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 $f(x)=-x$.
Best Answer
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.