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

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Answer 1 · 2016-12-23T14:46:01

$f(x)=x$

Making $x=y=0$ we have $f(0)=0$

Making $y=0$ we have

$f(x^2)=xf(x)$ but
$f((-x)^2)=f(x^2)=-xf(-x)$

so $f(x)$ is an odd function.

Making now $x = -y$ we have

$f(x^2-xf(x))=x f(0) = 0$

so supposing that

$f(x)=0 => x=0$

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

$x^2-xf(x)=x(x-f(x))=0$

considering $x ne 0$ we have

$f(x)=x$

Checking

$(x^2+y x)=x(x+y)$

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

$f(x) = x^(1/2)f(x^(1/2))= cdots = x^(1/2+1/4+cdots + 1/(2^n))f(x^(1/(2^n)))$

but $lim_(n->oo) x^(1/2^n) = 1$ and $lim_(n->oo)(1/2+1/4+cdots+1/(2^n))=1$ so

$f(x) = x f(1)$ now substituting this result into $f(x^2)=xf(x)$ we obtain

$f(x^2)=x^2f(1)->f(1)=1$

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