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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.

Mohsen Shahriari
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Vann
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  • 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. – Ninad Munshi Apr 13 '20 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 $f(\Bbb{R}^+) \subset \Bbb{R}^+$ – Ninad Munshi Apr 13 '20 at 15:17
  • $f(x)=\max{x,1,1-x}$ satisfies the functional equation and has range$[1,\infty)$. – Hagen von Eitzen Apr 13 '20 at 15:18
  • @HagenvonEitzen The $1-x$ portion doesn't work, but I like where you are thinking. – Ninad Munshi Apr 13 '20 at 15:21

1 Answers1

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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.

Hagen von Eitzen
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