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

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    $\begingroup$ Why? What have you tried? $\endgroup$ Jan 27, 2011 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, 2011 at 23:26
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    $\begingroup$ @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. $\endgroup$ Jan 27, 2011 at 23:36
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    $\begingroup$ 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, 2011 at 23:41

1 Answer 1

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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 $[m,M]$.

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