Hardy–Littlewood Tauberian theorem for Laplace transform

Question

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 three questions.

1. First, generally, what is the condition for the existence of an inverse Laplace transform?
2. Second, I am so doubious that this Tauberian theorem is true for $p=0$. The inverse Laplace transform of $1$ is $\delta(t)$: then, in this case is it true that $$ \dfrac{\omega_F(\tau \lambda)}{\omega_F(\tau)} \to 1 \implies \dfrac{F(tx)}{F(t)} \to H(t) $$ where $H(t)$ is the Heaviside step function, rather than converging to $1$?
3. Finally, the third question is: do the Tauberian theorem for the Laplace transform holds in the form $$ F(s) = \displaystyle\int^\infty_0 e^{-st} f(t) dt, $$ namely does it implies the asymptotic relation between $F$ and $f$?

Thanks!!!

Answer 1

Just to get a few points :) , let me repeat my comment : part of the answer is contained in Widder's book Laplace transform.