f(f(x)+y)=x+f(y) in R+

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#1 h Dec 30, 2020, 8:55 AM • 2 Y

Y by tiendung2006, Mango247

Find all functions $f : R_+ \to R_+$ such that $f(f(x)+y)=x+f(y)$ , for all $x, y \in R_+$

(Folklore)

PS

This post has been edited 1 time. Last edited by parmenides51, Dec 30, 2020, 8:56 AM

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#2 h Dec 30, 2020, 9:09 AM • 5 Y

Y by pco, A-Thought-Of-God, amar_04, srijonrick, Chryss

$x+f(y+z)=f(y+z+f(x))=f(f(f(z)+x)+y)=x+f(z)+f(y)$ Thus $f(x+y)=f(x)+f(y)$ and hence $f(x)=cx$ for all $x$ and constant $c$ . Thus $f(x)=x$ for all $x$ .

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#3 h Dec 31, 2020, 6:02 AM • 3 Y

Y by Mango247, Mango247, Mango247

I have this idea but I am not 100% sure if it is correct:
$f(f(x)+y)=x+f(y)$ , so letting $z$ equal $f(x)+y$ , $f(f(z))=z$ , implying $f(z)=f'(z)$ , and the only function satisfying this is $f(x)=x$ .
My problem comes from the fact that we assume that any $z\in R_+$ can be expressed in the form $f(x)+y$ . Would this be a false assumption, and is there anyway to show that it must be the case that $f(x)+y$ covers all $z$ ?

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#4 h Dec 31, 2020, 9:46 AM

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Lemma: $f$ is injective.
Proof: Assume $f(a)=f(b)$ . Give $x=a, y=c$ and $x=b, y=c$ . Then $a=b$ . So $f$ is injective.

$f(f(f(x))+y)=f(x)+f(y)=f(y)+f(x)=f(f(f(y))+x)$ and $f$ is injective, thus $f(f(x))-x=f(f(y))-y$ for all $x,y\in R_+$ . Then for all $x \in R_+$ we have $f(f(x))-x=k\Rightarrow f(f(x))=x+k$ .

$f(f(x)+y)=x+f(y)\Rightarrow f(f(f(x)+y))=f(x+f(y))\Rightarrow f(x)+y+k=f(x+f(y))=f(x)+y\Rightarrow k=0\Rightarrow f(f(x))=x$ .

$x=f(x), y=y$ gives us $f(x+y)=f(x)+f(y)$ .

Assume $f(a)>f(b)$ . $x=b, y=f(a)-f(b)$ gives us $a=b+f(f(a)-f(b))>b$ so $f$ is increasing.

With $f(x+y)=f(x)+f(y)$ and $f$ is increasing, we can say $f(x)=cx$ for all $x\in R_+$ and $c$ is constant. So $f(x)=x$ for all $x\in R_+$ .

This post has been edited 1 time. Last edited by BarisKoyuncu, Dec 31, 2020, 9:47 AM
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#5 h Sep 9, 2023, 1:44 AM

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parmenides51 wrote:

Find all functions $f : R_+ \to R_+$ such that $f(f(x)+y)=x+f(y)$ , for all $x, y \in R_+$

(Folklore)

PS

Solved with complex_afi
$\color{blue}\boxed{\textbf{Answer: }f\equiv x}$
$\color{blue}\boxed{\textbf{Proof:}}$
$\color{blue}\rule{24cm}{0.3pt}$
Cute
$f(f(x)+y)=x+f(y)...(\alpha)$ In $(\alpha) x\to y+f(x):$
$\Rightarrow f(f(f(x)+y)+y)=y+f(x)+f(y)$ By $(\alpha):$
$\Rightarrow f(x+y+f(y))=y+f(x)+f(y)$ By $(\alpha):$
$\Rightarrow f(x+y)+y=y+f(x)+f(y)$ $\Rightarrow f(x)+f(y)=f(x+y)$ And since it is in $\mathbb{R}^+$ we have that it is strictly increasing
$\Rightarrow f\equiv cx$ Replacing in $(\alpha):$
$\Rightarrow c(cx+y)=x+cy$ $\Rightarrow c=1$ $\Rightarrow \boxed{f\equiv x}_\blacksquare$ $\color{blue}\rule{24cm}{0.3pt}$

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#6 h Sep 10, 2023, 2:21 PM

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Here is my solution for this Problem:
Let $P(x,y)$ be the assertion: $f(f(x)+y)=x+f(y)$
$P(x+f(z),y)$ and with $P(z,x)$ and $P(x,y+z)$ we have $f(y+z)+x=f(f(x)+y+z)=f(f(x+f(z))+y)=x+f(y)+f(z)$

Then we will have $f(y+z)=f(y)+f(z)$ so that $f(x)=ax \quad \forall x>0$
Plugging this back in original equation, we get $a=1$ so $f(x)=x \quad \forall x>0$

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