# Contraction mapping theorem

• Feb 20th 2011, 04:23 PM
Danneedshelp
Contraction mapping theorem
Let $\displaystyle f$ be a function defined on all of $\displaystyle \mathbb{R}$, and assume there is a constant $\displaystyle c$ such that $\displaystyle 0<c<1$ and $\displaystyle |f(x)-f(y)|\leq\\k|x-y|$ for all $\displaystyle x,y\in\mathbb{R}$.

a) Show $\displaystyle f$ is continuous on $\displaystyle \mathbb{R}$.

Let $\displaystyle c$ be a real number and $\displaystyle \epsilon>0$. Then $\displaystyle |f(x)-f(c)|\leq\\k|x-c|$ by definition. So, let $\displaystyle \delta=\frac{\epsilon}{k}$. It follows that $\displaystyle |x-c|<\delta$ implies $\displaystyle |f(x)-f(c)|<k\frac{\epsilon}{k}=\epsilon$.

b) Pick some point $\displaystyle y_{1}\in\mathbb{R}$ and construct the sequence

$\displaystyle (y_{1},f(y_{1}),f(f(y_{1})),...)$.

In general, if $\displaystyle y_{n+1}=f(y_{n})$, show that the resulting sequence $\displaystyle (y_{n})$ is a Cauchy sequence. Hence, we may let $\displaystyle y=limy_{n}$.

I am not sure how to approach this one. Some help getting started would be appreciated. It seems almost trivial, but I am not sure how formalize it.
• Feb 20th 2011, 06:35 PM
choovuck
just write |f(f(x))-f(x)| < k|f(x)-x|, |f(f(f(x)))-f(f(x))| < k^2 |f(x)-x| and so on...
• Feb 20th 2011, 07:43 PM
Danneedshelp
So in general $\displaystyle |f(x)-f^{n}(x)|\leq\\k^{n}|f(x)-x|$. I am not sure how this proves the sequence is Cauchy. Why did you choose $\displaystyle f(x)-x$ in the expression $\displaystyle k^{i}|f(x)-x|$?

Do I do something like this?

$\displaystyle |f^{n}(x)-f^{m}(x)|=|f^{n}(x)-f(x)+f(x)-f^{m}(x)| \leq\\|f^{n}(x)-f(x)|+|f(x)-f^{m}(x)| \leq\\(k^{n}+k^{m})|f(x)-x|$

How do I control the size of $\displaystyle (k^{n}+k^{m})|f(x)-x|$?
• Feb 20th 2011, 08:10 PM
choovuck
Quote:

Originally Posted by Danneedshelp
So in general $\displaystyle |f(x)-f^{n}(x)|\leq\\k^{n}|f(x)-x|$. I am not sure how this proves the sequence is Cauchy. Why did you choose $\displaystyle f(x)-x$ in the expression $\displaystyle k^{i}|f(x)-x|$?

Do I do something like this?

$\displaystyle |f^{n}(x)-f^{m}(x)|=|f^{n}(x)-f(x)+f(x)-f^{m}(x)| \leq\\|f^{n}(x)-f(x)|+|f(x)-f^{m}(x)| \leq\\(k^{n}+k^{m})|f(x)-x|$

How do I control the size of $\displaystyle (k^{n}+k^{m})|f(x)-x|$?

first off, you get $\displaystyle |f^{n-1}(x)-f^{n}(x)|\leq\\k^{n-1}|f(x)-x|$, not $\displaystyle |f(x)-f^{n}(x)|\leq\\k^{n}|f(x)-x|$. then let m>n and use triangle inequality: $\displaystyle |f^m(x)-f^{n}(x)|\leq\\ \sum_{k=n}^{m-1} |f^{k+1}(x)-f^k(x)|<\ldots$ and u get geometric series. u can see this bound goes to zero if m and n go to infinity.