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!

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- Jun 29th 2010, 04:14 AMDeMathFunctional 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! - Jun 29th 2010, 05:48 AMNonCommAlg
it seems that $\displaystyle f_{\alpha}(x)=x^{\alpha x^{\beta}}, \ \alpha \in \mathbb{R},$ are all the solutions ...

- Jun 29th 2010, 05:52 AMshostakovich
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 !! - Jun 29th 2010, 05:56 AMNonCommAlg
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 $\displaystyle g(xy)=g(x)+g(y).$

- Jun 29th 2010, 06:01 AMDeMath
- Jun 29th 2010, 06:08 AMNonCommAlg