Lecture Notes in Advanced Calculus 1 (80315) Raz Kupferman Institute of Mathematics The Hebrew University


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1 Lecture Notes in Advanced Calculus 1 (80315) Raz Kupferman Institute of Mathematics The Hebrew University February 7, 2007
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3 Contents 1 Metric Spaces Basic definitions Point set topology in metric spaces Compactness Continuous functions on metric spaces Convergence of sequences of functions Completeness Contractive mapping theorem Baire s category theorem Paths and integration along paths Definitions Differentiable paths in R n Homotopy and connectedness Sets of measure zero and the Peano curve Integration of a vector field along a path Conservative fields Differential Calculus in R n Normed spaces Linear mappings between normed spaces Differentiability and derivatives
4 ii CONTENTS 3.4 Multivariate mean value and Taylor theorems Minima and maxima The inverse function theorem The implicit function theorem Lagrange multipliers Integration in R n 127
5 Chapter 1 Metric Spaces Lindenstrauss, Chapters IV and XI. 1.1 Basic definitions Discussion: say something about abstraction and generalization of concepts we ve met in previous courses. Definition 1.1 (Metric space) A metric space is a pair (X, d), where X is a nonempty set, and d : X X R + is a distance function, or, a metric. The metric d assigns to every pair of points x, y X a nonnegative number, d(x, y), which we will call the distance between x and y. The metric has to satisfy three defining properties: ➀ Symmetry: d(x, y) = d(y, x). ➁ Positivity: d(x, y) 0 with equality iff x = y. ➂ The triangle inequality, d(x, z) d(x, y) + d(y, z). Comment: It follows by induction that for every finite set of points (x i ) n i=1, d(x 1, x n ) d(x 1, x 2 ) + d(x 2, x 3 ) + + d(x n 1, x n ).
6 2 Chapter 1 Comment: Metric spaces were first introduced by Maurice Frechet in 1906, unifying work on function spaces by Cantor, Volterra, Arzelà, Hadamard, Ascoli, and other. Topological spaces were first introduced by Felix Haussdorf in 1914; their current formulation is a slight generalization from 1992 by Kazimierz Kuratowski. Examples: Both R and C are metric spaces when endowed with the distance function d(x, y) = x y. From now on, we will often refer to the metric space R, without explicit mention of the metric. Let F : R R be any strictly monotonic function (say increasing). Then, R endowed with the distance function d(x, y) = F(x) F(y) is a metric space. We only need to verify the triangle inequality: d(x, z) = F(x) F(z) F(x) F(y) + F(y) F(z) = d(x, y) + d(y, x). Note that the strict monotonicity is only needed to ensure the positivity of d. R n endowed with the Euclidean distance, n d(x, y) = (x i y i ) 2 is a metric space. Unless otherwise specified, R n will always be assumed to be endowed with the Euclidean metric. The surface of earth is a metric space, with d being the (shortest) distance along great circles. Any set X can be endowed with the discrete metric, 0 x = y d(x, y) = 1 x y. This is a particularly boring metric, since it provides no information about the structure of the space. We will often use it, however, as a clarifying example. i=1
7 Metric Spaces 3 Let (X, d) be an arbitrary metric space, and let Y X. Then the set Y with the function d restricted to Y is a metric space. We will call d Y the metric on Y induced by the metric on X. A subspace of a metric space always refers to a subset endowed with the induced metric. Exercise 1.1 (Ex. 3 in L.) Prove that the set of infinite sequences of real numbers, endowed with the function 1 x n y n d((x n ), (y n )) = 2 n 1 + x n y n is a metric space. n=1 Recall the definition of a normed space: a vector space X endowed with a function : X R (the norm), subject the three conditions of positivity, homogeneity, and the triangle inequality. Obviously, every normed space is a metric space, if we define d(x, y) = x y. Note, however, that normed spaces carry properties that are not pertinent to all metric spaces. Metric spaces need not be vector spaces. In addition, vector spaces, as metric spaces, are translational invariant, and homogeneous, where λ is a scalar. Examples: d(x z, y z) = d(x, y), d(λx, λy) = λ d(x, y), Both R and C are normed space, with x = x, and the resulting metric is the same as the one before. For X = R n and 1 < p <, the function p : X R defined by 1/p n x p = x i p i=1 is a norm, hence defines a metric. We give this family of metric spaces a name, l n p.
8 4 Chapter 1 The function : R n R defined by x = max 1 i n x i. is a norm. The corresponding metric space is denoted by l n. The set of infinite sequences (x i ) for which x p = x i p i=1 is finite is an example of an infinitedimensional normed space (hence an infinitedimensional metric space). We denote this space by l p. Consider C[0, 1], the set of continuous realvalued functions on [0, 1]. The function : C[0, 1] R defined by 1/p f = max 0 x 1 f (x) is a norm, therefore C[0, 1] is a metric space if endowed with the distance function d( f, g) = max f (x) g(x). 0 x 1 Exercise 1.2 Show that the functions p : R n R, defined by n x p = x i p i=1 1/p 1 < p < satisfy the definition of a norm. (Hint: the triangle inequality is proved through the Young inequality, followed by Hölder s inequality, followed by Minkowski s inequality. Prove all three inequalities.) Exercise 1.3 (Ex. 4 in L.) Let x = (x 1, x 2,..., x n ). Prove that the function ϕ(p) = x p defined for 1 < p < is monotonically decreasing, and that ϕ(p) x as p.
9 Metric Spaces Point set topology in metric spaces Throughout this section we will assume a metric space (X, d). Definition 1.2 A sphere of radius r centered at a X is the set of points, {x X : d(x, a) = r}. (Note that this set may be empty). An open ball of radius r centered at a is the set X a r whereas the closed ball is defined by B 0 (a, r) = {x X : d(x, a) < r}, B(a, r) = {x X : d(x, a) r}. Comment: In R 3 these definitions coincide with our intuitive pictures of spheres and balls. Note, however, that for a space endowed with the discrete metric we always have {a} r 1 B 0 (a, r) = X r > 1. Exercise 1.4 Draw the unit balls in R 2 for the metrics induced by the 1, 2, and norms. Definition 1.3 A set A X is called bounded if it is enclosed in some ball, it is called open if every point x A is the center of an open ball contained in A (note that we can replace open by closed, since if B 0 (a, r) A, then B(a, r/2) A). Comment: By definition, X itself is always an open set. By convention, the empty set is always considered to be open as well. A A A A bounded set Proposition 1.1 In every metric space (X, d), An open set ➀ The intersection of any finite number of open sets is open. ➁ The union of any collection (not necessarily countable) of open sets is open. ➂ An open ball is an open set.
