CHAPTER 5. COMPLETENESS 5.1. There are three basic properties about metric spaces which can be called infor- mally as three ‘C’s. They are completeness, compactness and connectedness. They will be the focus of the rest of this book. In the present chapter we study the first C: complete- ness. Let us briefly recall the following definitions from Chapter 3: a sequence xn in a { } metric space (X, ρ) is called a Cauchy sequence if, for each ε > 0, there exists N such that, for all m, n > N, ρ(xn, xm) < ε. or, according to our ad hoc definition, if there is a sequence of positive numbers αn decreasing to zero such that ρ(xn, xm) < αn for all m, n with m > n. A metric space is complete if all Cauchy sequences in it converge. According to Cauchy’s theorem, the real line R under the metric ρ(x, y) = x y | − | is complete. We have also seen that, more generally, the Euclidean space Rd is complete under the metric 2 2 ρ(x, y) = (x y ) + + (xd yd) 1 − 1 ··· − d for x = (x1, . , xd) and y = (y1, . , yd) in R . We will see more examples of complete metric spaces (actually, many of them are Banach spaces) in the future. Anyway, here we describe a “cheap way” to get a substantial supply of quick examples. Let (X, ρ) be an arbitrary metric space and let Y be any nonempty subset of X. Then we can equip Y with the metric ρY inherited from X: for all x, y Y , let ρY (x, y) = ρ(x, y); that is, ρY is ∈ just the restriction of ρ to Y Y . It is clear that ρY is a metric for Y , which is called the × metric on Y induced by ρ. The metric space (Y, ρY ) is called a subspace of X. Fact 1. With the above notation, if (X, ρ) is a complete metric space and if Y is a closed subset of X, then (Y, ρY ) is also a complete metric space. Proof: Let yn be a Cauchy sequence in Y . Then certainly it is a Cauchy sequence { } in X. Hence it has a limit in X. Since Y is a closed subset of X, this limit is in Y . This shows the convergence in Y of the Cauchy sequence yn . Q.E.D. { } Take any complete metric space, e.g. Rd. Then any nonempty closed set in it gives us an example of complete metric space. Notice the following strong form of converse to Fact 1; (it is strong because we do not have to assume the completeness of X.) Fact 2. A set Y of a metric space (X, ρ) is closed if (Y, ρY ) is complete. 1 The proof of this fact is left to you as an exercise (Exercise 4). 5.2. What can completeness do for you? In the present section we present the so called Banach’s principle for contractive mappings which hinges upon the completeness assumption. First we give: Definition. A mapping φ from a metric space (X, ρ) into itself is said to be a contractive mapping or simply a contraction if there exists a positive number α strictly less than 1 such that, for all x, y X, ρ(φ(x), φ(y)) αρ(x, y). ∈ ≤ Given a contractive mapping φ on X and take any point x X, we can define 0 ∈ a sequence xn in X by iteration: x = φ(x ), x = φ(x ) = φ(φ(x )), x = φ(x ) = { } 1 0 2 1 0 3 2 φ(φ(φ(x0))), etc. The question is, does this sequence converge? The answer is Yes, provided that the metric space X is complete. In that case it is easy to check that the limit z of this sequence is a fixed point of φ in the sense that φ(z) = z. Theorem (Banach’s Principle of Contractive Mappings). A contractive map- ping φ from a complete metric space (X, ρ) into itself has a unique fixed point in X, that is, a point z X satisfying φ(z) = z. Furthermore, for each x X, the sequence xn ∈ 0 ∈ { } defined iteratively by xn = φ(xn) (n 0) tends to this unique fixed point. + 1 ≥ Proof: Let x be an arbitrary point in X and define the sequence xn iteratively by 0 { } n 2 xn+ 1 = φ(xn). Denote by φ the composite of φ with itself n times. Thus φ (x) = φ(φ(x)), φ3(x) = φ(φ(φ(x))) etc. and in general φn+ 1(x) = φ(φn(x)). By induction, it is easy to show that, for all x, y X and for all positive integer n, ∈ ρ(φn(x), φn(y)) αnρ(x, y). ≤ Thus, for all positive