functional equation $f(xy)=\frac{f(x)f(y)}{f(1)}$

Question

I am looking for functions $f$ (in my case $\mathbb{N}^{*} \rightarrow \mathbb{R}^{+*}$) for which the ratio of the images only depends on the ratio of their antecedents :

$$\frac{f(n_1.n_0)}{f(n_2.n_0)}=\frac{f(n_1)}{f(n_2)} \forall (n_0,n_1,n_2) \in \mathbb{N}^{*3}$$

Which is equivalent to $$f(n_1.n_2)=\frac{f(n_1).f(n_2)}{f(1)} \forall (n_1,n_2) \in \mathbb{N}^{*2}$$

It is obvious that $f(n)=c.n^m$ ($c \in \mathbb{R}^{+*}$ and $m \in \mathbb{R}$) validate that equation.

Can we prove that they are the only solutions ?

Answer 1

From the initial identity, by induction

$$f(n)=f(p_0^{m_0}p_1^{m_1}\cdots p_k^{m_k})=\frac{f(p_0)^{m_0}f(p_1)^{m_1}\cdots f(p_k)^{m_k}}{f(1)^{m_0+m_1+\cdots m_k-1}}.$$

As the prime decomposition is unique, you can choose the $f(p_k)$ and $f(1)$ freely.

Answer 2

One can find a more general class of solutions. Denote by $\mathcal{P}$ the set of primes and take an arbitrary sequence of real numbers $(m_p)_{p\in \mathcal{P}}$, then for any $n\in \mathbb{N}^*$, uniquely write $n$ as, $$ n = p_1^{\alpha_1}...p_q^{\alpha_q} $$ where $q \geq 0$, $p_1, ..., p_q \in \mathcal{P}$ and $\alpha_1, ... ,\alpha_q \in \mathbb{N}^*$, possibly having a void product if $n = 1$.

Then define $f$ by $f(n) = p_1^{m_{p_1}\alpha_1}...p_q^{m_{p_q}\alpha_q}$, we easily check that $f(n_1.n_2) = f(n_1).f(n_2)$. In general $f$ cannot be written $f(n) = c.n^m$ for some real numbers $c >0$,$m$.

The next question: Is this more general class, the only class of solutions ?