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.

- First, generally, what is the condition for the existence of an inverse Laplace transform?
- 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 $1$?
- 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!!!