Given $a$ and $c$ positive numbers, is there a function $h$ such that $h(-ax)= -ch(x)$ for all $x$?

Question

Given $a>1$ and $c$ a positive number, is there a (no trivial) function $h$ such that $$h(-ax)= -ch(x)$$ for all $x$?.

I know that $h(x)=c^{\log_a|x|}$ satisfies $h(ax)=ch(x)$, however I am not sure how to deal with the $-a$ , $-c$ version.

Thanks in advance

Answer 1

Yes, such functions exist.

Presumably by non-trivial, you mean not the zero function.

It's clear that we must have $h(0)=0$, so we start with that declaration.

Next, define an equivalence relation on $\mathbb{R}{\setminus}\{0\}$ by $x\equiv y$ if $x=(-a)^ky$, for some integer $k$.

For each equivalence class, choose one representative $r$ and define $h(r)=1$.

That declaration uniquely determines $h(x)$ for any $x\in \mathbb{R}{\setminus}\{0\}$ with $x\equiv r$.

Thus, $h$ is defined for all $x\in\mathbb{R}$.

Answer 2

For such $h$, we have $$ h(a^2x)=c^2h(x).$$ In particular, $h(0)=-ch(0)$, so $h(0)=0$ or $c=-1$. Let $g\colon[0,1)\to \Bbb R$ be arbitrary and define $h$ as follows:

• if $x>0$, let $h(x)=(c^2)^{\lfloor \log_{a^2} x\rfloor }\cdot g(\log_{a^2} x \bmod 1)$
• if $x<0$, let $h(x)=-ch(-ax)$
• let $h(0)=0$ (or, if $c=-1$, take an arbitrary value)

These choices are both possible for arbitrary $g$, and on the other hand necessary by the conditions on $h$.

What if we want $h$ to be continuous? Then to begin with, $g$ must be continuous and $\lim_{g\to 1} g(x)=c^2g(0)$. Then

• If $|c|>1$, this makes $\lim_{x\to 0} h(x)=0$, so $h$ is continuous.

• If $c=-1$, the only globally continuous solutions have $g$ constant and so $h$ constant as well.

• If $c=+1$ or $|c|<1$, the only continuous solution is the zero function.