Chapter 1 Metric spaces 1.1 Metric and convergence We will begin with some basic concepts. Definition 1.1. (Metric space) Metric space is a set X, with a metric satisfying: 1. d(x, y) 0, d(x, y) = 0 x = y, 2. d(x, y) = d(y, x), 3. d(x, y) d(x, z) + d(z, y). d : X X R + = [0, + ) From now on, we will represent a metric space with (X, d). Here are some examples: Example 1.2. X = R n = (x 1, x 2,, x n ), d(x, y) = (x 1 y 1 ) 2 + (x 2 y 2 ) 2 + + (x n y n ) 2. Example 1.3. l p = (x 1, x 2,, x n, ), d(x, y) = ( (x i y i ) p ) 1 p, p 1. Example 1.4. X = C[a, b] = {f(x) f(x) is continuous in [a, b]}, d(f, g) = max f(x) g(x), x [a, b]. Example 1.5. X = C[a, b] = {f(x) f(x) is continuous in [a, b]}, d(f, g) = [a,b] f(x) g(x) dx. Definition 1.6. If there exists a seuence {x n } (X, d) and x (X, d), we will say x n x, iff d(x n, x) 0. Definition 1.7. Suppose that (X, d) is a metric space, A is an open subset of X, if x A, δ > 0, B δ (x) = {y d(x, y) < δ} A. A subset B X is a closed subset of X, if B c = {x X x / B} is open. 1
2 CHAPTER 1. METRIC SPACES Example 1.8. X = R, then the interval (0, 1) is an open subset, the interval [0, 1] is a closed subset. The interval [0, 1) is neither an open subset nor a closed subset. Proposition 1.9. We have following basic properties for open and closed sets: 1. Unions of open sets is open. 2. Finite intersection of open sets is open. 3. Intersection of closed sets is closed. 4. Finite union of closed sets is closed. Proof. According to the De Morgan s law, we have ( α A α ) c = α A c α and ( α A α ) c = α A c α. Hence we only need to prove the first two properties. Suppose that A = A α, A α is open. x A, α 0 I, such that x A α0. Since A α0 is α I open, δ > 0 such that B δ (x) A α0. Then B δ (x) A, i.e. A = A α is an open set. Suppose that B = n A i where A i (i = 1,, n) is an open set. x B, so for every i = 1, 2,, n, there exists δ i, such that B δi (x) A i, then take δ = min{δ 1, δ 2,, δ n }, we can derive that B δ (x) B, i.e. B = n A i is an open subset. Definition 1.10. The interior Å of a set A is the union of all open sets which are contained in A. The closure B of a set B is the intersection of all closed sets which contain B. Definition 1.11. If a set A satisfies Ā = X, then A is said to be dense in X. Definition 1.12. Metric space (X, d) is said to be separable, if there exists a countable dense subset A X. α I 1.2 Completeness Definition 1.13. (X, d) is a metric space, if lim d(x n, x m ) = 0, we say that {x n } n,m is a Cauchy seuence. The metric space (X, d) is called a complete metric space if every Cauchy seuence converges in X, i.e. x X, s.t. d(x n, x) 0, n. Example 1.14. R is a complete metric space and [0, 1] is also a complete metric space. Proof. First, let {x n } be a Cauchy seuence in [0, 1]. Then {x n } is a Cauchy seuence in R. Hence x n x R. Second, [0, 1] is closed x [0, 1] x n converges in [0, 1].
