# Proving surjectivity from $f\Bigl(\bigl(x+f(x)\bigr)^2\Bigr) = \bigl(x+f(x)\bigr)^2$ in a functional equation problem

Can anyone help me with this?

If, in a functional equation problem, I do some substitutions and found out that $$f\Bigl(\bigl(x+f(x)\bigr)^2\Bigr) = \bigl(x+f(x)\bigr)^2$$, in which $$f(x)$$ is a function that maps a real number to a real number $$(f:\mathbb{R}\to\mathbb{R})$$, can we imply that $$f(x)$$ is surjective over the positive reals?

Thanks for any help.

• At the very least we have that $f(\Bbb{R}^+) \subset \Bbb{R}^+$. I think a good route of investigation would be to ask if it's possible for $0$ to be neglected in the range. Commented Apr 13, 2020 at 15:08
• Here is some progress: The only polynomial solutions are $x$, $-x$ and $\frac{1}{2}-x$, but only one of them satisfies $f(\Bbb{R}^+) \subset \Bbb{R}^+$ Commented Apr 13, 2020 at 15:17
• $f(x)=\max\{x,1,1-x\}$ satisfies the functional equation and has range$[1,\infty)$. Commented Apr 13, 2020 at 15:18
• @HagenvonEitzen The $1-x$ portion doesn't work, but I like where you are thinking. Commented Apr 13, 2020 at 15:21

Define $$f\colon \Bbb R\to\Bbb R$$ as $$f(x)=\max\{x,1,1-x\}=\begin{cases}x&x\ge1\\1&0\le x\le 1\\1-x&x\le 0\end{cases}$$ Then $$f((x+f(x))^2) = (x+f(x))^2$$ holds for all $$x\in \Bbb R$$. Indeed, for $$x\le 0$$, we have $$f((x+f(x))^2)=f((x+(1-x))^2)=f(1)=1=(x+(1-x))^2.$$ For $$0\le x\le 1$$, we have $$f((x+f(x))^2)=f(\underbrace{(x+1)^2}_{\ge1})=(x+1)^2=(x+f(x))^2.$$ For $$x\ge 1$$, we have $$f((x+f(x))^2)=f(\underbrace{4x^2}_{>1})=4x^2=(x+f(x))^2.$$
Clearly, $$f(x)\ge1$$ for all $$x$$, so that the desired surjectivity over the positive reals does not hold.