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A function $ f: \mathbb{R} \rightarrow \mathbb{R} $ is called a contraction mapping if there exists a positive constant K < 1 such that

$ |f(x) - f(y)| \leq K |x-y| $

d) Suppose $f:\mathbb{R} \rightarrow \mathbb{R} $ is a contraction mpaping and for any $ x_0 \in \mathbb{R} $, consider the sequence $(x_n) $ defined recursively by $ x_n = f(x_{n-1}) $ for $ n \in \mathbb{N} $

define also $a_n = x_n - x_{n-1}$ so $x_ 0 + a_1 + a_2 + ... + a_n = x_n $ show that $\displaystyle \sum_{n=1}^\infty a_n$ converges absolutely and use this to conclude that the sequence $(x_n)$ converges

part d

Here is my attempt thus far

$ \displaystyle \sum_{n=1}^N |a_n| = |x_n - x_0| $

note that $ |x_2 - x_1| = |x_2 - x_0 -(x_1 -x_0)| \geq |x_2 -x_0| - |x_1 - x_0| $ so $ |x_2 - x_0| \leq |x_2 - x_1| + |x_1 - x_0| $ by a similar approach $ |x_3 - x_0| \leq |x_3 - x_2| + |x_2 - x_1| + |x_1 - x_0| $

then $ |x_n - x_0| \leq |x_1 - x_0| + |x_2 - x_1| + |x_3 - x_2| + ... + |x_{n} - x_{n-1}| $

now since $ |f(x_{n-1} - f(x_{n-2})| = |x_n - x_{n-1}| \leq K|x_{n-1} - x_{n-2}| \leq K^2|x_{n-2} - x_{n-3}| \leq ... $ we get that $ |x_n - x_0| \leq |x_1-x_0|(1 + K + K^2 +....+K^{n-1}) $ $ \displaystyle \lim_{N\rightarrow \infty} \sum_{n=1}^N|a_n| = \lim_{n\rightarrow \infty} |x_n - x_0| \leq \dfrac{|x_1 - x_0|}{1-K} $

now I don't know what to do - I feel as though I should conclude, as the limit is bounded so obviously converges to something.

I'm also not sure on how to go about to prove that the seqn x_n converges

any help thanks

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  • $\begingroup$ Please make your question self-contained. $\endgroup$ – Listing Nov 25 '13 at 18:36
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You're basically done. You've shown that for $S_N = \sum_{n=1}^N|a_n|$, the sequence $\{S_N\}$ is bounded. Since the terms in the series $\sum |a_n|$ are positive, you also know $\{S_N\}$ is monotone, and so $\lim_{N\to\infty} S_N$ exists. Thus $\sum_{n=1}^\infty |a_n|$ converges, which in turn implies $\sum_{n=1}^\infty a_n$ converges. Now conclude that $\{x_n\}$ converges by noting $x_N = x_0 + \sum_{n=1}^N a_n$ as described in the prompt (and as such $\{x_n\}$ converges to $x_0 + \sum_{n=1}^\infty a_n$ as $n \to \infty$).

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