Does the Laplace transform biject?

Question

Someone wrote on the Wikipedia article for the Laplace trasform that 'this transformation is essentially bijective for the majority of practical uses.'

Can someone provide a proof or counterexample that shows that the Laplace transform is not bijective over the domain of functions from $\mathbb{R}^+$ to $\mathbb{R}$?

Answer 1

Bijective from what space of functions to what space of functions?

For example, by the Paley-Wiener theorem the Laplace transform is a bijection from $L^2(0,\infty)$ to the functions $F$ analytic in the open right half plane whose restrictions to vertical lines in the right half plane have uniformly bounded $L^2$ norm.

Answer 2

It depends on you set of definition. If you ask wheter it is bijective from the functions defined on $\mathbb{R}$, compare the laplace transform of $$ f(x)=\begin{cases} \sin(x)&x\geq0\\ 0&x<0 \end{cases} $$

and $$ g(x)=\sin(x), \forall x. $$