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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.

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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

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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.

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