Unique Functional Equation with f and g

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

Unique Functional Equation with f and g

G H J

Reveal topic tags

functional equation

L

G H BBookmark kLocked kLocked NReply

Source: FKMO 2022 Problem 3

The post below has been deleted. Click to close.

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

47 posts

#1 h Apr 23, 2022, 11:49 AM • 2 Y

Y by doanquangdang, Rama12

A function $g \colon \mathbb{R} \to \mathbb{R}$ is given such that its range is a finite set. Find all functions $f \colon \mathbb{R} \to \mathbb{R}$ that satisfies $2f(x+g(y))=f(2g(x)+y)+f(x+3g(y))$ for all $x, y \in \mathbb{R}$ .

Z K Y

The post below has been deleted. Click to close.

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

96 posts

#2 h Apr 23, 2022, 2:14 PM • 5 Y

Y by Olympiadium, Acorn-SJ, JG666, Shiro2911, leonhard8128

Cute problem! Don't know why it's P3...

Let $P(x,y): 2f(x+g(y))=f(2g(x)+y)+f(x+3g(y)).$

step1) $f(\mathbb{R})$ is a finite set.
$P(0,y): 2f(g(y))=f(2g(0)+y)+f(3g(y)) \rightarrow f(2g(0)+y)=f(3g(y))-2f(g(y))$ .
since $g(y)$ can have finite number of values when $2g(0)+y$ can be any number in $\mathbb{R}$ , $f(\mathbb{R})$ is a finite set.

Let $f(\mathbb{R})={a_1,a_2, ...,a_n}$ $(a_1<a_2< ... <a_n)$ .

step2) If $f(a)=a_n$ , $f(a+2g(y))=a_n$ for all $y$ in $\mathbb{R}$ .
Just see $P(a-g(y),y)$ .

step3) $f(x)=constant.$
if $n\ge 2:$
There exist $b,a$ such that $f(b)=a_{n-1}$ and $f(a)=a_n$ .
$P(b-3g(a),a): 2f(b-2g(a))=f(2g(b-3g(a))+a)+f(b)=a_{n-1}+a_n$ which is a contradiction!

Z K Y

The post below has been deleted. Click to close.

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

1301 posts

#3 h Apr 25, 2022, 1:48 AM • 1 Y

Y by JG666

The answer is constant functions only, which obviously works.

Let $P(x,y)$ denote the assertion in the problem, $Im(g)=\{a_1, \cdots,a_n\}$ , and let $T_j=\{ y | g(y)=a_j\}$

Claim 1: $|Im(f)| \le |Im(g)|$

$P(x, t_j)$ where $t_j$ varies over all elements of $T_j$ gives $f(2g(x)+t_j)=2f(x+a_j)-f(x+3a_j)$ which is constant as $t_j$ varies over $T_j$ . Since $T_1 \cup T_2 \cup \cdots \cup T_n = \mathbb{R}$ , it follows that $Im(f) \subseteq \{ f(2g(x)+t_j) | 1\le j\le n\}$ so $Im(f)$ .

Let $Im(f)=\{ x_1,\cdots,x_m\}$ . WLOG $x_1=\max\{ x_1,\cdots,x_m\}, x_m=\max\{x_1,\cdots,x_m\}$ and $X_j=\{ x| f(x)=x_j\}$ . Now that $Im(f)$ is finite, if $f(x+g(y))$ is maximal, so are $f(x+3g(y))$ and $f(2g(x)+y)$

Select $x\in X_1-a_j$ . Since $x+a_j\in X_1, f(x+a_j)=x_1$ . By maximality of $x_1$ , it follows that $f(2g(x)+t_j)=x_1$ for all $t_j\in T_j$ (i.e. $2g(x)+T_j \subseteq X_1$ )

If $x+g(y)\in X_1$ then it must follow that $x+3g(y)\in X_1$ and $2g(x)+y\in X_1$ .

This means $X_1+2a_j \subseteq X_1$ because if $z\in X_1, P(z-g(y),y)$ gives $f(z+2g(y))=x_1$ . (1)

Now, select $x\in X_1, z\in X_m$ .

$P(x-g(y),y): 2g(x-g(y))+y \in X_1$ . By (1), $2g(x-g(y))+2g(z-g(y))+y\in X_1$

$P(z-g(y),y): 2g(z-g(y))+y \in X_m$ . By (1), $2g(x-g(y))+2g(z-g(y))+y\in X_m$

It follows that $x_1=x_m$ , so $f$ is constant.

This post has been edited 3 times. Last edited by CANBANKAN, Apr 25, 2022, 1:52 AM

Z K Y

The post below has been deleted. Click to close.

