Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove that for every uniform contraction function $f$ there exists a unique real $z$ such that $f(z)=z$. A function $f:\mathbb R\to\mathbb R$ is called a uniform contraction if there exists an $a$ in $(0,1)$ such that for all real $x$ and $y$, $|f(x)-f(y)|\leq a|x-y|$.

I proved $f$ is continuous. I let a sequence $\{x_n\}$ be defined as $x_{n+1}=f(x_n)$. I'm having trouble proving $\{x_n\}$ converges to $z$. From there I know how to complete the proof. Any help will be greatly appreciated.

share|cite|improve this question
Try to prove that $|x_{n+1} - x_n| \le a|x_n - x_{n-1}|$. Then try to show that $\{x_n\}$ is a Cauchy sequence. – Hans Engler Nov 16 '12 at 1:48

For, $N\in\mathbb{N}$, prove that $$ |f(x_m)-f(x_n)|<|x_0-f(x_0)|\frac{a^N}{1-a},\text{ for all }m>n>N. $$ This is the so called Contraction Mapping Theorem.

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.