Is the Laplace transform a linear operator?

Question

Many sources online states that the Laplace transform $\mathcal{L}: V \to W, f(t) \mapsto F(s)$ is a linear operator

For example:

1. http://www2.fiu.edu/~aladrog/PropLaplaceTransform.pdf (Wayback Machine)

2. http://www.saylor.org/site/wp-content/uploads/2013/04/ME401-1.2.2-LaplaceTransform.pdf

However, closely examining the definition of linear operator on Wikipedia, it says:

In mathematics, a linear map (also called a linear mapping, linear transformation or, in some contexts, linear function) is a mapping V → W between two modules (including vector spaces) that preserves (in the sense defined below) the operations of addition and scalar multiplication.

An important special case is when V = W, in which case the map is called a linear operator,[1] or an endomorphism of V.

Do we know that the Laplace transform is a endomorphism? It doesn't seem plausible given that we are taking a function in time domain, say $L_2[0, \infty)$ space, and mapping it into $H_2$ space

So should the Laplace transform be a linear operator?

Answer 1

As it is usually defined, the domain and range of the Laplace transformation are different spaces. The domain is usually a space of measurable functions defined on $[0,+\infty)$, and the range a space of holomorphic functions. One can however choose a space $V$ as the domain such that $\mathcal{L}(V) \subset V$ [for example the space of bounded holomorphic functions defined on the right half-plane], and then one can view $\mathcal{L}$ as an endomorphism of $V$.

But the main source of calling the Laplace transformation a linear operator is that not all people reserve the term "operator" for endomorphisms, many people call any [or possibly only continuous or closed] linear maps "operator". With that convention, the Laplace transformation is a linear operator in the more common settings.