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Find all the function that satisfy : $$f\left(\frac{xf(y)}{2}\right)+f\left(\frac{yf(x)}{2}\right)=4xy$$ I only find $f(0)=0$ but I can't prove $f(x)=2x$

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    $\begingroup$ $f(x)=-2x$ is another solution. $\endgroup$ Feb 25, 2013 at 14:42
  • $\begingroup$ Is $f$ a continuous function? And what is the domain, $\mathbb{R}$ or $\mathbb{C}$? $\endgroup$
    – Yimin
    Feb 25, 2013 at 15:57

1 Answer 1

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If your $f$ is differentiable at $x=0$. Then

Differentiate the equation for $x$ at $x=0$

$f'\left(\dfrac{xf(y)}{2}\right)\dfrac{f(y)}{2}+f'\left(\dfrac{f(x)y}{2}\right)\dfrac{y}{2}f'(x)=4y$.

Take $x=0$. Since $f(0)=0$. Say $A = f'(0)$.

$A\dfrac{f(y)}{2}+A\dfrac{y}{2}A=4y$

Which is

$f(y)= \dfrac{8y-A^2y}{A}$

$f'(y) = \dfrac{8-A^2}{A}$,take $y=0$, $f'(0)=A=\dfrac{8-A^2}{A}$.

$A=2$ or $A=-2$.

This is for smooth case. If not differentiable at $x=0$, then I have no idea.

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