Morphism which preserves convolution?

[mnb96]

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:
F(f\ast g) = F(f)\cdot F(g)

Is it possible to find another operator \mathcal{G} (different than the FT) for which the following is valid?
\mathcal{G}(f\ast g) = \mathcal{G}(f)+\mathcal{G}(g)

 

[matticus]

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?

 

[mnb96]

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:

\bar{x_i} = log(x_i)

x_i = e^{\bar{x_i}}

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?

 

[mnb96]

...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?