10 6 Chapter 1 ➃ A set is open iff it is the union of open balls. Proof : ➀ Let (A i ) n i=1 be a finite collection of open subsets of X and set A = n i=1 A i. If A = then we are done. Otherwise, for every x A we need to show that there exists an r > 0 such that B 0 (x, r) A. Since x belongs to each of the A i, which are all open in X, there exists a collection of positive numbers (r i ) n i=1 such that B 0 (x, r i ) A i, i = 1,..., n. Take, for example, r = min n i=1 r i, then B 0 (x, r) A i, i = 1,..., n, a d(a,x) x r i.e., B 0 (x, r) A. ➁ Every point x in the union A = α A α belongs to at least one open set A β, therefore there exists an r > 0 such that B 0 (x, r) A β A. ➂ Let B 0 (a, r) be an open ball, and let x B 0 (a, r). We claim that the open ball B 0 (x, r d(a, x)) in contained in B 0 (a, r). Indeed, if y B 0 (x, r d(a, x)), then by definition, d(y, x) < r d(a, x), i.e., d(a, y) d(a, x) + d(x, y) < r, so y B 0 (x, r d(a, x)) implies y B 0 (a, r). ➃ We have already seen that every open ball is open and that any union of open sets is open. Thus, it only remains to show that every open set can be represented as a union of open balls. By definition, to every point x in a open set A corresponds an open ball B 0 (x, r(x)) included in A. Clearly, A = x B 0 (x, r(x)) A. x A x A (2 hrs)
11 Metric Spaces 7 Comment: An infinite intersection of open sets is not necessarily open. Take for example X = R with the standard metric. Then for every integer n the set ( 1/n, 1/n) is open. On the other hand, ( 1 n, 1 ) = {0} n is not open. n=1 Definition 1.4 A set A X is called closed it is complement A c is open. The following proposition is a direct consequence of the previous proposition, using de Morgan s laws, Proposition 1.2 ➀ Every finite union of closed sets is closed. ➁ Every arbitrary intersection of closed sets is closed. Exercise 1.5 Show, by example, that it is not generally true that an infinite union of closed sets is closed. Comments: X and are both open and closed. For every x X the singleton {x} is a closed set. Every y x is the center of an open ball B(y, d(x, y)) {x} c. By the previous proposition, every finite collection of points is closed as well. Every closed ball B(a, r) is closed. Indeed, if y B c (a, r) then B 0 (y, d(y, a) r) B c (a, r). In a set endowed with the discrete metric, every subset is both open and closed. To prove it, it is sufficient to show that every singleton is both open and closed. We have already seen that singletons are always closed, regardless of the metric. They are also open since B 0 (x, 1/2) = {x}. A metric space endowed with the collection of its open subsets is an instance of a topological space (a set with a collection of subsets, the open sets, that are closed under finite intersection, arbitrary union, and contain the empty set and the whole space).
12 8 Chapter 1 Definition 1.5 (Limit of a sequence) A sequence (x n ) in a metric space (X, d) is called convergent if there exists a point x X such that lim d(x n, x) = 0. n (For every ɛ > 0 there exists an N ɛ such that for every n > N ɛ, d(x n, x) < ɛ.)the point x is called the limit of the sequence (x n ), and we write Comments: lim x n = x or x n x. n If a sequence converges then its limit is unique. Indeed, suppose that a sequence (x n ) has two limits x and y. Then, for every ɛ > 0 there exists a (sufficiently large) n such that d(x n, x) < ɛ and d(x n, y) < ɛ, from which we conclude that for every ɛ > 0, i.e., d(x, y) = 0. d(x, y) d(x, x n ) + d(x n, y) < 2ɛ, This notion of convergence coincides with that of convergence in topological spaces: (x n ) converges to x if for every open set A that contains x there exists an N A such that x n A for all n > N A (the sequence is eventually confined to every neighborhood of x). Exercise 1.6 Let (X, ρ) be a metric space and define for every x, y X, d(x, y) = ➀ Show that d is a metric on X. ρ(x, y) 1 + ρ(x, y). ➁ Show that convergence with respect to this new metric d is equivalent to convergence with respect to the original metric ρ. That is, prove that a sequence (x n ) is convergent in (X, d) if and only if it is convergent in (X, ρ).
13 Metric Spaces 9 Exercise 1.7 Let X = {0, 1} N (the space of infinite binary sequences), and define for every two sequences (x n ), (y n ) X, d 1 ((x n ), (y n )) = 2 n x n y n n=1 d 2 ((x n ), (y n )) = 2 min(i x i y i ). ➀ Show that these are indeed metrics. ➁ Show that these two metrics are equivalent. Proposition 1.3 A subset A X is closed iff the limit of every convergent sequence of points (x n ) is also in A. That is, A is closed iff for every (x n ) A such that there exists an x X for which x n x, x A. A Proof : Suppose first that A is closed (i.e., its complement is open), and let (x n ) be a sequence in A with limit x (a priori, not necessarily in A). We need to show that x A, or that x A c. If, by contradiction, x A c, then it is the center of an open ball B 0 (x, a) A c. This implies that d(x n, x) > a for all n, in contradiction with x being the limit of x n. Suppose then that A has the property that every convergent sequence in A has its limit in A. We need to show that A c is open. If this weren t the case, there would exists a point x A such that all the open balls B 0 (x, 1/n) have a nonempty intersection with A. For every n take a point x n A B 0 (x, 1/n). We thus obtain a sequence in A that converges to x A, hence, a contradiction. x Definition 1.6 Let A X be a set in a metric space. Its interior, A, is the union of all open subsets of A (it is the largest open subset of A). Its closure, Ā, is the intersection of all closed sets that include A (it is the smallest closed set that includes A). Comment: Obviously, A is open and Ā is closed. Example: Let X = R and A = Q. Then, A = and Ā = R. Exercise 1.8 Prove that for every A X (A ) c = A c.
