There are for example several canonical spaces to define the Fourier transform (i.e. Schwartz's space). Is there also a particularly suitable space to define the Laplace transform, so that the Laplace transform is at least bijective?


Solution 1:

The best answer I know of is here, given by Robert Israel: Does the Laplace transform biject?

In the first place, taking our cue from the unitarity of the Fourier transform, we should always try to check Hilbert spaces first, $L^2$ being our first choice. Then we can just look at what structure the transformation carries over. We take real functions to complex functions, and the $L^2$ condition gives analyticity because of the convergence of the integral. The link above tells you that there is a nice condition for bijectivity between two natural Hilbert spaces. There is a more general bijection, which is proven in very much the same way (neither of the two bijections is easy to prove) and can be found here: https://mathoverflow.net/questions/44713/when-i-can-safely-assume-that-a-function-is-a-laplace-transform-of-other-functio

If you are trying to find a single $S$ as source and target (say, by restricting the Laplace transform to real arguments), there may be some fixed point theorem in functional analysis that can prove its existence (?), but I doubt you will find a description for $S$. In any case, you will be shooting yourself in the foot, because when you restrict the Laplace transform to real arguments, you are forsaking all the analytic structure of the transform, you are making the inversion theorem an unfathomable curiosity, and of course you are breaking the links with the Fourier transform and the Mellin transform, which require integration over vertical lines in the complex plane. In short: don't do it!