10.1. METRIC SPACES 257

As to the sequence of iterates remaining in B where B is a ball as described, the inequalityabove in the case where m = 0 yields

d (x0, f n (x0))≤1

1− rd ( f (x0) ,x0)

and so, if the right side is less than δ , then the iterates remain in B. As to the fixed point be-ing unique, it is as before. If x,x′ are both fixed points in B, then d (x,x′)= d ( f (x) , f (x′))≤rd (x,x′) and so x = x′. ■

The contraction mapping theorem has an extremely useful generalization. In order toget a unique fixed point, it suffices to have some power of f a contraction map.

Theorem 10.1.43 Let f : (X ,d)→ (X ,d) have the property that for some n ∈ N, f n is acontraction map and let (X ,d) be a complete metric space. Then there is a unique fixedpoint for f . As in the earlier theorem the sequence of iterates { f n (x0)}∞

n=1 also convergesto the fixed point.

Proof: From Theorem 10.1.41 there is a unique fixed point for f n. Thus

f n (x) = x

Thenf n ( f (x)) = f n+1 (x) = f (x)

By uniqueness, f (x) = x.Now consider the sequence of iterates. Suppose it fails to converge to x. Then there is

ε > 0 and a subsequence nk such that

d ( f nk (x0) ,x)≥ ε

Now nk = pkn+ rk where rk is one of the numbers {0,1,2, · · · ,n−1}. It follows that thereexists one of these numbers which is repeated infinitely often. Call it r and let the furthersubsequence continue to be denoted as nk. Thus

d(

f pkn+r (x0) ,x)≥ ε

In other words,d ( f pkn ( f r (x0)) ,x)≥ ε

However, from Theorem 10.1.41, as k→∞, f pkn ( f r (x0))→ x which contradicts the aboveinequality. Hence the sequence of iterates converges to x, as it did for f a contraction map.■

Now with the above material on analysis, it is time to begin using the ideas from linearalgebra in this special case where the field of scalars is R or C.

10.1.8 Convergence Of FunctionsNext is to consider the meaning of convergence of sequences of functions. There are twomain ways of convergence of interest here, pointwise and uniform convergence.