14 10 Chapter 1 Proposition 1.4 The closure of a set A is the set of points that are the limit of some sequence in A, Proof : Define Ā = {x X : (x n ) A such that x n x}. B = {x X : (x n ) A such that x n x}. We need to show that both Ā B and B Ā. (1) We first show that B Ā. Let x B. Then it is the limit of a sequence (x n ) A. Since A Ā, this sequence is also a sequence in Ā. But Ā is closed, hence the limit x must be in Ā, i.e., B Ā. (2) We then show that Ā B. We first argue that B is closed; indeed, every point in its complement must have an open neighborhood that does not intersect A, i.e., a neighborhood that does not intersect B. On the other hand, every point in A is also in B (the trivial sequence). Thus, B is a closed set including A, and therefore Ā B. Definition 1.7 The boundary A of a set A is defined as A = Ā \ A. Comment: The boundary is always closed since it is the intersection of two closed sets, A = Ā (A ) c Exercise 1.9 (Ex. 16 in L.) Let A, B be subsets of a metric space (x, d). Prove that ➀ A B = A B. ➁ A B A B. ➂ If A is open then ( A) =. ➃ A is closed iff A A. ➄ A is open iff A A =.
15 Metric Spaces 11 Exercise 1.10 Let A be a set in a metric space (X, d). Show that its boundary A coincides with the set of points x, such that every open ball with center at x intersects both A and A c. A!A Consider now a metric space (X, d). Any Y X is a metric space with the induced metric. Note, however, that a set A Y can be open in the metric space (Y, d), but closed in the metric space (X, d) (e.g., the set Y itself is always open in (Y, d), but could be closed in (X, d)). Proposition 1.5 Let (Y, d) be a subspace of the metric space (X, d). A set A Y is open in (Y, d) iff it is the intersection of Y and a set that is open in (X, d). The same applies if we replace open by closed. Proof : (1) Let A Y be open in (Y, d). we can express it as follows, A = y Y {z Y : d(z, y) < r(y)} = y Y Y {z X : d(z, y) < r(y)} = Y ( y Y {z X : d(z, y) < r(y)} ). (2) Let A X be open in (X, d) and set B = A Y. We need to show that B is open in (Y, d). Take y B A. Because A is open, there exists an r such that {z X : d(z, y) < r} A. It follows that {z Y : d(z, y) < r} A Y = B, hence B is open in (Y, d). (4 hrs) Exercise 1.11 (Ex. 5 in L.) A metric space (X, d) is called separable if there exists a countable set {x n } X such that X = {x n }. Show that every subspace (Y, d) of a separable metric space (X, d) is separable. Exercise 1.12 (Ex. 6 in L.) (a) Prove that in a normed space, (B(a, r)) = B 0 (a, r) and B 0 (a, r) = B(a, r). (b) Prove that this is not necessarily true in a general metric space.
16 12 Chapter 1 Exercise 1.13 (Ex. 12 in L.) A point a in a subset A of a metric space X is called an accumulation point of A if in any open ball centered at a there exists at least one point of A other than a. Prove that a is an accumulation point of A iff there exists a sequence of distinct points in A that converges to a. Exercise 1.14 (Ex. 18 in L.) Prove that the only sets in R that are both open and closed are R and. Exercise 1.15 ( Ex. 26 in L.) A point x A X is called isolated in A if there exists a ball B(x, r) such that B(x, r) A = {x}. Prove that if A R is a nonempty, countable and closed, then it contains at least one isolated point. Exercise 1.16 ( Ex. 27 in L.) Prove that l, the space of infinite sequence with the metric induced by the norm x = sup x n n is not separable. Hint: use Cantor s diagonalization technique. 1.3 Compactness Definition 1.8 (Compactness) A metric space (X, d) is called compact if every sequence (x n ) has a convergent subsequence. A subset K X is called compact if the induced subspace (K, d) is compact (i.e., if every sequence in K has a subsequence that converges to a limit in K). Comment: Note that the compactness of a set is a property pertinent only to the set, whereas openness/closedness is property of a set in relation with the entire space. Exercise 1.17 (Ex. 17 in L.) Let (X, d) be a metric space and (x n ) a sequence that converges to a limit x. Prove that the set x {x n } is compact. Proposition 1.6 A finite union of compact sets is compact. Proof : Given a sequence (x n ) n i=1 K i = A, K i compact, it must have an infinite subsequence in at least one of the K i s. Since this K i is compact, there is a subsubsequence that converges in K i hence in A.
17 Metric Spaces 13 Examples: Every metric space that has a finite number of elements is compact. A set in a discrete space is compact iff it is finite (if it is infinite, a sequence that never repeats itself has no converging subsequence). The BolzanoWeierstraß theorem implies that a closed segment on the line is compact. Proposition 1.7 A compact set in a metric space is both bounded and closed. Proof : Let K be a compact set in a metric space (X, d). Closed By definition, every sequence (x n ) K has a subsequence that converges to a point in K. In particular, if (x n ) is a sequence that converges to a point x X, the point x must be in K. It follows from Proposition 1.3 that K is closed. Bounded Suppose, by contradiction, that K was not bounded. Given a point a X, no open ball B 0 (a, n) contains all of K. Thus, we can find a point Then, we can find a point and it follows that x 1 K such that d(x 1, a) > 1. x 2 K such that d(x 2, a) > 1 + d(x 1, a), d(x 2, x 1 ) + d(x 1, a) d(x 2, a) > 1 + d(x 1, a), i.e., d(x 2, x 1 ) > 1. We can further find a point x 3 K such that d(x 3, a) > 1 + d(x 2, a), whose distance from both x 1 and x 2 is greater than 1. We can thus generate a sequence (x n ) K, which has no converging subsequence, in contradiction with K being compact. a x 1 x 2
18 14 Chapter 1 a A Comment: People often identify compact with bounded and closed. In general, the implication is only onedirectional, although we will soon see situations in which there is indeed a correspondence between the two. A simple counter example is the segment (0, 1), which is closed (as a space) and bounded, but not compact. Proposition 1.8 If (X, d) is a compact metric space, then every closed subset is compact. Proof : This is quite obvious. Let (x n ) K X. Since X is compact, (x n ) has a subsequence (x nk ) that converges to some limit x X. But because K is closed, this limit must be in K, hence K is compact. Corollary 1.1 In a compact metric space closedness and compactness are the same. Theorem 1.1 (BolzanoWeirestraß in R m ) In R m (endowed with the Euclidean metric) a set is compact iff it is closed and bounded. Proof : We only need to show that closedness and boundedness imply compactness. Let K R m be closed and bounded. Thus, there exists an M > 0 such that max i=1,...,n xi < M x K. Let now (x n ) = (xn, 1..., xn m ) be a sequence in K. Consider the sequence formed by the first component, (xn). 1 By the BolzanoWeierstraß theorem, (x n ) has a subsequence such that the first component has a limit x 1. Consider now the second component: the subsequence has a subsubsequence such that the second component converges to a limit x 2. We repeat this procedure n times, until we obtain a subsequence which converges to a limit x = (x 1,..., x m ) componentbycomponent. Since K is closed, this limit is in K, hence K is compact. Exercise 1.18 (Ex. 7 in L.) Prove that in a metric space in which all the closed balls are compact, the compact sets coincide with the sets that are closed and bounded. Definition 1.9 Let (X, d) be a metric space with A, B X and a X. Then we define the distance between a point and a set, and the distance between two sets: d(a, A) = inf d(a, x) and d(a, B) = inf d(x, y). x A x A,y B
