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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!

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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.

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  • $\begingroup$ Thanks; that makes sense. Do you know if there are some conditions on K(S) under which its reciprocal is guaranteed to exist in a distributional sense? $\endgroup$ – Scott Hansen Nov 19 '19 at 15:24

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