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Methods Mathematical of Engineering Analysis
Erhan Cinlar ¸ Robert J. Vanderbei
February 2, 2000
Contents
Sets and Functions 1 Sets . . . . . . . . . . . . . . . . . . . . Subsets . . . . . . . . . . . . . . Set Operations . . . . . . . . . . Disjoint Sets . . . . . . . . . . . Products of Sets . . . . . . . . . 2 Functions and Sequences . . . . . . . . . Injections, Surjections, Bijections Sequences . . . . . . . . . . . . . 3 Countability . . . . . . . . . . . . . . . . 4 On the Real Line . . . . . . . . . . . . . Positive and Negative . . . . . . . Increasing, Decreasing . . . . . . Bounds . . . . . . . . . . . . . . Supremum and Inﬁmum . . . . . Limits . . . . . . . . . . . . . . . Convergence of Sequences . . . . 5 Series . . . . . . . . . . . . . . . . . . . Ratio Test, Root Test . . . . . . . Power Series . . . . . . . . . . . Absolute Convergence . . . . . . Rearrangements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1 2 2 3 3 4 4 5 6 8 9 9 9 9 10 11 14 16 17 18 19 23 23 23 24 25 26 26 26 29 30 30
Metric Spaces 6 Euclidean Spaces . . . . . . . . . . . . . . . . . . . . . . Inner Product and Norm . . . . . . . . . . . . . . Euclidean Distance . . . . . . . . . . . . . . . . . 7 Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . Usage . . . . . . . . . . . . . . . . . . . . . . . . Distances from Points to Sets and from Sets to Sets Balls . . . . . . . . . . . . . . . . . . . . . . . . 8 Open and Closed Sets . . . . . . . . . . . . . . . . . . . . Closed Sets . . . . . . . . . . . . . . . . . . . . . Interior, Closure, and Boundary . . . . . . . . . . i
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Open Subsets of the Real Line . . . . . . . Convergence . . . . . . . . . . . . . . . . . . . . Subsequences . . . . . . . . . . . . . . . . Convergence and Closed Sets . . . . . . . Completeness . . . . . . . . . . . . . . . . . . . . Cauchy Sequences . . . . . . . . . . . . . Complete Metric Spaces . . . . . . . . . . Compactness . . . . . . . . . . . . . . . . . . . . Compact Subspaces . . . . . . . . . . . . Cluster Points, Convergence, Completeness Compactness in Euclidean Spaces . . . . .
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31 34 35 36 37 37 38 40 40 41 42 45 45 46 46 47 48 48 50 51 53 54 56 57 57 58 60 60 63 63 64 69 69 70 70 71 71 71 76 77 78
Functions on Metric Spaces 12 Continuous Mappings . . . . . . . . . . Continuity and Open Sets . . . Continuity and Convergence . . Compositions . . . . . . . . . . RealValued Functions . . . . . Rn Valued Functions . . . . . . 13 Compactness and Uniform Continuity . Uniform Continuity . . . . . . . 14 Sequences of Functions . . . . . . . . . Cauchy Criterion . . . . . . . . Continuity of Limit Functions . 15 Spaces of Continuous Functions . . . . Convergence in C . . . . . . . . Lipschitz Continuous Functions Completeness . . . . . . . . . . Functionals . . . . . . . . . . .
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Differential and Integral Equations 16 Contraction Mappings . . . . . . . . . . . . . . . Fixed Point Theorem . . . . . . . . . . . . 17 Systems of Linear Equations . . . . . . . . . . . . Maximum Norm . . . . . . . . . . . . . . Manhattan Metric . . . . . . . . . . . . . . Euclidean Metric . . . . . . . . . . . . . . Conclusion . . . . . . . . . . . . . . . . . 18 Integral Equations . . . . . . . . . . . . . . . . . . Fredholm Equation . . . . . . . . . . . . . Volterra Equation . . . . . . . . . . . . . . Generalization of the Fixed Point Theorem 19 Differential Equations . . . . . . . . . . . . . . . . ii
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Convex Analysis 20 Convex Sets and Convex Functions . . . . . . . . . . . . . . . . . . . 21 Projection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Supporting Hyperplane Theorem . . . . . . . . . . . . . . . . . . . . Measure and Integration 23 Motivation . . . . . . . . . . . . . . . . . . . . . 24 Algebras . . . . . . . . . . . . . . . . . . . . . . Monotone Class Theorem . . . . . . . . 25 Measurable Spaces and Functions . . . . . . . . Measurable Functions . . . . . . . . . . Borel Functions . . . . . . . . . . . . . . Compositions of Functions . . . . . . . . Numerical Functions . . . . . . . . . . . Positive and Negative Parts of a Function Indicators and Simple Functions . . . . . Approximations by Simple Functions . . Limits of Sequences of Functions . . . . Monotone Classes of Functions . . . . . Notation . . . . . . . . . . . . . . . . . 26 Measures . . . . . . . . . . . . . . . . . . . . . Arithmetic of Measures . . . . . . . . . . Finite, σﬁnite, Σﬁnite measures . . . . Speciﬁcation of Measures . . . . . . . . Image of Measure . . . . . . . . . . . . Almost Everywhere . . . . . . . . . . . 27 Integration . . . . . . . . . . . . . . . . . . . . . Deﬁnition of the Integral . . . . . . . . . Integral over a Set . . . . . . . . . . . . Integrability . . . . . . . . . . . . . . . . Elementary Properties . . . . . . . . . . Monotone Convergence Theorem . . . . Linearity of Integration . . . . . . . . . . Fatou’s Lemma . . . . . . . . . . . . . . Dominated Convergence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
83 83 86 90 91 91 93 94 96 96 97 97 97 98 98 99 100 100 101 103 104 104 105 106 106 108 109 110 110 110 111 113 113 114
iii
Sets and Functions
This introductory chapter is devoted to general notions regarding sets, functions, sequences, and series. The aim is to introduce and review the basic notation, terminology, conventions, and elementary facts.
1
Sets
A set is a collection of some objects. Given a set, the objects that form it are called its elements. Given a set A, we write x ∈ A to mean that x is an element of A. To say that x ∈ A, we also use phrases like x is in A, x is a member of A, x belongs to A, and A includes x. To specify a set, one can either write down all its elements inside curly brackets (if this is feasible), or indicate the properties that distinguish its elements. For ex 10 14 19 15 20 21 2 3 4 5 6
SETS AND FUNCTIONS ···
i 1 2 3 4 5 6 . . .
2.5 Functional Inverses. Let f be a bijection from E onto F . Then, for each y in F there is a unique x in E such that f (x) = y. In other words, in the notation of (2.1), f −1 ({y}) = {x} for each y in F and some unique x in E. In this case, we drop some brackets and write f −1 (y) = x. The resulting function f −1 is a bijection from F onto E; it is called the functional inverse of f . This particular usage should not be confused with the general notation of f −1 . (Note that (2.1) deﬁnes a function f −1 form F into E, where F is the collection of all subsets of F and E is the collection of all subsets of E.)
3
Countability
Two sets A and B are said to have the same cardinality, and then we write A ∼ B, if there exists a bijection from A onto B. Obviously, having the same cardinality is an equivalence relation; it is 1. reﬂexive: A ∼ A, 2. symmetric: A ∼ B ⇒ B ∼ A, 3. transitive: A ∼ B and B ∼ C ⇒ A ∼ C. A set is said to be ﬁnite if it is empty or has the same cardinality as {1, 2, . . . , n} for some n in N; in the former case it has 0 elements, in the latter exactly n. It is said to be countable if it is ﬁnite or has the same cardinality as N; in the latter case it is said to have a countable inﬁnity of elements. In particular, N is countable. So are Z, N × N in view of exercises 2.3 and 2.4. Note that an inﬁnite set can have the same cardinality as one of its proper subsets. For instance, Z ∼ N, R+ ∼ (0, 1], R ∼ R+ ∼ (0, 1); see exercise 2.2 for the latter. Incidentally, R+ , R, etc. are uncountable, as we shall show shortly. A set is countable if and only if it can be injected into N, or equivalently, if and only if there is a surjection from N onto it. Thus, a set A is countable if and only if there is a sequence (xn ) whose range is A. The following lemma follows easily from these remarks.
3. COUNTABILITY
7
3.1 LEMMA. If A can be injected into B and B is countable, then A is countable. If A is countable and there is a surjection from A onto B, then B is countable.
3.2 THEOREM. The product of two countable sets is countable. PROOF. Let A and B be countable. If one of them is empty, then A × B is empty and there is nothing to prove. Suppose that neither is empty. Then, there exist injections f : A → N and g : B → N. For each pair (x, y) in A × B, let h(x, y) = (f (x), g(y)); then h is an injection from A × B into N × N. Since N × N is countable (see Exercise (2.4)), this implies via the preceding lemma that A × B is countable 2 3.3 COROLLARY. The set of all rational numbers is countable. PROOF. Recall that the set Q of all rationals consists of ratios m/n with m ∈ Z and n ∈ N. Thus, f (m, n) = m/n deﬁnes a surjection from Z × N onto Q. Since Z and N are countable, so is Z × N by the preceding theorem. Hence, Q is countable by Lemma 3.1. 2 3.4 THEOREM. The union of a countable collection of countable sets is countable. PROOF. Let I be a countable set, and let Ai be a countable set for each i in I. The claim is that A = i∈I Ai is countable. Now, there is a surjection fi : N → Ai for each i, and there is a surjection g : N → I; these follow from the countability of I and the Ai . Note that, then, h(m, n) = fg(m) (n) deﬁnes a surjection h from N × N onto A. Since N × N is countable, this implies via Lemma 3.1 that A is countable. 2 The following theorem exhibits an uncountable set. As a corollary, we show that R is uncountable. 3.5 THEOREM. Let E be the set of all sequences whose terms are the digits 0 and 1. Then, E is uncountable. PROOF. Let A be a countable subset of E. Let x1 , x2 , . . . be an enumeration of the elements of A, that is, A is the range of (xn ). Note that each xn is a sequence of zeros and ones, say xn = (xn,1 , xn,2 , . . .) where each term xn,m is either 0 or 1. We deﬁne a new sequence y = (yn ) by letting yn = 1 − xn,n . The sequence y differs from every one of the sequences x1 , x2 , . . . in at least one position. Thus, y is not in A but is in E. We have shown that if A ⊂ E and is countable, then there is a y ∈ E such that y ∈ A. If E were countable, the preceding would hold for A = E, which would be
8 absurd. Hence, E must be uncountable. 3.6 COROLLARY. The set of all real numbers is uncountable.
SETS AND FUNCTIONS 2
PROOF. It is enough to show that the interval [0, 1) is uncountable. For each x ∈ [0, 1), let 0.x1 x2 x3 · · · be the binary expansion of x (in case x is dyadic, say x = k/2n for some k and n in N, there are two such possible binary expansions, in which case we take the expansion with inﬁnitely many zeros), and we identify the binary expansion with the sequence (x1 , x2 , . . .) in the set E of the preceding theorem. Thus, to each x in [0, 1) there corresponds a unique element f (x) of E. In fact, f is a surjection onto the set E \ D where D denotes the set of all sequences of zeros and ones that are eventually all ones. It is easy to show that D is countable and hence that E \ D is uncountable. From this it follows that [0, 1) is uncountable. 2
Exercises:
3.1 Dyadics. A number is said to be dyadic if it has the form k/2n for some integers k and n in Z+ . Show that the set of all dyadic numbers is countable. Of course, every dyadic number is rational. 3.2 Let D denote the set of all sequences of zeros and ones that are eventually all ones. Show that D is countable. 3.3 Suppose that A is uncountable and that B is countable. Show that A \ B is uncountable. 3.4 Let x be a real number. For each n ∈ Z+ , let xn be the smallest dyadic number of the form k/2n that exceeds x. Show that x0 ≥ x1 ≥ x2 ≥ · · · and that xn > x for each n. Show that, for every > 0, there is an n such that xn − x < for all n ≥ n .
4
On the Real Line
The object is to review some facts and establish some terminology regarding the set ¯ R of all real numbers and the set R = [−∞, +∞] of all extended real numbers. The extended real number system consists of R and two extra symbols, −∞ and ∞. The ¯ relation < is extended to R by postulating that −∞ < x < +∞ for every real number ¯ x. The arithmetic operations are extended to R as follows: for each x ∈ R, x + ∞ = x − (−∞) = ∞ x + (−∞) = x − ∞ = −∞ x·∞ = ∞ −∞ if x > 0 if x < 0
4. ON THE REAL LINE x · (−∞) x/∞ = x/(−∞) ∞+∞ (−∞) + (−∞) ∞ · ∞ = (−∞) · (−∞) ∞ · (−∞) = = = = = = (−x) · ∞ 0 ∞ −∞ ∞ −∞.
9
The operations 0 · (±∞), (−∞) − (−∞), +∞/+∞, and −∞/−∞ are undeﬁned.
Positive and Negative
¯ We call x in R positive if x ≥ 0 and strictly positive if x > 0. By symmetry, then, x ¯ is negative if x ≤ 0 and strictly negative if x < 0. A function f : E → R is said to be positive if f (x) ≥ 0 for all x in E and strictly positive if f (x) > 0 for all x in E. Negative and strictly negative functions are deﬁned similarly. This usage is in accord with modern tendencies, though at variance with common usage1 .
Increasing, Decreasing
¯ ¯ A function f : R → R is said to be increasing if f (x) ≤ f (y) whenever x ≤ y. It is said to be strictly increasing if f (x) < f (y) whenever x < y. Decreasing and strictly decreasing functions are deﬁned similarly by reversing the inequalities. ¯ ¯ These notions carry over to functions f : E → R with E ⊂ R. In particular, since a ¯ sequence is a function on N, these notions apply to sequences in R. Thus, for example, ¯ is increasing if x1 ≤ x2 ≤ · · · and is strictly decreasing if x1 > x2 > · · ·. (xn ) ⊂ R
Bounds
¯ Let A ⊂ R. A real number b is called an upper bound for A provided that A ⊂ [−∞, b], and then A is said to be bounded above by b. Lower bounds and being bounded below are deﬁned similarly. The set A is said to be bounded if it is bounded above and below; that is, if A ⊂ [a, b] for some real interval [a, b]. ¯ These notions carry over to functions and sequences as follows. Given f : E → R, the function f is said to be bounded above, below, etc. according as its range is bounded above, below, etc. Thus, for instance, f is bounded if there exist real numbers a ≤ b such that a ≤ f (x) ≤ b for all x in E.
