TITLE: $f(\alpha x) = f(x)^{\beta}$ under different constraints
QUESTION [6 upvotes]: With $\alpha > 0,\, \beta \in \Bbb R^*,\, \alpha, \beta \neq 1$ and $f : \Bbb R \to \Bbb R_+^*$, let's consider the functional equation 
$$ f(\alpha x) = f(x)^{\beta} \tag{$\Xi$}$$ 
or equivalently $g(\alpha x) = \beta g(x)$ for $g = \ln f$.
The case where $\alpha = \sqrt2$, $\beta = 2$ and $f \in \mathcal C^2$ has already been solved here : Solving $(f(x))^2 = f(\sqrt{2}x)$ (the answer is $\exists \lambda\mid f(x) = e^{\lambda x^2}$).
What if we relax/change some of the constraints, for instance:

Keeping $f$ regular (say $\mathcal C^{\infty}$) but setting $\alpha, \beta$ generic
$f \in \mathcal C^0$
$f \in L^1$
(other ideas?)

REPLY [3 votes]: Summary at the bottom.

$$g(\alpha x)=\beta g(x)$$
$$\gamma = \frac{\ln |\beta|}{\ln\alpha}$$

If $\beta >0$, $\beta=\alpha^\gamma$ and $g(\alpha x)=\alpha ^\gamma g(x)$.
We define $h(x)$, for $x\neq 0$, as $h(x)= g(x)x^{-\gamma} $.
Then:
$$h(\alpha x)=g(\alpha x){\alpha^{-\gamma} x^{-\gamma}}=g(x){x^{-\gamma}}=h(x) $$
Now let $k_1(x)=h(\alpha^x)$ and $k_2(x)=h(-\alpha^x)$.
Then $k_1$ and $k_2$ can be any periodic functions, with a period of $1$.
Therefore:
$$
g(x)=\cases{x^\gamma k_1(\log_\alpha(x)) & if $x>0$ \cr x^\gamma k_2(\log_\alpha(-x)) & if $x<0$ }
$$

If $\beta<0$, $\beta=-\alpha^\gamma$. We use the same definition for $h$, $k_1$ and $k_2$, but now we have: $k_1(x+1)=-k_1(x)$ and $k_2(x+1)=-k_2(x)$.
So $k_1$ and $k_2$ can be any antiperiodic functions, with a period of $1$.
And:
$$g(x)=\cases{x^\gamma k_1(\log_\alpha(x)) & if $x>0$ \cr x^\gamma k_2(\log_\alpha(-x)) & if $x<0$ }$$

If $g \in \mathcal C^n$, then $k_1$, $k_2 \in \mathcal C^n$.
If $n\geq\gamma$, then in a neighborhood of $0^+$:
$$g(x)=\sum\limits_{k=0}^{n}\frac{g^{(k)}(0)}{k!}x^k + o(x^n)= x^\gamma k_1(\log_\alpha(x))$$
$$k_1(\log_\alpha(x))=\sum\limits_{k=0}^{n}\frac{g^{(k)}(0)}{k!}x^{k-\gamma} + o(x^{n-\gamma})$$
So $k_1(\log_\alpha(x))$ has a limit at $0^+$ (which could be infinite), so $k_1$ has a limit at $-\infty$. So it must be a constant (since it is periodic of period at most $2$). We can use the same reasoning for $k_2$. So
$g(x)=\cases{c_1x^\gamma & \text{if } x>0 \cr c_2x^\gamma & \text{if } x<0 } $
Therefore, all the derivatives of $g$ at $0$ of order smaller than $\gamma$ are $0$, and if $n>\gamma$, then $g$ is not $p=\lceil \gamma\rceil$ times differentiable near $0$. So $\gamma = n$ and $c_1=c_2$.

To sum it up:
In general:
$$g(x)=\cases{x^\gamma k_1(\log_\alpha(x)) & if $x>0$ \cr x^\gamma k_2(\log_\alpha(-x)) & if $x<0$ } $$
Where $k_1$ and $k_2$ are two periodic (if $\beta>0$) or antiperiodic (if $\beta<0$) functions, of period $1$.
If $g\in\mathcal C^{n}$, so are $k_1$ and $k_2$. And if $n\geq\gamma$, $n = \gamma$, and $g(x)=\lambda x^n$.
If there are any mistakes, please let me know.