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Consider the family of functions $g$ that satisfy the following conditions over a certain field or algebra:

$$ g(uv, w) =g(u,w) g(v,w) $$

over any elements $u, v, w$ of the given field or algebra

How does one determine or characterise this family of functions?

In the particular case that the family of function is analytic and the field is Abelian,

$$g(u,v)= \sum_{m,n} a_{m,n}u^m v^n $$ I get the following nilpotent condition on coefficients:

$$ \sum_{p+q=m} a_{p,n} a_{q,n} = a_{m,n} $$

for all m and n

I'm not sure how to interpret this condition other than it seems to be its own square

lurscher
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  • Maybe one should first determine the class of functions $f$ for which $f(uv) = f(u)f(v)$. See here: https://math.stackexchange.com/questions/43964/if-fxy-fxfy-then-show-that-fx-xt-for-some-t – amsmath Aug 30 '19 at 22:03
  • So, if the field is $\mathbb R$ and the functions $g(\cdot,w)$ are supposed to be continuous for every $w$, then the general solution is $g(u,w) = u_w^{f(w)}$, where $f$ is an arbitrary function of $w$ and $u_w$ is either $|u|$ or $u$, depending on $w$. – amsmath Aug 30 '19 at 22:34

1 Answers1

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The argument $w$ is merely an index. It's more instructive to write this as $$g_w(uv) =g_w(u) g_w(v) $$ Then we see that $g$ is just an arbitrary indexed family of homomorphisms of the multiplicative semigroup.

Matt Samuel
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  • But the question was how to characterize these. – amsmath Aug 30 '19 at 22:37
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    @ams That's hopeless in general. We need to pick a specific algebra to have any hope of characterizing them. – Matt Samuel Aug 30 '19 at 22:43
  • Sure, Matt. Even for $\mathbb R$ this is hopeless. With some more constraints though it is possible in special cases (see my comment(s)). – amsmath Aug 30 '19 at 22:45
  • it is interesting that the $w$ does not play a major part, the $g$ function is an attempt to find all binary functions that satisfy distributive property with multiplication, and your argument is pointing to it being equivalent to homomorphic property – lurscher Aug 30 '19 at 23:20