# Is Injectivity of the Generalized Laplace Transform Proveable?

• Pere Callahan
In summary, the conversation discusses a "generalized" Laplace transform and the difficulty in proving its injectivity. The operator L is defined for rational functions and it is clear that the zero function results in a zero Laplace transform. The speaker wonders if the opposite is also true, and if the zero function is the only one that yields a vanishing Laplace transform. They were able to prove this for a more restricted class of functions, but are unsure how to extend it to rational functions. They also consider the possibility of a weaker result, such as (Lf)(w) = 0 for many w implying that f must be rational, but question if this is true for all cases.
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.Pere

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 ...

## 1. What is a Generalized Laplace transform?

The Generalized Laplace transform is a mathematical tool used in engineering and physics to analyze systems that are described by differential equations. It is an extension of the traditional Laplace transform, which is used to solve linear differential equations with constant coefficients. The Generalized Laplace transform allows for the analysis of more complex systems with variable coefficients.

## 2. How does the Generalized Laplace transform work?

The Generalized Laplace transform works by converting a function of time into a function of a complex variable, s. This transformation allows for the analysis of the function in the complex plane, which can provide valuable insights into the behavior of the system. The transformed function can then be manipulated using algebraic techniques to solve for the original function in the time domain.

## 3. What are the applications of the Generalized Laplace transform?

The Generalized Laplace transform has a wide range of applications in physics and engineering. It is commonly used in control systems, signal processing, circuit analysis, and fluid dynamics. It is also used in the study of differential equations and can be applied to a variety of complex systems, including mechanical, electrical, and thermal systems.

## 4. Are there any limitations of the Generalized Laplace transform?

Like any mathematical tool, the Generalized Laplace transform has its limitations. It is most effective for linear, time-invariant systems and may not provide accurate solutions for nonlinear or time-varying systems. Additionally, the Generalized Laplace transform cannot be used for functions that do not have a finite number of discontinuities.

## 5. How is the Generalized Laplace transform different from the Fourier transform?

The Generalized Laplace transform and the Fourier transform are both mathematical tools used for analyzing systems described by differential equations. However, the Fourier transform only works for time-invariant systems, while the Generalized Laplace transform can handle systems with variable coefficients. Additionally, the Fourier transform produces a frequency-domain representation of a function, while the Generalized Laplace transform produces a complex-domain representation.

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