Number Theory: An Introduction to Mathematics

4 Metric Spaces 35

Ifx 1 ,x 2 ∈U,then

|fy(x 2 )−fy(x 1 )|=

∣

∣

∣

∣

∫ 1

0

f′(( 1 −t)x 1 +tx 2 )(x 2 −x 1 )dt

∣

∣

∣

∣

≤|x 2 −x 1 |/ 2.

It follows thatfyhas at most one fixed point inU. Since this holds for arbitraryy∈Rn,
the restriction ofφtoUis injective.
Suppose next thatη=φ(ξ)for someξ∈U. We wish to show that, ifyis nearη,
the mapfyhas a fixed point nearξ.
Chooser=r(ξ) >0 so that the closed ballBr={x∈Rn:|x−ξ|≤r}is
contained inU,andfixy∈Rnso that|y−η|<r/ 2 |A−^1 |.Then

|fy(ξ)−ξ|=|A−^1 (y−η)|

≤|A−^1 ||y−η|<r/ 2.

Hence if|x−ξ|≤r,then

|fy(x)−ξ|≤|fy(x)−fy(ξ)|+|fy(ξ)−ξ|

≤|x−ξ|/ 2 +r/ 2 ≤r.

Thusfy(Br)⊆Br. Also, ifx 1 ,x 2 ∈Br,then

|fy(x 2 )−fy(x 1 )|≤|x 2 −x 1 |/ 2.

ButBris a complete metric space, with the same metric asRn, since it is a closed
subset (ifxn∈Brandxn→xinRn,thenalsox∈Br). Consequently, by the con-
traction principle (Proposition 26),fyhas a fixed pointx∈Br.Thenφ(x)=y,which
proves (ii).
Suppose now thaty,η∈V.Theny=φ(x),η=φ(ξ)for uniquex,ξ∈U.Since

|fy(x)−fy(ξ)|≤|x−ξ|/ 2

and

fy(x)−fy(ξ)=x−ξ−A−^1 (y−η),

we have

|A−^1 (y−η)|≥|x−ξ|/ 2.

Thus

|x−ξ|≤ 2 |A−^1 ||y−η|.

IfF=φ′(ξ)andG=F−^1 ,then

ψ(y)−ψ(η)−G(y−η)=x−ξ−G(y−η)

=−G[φ(x)−φ(ξ)−F(x−ξ)].