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?

  • 2
    $\begingroup$ 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

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 $x \neq x'$.


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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