Hard FE with positive reals

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Hard FE with positive reals

G H J

Reveal topic tags

functional equation

L

G H BBookmark kLocked kLocked NReply

Source: Turkey Olympic Revenge 2023 Shortlist A4

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

185 posts

egxa

#1 h Jan 22, 2024, 7:22 AM • 1 Y

Y by Damiante

Find all functions $f:\mathbb{R^+}\to \mathbb{R^+}$ such that for all $x,y\in \mathbb{R^+}$
$f(xf(y)+y)=f(f(y))+yf(x)$
Proposed by Şevket Onur Yılmaz

This post has been edited 1 time. Last edited by egxa, Jan 22, 2024, 7:48 AM
Reason: typo

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1601 posts

#2 h Jan 22, 2024, 7:56 AM

Y by

solved with Malay

nice fe
solution

This post has been edited 1 time. Last edited by starchan, Jan 24, 2024, 6:08 PM
Reason: forgot to write co solver lmao

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

540 posts

#3 h Jan 22, 2024, 8:32 AM • 1 Y

Y by D_S

egxa wrote:

Find all functions $f:\mathbb{R^+}\to \mathbb{R^+}$ such that for all $x,y\in \mathbb{R^+}$
$P(x,y)$ : $f(xf(y)+y)=f(f(y))+yf(x)$
Proposed by Şevket Onur Yılmaz

* Claim: $f(x) \leq x$ for all $x>0$
Proof

* Claim: $f(1) = 1$
Proof

Now, $P(x,1)$ gives $f(x+1)=f(x)+1$ for all $x>0$ , hence by induction, $f(x+n)=f(x)+n$ for all $n \in \mathbb{N}$ and $x>0$ .
$P(x,n)$ $\implies$ $f(nx)=nf(x)$ for all $n \in \mathbb{N}$ and $x>0$ .

Suppose there exist $x>0$ such that $f(x)<x$ , let $r=\frac{p}{q}$ such that $f(x)<r<x$ .
We have $qf(x)=f(qx)=f(qx-p+p)=f(qx-p)+p>p$ $\implies$ $f(x)>r$ which contradicts $f(x)<r$ .
Therefore $\boxed{f(x)=x \quad \forall x >0}$ which satisfies the FE.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

659 posts

#4 h Feb 21, 2024, 8:54 PM • 3 Y

Y by Exposter, soryn, ehuseyinyigit

Cute.
Clearly, $f(x) \leq x$ else if we can set $x = \frac{f(y)-y}{f(y)}$ and get $f(x)=0$ contradiction!
Now $f(1)=1$ , Suppose $f(1) < 1$ then substitute $x = \frac{1}{1-f(1)}, y=1$ to get $f(f(1))=0$ contradiction!
By induction $f(x+n)=f(x)+n$ and $f(xn)=nf(x)$ , assume there exists $x$ s.t. $f(x) < x$ then choose rational $r = \frac{p}{q}$ b/w them $qf(x)=f(qx-p+p)=f(qx-p)+p > p$ so $f(x)>r$ contradiction! SO $f(x)=x$ is the only soln

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

311 posts

#5 h Feb 22, 2024, 2:34 AM • 1 Y

Y by ehuseyinyigit

Here is a simple solution using the cde method

Claim: $f(x) \le x$
Proof
Claim $f(1) = 1$
Proof
Claim $f(x+1) = f(x) + 1$
Proof
Claim $\boxed{f(x) = x}$
Proof (using cde method)

This post has been edited 3 times. Last edited by tadpoleloop, Feb 22, 2024, 2:38 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

