Is the convolution an invertible operation? I have a signal $f(x,y)$, which is discrete. I convolve this signal with a kernel $h(x,y)$:
$y(x,y) = f(x,y) \star h(x,y)$ (where $\star$ is the convolution operator)
Can I obtain $f(x,y)$ given only $y(x,y)$ and $h(x,y)$ ?
Note: Even though this may be a signal processing question I would like to know the answer (invertibility) from a Mathematical point of view.
 A: Consider the convolution theorem, $\operatorname{Four}\{f\star g\}=\operatorname{Four}\{f\}\operatorname{Four}\{g\}.$ It's clear the original frequency space representation of f can only be recovered by division if the zeros of $\operatorname{Four}\{g\}$ are a subset of the zeros of $\operatorname{Four}\{f \star g\}$.
But in general, convolution of functions is almost a ring (there's no exact identity element). The linear space of compactly supported distributions forms an actual ring under convolution, and so it has a group of units. These are distributions whose convolutions are always reversible. One could consider invertibility in a neighborhood, by localizing this ring, in the same way you consider x to be invertible to 1/x in a neighborhood not containing x=0.
A: If convolution still makes sense with improper integrals then I believe the following shows that the answer is "No".
Consider the 1d case, with
$ f(t) = \frac{1}{1+t^2}\quad\text{and}\quad h(t) = 1,$
so then
$y(t) = f \star h = \int_{-\infty}^{\infty}\ f(\tau)h(t-\tau)\ d\tau = \int_{-\infty}^{\infty}\ f(\tau)\ d\tau = \pi $
Assume if $f$ were recoverable from $y$ and $h$, then for any $g$ such that $g \star h = y$, we must have $g=f$.
Consider,
$g(t) = \sqrt{\pi}e^{-t^2} \neq f(t) \longrightarrow g \star h = \pi = y(t)$ 
Contradiction!
