Functional equation $f(ax)=bf(x)$

Question

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.

@Fundamental But this question is that $a$ is positive, while http://math.stackexchange.com/questions/1033898 may not. — doraemonpaul, Feb 01 '15 at 11:06

score 2 · Answer 1 · 2014-12-29T21:16:27.480

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 — user154729, Dec 29 '14 at 21:40

score 0 · Answer 2 · answered Jan 25 '15 at 11:16

0

$f(ax)=bf(x)$

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

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

$f(a^x)=\Theta(x)b^x$, 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

answered Jan 25 '15 at 11:16

doraemonpaul

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Functional equation $f(ax)=bf(x)$

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