Contraction mapping theorem

**Danneedshelp** · February 20th 2011, 04:23 PM

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

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

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

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

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

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

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.

**choovuck** · February 20th 2011, 06:35 PM

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...

**Danneedshelp** · February 20th 2011, 07:43 PM

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

Do I do something like this?

$|f^{n}(x)-f^{m}(x)|=|f^{n}(x)-f(x)+f(x)-f^{m}(x)|<br /> \leq\\|f^{n}(x)-f(x)|+|f(x)-f^{m}(x)|<br /> \leq\\(k^{n}+k^{m})|f(x)-x|$

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

**choovuck** · February 20th 2011, 08:10 PM

Originally Posted by Danneedshelp

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

Do I do something like this?

$|f^{n}(x)-f^{m}(x)|=|f^{n}(x)-f(x)+f(x)-f^{m}(x)|<br /> \leq\\|f^{n}(x)-f(x)|+|f(x)-f^{m}(x)|<br /> \leq\\(k^{n}+k^{m})|f(x)-x|$

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

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

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