8 Completeness
We recall the definition of a Cauchy sequence. Let (X, d) be a given metric space and let (xn) be a sequence of points of X. Then (xn) (xn) is a Cauchy sequence if for every ε> 0 there exists N N such that ∈ d(x ,x ) < ε for all n,m N. n m ≥ Properties of Cauchy sequences are summarized in the following propositions
Proposition 8.1. (i) If (xn) is a Cauchy sequence, then (xn) is bounded.
(ii) If (xn) is convergent, then (xn) is a Cauchy sequence.
(iii) If (xn) is Cauchy and it contains a convergent subsequence, then (xn) converges.
A Cauchy sequence need not converge. For example, consider the sequence (1/n) in the metric space ((0, 1), ). Clearly, the sequence is Cauchy in | · | (0, 1) but does not converge to any point of the interval.
Definition 8.2. A metric space (X, d) is called complete if every Cauchy sequence (xn) in X converges to some point of X. A subset A of X is called complete if A as a metric subspace of (X, d) is complete, that is, if every Cauchy sequence (xn) in A converges to a point in A. By the above example, not every metric space is complete; (0, 1) with the standard metric is not complete.
Theorem 8.3. The space R with the standard metric is complete.
Theorem 8.3 is a consequence of the Bolzano-Weierstrass theorem and Propositions 8.1 Recall that if (X , d ), 1 i m, are metric spaces and X = X1 . . . X , i i ≤ ≤ × × n then d(x,y)= max dj(xi,yi) 1≤j≤n where x = (x1,...,x ), y = (y1,...,y ) X, defines a metric on X. The n n ∈ pair (X, d) is called the product of (Xi, di).
Theorem 8.4. If (Xi, di) are complete metric spaces for i = 1,...,m, then the product (X, d) is a complete metric space.
49 1 m Proof. Let xn = (xn,...,xn ) and (xn) be a Cauchy sequence in (X, d). Then for a given ε > 0 there exists k such that d(x ,x ) < ε for all n,m k. n m ≥ Since d (xj ,xj ) d(x ,x ) < ε, j n m ≤ n m j it follows that xn is Cauchy in (Xj, dj) for j = 1,...m. Since (Xj, dj ) is { } j j j complete, for j = 1,...,m there exists x Xj such that xn x . Then 1 ∈ → setting x = (x ,...,xm), we see, in view of the above definition of d, that x x in X. n → Since 1 2 n / 2 max x1 y1 ,..., xn yn xi yi {| − | | − |} ≤ =1 | − | ! Xi √n max x1 y1 ,..., x y , ≤ {| − | | n − n|} it follows from Theorem 8.3 and Proposition 8.4 that Rn with the Euclidean metric is complete. ∞ Example 8.5. Denote by ℓ the set of all real sequences (xk) satisfying k xk < 1 =1 | | . Then ℓ1 is a vector space if addition and multiplication by a number are defined as∞ follows, P (xn) + (yn) = (xn + yn), α(xn) = (αxn).
If x = (xk) ℓ1, then ∈ ∞ x = (xk) := xk k k k k | | k=1 X∞ defines a norm on ℓ and d(x, y)= x y = xk yk defines a distance on 1 k − k k=1 | − | ℓ1. We claim that (ℓ1, d) is a complete metric space. P (n) Indeed, let (Xn) ℓ be a Cauchy sequence. Then with Xn(x ) we have ⊂ 1 k ∞ (n) (m) (n) (m) x x x x = Xn Xm . k − k ≤ l − l k − k l=1 X (n) Hence for every k 1, the sequence (xk ) is Cauchy in R and since R with ≥ (n) the standard metric is complete, the sequence (xk ) converges to some xk. Set X = (xk). We suspect that X is the limit in ℓ1 of the sequence (Xn). To see this we first show that X ℓ . Since (Xn) is Cauchy in ℓ , there is K such that ∈ 1 1 Xn Xm < 1 for all n,m K. In particular, k − k ≥ N ∞ ∞ (n) (n) (K) (K) xk xk xk + xk k ≤ k − k X=1 X=1 X=1 Xn XK + XK < 1+ XK ≤k − k k k k k 50 for every N 1. Fixing N and taking limit as n we get ≥ → ∞ N xk 1+ XK k | |≤ k k X=1 and taking limit as N we get → ∞ ∞ xk 1+ XK . k | |≤ k k X=1 So, X ℓ . Next we show that Xn X 0 as n . Given ε > 0, there is ∈ 1 k − k → → ∞ K such that Xn Xm < 1 for all n,m K. Consequently, for every N 1 we have k − k ≥ ≥ N (n) (m) xk xk Xn Xm < ε, for all n,m K. − ≤k − k ≥ k=1 X With n>K and N fixed, we let m to find that → ∞ N (n) xk xk ε. − ≤ k=1 X Since this is true for every N, ∞ (n) Xn X = x xk ε k − k k − ≤ k=1 X for n>K. Hence d(Xn,X) 0 and since X ℓ , the space ℓ is complete. → ∈ 1 1 A subspace of a complete metric space may not be complete. For example, R with the standard metric is complete but (0, 1) equipped with the same metric is not complete. Proposition 8.6. If (X, d) is a complete metric space and Y is a closed subspace of X, then (Y, d) is complete.
