Functional: f(x+y)+y<f^(3)(x)

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Functional: f(x+y)+y<f^(3)(x)

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Functional inequality

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Source: Benelux MO 2013 Q2

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445 posts

djb86

#1 h Apr 29, 2013, 6:30 PM • 3 Y

Y by biomathematics, Adventure10, Mango247

Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$f(x + y) + y \le f(f(f(x)))$
holds for all $x, y \in \mathbb{R}$ .

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23460 posts

pco

#2 h Apr 29, 2013, 6:49 PM • 14 Y

Y by nghiepdu-socap, shaiephraim, Uagu, andria, Higgsboson79, Amir Hossein, brokendiamond, MahdiTA, Jerry37284, sman96, Ajomoto, Adventure10, Assassino9931, and 1 other user

djb86 wrote:

Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$f(x + y) + y \le f(f(f(x)))$
holds for all $x, y \in \mathbb{R}$ .

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

$P(x,f(f(x))-x)$ $\implies$ $f(f(x))\le x$ and so $P(x,y)$ implies $Q(x,y)$ : $f(x+y)+y\le f(x)$

$Q(x,y-x)$ $\implies$ $f(y)+y\le f(x)+x$ and so $\boxed{f(x)=a-x}$ which indeed is a solution whatever is $a\in\mathbb R$

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207 posts

#3 h May 1, 2013, 7:58 AM • 2 Y

Y by Adventure10, Mango247

pco wrote:

$Q(x,y-x)$ $\implies$ $f(y)+y\le f(x)+x$ and so $\boxed{f(x)=a-x}$ which indeed is a solution whatever is $a\in\mathbb R$

That solution is very elegant, but can you explain this line? How can you conclude that fact?

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23460 posts

pco

#4 h May 1, 2013, 8:23 AM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

Nguyenhuyhoang wrote:

pco wrote:

$Q(x,y-x)$ $\implies$ $f(y)+y\le f(x)+x$ and so $\boxed{f(x)=a-x}$ which indeed is a solution whatever is $a\in\mathbb R$

That solution is very elegant, but can you explain this line? How can you conclude that fact?

$f(y)+y\le f(x)+x$
Swapping $x,y$ , we get $f(x)+x\le f(y)+y$

And so $f(x)+x=f(y)+y$ constant $=a$ for some real $a$ .

And so $f(x)=a-x$

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mathuz

#5 h May 1, 2013, 11:59 AM • 8 Y

Y by Egamberganov, Const41, Amir Hossein, Ajomoto, tony88, Adventure10, Mango247, and 1 other user

djb86 wrote:

Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$f(x + y) + y \le f(f(f(x)))$
holds for all $x, y \in \mathbb{R}$ .

(1) $f(f(f(x)))=f(x)$ :
because, $P(x,0)$ $\Rightarrow$ $f(x)\le f(f(f(x)))$ and
$P(f(x), f(f(f(x)))-f(x))$ $\Rightarrow$ $f(f(f(x)))\le f(x)$ .
hence, $f(x)\le f(f(f(x)))\le f(x)$ and $f(f(f(x)))=f(x).$
(2) $f(x)=c-x$ : ( $c=f(0)$ )
because, from (1) we have $f(x+y)+y\le f(x)$ .
$P(0,x)$ $\Rightarrow$ $f(x)+x\le f(0)$ and $P(x, -x)$ $\Rightarrow$ $f(0)\le f(x)+x$ .
hence, $f(x)+x=f(0)$ and $f(x)=c-x .$
Answer: $f(x)=c-x$ .

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55 posts

#6 h May 31, 2014, 9:27 AM • 2 Y

Y by Adventure10 and 1 other user

Easy and nice.

My Solution (almost same as others):

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74 posts

Jeje

#7 h Oct 8, 2014, 12:18 PM • 2 Y

Y by Adventure10, Mango247

I am sorry Bumping old post

f(f(x)) =< x imply that f(f(f(x))) =< f(x)

This mean it is increasing function isn't it????

Can anyone explain me why it was increasing function???

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327 posts

#8 h Oct 8, 2014, 12:23 PM • 2 Y

Y by Adventure10, Mango247

No, in their proof, it is stated that $f(f(x)) \le x$ is true for every $x \in \mathbb{R}$ , so we can change $x$ by $f(x)$ .

