f(x + f(y)) is equal to x + f(y) or f(f(x)) + y

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Source: Hong Kong TST - HKTST 2024 2.4

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#1 h Jul 20, 2024, 11:49 PM

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Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ satisfying the following condition: for any real numbers $x$ and $y$ , the number $f(x + f(y))$ is equal to $x + f(y)$ or $f(f(x)) + y$

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#2 h Jul 21, 2024, 7:40 AM

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parmenides51 wrote:

Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ satisfying the following condition: for any real numbers $x$ and $y$ , the number $f(x + f(y))$ is equal to $x + f(y)$ or $f(f(x)) + y$

Let $P(x,y)$ be the assertion $f(x+f(y))\in\{x+f(y),f(f(x))+y\}$

Let $U=\{x$ such that $f(x)\ne x\}$
Let $V=\mathbb R\setminus U=\{x$ such that $f(x)=x\}$

If $U=\emptyset$ , we get solution $f(x)=x\quad\forall x$
If $U\ne\emptyset$ , let $u\in U$ :

$P(u-f(x),x)$ $\implies$ new assertion $Q(u,x)$ : $f(u)=f(f(u-f(x)))+x$ $\forall u\in U,\forall x$
This implies $f(x)$ bijective and so $Q(u,f(u))$ , which is $f(f(u-f(f(u))))=0$ , implies $f(f(u))=u+a$ for some constant $a\ne 0$

And so $\forall x$ : either $f(f(x))=x$ (if $x\in V$ ), either $f(f(x))=x+a$ (if $x\in U$
And $Q(u,x)$ implies $f(u)\in\{u-f(x)+x,u-f(x)+x+a\}$ , $\forall x$

If $\exists v\in V$ (so that $f(v)=v$ ):
Setting $x=v$ in previous line, we get $f(u)\in\{u,u+a\}$ and so $f(u)=u+a$
But then $f(f(u))=u+a$ is $f(u+a)=u+a=f(u)$ , impossible since injective
And so $V=\emptyset$ and $f(f(x))=x+a$ $\forall x$

Then $Q(u,x)$ is $f(u)=u-f(x)+a+x$ $\forall x$ and so $f(x)=x+c$ , which indeed fits, whatever is $c\ne 0$

And solution (grouping the two cases) : $\boxed{f(x)=x+c\quad\forall x}$ , which indeed fits, whatever is $c\in\mathbb R$

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#3 h Sep 15, 2024, 4:39 PM

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The only function that satisfy are $\boxed{f(x)=x+a}$ for any real number $a$ , which can be directly verified to work. We now show that these are the only solutions. Let $P(x,y)$ denote the assertion. Let $f(0)=a$ .

Claim: $f$ is bijective
If $y_1\neq y_2$ and $f(y_1)=f(y_2)$ then comparing the assertions $P(x,y_1)$ with $P(x,y_2)$ gives that we must have $f(x+f(y_1))=x+f(y_1)$ which must mean $f$ is just the identity. Thus $f$ is injective. Then the assertion $P(x,0)$ gives that $f(x+a)=x+a$ or $f(x+a)=f(f(x))$ but as $f$ is injective the last one is equivalent to $f(x)=x+a$ , thus $x+a$ is in the domain of $f$ for all $x$ , making $f$ surjective.

The assertion $P(f(z)-a,0)$ gives that either $f(f(z))=f(z)$ or $f(f(z))=f(f(f(z)-a))$ but as $f$ is injective we must have either $f(z)=z$ or $f(z)=z+a$ . If $a=0$ then we are done so assume $a\neq 0$ . If $f(x)=x$ and $f(w)=w+a$ then chose $y$ such that $f(y)=w-x$ , yielding a contradiction for $P(x,y)$ . This finishes the proof.

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#4 h Mar 10, 2025, 4:53 PM

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We uploaded our solution https://calimath.org/pdf/HongKongTST2024-2.4.pdf on youtube https://youtu.be/90nKrPpO71Y.

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