# Contraction Functions

• Feb 22nd 2014, 02:58 PM
Aryth
Contraction Functions
I just learned about these yesterday, so I'm having a little trouble with this problem.

If T: [0,1] to [0,1] and if there is a real number alpha with 0<= alpha < 1 such that |T'(x)| <= alpha (0 <= x <= 1) where T' is the derivative of T, prove that T is a contraction on [0,1].

Any help would be appreciated!
• Feb 22nd 2014, 03:23 PM
Plato
Re: Contraction Functions
Quote:

Originally Posted by Aryth
If T: [0,1] to [0,1] and if there is a real number alpha with 0<= alpha < 1 such that |T'(x)| <= alpha (0 <= x <= 1) where T' is the derivative of T, prove that T is a contraction on [0,1].

Please review what you have posted. The reason being that is the usual definition of a Contraction Function.
We don't prove definitions.

What is the exact wording of the question?
• Feb 22nd 2014, 03:51 PM
Aryth
Re: Contraction Functions
Quote:

Originally Posted by Plato
Please review what you have posted. The reason being that is the usual definition of a Contraction Function.
We don't prove definitions.

What is the exact wording of the question?

That is the exact wording of the question. O.O

The definition I've been given is:

Let <M,p> be a metric space with a function T: M to M. We say that T is a contraction on M if there exists an alpha in R with 0 <= alpha < 1 such that p(Tx,Ty) <= alpha*p(x,y) (x,y in M) with alpha independent of x and y.
• Feb 22nd 2014, 04:19 PM
Plato
Re: Contraction Functions
Quote:

Originally Posted by Aryth
That is the exact wording of the question.
The definition I've been given is:
Let <M,p> be a metric space with a function T: M to M. We say that T is a contraction on M if there exists an alpha in R with 0 <= alpha < 1 such that p(Tx,Ty) <= alpha*p(x,y) (x,y in M) with alpha independent of x and y.

Then you are given that:
Quote:

Originally Posted by Aryth
If T: [0,1] to [0,1] and if there is a real number alpha with 0<= alpha < 1 such that |T'(x)| <= alpha (0 <= x <= 1) where T' is the derivative of T, prove that T is a contraction on [0,1].

So what is there to prove?
• Feb 22nd 2014, 07:04 PM
Aryth
Re: Contraction Functions
That's the problem, I suppose... I don't see how making the derivative bounded gives me that T is a contraction by definition...