Is this implied function unique? Let $f(p)>0$ be a function for $p\in(0,1)$ such that $$ f(p_{1}\times p_{2})=f(p_{1})\times f(p_{2})$$ for all $p_{1},p_{2}\in (0,1)$. Does $f(p)$ have to be of the form $f(p)=p^{\alpha}$ for some fixed $\alpha$?
 A: No. Here is a counterexample or, rather, an idea how to construct one (I did not check all details).Take $a_0=1/2$. Let $f(a_0^\alpha)=a_0^\alpha$ for every rational number $\alpha>0$. Note that the set $S_0$ of all positive rational powers of $1/2$ is a multiplicative subsemigroup of $(0,1)$. Suppose we already defined $f$ on some multiplicative semigroup $S$ of $(0,1)$. We say that such a subsemigroup $S$ is group-like  if for every $x,y\in S$ and $x^\alpha/y\in (0,1)$ for some rational $\alpha>0$, we have $x^\alpha/y\in S$. Clearly, $S_0$ is group-like. Suppose we have already defined $f$ on some group-like subsemigroup $S$ of $(0,1)$, and $x\in (0,1)\setminus S$. Consider any element $z$ of $(0,1)\setminus S$ and let $S'$ be the smallest group-like subsemigroup of $(0,1)$ containing $S,x$. Then we can define $f(x)=x^2$ and extend $f$ to the whole $S'$. By Zorn lemma, we can extend $f$ to the whole $(0,1)$. That $f(p)$ is not of the form $p^\alpha$ for any $\alpha$.
A: Yes. Observe that:
$$f(p^2)=f(p\times p)=f(p)\times f(p)=f^2(p)$$ and proceed on induction to see that:
$$f(p^{k+1})=f(p^k\times p)=f(p^k)\times f(p)=f^k(p)\times f(p)=f^{k+1}(p)$$
This general result $f(p^n)\equiv (f(p))^n$ implies that $f(p)$ is monomial.
To show this, suppose $f$ has $k$ terms, then $f^n$ has $k^n$ terms by the multinomial theorem, so $k=k^n\implies k=1$.
Of the monomial forms, only the exponential $p^\alpha$ has the desired multiplicativity property.
