Identical or Periodic?

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Identical or Periodic?

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Source: India EGMO TST 2023/4

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1184 posts

L567

#1 h Dec 10, 2022, 4:29 PM • 2 Y

Y by dangerousliri, ItsBesi

Let $f, g$ be functions $\mathbb{R} \rightarrow \mathbb{R}$ such that for all reals $x,y$ , $f(g(x) + y) = g(x + y)$ Prove that either $f$ is the identity function or $g$ is periodic.

Proposed by Pranjal Srivastava

This post has been edited 1 time. Last edited by L567, Dec 10, 2022, 7:00 PM

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L567

#2 h Dec 10, 2022, 4:36 PM

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If $g(x) = x+c$ for some constant $c$ , then $f(x+y+c) = x+y+c$ so it is the identity, so ignore this case.

Then, there exists some $k$ such that $p = g(k)-k-g(0) \neq 0$ . Then we have that $f(g(k) + y) = f(k+y) = f(g(0) + k+y)$ so $f$ is periodic with period $p$ .

Then $g(y+p) = f(g(0)+y+p) = f(g(0)+y) = g(y)$ so $g$ is also periodic, as desired. $\blacksquare$

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5608 posts

#4 h Dec 10, 2022, 5:07 PM

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

Notice that by $P(x,0)$ , $f(g(x)) = g(x)$ , so if $g$ is surjective, then $f$ is the identity function. Now we assume $g$ is not surjective.

$P(0,x): f(x+g(0)) = g(x)$ .

Case 1: $g$ is injective.
Then if $f(a) = f(b)$ , then $P(0,a-g(0))$ compared with $P(0,b-g(0))$ gives $g(a-g(0)) = g(b-g(0))$ , so $a=b$ .

The original FE can be rewritten as $f(g(x) + y)= f(x +y+g(0))$ , so using $f$ injective, we get $g(x) = x + g(0)$ , which is surjective, contradiction.

Case 2: $g$ is not injective.
Then $g(a) = g(b)$ for some $a\ne b$ . Comparing $P(a,x)$ and $P(b,x)$ gives $g(a+x) = g(b+x)$ for all reals $x$ , so $g(x) = g(x + (b-a))$ , which implies $g$ is periodic.

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#5 h Dec 10, 2022, 5:14 PM • 1 Y

Y by dangerousliri

Solution: The solution will be a casework whether $g$ is injective or not. We shall prove for all injective $g$ , $f$ must be the identity and $g$ must be periodic if $g$ is non-injective. Let $P(x,y)$ be the assertion to the Functional Equation.

Case 1: $g$ is an injective function.

$P(x,-g(x))$ gives $f(0) = g(x-g(x))$ . The final trick is to consider $P(0,-g(0))$ which gives you $f(0) = g(-g(0))$ . By injectivity you get $g(x) = x + g(0)$ . Putting this back into the original equation, we yield $f(x+y+g(0)) = x+y+g(0)$ which gives $f \equiv \text{id}$ as desired.

Case 2: $g$ is a non-injective function.

By assumption, there exists distinct $a,b$ in $\mathbb{R}$ such that $g(a) = g(b)$ . Consider $P(x,a)$ and $P(x,b)$ . This would give
$f(g(a) + y) = f(g(b) + y) = g(a+ y) = g(b+ y)$ This nicely rearranges to $g(y) = g(y+a-b)$ which shows $g$ is periodic function.

We have exhausted all cases so the solution is complete. $\blacksquare$

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#6 h Dec 10, 2022, 10:35 PM

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Let $P(x,y)$ be the given assertion.
$P(0,x-g(0))\implies f(x)=g(x-g(0))$
$P(x,y-x)\implies g(g(x)-x-g(0)+y)=g(y)$
Now either $g$ is periodic, or $g(x)=x+g(0)$ in which case $f(x+y+g(0))=x+y+g(0)$ and $f$ is the identity.

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Leo890

#7 h Dec 11, 2022, 12:14 AM

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Observe that $f(g(x))= g(x)$ and $f(g(0)+y)=f(g(y))$ .
Suppose that for each pair of real numbers $(x,p)$ we have $g(x) \neq f(g(x)+p) = g(x+p)$ .
Thus $f$ must be injective by below, so $g(y)= g(0)+y \implies f(g(0)+y)=f(g(y))= g(y) = g(0) + y$ , hence $f$ is the identity function.
If there exists such pair, then $g(x+p) = g(x) \implies g(x)+a = g(x+p)+a \implies f(g(x)+a) = f(g(x+p)+a) \implies g(a+x) = g(x+a+p)$ for any arbitrary real $a$ . Hence $g$ is periodic.

