Find all injective $f:\mathbb{R} \rightarrow \mathbb{R}$ satisfying $f(x+f(f(-y)))=f(x)+f(f(y))$

Question

Find $f:\mathbb{R} \rightarrow \mathbb{R}$ satisfying these two conditions: \begin{align} &(i) \ x\neq y \Rightarrow f(x) \neq f(y). \\ &(ii) \ f(x+f(f(-y)))=f(x)+f(f(y)). \end{align}

From (i), we get that function $f$ is an injective function.

My attempt: \begin{align} &\text{let } P(x, y): f(x+f(f(-y)))=f(x)+f(f(y)) \\ \ \\ &\text{let } f(0)=t. \\ \ \\ &P(x, 0): f(x+f(f(0)))=f(x)+f(f(0)). \\ & \Rightarrow f(x+f(t))=f(x)+f(t). \\ &x=0; \ f(f(t))=t+f(t). \\ &x=t; \ f(t+f(t))=2f(t). \\ &\Rightarrow f(f(f(t)))=2f(t) \\ \ \\ &P(0, y): f(f(f(-y)))=t+f(f(y)). \\ &y=t; \ f(f(f(-t)))=t+f(f(t)) \Rightarrow f(f(f(-t)))=2t+f(t). \\ &y=-t; \ f(f(f(t)))=t+f(f(-t)) \Rightarrow 2f(t)=t+f(f(-t)). \end{align}

• Still, I didn't use the injective one.
• My expectation is $f(x)=-x$.

Answer 1

It's straightforward to verify that for any constant $ a \in \mathbb R $, the function $ f : \mathbb R \to \mathbb R $ defined with $ f ( x ) = a - x $ for all $ x \in \mathbb R $, is injective and satisfies $$ f \Bigl ( x + f \bigl ( f ( - y ) \bigr ) \Bigr ) = f ( x ) + f \bigl ( f ( y ) \bigr ) \tag 0 \label 0 $$ for all $ x , y \in \mathbb R $. To prove that those are the only solutions, let $ a = f ( 0 ) $ and $ b = a - f ( a ) $. Putting $ x = b $ and $ y = 0 $ in \eqref{0} we get $ f ( b ) = 0 $. Then, setting $ x = b $ in \eqref{0} and using injecivity of $ f $ we have $$ f \bigl ( f ( - y ) \bigr ) = f ( y ) - b $$ for all $ y \in \mathbb R $. This lets us rewrite \eqref{0} as $$ f \bigl ( x + f ( y ) - b \bigr ) = f ( x ) + f ( - y ) - b \text , $$ which by substituting $ - y $ for $ y $ yields $$ f \bigl ( x + f ( - y ) - b \bigr ) = f ( x ) + f ( y ) - b \tag 1 \label 1 $$ for all $ x , y \in \mathbb R $. Interchanging $ x $ and $ y $ in \eqref{1} and comparing with \eqref{1} itself, we get $$ f \bigl ( x + f ( - y ) - b \bigr ) = f \bigl ( y + f ( - x ) - b \bigr ) \tag 2 \label 2 $$ for all $ x , y \in \mathbb R $. Substituting $ - x $ for $ x $ and $ 0 $ for $ y $ in \eqref{2} and using injectivity of $ f $, we can conclude $ f ( x ) = a - x $ for all $ x \in \mathbb R $, as desired.

Answer 2

In the following, I will write $f^{k}=f\circ f\circ\dots\circ f$.

Any function of the form $f(x)=a-x$, for $a\in\mathbb R$ is a solution. In particular, $f(0)=a$. It is obvious that such a function satisfies the equation, indeed $f_{a}^2(-y)=f_{a}(a+y)=a-(a+y)=-y$, hence $$f_{a}(x+f^2(-y))=f_{a}(x-y)=a-(x-y)=(a-x)+y=f_{a}(x)+f_{a}^2(y).$$

To prove that functions of this form are the only solutions, rewrite the condition $f(x+f^2(-y))=f(x)+f^{2}(y)$ as $f(x)=-f^{2}(y)+f(x+f^2(-y))$. For $x=0$ one obtains \begin{align} f^2(0)&=f\bigg(-f^{2}(y)+f^2\big(f(-y)\big)\bigg)=f\big(-f^2(y)\big)+f^2\big(-f(-y)\big). \end{align} Apply $f$ again, hence \begin{align} f^3(0)&=f(f^2(0))=f\bigg(f\big(-f^2(y)\big)+f^2\big(-f(-y)\big)\bigg)=\\ &=f^2\big(-f^2(y)\big)+f^2\big(f(-y)\big)=f^2\big(-f^2(y)\big)+f^3(-y). \end{align} If $y=0$, the equation states $f^2(-f^2(0))=0$.

It follows that \begin{align} f(x)=f(x+0)=f\bigg(x+f^2\big(-f^2(0)\big)\bigg)=f(x)+f^2\big(f^2(0)\big). \end{align} Hence $f^2(f^2(0))=0=f^2(-f^2(0))$, by injectivity $f^2(0)=-f^2(0)$, namely $f^2(0)=0$.

It follows that \begin{align} f^4(-y)=f^2\big(0+f^2(-y)\big)=f\big(f(0)+f^2(y)\big)=f^2(0)+f^2(-y)=f^2(-y), \end{align} which implies by injectivity that $f^2(-y)=-y$.

Finally \begin{align} f(0)=f(y-y)=f\big(y+f^2(-y)\big)=f(y)+f^2(y)=f(y)+y, \end{align} that is $f(y)=f(0)-y$, which ends the proof.