Proof. Let (xn) be a Cauchy sequence of points in Y . Then (xn) also satisfies the Cauchy condition in X, and since (X, d) is complete, there exists x X such that ∈ xn x. But Y is also closed, so x Y showing that Y is complete. → ∈ Proposition 8.7. If (X, d) is a metric space, Y X and (Y, d) is complete, ⊂ then Y is closed.
Proof. Let (xn) be a sequence of points in Y such that xn x. We have to show → that x Y . Since (xn) converges in X, it satisfies the Cauchy condition in X and so, it also∈ satisfies the Cauchy condition in Y . Since (Y, d) is complete, it converges to some point in Y , say to y Y . Since any sequence can have at most one limit, x = y. So x Y and Y is closed.∈ ∈
51 Let (X, d) and (Y, ρ) be metric spaces. A function f : X Y is said → to be bounded if the image f(X) is contained in a bounded subset of Y . Denote by B(X,Y ) the set of all functions f : X Y which are bounded → and by C (X,Y ) the space of bounded continuous function f : X Y . If b → Y = R, we simply write B(X) and Cb(X) instead of B(X,Y ) and Cb(X, R), respectively. We have C (X,Y ) B(X,Y ). For f, g B(X,Y ), we set b ⊂ ∈ D(f, g) := sup ρ(f(x), g(x)) x X , { | ∈ } where ρ denotes the metric on Y . The metric on Cb(X,Y ) is defined in the same way.
Theorem 8.8. Suppose that (Y, ρ) is a complete metric space. Then the spaces B(X,Y ) and (Cb(X,Y ), D) are complete. Proof. The verification that D is a metric is left as an exercise. It suffices to show that B(X, Y ) is a complete and that Cb(X, Y ) is a closed subset of B(X, Y ). Let fn be a Cauchy sequence in B(X, Y ). Then (fn(x)) is a Cauchy sequence in (Y,ρ) { } for every x X. Since, by assumption, (Y,ρ) is complete, the sequence (fn(x)) ∈ converges in Y . Define f(x) = limn→∞ fn(x). Given ε> 0, there is N such that
ρ(fn(x),fm(x)) <ε for n,m N. ≥ Fix n N and let m tend to . Since the function z ρ(fn(x),z) is continuous, it follows≥ that ∞ → ρ(fn(x),f(x)) ε for n N (1) ≤ ≥ which implies that D(fn,f) ε for n N. ≤ ≥ It remains to show that f is bounded. Since fN is bounded, there are y Y and ∈ r > 0 such that fN (X) Br(y). From (1) and the triangle inequality, we obtain ⊂ ρ(f(x),y) ρ(f(x),fN (x)) + ρ(fN (x),y) < ε + r showing that f(X) Br+ε(y) and that f≤is bounded. Hence the space (B(X, Y ),D) is complete as claimed.⊂ Next we shall show that Cb(X, Y ) is a closed subset of B(X, Y ). Take any sequence (fn) Cb(X, Y ) such that D(fn,f) 0 as n . We have to show that f is continuous.⊂ Fix x X and let ε> 0.→ Then there→ is∞N N such that 0 ∈ ∈ D(fn,f) < ε/3 for all n N. ≥ In particular, in view of definition of D,
ρ(fN (x),f(x)) < ε/3 for all n N. ≥
Since fN is continuous at x0, there is δ > 0 such that
ρ(fN (x),fN (x0)) < ε/3 for all x satisfying d(x0, x) <δ.
