Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
3  
Why? What have you tried? –  Matthew Conroy Jan 27 '11 at 23:20
    
$f(x) = 1-x$, $\forall x \in [0,1]$ or $f(x) = 0$, $\forall x \in [0,1]$ –  user17762 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-Alvarez Jan 27 '11 at 23:36
2  
They'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 Answer 1

up vote 6 down vote accepted

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 $[m,M]$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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