Supremum and Inﬁmum
¯ If A ⊂ R is bounded above, then it has a least upper bound, that is, an upper bound b such that no number less than b is an upper bound; we call that least upper bound the supremum of A. If A is not bounded above, we deﬁne the supremum to be +∞. The
1 Often used concepts should have the simpler names. Mindbending double negatives should be avoided, and as much as possible, the mathematical usage should not conﬂict with the ordinary language.
10
SETS AND FUNCTIONS
inﬁmum of A is deﬁned similarly; it is −∞ if A has no lower bound and is the greatest lower bound otherwise. We let inf A, sup A
denote the inﬁmum and supremum of A, respectively. For example, inf{1, 1/2, 1/3, . . .} = 0, inf(a, b] = inf[a, b] = a, sup{1, 1/2, 1/3, . . .} = 1, sup(a, b) = sup(a, b] = b.
In particular, inf ∅ = +∞ and sup ∅ = −∞. If A is ﬁnite, then inf A is the smallest element of A, and sup A is the largest. Even when A in inﬁnite, it is possible that inf A is an element of A, in which case it is called the minimum of A. Similarly, if sup A is an element of A, then it is also called the maximum of A. ¯ If f : E → R, it is customary to write
x∈D
inf f (x) = inf{f (x) : x ∈ D}
and call it the inﬁmum (or maximum) of f over D ⊂ E, and similarly with the supre¯ mum. In the case of sequences (xn ) ⊂ R, inf xn , sup xn
denote, respectively, the inﬁmum and supremum of the range of (xn ). Other such notations are generally selfexplanatory; for example,
n≥k
inf xn = inf{xk , xk+1 , . . .},
sup xnk = sup{xn1 , xn2 , . . .}.
k≥1
Limits
¯ If (xn ) is an increasing sequence in R, then sup xn is also called the limit of (xn ) and is denoted by lim xn . If it is a decreasing sequence, then inf xn is called the limit of (xn ) and again denoted by lim xn . ¯ Let (xn ) ⊂ R be an arbitrary sequence. Then 4.1 xm = inf xn ,
n≥m
xm = sup xn , ¯
n≥m
m ∈ N,
deﬁne two sequences; (xn ) is increasing, and (¯n ) is decreasing. Their limits are called x the limit inferior and the limit superior, respectively, of the sequence (xn ): 4.2 4.3 lim inf xn = lim xn = sup inf xn ,
m n≥m m n≥m
lim sup xn = lim xn = inf sup xn , ¯
Figure 1 is worthy of careful study. Note that, in general, 4.4 −∞ ≤ lim inf xn ≤ lim sup xn ≤ +∞.
If lim inf xn = lim sup xn , then the common value is called the limit of (xn ) and is denoted by lim xn . Otherwise, if limits inferior and superior are not equal, the sequence (xn ) does not have a limit.
4. ON THE REAL LINE
11
Figure 1: Lim Sup and Lim Inf. The pairs (n, xn ) are connected by the solid lines for clarity. The pairs (n, xn ) form the lower dotted line and (n, xn ) the upper dotted line. ¯
Convergence of Sequences
A sequence (xn ) of real numbers is said to be convergent if lim xn exists and is a real number. An examination of Figure 1 shows that this is equivalent to the classical deﬁnition of convergence: (xn ) converges to x if for every > 0, there is an n such that xn −x < for all n ≥ n . The phrase “there is n ... for all n ≥ n ” can be expressed in more geometric terms by phrases like “the number of terms outside (x − , x + ) is ﬁnite,” or “all but ﬁnitely many terms are in (x − , x + ),” or “xn − x < for all n large enough.” The following is a summary of the relations between convergence and algebraic operations. The proof will be omitted. 4.5 THEOREM. Let (xn ) and (yn ) be convergent sequences with limits x and y respectively. Then, 1. lim cxn = cx, 2. lim(xn + yn ) = x + y, 3. lim xn yn = xy, 4. lim xn /yn = x/y provided that yn , y = 0. In practice, we do not have the sequence laid out before us. Instead, some rule is given for generating the sequence and the object is to show that the resulting sequence will converge. For instance, a function may be speciﬁed somehow and a procedure described to ﬁnd its maximum; starting from some point, the procedure will give the
12
SETS AND FUNCTIONS
successive points x1 , x2 , . . . which are meant to form the sequence that converges to the point x where the maximum is achieved. Often, to ﬁnd the limit of (xn ), one starts with a search for sequences that bound (xn ) from above and below and whose limits can be computed easily: suppose that yn ≤ xn ≤ zn for all n, lim yn = lim zn ,
then lim xn exists and is equal to the limit of the other two. The art involved is in ﬁnding such sequences (yn ) and (zn ). 4.6 EXAMPLE. This example illustrates the technique mentioned above. We want to show that (n1/n ) converges. Note that n1/n ≥ 1 always, and put xn = n1/n − 1, and consider the sequnce (xn ). Now, (1 + xn )n = n, and by the binomial theorem (a + b)n = an + nan−1 b + ≥ for a, b ≥ 0 and n ≥ 2. So, n = (1 + xn )n ≥ or 0 ≤ xn ≤ It follows that lim xn = 0, and hence lim n1/n = 1. 2 . n−1 n(n − 1) 2 xn , 2 n(n − 1) n−2 2 a b + · · · + bn 2
n(n − 1) n−2 2 a b 2
Exercises:
4.1 Show that if A ⊃ B then inf A ≤ inf B ≤ sup B ≤ sup A. Use this to show that, if A1 ⊃ A2 ⊃ · · ·, then inf A1 ≤ inf A2 ≤ · · · ≤ inf An ≤ · · · ≤ ≤ sup An ≤ · · · ≤ sup A2 ≤ sup A1 . Use this to show that (xn ) is increasing, (¯n ) is decreasing, and lim xn ≤ x lim xn (see (4.1) – (4.3) for deﬁnitions). ¯ ¯ 4.2 Show that sup(−xn ) = − inf xn for any sequence (xn ) in R. Conclude that lim sup(−xn ) = − lim inf xn .
4. ON THE REAL LINE 4.3 Cauchy Criterion. Sequence (xn ) is convergent if and only if for every > 0 there is an n such that xm − xn  ≤ for all m ≥ n ≥ n . Prove this by examining Figure 1 on the deﬁnition of the limit. 4.4 Monotone Sequences. If (xn ) is increasing, then lim xn exists (but could be +∞). Thus, such a sequence converges if and only if it is bounded above. Show this. State the version of this for decreasing sequences. 4.5 Iterative Sequences. Often, xn+1 is obtained from xn via some rule, that is, xn+1 = f (xn ) for some function f . If (xn ) is so obtained from some function f , it is said to be iterative. If (xn ) is such and f is continuous and lim xn = x exists, then x = f (x). This works well for identifying the limit especially when f is simple and x = f (x) has only one solution. In general, with complicated functions f , the reverse is true: To ﬁnd x satisfying x = f (x), one starts at some point x0 , computes x1 = f (x0 ), x2 = f (x1 ), ..., and tries to show that x = lim xn exists and satisﬁes x = f (x). 4.6 Domination. A sequence (xn ) is said to be dominated by a sequence (yn ) if xn ≤ yn for each n. Show that, if so 1. inf xn ≤ inf yn , 2. sup xn ≤ sup yn , 3. lim inf xn ≤ lim inf yn , 4. lim sup xn ≤ lim sup yn . In particular, if the limits exist, lim xn ≤ lim yn . Incidentally, (xn ) deﬁned by (4.1) is the maximal increasing sequence dominated by (xn ), and (¯n ) is the minimal decreasing sequence domix nating (xn ). 4.7 Comparisons. Let (xn ) be a positive sequence. Then, (xn ) converges to 0 if and only if it is dominated by a sequence (yn ) with lim sup yn = 0. Show this. Favorite sequences (yn ) used in this role are given by yn = 1/n, yn = rn for some ﬁxed number r ∈ (0, 1), and yn = np rn with p ∈ (−∞, +∞) and r ∈ (0, 1). 4.8 Existence of Least Upper Bounds. Let A be a nonempty subset of R and let B = {b : b is an upper bound of A}. Assuming that B is nonempty, show that B has a minimum element.
13
14
SETS AND FUNCTIONS
5
Series
n
Given a sequence (xn ) ⊂ R, the sequence (sn ) deﬁned by 5.1 sn =
i=1
xi
is called the sequence of partial sums of (xn ), and the symbolic expression 5.2 xn
is called the series associated with (xn ). The series is said to converge to s, and then we write
∞
5.3
1
xn = s
if and only if the sequence (sn ) converges to s. Sometimes, we write x1 + x2 + · · · for the series (5.2). Sometimes, for convenience ∞ ∞ of notation, we shall consider series of the form 0 or m , depending on the index set. Here are a few examples:
∞
xn x n! n=0
∞ n=0 ∞ n
=
1 1−x
for x ∈ (−1, 1),
= ex = = π2 , 6
for x ∈ R,
1 n2 n=1
∞
xn
n=m
xm 1−x
for x ∈ (−1, 1).
The following result is obtained by applying the Cauchy Criterion (Exercise 4.3) to the sequence of partial sums. 5.4 THEOREM. The series n such that 5.5 for all m ≥ n ≥ n . In particular, taking m = n in (5.5) we obtain xn  ≤ . Thus we have obtained the following: 5.6 COROLLARY. If xn converges, then lim xn = 0. xn converges if and only if for every > 0 there is an
m

i=n
xi  ≤
5. SERIES The converse is not true. For example, lim 1/n = 0 but 1/n is divergent. In the case of series with positive terms, partial sums form an increasing sequence, and hence, the following holds (see Exercise 4.4): 5.7 PROPOSITION. Suppose that the xn are positive. Then only if the sequence of partial sums is bounded. xn converges if and
15
In many cases, we encounter series whose terms are positive and decreasing. The following theorem due to Cauchy is helpful in such cases, especially if the terms involve powers. Note the way a rather thin sequence determines the convergence or divergence of the whole series. 5.8 THEOREM. Suppose that (xn ) is decreasing and positive. Then if and only if the series x1 + 2x2 + 4x4 + 8x8 + · · · converges. PROOF. Let sn = x1 + · · · + xn as usual and put tk = x1 + 2x2 + · · · + 2k x2k . Now, for n ≤ 2k , since x1 ≥ x2 ≥ · · · ≥ 0, sn ≤ x1 + (x2 + x3 ) + (x4 + · · · x7 ) + · · · + (x2k + · · · + x2k+1 −1 ) ≤ x1 + 2x2 + 4x4 + · · · + 2k x2k = tk , xn converges
and for n ≥ 2k , sn ≥ x1 + x2 + (x3 + x4 ) + (x5 + · · · x8 ) + · · · + (x2k−1 +1 + · · · + x2k ) 1 ≥ x1 + x2 + 2x4 + · · · + 2k−1 x2k 2 1 = tk . 2
Thus, the sequences (sn ) and (tn ) are either both bounded or both unbounded, which completes the proof via Proposition 5.7 2
5.9 EXAMPLE. 1/np converges if p > 1 and diverges if p ≤ 1. For p ≤ 0, the claim is trivial to see. For p > 0, the terms xn = 1/np form a decreasing positive sequnce, and thus, the preceding theorem applies. Now,
∞
2k x2k =
k=0
(21−p )k ,
which converges if 21−p < 1 and diverges otherwise. Since 21−p < 1 if and only if p > 1, we are done.
16
SETS AND FUNCTIONS
5.10 EXAMPLE. The series
∞
2
1 n(log n)p
converges if p ∈ (1, ∞) and diverges otherwise. Here we start the series with n = 2 since log 1 = 0. Since the logarithm function is monotone increasing, Theorem 5.8 applies. Now, xn = 1/n(log n)p and so
∞ ∞
2k x2k =
k=1 1
2k
1 1 = k (log 2k )p 2 (log 2)p
∞
1
1 , kp
which converges if and only if p > 1 in view of the preceding example.
Ratio Test, Root Test
The ratio test ties the convergence of large n; it is highly useful. 5.11 THEOREM. 1. If lim sup xn+1 /xn  < 1, then 2. If lim inf xn+1 /xn  > 1, then xn converges. xn diverges. xn to the behavior of the ratios xn+1 /xn for
PROOF. (1) If lim sup xn+1 /xn  < 1, then there is a number r ∈ [0, 1) and an integer n0 such that xn+1 /xn  ≤ r for all n ≥ n0 . Thus xn0 +k  ≤ xn0 rk for all k ≥ 0, and therefore, for m > n > n0 ,
m ∞ ∞

i=n
xi  ≤
i=n
xi  ≤ xn0 
i=n
ri−n0 = xn0 
rn−n0 . 1−r
Given > 0 choose n so that xn0 rn −n0 /(1 − r) < . Then Cauchy’s criterion works with this n and xn converges. (2) If lim inf xn+1 /xn  > 1 then there is an integer n0 such that xn+1  ≥ xn  for all n ≥ n0 . Hence, xn  ≥ xn0  for all n ≥ n0 which shows that (xn ) does not converge to 0 as it must in order for xn to converge (see Corollary 5.6). 2 The preceding test gives no information in cases where lim inf xn+1 /xn  ≤ 1 ≤ lim sup xn+1 /xn .
5. SERIES For instance, for the two series 1/n and 1/n2 , both the lim inf and the lim sup are equal to 1, but the ﬁrst series diverges whereas the second converges. Also, the series 11 1 1 1 1 1 1 5.12 ++ + 2 + 3 + 3 + 4 + 4 + ··· 2 3 22 3 2 3 2 3 obviously converges to 3/2; yet, the ratio test is miserably inconclusive: lim inf xn+1 = lim xn xn+1 = lim xn 2 3 3 2
n
17
=0
n
lim sup
= ∞.
The following test, called the root test, is a stronger test — if the ratio test works, so does the root test. But the root test works in some situations where the ratio test fails; for example, the root test works for the series (5.12). 5.13 THEOREM. Let a = lim sup xn 1/n . Then diverges if a > 1. xn converges if a < 1, and
PROOF. Suppose that a < 1. Then, there is a b ∈ (a, 1) such that xn 1/n ≤ b for all n ≥ n0 , where n0 is some integer. Then, xn  ≤ bn for all n ≥ n0 , and comparing xn with the geometric series bn shows that xn converges. Suppose that a > 1. Then, a subsequence of xn  must converge to a > 1, which means that xn  ≥ 1 for inﬁnitely many n. So, (xn ) does not converge to zero, and hence, xn cannot converge. 2
Power Series
Given a sequence (cn ) of complex numbers, the series
∞
5.14
0
cn z n
is called a power series. The numbers c0 , c1 , . . . are called the coefﬁcients of the power series; here z is a complex number. In general, the series will converge or diverge, depending on the choice of z. As the following theorem shows, there is a number r ∈ [0, ∞], called the radius of convergence, such that the series converges if z < r and diverges if z > r. The behavior for z = r is much more complicated and cannot be described easily. 5.15 THEOREM. Let a = lim sup cn 1/n and r = 1/a. 1. If z < r, then cn z n converges.
