# 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.

• 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. Apr 26, 2013 at 16:42
• @Bjorn: consider noise = 0 Apr 26, 2013 at 16:57
• 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. Feb 15, 2017 at 12:28

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.

• 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! Apr 26, 2013 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? Apr 26, 2013 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. Apr 26, 2013 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. Apr 26, 2013 at 15:40
• @MichaelC.Grant: yes, I have searched before to post this answer too Apr 26, 2013 at 16:57

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,

$$g(t) = \sqrt{\pi}e^{-t^2} \neq f(t) \longrightarrow g \star h = \pi = y(t)$$

• 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. Aug 6, 2019 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 $h$? Dec 8, 2019 at 9:30