CHAPTER 5. COMPLETENESS 5.1. There Are Three Basic Properties

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 mapping 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, realvalued 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. By the way the norm fn fm ≡ − ≤ ≥ − is deﬁned, this tells us that

fn(x) fm(x) αn x X and n, m with n m. ( ) | − | ≤ ∀ ∈ ∀ ≤ ∗

Hence, for each x in X, fn(x) is a Cauchy sequence of scalars and, by Cauchy’s Criterion, { } the limit limn fn(x) exists. Denote this limit by f(x). As x is an arbitrary point in X, we get a function f on X. Now we go back to the inequality in ( ). In this inequality, ﬁx n ∗ and x, but let m . This will gives us fn(x) f(x) αn, which holds for arbitrary x → ∞ | − | ≤ and n. The last assertion allows us to conclude fn f ∞ αn, provided we have checked − ≤ one small item: f ℓ∞ (X), i.e. f is a bounded function. Notice that, for all x X, ∈ ∈ f(x) = (f(x) f (x)) + f (x) α + f , | | | − 1 1 | ≤ 1 1 4 which tells us that f is indeed bounded. Therefore the Cauchy sequence fn in ℓ∞ (X) { } converges to f in supnorm.

5.5. Given a metric space (X, ρ), which may be incomplete, we have a “big” complete space b(X). It would not be too wild to think about the possibility of embedding X in C this complete space, in order to get a completion of X. Let’s try to accomplish this.

Notice that, ﬁxing any point a in X, the function fa deﬁned by fa(x) = ρ(x, a) is continuous. To verify this, all we need is to establish the following elementary inequality:

ρ(x, a) ρ(y, a) ρ(x, y). ( ) | − | ≤ †

In fact, the left hand side of the above inequality is either ρ(x, a) ρ(y, a) or ρ(y, a) ρ(x, a), − − and in both cases ( ) follows from the triangular inequality. †

Rewrite ( ) as ρ(x, a) ρ(x, b) ρ(a, b). Thus fa(x) fb(x) ρ(a, b) for all x X † | − | ≤ | − | ≤ ∈ and hence fa fb ∞ ρ(a, b). On the other hand, we have fa fb ∞ fa(a) fb(a) = − ≤ − ≥ | − | ρ(a, a) ρ(a, b) = ρ(a, b). Hence | − |

fa fb ∞ = ρ(a, b). (⋆) −

Now we would like to deﬁne a map Φ form X to b(X) be putting Φ(a) = fa for each C a X. The identity (⋆) tells us that Φ is an isometric embedding, i.e. the distance ∈ between points are preserved under Φ. The closure of its range in b(X), namely Φ(X), is C a complete metric space by itself (its completeness is “inherited” from b(X)) and hence C may be served as the completion of X. Everything seems to be ﬁne, except one: there is

no guarantee that fa is bounded to ensure Φ(a) = fa b(X). In other words, the map Φ ∈ C may be “illdefined”. This argument here only works when the space X is bounded, i.e., there exists M > 0 such that ρ(x, y) M for all x, y X. ≤ ∈ When the boundedness assumption on X is dropped, we have to modify our argument by the following little trick. Take an arbitrary point p in X and fix it. Define Ψ : X → b(X) by Ψ(a) = fa fp, which is a bounded function in view of (⋆). We have seen that C − fa fp is also continuous. Therefore the map Ψ is welldefined. Finally, Ψ is an isometry; − indeed, for all a, b X, ∈

Ψ(a) Ψ(b) = (fa fp) (fb fp) = fa fb = ρ(a, b), − − − − − 5 by using (⋆) again. Thus Ψ is an isometric embedding of X into the complete metric space

b(X) and hence Ψ(X) can be regarded as the completion of X. C 5.6. The existence of the completion of a metric space established in the last section is just a general abstract principle. In the concrete situations we often have to describe how an element in the completion looks like. For example, an important tool in partial diﬀerential equations called Sobolev Spaces, is nothing but completions of some spaces of smooth functions under suitable norms. But in solving many long standing basic problems about existence and regularity of solutions, knowing some concrete discription is a ‘must’.

Another example is from number theory: the ring Zp of padic integers, whih can be consider the completion of the ring Z of usual integers under the padic metric. In order to do padic analysis (such as proving or applying Hensel’s lemma) we must know a concrete way to describe elements in this completion, which are called padic integers. In what follows we are going to brieﬂy describe this example, which looks quite esoteric for ﬁrst timers. Although details are not given and our description here is incomplete, it gives some basic idea.

n Let us take ﬁx a prime number p. Then any nonzero integer a can be factorized as p a0, n where a is not divisible by p. We deﬁne the ‘size’ or the (p-adic) norm a p of a p a 0 | | ≡ 0 −n to be p . When a = 0, we simply put a p = 0. (The more factors of p can be extracted | | from a, the smaller of its size a p is. Notice that a p 1 for all nonzero integer a, and | | | | ≤ m a p < 1 occurs exactly when a 0 (mod p).) Take two nonzero integers, say a = p a | | ≡ 0 n m+ n and b = p b0, where a0 and b0 are not divisible by p. Then we have ab = p a0b0 −(m+ n) −m −n where a b is not divisible by p and hence ab p = p = p p = a p b p. Next, 0 0 | | | | | | let us assume that m n, in other words, a p b p, or b p = max a p, b p . Then ≥ | | ≤ | | | | {| | | | } m = n + ℓ for some integer ℓ 0. Thus a + b = pma + pnb = pn(a + pℓb ) so that ≥ 0 0 0 0 −n −n a + b p p = a p = max a p, b p ; (of course, in case ℓ > 0, we have a + b p = p ). | | ≤ | | {| | | | } | | We summarize what we have here:

a + b p max a p, b p , ab p = a p b p. | | ≤ {| | | | } | | | | | |

The padic distance of two integers a and b is deﬁned to be disp(a, b) = a b p. It is | − | quite straightforward to check that disp is a metric for Z. In fact, the triangular inequality

disp(a, c) disp(a, b) + disp(b, c) can be replaced by the following stronger ultrametric ≤ inequality:

disp(a, c) max disp(a, b), disp(b, c) . ≤ { } 6 The completion Zp of Z can be described as the set of all sums

∞ n anp n= 0 where an (for all n 0) are integers satisfying 0 an p 1. It is not hard to see that ≥ ≤ ≤ − n k the sequence sn of the partial sums sn = akp is a Cauchy sequence with respect { } k= 1 ∞ n to the metric disp and hence has a limit in Zp, giving us the sum anp . It is true but n= 0 ∞ n hightly nontrivial that every elemnt in Zp can be written in a unique way as anp n= 0 with 0 an p 1. For example, ≤ ≤ − ∞ 1 = (p 1)pn. Do you know why? − n= 0 −

Zp is among the ﬁrst examples of so called local rings, a concept important for both theories of algebraic numbers and algebraic geometry.

The space (Zp, disp) of padic integers is an ultrametric space. Ultrametric spaces have many strange properties. For examples:

1. Every closed ball is open as well. 2. Every point in an open ball is a center of the ball. 3. Given two open balls, either they are disjoint, or one is contained in another.

The details are left as an exercise.