18 2. If z > r, then cn z n diverges.
SETS AND FUNCTIONS
PROOF. Put xn = cn z n and apply the root test with lim sup xn 1/n = z lim sup cn 1/n = az = z . r 2
5.16 EXAMPLE. 1. 2. 3. 4. z n /n! = ez and r = ∞. z n converges for z < 1 and diverges for z ≥ 1; r = 1. z n /n2 converges for z ≤ 1 and diverges for z > 1; r = 1. z n /n converges for z < 1 and diverges for z > 1; r = 1; for z = 1 the series diverges, but for z = 1 but z = 1 it converges.
Absolute Convergence
The series xn is said to converge absolutely if xn  is convergent. If the xn are all positive numbers, then absolute convergence is the same as convergence. Using Cauchy’s criterion (see Theorem 5.4) on both sides of
m m

i=n
xi  ≤
i=n
xi 
shows that if for example,
xn converges absolutely then it converges. But the converse is not true: (−1)n /n
converges but is not absolutely convergent. The comparison tests above, as well as the root and ratio tests, are in fact tests for absolute convergence. If a series is not absolutely convergent, one has to study the sequence of partial sums to determine whether the series converges at all.
5. SERIES
19
Rearrangements
Let (k1 , k2 , . . .) be a sequence in which every integer n ≥ 1 appears once and only once, that is, n → kn is a bijection from N onto N. If yn = xkn , n ∈ N,
for such a sequence (kn ), then we say that (yn ) is a rearrangement of (xn ). Let (yn ) be a rearrangement of (xn ). In general, the series yn and xn are quite different. However, if xn is absolutely convergent, then so is yn and it converges to the same number as does xn . The converse is also true: if every rearrangement of the series xn converges, then the series xn is absolutely convergent and all its rearrangements converge (to the same sum). On the other hand, if xn is not absolutely convergent, its various rearrangements may converge or diverge, and in the case of convergence, the sum generally depends on the rearrangement chosen. For instance, 1− 111111 + − + − + − ··· 234567
is convergent, but not absolutely so. Its rearrangement 1+ 111111 − + + − + + ··· 325749
(with + + − + + − + + − pattern) is again convergent, but not to the same sum. In fact, the following theorem due to Riemann shows that one can create rearrangements that are as bizarre as one wants. 5.17 THEOREM. Let xn be convergent but not absolutely. Then, for any two ¯ numbers a ≤ b in R there is a rearrangement yn of xn such that
n n
lim inf
1
yi = a,
lim sup
1
yi = b.
We omit the proof. Note that, in particular, taking a = b we can ﬁnd a rearrangement yn with sum a, no matter what a is.
Exercises:
5.1 Determine the convergence or divergence of the following: √ √ 1. ( n + 1 − n) √ √ 2. ( n + 1 − n)/n √ 3. (sin n)/(n n) 4. (−1)n n/(n2 + 1).
In case of convergence, indicate whether it is absolute convergence.
20 5.2 Show that if xn converges then so does √ xn /n .
SETS AND FUNCTIONS
5.3 Show that if xn converges and (yn ) is bounded and monotone (either increasing or decreasing), then xn yn converges. 5.4 Find the radius of convergence of each of the following power series: 1. 2. 3. 4. n2 z n , 2n z n /n!, 2n z n /n2 , n3 z n /3n . c2n z 2n ,
5.5 Suppose that f (z) = cn z n . Express the sum of the even terms, and the sum of the odd terms, c2n+1 z 2n+1 , in terms of f . 5.6 Suppose that f (z) = cn z n . Express c3n z 3n in terms of f .
5.7 Rearrangements. Let xn be a series that converges absolutely. Prove that every rearrangement of xn converges, and that they all converge to the same sum. 5.8 Riemann’s Theorem. Prove Riemann’s theorem 5.17 by ﬁlling in the details in the following outline: 1. Let (x+ ) denote the subsequence consisting of the positive elements n of (xn ) and let (x− ) denote the subsequence of negative elements of n (xn ). Both of these sequences must be inﬁnite. 2. Both sequences (x+ ) and (x− ) converge to zero. n n 3. Both series x+ and n x− diverge. n 4. Suppose that a, b ∈ R and deﬁne a rearrangement as follows: start with the positive elements and choose elements from this set until the partial sum exceeds b. Then, choose elements from the set of negative elements until the partial sum is less than a. Then, choose elements from the set of positive elements until the partial sum exceeds b. Continue this proceedure of alternating between elements of the positive and negative sets indeﬁnitely. 5. Prove that the procedure described above can be continued ad inﬁnitum. 6. Prove that this rearrangement has the properties stated in Riemann’s theorem. 7. Extend the above arguments to the case where a, b = ±∞. 5.9 Poisson distribution. Let pn = e−λ λn /n! where λ is a positive real. Show that 1. pn > 0,
5. SERIES 2. 3.
∞ n=0 ∞ n=0
21 pn = 1, npn = λ.
∞
5.10 Borel Summability. Consider a series n=0 xn with partial sums sn = n i=0 xi . We say that the series is Borel summable if
∞ λ→∞
lim
sn pn
n=0
converges, where pn are the Poisson probabilities deﬁned in Exercise 5.9. ∞ For what values of z is the geometric series n=0 z n Borel summable?
22
SETS AND FUNCTIONS
Metric Spaces
Basic questions of analysis on the real line are tied to the notions of closeness and distances between points. The same issue of closeness comes up in more complicated settings, for instance, like when we try to approximate a function by a simpler function. Our aim is to introduce the idea of distance in general, so that we can talk of the distance between two functions with the same conceptual ease as when we talk of the distance between two points in the plane. After that, we discuss the main issues: convergence, continuity, approximations. All along, there will be examples of different spaces and different ways of measuring distances.
6
Euclidean Spaces
This section is to review the space Rn together with its Euclidean distance. Recall that each element of Rn is an ntuple x = (x1 , . . . , xn ), where the xi are real numbers. The elements of Rn are called points or vectors, and we are familiar with the operations like addition of vectors and multiplication by scalars.
Inner Product and Norm
For x and y in Rn , their inner product x · y is the number
n
6.1
x·y =
1
xi yi .
If we regard x and y as column vectors, then x · y = xT y. For x in Rn , the norm of x is deﬁned to be the positive number 6.2 x= √ x·x=
n 1
x2 . i
The norm satisﬁes the following: 6.3 6.4 6.5 x ≥ 0 for every x in Rn , x = 0 if and only if x = 0, x + y ≤ x + y for all x and y in Rn . 23
24
METRIC SPACES
Of these, 6.3 and 6.4 are obvious, and 6.5 is immediate from the following, which is called the Schwartz inequality. 6.6 PROPOSITION. x · y ≤ x y for all x and y in Rn . PROOF. Consider the function f (λ) = λy − x 2 = λ2 y 2 − 2λ(x · y) + x 2 . This function is clearly positive and quadratic and its minimum occurs at x·y . λ= y2 For this value of λ we have 0 ≤ f( x·y (x · y)2 )=− +x 2 y y2
2
from which Schwartz’s inequality follows immediately.
2
Euclidean Distance
For x and y in Rn , the Euclidean distance between x and y is deﬁned to be the number x − y . It follows from the properties given above that, for all x, y, z in Rn , 1. x − y ≥ 0, 2. x − y = y − x , 3. x − y = 0 if and only if x = y, 4. x − y + y − z ≥ x − z . The last is called the triangle inequality: on R2 , if the points x, y, z are the vertices of a triangle, this is simply the wellknown fact that the sum of the lengths of two sides is greater than or equal to the length of the third side. The set Rn together with the Euclidean distance is called ndimensional Euclidean space. The Euclidean spaces are important examples of metric spaces.
Exercises:
6.1 Show that the mapping (x, y) → x · y from Rn × Rn into R is a linear transformation in x and is a linear transformation in y (and therefore is said to be bilinear). Conclude that (x + y) · (x + y) = x · x + 2x · y + y · y. Use this and the Schwartz inequality to prove (6.5).
7. METRIC SPACES 6.2 Show that x + y 2 + x − y 2 = 2 x 2 + 2 y 2 . Interpret this in geometric terms, on R2 , as a statement about parallelograms. 6.3 Points x and y are said to be orthogonal if x · y = 0. Show that this is equivalent to saying that the lines connecting the origin to x and y are perpendicular. In general, letting α be the angle between the lines through x and y, we have x · y = x y cos α.
25
7
Metric Spaces
Let E be a set. A metric on E is a function d : E ×E → R+ that satisﬁes the following for all x, y, z in E: 1. d(x, y) = d(y, x), 2. d(x, y) = 0 if and only if x = y, 3. d(x, y) + d(y, z) ≥ d(x, z). A metric space is a pair (E, d) where E is a set and d is a metric on E. In this context, we think of E as a space, call the elements of E points, and refer to d(x, y) as the distance from x to y. EXAMPLES. 7.1 Euclidean spaces. Consider Rn with the Euclidean distance d(x, y) = x − y on it. It follows from (1)–(4) that d is a metric on Rn . Thus, (Rn , d) is a metric space and is called ndimensional Euclidean space. 7.2 Manhattan metric. On Rn deﬁne a metric d by
n
d(x, y) =
1
xi − yi .
This d is called the Manhattan metric, or l1 metric, on Rn , and (Rn , d) is a metric space again. Note that for n > 1 this metric is different from the Euclidean metric of the preceding example. 7.3 Space C. Let C denote the set of all continuous functions from the interval [0, 1] into R. For x and y in C, let d(x, y) = sup x(t) − y(t).
0≤t≤1
26
METRIC SPACES
It is clear that d(x, y) is a positive real number, that d(x, y) = d(y, x), and that d(x, y) = 0 if and only if x = y. As for the triangle inequality, we note that x(t) − z(t) ≤ x(t) − y(t) + y(t) − z(t) ≤ d(x, y) + d(y, z) for every t in [0, 1], from which we have d(x, y) + d(y, z) ≥ d(x, z). Thus, d is a metric on C, and (C, d) is a metric space. This metric space is important in analysis.
Usage
In the literature, it is common practice to call E a metric space if (E, d) is a metric space for some metric d. If there is only one metric under consideration, this is harmless and saves time. For instance, the phrase “Euclidean space Rn ” refers to (Rn , d) where d is the Euclidean metric. For a while at least, we shall indicate the metric involved in each case in order to avoid all possible confusion.
Distances from Points to Sets and from Sets to Sets
Let (E, d) be a metric space. For x in E and A ⊂ E, let 7.4 d(x, A) = inf{d(x, y) : y ∈ A};
this is called the distance from the point x to the set A. For A ⊂ E and B ⊂ E, the distance from A to B is deﬁned by 7.5 d(A, B) = inf{d(x, y) : x ∈ A, y ∈ B}.
The diameter of a set A ⊂ E is deﬁned to be 7.6 diam A = sup{d(x, y) : x ∈ A, y ∈ A}.
A set is said to be bounded if its diameter is ﬁnite.
Balls
Let (E, d) be a metric space. For x in E and r in (0, ∞), 7.7 B(x, r) = {y ∈ E : d(x, y) < r}
is called the open ball with center x and radius r, and 7.8 ¯ B(x, r) = {y ∈ E : d(x, y) ≤ r}
is the corresponding closed ball. For example, if E = R3 and d is the usual Euclidean metric, then B(x, r) becomes ¯ the set of all points inside the sphere with center x and radius r, and B(x, r) is the set of all points inside or on that sphere.
Exercises and Complements:
7. METRIC SPACES 7.1 Discrete metric. Let E be an arbitrary set. Deﬁne d(x, y) = 1 0 if x = y, if x = y.
27
Show that this d is a metric on E. It is called the discrete metric on E. 7.2 Metrics on Rn . For each number p ≥ 1,
n
dp (x, y) = (
1
xi − yi p )1/p
deﬁnes a metric dp on Rn . Note that d1 is the Manhatten metric, and d2 is the Euclidean metric. Finally, d∞ (x, y) = sup xi − yi 
1≤i≤n
is again a metric on Rn . Show this. 7.3 Equivalent Metrics. Two metrics d and d are equivalent if there exist strictly positive constants c1 and c2 such that for all x, y: c1 d (x, y) ≤ d(x, y) ≤ c2 d (x, y). Show that d1 , d2 , and d∞ are all equivalent to each other. 7.4 Weighted Metrics on Rn . The metrics introduced in the preceding exercise treat all components of x−y equally. This is reasonable if Rn is thought of geometrically and the selection of a coordinate system is unimportant. On the other hand, if x = (x1 , . . . , xn ) stands for a shopping list that requires buying x1 units of product one, and x2 units of product two, and so on, then it would make much better sense to deﬁne the distance between two shopping lists x and y by
n
d(x, y) =
1
wi xi − yi 
where x1 , . . . , wn are ﬁxed, strictly positive numbers, with wi being the value of one unit of product i. Show that this d is indeed a metric. More generally, paralleling the metrics introduced in the previous exercise,
n
dp (x, y) = (
i
wi xi − yi p )1/p ,
x, y ∈ Rn ,
is a metric on Rn for each p ≥ 1 and each ﬁxed, strictly positive vector w (the latter means w1 > 0, . . . , wn > 0).
28
METRIC SPACES
7.5 l2 Spaces. Instead of Rn , now consider the space R∞ of all inﬁnite sequences in R, that is, each x in R∞ is a sequence x = (x1 , x2 , . . .) of real numbers. In analogy with the d2 metrics introduced on Rn in Exercises 7.2 and 7.4, we deﬁne
∞
d2 (x, y) = (
1
xi − yi 2 )1/2 .
