If $k\in\Bbb R$ is fixed, find all $f:\Bbb R\to\Bbb R$ that satisfy $f(f(x))=k$ for all real $x$.
If $k\ge 0$, $f(x)=|k+g(x)-g(|x|)|$ is a solution for any $g:\Bbb R\to\Bbb R$.
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user263326
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This only works for nonnegative $k$. Moreover, you have not asked a question. What do you want to know? – Hrodelbert Sep 02 '15 at 08:06
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@Hrodelbert. Clearly, he wants to find all $f:\Bbb R\to\Bbb R$ that satisfy $f(f(x))=k$ for all real $x$. – wythagoras Sep 02 '15 at 08:07
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@wythagoras In my interpretation, the first line is an exercise from somewhere (since it is written using an imperative), the second is a partial answer. If he wants us to solve it, he should say so and give some more background. – Hrodelbert Sep 02 '15 at 08:51
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It's a problem I've come up with myself. – user263326 Sep 02 '15 at 12:23
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Choose arbitrary subsets $A,B\subseteq\mathbb R$ such that $A\cup B=\mathbb R$ and $A\cap B=\emptyset$. We may assume without loss of generality that $k\in B$. Let $g:A\to B$ be an arbitrary function. Then define $$f(x)=\begin{cases}k;&x\in B,\\ g(x);&x\in A.\end{cases}\tag{$\ast$}$$ This clearly satisfies the requirements.
Conversely, if $f:\mathbb R\to\mathbb R$ is any function that satisfies $f(f(x))=k$, we may choose $B=f(\mathbb R)$ and $A=\mathbb R\setminus B$ which shows that $f$ is of the form given above.
So $(*)$ in fact gives all possible solutions.
Dejan Govc
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