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.

If your text material give a different definition please post it.

**What is the exact wording of the question?**

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.

If your text material give a different definition please post it.

**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.

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?**

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...