1.2. COMPLETENESS 3 Example 1.15. (0, 1) is not a complete metric space. Proof. Take x n = 1 n, x n 0 / (0, 1), but x n is a Cauchy seuence in (0, 1), then we will say that (0, 1) is not a complete metric space. We have the following theorem, whose proof is one of our exercises. Theorem 1.16. Given a subset Y of a complete metric space (X, d), the metric space (Y, d) is complete iff Y is a closed subset. Definition 1.17. Let X be a metric space. A set E X is said to be nowhere dense if its closure E has an empty interior. The sets of the first category in X are that are countable unions of nowhere dense sets. Any subset of X that is not of the first category is said to be of the second category in X. Proposition 1.18. We have the following properties on the category. (a) If A B and B is of the first category in X, so is A. (b) Any countable union of sets of the first category is of the first category. (c) Any closed set E X whose interior is empty is of the first category in X. Theorem 1.19. (Baire Category Theorem) If (X, d) is a complete metric space, then the intersection of every countable collection of dense open subsets of X is dense in X. Corollary 1.20. A complete metric space is of the second category. Proof of the corollary. Suppose not. Let {E i } be a countable collection of nowhere dense subset of X and X = i E i. Denote V i = X \ E i. x X, if x / V i, then x E i. Since E i is nowhere dense, E i is nowhere dense. Hence n N, x n E i c = Vi such that x n B 1/n (x). Hence x V i. Thus each V i is dense. By Baire category Theorem, V i. Therefore, i E i = ( i V i ) c X. Hence X cannot be the union of countable nowhere dense sets. Therefore, X is of the second category. Proof of Baire Category Theorem. Suppose that V 1, V 2,... are dense open subsets of X. Let U 0 be an arbitrary nonempty open set in X, If n 1 and an open set U n 1 has been chosen, then there exists an open set U n V n U n 1 where we use the fact that V n is dense and U n may be taken to be a ball of radius less than 1 n. Put K = Note that the centers of the balls U n form a Cauchy seuence which converges to some point of K, and so K. Our construction shows that K U 0 and K V n for each n. Hence U 0 intersect V n. U n
4 CHAPTER 1. METRIC SPACES Definition 1.21. (X, d X ), (Y, d Y ) are two metric spaces. The mapping T : X Y is called an isometry if d Y (T x 1, T x 2 ) = d X (x 1, x 2 ) If it is also onto, then it is called isometric-isomorphism, we represent this by X = T (X). As we mentioned in the above, not all the metric spaces are complete metric spaces(see the example 1.15), however, by the definition of isometry, we can define the completion of a metric space. Definition 1.22. A metric space ( X, d) is a completion of (X, d) if the following conditions are satisfied: (a) there is an isometry σ : X X; (b) σ(x) is dense in X, i.e. σ(x) = X; (c) ( X, d) is complete. Now we introduce the main theorem of the completion of metric spaces. Theorem 1.23. Every metric space has a completion. Example 1.24. Q represents the set of rational numbers, so R is the completion of Q. Example 1.25. Let Then C[0, 1] = {f(x) f(x) is continuous in [0, 1]} with metric d 1 (f, g) = L 1 [0, 1] = {f(x) is the completion of the metric space (C[0, 1], d 1 ). [0,1] f(x) dx < } 1 0 f(x) g(x) dx. 1.3 Compactness Compactness is one of the most important concepts in analysis. We will define compact sets in a metric space by means of seuence. Definition 1.26. K X is seuentially compact if every seuence of K has a convergent subseuence whose limit is also in K. Example 1.27. X = R, K = [0, 1] is a seuentially compact subset, but K = (0, 1) is not a seuentially compact subset. By the well known Bolzano-Weistrass theorem which we have already learnt in the mathematical analysis, the following theorem holds