19 Metric Spaces 15 Proposition 1.9 Let A, B X be nonempty, closed disjoint sets, such that at least one of them is compact. Then d(a, B) > 0. Comment: A counterexample (an example that highlights the importance of the conditions): let X = R 2 with A = {(x, 1/x) : x > 0} and B = {(x, 0) : x 0}. The sets A, B are nonempty, disjoint and closed (why?). Yet, d(a, B) = 0. There is no contradiction because neither A nor B is compact. Proof : Suppose w.l.o.g. that A were compact, i.e., every sequence in A has a subsequence that converges to a point in A. Assume, by contradiction, that d(a, B) = 0. Then, there exists sequences (x n ) A, (y n ) B, such that d(x n, y n ) 0. The sequence (x n ) has a subsequence (x nk ) with limit x A, Now, d(x, y nk ) d(x, x nk ) + d(x nk, y nk ) 0, from which follows at once that y nk x. But since B is closed, x B, contradicting the disjointness of A and B. Corollary 1.2 If A is open in (X, d) and K A is compact, then A c is closed and disjoint from K, from which follows that d(k, A c ) > 0. Definition 1.10 Let (X, d) be a metric space. A covering of X is a collection of subsets {A α }, whose union coincides with X. A covering is called a finite covering if the number of sets is finite. It is called an open covering if all the A α are open sets (in X). Let {A α } be a covering of X. One may ask whether we could still cover X with a smaller collection of subsets. Consider for example the case where X = R; the countable collection of subsets {(n 1, n + 1) : n Z} is a covering of R. The omission of any of those (n 1, n + 1) would fail to cover the point x = n. As the following theorem shows, compactness can be defined alternatively as every open covering can be reduced to a finite subcovering. Theorem 1.2 (HeineBorel) The following three properties of a metric space (X, d) are equivalent: ➀ X is compact. A (6 hrs) K d!k,a c "
20 16 Chapter 1 ➁ Every open covering of X has a finite subcovering. ➂ Every collection of closed sets {F α } α I has the property that if the intersection of every finite subcollection is not empty, then the intersection of the entire collection is not empty. Proof : The equivalence of properties 2 and 3 follows from the identity, ( α F α ) c = α F c α. Note that the Fα c are a collection of open sets. Now, {F α } α I : Fα c = X exists a finite J I s.t. α I is equivalent to {F α } α I : α I F c α c = exists a finite J I s.t. which in turn is equivalent to {F α } α I : F α = exists a finite J I s.t. α I Fα c = X α J α J F c α c F α =. Finally, this is equivalent to {F α } α I : for all finite J I, F α. F α α J α I α J =, We next show that 3 implies 1. Let (x n ) be a sequence, and consider the (not necessarily closed) sets A n = {x n, x n+1,... }. This is a decreasing sequence of sets that has the property that every finite subcollection has a nonempty intersection. By assumption, the countable intersection, n=1 A n is not empty: there exists a point x {x n, x n+1,... }. n=1
21 Metric Spaces 17 This point x belongs to each of the closed sets A k, and therefore in each of the A k there exists a point x nk whose distance from x is less than 1/k. Thus, we can construct a subsequence (x nk ) that converges to x, hence X is compact. It remains to show that 1 implies 2, for which we need two lemmas. Lemma 1.1 Let (X, d) be a compact metric space, and F an open covering of X. Then there exists an ɛ > 0 such that every open ball of radius ɛ (or less) is contained in at least one of the sets in F. Comment: Given an open covering, the supremum over all such ɛ is called the Lebesgue number of the covering. Proof : Suppose that the desired property does not hold, i.e., we can construct a sequence of open balls B 0 (x n, 1/n), that none of them is contained in any of the sets in F. Since X is compact, the sequence (x n ) has a convergent subsequence x nk x. Since F is a covering, there exists an open set A F such that x A. Moreover, since A is open, there exists an open ball B 0 (x, r) contained in A. Since x is a partial limit of (x n ), there exists infinitely many n for which d(x n, x) < r/2, and at least one for which 1/n < r/2. Thus, the set B 0 (x n, 1/n) is a subset of A F in contradiction with our assumption. Lemma 1.2 Let (X, d) be a compact metric space. Then for every ɛ > 0, X can be covered by a finite set of open balls of radius ɛ. X 1 2 Proof : Let ɛ > 0 be given and let x 0 X be an arbitrary point. If B 0 (x 0, ɛ) = X then we are done. Otherwise, there exists a point x 1 X \ B 0 (x 0, ɛ). If B 0 (x 0, ɛ) B 0 (x 1, ɛ) = X, then we are done, otherwise, etc. Thus, either the lemma is correct, or we can construct an infinite sequence of point (x n ) that are pairwise at a distance greater than ɛ. Such a sequence does not have a convergent subsequence, in contradiction to X being compact. Completion of the HeineBorel theorem: Let X be compact and let F be an open covering. By the first lemma there exists an ɛ > 0 such that every open ball of
22 18 Chapter 1 radius ɛ is contained in an element of F. By the second lemma, we can cover X with a finite number of such open balls. This finite collection of open balls is contained in a finite collection of sets in F, which is therefore a covering. Comment: The term compactness was introduced by Maurice Fréchet in It was primarily introduced in the context of metric spaces in the sense of sequential compactness. The covering compact definition has become more prominent because it allows us to consider general topological spaces, and many of the old results about metric spaces can be generalized to this setting. This generalization is particularly useful in the study of function spaces, many of which are not metric spaces. One of the main reasons for studying compact spaces is because they are in some ways very similar to finite sets. In other words, there are many results which are easy to show for finite sets, the proofs of which carry over with minimal change to compact spaces. Comment: A counterexample : the space X = (0, 1) is not compact, yet for every ɛ > 0 we can cover (0, 1) with a finite number of segments of size ɛ. There is no contradiction. We can form an open covering of (0, 1), F = {(1/(n + 2), 1/n) : n N}, which does not have a finite subcovering. The HeineBorel theorem refers to compact metric spaces. An analogous theorem applies for compact sets in metric spaces: Theorem 1.3 (HeineBorel for sets) A set K in a metric space X is compact iff for every collection of open sets {A α } α I, such that K α I A α, there exists a finite subcollection {A α } α J, such that K α J A α. Exercise 1.19 Prove it. Definition 1.11 A set A X is called dense in X if Ā = X (every nonempty open set in X intersects A). A metric space is called separable if is contains a dense countable subset. Example: R is separable because it contains the countable dense set Q. So is any R n, which contains the countable dense set Q n.
