Functional equation $f(ax)=bf(x)$ [duplicate]

What are all the solutions to the functional equation $f(ax)=bf(x)$, where $a,b>0$, and $f$ is continuous, strictly monotone and increasing, and $x$ ranges over the reals? references? proof?

Thanks

Additional details following the first response:

It is easy to see that the function $f(x)=c\cdot x^\alpha$, with $\alpha =\log b/\log a$ and any constant $c$ is a solution for the functional equation. Also, the $c$ can be different for the positives and the negatives. So, if $x^\alpha$ is not monotone itself, we can create a monotone solution by gluing together a positive $c$ for the positive side with a negative $c$ for the negative side. Is this correct?

My question is if these are all the solutions, and if this is appears in the literature.

Thanks.

marked as duplicate by Jonas Meyer, Dan Rust, graydad, Johanna, user149792Mar 2 '15 at 2:03

• ... where $2$ can be replaced by any $b$. – user147263 Dec 29 '14 at 21:28
• @Fundamental But this question is that $a$ is positive, while math.stackexchange.com/questions/1033898 may not. – doraemonpaul Feb 1 '15 at 11:06

It is highly appreciated if you tell us what you tried, what you think about the solution, etc.. So since you didn't do that I'll give you a solution for only a piece of the question, and I hope you can complete by your own. Note that my solution will not take advantage of $f$ being strictly monotone (or monotone at all), so the full solution should use that.

Assume $a>1$ and $b=1$. I'll prove $f=const$.

Assume there are $x_1,\:x_2\in\mathbb R$ with $f(x_1)\not=f(x_2)$. Denote $a_n=x_1/a^n$ and $b_n=x_2/a^n$. It's easy to see $\lim a_n=\lim b_n=0$, so by continuity at $x_0=0$: $\lim f(a_n)=\lim f(b_n)$.

Since for all $n\in\mathbb N$ we have $f(a_n)=f(x_1),\:f(b_n)=f(x_2)$ we get $f(x_1)\not=f(x_2)$, by contradiction to the assumption that $f$ is not constant.

It's easy to see that every constant function ineed makes $f(ax)=f(x)$.

Please complete it by your own for a general case. Use this site to prove the full thing if you need, but try to be more specific with your requests.

• Thanks. I added more explanations. – user154729 Dec 29 '14 at 21:40

$f(ax)=bf(x)$

$f(aa^x)=bf(a^x)$

$f(a^{x+1})=bf(a^x)$

$f(a^x)=\Theta(x)b^x$, where $\Theta(x)$ is an arbitrary periodic function with unit period

$f(x)=\Theta(\log_ax)x^{\log_ab}$, where $\Theta(x)$ is an arbitrary periodic function with unit period