1.3. COMPACTNESS 5 Theorem 1.28. K R n is seuentially compact iff it is bounded and closed. In the following, we will find an euivalent criterion for the seuentially compactness of a metric space that is easier to verify. First, we introduce the definition of ε-net of a subset Ω. Definition 1.29. Let Ω X, we say that F X is an ε-net of Ω if Ω reuired to be a subset of Ω). x F B ε (x) (F is not Definition 1.30. A subset of a metric space is totally bounded if it has a finite ε-net for any ε > 0. Theorem 1.31. A subset K X as a metric space is seuentially compact iff it is complete and totally bounded. Proof. Step 1. K is seuentially compact K is complete and totally bounded. If {x n } is a Cauchy seuence in K {x n } has convergent subseuence x nk x n x K K is complete. x K ε > 0, choose any x 1 K, if K B ε (x 1 ), then we can end the proof. If it is not true, then x 2 K\B ε (x 1 ), we continue this step, if it can be ended in finite steps, i.e. K n B ε (x i ). If it can t be ended in finite steps, we can find a seuence {x i } + with d(x i, x j ) > ε, i, j, if j > i, then x j / i k=1 B ε (x k ), since K is seuentially compact x nk x, this is a contradiction with d(x nk, x nl ) > ε, n k n l, then we can find finite many balls such that K n B ε (x i ), i.e. K is totally bounded. Step 2. K is totally bounded and complete K is seuentially compact. Let {x n } be a seuence in K, for ε n = 1 2, a finite ε n n -net, {x n 1,, xn l n }, such that K ln B 1 (xn 2 n i ), when n = 1, K l 1 B 1 (x 1 i ), because there are only finite many balls that 2 contain {x n } + x1 i 1, s.t. B 1 (x 1 i 2 1 ) contains infinite many {x n }, we denote it by {x n 1 }, k k = 1, 2, When n = 2, ε = 1 4, K l 2 B 1 4 {x n 1 k }, we denote it by {x n 2 k }, which is a subseuence of {x n 1 k }. (x 2 i ); x2 i 2 {x n 1 }, s.t. B 1 (x 2 k i 4 2 ) contains infinite many of Repeat the step again and again, we can see that for every 1, d(x n k, x n m ) < 1 2 + 1 2 = 1, pick one element from each {x 2 1 n } diagonally, we will get a new subseuence {x k n {x n } is a subseuence of {x n } with the property that d(x n, x n p p ) < 1 }, then 2 1, for any p, i.e. {x n } is a Cauchy seuence, by the condition that K is complete, then {x n } x K, i.e. K is seuentially compact. The proof of the following lemma is an exercise. Lemma 1.32. A seuentially compact metric space is separable.
6 CHAPTER 1. METRIC SPACES Then we will introduce the definition of compactness and claim the euivalence between compactness and seuentially compactness. Definition 1.33. Let {A α } be a collection of open sets, if U α A α, then A α is called an open cover of U. Definition 1.34. A subset K X is compact if every open cover has a finite sub-cover. Theorem 1.35. A subset K of a metric space is compact iff it is seuentially compact. Proof. We first show that K is seuentially compact as long as K is compact. Step 1. Every seuence has a convergent subseuence. If {x n } does not exist a convergent subseuence, denote Then A n is closed. On the other hand, A n = {x 1, x 2,, x n 1, x n+1, } (X \ A n ) = X \ A n = X \ = X K. Since K is compact, one has N (X \ A n) K. Therefore, we have X \ {x n } n=n+1 K. But {x n } n=n+1 K. There is a contradiction. Step 2. K is closed. x 0 X \ K, we have K x K B 1 2 d(x,x 0) (x) Since K is compact, there is a finite open cover for K. Assume that K n B 1 2 d(x i,x 0 ) (x i) Let δ = 1 4 min 1 i n d(x i, x 0 ). Then for any x B δ (x 0 ), we have d(x k, x) d(x k, x 0 ) d(x 0, x) d(x k, x 0 ) 1 4 d(x k, x 0 ) = 3 4 d(x k, x 0 ) This implies that x / B 1 2 d(x i,x 0 ) (x i) for all i = 1,, n. Hence B δ (x 0 ) K =. Therefore, K is a closed set. Step 3. Seuentially compact must be compact. Suppose not, there exists open cover {A α } which does not have a finite cover. Since K is seuentially compact, n N, an 1 2n -net Therefore, we have Ñ n = {y (n) 1, y(n) 2,, y(n) k(n) } K k(n) B 1 (yi n ). 2n Hence, for i = 1,, k(n), one can choose x n i B 1 (yi n ) K such that 2n K k(n) B 1 n (x n i ).
