18.950 Handout 2. The Contraction Mapping Theorem.

18.950 Handout 2. The Contraction Mapping Theorem.

Here we prove a very useful fixed point theorem called the contraction mapping theorem. Later we will apply this theorem to prove existence and uniqueness of solutions to ODE’s, and also to prove inverse and implicit func­

tion theorems.

Definition 1. Let f : X → X be a map of a metric space to itself. A point a ∈ X is called a fixed point of f if f (a) = a.

Recall that a metric space (X, d) is said to be complete if every Cauchy sequence in X converges to a point in X.

Exercise. For any positive integers n, m, a set Ω ⊆ Rⁿ , a positive number R and a point y₀ ∈ R^m, let C (Ω; B_R(y₀)) denote the collection of continuous maps f : Ω → B_R(y₀). Then C (Ω, B_R(y₀)) is a complete metric space with the uniform or sup metric defined by

d(f, g) = �f − g�_sup = sup f (x) − g(x)

x∈Ω

| |

for f, g ∈ C (Ω; B_R(y₀)).

Definition 2. Let (X, dX) and (Y, dY ) be metric spaces. A map φ : X →Y is called a contraction if there exists a positive number c < 1 such that

dY (φ(x), φ(y)) ≤ cdX (x, y) for all x, y ∈ X.

A trivial example is the map φ(x) = 1₂x for x ∈ X = Rⁿ .

Theorem 1 (Contraction mapping theorem). Let (X, d) be a complete metric space. If φ : X → X is a contraction, then φ has a unique fixed point.

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Proof. By definition of contraction, there exists a number c ∈ (0, 1) such that

d(φ(x), φ(y)) ≤ cd(x, y). (1)

Let a₀ ∈ X be an arbitrary point, and define a sequence a_n inductively by setting an+1 = φ(an). We claim that an is Cauchy. To see this, first note that for any n ≥ 1, we have by (1) that d(a_n+1, a_n) = d(φ(a_n), φ(a_n−1)) ≤ cd(a_n, a_n−1), and so we can check easily by induction that

d(a_n+1, a_n) ≤ c ⁿd(a₁, a₀) (2) for all n ≥ 1. This and the triangle inequality then gives that for m > n ≥ 1,

d(a_m, a_n) ≤ d(a_m, a_m−1) + d(a_m−1, a_m−2) + . . . + d(a_n+1, a_n)

≤ (c ^m−1 + c ^m−2 + . . . + c )d(an ₁, a₀) cn

d(a₁, a₀). (3)

≤ 1 − c

This shows that d(a_m, a_n) → 0 as n, m → ∞, i.e. that {a_n} is Cauchy as claimed.

Since (X, d) is complete, there exists a ∈ X such that a_n → a. Being a contraction, φ is continuous, and hence

φ(a) = lim φ(a_n) = lim a_n+1 = a.

n→∞ n→∞

Thus a is a fixed point of φ.

If b ∈ X is also a fixed point of φ, then

d(a, b) = d(φ(a), φ(b)) ≤ cd(a, b)

which implies, since c < 1, that d(a, b) = 0 and hence that a = b. Thus the fixed point is unique.

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