52 Hence, if d(x0, x) <δ, then
ρ(f(x),f(x )) ρ(f(x),fN (x)) + ρ(fN (x),fN (x )) + ρ(f(x) ,fN (x )) 0 ≤ 0 0 0 2D(fN ,f)+ ρ(fN (x),fN (x )) < 2(ε/2)+ ε/3= ε. ≤ 0 This means that f is continuous at x0 and since x0 was an arbitrary point, f is continuous, i.e., f Cb(X, Y ) as claimed. ∈ 8.0.1 Banach Fixed Point Theorem Definition 8.9. Let (X, d) be a metric space. A map f : X X is called → a contraction if there is constant c (0, 1) such that ∈ d(f(x),f(y)) cd(x,y) ≤ for all x,y X. ∈ Theorem 8.10 (Banach Fixed Point Theorem). Let (X, d) be a com- plete metric space and f : X X a contraction. Then there exists exactly → one point u X such that f(u) = u. Moreover, for every x X, the ∈ ∈ sequence (f n(x)) converges to u. Proof. Claim: If a,b X, then ∈ 1 d(a,b) [d(a,f(a)) + d(b,f(b))]. (2) ≤ 1 c − Indeed, d(a,b) d(a,f(a)) + d(f(a),f(b)) + d(f(b),b) ≤ d(a,f(a)) + cd(a,b)) + d(f(b),b) ≤ from which we conclude (2) Using (2), the map f can have at most one fixed point since if f(a) = a and f(b) = b, then the right side of (2) is equal to 0 implying that d(a,b) = 0. To prove existence of a fixed point, take any x X. Since f is a contraction with the constant c, the n-fold composition f n is a contraction∈ with the constant cn. Using this and (2) with a = f n(x) and b = f n(b), and noticing that f(a)= f n+1(x)= f n(f(x)), f(b)= f m(f(x)), we conclude 1 d(f n(x),f m(x)) = d(a,b) d(a,f(a)) + d(b),b)] ≤ 1 c − 1 n n m m = d(f (x)f (f(x))) + d(f (x),f (f(x))] 1 c n− m c + c d(x, f(x)) ≤ 1 c − This implies that the sequence (f n(x)) is Cauchy In view of the completeness of X, there is u such that f n(x) u. In view of the continuity of f, f(f n(x)) converges to f(u). On the other hand→f(f n(x)) = f n+1(x) converges to u and so, f(u) = u as required.
53 8.0.2 Application of Banach Fixed Point Theorem Here is an application of the Banach fixed point theorem to the local exis- tence of solutions of ordinary differential equations.
Theorem 8.11 (Picard’s Theorem). Let U be an open subset of R2 and let f : U R be a continuous function which satisfies the Lipschitz condition with respect→ to the second variable, that is,
f(x,y1) f(x,y2) α y1 y2 | − |≤ | − | for all (x,y1), (x,y2) U, and some α > 0. Then for a given (x0,y0) U ∈ ∈ there is δ > 0 so that the differential equation
y′(x)= f(x,y(x)) has a unique solution y : [x0 δ, x0 + δ] R such that y(x0)= y0. − → Proof. Note that it is enough to show that there are δ > 0 and a unique function y : [x δ, x + δ] R such that 0 − 0 → x
y(x)= y0 + f(t,y(t))dt. x Z 0 Fix (x ,y ) U. Then we find δ > 0 and b > 0 such that if I = [x δ, x + δ] 0 0 ∈ 0 − 0 and J = [y0 b,y0 + b], then I J U. Since f is continuous and I J is closed and bounded,− f is bounded on ×I ⊂J. That is, f(x, y) M for some× M and all (x, y) U. Replacing δ by a smaller× number we| may| ≤assume that αδ < 1 and αM
For x x x + δ, 0 ≤ ≤ 0 x x
(Tg)(x) y0 = f(t,g(t))dt f(t,g(t)) dt M x x0 The same inequality holds for x0 δ x x0, and so Tg X for any g X. Since f is Lipschitz with respect to− the≤ second≤ variable, we obtain∈ for g,h X∈and ∈ 54 x [x , x + δ], ∈ 0 0 x (Tg)(x) (Th)(x) = [f(t,g(t)) f(t,h(t))] dt | − | x − Z 0 x f(t,g(t)) f(t,h(t)) dt ≤ x | − | Z 0 α x x d(g,h) < αδd(g,h). ≤ | − 0| Similarly, (Tg)(x) (Th)(x) α x x0 d(g,h) < αδd(g,h) for x [x0 δ, x0]. Since αδ <| 1, T is a− contraction| ≤ and| in− view| of Banach’s fixed point theorem∈ − there exists a unique continuous function y : I J such that → x y(x) = (Ty)(x)= y0 + f(t,y(t))dt. x Z 0 55