1.4. CONTINUOUS FUNCTIONS 7 Denote N n = {x (n) 1, x(n) 2,, xn k(n) } Therefore, n N, x n N n such that B 1/n (x n ) cannot be covered by finite sets in A α. Since K is seuentially compact, there exists a subseuence {x nk } such that x nk x 0 A α0 Since A α0 is open, there exists a δ > 0 such that B δ (x 0 ) A α0. If n k is sufficiently large, we have d(x nk, x 0 ) < δ/2. Therefore, This yields a contradiction. x B 1 n k (x nk ) B δ (x 0 ) Definition 1.36. A subset K of a metric space is pre-compact if its closure K is compact. 1.4 Continuous functions Definition 1.37. (Continuous functions) Given two metric spaces (X, d X ) and (Y, d Y ), we say that f : X Y is continuous, if x X lim n f(x n) = f(x), whenever x n x. Proposition 1.38. f : X Y is continuous at x 0 if and only if ɛ > 0, δ > 0, one has d Y (f(x), f(x 0 )) < ɛ provided that d X (x, x 0 ) < δ. Proof. The sufficient part is trivial. Now let s prove the necessary part. Suppose not, then ɛ 0, n, x n such that d X (x n, x 0 ) < 1/n, d Y (f(x n ), f(x 0 )) ɛ 0. In fact, we have the following characterization for the continuous functions. Proposition 1.39. f is continuous function from (X, d X ) to (Y, d Y ) iff f 1 (U) is an open set for any open set U X. We can also define so called lower and upper semicontinuous functions from a metric space (X, d) to R. Definition 1.40. A function f : X R is called lower (upper) semicontinuous if lim inf n f(x n) f(x) and lim sup f(x n ) f(x), n respectively, whenever x n x as n. We will denote the class of all continuous functions from X to Y by C(X, Y ), i.e. C(X, Y ) = {f : X Y f is continuous}
8 CHAPTER 1. METRIC SPACES Example 1.41. We define discrete metric space(x, d D ) as: { 0, x = y d D (x, y) = 1, x y Then, we can say that C((X, d D ), Y ) = all maps from X to Y. We can also have the notation as uniform continuity. Definition 1.42. Given f C(X, Y ), we say that f is uniformly continuous, if ε > 0, δ > 0, for any x 1, x 2 X satisfying d X (x 1, x 2 ) < δ, one has d Y (f(x 1 ), f(x 2 )) < ε. The proof of the following two theorems is similar to the ones of mathematical analysis. We leave them as exercises. Theorem 1.43. f C(K, R), K is compact, then f is uniformly continuous. Theorem 1.44. f C(K, R), K is compact, then f is bounded and attains its maximum and minimum. We will discuss the compactness of a special metric space that is composed of the continuous functions. C B (X) = {f : X R f is continuous and bounded} with metric d(f, g) = sup f(x) g(x) x X Then we have the following theorem. Theorem 1.45. The metric space (C B (X), d) is complete. When X is compact, we just write the above metric space C(X) for simplicity. Definition 1.46. F C(X, Y ) is eui-continuous if for any x X, ε > 0, δ(ε, x) such that y X with d X (x, y) < δ, one has d Y (f(x), f(y)) ε for any f F. If δ does not depend on x, then we say that F is uniformly euicontinuous. Theorem 1.47. An eui-continuous family F C(K, Y ) of functions from a compact metric space K to a complete metric space Y is uniformly-eui-continuous. Proof. We prove the theorem by contradiction argument. Step 1. Suppose the theorem is not true, ε 0 > 0, for δ n = 1 n > 0, x n, y n K, f n F, such that d(f n (x n ), f n (y n )) ε 0 with d(x n, y n ) 1 n. Since K is compact, {x n } has a convergent subseuence x nk x K.
1.4. CONTINUOUS FUNCTIONS 9 Step 2. By the condition F is eui-continuous at x, for ε = ε 0 10 d(x, x) < 1 N, then d(f(x), f(x)) < ε 0 10. For k sufficiently large, we have > 0, N, such that if ε 0 d(f nk (x nk ), f nk (y nk )) d(f nk (x nk ), f nk (x)) + d(f nk (y nk ), f nk (x)) ε 0 10 + ε 0 10 <ε 0. This is a contradiction. Therefore, F is uniformly-eui-continuous. Here, we will state an important theorem which describes the compact criteria for C(K). Theorem 1.48. (Arzela-Ascoli theorem) Let K be a compact metric space. A subset F of C(K) is compact iff it is closed, bounded, and eui-continuous. Proof. Step 1. compact set is closed, bounded, and eui-continuous. Note that a compact set is seuentially compact, hence it is complete and totally bounded. Since F is compact, then F is complete, i.e. F is a closed subset of a complete metric space C(K) and it is totally bounded. What left is to show that F is eui-continuous. In fact, because F is totally bounded. ε > 0, ε a finite 100 -net, i.e. f 1, f 2,, f N, such that F N B ε (f i), i.e. for any f F, f i0, 100 i 0 {1, 2, N}, s.t. f(x) f i0 (x) < ε, x K. 100 On the other hand, x K, δ > 0, if d(x, y) < δ, one has Hence for any f F, one has f i (x) f i (y) < ε (i = 1, 2,, N). 