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Given $f : [a,b] \to [a,b]$ and $|f(x)-f(y)| < |x-y|$ for all $x,y \in [a,b]$ with $x\neq y$, I have shown so far that there exists a unique fixed point in $[a,b]$ and that fixed point is attracting. How can I now show that the basin of attraction contains $(a,b)$?

Riemann
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user278689
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1 Answers1

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Let $\hat{x}$ be the unique fixed point. Let $x_0 \in [a,b]$ and $x_{n+1} = f(x_n)$. It is clear that $|x_{n+1}-\hat{x}| < |x_n-\hat{x}|$ as long as $x_n \neq \hat{x}$.

Let $\delta = \inf_n |x_n-\hat{x}|$. If $\delta=0$ we are finished.

Suppose $\delta>0$. Let $\lambda = \max_{|x-\hat{x}| \ge \delta} { |f(x)-\hat{x}| \over |x- \hat{x}| } $ (the $\max$ exists because it is the $\sup $ of a continuous function over a compact domain), and note that $\lambda <1$. Since $|x_n-\hat{x}| \ge \delta$ for all $n$, we have $|x_{n+1}-\hat{x}| \le \lambda |x_n-\hat{x}|$ which contradicts $\delta >0$ for sufficiently large $n$. Hence $\delta =0$.

copper.hat
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