1. Functional equation

Solve the functional equation:

$\displaystyle f(xy)=[f(x)]^{y^\beta}[f(y)]^{x^\beta}$, where $\displaystyle x>0,~f(x)>0,~\beta\in\mathbb{R}$.

P.S. Good luck!

2. it seems that $\displaystyle f_{\alpha}(x)=x^{\alpha x^{\beta}}, \ \alpha \in \mathbb{R},$ are all the solutions ...

3. let b=beta
taking the logarithm of the both sides , we get :

$\displaystyle ln(f(xy))= y^bln(f(x))+x^bln(f(y))$ , let $\displaystyle g(x)=ln(f(x))$

we get : $\displaystyle g(xy)=y^bg(x)+x^bg(y)$

dividing both sides by $\displaystyle x^by^b$:
$\displaystyle \frac{g(xy)}{x^by^b} = \frac{g(x)}{x^b} + \frac{ g(y)}{y^b}$

let : $\displaystyle h(x)= e^{\frac{g(xy)}{x^by^b}}$
we have : $\displaystyle h(xy)=h(x)+h(y)$ which is equivalent to : $\displaystyle k(xy)=k(x)k(y)$ , if i remember well , the solution of the Last equation is $\displaystyle x^a$ for some real a , a difficult result of Erdos !!

4. you'd need some sort of continuity condition i guess. anyway, just put $\displaystyle g(x)=x^{-\beta} \ln f(x)$ to end up with the simple equation $\displaystyle g(xy)=g(x)+g(y).$

5. Originally Posted by NonCommAlg
it seems that $\displaystyle f_{\alpha}(x)=x^{\alpha x^{\beta}}, \ \alpha \in \mathbb{R},$ are all the solutions ...
Yes, but it would be even more interesting see your full solution, dear NonCommAlg.

6. Originally Posted by DeMath
Yes, but it would be even more interesting see your full solution, dear NonCommAlg.
i've already posted it. see my second response.