integers m and n with m > n, we have n m ρ(xn, xm) = ρ(φ (x0), φ (x0)) ρ(φn(x ), φn+ 1(x )) + ρ(φn+ 1(x ), φn+ 2(x )) + + ρ(φm−1(x ), φm(x )) ≤ 0 0 0 0 ··· 0 0 n n+ 1 m−1 α ρ(x , φ(x )) + α ρ(x , φ(x )) + + α ρ(x , φ(x )) βn ≤ 0 0 0 0 ··· 0 0 ≤ ∞ k where βn ρ(x0, x1) α are positive numbers decreasing to zero as n . This ≡ k= n → ∞ shows that xn is a Cauchy sequence in X. The completeness of X tells us that xn { } { } converges, say to z. A contractive mapping is clearly a continuous mapping and hence we are allowed to let n in the identity φ(xn) = xn to conclude φ(z) = z. → ∞ + 1 2 To show the “uniqueness” part, let z1 and z2 be fixed points of the contraction φ: φ(z1) = z1 and φ(z2) = z2. Then ρ(z , z ) = ρ(φ(z ), φ(z )) αρ(z , z ). 1 2 1 2 ≤ 1 2 Since 0 < α < 1, we must have ρ(z1, z2) = 0 and hence z1 = z2. Q.E.D. 5.3. Banach’s contraction principle gives applications to local solvability of initial value problem in ordinary differential equations under the Lipschitz condition, the inverse function theorem in mathematical analysis of smooth functions with several variables, and a basic fact for constructing fractals, such as the Cantor set. These applications need substantial background for their descriptions. Here we briefly present an application to justify Newton’s method in a special case as well as an application to solvability of Bellman’s equation in dynamical programming. Example. Let a be any positive number greater than 1 and let X = [√a, ) x R: x √a , ∞ ≡ { ∈ ≥ } which is a closed subset R of the real line R and hence is a complete metric space in its own right, (according to Fact 2 in 5.1). Consider the map φ on X given by § 1 a φ(x) = x + . 2 x Here we ask the reader to check that φ indeed maps X into X itself, as well as the inequality φ(x) φ(y) α x y for x, y X, where α = 1/2. So, by Banach’s | − | ≤ | − | ∈ contraction principle, φ has a unique fixed point, say z. We can find this unique fixed point of φ, say z. By solving the equation φ(z) = z, we can find this fixed point, which turns out to be z = √a. Take any convenient number x0 in X, for example, x0 = a. Then the iteration xn = φ(xn) with x = a will give us a sequence xn of numbers + 1 0 { } approximating √a. When a = 2, xn are rational numbers approximating the irrational number √2. In this example the recipe for the contractive map φ comes from Newton’s method by applying f(x) φ(x) = x − f ′(x) to the function f(x) = x2 a. The reader is asked to check this. − 3 By using Lagrange’s mean value theorem and Banach’s principle of contractive map- pings, we can prove the following: Proposition. If φ is a differentiable function on a closed interval I such that φ(x) I ∈ for all x I and there exists a positive α < 1 such that φ′(x) α, then there is a unique ∈ | | ≤ point a in I such that φ(a) = a. We leave the proof of the above proposition to those readers who know Lagrange’s mean value theorem. Notice that, for the function φ given in the above example, we have 1 a φ′(x) = 1 2 − x2 and hence 0 φ′(x) 1/2 for all x [√a, ). ≤ ≤ ∈ ∞ 5.4. Recall that, given a metric space X, b(X) stands for the space of all bounded, C continuous, real-valued functions on X, and the norm f of f b(X) is given by ∈ C f = sup ∈ f(x) . Also, the recipe ρ(f, g) = f g defined a metric ρ on b(X). x X | | − C Next we recall that, given any nonempty set X (not necessary a metric space), the space consisting of all bounded functions on X is denoted by ℓ∞ (X). Proposition. With the notation as above, we have: (a) as a metric space, ℓ∞ (X) is complete, and (b) b(X) is closed in ℓ∞ (X) and hence is a complete space by itself. C Proof: Part (b) is just a repetition of a proposition in 4.6, as well as Fact 1 in our § previous section. So it is enough to establish part (a). Let fn be a Cauchy sequence { } in ℓ∞ (X). Then there is a sequence of positive numbers αn decreasing to zero such that ρ(fn, fm) fn fm ∞ αn for all n, m with m n.