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Let $f:[0,1] \rightarrow [0,1]$ be such that $|f(x)-f(x')| <|x-x'|$ for all $x,x'\in [0,1]$ with $x \not= x'$. Show that there is a unique point $t\in [0,1]$ such that $f(t)=t$.

I noticed that $f$ is a Lipschitz function so that $f$ is a uniformly continuous function on $[0,1]$. By the intermediate value theorem, there exists at least one $t \in [0,1]$, such that $f(t)=t$.

To show uniqueness, how would like to show that $f$ is strictly monotonic on $[0,1]$. That's where I'm stuck... I tried using sequences : $|x_{n+1} > x_n|$ (or $<$). Is that the way to go?

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    $\begingroup$ Suppose that $x$ and $x'$ are two fixed points. Feed them into your condition on $f$. $\endgroup$ – Stephen Montgomery-Smith Nov 18 '13 at 1:19
  • $\begingroup$ @StephenMontgomery-Smith, I don't understand your hint. You want me to feed them into $|f(x)-f(x')| < |x-x'|$? Is the goal to show monotonicity or $f(t)=t$? $\endgroup$ – Justin D. Nov 18 '13 at 1:21
  • $\begingroup$ The intermediate value theorem does not quite do the trick. You must consider the cases where f(x) > x on (0,1) and similarly where f(x) < x. If you consider what happens with the function f(x) = $x^2$ you will get a clue how this strong continuity helps you.,? $\endgroup$ – Betty Mock Nov 18 '13 at 1:21
  • $\begingroup$ @JustinD. The goal is to show $x = x'$. Since $x$ is a fixed point, you know $f(x) = x$. Similarly with $x'$. $\endgroup$ – Stephen Montgomery-Smith Nov 18 '13 at 1:23
  • $\begingroup$ @BettyMock, I am actually using Bolzano's Intermediate Value Theorem which I think does not require the endpoints to be of different signs $\endgroup$ – Justin D. Nov 18 '13 at 1:23
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Suppose $t$ and $t'$ are distinct fixed points of $f$. Then $f(t) = t$ and $f(t') = t'$ so $|f(t) - f(t')| = |t - t'|$, contradicting $|f(x) - f(x')| < |x - x'|$ for all $x, x' \in [0, 1]$ with $x \neq x'$.

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Let arbitrary $x_1\in [0,1]$.

And suppose that $|f(x)-f(x')| <a|x-x'|$ for $a<1$.

Define $x_2=f(x_1),...,x_{n+1}=f(x_n)$.

Then $x_n$ is Cauchy sequence,because $|x_{n+1}-x_n|=|f(x_n)-f(x_{n-1})|\leq a|x_n-x_{n-1}|=a|f(x_{n-1}-f(x_{n-1}|\leq a^2|x_{n-1}-x_{n-2}|\leq ...a^{n-1}|x_2-x_1|$.

Let now $m>n $,$m,n\Bbb N$. We have that $|x_m-x_n|\leq |x_m-x_{m-1}|+|x_{m-1}-x_{m-2}|+...+|x_{n+1}-x_n|\leq a^{m-2}|x_2-x_1|+a^{m-3}|x_2-x_1|+...+a^{n-1}|x_2-x_1|=|x_2-x_1|a^{n-1}[1+a+...+a^{m-n-1}]\leq \frac {|x_2-x_1|}{a(1-a)}a^n$ with $M=\frac {|x_2-x_1|}{a(1-a)}$.

Let $ε>0$. Then there is a $n_0\in \Bbb N:Ma^{n_0}<ε$. Then $\forall m>n>n_0:|x_m-x_n|\leq Ma^n\leq Ma^{n_0}<ε$.

Because $[0,1]$ is complete we have that $x_n$ converges to a $x_0\in [0,1]$.

So, $x_n\to x_0=>x_{n+1}\to x_0<=>f(x_{n+1})\to f(x_0)$ and thus $x_0=f(x_0)$.

$f$ cannot have two fixed points $x,y$ because $<|x-y|=|f(x)-f(y)|\leq a|x-y|=>1\leq a$ which is false.

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  • $\begingroup$ @Justin D. ,check my proof please.thank you $\endgroup$ – Haha Nov 24 '13 at 12:58

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