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Jeje

#9 h Oct 8, 2014, 12:59 PM • 1 Y

Y by Adventure10

Oh yeahhhh i am sorry too stupid argumentt

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6012 posts

MSTang

#10 h Apr 26, 2016, 3:15 PM • 3 Y

Y by TooPicky07, Adventure10, Mango247

~~Take $y=0$ to get $f(x) \le f(f(f(x)))$ for all $x$ .~~ Take $y = f(f(x)) - x$ to get $f(f(f(x))) + f(f(x)) - x \le f(f(f(x)))$ , or $f(f(x)) \le x$ for all $x$ . Replacing $x$ with $f(x)$ in this inequality, we get $f(f(f(x))) \le f(x)$ for all $x$ . ~~Thus, $f(x) = f(f(f(x)))$ for all $x$ .~~ Then the given inequality becomes $f(x+y) + y \le f(x)$ for all $x$ , $y$ .

Take $x=0$ to get $f(y) + y \le f(0)$ for all $y$ , and take $y=-x$ to get $f(0) - x \le f(x) \implies f(0) \le f(x) + x$ for all $x$ . Thus, $f(0) = f(x) + x$ for all $x$ , so $f(x) = a-x$ where $a=f(0)$ is a constant. It is easy to check that all functions of this form satisfy the original functional inequality, so the solutions are $f(x) = a-x$ for any constant $a$ .

(Strikethrough edits made after reading first solution in this thread

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This post has been edited 1 time. Last edited by MSTang, Apr 26, 2016, 3:16 PM

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#11 h Apr 3, 2017, 8:38 PM • 2 Y

Y by Adventure10, Mango247

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#12 h Apr 9, 2017, 1:06 AM • 1 Y

Y by Adventure10

$(x, f(f(x))-x)$ yields $f(f(x)) \le x$ . $(x, 0)$ yields $f(f(f(x))) \ge f(x)$ . Combining the two gives $f(f(f(x))) = f(x)$ . Thus, $f(x + y) + y \le f(x)$ . Now, "swap" the roles of $x+y$ and $x$ to get $f(x) - y \le f(x + y)$ . Thus, $f(x + y) = f(x) - y$ . Letting $x = 0$ , $f(y) = f(0) - y$ . Indeed all such functions work. Yay.

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1179 posts

#13 h Jan 5, 2019, 6:04 PM • 1 Y

Y by Adventure10

Nice problem.

Taking $y=f(f(x))-x$ , we see $f(f(x))\le x$ . Thus, the original FE can be rewritten as $f(x+y)+y\le f^2(f(x))\le f(x)$ Swapping the sign on $y$ , we see $f(x-y)-y\le f(x)$ . Replacing $x$ with $x-y$ , we get $f(x)+y\le f(x-y)$ . It follows, $f(x)=f(x-y)-y$ . Taking $x=0$ , we get that $f(-y)=y+f(0)$ , yielding the family of solutions $\boxed{f=-x+k}$ .

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ZETA_in_olympiad

2211 posts

ZETA_in_olympiad

#14 h Jul 30, 2022, 1:31 PM • 1 Y

Y by panche

$P(x,y)\implies f(x + y) + y \leqslant f(f(f(x)))$
$P(x,f(f(x))-x)\implies f(f(x))\leqslant x$
$P(0,x)\implies f(x)\leqslant f(0)-x$
$P(x,-x)\implies f(x)\geqslant f(0)-x$
$\therefore f(x)=c-x \; \text{(works)}$

This post has been edited 1 time. Last edited by ZETA_in_olympiad, Jul 30, 2022, 1:34 PM

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#15 h Feb 22, 2025, 12:21 AM

Y by

Let us denote $f^k(x)$ as the $k$ -th composition of $f(x)$ . We can take $P(x, f(f(x)) - x)$ to obtain $f(f(x)) \leq x$ . From here we use $P(0,0)$ to obtain $f^3(x) \geq f(x)$ . Combining both $f(f(x)) \leq x$ and $f^3(x) \geq f(x)$ we obtain $f^3(x) = f(x)$ and hence
$f(x+y) + y \leq f(x)$ We substitute $P(x,-x)$ to obtain
$f(x) \geq -x + f(0)$ We substitute $P(f(0),y-f(0))$ to obtain
$f(y) + y - f(0) \leq f^2(0)$ Since we know $f(f(x)) \leq x$ we obtain
$f(y) + y - f(0) \leq f^2(0) \leq 0 \Longrightarrow f(y) \leq -y + f(0)$ Thus we must have $f(x) = -x+f(0)$ , which does in fact work.

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#16 h Mar 20, 2025, 10:06 AM

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Ditto as pco, so won't elaborate.

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