662 posts

#6 h Oct 23, 2024, 7:58 PM

Y by

Incredible problem once again!
$f(xf(y)+y)=f(f(y))+yf(x)$ Claim: $f(x)\leq x$ .
Proof: $f(xf(y)+y)=f(f(y))+yf(x)>f(f(y))$ hence $xf(y)+y\neq f(y)$ which implies $\frac{f(y)-y}{f(y)}$ is never a positive real. Thus, $y\geq f(y)$ . $\square$
Claim: $f(1)=1$ .
Proof: Plug $y=1$ to get $f(xf(1)+1)=f(f(1))+f(x)>f(x)$ hence $xf(1)+1\neq x$ which yields $\frac{1}{1-f(1)}$ is not a positive real. We can see that $f(1)\geq 1$ and we utilise $f(x)\leq x$ to observe that $f(1)=1$ . $\square$
Claim: $f(x+n)=f(x)+n$ .
Proof: Plugging $y=1$ gives $f(x+1)=1+f(x)$ and by inducting we get $f(x+n)=f(x)+n$ as expected. $\square$
Claim: $f(x)=x$ for all $x<1$ .
Proof: Choose $y<1$ . Plugging $x=\frac{1-y}{f(y)}$ implies
$1=f(f(y))+yf(\frac{1-y}{f(y)})\iff \frac{1-y}{f(y)}\geq f(\frac{1-y}{f(y)})=\frac{1-f(f(y))}{y}\geq \frac{1-f(y)}{y}$ Thus, $y-y^2\geq f(y)-f(y)^2\iff (y-f(y))(1-y-f(y))\geq 0$ . Hence we conclude that $f(y)=y$ or $f(y)\leq 1-y$ .
$xf(y)+y\geq f(xf(y)+y)=f(f(y))+yf(x)$ Suppose that $f(a)\neq a$ for $a\in (0,1)$ . Plug $y=a$ to see that
$x(1-a)+a\geq xf(a)+a= f(f(a))+af(x)>af(x)\implies f(x)<\frac{x(1-a)}{a}+1$ For any $k$ positive integer, we have $f(k)=k$ by induction with base case $f(1)=1$ . Subsequently,
$k<\frac{k(1-a)}{a}+1\iff k(1-\frac{1-a}{a})<1\iff \frac{1}{k}>1-\frac{1-a}{a}$ While $k$ goes to positive infinity, $LHS$ diverges to $0$ hence $1\leq \frac{1-a}{a}$ which is equavilent to $\frac{1}{2}\geq a$ . This gives that $f(x)=x$ for all $x\in (\frac{1}{2},1]$ . Now suppose that $a\in (0,\frac{1}{2})$ and $f(a)\neq a$ and $f(a)\leq 1-a$ . Choose $y=\frac{1}{2}+\epsilon,x=a$ where $\{a\}<\frac{1}{2}$ and $\epsilon<\frac{1-a}{2(a+1)}$
$f(\frac{a+1}{2}+a\epsilon+\epsilon)=\frac{1}{2}+\epsilon+(\frac{1}{2}+\epsilon)f(a)$ Since $\frac{1}{2}<\frac{a+1}{2}+a\epsilon+ \epsilon<1$ , we see that $f(\frac{a+1}{2}+a\epsilon+\epsilon)=\frac{a+1}{2}+a\epsilon+\epsilon$ which implies $f(a)=a$ hence $f(x)=x$ for all $x\in (0,\frac{1}{2}]$ . By combining these, we conclude that $f(x)=x$ for all $x\in (0,1]$ and by using $f(x+n)=f(x)+n$ , we get that $f(x)=x$ for all positive reals as desired. $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

11100 posts

#7 h Yesterday at 4:04 PM

Y by

egxa wrote:

Find all functions $f:\mathbb{R^+}\to \mathbb{R^+}$ such that for all $x,y\in \mathbb{R^+}$
$f(xf(y)+y)=f(f(y))+yf(x)$
Proposed by Şevket Onur Yılmaz

Claim: $f(1)=1$
If $f(1)<1$ then:
$P\left(\frac1{1-f(1)},1\right)\Rightarrow f\left(\frac1{1-f(1)}\right)=f(f(1))+f\left(\frac1{1-f(1)}\right)\Rightarrow f(f(1))=0,$
contradiction. On the other hand, if $f(1)>1$ then:
$P\left(\frac{f(1)-1}{f(1)},1\right)\Rightarrow f(f(1))=f(f(1))+f\left(\frac{f(1)-1}{f(1)}\right)\Rightarrow f\left(\frac{f(1)-1}{f(1)}\right)=0,$
contradiction. Therefore we must have $f(1)=1$ .