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

1301 posts

#4 h Jun 7, 2022, 6:19 PM • 1 Y

Y by JG666

Clarification:

CANBANKAN wrote:

Select $x\in X_1-a_j$ . Since $x+a_j\in X_1, f(x+a_j)=x_1$ . By maximality of $x_1$ , it follows that $f(2g(x)+t_j)=x_1$ for all $t_j\in T_j$ (i.e. $2g(x)+T_j \subseteq X_1$ )

If $x+g(y)\in X_1$ then it must follow that $x+3g(y)\in X_1$ and $2g(x)+y\in X_1$ .

This means $X_1+2a_j \subseteq X_1$ because if $z\in X_1, P(z-g(y),y)$ gives $f(z+2g(y))=x_1$ . (1)

Here for a set $X$ and a real $c$ , $X+c$ doesn't mean $X\cup \{c\}$ but $\{a+c| a\in X\}$

Similarly, $X-c$ doesn't mean $X\backslash \{c\}$ but $\{a-c| a\in X\}$ .

Sorry for the possible confusion, and hopefully this clears things up. In the future, when I use notation, I will define them.

This post has been edited 1 time. Last edited by CANBANKAN, Jun 7, 2022, 6:19 PM

Z K Y

The post below has been deleted. Click to close.

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

52 posts

#5 h Jun 9, 2022, 3:52 PM • 1 Y

Y by hakN

We see that f has finitely many values by $P(0,x)$ as also done by above, we define $\alpha, \beta$ such that $f(\alpha)=a, f(\beta)=b$ where $a,b$ are respectively the maximum and minimum values that f can take.
$P(\alpha - g(\beta - g(\alpha)), \beta - g(\alpha)) \implies f( \alpha + 2g(\beta - g(\alpha)))=a$ $P(\beta - g(\alpha), \alpha) \implies f(\alpha + 2g(\beta - g(\alpha))) = b$ so $\alpha = \beta$ , $f(x)=c$ for some real $c$ .

This post has been edited 2 times. Last edited by Cookierookie, Jun 9, 2022, 4:02 PM
Reason: Edit

Z K Y

The post below has been deleted. Click to close.

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

67 posts

#6 h Jun 10, 2022, 9:53 PM

Y by

Slightly different solution:
Claim
$f$ has finite range.
Proof
Notice $2f(x+g(y))-f(x+3g(y))=f(2g(x)+y)$ , move $y$ over all reals, the LHS takes finitely many values, and the RHS is $f$ of any real number.

As a corrolary of this, $f$ has an absolute maximum. Say $k$ is cool if $f(k)$ achieves this maximum. The idea is that if $x+g(y)$ is cool, then $x+3g(y)$ and $2g(x)+y$ are also cool.
Claim
For all real $y$ , there's an $x$ such that $2g(x)+y$ is cool.
Proof
Basically choose the $x$ such that $x+g(y)$ is cool.

Claim
If $k$ is cool, then $k+2g(y)$ is also cool for any $y\in\mathbb{R}$ .
Proof
Plug $x=k-g(y)$ in to the equation.

Now just notice that combining the two claims, we get that if $s$ is the sum of the elements of $g(\mathbb{R})$ , then $y+2s$ is cool for all $y$ , thus all reals are cool and $f$ is constant, which indeed works.

Z K Y

The post below has been deleted. Click to close.

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

27 posts

Ruy

#7 h Sep 19, 2022, 9:34 PM

Y by

Pretty much concise solution

This post has been edited 1 time. Last edited by Ruy, Sep 19, 2022, 9:42 PM
Reason: typo

Z K Y

The post below has been deleted. Click to close.

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

ZETA_in_olympiad

2211 posts

ZETA_in_olympiad

#8 h Oct 7, 2022, 4:59 PM • 1 Y

Y by Mango247

Easy to see that $f$ takes finitely many values. Let $f(X)$ be the maximum value. Setting $x=X-g(x-3g(X))$ and $y=x-3g(X),$ we deduce $f(X+2g(x-3g(X)))=f(X).$ And setting $x=x-3g(X)$ and $y=X,$ we get $f(x)=f(X)$ for all $x.$ Any constant function works.

Z K Y

The post below has been deleted. Click to close.

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

ZETA_in_olympiad

2211 posts

ZETA_in_olympiad

#9 h Oct 7, 2022, 5:09 PM

Y by

CANBANKAN wrote:

$x_1=\max\{ x_1,\cdots,x_m\}, x_m=\max\{x_1,\cdots,x_m\}$

Typo, one should be min

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a