Find all continuous $f:[0,1] \rightarrow [0,1]$ such that $f(1-f(x))=f(x)$.
Asked
Active
Viewed 263 times
3
-
3Why? What have you tried? – Matthew Conroy Jan 27 '11 at 23:20
-
$f(x) = 1-x$, $\forall x \in [0,1]$ or $f(x) = 0$, $\forall x \in [0,1]$ – Jan 27 '11 at 23:26
-
2@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. – Mariano Suárez-Álvarez Jan 27 '11 at 23:36
-
2They're fun problems. As long as it doesn't become excessive I don't see why there has to be an explanation. – Zarrax Jan 27 '11 at 23:41
1 Answers
7
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 $[m,M]$.
Ofir
- 8,015