question on contraction mapping?

Question

If T is a contraction mapping, how could i show that T^n (i.e. T raise to the power n) is also a contraction mapping???
please help me out??? I hav to submit the proof tommorow...

Answer 1

I think it can be done by induction:

We know T is a contraction, so
|T(x) - T(y)| <= a|x - y|, where a<1.

We want to show
|T^n(x) - T^n(y)| <= b|x - y| where b<1.

Case n=1: This is true by assumption.

Now assume that T^(n-1) is a contraction, so
|T^(n-1)(x) - T^(n-1)(y)| <= c|x-y| where c<1. Then
|T^n(x) - T^n(y)| = |T(T^(n-1))(x) - T(T^(n-1))(y)|
<= a|T^(n-1)(x) - T^(n-1)(y)| <= ac|x-y|.

And since a<1,c<1, we have ac<1 and so T^n is a contraction.

I hope that you can follow all of this and it helps.

Answer 2

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Answer 3

by definitioan a contraction mappinfg is a mapping

| Tx - Ty|<= k|x-y|

for n = 2
| TTx - TTy |
set x' = Tx and y'=Ty
then
| TTx - TTy | = | Tx' - Ty' | <= k|x'-y'|

etc induction on n : the last <= is because T is a contractive mapping.