This d2 satisﬁes all the conditions for a metric except that d2 (x, y) can be ∞ for some x and y. To remedy the latter, we let E be the set of all x in R∞ with
∞
x2 < ∞. i
1
Then, by an easy generalization of the Schwartz inequality, it follows that d2 (x, y) < ∞ for all x and y in E. Thus, (E, d2 ) is a metric space. It is generally denoted by l2 . 7.6 Metrics on C. Consider the set C of all continuous functions from [0, 1] into R. The interval [0, 1] can be replaced by any bounded interval [a, b], in which case one writes C([a, b]). A number of metrics can be deﬁned on C in analogy with those in Exercise 7.2. The analogy is provided by the following observation: every x in Rn can be thought of as a function 12 x from { n , n , . . . , n } into R, namely, the function x with x(t) = xi for n 12 t = i/n. Thus, replacing the set { n n , . . . , n } with the interval [0, 1] and n replacing the summation by integration, we obtain
1
dp (x, y) = (
0
x(t) − y(t)p dt)1/p
for all x and y in C. Since any continuous function on [0, 1] is bounded, the integral here is ﬁnite and it is easy to check the conditions for this dp to be a metric, except perhaps for the triangle inequality. So, for each p ≥ 1, this dp is a metric on C. Incidentally, the metric of Example 7.3 can be denoted by d∞ in analogy with d∞ in Exercise 7.2. 7.7 Open Balls. Let E = R2 . Describe the open ball B(x, r), for ﬁxed x and r, under each of the following metrics: 1. d2 of Exercise 7.2. 2. d1 of Exercise 7.2. 3. d∞ of Exercise 7.2. 4. d2 of Exercise 7.4 with w1 = 1 and w2 = 5. 7.8 Open Balls in C. For the metric space of Example 7.3, describe B(x, r) for a ﬁxed function x and ﬁxed number r > 0. Draw pictures!
8. OPEN AND CLOSED SETS 7.9 Product Spaces. Let (E1 , d1 ) and (E2 , d2 ) be arbitrary metric spaces. Let E = E1 × E2 and deﬁne, for x = (x1 , x2 ) in E and y = (y1 , y2 ) in E, d(x, y) = [d1 (x1 , y1 )2 + d2 (x2 , y2 )2 ]1/2 . Show that d is a metric on E. The metric space (E, d) is called the product of the metric spaces (E1 , d1 ) and (E2 , d2 ).
29
8
Open and Closed Sets
Let (E, d) be a metric space. All points mentioned below are points of E, all sets are subsets of E. Recall the deﬁnition 7.7 of the open ball B(x, r) with center x and radius r. 8.1 DEFINITION. A set A is said to be open if for every x in A there is an r > 0 such that B(x, r) ⊂ A. A set is said to be closed if its complement is open. For example, if E = R with the usual distance, the intervals (a, b), (−∞, b), (a, ∞) are open, the intervals [a, b], (−∞, b], [a, ∞) are closed, and the interval (a, b] is neither open nor closed. 8.2 PROPOSITION. Every open ball is open. PROOF. Fix x and r. To show that B(x, r) is open, we need to show that for every y in B(x, r) there is a q > 0 such that B(y, q) ⊂ B(x, r). This is accomplished by picking q = r − d(x, y). Since y is in B(x, r), we have d(x, y) < r and, hence, q > 0. And, every point of B(y, q) is a point of B(x, r), because z ∈ B(y, q) means d(z, y) < q which implies that d(z, x) ≤ d(z, y) + d(y, x) < q + d(y, x) = r. 2
8.3 THEOREM. The sets ∅ and E are open. The intersection of a ﬁnite number of open sets is open. The union of an arbitrary collection of open sets is open. PROOF. The ﬁrst assertion is trivial from the deﬁnition. We prove the second assertion for the intersection of two open sets. The general case follows from the repeated aplication of the case for two. Let A and B be open. Let x ∈ A ∩ B. Since A is open and x is in A, there is p > 0 such that B(x, p) ⊂ A.
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Similarly, there is a q > 0 such that B(x, q) ⊂ B. Let r = p ∧ q, the smaller of p and q. Then, B(x, r) ⊂ B(x, p) ⊂ A and B(x, r) ⊂ B(x, q) ⊂ B. Hence, B(x, r) ⊂ A ∩ B. So, A ∩ B is open. For the last assertion, let {Ai : i ∈ I} be an arbitrary collection of open sets. We want to show that A = ∪i Ai is open. Let x be in A. Then, x ∈ Ai for some i ∈ I. Since Ai is open, there is an r > 0 such that B(x, r) ⊂ A. Since Ai ⊂ A, this shows that B(x, r) ⊂ A. So, A is open. 2 The following characterization is immediate from the preceding theorem together with Proposition 8.2. 8.4 PROPOSITION. A set is open if and only if it is the union of a collection of open balls. PROOF. If A is the union of a collection of open balls, then A must be open in view of 8.2 and 8.3. To show the converse, let A be open. Then, for every x in A, there is an open ball Ax = B(x, r(x)) contained in A. Obviously, the union of all these Ax is exactly A. 2
Closed Sets
Recall that a subset of E is closed if and only if its complement is open. Thus, the following theorem is immediate from Theorem 8.3 above and the fact that the complement of a union is the intersection of complements and vice versa. 8.5 THEOREM. The sets ∅ and E are closed. The union of ﬁnitely many closed sets is closed. The intersection of an arbitrary collection of closed sets is closed. Every closed ball is closed. This last observation can be proved along the lines of ¯ 8.2: if y ∈ E \ B(x, r) then d(y, x) > r, and picking p = d(x, y) − r > 0 we see that ¯ ¯ B(y, p) ⊂ E \ B(x, r), which proves that E \ B(x, r) is open. In particular, for each x in E, the singleton {x} is closed. It follows from this and the preceding theorem that every ﬁnite set is closed.
Interior, Closure, and Boundary
Let A be a subset of E. The collection of all closed sets containing A is not empty ¯ (since E belongs to that collection.) The intersection A of that collection is a closed ¯ set by the last theorem. Clearly, A is the smallest closed set that contains A, that is, if ¯ ¯ B ⊃ A and B is closed then B ⊃ A. The set A is called the closure of A. We deﬁne the interior of A similarly as the largest open set contained in A, and we denote it by A◦ . In other words, A◦ is the union of all open sets contained in A. Note
8. OPEN AND CLOSED SETS that 8.6
31
¯ A◦ ⊂ A ⊂ A.
¯ We deﬁne the boundary of A to be the set ∂A = A \ A◦ . For example, if A is the open ball B(x, r) in the Euclidean space E = Rn , the ◦ ¯ ¯ A = A, A = B(x, r), and ∂A is the sphere of radius r centered at x. If E = R with ¯ the usual metric, and if A = (a, b], then A = [a, b] and A◦ = (a, b) and ∂A = {a, b}. The following seems self evident. 8.7 PROPOSITION. A set is closed if and only if it is equal to its closure. A set is open if and only if it is equal to its interior.
Open Subsets of the Real Line
We take E = R with the usual distance. Then, every open ball is an open interval, and according to Proposition 8.4, every open set is the union of a collection of open balls. The following sharpens the picture by taking into account the special nature of the real line. 8.8 THEOREM. A subset of R is open if and only if it is the union of a countable collection of disjoint open intervals. PROOF. The “if” part is immediate from Proposition 8.4 and the fact that every open ball is an interval in this case. To prove the “only if” part, let A be an open subset of R. Recall that the set Q of all rationals is countable. For each q in Q ∩ A, let aq = sup{y ≤ q : y ∈ A}, Then, B=
q∈Q∩A
bq = inf{y ≥ q : y ∈ A}.
(aq , bq )
is the union of a countable collection of open intervals. We show next that A = B by showing that A ⊂ B and B ⊂ A. Let x be in A. Since A is open, there is a ball B(x, r) contained in A. Take a rational number q in this ball. Clearly, B(x, r) ⊂ (aq , bq ). Thus, x is in B. Since this is true for every x in A, we have that A ⊂ B. Fix q ∈ Q ∩ A. Clearly, (aq , bq ) ⊂ A. Hence, B ⊂ A. We have shown that A = B, and B has the desired form except that the intervals (aq , bq ) are not necessarily disjoint. Note that if r ∈ (aq , bq ) then (ar , br ) = (aq , bq ) and q ∈ (ar , br ). Let us write q ≈ r if and only if (aq , bq ) = (ar , br ). This deﬁnes an equivalence relation on the set Q ∩ A. Thus, by picking exactly one q from each
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METRIC SPACES
0
Figure 2: The set D = ∪Dq .
1
equivalence class, we can form a set I ⊂ Q ∩ A such that (aq , bq ) ∩ (ar , br ) = ∅ for all distinct q and r in I, and A=B=
q∈I
(aq , bq ). 2
8.9 EXAMPLE. The Cantor Set. Start with the unit interval B = [0, 1]. To each q in the set I = {1/2; 1/4, 3/4; 1/8, 3/8, 5/8, 7/8; 1/16, 3/16, . . . , 15/16; . . .} we associate an open interval Dq in the following fashion: D1/2 is the open interval (1/3, 2/3) which is the middle third of B. Deleting it from B leaves two closed intervals, [0, 1/3] and [1/3, 1]. Let D1/4 be the interval (1/9, 2/9), which is the middle third of [0, 1/3], and let D3/4 be (7/9, 8/9), which is the middle third of [2/3, 1]. Deleting those middle thirds, we are left with four closed intervals of length 1/9 each. Let D1/8 , D3/8 , D5/8 , D7/8 be the open intervals that make up the middle thirds of those closed intervals. Delete the middle thirds, and continue in this manner (see Figure 2). Then, D=
q∈I
Dq
is the union of the countably many disjoint open intervals Dq , q ∈ I. It is an example of a nontrivial open set. Incidentally, note that the lengths of the Dq sum to 1 11 1 1 1 1 +( + )+( + + + ) + · · · = 1. 3 99 27 27 27 27 Thus, the “length” of D is 1. But the set C = B \ D is not empty. The set C = B \ D is called the Cantor set. It is obviously a closed set. The construction above shows that C is obtained by starting with B and deleting the middle third of every interval we can ﬁnd. Thus, there is no open interval contained in C. That is, there are no open balls in C. Hence, the interior of C must be empty, and C is pure boundary: ¯ C ◦ = ∅, C = C, ∂C = C. Also, since the length of D is equal to the length of B, the length of C = B \ D must be 0. In summary, the Cantor set is very thin.
8. OPEN AND CLOSED SETS
33
y=f(x)
x=g(y)
Figure 3: The cantor function. Nevertheless, C has at least as many points as the interval [0, 1]. We prove this next by showing, via construction, that there exists an injection g from [0, 1] into C. To this end, we start by deﬁning an increasing function f from D into [0, 1] by letting f (x) = q, if x ∈ Dq . Then, we deﬁne the function g on [0, 1] by setting g(1) = 1 and g(y) = inf{x ∈ D : f (x) > y}, 0 ≤ y < 1.
We show ﬁrst that g(y) ∈ C for every y. This is obvious for y = 1. Let y ∈ [0, 1); note that g(y) is the inﬁmum of the union of all intervals Dq with q > y; clearly, that inﬁmum cannot belong to D; so g(y) must belong to C (since it is obvious that g(y) ∈ B). Finally, we show that g : [0, 1] → C is an injection by showing that if y < z, then g(y) < g(z). Fix y < z. Note that there is at least one q in I such that y < q < z, and the corresponding set Dq is contained in {x ∈ D : f (x) > y} but not in {x ∈ D : f (x) > z}. It follows that the number g(y) is to the left of the interval Dq whereas g(z) is to the right. So, g(y) < g(z) if y < z. Hence, g : [0, 1] → C is an injection.
Exercises and Complements:
8.1 Let (E, d) be a metric space. Show that ¯ A = {x ∈ E : d(x, A) = 0} A◦ = {x ∈ E : d(x, Ac ) > 0} ∂A = {x ∈ E : d(x, A) = 0 and d(x, Ac ) = 0}.
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METRIC SPACES
8.2 Let (E, d) be a metric space. Fix A ⊂ E. Show that A = {x ∈ E : d(x, A) < } is an open set containing A for each > 0. Show that ¯ A = ∩ >0 A . 8.3 Boundedness. Let (E, d) be a metric space. Show that a subset A of E is bounded if and only if it is contained in some ball, that is, if and only if A ⊂ B(x, r) for some x and r. 8.4 Take E = R and d the usual metric. Let A ⊂ E. Show that if A is closed and bounded above, then sup A belongs to A (that is, A has a maximum). Similarly, if A is closed and bounded below, then it has a minimum. Show that an open set A cannot have a minimum, that is, inf A cannot belong to A. 8.5 Let D be the open set of Example 8.9. Find its interior and boundary. ¯ 8.6 Denseness. A set D is said to be dense in E if D = E. Let D be dense in E. Show that every x in E is at 0 distance from D. Thus, every open ball has at least one point of D. Show that the set Q of all rationals is dense in R, the set of all pairs of rationals is dense in R2 , etc. 8.7 Separability. The metric space E is said to be separable if there exists a countable set D that is dense in E. So, for example, the Euclidean spaces R, R2 , R3 , ... are separable. 8.8 Discrete metric spaces. Let E be arbitrary and suppose that d is the discrete metric (see (7.1) for it) on E. Show that each subset A is both open and closed. For r ≤ 1, every open ball B(x, r) consists of exactly the ¯ point x. Note that B(x, 1) = {x}, B(x, 1) = E for every x (Moral: ¯ r) is not necessarily the closure of B(x, r)). If E is countable, then B(x, it is separable (trivially). If E is uncountable, it is not separable. Show this.
9
Convergence
Let (E, d) be a metric space. Our goal is to discuss the notion of convergence for a sequence of points in E. We do so by employing the concept of convergence in R, for which we refer to Section 4 of Chapter . 9.1 DEFINITION. A sequence (xn ) in E is said to be convergent in E if there exists a point x in E such that lim d(xn , x) = 0. And, then, (xn ) is said to converge to x, the point x is called the limit of (xn ), and the notation x = lim xn is used to indicate it.
REMARK: The preceding deﬁnition includes, implicit in it, the fact that a convergent
9. CONVERGENCE sequence has exactly one limit. To see this, suppose that (xn ) converges to x and to y, that is, lim d(xn , x) = 0 and lim d(xn , y) = 0. Then, 0 ≤ d(x, y) ≤ d(x, xn ) + d(xn , y) by the triangle inequality, and the right side converges to zero. Thus, d(x, y) = 0, which means that x = y. The following brings together a number of rewordings of convergence. Each is a slight alteration of the others. No proof seems needed. 9.2 THEOREM. The following statements are equivalent: 1. (xn ) converges to x. 2. For every > 0 there is an n such that d(xn , x) < for all n ≥ n . 3. The set {n : d(xn , x) ≥ } is ﬁnite for each > 0. 4. For every > 0, the ball B(x, ) includes all but a ﬁnite number of the terms xn .
