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December 2nd, 2016, 06:47 AM  #1 
Newbie Joined: Dec 2016 From: Montreal Posts: 4 Thanks: 0  Banach's Contraction Mapping Theorem
As I understand the theorem, we Let (X,d) be a metric space and let f : X →X be a mapping. Then, a point x ∈ X is a fixed point of f if x = f (x), and f is called a contraction if there exists a fixed constant h < 1 such that d( f (x), f (y) ) ≤ hd(x,y), for all x,y ∈ X. I'm a little confused on how to approach problems using Banach's Contraction theorem though. For example, if f(x) = 3x − 4, g(x) = (1/2)sin x and q(x) = e^(x^2) d( f (x), f (y) ) = (3x4)  (3y4) ≤ hd(x,y) =  3x  3y  ≤ hd(x,y) = 3 x  y  ≤ hd(x,y) Would this be the proper approach f(x) = 3x − 4? 
December 2nd, 2016, 03:54 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,629 Thanks: 622 
Proper approach for what? Since h < 1, f(x) does not satisfy.

December 3rd, 2016, 08:29 AM  #3  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894  Quote:
Then it would be obvious that d(f(x), f(y))= 3x y= 3 d(x, y). The "h" is at least 3 so this is not a "contraction map".  

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banach, contraction, mapping, theorem 
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