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!

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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].

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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?

Quote:

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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?

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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.

Quote:

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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.

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Then you are given that:

Quote:

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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].

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So what is there to prove?

That's the problem, I suppose... I don't see how making the derivative bounded gives me that T is a contraction by definition...