$P(x,1)\Rightarrow f(x+1)=f(x)+1\Rightarrow f(x+n)=f(x)+n$ for any $0\le n\in\mathbb Z$ by induction. In particular, since $f(1)=1$ we have $f(n)=n$ for all $n\in\mathbb N$ .

$P(y+1,x)\Rightarrow f(yf(x)+x+f(x))=f(f(x))+xf(y)+x$
$P(y,x)\Rightarrow f(f(x))+xf(y)=f(yf(x)+x)$
Comparing these last two, we get that $f(yf(x)+x+f(x))=f(yf(x)+x)+x$ , or $f(y+x+f(x))=f(y+x)+x$ . Call this assertion $Q(x,y)$ .

$Q(x,1)\Rightarrow f(1+x+f(x))=f(x+1)+x\Rightarrow f(x+f(x))=x+f(x)$
$P(1,x)\Rightarrow f(x+f(x))=x+f(f(x))\Rightarrow f(f(x))=f(x)\forall x$
Fix $x$ and let $n\gg\lceil x\rceil+1$ be some large integer:
$Q(x,n-x)\Rightarrow f(n+f(x))=f(n)+x\Rightarrow f(f(x))=x\Rightarrow\boxed{f(x)=x}$
which is indeed a solution.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1247 posts

ezpotd

#8 h Yesterday at 4:55 PM

Y by

We claim the answer is only $f(x) = x$ for all $x$ . It is trivial to verify these solutions work.

Let $P$ denote the assertion, taking $P(x,1)$ gives $f(xf(1) + 1) = f(f(1)) + f(x)$ , thus we have $f(1) \ge 1$ , otherwise setting $x = \frac{1}{1 - f(1)}$ gives $f(f(1)) = 0$ , impossible. Now if $y < f(y)$ , we can take $P(1 - \frac{y}{f(y)},y)$ to get $yf(x) = 0$ , impossible, so $f(y) \le y$ . This also forces $f(1) = 1$ . Setting $P(x,1)$ then gives $f(x +1) = f(x) + 1$ for all $x$ , so $f(n) = n$ over positive integers. Then we have $f(f(y)) + yf(x) = f(xf(y) + y) \le xf(y) + y$ , giving $xf(y) - yf(x) \ge f(f(y)) - y$ . Taking $x$ to be an arbitrarily large integer, we can then write $x(f(y) - y) \ge f(f(y)) - y$ , which forces $f(y) - y = 0$ otherwise sufficiently large $x$ would break the inequality. Thus $f$ is the identity for all $y$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

5532 posts

#9 h Yesterday at 5:24 PM

Y by

The only solution is $f(x) = x$ , which clearly works. Now we show it's the only one.

Let $P(x,y)$ be the given assertion.

Claim: $f(y) \le y$ for each $y$ .
Proof: Suppose $f(y) > y$ for some $y$ . Setting $x = \frac{f(y) - y}{f(y)}$ in the equation gives $y f(x) = 0$ , a contradiction. $\square$

Claim: $f(1) = 1$ .
Proof: If not, set $x$ so that $xf(1) + 1 = x$ (this is possible because we have $f(1) < 1$ ). We get $f(f(1)) = 0$ , absurd. $\square$

$P(x, 1): f(x+1) = f(x) + 1$ .

$P(x+1, y): f((xf(y) + y) + f(y)) = f(f(y)) + yf(x) + y$ .

Comparing to original FE gives $f((xf(y) + y) + f(y)) - f(xf(y) + y) = y$ This means that for all $x > y$ , $f(x+f(y)) = f(x) + y$ , so $f(x) + y \le x + f(y)\implies x - f(x) \ge y - f(y)$ , so $x - f(x)$ is nonnegative and nondecreasing. Since $f(1) = 1$ , setting $x = 1$ here gives $y - f(y)\le 0$ for all $y < 1$ , so $f(y) = y \forall y < 1$ . Now, combined with the fact that $f(x+1) = f(x) + 1$ , we get that $f$ must be the identity.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a