Does $g(x) \sim c g(bx)$ imply $g(x) \sim K x^{-\log_b(c)}$

Question

define $f(x)\sim g(x)$ for $x \rightarrow x_0$ if and only if $\frac{f(x)}{g(x)} \rightarrow 1$ for $x\rightarrow x_0$

for fixed $b>0$ and fixed $c>0$ the relation $g(x) \sim c \ g(bx)$ holds for $x\rightarrow 0$.

Does this relation imply $g(x)\sim k x^{-log_b(c)}$ and how can I prove that?

The paper 'Forecasting Multifractal Volatility' by Calvet and Fisher claims this on page 53 without any proof, where g is defined as the q-th moment of a random variable $\Omega(x)$: $g(x):=\mathbb{E}(\Omega(x)^q)$

$\Omega(x)$ is positive, satisfies $c^q\mathbb{E}(\Omega(cx)^q)\leq\mathbb{E}(\Omega(x)^q)\leq n^{\max(0,1-q)}\mathbb{E}(\Omega(x/n)^q)$

Due to the advice of homogenous functions I figured out this way: $g(x)\sim c \ g(bx) \sim c^k g(b^k \ x)$ for any $k\in \mathbb{N}$ setting $k(x):=\lceil -\log_b(x)\rceil$ gives $g(x) \sim c^{\lceil -\log_b(x)\rceil} g(b^{\lceil -\log_b(x)\rceil}) \sim x^{-\log_b(c)} g(1)$ However I'm pretty sure that setting k like that is wrong, since k suddenly deppends on x which leads to wrong limits?

Thank you for any advice

Answer 1

No, unless you require some extreme sort of regularity.

For example, $$ g(x) = x\cos\big(2\pi \log_2 x\big),\quad x>0 $$ satisfies the equation $$ g(x) = \frac{1}{2}\;g(2x) $$ (not just asymptotically, but exactly).

And the same is true for any other function of period $1$ in place of $\cos(2\pi \theta)$. For example, you could choose a positive periodic function.

The claim in your statement was that the solution is $Kx$, where $K$ is a constant. But in fact, $K$ need not be a strict constant, just a "wavering constant" (to quote Dale Alspach).