35
9.3 COROLLARY. Every convergent sequence is bounded. PROOF. Let (xn ) be convergent and x its limit. In view of the equivalence of 1 and 4 in Theorem 9.2, B(x, 1) includes all but a ﬁnite number of the terms xn . Let r be the maximum of the distances from x to those terms xn outside B(x, 1), if there are any; otherwise, set r = 1. Clearly r < ∞ and B(x, r) contains (xn ), which means that (xn ) is bounded. 2
Subsequences
It follows from Theorem 9.2 that we may remove a ﬁnite number of terms, or rearrange the terms, without affecting the convergence. The following generalizes this. 9.4 PROPOSITION. If a sequence converges to x, then every subsequence of it converges to the same x. PROOF. Let (xn ) be a sequence with limit x. Let (yn ) be a subsequence of it, that is, yn = xkn for some k1 < k2 < · · ·. Now, by Theorem 9.2, for every > 0 the ball B(x, ) includes all the terms xn except for some ﬁnite number of them; therefore the same must be true for the terms yn . So, by Theorem 9.2, the subsequence (yn ) converges to x. 2
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METRIC SPACES
Convergence and Closed Sets
Think of a particle that moves in E by jumps: ﬁrst it is at x1 , then at x2 , then at x3 , and so on. The following gives meaning to the term “closed set” if you think of sequences in this fashion. 9.5 THEOREM. A set is closed if and only if it includes the limit of every sequence in it. PROOF. “Only if” part. Suppose that A is a closed set and that (xn ) is a sequence in A with limit x. We show that, then, x must belong to A. For, otherwise, if x were in Ac , there would exist an > 0 such that B(x, ) ⊂ Ac since Ac is open and B(x, ) would include inﬁnitely many terms since x is the limit, which would contradict the fact that all the xn are in A. “If” part. We show that if A is not closed then there is a sequence (xn ) in A that converges to some point x in Ac . Suppose that A is not closed. Then Ac is not open. Thus, there exists an x in Ac such that B(x, r)∩A has at least one point for each r > 0. Hence, for each n in N, there is an xn in A such that d(xn , x) < 1/n. Obviously, (xn ) is in A and converges to x which is not in A. 2
Exercises:
9.1 Discrete metric spaces. Suppose that d is the discrete metric on E. Show that (xn ) is convergent if and only if it is ultimately stationary, that is, if and only if it has the form (x1 , x2 , . . . , xn , x, x, x, . . .) for some n. 9.2 Let (E, d) be arbitrary. Show that if (xn ) converges to x and (yn ) converges to y, then d(xn , yn ) converges to d(x, y). Hint: ﬁrst show that, for arbitrary x, y, z in E, d(x, y) − d(x, z) ≤ d(y, z). Use this to write d(xn , yn ) − d(x, y) ≤ d(xn , yn ) − d(xn , y) +d(xn , y) − d(x, y) ≤ d(yn , y) + d(xn , x), and take limits. 9.3 Show that if (xn ) converges to x, then d(xn , A) converges to d(x, A) for each ﬁxed subset A of E.
10. COMPLETENESS
37
10
Completeness
Let (E, d) be a metric space. Recall that a sequence (xn ) in E is convergent if there is an x in E such that lim d(xn , x) = 0. This deﬁnition has two shortcomings. First, starting with (xn ), we rarely have a candidate x for the limit. Second, often we are not interested in computing the limit itself; it is generally sufﬁcient to know that the limit exists and has such and such properties. This section is aimed at rectifying these shortcomings.
Cauchy Sequences
10.1 DEFINITION. A sequence (xn ) in E is said to be Cauchy if for every > 0 there is an n such that d(xm , xn ) < for all m > n ≥ n . The following is nearly a restatement of this deﬁnition in slightly more geometric terms. 10.2 LEMMA. A sequence (xn ) is Cauchy if and only if for every > 0 there is a ball of radius that contains all but ﬁnitely many of the terms xn . PROOF. Suppose that (xn ) is Cauchy. Let > 0. Then, there is n such that d(xm , xn ) < for all m > n ≥ n . Thus, in particular, the ball B(xn , ) contains all the terms except possibly x1 , . . . , xn −1 . This proves the necessity of the condition. Conversely, suppose that for every > 0 there is a ball B(x, ) with some x as its center such that all but a ﬁnite number of the terms are in the ball. Given > 0, now pick x so that B(x, /2) contains all the xn except perhaps ﬁnitely many, that is, there is n such that xn ∈ B(x, /2) for all n ≥ n . Now, if m > n ≥ n , then d(xm , xn ) ≤ d(xm , x) + d(x, xn ) < /2 + /2 = . Hence, (xn ) is Cauchy. This proves the sufﬁciency. 2
10.3 THEOREM. 1. Every convergent sequence is Cauchy. 2. Every Cauchy sequence is bounded. 3. Every subsequence of a Cauchy sequence is Cauchy.
38
METRIC SPACES
PROOF. The ﬁrst claim is immediate from the preceding lemma and Theorem 9.2. The second claim is proved, via the preceding lemma, by following the proof of Corollary 9.3. The last claim is immediate from the preceding lemma. 2 The following shows that if a sequence is Cauchy and you can ﬁnd a subsequence of it that converges to some point x, then the original sequence converges to x. 10.4 PROPOSITION. A Cauchy sequence that has a convergent subsequence is itself convergent. PROOF. Let (xn ) be Cauchy. Let x be the limit of a convergent subsequence of it. Pick > 0. By Lemma 10.2, there is a ball B(y, ) that contains all but a ﬁnite number of the xn . That ball B(y, ) must contain all but a ﬁnite number of the subsequence as ¯ ¯ well. Thus, x must be in B(y, ). Then, B(x, 3 ) contains B(y, ) and hence contains all but a ﬁnite number of the xn . Thus, (xn ) is convergent and x = lim xn in view of Theorem 9.2. 2
Complete Metric Spaces
All the results above suggest that all Cauchy sequences should be convergent, which is in fact what we hope for. Unfortunately, this is not true in general. Here is an example. Suppose that E √ Q, the set of all rationals, with the metric it inherits from the = real line. Let x = 2, which is not a rational number, and let (xn ) be a sequnce in Q that converges to x in the sense of convergence in R: for instance, pick xn to be a rational number in the interval (x, x + 1/n) for each n. Over the metric space Q, the sequence (xn ) is Cauchy, but fails to be convergent in Q simply because x is not in Q. The problem here is not with the Cauchy sequence, but with the space Q. The space Q has holes in it! The following introduces the extra notion we want. 10.5 DEFINITION. The metric space (E, d) is said to be complete if every Cauchy sequence in E converges to a point of E. The following is immediate from Theorem 9.5. 10.6 PROPOSITION. If (E, d) is complete and D ⊂ E is closed, then (D, d) is a complete metric space. The following shows that familiar spaces are complete. Other examples are listed in exercises.
10. COMPLETENESS 10.7 THEOREM. Every Euclidean space is complete. PROOF. We start with the onedimensional Euclidean space, namely R. Let (xn ) ⊂ R be Cauchy. Then, for every > 0 there is a ball of radius (namely an open interval of length 2 ) that contains all but ﬁnitely many of the xn . Therefore, the numbers x = lim inf xn and y = lim sup xn must belong to that ball, which means that 0 ≤ y − x < 2 . Since this is true for every > 0, we must have x = y, that is, (xn ) is convergent. This proves that R is complete. Now, ﬁx k ≥ 2 and consider the Euclidean space Rk . We write x = (a, b, . . . , c) for each x in Rk for simplicity of notation (in other words, the coordinates of x are a, b, . . . , c). Consider a Cauchy sequence of points xn = (an , bn , . . . , cn ) in Rk . Given > 0, then, for all m and n large enough, we have d(xm , xn ) = (am − an 2 + bm − bn 2 + · · · + cm − cn 2 )1/2 < , which shows that am − an  < , bm − bn  < , . . . , cm − cn  < .
39
In other words, the sequences (an ), (bn ), ..., (cn ) in R are Cauchy. We have just shown that R is complete. So, these sequences must be convergent in R, say, with limits a, b, . . . , c respectively. Now, let x = (a, b, . . . , c) and note that d(xn , x)2 = an − a2 + bn − b2 + · · · + cn − c2 converges to 0. Hence, lim d(xn , x) = 0, and (xn ) is convergent. This completes the proof that Rk is complete. 2
Exercises and Complements:
10.1 Show that the following metric spaces are complete: 1. E = R2 with the Manhattan metric d. 2. E arbitrary, d is the discrete metric. In fact, each Rk is a complete metric space with any one of the metrics dp introduced in Exercises 7.2 and 7.4. 10.2 Show that the space l2 introduced in Exercise 7.5 is complete. Incidently, so is the space C of Example 7.3 and Exercise 7.6. 10.3 Two Cauchy sequences (xn ) and (yn ) are said to be equivalent if their merger (x1 , y1 , x2 , y2 , . . .) is Cauchy. In this case, we write (xn ) ≡ (yn ). Show that this deﬁnes an requivalence relation. That is, 1. (xn ) ≡ (xn ) 2. (xn ) ≡ (yn ) implies that (yn ) ≡ (xn ) 3. (xn ) ≡ (yn ), (yn ) ≡ (zn ) implies that (xn ) ≡ (zn ).
40
METRIC SPACES
11
Compactness
Let (E, d) be a metric space. It will be convenient to refer to E as a metric space, without mentioning d. We shall use the picturesque phrase “the collection {Ai : i ∈ I} covers B” to mean that ∪i∈I Ai ⊃ B. 11.1 DEFINITION. A set C ⊂ E is said to be compact if every collection of open sets that covers C has a ﬁnite subcollection that covers C. The metric space (E, d) is said to be compact if E is so. We shall show that, for many metric spaces, compact sets are precisely the sets that are bounded and closed. The following are aimed in that direction. The proofs are excessively detailed in order to facilitate understanding. 11.2 PROPOSITION. Every compact set is bounded. PROOF. Let C be compact. For each x in C, let Bx be a ball of radius 1 centered at x. Obviously, then, the collection {Bx : x ∈ C} of open sets covers C. Hence, there must be a ﬁnite subcollection, say of sets Bx1 , . . . , Bxn , that covers C. Since the union of balls Bx1 , . . . , Bxn must be bounded, this implies that C is bounded as well. 2
11.3 PROPOSITION. Every closed subset of a compact set is compact. PROOF. Let D be compact. Let C ⊂ D be closed. Fix a collection of open sets that covers C. Adding the open set E \ C to that collection, we obtain a collection of open sets that covers D. Since D is compact, the latter collection has a ﬁnite subcollection that still covers D. Removing E \ C from that subcollection (if it were in), we obtain a ﬁnite subcollection of the original collection that covers C. Thus, C must be compact. 2
Compact Subspaces
Recall that every subset D of E can be regarded as a metric space by itself, with the metric it inherits from E. Whether D is open or not as a subset of E, it is open automatically when it is regarded as a metric space. The concept of compactness does not suffer from such foolishness. 11.4 PROPOSITION. A set D is compact as a metric space if and only if it is compact as a subset of E.
11. COMPACTNESS PROOF. A subset of D is an open ball in the space D if and only if it has the form B ∩ D for some open ball B of the space E. Since an open set is the union of all the open balls it contains, it follows that A is an open subset of the space D if and only if A = B ∩D for some open subset B of the space E. Now, the deﬁnition of compactness does the rest. 2
41
Cluster Points, Convergence, Completeness
This is to look into the connections between compactness and convergence. 11.5 DEFINITION. A point x in E is called a cluster point 2 of a subset A of E provided that every open ball centered at x contains inﬁnitely many points of A.
11.6 THEOREM. Every inﬁnite subset of a compact set has at least one cluster point in that compact set. PROOF. We shall show that if C is compact, and A ⊂ C, and A has no cluster point in C, then A is ﬁnite. Let A and C be such. Since no x in C is a cluster point of A, for every x in C there is an open ball B(x, r) that contains only ﬁnitely many points of A. Those open balls cover C obviously. Since C is compact, there must be a ﬁnte number of them that cover C and, therefore, A. Since each one of those ﬁnitely many balls has a ﬁnte number of points of A, the total number of points in A must be ﬁnite. 2 The following is the way compactness helps in discussing convergence. In particular, together with Proposition 10.4, it shows that every Cauchy sequence in a compact set is convergent. 11.7 THEOREM. Every sequence in a compact set has a subsequence that converges to some point of that set. PROOF. Let C be compact. Let (xn ) ⊂ C. If the set A = {x1 , x2 , . . .} is ﬁnite, then at least one point of A, say x, appears inﬁnitely often in the sequence, and hence (x, x, . . .) is a subsequence, which obviously converges to x ∈ A ⊂ C. Now suppose that A is inﬁnite. By the preceding theorem, then A has a cluster point x in C. Since each one of the balls B(x, 1/n), n = 1, 2, . . ., has inﬁnitely many points in C, we may pick k1 so that xk1 is in B(x, 1), pick k2 > k1 so that xk2 is in B(x, 1/2), pick k3 > k2 so that xk3 is in B(x, 1/3), and so on. Obviously, (xkn ) converges to x. 2
2 Other
terms in common use include limit point, adherence point, point of accumulation, etc.
42 11.8 COROLLARY. Every compact set is closed.
METRIC SPACES
PROOF. Let C be compact. The preceding theorem implies that every convergent sequence in C converges to a point of C. Thus, C is closed by Theorem 9.5. 2
11.9 COROLLARY. Every compact metric space is complete. Every Cauchy sequence in a compact metric space is convergent. PROOF. The second statement is immediate from Theorem 11.7 and Proposition 10.4. The ﬁrst follows from the second by the deﬁnition of completeness. 2
Compactness in Euclidean Spaces
We have seen that, for an arbitrary metric space, every compact set is bounded and closed (Proposition 11.2 and Corollary 11.8). In the case of Euclidean spaces, the converse is true as well. This is called the HeineBorel Theorem. 11.10 THEOREM. A subset of a Euclidean space is compact if and only if it is bounded and closed. We start by listing an auxiliary result that is trivial at least for R, R2 , R3 . We omit its proof. 11.11 LEMMA. Let B be a bounded subset of a Euclidean space E. Then, for every > 0 there is a ﬁnite collection of closed balls of radius that covers B. Here is the proof of Theorem 11.10. PROOF. As mentioned above, 11.2 and 11.8 prove the necessity part. We now prove the sufﬁciency of the condition. Let E be a Euclidean space and let C be a closed and bounded subset of E. Suppose that C is not compact. Then, there is a collection {Ai : i ∈ I} of open sets that covers C but is such that 11.12 no ﬁnite subcollection {Ai : i ∈ I} covers C.
