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Find all the function satisfy: $$f(f(x)+y)=2x+f(f(y)-x), \forall x , y \in \mathbb{R}$$

I have tried that: Let $x:=-x $ we have : $$f(f(y)+x)=2x+f(f(-x)+y) ,(1) $$ Then in $(1)$ $x:=y;y:=x$ we have : $$f(f(x)+y)=2y+f(f(-y)+x)$$ So : $$2x+f(f(y)-x)=2y+f(f(-y)+x), (2)$$ Then let :$2x=f(y)-f(-y)$ we have : $$f(y)-f(-y)=2y$$ And stuck !

Edit: This is IMO2002 Shortlisted Problem A1 (CZE).

Sil
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Haruboy15
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3 Answers3

12

For convenience, let $P(x, y)$ represent the equation $f(f(x)+y)=2x+f(f(y)-x)$.

$P(x, -f(x))$ gives $f(f(-f(x))-x)=f(0)-2x$. Thus $f(x)$ is surjective.

Suppose $f(a)=f(b)$ for some $a, b$. Then $P(x, a)$ and $P(x, b)$ give $$f(f(x)+a)=2x+f(f(a)-x)=2x+f(f(b)-x)=f(f(x)+b)$$ Since $f(x)$ is surjective, $f(x+a)=f(x+b) \, \forall x \in \mathbb{R}$.

Now using the above equation and $P(a, y), P(b, y)$ give

\begin{align} 2a=f(f(a)+y)-f(f(y)-a) & =f(f(a)+y)-f(f(y)-(a+b)+b) \\ &=f(f(a)+y)-f(f(y)-(a+b)+a) \\ & =f(f(b)+y)-f(f(y)-b) \\ & =2b \end{align}

Thus $f(x)$ is also injective.

Finally, $P(0, y)$ gives $f(y+f(0))=f(f(y))$, so since $f(x)$ is injective, $f(y)=y+f(0) \, \forall y \in \mathbb{R}$. Checking, this is indeed a family of solutions.

Ivan Loh
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Purely from a calculus perspective:

Taking the derivative of the equation with respect to $x$, we have $$ f'(f(x)+y)f'(x) = 2-f'(f(y)-x) $$ Taking the derivative with respect to $y$ instead, we have $$ f'(f(x)+y) = f'(f(y)-x)f'(y) $$ Combining these, we can note that

$$ f'(f(y)-x)(f'(x)f'(y)+1) = 2 $$ When $x=f(y)$, this becomes $$ f'(f(y))f'(y) = \frac{2}{f'(0)}-1 $$ Which we may integrate to get

$$ f(f(y)) = \left(\frac{2}{f'(0)}-1\right)y+C $$ From the original equation, when $x=0$, we have $$ f(f(0)+y) = f(f(y)) $$ So we have $$ f(f(0)+y) = \left(\frac{2}{f'(0)}-1\right)y+C $$ Or $$ f(y) = \left(\frac{2}{f'(0)}-1\right)y+\hat C $$ Now, taking the derivative with respect to $y$ at $y=0$, we have $$ f'(0) = \left(\frac{2}{f'(0)}-1\right) $$ Which tells us that $f'(0)=1$ or $f'(0)=-2$. But we also have $f'(y)=f'(0)$, and so $$ f'(0)^2 = \frac{2}{f'(0)}-1 $$ Which immediately tells us that $f'(0)=1$. Therefore, $$ f(x) = x + \hat C $$ And this is the only solution family to the equation.

Glen O
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Since this is an IMO2002 SL problem, here is a solution from The IMO Compendium A Collection of Problems Suggested for The International Mathematical Olympiads: 1959-2009:

Notice that $f$ is surjective. Indeed, setting $y=-f(x)$ gives $f(f(-f(x))-x)=f(0)-2x$, where the right-hand expression can take any real value.

In particular, there exists $x_0$ for which $f(x_0)=0$. Now setting $x=x_0$ in the functional equation yields $f(y)=2x_0+f(f(y)-x_0)$, so we obtain $$ f(z)=z-x_0\mbox{ for }z=f(y)-x_0. $$ Since $f$ is surjective, $z$ takes all real values. Hence for all $z$, $f(z)=z+c$ for some constant $c$, and this is indeed a solution.

Sil
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