$ is continous on $$(0,+\infty)$$.

I need to prove the same result of continuity when $$f\in L^2(0,+\infty)$$. I've tried to use the theorem of continuity of parametric integrals, but I don't know how to bounded above the integrand indepently from the variable $$s$$. I've also tried the Hölder's inequality, but it hasn't worked.

Does anyone have an advice?

Use Cauchy-Schwarz. For $$f\in L^2(0,\infty)$$ and $$\Re(s)>0$$ $$F(s)=\int_0^\infty f(t)e^{-st}dt$$ converges and is continuous. The continuity follows from $$|F(s)-F(u)|\le \|f\|_{L^2(0,\infty)} \|e^{-st}-e^{-ut}\|_{L^2(0,\infty)}$$ The convergence follows from $$|\int_A^B f(t)e^{-st}dt|\le \|f\|_{L^2(0,\infty)}\|e^{-st}\|_{L^2(A,B)}$$
For the same reason $$f(t)e^{-\epsilon t}$$ is $$L^1(0,\infty)$$ for all $$\epsilon>0$$.
• I think I got it. But, instead of bounding the integral, I use the Cauchy-Schwarz inequality with the integrand and prove that $L(f)\in\mathcal{C}^0([\varepsilon,+\infty))$ for all $\varepsilon>0$. Nov 30, 2021 at 16:15