23 Metric Spaces 19 Proposition 1.10 A compact metric space is separable. Proof : If X is compact, then it can be covered with a finite number of open balls of size 1. The centers of these balls form the first elements of a countable set. The next elements are the finite number of points that are the centers of open balls of size 1/2 that cover X. Thus, we proceed, forming a countable set of points in X. It is obvious that this set is dense in X. Exercise 1.20 (Ex. 8 in L.) Let (X, d) be a metric space in which the closedandbounded sets are compact (like R n ). Prove that for any a X and F X closed, there exists a point f F such that d(a, f ) = d(a, F). Exercise 1.21 (Ex. 20 in L.) Show that the finite product space of compact spaces is compact: Let (X 1, d 1 ),..., (X n, d n ) be metric spaces and consider the product space and X = n { (x1,..., x n ) : x j X j j }, i=1 d(x, y) = max j=1,...,n d i(x i, y i ). (1) Prove that (X, d) is a metric space. (2) Prove that if the (X i, d i ) are compact then (X, d) is compact. 1.4 Continuous functions on metric spaces Metric spaces are an abstraction that generalizes the real line. Abstractions, such as metric spaces, groups, rings, etc., never live on their own. They are always associated with mappings (morphisms, in the language of categories) between instances of the same abstraction, which preserve certain properties (e.g., groups and homomorphisms). In the first calculus courses we learned about continuous mappings (i.e., functions) R R. This concept will be generalized into continuous mappings between metric spaces. X Y x " f!x"!
24 20 Chapter 1 Definition 1.12 Let (X, d) and (Y, ρ) be two metric spaces. A mapping f : X Y is said to be continuous at a point x X if for every ɛ > 0 there exists a δ > 0 such that d(x, x) δ ρ( f (x ), f (x)) ɛ. That is, f is continuous at x if for every ball B( f (x), ɛ) Y there exists a ball B(x, δ) X whose image under f is contained in B( f (x), ɛ). A function is called continuous if it is continuous at all points. Examples: Let (X, d) be an arbitrary metric space and a X an arbitrary point. The function x d(x, a) is a continuous mapping f : (X, d) (R, ) because for every ɛ > 0 take δ = ɛ. Then d(x, z) ɛ implies that f (x) f (z) = d(x, a) d(z, a) d(x, z) ɛ. Every function defined on a discrete space is continuous because we can always take δ = 1/2. Every function f : X R n is continuous iff each of its component is a continuous function X R. (8 hrs) Proposition 1.11 (Sequential continuity) Let f : (X, d) (Y, ρ) be a function between two metric spaces. Then f is continuous at a iff for every sequence (x n ) in X lim x n = a lim f (x n ) = f (a). n n Proof : Suppose first that f is continuous at a and let (x n ) be a sequence in X that converges to a. Since f is continuous, there exists for every ɛ > 0 a δ > 0 such that x B(a, δ) implies f (x) B( f (a), ɛ). Because x n a, there exists an N δ such that x n B(a, δ) for all n > N δ, hence f (x n ) B( f (a), ɛ) for all n > N δ, from which follows that f (x n ) f (a). Suppose next that for every sequence (x n ) that converges to a the sequence f (x n ) converges to f (a). Suppose that f were not continuous: then there exists an ɛ >
25 Metric Spaces 21 0 such that for every δ > 0 there exists an x n such that d(x n, a) δ and yet ρ( f (x n ), f (a)) > ɛ, in contradiction with the assumption. The following theorem implies alternative definitions to the continuity of functions between metric spaces. In fact, these definitions are more general in the sense that they remain valid in (nonmetric) topological spaces (i.e., spaces that are only endowed with a structure of open sets). Theorem 1.4 Let f : (X, d) (Y, ρ) be a function between two metric spaces. Then the following three statements are equivalent: ➀ f is continuous. ➁ The preimage of every set open in Y is a set open in X. ➂ The preimage of every set closed in Y is a set closed in X. Proof : For every set A Y, its preimage under f is defined as f 1 (A) = {x X : f (x) A}. The equivalence between 2 and 3 follows from the complementation property of the preimage: f 1 (A c ) = ( f 1 (A)) c. We now show that 1 implies 2. Let A Y be an open set, and let x f 1 (A). Since A is open and f (x) A, there exists a ball B( f (x), ɛ) A. Since f is continuous, there exists a ball B(x, δ) whose image under f is a subset of B( f (x), ɛ), i.e., B(x, δ) f 1 (A), which implies that f 1 (A) is open. It remains to show that 2 implies 1. Let x X be an arbitrary point; by assumption, for every ɛ > 0 the preimage of the set B 0 ( f (x), ɛ) is open in X, and includes the point x, i.e., there exists an open ball B 0 (x, δ) f 1 (B 0 ( f (x), ɛ)), which implies continuity at x. Comment: A counterexample : continuous function not necessarily map open sets into open sets. For example the function f : R R defined by f : x 0 maps the open set (0, 1) into the closed set {0}. Definition 1.13 (Function composition) Let f : (X 1, d 1 ) (X 2, d 2 ) and g : (X 2, d 2 ) (X 3, d 3 ) be functions between metric spaces. Their composition g f is a function (X 1, d 1 ) (X 3, d 3 ) defined by (g f )(x) = g( f (x)). x f!1 "A# f"x# A
