# Find all real functions f from RR->RR satisfying the relation f(x²+yf(x))=xf(x+y) ?

Dec 23, 2016

$f \left(x\right) = x$

#### Explanation:

Making $x = y = 0$ we have $f \left(0\right) = 0$

Making $y = 0$ we have

$f \left({x}^{2}\right) = x f \left(x\right)$ but
$f \left({\left(- x\right)}^{2}\right) = f \left({x}^{2}\right) = - x f \left(- x\right)$

so $f \left(x\right)$ is an odd function.

Making now $x = - y$ we have

$f \left({x}^{2} - x f \left(x\right)\right) = x f \left(0\right) = 0$

so supposing that

$f \left(x\right) = 0 \implies x = 0$

(remember that $f \left(x\right)$ is an odd function) we have

${x}^{2} - x f \left(x\right) = x \left(x - f \left(x\right)\right) = 0$

considering $x \ne 0$ we have

$f \left(x\right) = x$

Checking

$\left({x}^{2} + y x\right) = x \left(x + y\right)$

Note. Another way of proof for $f \left(x\right) = x$ is by doing for $x \ge 0$

$f \left(x\right) = {x}^{\frac{1}{2}} f \left({x}^{\frac{1}{2}}\right) = \cdots = {x}^{\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{{2}^{n}}} f \left({x}^{\frac{1}{{2}^{n}}}\right)$

but ${\lim}_{n \to \infty} {x}^{\frac{1}{2} ^ n} = 1$ and ${\lim}_{n \to \infty} \left(\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{{2}^{n}}\right) = 1$ so

$f \left(x\right) = x f \left(1\right)$ now substituting this result into $f \left({x}^{2}\right) = x f \left(x\right)$ we obtain

$f \left({x}^{2}\right) = {x}^{2} f \left(1\right) \to f \left(1\right) = 1$