# Does the Laplace transform biject?

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 $\mathbb{R}$?

#### Solutions Collecting From Web of "Does the Laplace transform biject?"

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.

For “the majority of practical uses” it is important that the Laplace transform ${\cal L}$ is injective. This means that when you have determined a function $s\mapsto F(s)$ that suits your needs, there is at most one process $t\mapsto f(t)$ such that $F$ is its Laplace transform. You can then look up this unique $f$ in a catalogue of Laplace transforms.

This injectivity of ${\cal L}$ is the content of Lerch’s theorem and is in fact an essential pillar of the “Laplace doctrine”. The theorem is proven first for special cases where we have an inversion formula, and then extended to the general case.

The difference between “injectivity” and “bijectivity” here is that we don’t have a simple description of the space of all Laplace transforms $F$. But we don’t need to know all animals when we want to analyze a zebra. Lerch’s theorem tells us that it has a unique pair of parents.

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