Hello,

Let X1 be in E. For n>=1, let Xn+1 = f(Xn)

show {Xn} is contractive

I have attempted several things:

1. |f(X2) - f(X1)|

= |X3-X2|

<= |X2-X1| + |X1 - X3| (by triangle inequality)

< |X2-X1|

By definition, a sequence is contractive if |Pn+1 -Pn|<= b|Pn-Pn-1| or |Pn+1-Pn|<= b^n-1|p2-p1|

So, I know |X3-X2| < |X2-X1|, but how do i find the b?

2. Show {Xn} is cauchy

to show that, I have to find an N s.t. |Pn -Pm|< epsilon for all n,m>=N.

So, |f(Xn-1)-f(Xm-1)|

= |Xn-Xm|

<= |Xn -p| + |p-Xm| (let |Xn - p| < epsilon/2 and |Xm -p| < epsilon/2)

< epsilon/2 + epsilon/2

=epsilon

But what is N?

Would either work?