Let $f:\mathbb R\to \mathbb R$ is strickly increasing function . And $a< f(a)< f(b)< b\ \ $for $ \ a,b\in \mathbb R$. How to show that $\exists \zeta\in (a,b),f(\zeta)=\zeta $ ?
Question about increasing function
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0@Didier Ahhh - yes, that was just a thinko on my part. I was thinking that without continuity we could just hide the 'crossover point' at a discontinuity, but of course that doesn't work here. Thank you! – 2011-10-10
2 Answers
Hint: consider $\zeta=\sup Z$ with $Z=\{x\in[a,b]\mid\forall y\in[a,x],f(y)>y\}$. Then $\zeta (why?), $f(\zeta)>\zeta$ is impossible (why?) and $f(\zeta)<\zeta$ is impossible as well (why?). Ergo.
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0thx,that's nice. – 2011-10-11
consider this set $A=\{x \in (a,b) / f(x)\geq x\}$
$A \neq \varnothing $ because $a \in A$
$b \notin A$
let $a_{sup}$ be the superior bond of A
consider $B= \{x \in (a_{sup},b) / f(x)\leq x\}$
and $b_{inf}$ it inferior bond
let $\zeta=\frac{b_{inf}+a_{sup}}{2}$
(i) $b_{inf}=a_{sup}=\zeta$
suppose $b_{inf} \neq a_{sup}$
then $b_{inf} > a_{sup}$
and $b_{inf}\geq f(b_{inf})>f(\zeta)>f(a_{sup})\geq a_{sup}$
( $b_{inf}\geq f(b_{inf})$ because $b_{inf}$ is not in A; same to $f(a_{sup})\geq a_{sup}$)
since $A \cup B = (a,b)$ because we always have $\forall x \in (a, b)$ ether $f(x)\geq x$ or $f(x)\leq x$
then $\zeta$ is in A or in B
in both case we get $\zeta\geq b_{inf}$ or $\zeta\leq a_{sup}$
witch is contradictory
(ii)$f(\zeta)=\zeta$
$\zeta \in A \bigcap B$ so $f(\zeta)=\zeta$
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0yes, you are right; i looking to fixe that – 2011-10-10