# contraction mapping theorem

• Dec 21st 2012, 05:46 AM
Plato13
contraction mapping theorem
I need to show f:[0,1]->[0,1] is a contraction mapping. http://latex.codecogs.com/png.latex?...|x^{0.5}-g(x)| where g is continous map form [0,1] to itself
• Dec 21st 2012, 06:42 AM
HallsofIvy
Re: contraction mapping theorem
The definition of "contraction mapping" is that |f(a)- f(b)|< |a- b|. A "strict contraction" would have |f(a)- f(b)|<= c|a- b| where c< 1. Can you show either of those? Use the fact that |sin(x)|<= 1 of course.
• Dec 21st 2012, 07:15 AM
Plato13
Re: contraction mapping theorem
Yes I'm looking for the latter strict contraction as you say. ok so |F(x)-F(y)|=|sinx|||x^{0.5}-g(x)|-|y^{0.5}-g(y)| which is less than or equal to [x^{0.5}-g(x)|+|y^0.5-g(y)| by triangle inequality and |sinx|<=1. Hmm tex doesn't seem to be working. Hope you can read it
• Dec 21st 2012, 07:19 AM
emakarov
Re: contraction mapping theorem
It is not enough to use the fact that |sin(x)| <= 1. Suppose g(x) = 0. Then sqrt(x) is not contracting.
• Dec 21st 2012, 08:13 AM
Plato13
Re: contraction mapping theorem
can you help? Maybe use domain and co-domain of the functions?
• Dec 21st 2012, 09:05 AM
emakarov
Re: contraction mapping theorem
Let g(x) = 0, i.e., f(x) = sin(x) * sqrt(x). Then f'(x) exceeds 1 between 0.7 and 0.8. By the mean value theorem, f(0.8) - f(0.7) > 0.1, so f(x) is not a contraction.

https://s3.amazonaws.com/grapher/exports/inwamx1sxw.png
• Dec 21st 2012, 09:23 AM
Plato13
Re: contraction mapping theorem
look at post 2.
• Dec 21st 2012, 09:26 AM
emakarov
Re: contraction mapping theorem
Quote:

Originally Posted by Plato13
look at post 2.

I have looked. What are you saying?