(a) Let = 1/2. By the preceding lemma, we can ﬁnd a ﬁnite number m of closed balls B1 , . . . , Bm of radius that cover C. Then, C = (C ∩ B1 ) ∪ · · · ∪ (C ∩ Bm ). In view of (11.12), at least one of C ∩ B1 , . . . , C ∩ Bm cannot ever be covered by a ﬁnite subcollection of the Ai ; let that one be denoted by C1 . Now, C1 is closed, its diameter is at most 2 = 1 (since the Bk have diameter 1), and (11.12) is true for C1 .
11. COMPACTNESS (b) Applying the arguments of the preceding paragraph with = 1/4 to the set C1 we get a new set C2 ⊂ C1 that is closed, has diameter at most 1/2, and (11.12) holds for C2 . Repeating this with = 1/6, 1/8, 1/10, . . . we obtain further sets C3 , C4 , C5 , . . . with the same properties but with diameters at most 1/3, 1/4, 1/5, . . .. Clearly C1 ⊃ C2 ⊃ C3 ⊃ · · ·. (c) Since (11.12) holds for each Cn , it must be that no Cn is empty (covering an empty set takes no effort). Thus, we may pick x1 from C1 , x2 from C2 , and so on to obtain a sequence (xn ). (d) This sequence is Cauchy: given > 0 choose n so that 1/2n < , and then xn , xn+1 , . . . are all in a ball of radius since all these terms are in Cn which has diameter less than 1/n. Since E is Euclidean, it is complete (see Theorem 10.7), which means that every Cauchy sequence converges. Hence, the sequence (xn ) converges to some point x0 in E. Since, for each n, (xm : m ≥ n) ⊂ Cn and Cn is closed, the limit x0 belongs to Cn by Theorem 9.5. (e) Since the Ai cover C, there must exist an i in I such that x0 is in Ai . Fix that i. Since Ai is open, there is an > 0 such that B(x0 , ) ⊂ Ai . Now choose n large enough that 1/n < /2. Since, x0 ∈ Cn and diam Cn ≤ 1/n < /2, we see that Cn ⊂ B(x0 , ). In other words, Ai covers Cn . This contradicts the earlier assertion that (11.12) holds for all Cn . This completes the proof. 2
43
Exercises:
11.1 Supremums. Let A be a nonempty subset of R. Suppose that A is bounded above but has no greatest element. Show that, then, sup A is a cluster point of A. 11.2 Show that the union of a ﬁnite number of compact sets is again compact. 11.3 Give an example of an inﬁnite subset of R that has no cluster points. Give an example of one with exaclty two cluster points. Identify the cluster points of the set A = {x ∈ R : x = 1 1 + for some m, n in N}. mn
11.4 Sequences in R. By the HeineBorel theorem, every closed interval [a, b] ⊂ R is compact. Thus, every bounded sequence in R has a convergent subsequence (cf. Theorem 11.7). Another consequence is the following useful result: Let (xn ) be a bounded sequence in R. Suppose that all convergent subsequences of it have the same limit x. Then, (xn ) converges to x.
44 Prove this by following the steps below.
METRIC SPACES
(a) Show that x = lim inf xn and x = lim sup xn are cluster points of ¯ (xn ). (b) Show that there is a subsequence of (xn ) that converges to x. Similarly, then, there is a subsequence that converges to x. ¯ (c) By the hypothesis that all convergent subsequences have the same limit, we conclude that x = x, which means that lim xn exists (and is in R since ¯ (xn ) is bounded).
Functions on Metric Spaces
Elementary analysis is mostly about functions from R into R, or functions from Rn into R, or, somewhat more generally, functions from Rn into Rm . Our aim is to consider functions from one metric space to another. Replacing Euclidean spaces by metric spaces introduces no new difﬁculties and is immensely useful for dealing with various problems concerning differential and integral equations. For mappings from a metric space to another we employ either notations like T, S, U or notations like f, g, h. Generally, the transformation notation is cleaner: we write T x for the image of x under T and T −1 B for the inverse image of B, which become f (x) and f −1 (B) in the standard function notation.
12
Continuous Mappings
Throughout this section, E, E , etc. will be metric spaces with corresponding metrics d, d , etc. Given a mapping T from E into E , we write T x for the image of the point x of E and T −1 B for the inverse image of the subset B of E . On a ﬁrst reading, the reader may wish to take E = R and d (x, y) = x − y as usual. 12.1 DEFINITION. A mapping T : E → E is said to be continuous at the point x of E provided that for every > 0 there is a δ > 0 such that y ∈ E, d(x, y) < δ ⇒ d (T x, T y) < .
The mapping T is said to be continuous if it is continuous at every x of E.
REMARKS: (a) In the deﬁnition, δ is allowed to depend on and x. (b) When E = E = R with the usual metric, the preceding is the classical deﬁnition of continuity. (c) The condition for T to be continuous at x can be rephrased in more geometric terms as follows: for every > 0 there is a δ > 0 such that T maps the open ball B(x, δ) of E into the open ball B (T x, ) of E . Here, B(x, δ) = {y ∈ E : d(x, y) < δ}, B (T x, ) = {y ∈ E : d (T x, y) < }. 45
46
FUNCTIONS ON METRIC SPACES
Continuity and Open Sets
12.2 THEOREM. A mapping T : E → E is continuous if and only if T −1 B is an open subset of E for every open subset B of E . PROOF. Suppose that T is continuous. Let B ⊂ E be open. We want to show that, then, A = T −1 B is open, that is, for every x in A there is δ > 0 such that B(x, δ) ⊂ A. To this end, ﬁx x in A, note that y = T x is in B, and therefore, there is > 0 such that B (y, ) ⊂ B (since B is open). By the continuity of T , for that , there is a δ > 0 such that T maps B(x, δ) into B (y, ). Since B (y, ) ⊂ B, we have B(x, δ) ⊂ A as needed. Suppose that T −1 B is open in E for every open subset B of E . Let x in E be arbitrary. We want to show that, then, T is continuous at x. To this end, ﬁx > 0. Since B (T x, ) is open, its inverse image is open, that is A = T −1 B (T x, ) is an open subset of E. Note that x is in A; therefore, there is a δ > 0 such that B(x, δ) ⊂ A, and then T maps B(x, δ) into B (T x, ). So, T is continuous at x. 2
Continuity and Convergence
If (xn ) is a sequence in E, we write xn → x to mean that (xn ) converges to x in E in the metric d, that is, d(xn , x) → 0. Similarly, we write yn → y to mean that the sequence (yn ) in E converges to y in the metric d . The following is probably the most useful characterization of continuity. 12.3 THEOREM. A mapping T : E → E is continuous at the point x of E if and only if (xn ) ⊂ E, xn → x
d d d
⇒
T xn → T x.
d
PROOF. Suppose that T is continuous at x. Let (xn ) ⊂ E be such that xn → x. We want to show that, then, T xn → T x, which is equivalent to showing that for every > 0 the ball B (T x, ) contains all but ﬁnitely many of the points T xn . To this end, ﬁx > 0. By the continuity of T at x, there is δ > 0 such that T maps B(x, δ) into B (T x, ). Since xn ∈ B(x, δ) for all but ﬁnitely many n, it follows that T xn ∈ B (T x, ) for all but ﬁnitely many n, which is as desired. Suppose that T is not continuous at x. Then, there is > 0 such that for every δ > 0 there is y in E such that d(x, y) < δ and d (T x, T y) ≥ . Thus, for that ,
d
d
12. CONTINUOUS MAPPINGS taking δ = 1, 1/2, 1/3, . . . we can pick y = x1 , x2 , x3 , . . . such that d(xn , x) < 1/n and d (T xn , T x) ≥ . Hence, there is a sequence (xn ) ⊂ E such that xn → x but (T xn ) does not converge to T x. 2
d
47
Compositions
The following result is recalled best by the phrase “a continuous function of a continuous function is continuous”. 12.4 THEOREM. If T : E → E is continuous at x ∈ E and S : E → E is continuous at T x ∈ E , then S ◦ T : E → E is continuous at x ∈ E. If T is continuous and S is continuous, then S ◦ T is continuous. PROOF. The second assertion is immediate from the ﬁrst. To show the ﬁrst, let (xn ) ⊂ E be such that xn → x. If T is continuous at x, the T xn → T x by the last theorem; and if S is continuous at T x, this in turn implies that S(T xn ) → S(T x) by the last theorem again, which means that S ◦ T is continuous at x. 2
d d d
EXAMPLES. 12.5 Constants. Let T : E → E be deﬁned by T x = b where b in E is ﬁxed. This T is continuous. 12.6 Identity. Let T : E → E be deﬁned by T x = x. This T is continuous, as is easy to see from Theorem 12.2 or 12.3. 12.7 Restrictions. Let T : E → E be continuous. For D ⊂ E, the restriction of T to D is the mapping S : D → E deﬁned by putting Sx = T x for each x ∈ D. Obviously, the continuity of T implies that of S. 12.8 Discontinuity. Let f : R → R be deﬁned by setting f (x) = 1 if x is rational and f (x) = 0 if x is irrational. This function is discontinuous at every x ∈ R. To see it, ﬁx x in R. For every δ > 0, the ball B(x, δ) has inﬁnitely many rationals and inﬁnitely many irrationals. Thus, it is impossible to satisfy the condition for continuity at x (for any < 1). 12.9 Lipschitz continuity. A mapping T : E → E is said to satisfy a Lipschitz condition if there exists a constant K ∈ (0, ∞) such that d (T x, T y) ≤ Kd(x, y)
48
FUNCTIONS ON METRIC SPACES
for all x, y in E. Every such mapping is continuous: given > 0, choose δ = /K no matter what x is. 12.10 Coordinate mappings. Let E = Rn , the ndimensional Euclidean space, ﬁx i in {1, . . . , n}, and deﬁne Pi : Rn → R by Pi x = xi , the ith coordinate of x. Then, Pi satisﬁes the Lipschitz condition above with K = 1 and, thus, is continuous.
RealValued Functions
Functions f from a metric space E into R can be combined through arithmetic operations to obtain new functions. For instance, f + g is the function whose value at x is f (x) + g(x). In deﬁning f /g, however, one must exercise some caution at points x where g(x) = 0. It is best to limit the deﬁnition of f /g to the set {x ∈ E : g(x) = 0}. The following is immediate from Theorem 12.3. 12.11 PROPOSITION. If f : E → R and g : E → R are continuous, then so are f + g, f − g, f · g, f /g except that, in the last case, f /g should be treated as a function on {x : g(x) = 0}.
Rn Valued Functions
These are functions from a metric space E into the Euclidean space Rn (with the Euclidean distance). The following reduces the notion of continuity for such mappings to the case of realvalued functions. We use the projection mappings Pi introduced in Example 12.10: Pi x is the ith coordinate of the vector x in Rn . 12.12 PROPOSITION. A mapping T : E → Rn is continuous if and only if the mappings P1 ◦ T, . . . , Pn ◦ T from E into R are continuous. PROOF. Let T be continuous. Then, Pi ◦ T is continuous for each i because a continuous function of a continuous function is continuous. Suppose that P1 ◦T, . . . , Pn ◦T are continuous. To show that, then, T is continuous, we start by observing that
n
12.13
u−v =
1
Pi u − Pi v2 ,
u, v ∈ Rn .
Now, ﬁx x ∈ E and > 0. Using the deﬁnition of continuity for Pi ◦ T at x with √ i = / n, we ﬁnd δi > 0 such that √ d(x, y) < δi ⇒ Pi T x − Pi T y < / n.
12. CONTINUOUS MAPPINGS Let δ = min{δ1 , . . . , δn }. Then δ > 0 and d(x, y) < δ √ ⇒ Pi T x − Pi T y < / n for each i ⇒ Tx − Ty < 2
49
in view of 12.13 used with u = T x and v = T y.
Exercises:
12.1 Continuity of metrics. Recall the deﬁnition of the product space E × E from Exercise 7.9 in Chapter with (E1 , d1 ) = (E2 , d2 ) = (E, d). Show that d : E × E → R+ is continuous. 12.2 Continuity of pairs. Let f : E → E and g : E → E be continuous. Deﬁne h : E → E × E by h(x) = (f (x), g(x)). Show that h is continuous. 12.3 Closed sets. If T : E → E is continuous, then T −1 B is a closed subset of E for every closed subset B of E . Show. For f : E → R continuous, show that the sets {x ∈ E : f (x) ≤ b}, {x ∈ E : f (x) = b}, {x ∈ E : f (x) ≥ b} are closed in E. 12.4 Indicators. For A ⊂ E let 1A be the indicator of A, that is, 1A (x) = 1 if x ∈ A and 1A (x) = 0 if x ∈ A. Show that 1A is continuous at all points x ∈ E except for x ∈ ∂A. 12.5 Left and Right Continuity. Let f : R → E . Order properties of the real line enable us to reﬁne the notion of continuity as follows. The function f is said to be rightcontinuous at x ∈ R provided that f (xn ) → f (x) for every decreasing sequence (xn ) ⊂ R with limit x. Similarly, f is said to be leftcontinuous at x if f (xn ) → f (x) for every increasing sequence (xn ) with limit x. Show that f is continuous at x if and only if it is both rightcontinuous and leftcontinuous at x. 12.6 Functional inverses. Let f : R+ → R+ be a continuous and strictly increasing bijection. Let f −1 (y) be that point x for which f (x) = y. Show that the function f −1 is continuous and strictly increasing. 12.7 Legendre Transforms. A function f : R → R is called convex if f (px + qy) ≤ pf (x) + qf (y) for all x, y ∈ R and all p, q ∈ (0, 1) satisfying p + q = 1. The Legendre transform of a convex function f is the function g : R → R deﬁned by g(y) = max(xy − f (x)).
x d d
50 Show that g is convex and that
FUNCTIONS ON METRIC SPACES
f (x) = max(xy − g(y)).
y
State any extra “smoothness” assumptions you might need. 12.8 Sections. Let f : E1 × E2 → R be continuous. Show that, for each y in E2 , the mapping x → f (x, y) from E1 into R is continuous. Similarly, y → f (x, y) is continuous for each x. Unfortunately, the converse does not hold: it is possible to have x → f (x, y) continous for each y and y → f (x, y) continuous for each x even though f is not continuous. Give an example of such a function.
