256 CHAPTER 10. NORMED LINEAR SPACES

Also note that this estimate yields

d (x0, f n (x0))≤d (x0, f (x0))

1− r

Now d (x0,x)≤ d (x0, f n (x0))+d ( f n (x0) ,x) and so

d (x0,x)−d ( f n (x0) ,x)≤d (x0, f (x0))

1− r

Letting n→ ∞, it follows that

d (x0,x)≤d (x0, f (x0))

1− r

It only remains to verify that there is only one fixed point. Suppose then that x,x′ aretwo. Then

d(x,x′)= d

(f (x) , f

(x′))≤ rd

(x′,x)

and so d (x,x′) = 0 because r < 1. ■The above is the usual formulation of this important theorem, but we actually proved a

better result.

Corollary 10.1.42 Let B be a closed subset of the complete metric space (X ,d) and letf : B→ X be a contraction map

d ( f (x) , f (x̂))≤ rd (x, x̂) , r < 1.

Also suppose there exists x0 ∈ B such that the sequence of iterates { f n (x0)}∞

n=1 remains inB. Then f has a unique fixed point in B which is the limit of the sequence of iterates. Thisis a point x ∈ B such that f (x) = x. In the case that B = B(x0,δ ), the sequence of iteratessatisfies the inequality

d ( f n (x0) ,x0)≤d (x0, f (x0))

1− rand so it will remain in B if

d (x0, f (x0))

1− r< δ .

Proof: By assumption, the sequence of iterates stays in B. Then, as in the proof of thepreceding theorem, for m < n, it follows from the triangle inequality,

d ( f m (x0) , f n (x0)) ≤n−1

∑k=m

d(

f k+1 (x0) , f k (x0))

≤∞

∑k=m

rkd ( f (x0) ,x0) =rm

1− rd ( f (x0) ,x0)

Hence the sequence of iterates is Cauchy and must converge to a point x in X . However, Bis closed and so it must be the case that x ∈ B. Then as before,

x = limn→∞

f n (x0) = limn→∞

f n+1 (x0) = f(

limn→∞

f n (x0))= f (x)