# Generalized Laplace transform

1. Dec 6, 2009

### Pere Callahan

"Generalized" Laplace transform

Hello,

I'm having trouble proving injectivity of what might be called a "generalized" Laplace transform (not the one by Varma).

Let f be a rational function and C be a fixed closed contour in the complex plane, (such that C contains not pole of f):
The operator L is then defined by
$$(Lf)(w) = \int_C{e^{zw}f(z)dz}.$$

One could define L also for a more general class of functions, but for the moment rational ones suffice. It is clear that when f is the zero function, Lf is the zero function as well. I'm wondering about the converse; for the ordinary Laplace transform we know that the zero function is in fact the only function yielding a vanishing Laplace transform. I think that should also be true in this case but I have been unable to prove it.
When we restrict the class of functions in the domain even further so as to only contain "step functions" with N steps i was able to prove that (Lf)(w)=0 for N real values of w implies that f is the zero function by using the linear independence of exponentials of different frequencies.
I would appreciate any help in doing the step () from step functions to rational functions.

Best,
Pere

2. Dec 7, 2009

### Pere Callahan

Re: "Generalized" Laplace transform

Thinking again, it seems to me that Cauchy's integral theorem implies that Lf is identical zero for any holomorphic function f, in particular for poynomials.
This does not necessarily contradict what i wrote about step functions, becuase step functions are certainly not holomorphic...

Can one then prove something weaker for example that (Lf)(w) = 0 for sufficiently many w implies that f must be rational? Probably f(z) = 1/z^2 would be a counter example to this ...