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If I have a signal $f(x,y)$ (discrete) and I convolve this signal with a kernal $h(x,y)$:

$y(x,y) = f(x,y) \star h(x,y)$ (where $\star$ is the convolution operator)

Can I obtain $f(x,y)$ given only $y(x,y)$ and $h(x,y)$ ?

Note: even tho this may be a signal processing question I would like to know the answer (invertibility) from the Math point of view.

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  • $\begingroup$ Just a note about the practical implications: the process known as deconvolution exists and tends to be sensitive to noise unless special care is taken and something is known about the signal and/or noise. $\endgroup$ – Bjorn Roche Apr 26 '13 at 16:42
  • $\begingroup$ @Bjorn: consider noise = 0 $\endgroup$ – dynamic Apr 26 '13 at 16:57
  • $\begingroup$ I don't think it matters whether we look at the 1-d or n-d case, so the question could be simplified a bit. $\endgroup$ – masterxilo Feb 15 '17 at 12:28
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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.

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    $\begingroup$ Actually, the Fourier argument is not correct. Just because the Fourier transform of $g$ may have one or more zeros does not mean the convolution cannot be inverted. After all, the Fourier transform may have more frequencies than $g$ has degrees of freedom. In the continuous-time case in particular, the convolution may be invertible as long as the zeros are isolated. The inverse Fourier transform of $\textrm{Four}\{f\star g\}/\textrm{Four}\{g\}$ might still exist! $\endgroup$ – Michael Grant Apr 26 '13 at 14:18
  • $\begingroup$ That's wonderful. You mean the poles are isolated though, right? Because I accounted for the usual removable singularities. Could you explain in an answer how the inverse fourier transform of the quotient might still exist and be equal to the original function? $\endgroup$ – Loki Clock Apr 26 '13 at 14:23
  • $\begingroup$ You know, I might have bitten off more than I can chew by making the claim about the inverse Fourier transform of the quotient. But even if we walk that back I don't think the Fourier argument alone is enough to argue non-invertibility. $\endgroup$ – Michael Grant Apr 26 '13 at 15:23
  • $\begingroup$ Well, what I mean by the original f in frequency space is the function's Fourier transform as it was. So there's going to be some change to it if there are non-removable singularities. This is not to argue non-invertibility in general, just that this division technique wouldn't restore the original frequency-space representation. I will change it so as not to confuse the function's possible frequency space representations with the function. $\endgroup$ – Loki Clock Apr 26 '13 at 15:40
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    $\begingroup$ @MichaelC.Grant: yes, I have searched before to post this answer too $\endgroup$ – dynamic Apr 26 '13 at 16:57
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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!

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  • $\begingroup$ This answer provides conditions that describe when we can recover $f$ for compactly supported $f$. If you truncate the $f$ in your example, I think you should be able to recover it. $\endgroup$ – alexpghayes Aug 6 '19 at 22:24
  • $\begingroup$ A tangential question: is there a name for a class of functions (in this answer $f$ and $g$) which all result in the same value when convolved with a given function $h$? $\endgroup$ – syockit Dec 8 '19 at 9:30

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