# When is the reciprocal of a Laplace transform also a Laplace transform?

In the literature for a problem I am studying, there are classes of models of the form $$f*k=g$$ and some of the form $$f = g*k^{'}$$.

I am interested in whether for every $$k$$, or for some "nice" class of $$k$$s, there always exists a well-defined $$k'$$. Formally, of course, we may take the Laplace transform of both equations, and it follows that $$K^{'}(s)=\frac{1}{K(s)}$$. However, not every function in the s-domain is a valid Laplace transform.

Clearly it is true for $$K(s)=1$$, i.e., when one takes the transform of the Dirac delta function. However, if we consider the integral definition $$\mathscr{L}\{f\}\equiv\int_0^\infty f(t)e^{-st}dt$$ It appears that the magnitude of the transform ought to decrease as $$Re(s)\rightarrow\infty$$, so for instance $$K(s)=1/s$$ is a valid transform, but $$\frac{1}{K(s)}=s$$ is not.

On the other hand, considering the discrete analaog of the convolution problem ($$\mathbf{f}$$ and $$\mathbf{g}$$ as equal-length vectors, and $$\mathbf{K}$$ as a triangular Toeplitz matrix): you can show that the inverse of a triangular Toeplitz is itself triangular Toeplitz, so if $$\mathbf{Kf=g}$$, we should also be able to write $$\mathbf{f=K^{-1}g}$$, where $$\mathbf{K^{-1}}$$ also represents a convolution operation.

So it seems I believe two contradictory things. I am hoping someone with deeper expertise could resolve the difficulty!

The Laplace transform of an exponentially bounded function does go to $$0$$ as $$\text{Re}(s) \to \infty$$, so its reciprocal can't be the Laplace transform of such a function. On the other hand, it might be the Laplace transform of a distribution. Thus $$s$$ is the Laplace transform of the derivative of the Dirac delta.