Functional equation unique solution logarithm

Question

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

Answer 1

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.