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It's a well known result that, with some additional assumptions (often continuity or monotonicity), all functions $f : \mathbb{R}_{> 0} \to \mathbb{R}$ satisfying $f(ab) = f(a) + f(b)$ are constant multiples of $\log$.

It occurs to me that I've never seen a function which shows the necessity of the additional assumptions. Is there a function which satisfies $f(ab) = f(a) + f(b)$ but is not a constant multiple of $\log$?

Charles Hudgins
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    Check out https://math.stackexchange.com/questions/423492 or https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation . – kimchi lover Jan 27 '20 at 03:22

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There are (many) exotic solutions $\alpha\colon\mathbb{R}\to\mathbb{R}$ of the functional equation $\alpha(x+y)=\alpha(x)+\alpha(y)$. Indeed, $\alpha$ is a solutions iff it is linear with respect to the structure of a rational vector space. The exotic ones are exactly those sich are not multiples of the identity.

So $f:=\alpha\circ\log$ is an exotic logarithmic function if $\alpha$ is as above.

Edit: It can be seen quite easily that $f$ satisfies the equation (1) $f(ab)=f(a)+f(b)$ if $\alpha:=f\circ\exp$ satisfies (2) $\alpha(x+y)=\alpha(x)+\alpha(y)$. Thus $f$ solves (1) iff $f=\alpha\circ\log$ with some $\alpha$ satisfying (2).

Jens Schwaiger
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  • This combined with kimchi lover's comment answers my question. Thank you. – Charles Hudgins Jan 27 '20 at 03:44
  • Based on the links, it seems like there are no explicit constructions of such an $\alpha$. Is that right? Is it known that such $\alpha$ can only be exhibited non-constructively? – Charles Hudgins Jan 27 '20 at 03:48
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    The construction of $\alpha$ depends in the existence of a base of $\mathbb{R}$ considered as a vector space over the field of rational numbers. – Jens Schwaiger Jan 27 '20 at 03:57
  • Solovay showed that it is consistent with Zermelo-Frankel set theory (without the Axiom of Choice) that all subsets of $\mathbb R$ are Lebesgue measurable. These "exotic logarithms" are non-measurable functions. Thus without using some form of the Axiom of Choice, either you can't construct $f$ or you can't prove it satisfies the requirements. – Robert Israel Jan 27 '20 at 05:35