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!
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.