1
$\begingroup$

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 $t\geq 0$ ?

$\endgroup$
1
$\begingroup$

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 $t \in (1,2)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.