Sign up ×
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.

Prove/ Disprove:

  1. Let $f:(0,1)\to(0,1)$ be such that $|f(x)-f(y)|\leq 0.5|x-y|$ for all $x ,y.$ Then f has a fixed point.

    2.Let $f:\mathbb R\to\mathbb R$ be continuous and periodic with period $T>0.$Then there exists a point $x_0\in\mathbb R$ such that


share|cite|improve this question

3 Answers 3

up vote 3 down vote accepted
  • Let $f(x)=\frac{x}{2}$ so $f$ hasn't a fixed point in $(0,1)$.

  • Let $g(x)=f(x+T/2)-f(x)$ then $g$ is continuous and $g(0)g(T/2)\le0$ so use the intermediate value theorem to conclude.

share|cite|improve this answer
Good points. Applicable ones. +1 –  Babak S. Jan 22 '14 at 14:29

The counterexample for $1$ is the function $f(x) = x/4$ which would only have a fixed point if $0$ was included in the interval.

share|cite|improve this answer
All the counterexamples are from geometrical visualisations. But to write concrete proof (probably method of contradiction will be fruitful) can someone give any idea? –  user121418 Jan 22 '14 at 13:36
$f(x) = x/4$ suffices the condition $|f(x) - f(y)|\leq 0.5 |x - y|$ and does not have a fixed point on $(0,1)$. What more do you need, there's no geometry here! –  5xum Jan 22 '14 at 13:38
5xum See the counterexamples $x/2,x/4,0$ are all the lines in $\mathbb R^2$ with slopes less than the slope of $y=x$ satisfying the given criteria and it is also obvious that all the lines cut $y=x$ line only at (0,0). This is the picture in background but not a formal proof. –  user121418 Jan 22 '14 at 13:47
No, but you can easily MAKE a formal proof. Like this: Take any pair of $x,y\in\mathbb R$. For this pair, $f(x) - f(y) = (x-y)/4$, meaning that $|f(x) - f(x)| = 0.25 |x - y| < 0.5 |x-y|$. If you cannot prove that $x/4$ does not have a fixed point on $(0,1)$, you should do some more studying before trying to solve the task in your question. –  5xum Jan 22 '14 at 13:53

Or use the trivial function $f(x) = 0$.

share|cite|improve this answer

Your Answer


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.