100 f(x) f(y) f(x) f i (x) + f i (x) f i (y) + f(y) f i (y) ε 100 + ε 100 + ε 100 <ε, (1.1) i.e. F is eui-continuous. Step 2. closed, bounded and eui-continuous set is compact. Suppose that F is a bounded, euicontinuous subset of C(K). We will show that every seuence {f n } in F has a convergent subseuence. There is a countable dense set {x 1, x 2, x 3...} in the compact domain K. We choose a subseuence {f 1,n } of {f n } such that the seuence of values {f 1,n (x 1 )} converges in R, Such a subseuence exists because {f n (x 1 )} is bounded in R, since F is bounded in C(K). We choose a subseuence {f 2,n } of {f 1,n } such that {f 2,n (x 2 )} converges, which exists for the same reason. Repeating this procedure, we obtain seuences {f k,n } for k = 1, 2,... such that {f k,n } is a subseuence of {f k 1,n }, and {f k,n (x k )} converges as n. Finally, we define a
10 CHAPTER 1. METRIC SPACES diagonal subseuence {g k } by g k = f k,k. By construction, the seuence {g k } is a subseuence of {f n } with the property that {g k (x i )} converges in R as k for all x i in a dense subset of K. So far, we have only used the boundedness of F. The euicontinuity of F is needed to ensure the uniform convergence of {g k }. Let ɛ > 0. Since F is euicontinuous and K is compact, therefore, F is uniformly euicontinuous. Conseuently, ɛ > 0, there is a δ > 0 such that d(x, y) < δ implies g k (x) g k (y) < ɛ 3. Since {x i } is dense in K, we have K B δ (x i ). Since K is compact, there is a finite subset of {x i }, which we denote by {x 1,..., x n }, such that n K B δ (x i ). The seuence {g k (x i )} k=1 is convergent for each i = 1,, n, and hence is a Cauchy seuence for each i = 1,, n, so there is an N such that g j (x i ) g k (x i ) < ɛ 3. for all j, k N and i = 1,..., n. For any x K, there is an i such that x B δ (x i ). Then, for j, k N, we have g j (x) g k (x) g j (x) g j (x i ) + g j (x i ) g k (x i ) + g k (x i ) g k (x) < ɛ. It follow that {g k } is a Cauchy seuence in C(K). Since F is closed set in the complete space C(K), it converges to a limit in F. Hence F is compact. Now we take an example to see the applications of Arzela-Ascoli theorem. Example 1.49. Prove that F = {f f(x) = a n sin(nπx) with n a n 1 for x [0, 1]} is a compact set in C[0, 1]. Proof. Step 1. It is easy to see that F is bounded in C[0, 1]. f C[0, 1]. Suppose that f k is represented as Let {f k } F converge to f k = a k,n sin(nπx).
1.5. CONTRACTION MAPPING THEOREM 11 Then we have a k,n = 1 0 f k (x) sin(nπx)dx. Since {f k } converges to f uniformly on [0, 1], one has lim a k,n = k Therefore, for any N N, we have This implies that 1 0 N n a n = lim f(x) sin(nπx)dx := a n. k n a n 1. Hence f F. This yields that F is closed. N n a k,n 1. Step 2. By the mean value theorem, for any x < y R there is a x < ξ < y with sin x sin y = (cos ξ)(x y). Hence, for all x, y R we have sin x sin y x y. Thus, every f F satisfies f(x) f(y) a n sin(nπx) sin(nπy) πn a n x y π x y. Therefore, given ɛ > 0, we can pick δ = ɛ/π, and then x y < δ implies f(x) f(y) < ɛ for all f F. From the Arzleà-Ascoli theorem, F is a compact subset of C([0, 1]). In the above example, we can see that a special class of functions f, Lipschitz functions, which satisfy d(f(x), f(y)) Ld(x, y) must be uniformly eui-continuous. We will study a special class of Lipschitz mapping in the next section. 1.5 Contraction mapping theorem Definition 1.50. (Contraction map) Let(X, d) be a metric space. A map T : X X is a contraction map if there exists a fixed θ [0, 1), such that d(t x, T y) θd(x, y), x, y X. (1.2) Now let us state a fixed point theorem of the contraction map which can be a useful tool to prove the existence of solutions of the euations.
