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 further that $F(s)$ can be extended analytically to the closed half plane $\{ Re(s)\ge 0 \}$ (i.e., no singularities on the imaginary axis). How can we prove that $f(t) \to 0$ as $t\to \infty$?
Attempt: If $F(s)$ is a rational function, this follows from the Final Value Theorem. But I couldn't find a proof of the Theorem.
I am thinking of using the inverse Laplace transform $$ f(t) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty}F(s)e^{st}ds $$ for $c>0$, then push $c\to 0$, and then apply Riemann-Lebesgue lemma. But I am stuck on proving the function $y\to F(iy)$ belongs to $L^1$ or $L^2$.
Thank you in advance.