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As stated in the title; Let $f: \mathbb R \to \mathbb R $ be a continuous function satisfying $f(f(x))$ = $f(x)$ then

(a) $f$ must be constant

(b) $f(x) = x$ for all $x$ in range of $f$

(c) $f$ must be a non constant polynomial

(d) There is no such function

By randomly trying different functions I discovered that $f(x) = x$ and $f(x) = 1-x$ satisfy given property . So using this option (c) seems to be correct.

But , My question is that How can I make sure that these are the only functions that hold this property ? and if there are any other function (other than these two) then how should I find them .

Thank you

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    $\begingroup$ Note: the header question says $f(f(x))=x$, but in the body of the post you say $f(f(x))=f(x)$. Which did you intend? $\endgroup$ – lulu Jun 12 at 17:00
  • $\begingroup$ @lulu sorry,it was a typo i have edited the question now. $\endgroup$ – sat091 Jun 12 at 17:07
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    $\begingroup$ Are you sure? Given $f(x)=1-x$ we have $f\circ f(x)=f(1-x)=1-(1-x)=x\neq f(x)$. $\endgroup$ – lulu Jun 12 at 17:12
  • $\begingroup$ constant functions also meet the condition $\endgroup$ – Red shoes Jun 12 at 17:12
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    $\begingroup$ $|x|$ satisfies the given conditions and isn't a polynomial. $\endgroup$ – lulu Jun 12 at 17:13
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The answer should be $b$. Suppose not. That means there exists $f$ and there exists $y_0$ such that we have some $x_0$ with $f(x_0)=y_0$ and we have $f(y_0)\neq y_0$.

Then, $f(x_0)=y_0$, but $f(f(x_0))=f(y_0)\neq y_0$, which contradicts the assumtion that $f(f(x))=f(x)$ for all $x$.

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  • $\begingroup$ What you're arguing for, it's (b), not (c). $\endgroup$ – Adam Latosiński Jun 12 at 17:24
  • $\begingroup$ Ah thanks, excuse the mistake. $\endgroup$ – EBP Jun 12 at 17:27
  • $\begingroup$ @EBP but what about $f(x) = \mid x\mid $? $\endgroup$ – sat091 Jun 12 at 17:36
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    $\begingroup$ This statement holds for $f(x)=|x|$. The image of $f(x)$ is $[0,\infty)$, and this is exactly the domain where $f(x)$ itself is the identity. $\endgroup$ – EBP Jun 12 at 17:40
  • $\begingroup$ thanks for the clarification ! $\endgroup$ – sat091 Jun 12 at 17:48
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If $x$ is in the range of $f$ means $x=f(y)$ for some $y$. Since $f(f(y))=f(y)$ you have $f(x)=x$.

Just to see how arbitrary your function may be I give the following example satisfying your conditions:

$$f(x)=\left\{ \begin{array}{cc} x,&|x|\leq 1\\ \sin(\pi x/2),& |x|>1 \end{array}\right. $$

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