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?


1 Answer 1


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 $\epsilon>0$.

  • $\begingroup$ 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 $\varepsilon>0$. $\endgroup$ Nov 30, 2021 at 16:15
  • $\begingroup$ What do you mean? $\endgroup$
    – reuns
    Nov 30, 2021 at 18:35

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