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

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 3
    $\begingroup$ $|x|$ satisfies the given conditions and isn't a polynomial. $\endgroup$ – lulu Jun 12 at 17:13

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 $x$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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:

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.