Functional equation

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Dadgarnia

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Dadgarnia

#1 h Apr 15, 2018, 12:15 PM • 10 Y

Y by anantmudgal09, tenplusten, Mathuzb, vsathiam, Davi-8191, Amir Hossein, Muaaz.SY, Adventure10, Mango247, jasperE3

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ that satisfy the following conditions:
a. $x+f(y+f(x))=y+f(x+f(y)) \quad \forall x,y \in \mathbb{R}$
b. The set $I=\left\{\frac{f(x)-f(y)}{x-y}\mid x,y\in \mathbb{R},x\neq y \right\}$ is an interval.

Proposed by Navid Safaei

This post has been edited 1 time. Last edited by Dadgarnia, Apr 15, 2018, 12:15 PM

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pco

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pco

#2 h Apr 15, 2018, 12:19 PM • 4 Y

Y by Tawan, Muaaz.SY, Adventure10, Mango247

Dadgarnia wrote:

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ that satisfy the following conditions:
a. $x+f(y+f(x))=y+f(x+f(y)) \quad \forall x,y \in \mathbb{R}$
b. The set $I=\left\{\frac{f(x)-f(y)}{x-y}\mid x,y\in \mathbb{R},x\neq y \right\}$ is an interval.

Proposed by Navid Safaei

When you write "an interval", do you mean a "bounded interval" (which implies continuity for example)? or is $[0,+\infty)$ for example considered as an "interval" in your problem statement ?

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Dadgarnia

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Dadgarnia

#3 h Apr 15, 2018, 12:27 PM • 2 Y

Y by Muaaz.SY, Adventure10

pco wrote:

Dadgarnia wrote:

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ that satisfy the following conditions:
a. $x+f(y+f(x))=y+f(x+f(y)) \quad \forall x,y \in \mathbb{R}$
b. The set $I=\left\{\frac{f(x)-f(y)}{x-y}\mid x,y\in \mathbb{R},x\neq y \right\}$ is an interval.

Proposed by Navid Safaei

When you write "an interval", do you mean a "bounded interval" (which implies continuity for example)? or is $[0,+\infty)$ for example considered as an "interval" in your problem statement ?

Yes this implies continuity but this is not necessarily a bounded interval.

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pco

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pco

#4 h Apr 15, 2018, 1:01 PM • 11 Y

Y by rightways, atmchallenge, naw.ngs, Amir Hossein, Promi, MrOrange, Muaaz.SY, Siddharth03, Adventure10, Mango247, MS_asdfgzxcvb

Dadgarnia wrote:

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ that satisfy the following conditions:
a. $x+f(y+f(x))=y+f(x+f(y)) \quad \forall x,y \in \mathbb{R}$
b. The set $I=\left\{\frac{f(x)-f(y)}{x-y}\mid x,y\in \mathbb{R},x\neq y \right\}$ is an interval.

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

If $1\in I$ , then $\exists x\ne y$ such that $f(x)-f(y)=x-y$
And so $y+f(x)=x+f(y)$ and then $P(x,y)$ implies $x=y$ , contradiction.
So $1\notin I$

It $t\ne 1\in I$ , then let $x\ne y$ such that $\frac{f(x)-f(y)}{x-y}=t$
We have $x+f(y)\ne y+f(x)$ and
$A=\frac{f(x+f(y))-f(y+f(x))}{(x+f(y))-(y+f(x))}=\frac{x-y}{(x-y)-(f(x)-f(y)}=\frac 1{1-t}$

So $t\in I$ implies $\frac 1{1-t}\in I$ which unfortunately is impossible :
If $t>1$ , then $\frac 1{1-t}<0$ and we can not have both in I (else $1\in I$ )

If $t<1$ , then we need $\frac 1{1-t}<1$ which means $t<0$ but then $\frac 1{1-t}>0$ and replacing $t$ by $\frac 1{1-t}$ gives contradiction.

