Morphism which preserves convolution?

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Discussion Overview

The discussion revolves around the existence of a morphism that preserves convolution with respect to pointwise addition, similar to how the Discrete Fourier Transform preserves convolution with respect to pointwise multiplication. Participants explore potential operators that could fulfill this requirement.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant inquires about the possibility of finding an operator \mathcal{G} such that \mathcal{G}(f\ast g) = \mathcal{G}(f)+\mathcal{G}(g).
  • Another participant suggests defining \mathcal{G}(f) = ln(F(f)), proposing that this could lead to the desired property.
  • A later reply acknowledges the suggestion and discusses the implications of applying a logarithmic transformation to the Discrete Fourier Transform, suggesting it could be an isomorphism under certain conditions.
  • Concerns are raised regarding the validity of the logarithmic transformation when considering the Fourier transform's output, particularly with respect to negative and zero values, which complicates the isomorphism.

Areas of Agreement / Disagreement

Participants express differing views on the applicability of the logarithmic transformation as an isomorphism, particularly concerning its limitations with negative and zero values in the context of the Fourier transform. The discussion remains unresolved regarding the existence of a suitable morphism.

Contextual Notes

Limitations include the dependence on the domain of the logarithmic transformation and the conditions under which the Fourier transform outputs non-zero components. These factors contribute to the complexity of establishing a valid isomorphism.

mnb96
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Hello,
I was wondering if there exists a (iso)morphism which preserves the operation of convolution, in respect to the pointwise-addition operation.
For example: it is well known that the Discrete Fourier Transform is a morphism which preserves convolution in respect to pointwise-multiplication:
[tex]F(f\ast g) = F(f)\cdot F(g)[/tex]

Is it possible to find another operator [tex]\mathcal{G}[/tex] (different than the FT) for which the following is valid?
[tex]\mathcal{G}(f\ast g) = \mathcal{G}(f)+\mathcal{G}(g)[/tex]
 
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Just looking I would say if you defined G(f) = ln(F(f)), then you would have G(f*g) = ln(F(f)F(g)) = ln(F(f)) + ln(F(g)) = G(f) + G(g).

Right?
 
Matticus thanks for your reply!
I am quite amazed for the fact that I didn't figure out that before. In fact, I think your suggestion makes perfectly sense.
Moreover, if we restrict ourselves to the discrete domain, taking the logarithm of the DFT would be equivalent to applying a coordinate transformation of the kind:

[tex]\bar{x_i} = log(x_i)[/tex]

[tex]x_i = e^{\bar{x_i}}[/tex]

and if I am not wrong, the Jacobian of such a transformation is never 0. This would imply that it is a one-to-one mapping, and so an isomorphism. I guess this observation could be applied as well to continuous functions, but I don't know how.
Does it make sense?
 
...uhm, actually there are problems is one is looking for an isomorphism. The isomorphism I mentioned with the log transformation is perfectly valid only in the domain of real positive numbers. Now, the Fourier transform outputs complex numbers, including real negative numbers and zero, so two problems arise:

1) the logarithm of a negative number can have multiple values
2) the logarithm of zero is not defined

The first problem can be solved by considering only the principal first branch of the negative logarithm, and that's fine.

But how can one ensure that the Fourier transform has always non-zero components?
 

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