13
Compactness and Uniform Continuity
As before, E, E , etc. are metric spaces with metrics d, d , etc. This section is on the effect of compactness on continuity. 13.1 THEOREM. Let T : E → E be continuous. If E is compact, then the range of T is a compact subset of E . PROOF. Let D ⊂ E be the range of T . Assuming that E is compact, we need to show that D is compact. Let {Bi : i ∈ I} be a collection of open subsets of E that covers D. Then, the continuity of T implies via Theorem 12.2 that the sets Ai = T −1 Bi , i ∈ I, are open. Moreover, {Ai : i ∈ I} covers E: if x is in E then T x is in D, and hence, T x is in Bi for some i, which implies that x is in the corresponding Ai . Now the compactness of E implies that there exists a ﬁnte set J ⊂ I such that {Ai : i ∈ J} covers E. Thus, if x ∈ E, then x ∈ Ai for some i in J and therefore T x ∈ Bi for some i in J. That is, {Bi : i ∈ J} covers D. So, D must be compact. 2 Recall that every compact set is closed and bounded. Thus, if f : E → R is continuous and E is compact, then the range of f is bounded and closed, which implies that f attains a maximum and a minimum, that is, there are x0 and x1 , such that f (x0 ) ≤ f (x) ≤ f (x1 ) for all x ∈ E (see Exercise 11.1 in Chapter to the effect that if D ⊂ R is closed and bounded then inf A and sup A belong to D). We have thus shown the following: 13.2 COROLLARY. Let E be compact and f : E → R continuous. Then, f is bounded and attains a maximum and a minimum. The conclusion fails if E is not compact. For instance, f (x) = x on E = (0, 1) is bounded but has neither a maximum nor a minimum. Also, f (x) = 1/x on E = (0, 1) is not bounded and has neither a maximum nor a minimum.
13. COMPACTNESS AND UNIFORM CONTINUITY
51
Uniform Continuity
Recall the deﬁnition of continuity: T : E → E is continuous provided that for every x in E and every > 0 there is a δ > 0 (depending on x and ) such that d(x, y) < δ implies d (T x, T y) < for all y in E. The importance of the following is to remove the dependence δ of on x. 13.3 DEFINITION. A mapping T : E → E is said to be uniformly continuous provided that for every > 0 there is a δ > 0 such that x, y ∈ E, d(x, y) < δ ⇒ d (T x, T y) < .
Obviously, every uniformly continuous function is continuous. The converse is false. For example, the function f : (0, 1) → R deﬁned by f (x) = 1/x is continuous but not uniformly so. The failure here is not due to the unboundedness of f . For instance, the function f : (0, 1) → [−1, 1] deﬁned by f (x) = sin 1/x is continuous but not uniformly so. The mappings of Examples 12.5, 12.6, 12.9, and 12.10 are uniformly continuous. In fact, they are all special cases of 12.9 on Lipschitz continuity. Being Lipschitz almost encapsulates the notion of uniform continuity 13.4 PROPOSITION. Let T : E → E be Lipschitz continuous. Then T is uniformly continuous.
PROOF. Fix > 0 and choose δ = /K. This δ works and is independent of x.
2
(Exercise 13.6 provides an “almost converse” to this result). The following shows the important role of compactness on uniform continuity. 13.5 THEOREM. Let T : E → E be continuous. If E is compact, then T is uniformly continuous.
PROOF. Fix > 0. We search for δ > 0 that will fulﬁll the condition for uniform continuity. Since T is continuous, for each x in E there is δ(x) > 0 such that 13.6 d(x, y) < δ(x) ⇒ d (T x, T y) < /2.
The collection of open balls B(x, δ(x)/2), x ∈ E, covers E. Since E is compact, there must exist a ﬁnite number of them, say those corresponding to x1 , . . . , xn , that cover E. Deﬁne 1 δ = min{δ(x1 ), . . . , δ(xn )}. 2
52
FUNCTIONS ON METRIC SPACES
Then, δ > 0 and it remains to show that this δ works. Let x, y in E be arbitrary and suppose that d(x, y) < δ. By the way the x1 , . . . , xn are chosen, there is an i such that x is in B(xi , δ(xi )/2), that is, d(x, xi ) < Moreover, for the same i, 1 d(y, xi ) ≤ d(y, x) + d(x, xi ) ≤ δ + δ(xi ) ≤ δ(xi ). 2 Thus, d(x, xi ) < δ(xi ) and d(y, xi ) < δ(xi ), which by 13.6 imply that d (T x, T y) < /2, and d (T y, T xi ) < /2. Thus, d (T x, T y) < by the triangle inequality. 2 1 δ(xi ). 2
Exercises:
13.1 Metrics. Show that, for ﬁxed x0 in E, the function x → d(x, x0 ) from E into R+ is uniformly continuous. 13.2 Compositions. Let T : E → E and S : E → E be uniformly continuous. Show that, then, S ◦ T : E → E is uniformly continuous. 13.3 Homeomorphisms. Recall that for a bijection f : E → E we deﬁne the functional inverse f −1 by setting f −1 (y) = x if and only if f (x) = y. A homeomorphism from E onto E is a bijection that is continuous and whose functional inverse is also continuous. Incidentally, two spaces E and E are said to be homeomorphic if there exists a homeomorphism from one to the other. Compactness helps in checking for homeomorphisms. Show that if f : E → E is a continuous bijection and E is compact, then f is a homeomorphism. 13.4 Extensions. Let D be dense in E (see Exercise 8.6 in Chapter for the deﬁnition). Note that this means that every point of E \ D is a cluster point of D. Suppose that f : D → R is uniformly continuous. Show ¯ that, then, there exists a unique continuous function f : E → R such that ¯ ¯ f (x) = f (x) for all x in D. Then, f is called the continuous extension of f onto E. 13.5 Cantor function. Let E = [0, 1], and C be the Cantor set, and D = E \ C; see Example 8.9 in Chapter . Note that D is dense in E, since C has no open intervals contained in it. Show that the function f constructed in 8.9 of Chapter is a uniformly continuous function from D into [0, 1]. By the preceding exercise, then,
14. SEQUENCES OF FUNCTIONS ¯ ¯ f has a continuous extension f onto E = [0, 1]. In fact, f is uniformly continuous (why?). ¯ The function f is called the Cantor function. It is increasing and continu¯ ous. Its derivative exists at every x in D and is equal to 0. So, although f increases from 0 to 1 in a continuous fashion, all its increase is on the set C, and C has “length” 0. 13.6 Lipschitz Continuity. A mapping T : Rn → R is uniformly continuous if and only if for every > 0 there exists K such that T x − T y ≤ K · x − y + for all x and y in Rn . Prove this. Hints: (a) The “if” part is easy. Choose δ= /2 . K /2
53
(b) For the “only if” part: ﬁx > 0 and x and y; choose a chain of points x = x0 , x1 , x2 , . . . , xm = y with distances xi − xi+1 < δ; ask, how many such points do we need, and note that
m
T x − T y ≤
1
T xi − T xi+1  ≤ n ;
ﬁgure out m needed and then what K should be.
14
Sequences of Functions
Let E and E be metric spaces with respective metrics d and d . Let (Tn ) be a sequence of mappings from E into E . 14.1 DEFINITION. The sequence (Tn ) is said to converge pointwise to a mapping T : E → E provided that the sequence (Tn x) converges to T x in E for each point x in E. In other words, for each x in E, we must have 14.2 lim d (Tn x, T x) = 0,
n
that is, for every > 0 there must be an n ,x such that d (Tn x, T x) < for all n ≥ n ,x . If n ,x can be chosen to be free of x, we obtain the following stronger concept of convergence:
54
FUNCTIONS ON METRIC SPACES
f (x) n
f
1 f 2 f
3 x 1
Figure 4: Here (fn ) converges to f , where f (x) = 0 for x < 1 and f (x) = 1 for x ≥ 1. Convergence is pointwise but not uniform.
14.3 DEFINITION. The sequence (Tn ) is said to converge uniformly to a mapping T provided that lim sup d (Tn x, T x) = 0.
n x∈E
Obviously, uniform convergence of (Tn ) implies pointwise convergence (and the limit T is the same). That the converse is generally false can be seen from Figures 4 and 5 below: here the functions fn : [0, ∞) → [0, 1] converge pointwise, but not uniformly.
Cauchy Criterion
As with sequences of points, it is important to have a criterion for the uniform convergence of (Tn ) expressed in terms of the Tn themselves. The following Cauchy criterion does this: 14.4 THEOREM. Suppose that E is complete. Then, (Tn ) is uniformly convergent if and only if for every > 0 there is an n with 14.5 sup d (Tn x, Tm x) <
x
for all m > n ≥ n .
14. SEQUENCES OF FUNCTIONS
55
f (x) n
f
n x n
n−1
Figure 5: These fn converge to f = 0 pointwise, but not uniformly.
f (x) n
f f f 3 2
1
x
Figure 6: These fn converge to 0 uniformly (and hence pointwise).
56
FUNCTIONS ON METRIC SPACES
PROOF. Suppose that (Tn ) converges uniformly, say, to T . Then, for every > 0, there is an n such that d (Tn x, T x) < /2 for all n ≥ n . Thus, for m, n ≥ n , d (Tn x, Tm x) ≤ d (Tn x, T x) + d (T x, Tm x) < /2 + /2 = for all x. So, (Tn ) is Cauchy (for every > 0 there is n such that 14.5 holds). Let (Tn ) be Cauchy. Then, in particular, for each x in E the sequence (Tn x) in E is Cauchy. Since E is complete, this implies that (Tn x) converges to some point of E , call it T x. This deﬁnes a mapping T : E → E . We want to show that (Tn ) converges to T uniformly. Since (Tn ) is Cauchy, for every > 0 there is an n such that d (Tn x, Tm x) < for all m, n ≥ n for all x. Now, let m → ∞; then, (Tm x) converges to T x and the continuity of y → d (Tn x, y) implies that d (Tn x, Tm x) → d (Tn x, T x). Thus, as we needed to show, for > 0 there is an n with d (Tn x, T x) < for all n ≥ n and all x ∈ E. 2
Continuity of Limit Functions
As can be seen from Figure 4, the pointwise limit of a sequence of continuous functions is not necessarily continuous. In fact, the primary use of uniform convergence is to ensure the continuity of the limit function. 14.6 THEOREM. Suppose that each Tn is continuous and (Tn ) converges to T uniformly. Then, T is continuous. PROOF. Fix x in E. Note that for all n and y d (T x, T y) ≤ d (T x, Tn x) + d (Tn x, Tn y) + d (Tn y, T y). Given > 0, there is an n such that the ﬁrst and third terms on the right side are less than /3 each for n = n ; This comes from the uniform convergence of (Tn ) of T . Moreover, the continuity of Tn at the point x implies the existence of δ = δ ,x such that the second term on the right with n = n is less than /3 for all y ∈ B(x, δ). Hence, for every > 0 there is a δ = δ ,x such that d(x, y) < δ implies that d (T x, T y) < for all y; that is, T is continuous at x. 2
Exercises:
14.1 Let 0 ≤ a < b < 1. Let fn : [a, b] → R+ be deﬁned by fn (x) = xn . Show that (fn ) converges uniformly to f = 0.
15. SPACES OF CONTINUOUS FUNCTIONS 14.2 Let Tn : [0, 1] → [0, 1] be deﬁned by Tn x = xn (1 − x). Show that (Tn ) is uniformly convergent. 14.3 Let f : R → R be uniformly continuous. Deﬁne fn (x) = f (x + 1/n). Show that (fn ) converges uniformly to f . 14.4 Let (fn ) be deﬁned as a sequence of functions from R+ into R+ by √ √ √ f1 (x) = x, f2 (x) = x + x, f3 (x) = x + x + x, . . . Show that (fn ) is convergent and ﬁnd the limit function.
57
15
Spaces of Continuous Functions
Throughout this section (E, d) will be a compact metric space, and all functions are from E into R. On a ﬁrst reading, the reader should take E = [a, b], a closed interval. Our aim is to illustrate the uses of the foregoing concepts in the analysis of the function space C(E, R) of all continuous functions from E into R. For brevity, we write C for C(E, R). The set C is a vector space: if f and g are in C then so is af + bg for each a in R and b in R. Moreover, various arithmetic operations are welldeﬁned on C: f + g, f − g, f · g, and f /g all belong to C if f and g are in C, except that in the case of f /g one must worry about g(x) = 0. Although each point of C is a function, in many respects C is like a Euclidean space. We may, for instance, deﬁne a norm of C as follows. Let f ∈ C. Being a continuous function on a compact metric space, f is bounded and attains its maximum and minimum. It follows that 15.1 f = max f (x)
x∈E
is a welldeﬁned positive realnumber; it is called the norm of f . It is indeed a norm: 15.2 15.3 15.4 f ≥ 0; f = 0 if and only if f = 0; cf = c · f ; f +g ≤ f + g .
As with Euclidean spaces, we may use the norm above to deﬁne a metric on C. We deﬁne the distance between f and g to be 15.5 d(f, g) = f − g .
Convergence in C
The following shows that the convergence in the metric space C is equivalent to the uniform convergence of functions on E.
58
FUNCTIONS ON METRIC SPACES
15.6 THEOREM. A sequence (fn ) in C is convergent if and only if the sequence of functions fn : E → R is uniformly convergent. PROOF. The deﬁnition of convergence for a sequence of points in a metric space and the deﬁnition of uniform convergence for a sequence of functions fn : E → R are such that the claim is simply that lim d(fn , f ) = 0
n
⇔
lim sup fn (x) − f (x) = 0.
n x∈E
But this is obvious in view of 15.5 and 15.1.
2
Conceptually, then, the somewhat complex concept of uniform convergence of a sequence of functions is equivalent to the simpler concept of convergence of a sequence in a metric space.
Lipschitz Continuous Functions
A function f ∈ C is said to be Lipschitz continuous if there exists a constant K such that 15.7 f (x) − f (y) ≤ K · d(x, y) for all x, y ∈ E. Let BK be the set of all f in C satisfying 15.7. Then, clearly, the set of all Lipschitz continuous functions is exactly the union of the BK ’s. If E = [a, b], f is differentiable, and the derivative f is bounded (that is, there exists a K such that f (x) ≤ K for all x ∈ [a, b]), then f is Lipschitz continuous. Consider a ﬁxed K and let AK denote the set of all differentiable functions f whose derivatives f are continuous and bounded by K. The set AK is not closed, which can be seen from Figure 7 where (fn ) ⊂ AK , (fn ) converges to f in C, but f is not in AK . In fact, the closure of AK is precisely BK . We leave this without proof. Instead, we show the following partial result with general E. 15.8 PROPOSITION. BK is a closed subset of C. PROOF. We use the characterization Theorem 9.5 from Chapter . Let (fn ) ⊂ BK converge to the point f in C. We need to show that f is in BK . Now, for arbitrary x and y in E, f (x) − f (y) ≤ f (x) − fn (x) + fn (x) − fn (y) + fn (y) − f (y) ≤ f − fn + Kd(x, y) + fn − f for all n. Since fn − f → 0, this shows that f satisﬁes 15.7. 2
As mentioned above, the set of all Lipschitz continuous functions coincides exactly with ∪K BK . Even though each BK is closed, the union is not. This fact can be seen from the sequence of functions shown in Figure 8. In fact, its closure is precisely C, that is, every f in C is the limit of a sequence of Lipschitz continuous functions.