So no such I.
So $\boxed{\text{no such function}}$

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tenplusten

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tenplusten

#5 h Apr 15, 2018, 3:15 PM • 3 Y

Y by Muaaz.SY, Adventure10, Mango247

Can anyone explain me what does "being interval" mean i.e if some set is interval what do we know about that set?
Thanks!

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pco

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pco

#6 h Apr 15, 2018, 3:17 PM • 3 Y

Y by Muaaz.SY, Adventure10, Mango247

tenplusten wrote:

Can anyone explain me what does "being interval" mean i.e if some set is interval what do we know about that set?
Thanks!

The only property of "interval" I used in my previous post is : $\forall x<y\in I$ , then $[x,y]\subseteq I$

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tenplusten

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tenplusten

#7 h Apr 15, 2018, 3:23 PM • 2 Y

Y by Adventure10, Mango247

pco wrote:

$\forall x<y\in I$

Does $x$ also belong to $I$ ,or just $y$ ?

pco wrote:

$[x,y]\subseteq I$

Does it mean any number in the interval $[x,y]$ belongs to $I$ .

Sorry if above questions are stupid.
Thanks!

This post has been edited 1 time. Last edited by tenplusten, Apr 15, 2018, 3:23 PM
Reason: also* not alsp

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pco

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pco

#8 h Apr 15, 2018, 3:29 PM • 4 Y

Y by tenplusten, Tawan, Adventure10, Mango247

What I wrote : $x,y\in I$ and $x<y$ $\implies$ $z\in I$ $\forall z$ such that $x\le z\le y$

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tenplusten

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tenplusten

#9 h Apr 15, 2018, 3:31 PM • 2 Y

Y by Adventure10, Mango247

Lol.I now see that first question is quite stupid.
Thanks for help.

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Ali3085

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Ali3085

#11 h Jun 21, 2020, 6:02 PM • 3 Y

Y by Inconsistent, Muaaz.SY, Mango247

let $P(x,y)$ be the assertion $x+f(y+f(x))=y+f(x+f(y)) \quad \forall x,y \in \mathbb{R}$ claim: $f$ is injective
proof:
note that if $1 \notin I$
we have $\frac{f(y+f(x))-f(x+f(y))}{y-x+f(x)-f(y)}=\frac{y-x}{y-x+f(x)-f(y)}$ so we have
$\frac{y-x}{y-x+f(x)-f(y)} \in I \forall x \neq y$
if $y>x$ then $f(y)>f(x) \forall y>x$ or $f(x)>f(y) \forall y>x$ else we will have $a>1,b<1 : a,b\in I \implies 1 \in I$ contradiction
$\blacksquare$
now let $t \in I$
$P(x,ty) \implies x+f(xt+f(y))=yt+f(x+f(yt))$ let it $Q(x,y)$
from $P(x,0)$ and $Q(x,0)$
$x+f(xt+c)=f(x+c)=x+f(f(x))$ so $f(x)=xt+c$
if we plug it into the original one, we will reach a contradiction
so
there's no such function
and we win

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Mr.C

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Mr.C

#12 h Sep 3, 2020, 6:46 PM

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solution

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jasperE3

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jasperE3

#13 h Mar 30, 2025, 5:17 PM

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Note that $1\notin I$ otherwise if $\frac{f(u)-f(v)}{u-v}=1$ for $u\ne v$ we get a contradiction by $P(u,v)$ .

So if $1\ne k\in I$ with $k=\frac{f(x)-f(y)}{x-y}$ we have:
$I\ni\frac{f(x+f(y))-f(y+f(x))}{x-y+f(y)-f(x)}=\frac{x-y}{x-y+f(y)-f(x)}=\frac1{1-k}$ But we must have either $k<1<\frac1{1-k}$ or $\frac1{1-k}<1<k$ so since $I$ is an interval this requires $1\in I$ , contradiction.

So there is no such function.

This post has been edited 1 time. Last edited by jasperE3, Mar 30, 2025, 5:18 PM

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