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#1 h Saturday at 3:51 PM

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Find all the functions f: R+ to R+ which satisfy the functional equation:
f(2f(x)+f(y))=2x+y (x,y R+)

This post has been edited 1 time. Last edited by nguyenvana, Yesterday at 8:53 AM
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#2 h Saturday at 5:50 PM • 1 Y

Y by nguyenvana

nguyenvana wrote:

Find all the functions f: R+ to R+ which satisfy the functional equation:
f(2f(x)+f(y)+xy)=xy+2x+y (x,y R+)

Let $P(x,y)$ be assertion $f(2f(x)+f(y)+xy)=xy+2x+y$

$f(x)$ is obviously surjective
If $f(a)=f(b)$ , comparaison of $P(a,b)$ with $P(b,a)$ implies $a=b$ and so $f(x)$ is injective.

$P(2,x)$ $\implies$ $f(2f(2)+f(x)+2x)=3x+4$
$P(x+1,1)$ $\implies$ $f(2f(x+1)+f(1)+x+1)=3x+4$
And so, since injective : $2f(x+1)=f(x)+x+2f(2)-f(1)-1$ , and so $f(x+1)=\frac{f(x)+x}2+a$ for some constant $a$
So $f(x+2)=\frac{f(x+1)+x+1}2+a$ $=\frac{f(x)+3x}4+\frac{3a+1}2$

$P(x,2)$ $\implies$ $f(2f(x)+2x+f(2))=4x+2$
$P(1,2x)$ $\implies$ $f(f(2x)+2x+2f(1))=4x+2$

And so, since injective : $f(2x)+2x+2f(1)=2f(x)+2x+f(2)$ and so $f(2x)=2f(x)+b$ for some constant $b$

So $f(2(x+1))=2f(x+1)+b$ $=f(x)+x+2a+b$
But $f(2(x+1))=f(2x+2)=\frac{f(2x)+6x}4+\frac{3a+1}2$ $=\frac{2f(x)+b+6x}4+\frac{3a+1}2$
And so $f(x)+x+2a+b=\frac{2f(x)+b+6x}4+\frac{3a+1}2$
And so $f(x)=x+1-a-\frac{3b}2$ $\forall x>0$

Plugging $f(x)=x+u$ in original equation, we get $u=0$ and
$\boxed{f(x)=x\quad\forall x>0}$ , which indeed fits

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