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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.

dynamic
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2 Answers2

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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.

Loki Clock
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    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! – Michael Grant Apr 26 '13 at 14:18
  • 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? – Loki Clock Apr 26 '13 at 14:23
  • 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. – Michael Grant Apr 26 '13 at 15:23
  • 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. – Loki Clock Apr 26 '13 at 15:40
  • Agreed. But that's not the ultimate objective. The objective is to find a function $f$ that satisfies the original, product equation. That is not necessarily the same as asking if you can perform the quotient. For instance, consider the linear system $Ax=b$, where $A$ is singular. It is true that you cannot evaluate $A^{-1}b$ in that case. But you cannot immediately conclude that no solution for $x$ exists, because it is still possible for $b$ to be in the range space of $A$. – Michael Grant Apr 26 '13 at 15:42
  • But achieving the ultimate objective is not the same as achieving an inverse operation to the convolution, or in my example to the multiplication of fourier transforms. One still can't reverse the multiplication by Four{g} or by A. After all he wants f back, not some f' that will give the same convolution with g. – Loki Clock Apr 26 '13 at 16:05
  • michael if you want post a answer – dynamic Apr 26 '13 at 16:43
  • @LokiClock, actually, if $b$ is in the range space of $A$, then using the pseudoinverse $A^\dagger b$ will recover it. – Michael Grant Apr 26 '13 at 16:56
  • @yes123, I wish I were confident enough to post an answer. I've been looking at Google for resources on deconvolution and haven't found a clear resource on invertibility conditions. – Michael Grant Apr 26 '13 at 16:57
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    @MichaelC.Grant: yes, I have searched before to post this answer too – 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!

Alec Jacobson
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    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. – alexpghayes Aug 06 '19 at 22:24
  • 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 $h$? – syockit Dec 08 '19 at 09:30