This question is a follow up question on Decay of a function from its Laplace transform.
Let $f:[0, \infty) \to \mathbb{R}$ be a continuous function whose Laplace transform $$ F(s) = \int_0^\infty f(t)e^{-st} dt $$ is convergent for all $Re(s) >0$. Assume $F(s)$ can be extended analytically to an open set containing the closed half plane $\{ Re(s)\ge 0 \}$ What can we say about the growth of $f(t)$ as $t\to \infty$? What I wish to prove is that $f$ has polynomial growth $f(t) = O(t^k)$ for some $k\ge 0$, and I think we should just get the bounded case $f(t)=O(1)$, but I couldn't prove it.
Attempt:
I tried to use the inverse Laplace transform $$ f(t) = \frac{1}{2\pi i} \int_{x-i\infty}^{x+i\infty}F(s)e^{st}ds $$ for $x>0$, then push $x\to 0$. Using Mean Value Theorem and continuity of derivative, I showed that as a function of $y$, $F(x+iy)\to F(0+iy)$ in $L^1(-\lambda, \lambda)$ or $L^2(-\lambda, \lambda)$, where $\lambda>0$ is arbitrary. I think shifting the integration line to $x=0$ should tell something about $f(t)$ (for example, if we can prove $L^1$, Riemann-Lebesgue lemma would say $f(t)\to 0$). But I am not sure how to proceed.
Thank you in advance.