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This is what I have first asked in SE but I think it is more suitable for here. I am interested in the set of all continuous functions $f: (0, \infty) \longrightarrow \Bbb{R}$ with the following property:

For any irrational positive number $\alpha$ there exists a real number $b_\alpha$ such that for any $x > 0 $ we have $f(\alpha x) = b_\alpha f(x)$.

Any reference or suggestions would be helpful.

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    $\begingroup$ What kind of information would you like? Where did this come up? It will be hard to say anything unless you provide more information. $\endgroup$ Jan 22, 2014 at 23:00

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$f$ must be a power function, $f(x)= cx^e$. Proof:

Assume $f(x) \neq 0$ for some $x$, Then $b_\alpha= f(\alpha x)/f(x)$ for all $\alpha$, hence $b_\alpha$ is a continuous function $(0,\infty)-\mathbb Q^{>0} \to \mathbb R$.

$b_{\alpha \beta} = f(\alpha \beta x)/f(x) = b_\alpha f(\beta x)/f(x)= b_\alpha b_\beta$. Hence $b_\alpha$ is a continuous group homomorphism.

We can adapt the usual proof that any continuous group homomorphism $\mathbb R \to \mathbb R$ is linear to a proof that $b_\alpha$ is a power function $x \to x^c$. Hence $f(x) = c x^e$ for some constant $c$.

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  • $\begingroup$ what is the domain of $b_\alpha$? $\endgroup$
    – user40021
    Jan 23, 2014 at 14:21
  • $\begingroup$ Irrational positive numbers. But we can easily extend it to all positive numbers. $\endgroup$
    – Will Sawin
    Jan 23, 2014 at 15:50

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