TITLE: Why isn't the parallel between the Fourier transform and the Laplace transform complete?
QUESTION [6 upvotes]: I mean the question in the following sense.  For Fourier, we can do it on compact intervals and then we get a sequence of coefficients.  We can do it continuum-style, and then we get a superposition of waves of continuum-varying frequencies.  We can even do it abstractly on compact groups.
So why is the only Laplace transform (at least that I've ever heard about) on $[0, \infty)$?

REPLY [5 votes]: As AD. said above, for LCA semigroups one can construct the semigroup algebra just as in the group case and apply Gelfand theory. 
It might be worth noting that the characters of the resulting Banach algebra change when we lose the group structure. For a LCA group $G$, the character space (the Pontryagin dual) consist of group homomorphisms $\phi: G \rightarrow S^1$. In the case $G = \mathbb{R}$, a typical character is $\phi_t(x) = e^{itx}$. The Gelfand-Fourier transform $f \mapsto \hat{f}$ is then given by
$$
\hat{f} (\phi_t) = \int f(x) \phi_t(x)dx.
$$
This is why $e^{itx}$ appears in the Fourier transform. When we just have a semigroup $A$, the characters are now semigroup maps $\psi: A \rightarrow \mathbb{R}_+^{\times}$. For $A = \mathbb{R}_+$, you get maps $\{\psi_{-s}(x) = e^{-sx}\}$, which appear in the Laplace transform.
The two-sided Laplace transform can be generalized to LCA groups a la Gelfand also, by relaxing the definition of a character. The circle is replaced by $\mathbb{C}^{\times}$. These are called "generalized characters".