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does not satisfy the letter of your criteria
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David Treumann
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In a finite abelian group, I think we have the relation

f*g = sum over pairs (x,y) in the group of ([y]*([x].f)).([x]*g)

Here [x] denotes the indicator function that takes the value 1 at x and zero elsewhere. On the trivial group this says f*g = f.g, so I don't think it reduces to 0 = 0. The relation can be deduced from the fact that ([u]*h)(v) = h(v-u) for all h, u, and v, and that summing the terms over just y gives f(x).([x]*g) = f(x).g(blank - x), and the definition of convolution. Maybe this is cheating but it follows the letter of your criteria.

In a finite abelian group, I think we have the relation

f*g = sum over pairs (x,y) in the group of ([y]*([x].f)).([x]*g)

Here [x] denotes the indicator function that takes the value 1 at x and zero elsewhere. On the trivial group this says f*g = f.g, so I don't think it reduces to 0 = 0. The relation can be deduced from the fact that ([u]*h)(v) = h(v-u) for all h, u, and v, and that summing the terms over just y gives f(x).([x]*g) = f(x).g(blank - x), and the definition of convolution. Maybe this is cheating but it follows the letter of your criteria.

In a finite abelian group, I think we have the relation

f*g = sum over pairs (x,y) in the group of ([y]*([x].f)).([x]*g)

Here [x] denotes the indicator function that takes the value 1 at x and zero elsewhere. On the trivial group this says f*g = f.g, so I don't think it reduces to 0 = 0. The relation can be deduced from the fact that ([u]*h)(v) = h(v-u) for all h, u, and v, and that summing the terms over just y gives f(x).([x]*g) = f(x).g(blank - x), and the definition of convolution.

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David Treumann
  • 4.9k
  • 26
  • 36

In a finite abelian group, I think we have the relation

f*g = sum over pairs (x,y) in the group of ([y]*([x].f)).([x]*g)

Here [x] denotes the indicator function that takes the value 1 at x and zero elsewhere. On the trivial group this says f*g = f.g, so I don't think it reduces to 0 = 0. The relation can be deduced from the fact that ([u]*h)(v) = h(v-u) for all h, u, and v, and that summing the terms over just y gives f(x).([x]*g) = f(x).g(blank - x), and the definition of convolution. Maybe this is cheating but it follows the letter of your criteria.