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Contraction Mapping Theorem. Prove $\{ y_{1},f(y_{1}),f(f(y_{1})),\ \ldots) \} $ is Cauchy. (Abbott p 114 q4.3.9b)

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Let $f$ be a function defined on all of $R$. Assume there is a constant $c$ such that $0< c <1$ and $ |f(x)\ -f(y)\leq c|x-y|$ for all $x,\ y\in R$.

(a) Show that $f$ is continuous on $R$ for all $c \in R$.

(b) Pick some point $y_{1}\in R$ and construct the sequence $ (y_{1},f(y_{1}),f(f(y_{1})),\ \ldots)\ $In general, if $y_{n+1}=f(y_{n})$ , show that the resulting sequence $(y_{n})$ is Cauchy. Hence let $y=\displaystyle \lim y_{n}$.

Solution (b) Observe that for any fixed $n\in N$, $|y_{m+1}-y_{m+2}|=|f(y_m) - f(y_{m + 1})| \leq c|y_{m}-y_{m+1}|. $ This idea can be extended inductively to conclude that

$\begin{align} |y_{m+1}-y_{m+2}| & \leq c|y_{m}-y_{m+1}|\\ & \leq c^2|y_{m-1}-y_{m}| \\ & \leq ... \leq c^{m}|y_{1}-y_{2}| \end{align}$.

1. Why start with $|y_{m+1}-y_{m+2}|$?

2. Intuitively, how do we know $|y_{m+1}-y_{m+2}| \le ... \le c^{m}...|y_{1}-y_{2}|$?
From $|y_{m+1}-y_{m+2}|$ to $...|y_{1}-y_{2}|$, how do we know we need $c^m$ and not c^{another exponent}? Not questioning about proofs or rigour in this question 2.

$0 <c<1$ implies $\displaystyle \sum_{n=1}^{\infty}c^{n}$ converges. This will invoke $(y_{n})$ is Cauchy. How? Note for $m <n$:

$3.$ Why need $m < n$? What does this signify?

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4. What's the modus operandi of all these (fey) steps? I know the first inequality is Triangle Inequality and the second is my question 2. But how to presage to do this?

$5.$ Geometric series formula engenders $1 + c + .. + c^{\color{magenta}{n - m} - 1} = \dfrac{1(1 - c^{\color{magenta}{n - m}})}{1 - c}$. Why $1/(1 - c)$?

Let $\epsilon >0$, and choose $N\in \mathbb{N}$ large enough so that $c^{\color{red}{N} - 1}|y_1 - y_2| \dfrac{1}{1 - c}< \epsilon$.

6. Don't we want the RHS term in $\color{magenta}{♫} < \epsilon$? So the red $c^{\color{red}{N} - 1}$ should be $c^{m - 1}$?

Then the overhead shows: $n>m\geq N$ implies $|y_{m}-y_{n}|<\epsilon$. This $\iff (y_{n})$ is Cauchy.


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