12 CHAPTER 1. METRIC SPACES Theorem 1.51. (Contraction map theorem) Let (X, d) be a complete metric space, T : X X is a contraction map satisfying (1.2), then T has a uniue fixed point x X such that T x = x. Proof. Step 1. (Existence). Pick any x 0 X, and define It follows from (1.2) that one has x n = T x n 1 for n 1. d(x n+1, x n ) = d(t (x n ), T (x n 1 )) θd(x n, x n 1 ) θ 2 d(x n 1, x n 2 ) θ n d(x 1, x 0 ). Then for m = n + k > n d(x m, x n ) d(x n+k, x n+k 1 ) + d(x n+k 1, x n+k 2 ) + + d(x n+1, x n ) θ n d(x 1, x 0 )(θ k 1 + + 1) θn 1 θ d(x 1, x 0 ). Thus {x n } is a Cauchy seuence. Then we take limit on x n+1 = T (x n ) x = T (x), where x = lim x n. n + Step 2. (Uniueness). If z such that T (z) = z, then Therefore, we have d(x, z) = 0. Hence x = z. d(x, z) = d(t (x), T (z)) θd(x, z). (1.3) Then let us see two examples of the applications of the above fixed point theorem of contraction map. Example 1.52. Consider the initial value problem for the ordinary differential euation, { u (t) = f(t, u), u(t 0 ) = u 0, where f(t, u) is continuous on [t 0 h, t 0 + h] [u 0 b, u 0 + b](then M > 0, s.t. f(t, u) M), what s more, f(t, u) is Lipchitz continuous with respect to u, i.e. L > 0, s.t. f(t, u 1 ) f(t, u 2 ) L u 1 u 2. Then we can use the contraction mapping theorem to prove the existence and uniueness of a solution for the problem (1.4) on [t 0 ɛ, t 0 + ɛ] for some ɛ > 0. Proof. Step 1. Define X = {u C[t 0 ɛ, t 0 + ɛ] u(t 0 ) = u 0, u(t) u 0 b for t [t 0 ɛ, t 0 + ɛ]}, where ɛ (0, h] is to be determined. Define T : X X as t (T u)(t) = u 0 + f(τ, u(τ))dτ. t 0 (1.4)
1.5. CONTRACTION MAPPING THEOREM 13 We claim that T maps X into itself. It is easy to see that for any u X, we have T u C[t 0 ɛ, t 0 + ɛ] and (T u)(t 0 ) = u 0. Furthermore, we have t T u(t) u 0 = f(τ, u(τ))dτ M t t 0 Mɛ. t 0 Therefore, if ɛ M/b, then T maps X to itself. Step 2. If ɛ is sufficiently small, then T is a contraction map. Indeed, t t d(t u, T v) = (f(τ, u(τ)) f(τ, v(τ)))dτ L u(τ) v(τ) dτ L t t 0 d(u, v). t 0 t 0 Hence if then we have { ɛ min h, M } b, 1, 2L d(t u, T v) 1 d(u, v). 2 So T is a contraction map. Step 3. By Contraction map theorem, u, such that T (u) = u, i.e., t ū = u 0 + f(τ, ū(τ))dτ. t 0 This is exact the solution of the problem (1.4). Furthermore, the contraction map theorem also gives the uniueness of the solution. Example 1.53. Let K be a convex and compact subset of R n, T : K K satisfies the condition T (s) T (t) s t, show that T has at least one fixed point(be careful: the map T here may not be a contraction map, so we can not use the contraction map theorem directly.) Proof. Pick up any x K. Define T n : K K as follows We claim that T n is a contraction map. In fact This means that n, x n, such that T n x = (1 1 n )T x + 1 n T x. T n (x) T n (y) (1 1 n ) T x T y (1 1 ) x y. n (1 1 n )T x n + 1 n T x = T n(x n ) = x n. (1.5) Since K is compact, then {x nj } and x K, s.t. {x nj } x, then take limit on both sides of (1.5), we can derive that T x = x. So T has at least one fixed point.