Find all continuous $f:[0,1] \rightarrow [0,1]$ such that $f(1-f(x))=f(x)$.

  • 3
    $\begingroup$ Why? What have you tried? $\endgroup$ – Matthew Conroy Jan 27 '11 at 23:20
  • $\begingroup$ $f(x) = 1-x$, $\forall x \in [0,1]$ or $f(x) = 0$, $\forall x \in [0,1]$ $\endgroup$ – user17762 Jan 27 '11 at 23:26
  • 2
    $\begingroup$ @bobokinks, it really looks like you are posting random functional equations! :) If you explained why you want to know the answer, what you have tried, why you expect that there is a sensible/interesting answer, &c, it'd be nice. $\endgroup$ – Mariano Suárez-Álvarez Jan 27 '11 at 23:36
  • 2
    $\begingroup$ They're fun problems. As long as it doesn't become excessive I don't see why there has to be an explanation. $\endgroup$ – Zarrax Jan 27 '11 at 23:41

Let $m,M$ be the minimum and maximum $f$ achieves on $[0,1]$ (there are such since f is continuous). From the intermediate value theorem, for each $m\leq y \leq M$ there is an $x\in [0,1]$ such that $f(x)=y$, so $f(1-y)=f(1-f(x))=f(x)=y$. This shows that if $m \leq y\leq M$ then $f(1-y)=y$.

for the $ [0,1] - [1-M, 1-m] $ you can extend $f$ any way you want as long as its range is in $[m,M]$.


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.