26 22 Chapter 1 Proposition 1.12 (Composition of continuous mappings) The composition of continuous mappings is a continuous mapping. Proof : We can use either interpretation of continuity. The proof is immediate if we use the topological interpretation and observe that (g f ) 1 = f 1 g 1. Corollary 1.3 For continuous functions f, g : X R, the functions f ± g, and f g are continuous. Furthermore, f /g is continuous at all points where g 0. Proof : The function x ( f (x), g(x)) is a continuous mapping from R to R 2, because, by the separate continuity of f, g, there exists for every ɛ > 0 a δ > 0, such that x a δ implies. f (x) f (a) 2 + g(x) g(a) 2 ɛ. Furthermore, the function (x, y) x+y is a continuous function from R 2 to R because for every ɛ > 0 we can take δ = ɛ/2 and x a 2 + y b 2 ɛ/2 implies (x + y) (a + b) x a + y b ɛ. Finally the mapping x f (x) + g(x) is a composition of these two continuous mapping. The same argument holds for subtraction and multiplication. To prove the continuity of division, it is sufficient to show that the mapping x 1/x is continuous at all points where x 0 (which we know from elementary calculus). Proposition 1.13 (Continuity and compactness) Continuous functions between metric spaces map compact sets into compact sets. Proof : Let f : X Y and K X a compact set. Take a sequence (y n ) in f (K), and consider a corresponding sequence (x n ) in K such that f (x n ) = y n (it is not necessarily unique). Since K is compact, (x n ) has a subsequence that converges to a limit x K. Because f is continuous, the image of this subsequence converges to f (x) f (K), hence f (K) is compact. Corollary 1.4 Recall that in compact spaces, the closed sets coincide with the compact sets: thus, if f : X Y and X is compact, the image of every closed set is closed. By complementation, the image of every open set is open.
27 Metric Spaces 23 The following theorem is a generalization of a wellknown theorem in elementary calculus: Proposition 1.14 (Extremal values in compact spaces) A continuous function from a compact metric space into R assumes a minimum and a maximum. Proof : Let f : X R with X compact. By the previous proposition, the set f (X) R is compact, hence bounded and closed. Since it is bounded, it has a finite supremum, and since it is closed this supremum must belong to the set, i.e., it is a maximum. Discussion: Note the advantage of working with abstract concepts: we prove powerful theorems without having to deal with problemspecific details. Definition 1.14 (Uniform continuity) A mapping f : (X, d) (Y, ρ) is called uniformly continuous if there exists for every ɛ > 0 a δ > 0 such that for every a, b X, d(a, b) δ implies ρ( f (a), f (b)) ɛ. That is, it is uniformly continuous if it is continuous, and the same δ can be chosen for all points. The function δ(ɛ) is called the continuity modulus of the function. Definition 1.15 (Hölder and Lipschitz continuity) A mapping f : (X, d) (Y, ρ) is called Hölder continuous of order α if there exists a real number k > 0 such that for every a, b X, ρ( f (a), f (b)) k [d(a, b)] α. The mapping is called Lipschitz continuous if it is Hölder continuous of order α = 1. Exercise 1.22 (Ex. 9 in L.) Let F X be a closed subset in a metric space. Prove that the function X R defined by x d(x, F) is Lipschitz continuous. Theorem 1.5 A continuous function on a compact metric space is uniformly continuous. Comment: Note that we can always replace this statement by a continuous function defined on a compact set in a metric space.
28 24 Chapter 1 Proof : Let f : (X, d) (Y, ρ) be continuous with X compact. Suppose it isn t uniformly continuous. Then, there exists an ɛ > 0 such that for all δ = 1/n there exist (x n, y n ) such that d(x n, y n ) 1/n and yet ρ( f (x n ), f (y n )) > ɛ. (10 hrs) Since X is compact, there exists a subsequence (x nk, y nk ) such that x n x and y n y. Because d(x n, y n ) 1/n it follows that x = y. Recall that the distance function is continuous in each of its arguments: it follows that the mapping (x, y) ρ( f (x), f (y)) is continuous i.e., ρ( f (x n ), f (y n )) ρ( f (x), f (x)) = 0, thus a contradiction. Definition 1.16 (Homeomorphism) A mapping f : (X, d) (Y, ρ) between metric spaces is called a homeomorphism if it is onetoone, onto, and both f and f 1 are continuous, i.e., A X is open iff f (A) Y is open. Two metric spaces are called homeomorphic of there exists a homeomorphism between them. Comment: It is easy to see that homeomorphism is an equivalence relation between metric spaces: it is symmetric, reflexive and transitive. In fact, homeomorphism is a topological concept (only requires the definition of open sets), and is also called topological equivalence. Note that this equivalence is with respect to all topological properties (openness, closedness, compactness), but not necessarily with respect to metric properties, such as boundedness. Indeed, a homeomorphism maps open sets into open sets, closed sets into closed sets, and compact sets into compact sets. Examples: The identity map from (R n, p ) to (R n, q ) is a homeomorphism, because open balls with respect to the p metric are open sets (although not balls) with respect to the q metric. Every two open segments (a, b) and (c, d) on the line are homeomorphic, because the linear mapping x c+(d c)(x a)/(b a) is a homeomorphism. Every open segment (a, b) is homeomorphic to the entire real line because it is homeomorphic to the open segment ( π/2, π/2) and the latter is homeomorphic to the real line though the homeomorphism x tan x.
