Solution of the equation $f(ax) = bf(x)$

Question

Given the equation $f(ax) = bf(x)$, with $a, b > 0$, demonstrate that the solution is: $$f(x) = g(\log x)x^{\frac{\log b}{\log a}}$$

where $g(x) = g(x + \log a)$ is an arbitrary periodic function with period $\log(a)$.

By the method of induction I arrived in the particular solution:

$f(x) = Cx^{\frac{\log b}{\log a}}$, where $C$ is an arbitrary constant (case $g(x) =$ constant).

But I could not demonstrate how to get in the generic solution with the associated periodic function $g(x)$. Can someone help me?

Answer 1

Assuming $x > 0$

$$ f(a^{\log_a(ax)})-bf(a^{\log_a x}) = 0 $$

or

$$ F(\log_a x + 1)-bF(\log_a x) = 0 $$

or calling $u = \log_a x$

$$ F(u+1)-b F(u) = 0 $$

This is a difference functional equation with solution

$$ F(u) = \Phi(u)b^u $$

where $\Phi(u)$ is any periodic function with period $1$

hence

$$ f(x) = \Phi(\log_a x)b^{\log_a x}=\Phi\left(\frac{\ln x}{\ln a}\right)b^{\frac{\log_a x}{\log_b x}\log_b x} = g(\ln x)x^{\frac{\ln b}{\ln a}} $$

NOTE

Regarding the periodicity of $\Phi(u)$ we have

$$ \Phi(u) = \Phi(u+1)\to \Phi(\log_a x) = \Phi(\log_a x+1) \to \Phi\left(\frac{\ln x}{\ln a}\right) = \Phi\left(\frac{\ln x}{\ln a}+1\right) $$

and finally

$$ \Phi(\ln x) = \Phi(\ln x+\ln a) $$

Answer 2

Hints can be useful if you are willingful to hear.

• Let $f$ be a solution of the equation, then define $g(x) = \frac{f(e^x)}{e^{Ax}}$, where $A = \frac{\log(b)}{\log(a)} = \log_a(b)$.

  $g(x)$ is periodic of period $\log(a)$: $$g(x+\log(a)) = \frac{f(e^{x+\log(a)})}{e^{Ax+A\log(a)}} = \frac{f(ae^{x})}{e^{Ax+\log(b)}} = \frac{bf(e^{x})}{be^{Ax}} = g(x)$$

So every solution can be written as $f(x) = g(\log(x))x^A$.

• Viceversa, an $f$ of the kind $f(x) = g(\log(x))x^A$ is a solution of the equation:

$$f(ax) = g(\log(ax))(ax)^A = g(\log(x)+\log(a))x^Aa^A = a^A f(x) = bf(x)$$

So all the $f(x) = g(\log(x))x^A$ are solutions.