$\lim_{t\to \infty}k_1(t)=\lim_{\lambda \to 0}\lambda \widehat{k_1}(\lambda)$?

Question

Can we conclude that $$\displaystyle \lim_{t\to \infty}k_1(t)=\lim_{\lambda \to 0}\lambda \widehat{k_1}(\lambda)?$$ Here $\widehat{k_1}$ is the Laplace transform of a function $k_1:(0,\infty)\to \mathbb{R}$ nonnegative, left-continuous, nonincreasing. It is a question that appear while I have reading the book of Jan Prüss (Evolutionary integral equations and applications, pg. 93), exist more assumptions about $k_1$. I am trying use integration by parts, but i cant answer.

Answer 1

Yes. $\lambda \hat {k_1}(\lambda) =\lambda \int_0^{\infty} e^{-\lambda t } k_1(t)dt=\int_0^{\infty} e^{-s} k_1(\frac s {\lambda}) ds$. We can aply Monotone Convergence Theorem to conclude that $\lambda \hat {k_1}(\lambda) \to k_1(\infty)$ as $ \lambda$ decreases to $0$.