29 Metric Spaces 25 Proposition 1.15 If f is a continuous, onetoone mapping from a compact metric space (X, d) onto an (arbitrary) metric space (Y, ρ), then f is a homeomorphism, i.e., f 1 is continuous. Proof : Let K X be a closed set. Since it is a closed subset of a compact space, it is compact, and by Proposition 1.13 f (K) is compact and, in particular, closed. By Theorem 1.4, if f maps closed sets into closed sets then f 1 is continuous. Discussion: How does one show that two metric spaces are not homeomorphic? In principle, we should show that there does not exist a homeomorphism between them. We can t of course check all mappings. The idea is to show that the two spaces differ in a topological invariant a property that is preserved by homeomorphisms. Examples: R is not homeomorphic to a discrete space because in a discrete space all compact sets are finite, whereas this not true in R. Thus, the image of [0, 1] in the discrete space cannot be compact. (0, 1) is not homeomorphic to [0, 1] because the first is not compact whereas the second is. The unit square [0, 1] [0, 1] is not homeomorphic to [0, 1]. Both spaces are compact, and it requires more subtle arguments to show it. R n is not homeomorphic to R m for m n. This is a very deep theorem (invariance of domain, Brouwer 1912), which is hard to prove and beyond the scope of this course (requires advanced topology). Exercise 1.23 (Ex. 13 in L.) Let F, G be sets in a metric space X and let Y be a metric space. Given two functions f : F Y and g : g Y such that f = g on F G, we define the function f (x) x F ( f g)(x) = g(x) x G.
30 26 Chapter 1 ➀ Prove that if F, G are closed in X and f, g are continuous, then f g is continuous. ➁ Prove that this is also correct if F, G are open in X. ➂ Show that this is not necessarily the case if F is open and G is closed. Exercise 1.24 (Ex. 14 in L.) Let f : (0, 1) R be continuous. Show that f can be extended into a continuous function on [0, 1] iff f is uniformly continuous on (0, 1). Exercise 1.25 (Ex. isometry if 23 in L.) A function f : (X, d) (Y, ρ) is called an ρ( f (x), f (y)) = d(x, y) x, y X. ➀ Show that every isometry from X onto Y is an homeomorphism. Give an example to show that the converse is not true. ➁ Prove that every isometry from R onto itself is of the form x ±x + a. Exercise 1.26 (Ex. 24 in L.) Show that every isometry from a compact space into itself is onto. Exercise 1.27 (Ex. 19 in L.) Let A X. The indicator function of the the set A is defined as 1 x A χ A (x) = 0 x A. Show that the points of discontinuity of χ A coincide with A. Exercise 1.28 (Ex. 30 in L.) Prove that the following spaces are homeomorphic (using the natural metrics): ➀ R 2. ➁ The unit disc: D 2 = { (x, y) : x 2 + y 2 < 1 }. ➂ The unit square: Sq 2 = {(x, y) : max( x, y ) < 1}. ➃ The unit sphere in R 3 less one point, e.g. S 2 \ = { (x, y, z) : x 2 + y 2 + z 2 = 1 } \ {(0, 0, 1)}.
31 Metric Spaces Convergence of sequences of functions This section deals with sequences of functions between metric spaces. Definition 1.17 (Pointwise and uniform convergence) A sequence ( f n ) of functions from (X, d) to (Y, ρ) is said to converge pointwise to a limit function f, if for every x X, lim n f n(x) = f (x), i.e., if for every x X and ɛ > 0 there exists an N x,ɛ such that ρ( f n (x), f (x)) ɛ for all n > N x,ɛ. It is said to converge uniformly if N can be chosen independently of x. Comments: Pointwise convergence means that lim ρ( f n(x), f (x)) = 0 n x, whereas uniform convergence means lim sup n x X ρ( f n (x), f (x)) = 0. A sequence of functions ( f n ) that converges uniformly to f satisfies the Cauchy criterion for uniform convergence, lim sup ρ( f n (x), f m (x)) = 0. n,m x X (This is because ρ( f n (x), f m (x)) ρ( f n (x), f (x)) + ρ( f m (x), f (x)).) It can be shown (see exercise below) that for functions X R the Cauchy criterion is also a sufficient condition for uniform convergence. Exercise 1.29 Prove that for functions X R the Cauchy criterion is a necessary and sufficient conditions for uniform convergence. Examples:
32 28 Chapter 1 The functions between the discrete spaces N and {0, 1} defined by f n : j δ j,n converge to the constant function f : j 0 pointwise by not uniformly. In fact, it is easy to see that a sequence of functions f n into a discrete space converges uniformly to f iff f n coincides with f eventually. The sequence of functions f n : [0, 1) [0, 1) (with the standard metric) defined by f n (x) = x n converges to the function f, defined by f (x) = 0, pointwise, but not uniformly, because for every n, sup f n (x) f (x) = sup x n 0 = 1. 0 x<1 0 x<1! 1/! Consider the following sequence of functions f n : R R: n 2 x 0 < x < 1/n f n (x) = 2n n 2 x 1/n < x < 2/n 0 otherwise. It converges pointwise to the function f = 0, but not only it does not converges uniformly, we even have lim sup n x R f n (x) f (x) = lim n n =. Consider the functions f n (x) = x n on the closed interval [0, 1]. Although each of the f n is continuous, this sequence converges pointwise to the discontinuous function 0 0 x < 1 f (x) = 1 x = 1. The following theorem shows that this can t happen if the convergence is uniform. Theorem 1.6 Let ( f n ) be a sequence of functions from (X, d) to (Y, ρ), which converge uniformly to a function f. If all the f n are continuous, then the limit is continuous. Moreover, if all the f n are uniformly continuous on X, then f is uniformly continuous on X.
