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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 $ ?

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What you've given isn't enough - you also need that $f$ is continuous. With that, consider the function $g(x) = f(x)-x$ and use the intermediate value theorem... –  Steven Stadnicki Oct 10 '11 at 5:10
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@Steven, did you spot the hypothesis that $f$ must be increasing? –  Did Oct 10 '11 at 5:11
    
@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! –  Steven Stadnicki Oct 10 '11 at 5:18
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2 Answers 2

up vote 4 down vote accepted

Hint: consider $\zeta=\sup Z$ with $Z=\{x\in[a,b]\mid\forall y\in[a,x],f(y)>y\}$. Then $\zeta<b$ (why?), $f(\zeta)>\zeta$ is impossible (why?) and $f(\zeta)<\zeta$ is impossible as well (why?). Ergo.

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"∀y∈[a,x],f(y)>y". why such a interval exists. –  Leitingok Oct 10 '11 at 6:11
    
Try $[a,f(a)]$. –  Did Oct 10 '11 at 6:15
    
thx,that's nice. –  Leitingok Oct 11 '11 at 3:10
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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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why $a_{max}$ exists ? –  Leitingok Oct 10 '11 at 6:09
    
Sorry but $b_{min}=a_{max}$ is not always true. –  Did Oct 10 '11 at 6:17
    
@Leitingok i changed max to sup (witch i meant) –  Hassan Oct 10 '11 at 7:10
    
@DidierPiau why? –  Hassan Oct 10 '11 at 7:10
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Usually the burden of the proof lies on the guy who pretends that something holds, but nevermind. Imagine a continuous S-shaped graph crossing the first diagonal three times on $[a,b]$, at the points $u$, $v$ and $w$ say, with $a<u<v<w<b$. Then $a_{max}=w$ and $b_{min}=u$ hence $b_{min}<a_{max}$. –  Did Oct 10 '11 at 7:39
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