When is a function of a product the product of functions? I am trying to figure out what are all functions, $f, g$ that satisfy the following:
$f(ab) = g(a)g(b) \forall a,b$ 
In other words, a function of a product of arguments is always the product of a (possibly different) function of both of the arguments separately?
For example, I know that $f(x) = kx^n, g(x) = \sqrt{k}x^n$ for all positive $k$ and real $n$ holds true. Is there a way to find other such functions, or prove that these are the only such functions? (As pointed out, the functions should also be continuous and R $\rightarrow$ R.
 A: Hint: $\;b=1$ gives $f(a)=g(a)\,g(1)=\lambda \,g(a)\,$ with $\lambda = g(1)$.


*

*If $\lambda=0$ then $f(x)=0\,$, which in turn implies $g(x)=0\,$.

*Otherwise, substituting back into the original equation: $$
\lambda\,g(ab) = g(a)\,g(b) \;\;\iff\;\; \frac{g(ab)}{\lambda} = \frac{g(a)}{\lambda}\,\frac{g(b)}{\lambda} $$ Let $h(x)=g(x)\,/\,\lambda$ and the equation becomes $\,h(ab)=h(a)\,h(b)\,$. The latter is related to Cauchy's functional equation and, for continuous $h\,$, the solutions on $\mathbb{R}^+$ are $h(x)=x^n$ which then give $g(x)=\lambda \,x^n$ then $f(x)=\lambda^2 x^n$ for $x \ge 0$  (edit: read on for the complete answer).

[ EDIT ]  As pointed out in OP's comment, the above only addresses the case $x \ge 0\,$. The following completes the answer for the entire $\mathbb{R}\,$.


*

*If $h(b)=0$ for some $b \ne 0$, then $h(a)=h(\frac{a}{b}\,b)=h(\frac{a}{b})\,h(b)=h(\frac{a}{b}) \cdot 0=0\,$, which again gives the trivial solution $h(x)=g(x)=f(x)=0\,$.

*Otherwise, $h(b) \ne 0$ for $b \ne 0\,$, therefore by continuity $h(x)$ can not change sign on either $(-\infty,0)$ or $(0,\infty)$. (Since $a \ge 0 \implies h(a)=h(\sqrt{a}\,\sqrt{a})=\big(h(\sqrt{a})\big)^2 \ge 0\,$ it follows that $h$ is non-negative on $\mathbb{R}^+$, which is consistent with the first part of the proof).
Since $h(-a)=h(-1)h(a)\,$ then $h(a^2)=h\big((-a)^2\big)=\big(h(-1)\big)^2\,\big(h(a)\big)^2=\big(h(-1)\big)^2\,h(a^2)\,$ so $\big(h(-1)\big)^2=1\,$ or $h(-1)=\pm 1$. Therefore $h(x)$ must be either odd or even. Given the previously determined $h(x)=x^n$ solution on $\mathbb{R}^+\,$, this means either $h(x)=|x|^n$ or $h(x)=\operatorname{sgn}(x)\,|x|^n $ on $\mathbb{R}$ so in the end the two solutions are $g(x) = \lambda |x|^n\,$ and $\,g(x)=\lambda \operatorname{sgn}(x)\,|x|^n\,$ then $f(x) = \lambda g(x)$.