33 Metric Spaces 29 Proof : Suppose that all the f n are continuous at a point x X and let (x m ) be a sequence that converges to x. For every n, m, ρ( f (x m ), f (x)) ρ( f (x m ), f n (x m )) + ρ( f n (x m ), f n (x)) + ρ( f n (x), f (x)). Let ɛ > 0 be given. Because the f n converge to f uniformly, we can choose n sufficiently large such that the first and third term are less than ɛ/3, irrespective of m. Having fixed n, there exists an N such that for all m > N, the second term is less than ɛ/3 (since f n is continuous). To conclude, for all ɛ > 0 there exists an N ɛ such that for all m > N ɛ, ρ( f (x m ), f (x)) ɛ, i.e., f is continuous. Assume next that the f n are uniformly continuous. For x, y X we write ρ( f (x), f (y)) ρ( f (x), f n (x)) + ρ( f n (x), f n (y)) + ρ( f n (y), f (y)). Since f n converges uniformly to f, given ɛ > 0, there exists an n such that the first and third terms are less than ɛ/3. Since this f n is uniformly continuous, there exists a δ > 0 such that d(x, y) δ implies ρ( f n (x), f n (y)) ɛ/3, i.e., implies ρ( f (x), f (y)) ɛ. Proposition 1.16 If ( f n ) are continuous and converge uniformly to f, then for every sequence (x n ) in X, x n x implies f n (x n ) f (x). Comment: A counter example : the sequence f n (x) = x n defined on [0, 1] converges pointwise to the function 0 0 x < 1 f (x) = 1 x = 1. Consider the sequence x n = 1 1/n, which converges to 1. Then, ( lim f n(x n ) = 1 1 ) n = 1 f (1). n n e Proof : We have ρ( f n (x n ), f (x)) ρ( f n (x n ), f (x n )) + ρ( f (x n ), f (x)) sup ρ( f n (y), f (y)) + ρ( f (x n ), f (x)). y X
34 30 Chapter 1 The first term on the right hand side converges to zero by the uniform convergence of f n, whereas the second term converges to zero by the continuity of f (which results from the previous proposition). Definition 1.18 Let ( f n ) be a sequence of functions from a metric space (X, d) into R (we need the target to be a normed space). The series n=1 f n is said to converge uniformly to f if the sequence of partial sums converges uniformly to f. Theorem 1.7 (Weierstraß Mtest) Let ( f n ) be a sequence of functions from a metric space (X, d) into R. Suppose that there exists a sequence of real numbers (M n ) such that sup f n (x) M n, x X and that the series n=1 M n converges. Then, the series n=1 f n converges uniformly on X. Example: The figure below shows the functions S n (x) = n k=1 sin kx k 1.1 for n = 2, 20, 200, By the Weierstraß Mtest, the sequence S n converges uniformly on [0, 2π], and since each of the partial sums is a uniformly continuous function, the limit is uniformly continuous n= n= n= n=2000
35 Metric Spaces 31 Proof : For every x the series S n (x) = n f n (x) converges absolutely (satisfies Cauchy s criterion), which defines a limiting function S (x). It remains to show that the convergence is uniform. Note that for every n, i.e., S n (x) S (x) = S n (x) S (x) k=n+1 k=n+1 f k (x) f k (x), k=n+1 and since the right hand side does not depend on x, sup S n (x) S (x) M k. x X It remains to let n. k=n+1 M k, Comment: In particular, if all the f n are continuous, so is the series S. (12 hrs) Proposition 1.17 Let f n : [a, b] R be a sequence of continuous functions that converges uniformly on [a, b] to a limit f. Then, b lim n a f n (x) dx = b a f (x) dx. Comment: We restrict ourselves to functions R R because we only know what are integrals of such functions. Proof : The integrals exists because all the f n and their limit f are continuous (Theorem 1.6), hence Riemann integrable. Now, for every n, b a f (x) dx b a f n (x) dx = b a [ f (x) f n (x)] dx. To prove the theorem, we need to show that the right hand side tends to zero: b b [ f (x) f n (x)] dx f (x) f n (x) dx b a sup a y b a f (y) f n (y) dx = (b a) sup f (y) f n (y), a y b a
36 32 Chapter 1 and since f n converges to f uniformly, the right hand side tends to zero. Theorem 1.6 provides a criterion for the limit of a sequence of continuous functions to be continuous (respectively, uniformly continuous). What about differentiability? Under what conditions is the limit of a sequence of differentiable functions differentiable? Theorem 1.8 Let f n : (a, b) R be a sequence of differentiable functions such that (i) the sequence f n (x) converges at at least one point c (a, b), and (ii) the sequence of derivatives, f n converges uniformly on (a, b) to a limit function g. Then, ➀ The sequence f n converges uniformly on (a, b) to a limit f. ➁ f is differentiable with f = g. Comments: We need the first condition, because the nonconvergent sequence f n (x) = n satisfies the second condition with g = 0. Since we have not assumed the f n to be continuously differentiable, we can t assume g to be continuous, and not even integrable. This prevents us from defining f (x) = x g(y) dy, and then show that f a n converges to f uniformly. This will cost us in technical complications. Providing a simpler proof for the case where the f n are continuously differentiable is left as an exercise (see below). Proof : The fact that the target space is R allows us to prove uniform convergence based on Cauchy s criterion. For every m, n, we use the mean value theorem to get f m (x) f n (x) ( f m f n )(x) ( f m f n )(c) + f m (c) f n (c) where ξ is between x and c. Hence, sup x = x c ( f m f n)(ξ) + f m (c) f n (c), f m (x) f n (x) (b a) sup f m(x) f n(x) + f m (c) f n (c). x The first term on the right hand side converges to zero, as m, n, by Cauchy s criterion for uniform convergence for the sequence of derivatives. The second
37 Metric Spaces 33 term converges to zero by the first assumption. Hence, the functions f n converges uniformly (to a limiting function f ) by Cauchy s criterion. It remains to show that f = g. Let x be an arbitrary (but fixed!) point, and set 1 [ f (x + h) f (x)] h 0 h ϕ(h) = g(x) h = 0. We will be done if we prove that ϕ is continuous at zero. How can we show that a function is continuous? For example, by showing that it is the uniform limit of a sequence of continuous functions. Define the sequence of functions 1 ϕ n (h) = [ f h n(x + h) f n (x)] h 0 f n(x) h = 0. These functions are, by definition, continuous at zero, and we already know that they converge pointwise to ϕ. To show that they converge uniformly, we show that they satisfy Cauchy s criterion: if h 0 then whereas for h = 0, ϕ n (h) ϕ m (h) = 1 h ( f n f m )(x + h) ( f n f m )(x) sup f n(y) f m(y), y X ϕ n (h) ϕ m (h) = f n(0) f m(0) sup f n(y) f m(y). y X Letting n, m, and using the fact that the right hand sides do not depend on h and that ( f n) converge uniformly, we have that ϕ n converges to ϕ uniformly, which concludes the proof. Exercise 1.30 (Ex. 31 in L.) Let f n : (a, b) R be a sequence of continuously differentiable functions such that (i) the sequence f n (x) converges at at least one point c (a, b), and (ii) the sequence of derivatives, f n converges uniformly on (a, b) to a limit function g. Prove that x lim ( f n(x) f n (c)) = lim f n n n(y) dy = c x c g(y) dy. Conclude then that ( f n ) has a limit, which we denote by f, and that f = g.
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