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December 2nd, 2016, 06:47 AM   #1
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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) ) = |(3x-4) - (3y-4)| ≤ hd(x,y)
= | 3x - 3y | ≤ hd(x,y)
= 3| x - y | ≤ hd(x,y)

Would this be the proper approach f(x) = 3x − 4?
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December 2nd, 2016, 03:54 PM   #2
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Proper approach for what? Since h < 1, f(x) does not satisfy.
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December 3rd, 2016, 08:29 AM   #3
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Quote:
Originally Posted by zactops View Post
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) ) = |(3x-4) - (3y-4)| ≤ hd(x,y)
= | 3x - 3y | ≤ hd(x,y)
= 3| x - y | ≤ hd(x,y)

Would this be the proper approach f(x) = 3x − 4?
Since you write d(f(x), f(y)) as |f(x)- f(y)| why not write d(x, y) as |x- y|

Then it would be obvious that d(f(x), f(y))= 3|x- y|= 3 d(x, y). The "h" is at least 3 so this is not a "contraction map".
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