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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!!!

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  • $\begingroup$ I can not find this book, could you provide the link of this? Thanks! $\endgroup$
    – mnmn1993
    Commented Jul 25, 2022 at 2:00
  • $\begingroup$ Actually, I have read several books about Laplace Transform, but I can not find the answers. $\endgroup$
    – mnmn1993
    Commented Jul 25, 2022 at 2:01
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    $\begingroup$ Sorry, my memory failed. The author is actually Widder. $\endgroup$ Commented Jul 25, 2022 at 6:42
  • $\begingroup$ Thank you so much for the help! This book settled question 2. But I can not find answers for questions 1 and 3. For question 3, I have read through chapter 5 about the tauberian theorem but I do not find out any useful stuff. For question 1, should the condition be the one tgat the mellin inverse integral is convergent? $\endgroup$
    – mnmn1993
    Commented Jul 26, 2022 at 7:54

1 Answer 1

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

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  • $\begingroup$ I finally get the results from the book, thanks! $\endgroup$
    – mnmn1993
    Commented Jul 31, 2022 at 14:06
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    $\begingroup$ Can you perhaps elaborate? (As it stands, this answer is not much more than a link-only answer.) $\endgroup$
    – Stefan Kohl
    Commented Jul 31, 2022 at 16:59

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