# Laplace transform, Inverse Laplace transform

Let $(\mathcal{L}f)(s)$ be the Laplace transform of a piecewise continuous function $f(t)$ defined for $t\geq 0$. If $(\mathcal{L}f)(s)\geq 0$ for all $s\in\mathbb{R^+}$ does this imply that $f(t)\geq 0$ for all $t\geq 0$ ?

No. $e^{-st}$ is always a decreasing function of $t$ when $s > 0$, so counterexamples are easy to construct.
Counterexample: let $u(t)$ denote the unit step function, i.e. $$u(t) = \begin{cases} 0 & t < 0\\ 1 & t \geq 0 \end{cases}$$ Let $f(t) = u(t) - 2u(t-1) + u(t - 2)$. Confirm that $(\mathcal L f)(s) \geq 0$ for all $s \in \Bbb R^+$, but $f(t) < 0$ for $t \in (1,2)$.