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I was wondering if there might be a way to prove that $f(x) = c \mathrm{log} x$ with $c=const$ is the unique solution to $f(x y) = f(x) + f(y)$. Any ideas would be appreciated!

Best regards

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  • $\begingroup$ how about $f(x)$ defined as follows: $f(0)=1, \: f(x)=0, \: x \ne 0$ $\endgroup$ – Vasya Mar 27 '18 at 17:34
  • $\begingroup$ It wouldn't be continuous... which OP probably isn't looking for. $\endgroup$ – TheSimpliFire Mar 27 '18 at 17:50
  • $\begingroup$ Have you tried searching this on Google? There are plenty of proofs. $\endgroup$ – TheSimpliFire Mar 27 '18 at 17:53
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Let $g(x) = f(e^x)$. Then we have that $$ g(x + y) = f(e^{x + y}) = f(e^x e^y) =f(e^x) + f(e^y) = g(x) + g(y). $$

This is called the Cauchy functional equation. The only continuous solutions are given by $g(x) = cx$ for some constant $c$. This corresponds to the solution $f(x) = c \log x$ to the original equation. There are many pathological non-continuous solutions though.

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