The Hardy–Littlewood Tauberian theorem for Laplace transform in Chapter XIII in "An Introduction to Probability Theory and Its Applications" by Feller reads as follows
Let $F : [0,\infty) \to \mathbb{R}$ of bounded variation, $p \geq 0$ be real number and $$\omega_F(s) = \int^\infty_0 e^{-st} d F(t).$$ Then each of the relations $$ \dfrac{\omega_F(\tau \lambda)}{\omega_F(\tau)} \to \lambda^{-p}\hspace{15pt} \text{as $\tau \to 0$}.$$ $$ \dfrac{F(tx)}{F(t)} \to x^{p} \hspace{15pt} \text{as $t \to \infty$}.$$ implies the other as well as $$ \omega_F(1/t) \sim F(t) \Gamma(p+1) \hspace{15pt} \text{as $t \to \infty$}.$$
I have two questions. Firstly, I am so doubt that if the statement is true for $p=0$. It seems that the inverse Laplace transform of $1$ is $\delta(t)$. In this case is it true that $\dfrac{\omega_F(\tau \lambda)}{\omega_F(\tau)} \to 1$ implies $\dfrac{F(tx)}{F(t)} $ converging to the Heaviside step function rather $1$?
The second question is that do we have the Tauberian theorem for the Laplace transform in form of $F(s) = \displaystyle\int^\infty_0 e^{-st} f(t) dt$, namely the asymptotic relation of $F$ and $f$. Thanks!!!