Functional equation

**DeMath** · June 29th 2010, 04:14 AM

Solve the functional equation:

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

P.S. Good luck!

**NonCommAlg** · June 29th 2010, 05:48 AM

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

**shostakovich** · June 29th 2010, 05:52 AM

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

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

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

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

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

**NonCommAlg** · June 29th 2010, 05:56 AM

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

**DeMath** · June 29th 2010, 06:01 AM

Originally Posted by NonCommAlg

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

Yes, but it would be even more interesting see your full solution, dear NonCommAlg.

**NonCommAlg** · June 29th 2010, 06:08 AM

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.

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