Functional equation

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#1 h Yesterday at 11:18 AM • 1 Y

Y by Drago119

Find all functions f:R R satisfying the equation

f(f(x)y) + f(x+ f(y)) = x f(y) + f(x+y)

for all x,y real numbers

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pco

#2 h Yesterday at 3:27 PM • 1 Y

Y by Ecrin_eren

Ecrin_eren wrote:

Find all functions f:R R satisfying the equation

f(f(x)y) + f(x+ f(y)) = x f(y) + f(x+y)

for all x,y real numbers

Let $P(x,y)$ be the assertion $f(f(x)y)+f(x+f(y))=xf(y)+f(x+y)$
Let $a=f(0)$

$P(0,0)$ $\implies$ $f(a)=0$
$P(a,a)$ $\implies$ $f(2a)=a$
$P(a,0)$ $\implies$ $f(2a)=a^2-a$
And so $a^2-a=a$ and so $a\in\{0,2\}$

If $a=2$ :
$P(x,0)$ $\implies$ $2+f(x+2)=2x+f(x)$
$P(x,2)$ $\implies$ $f(2f(x))+f(x)=f(x+2)$
Adding : $2+f(2f(x))=2x$ and so $f(x)$ is bijective
But $P(2,0)$ $\implies$ $f(4)=2=f(0)$ , impossible since injective

So $a=0$
$P(0,x)$ $\implies$ $f(f(x))=f(x)$
$P(x,f(y))$ $\implies$ $f(f(x)f(y))=xf(y)$ and so :
Either $\boxed{\text{S1 : }f(x)=0\quad\forall x}$ , which indeed fits
Either $\exists y$ such that $f(y)\ne 0$ and last line implies surjectivity.
And then surjectivity and $f(f(x))=f(x)$ imply $\boxed{\text{S2 : }f(x)=x\quad\forall x}$ , which indeed fits too

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#3 h Yesterday at 5:40 PM

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how does surjectivity imply $f(f(x))=f(x)$ ?

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Burmf

#4 h Yesterday at 5:48 PM • 1 Y

Y by pco

$P(0,x)$ implies $f(f(x)) = f(x)$ . Then since f is surjective $R = Im(f)$

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