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#8 h Jan 11, 2024, 4:53 PM

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Let us assume that g is not periodic. We have to prove that in this case, f is the identity function.
Let $P(x, y)$ denote the given assertion.
$P(0, y)$ gives us $f(y+g(0)) = g(y) \implies f(y) = g(y-g(0))$ .
Substituting this in the original equation,
$g(y + g(x) - g(0)) = g(y+x)$ .
Note that if for some $r$ , if $g(r) - g(0) \ne r$ , then $|g(r) - g(0) - r|$ is a multiple of the period of $g$ , so $g$ becomes periodic.
Therefore for all r, $g(r) - g(0) = r \implies g(r) = r + g(0)$ . Now $f(r) = g(r-g(0)) = r - g(0) + g(0) = r$ , which yields that $f$ is indeed the identity function.

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Jndd

#9 h Jul 25, 2024, 9:00 PM

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Plug in $y=0$ to get $f(g(x))=g(x)$ , which gives $f(g(x)+y)=g(x+y)=f(g(x+y)).$ We will do cases on whether $f$ is injective. If it is, then we have $g(x)+y=g(x+y)$ , and plugging in $x=0$ gives us $g(0)+y=g(y)$ . Thus, $g(x)=x+f(0)$ for all $x$ , which gives $f(g(x)+y)=f(x+y+f(0))=g(x+y)=x+y+f(0).$ Thus, $f$ must be the identity.

If $f$ is not injective, then $f$ clearly cannot be the identity. There must be some $a\neq b$ such that $f(a)=f(b)$ . For each $x$ , there is some $y_1$ and $y_2$ such that $g(x)+y_1=a$ and $g(x)+y_2=b$ , so $g(x+y_1)=f(a)=f(b)=g(x+y_2),$ giving $g(x)=g(x+(y_2-y_1)) = g(x+(b-a))$ for all $x$ . Thus, $g$ is periodic.

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#10 h Jul 26, 2024, 8:32 PM

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Suppose that $g$ is not periodic then let $P(x,y)$ the assertion of the F.E.
$P(x,y-g(x))$ gives $f(y)=g(y+x-g(x))$ , now let $x$ vary among all reals to get that $g(x)-x$ is constant therefore $g(x)=x+c$ which by replacing means $f(x+y+c)=x+y+c$ or $f= \text{id}$ thus we are done :cool:

:cool:

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#11 h Oct 25, 2024, 10:23 AM • 2 Y

Y by pb_ana, Kaus_sgr

Nice and easy FE

We will treat two cases: $g$ is injective and $g$ is not injective.

Case 1: $g$ is injective
plugging in $y=-g(x)$ we get: $f(0)=g(x-g(x))$ , so, by the injectivity of $g$ , $x-g(x)$ is equal to a constant $c$ , $\forall x \in \mathbb{R}$ .
Therefore $g$ is also surjective, so setting $y=0$ we get $f(g(x))=g(x)$ , therefore $f$ is the identity function.

Case 2: $g$ is not injective
Then, there exist some $a, b \in \mathbb{R}$ such that $a \neq b$ but $f(a)=f(b)$ .
Therefore, setting $x=a$ and then $x=b$ we get that $g(a+y)=g(b+y)$ , $\forall y \in \mathbb{R}$ . Therefore, $g$ is periodic with period $(a-b)$ , as desired

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#12 h Mar 14, 2025, 12:53 AM

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This post has been edited 2 times. Last edited by Marcus_Zhang, Mar 14, 2025, 12:54 AM

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#13 h Mar 16, 2025, 4:21 PM

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First, suppose that $g(x)$ is not injective. Then there is some $a, a+T$ such that $g(a)=g(a+T)$ where $T \neq 0.$ Then $g(a+T+y)=f(g(a+T)+y)=f(g(a)+y)=g(a+y).$ Therefore since $y$ can vary $g(x)$ is periodic.

Now, suppose that $g(x)$ is injective. Then $y=-g(x) \implies f(0)=g(x-g(x)).$ Hence $x-g(x)=c$ for all real $x$ and a constant $c,$ therefore $f(x+y+c)=x+y+c$ so $f(x)$ is the identity function. QED

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#14 h Apr 25, 2025, 8:05 AM

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We will do casework on if $g$ is injective or not.

If $g$ is injective, we plug in $y = -g(x)$ to see that $f(0) = g(x-g(x)),$ and by plugging in $x = 0$ to that, we have that $f(0) = g(x - g(x)) = g(- g(0)),$ and hence $g(x) = x + g(0).$ Thus, we have that $f$ is identity.

If $g$ is not injective, let $a, b$ be values such that $g(a) = g(b).$ By plugging in $x = a, x = b,$ we see that $g(a + y) = f(g(a) + y) = f(g(a) + y) = g(b + y).$ Thus, we see that $g$ is periodic, with frequ3ence of $|a - b|.$

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