15. SPACES OF CONTINUOUS FUNCTIONS
59
f
¡
f2 f1
a
b
Figure 7: A sequence of differentiable functions whose derivatives are bounded but whose limit is not differentiable.
f
f2 f1
a
b
Figure 8: A sequence of Lipschitz continuous functions converging to a continuous function that is not Lipschitz.
60
FUNCTIONS ON METRIC SPACES
Completeness
The space C is not bounded. Therefore it cannot be compact. But, at least, it is complete. 15.9 THEOREM. The space C is complete. PROOF. Let (fn ) ⊂ C be Cauchy, that is, for every > 0 there is an n such that fn − fm ≤ for all m > n ≥ n . This is equivalent to the condition 14.5 (here E = R which is complete). Thus, by Theorem 14.4, (fn ) is uniformly convergent as a sequence of functions on E. But, by Theorem 15.6, uniform convergence is equivalent to convergence in C. So, (fn ) is convergent in C. 2
Functionals
Since C is a metric space, we may speak of functions deﬁned on C as we speak of functions deﬁned on E. For linguistic clarity, a function from C into R is called a functional. Here are some examples of functionals: for f ∈ C, 15.10 15.11 15.12 M (f ) = max f (x)
x∈E
Px (f ) = f (x),
x ∈ E ﬁxed
F (f ) = φ(f (x1 ), . . . , f (xk )),
where φ : Rk → R is ﬁxed and x1 , . . . , xk are ﬁxed in E. Here are some further examples of functionals, in the particular case where E = [a, b]:
b
15.13 15.14
L(f ) =
a b
f (x)dx, φ(x)f (x)dx,
a
Lφ (f ) =
where φ ∈ C is some ﬁxed function. The functional M is uniformly continuous; in fact, it is Lipschitz continuous with Lipschitz constant K = 1: M (f ) − M (g) =  max f (x) − max g(x)
x x
≤ max f (x) − g(x)
x
= f −g = d(f, g). Even easier is the Lipschitz continuity of the coordinate mapping Px : Px (f ) − Px (g) = f (x) − g(x) ≤ f − g .
15. SPACES OF CONTINUOUS FUNCTIONS Assuming that the function φ : Rk → R is continuous, the function F is continuous: if fn − f → 0, then the sequence of points (fn (x1 ), . . . , fn (xk )) ∈ Rk converges to the point (f (x1 ), . . . , f (xk )) ∈ Rk as n → ∞, and the continuity of φ implies that F (fn ) → F (f ). The functional L is a linear transformation from C into R. It is uniformly continuous; in fact, it is Lipschitz continuous with Lipschitz constant K = b − a. So is Lφ b with Lipschitz constant K = a φ(x)dx.
61
Exercises:
15.1 If f and g are two continuous functions on a compact metric space, show that  max f (x) − max g(x) ≤ max f (x) − g(x).
x x x
62
FUNCTIONS ON METRIC SPACES
Differential and Integral Equations
The aim of this chapter is to discuss several applications of metric space ideas to some classical problems of engineering analysis. We shall start with one theorem, the ﬁxed point theorem for contractions on a metric space, and show how various problems can be beaten to submission with it.
16
Contraction Mappings
The aim of this section is to prepare the stage for some applications to differential and integral equations encountered frequently in engineering. Throughout, E is a metric space with some metric d. We shall use the term “transformation on E” to mean a mapping from E into E. If T is a transformation on E, then the image T x of x is a point in E, and the image of T x is T (T x), for which we will write T 2 x. In other words, we are writing T 2 for T ◦ T . Similarly, we deﬁne further iterates by T n+1 x = T (T n x), x ∈ E, n ≥ 0,
with T 0 x = x for all x. So, T 0 is the identity, T 1 is T , etc. Given a point x in E, if we write x0 = x, x1 = T x, x2 = T 2 x, x3 = T 3 x, . . . , we obtain a sequence (xn ) in E; this sequence is called the orbit starting at x. One should think of xn = T n x as the position at time n of a particle that starts at x and moves successively to T x, T 2 x, . . . . 16.1 DEFINITION. A transformation T on E is said to be a contraction if it is Lipschitz continuous with some Lipschitz constant α < 1. In other words, T is a contraction of E if there exists a constant α ∈ [0, 1) such that 16.2 d(T x, T y) ≤ αd(x, y) 63 for all x, y ∈ E.
64
DIFFERENTIAL AND INTEGRAL EQUATIONS
E Tx x
T3x T2x
Figure 9: The orbit of x under the map T .
Fixed Point Theorem
A point x is said to be a ﬁxed point of a transformation T if T x = x. Figure 10 shows a transformation T on E = [0, 1]; there, x∗ is the unique ﬁxed point of T , and the orbit (T n x0 ) of x0 converges to the ﬁxed point x∗ . The following theorem shows that every contraction of a complete metric space has a unique ﬁxed point. Its proof shows how to obtain the ﬁxed point by a method of successive approximations. 16.3 THEOREM. Suppose that E is complete. Let T be a contraction on E. Then, T has a unique ﬁxed point and for each point x0 in E, the orbit (T n x0 ) converges to that ﬁxed point. PROOF. Fix x0 in E and let (x0 , x1 , x2 , . . .) be its orbit. We show ﬁrst that this sequence is Cauchy. Indeed, suppose that m < n. Then xm = T m x0 and xn = T n x0 = T m T n−m x0 = T m xn−m . Hence, since d(T m x, T m y) ≤ αm d(x, y) in view of 16.2, we have d(xm , xn ) ≤ αm d(x0 , xn−m ) ≤ αm [d(x0 , x1 ) + d(x1 , x2 ) + · · · + d(xn−m−1 , xn−m )]. Now note that d(xi , xi+1 ) = d(T i x0 , T i x1 ) ≤ αi d(x0 , x1 ). Thus, d(xm , xn ) ≤ αm d(x0 , x1 )[1 + α + α2 + · · · + αn−m−1 ] 1 − αn−m = αm d(x0 , x1 ) 1−α m d(x0 , x1 ) ≤α . 1−α Since α < 1, the right side goes to 0 as m → ∞. Hence, the sequence (xn ) is Cauchy.
16. CONTRACTION MAPPINGS
65
y
y=x y=Tx
x* x2 x1
x0 1
x
Figure 10: A contraction on [0, 1]. Since E is complete, the sequence (xn ) must converge to some point x in E. Then, by the continuity of T , T x = T (lim xn ) = lim T xn = lim xn+1 = x, that is, x is a ﬁxed point. To complete the proof, we now show that the ﬁxed point is unique. To this end, let y be another ﬁxed point. Then, Tx = x and T y = y,
and hence, by the contraction condition, d(x, y) = d(T x, T y) ≤ αd(x, y). Since α < 1, this is possible only if d(x, y) = 0, that is, x = y. 2 The preceding theorem can be used to prove existence and uniqueness of solutions to a wide variety of equations. Besides showing that T x = x has a solution, the proof gives a practical method for arriving at it. Indeed, start from an arbitrary point x0 and successively compute x1 = T x, x2 = T x1 , x3 = T x2 , . . . . The xn get close to x (geometrically fast): d(xn+1 , x) = d(T xn , T x) ≤ αd(xn , x), which shows that 16.4 d(xn , x) ≤ αn d(x0 , x).
Exercises:
66
DIFFERENTIAL AND INTEGRAL EQUATIONS
y
y=x y=Tx
0
x0
Figure 11: Exercise 16.1.
1
x
16.1 For the transformation T : [0, 1] → [0, 1] shown in Figure 11 ﬁnd the orbit of the point x0 indicated. 16.2 For the transformation T : [0, 1] → [0, 1] given by T x = 0.3 + 0.2x + 0.5x3 , Figure 12 shows that there are exactly two ﬁxed points. Find them. Show that, for arbitrary x0 = 1, the orbit of x0 converges to the smaller ﬁxed point x∗ . 16.3 Branching processes. In a chain reaction, each particle gives rise to a random number of new particles. Each of these new particles act independently and produces random numbers of newer particles. And this continues indeﬁnitely. Let pk be the probability that a particle produces k particles; here p0 , p1 , p2 , . . . are positive numbers with pk = 1. Starting with one particle, we now consider the probability that the chain reaction ﬁzzles out, that is, the population of particles becomes extinct. Let xn be the probability that the nth generation is extinct already. Note that the (n + 1)th generation consists of particles that are nth generation offspring of the individuals of the ﬁrst generation. In order for the population to be extinct at or before the (n + 1)th generation, populations initiated by the
16. CONTRACTION MAPPINGS
67
y
y=x
y=Tx
0
x*
x**=1
x
Figure 12: Exercise 16.2.
0
1
2 3 generations
Figure 13: Exercise 16.3.
4
68
DIFFERENTIAL AND INTEGRAL EQUATIONS particles of the ﬁrst generation must all become extinct. Thus,
∞
xn+1 =
k=0
pk (xn )k .
In other words, xn+1 = T xn where T : [0, 1] → [0, 1] is deﬁned by
∞
Tx =
k=0
pk xk ,
x ∈ [0, 1].
Now, the probability x∗ of eventual extinction for the population is the limit of xn , and thus satisﬁes x∗ = T x∗ . (a) Show that x1 = p0 . Show that the sequence (xn ) increases to the extinction probability x∗ . (b) Assume that p0 > 0. If p0 + p1 = 1 (so that p2 = p3 = · · · = 0) show that x∗ = 1. (c) Show that the mapping x → T x is increasing and convex. (d) Let a = k=1 pk k, that is, a is the expected number of particles produced by one particle. Show that if a ≤ 1, then x = T x has only one solution and the ﬁxed point is x∗ = 1. (e) Suppose that a > 1. Then, show that x = T x has exactly two solutions. One solution is 1, the other is the extinction probability x∗ . Show this by examining the graph of T and using (a). 16.4 Let T : [0, 1] → [0, 1] be deﬁned by T x = 4x(1 − x). Show that T has exactly two ﬁxed points. Compute them. Give an example of an orbit that converges to the ﬁxed point x∗ = 0. Note the highly chaotic nature of the orbits. 16.5 Let T : [0, 1] → [0, 1] be deﬁned by T x = 2x (mod 1), that is, T x = 2x if 2x < 1 and T x = 2x − 1 if 2x ≥ 1. The only ﬁxed point is x∗ = 0. Incidentally, if x = 0.ω1 ω2 ω3 · · · is the binary representation of x then T x = 0.ω2 ω3 ω4 · · · and T 2 x = 0.ω3 ω4 ω5 · · ·, etc. Note the highly chaotic nature of the orbits by plotting (T n x). 16.6 Let T : Rn → Rn be a linear transformation, say T x = Ax where A is some n × n matrix. Give a condition on A that guarantees T to be a contraction (with the Euclidean metric on Rn ).
∞
17. SYSTEMS OF LINEAR EQUATIONS 16.7 Let T x = Ax + b where A is n × n matrix and b is a ﬁxed vector in Rn . Consider E = Rn with the weighted Manhattan metric d(x, y) = n i=1 wi · xi − yi  where the weights w1 , . . . , wn are strictly positive. Show that, to assume that T is a contraction of this metric space E, it is sufﬁcient to have
n
69
wi aij  < wj ,
i=1
j = 1, . . . , n.
17
Systems of Linear Equations
In this section we discuss the use of the ﬁxed point theorem in solving systems of linear equations. As a byproduct, we get a chance to discuss the importance of choosing the right metric for a particular application. Let E = Rn ; we do not specify the metric just yet. Fix b ∈ Rn and consider the system of linear equations
n
17.1
xi =
j=1
aij xj + bi ,
i = 1, . . . , n,
where the aij are real numbers. Writing A for the n × n matrix of elements aij , the system 17.1 is equivalent to 17.2 x = Ax + b. In other words, the problem is to ﬁnd the ﬁxed point of the transformation T : Rn → Rn deﬁned by 17.3 T x = Ax + b. If T is a contraction, then we can use Theorem 16.3 and obtain the unique solution of T x = x by the method of successive approximations. The conditions under which T is a contraction depend on the choice of metric on E = Rn . We discuss three cases.
Maximum Norm
Suppose that d is the metric associated with the maximum norm: d(x, y) = max xi − yi .
1≤i≤n
Then, since T x − T y = Ax − Ay = A(x − y),
n
d(T x, T y) =
max 
i j=1
aij (xj − yj )
70
DIFFERENTIAL AND INTEGRAL EQUATIONS ≤ max
i j
aij  · xj − yj  aij  max xk − yk 
j k
≤ max
i
= (max
i j
aij )d(x, y).
Thus, the contraction condition 16.2 is satisﬁed if 17.4 α = max
i j
aij  < 1.
Manhattan Metric
Suppose that d is the Manhattan metric:
n
d(x, y) =
i=1
xi − yi .
Then, d(T x, T y) =
i

j
aij (xj − yj ) aij  · xj − yj 
≤
i j j
≤ (max
i
aij )d(x, y),
and the contraction condition is satisﬁed if 17.5 α = max
j i
aij  < 1.
Euclidean Metric
Suppose that d is the ordinary Euclidean distance. Then, d(T x, T y)2 =
i
2 aij (xj − yj )
j
a2 ij
j j
(xj − yj )2
≤
i
=(
i j
a2 )d(x, y)2 , ij
18. INTEGRAL EQUATIONS where we used Schwartz’s inequality at the second step. Thus, the contraction condition 16.2 is satisﬁed if 17.6 α= a2 < 1. ij
i j
71
Conclusion
Under each of the metrics discussed, Rn is a complete metric space. Hence, if at least one of the conditions 17.4–17.6 holds, Theorem 16.3 applies to show that there exists a unique solution to 17.1. The sequence of successive approximations x(0) , x(1) , . . . (whose limit is the ﬁxed point x) has the following form: 17.7 x(k+1) = Ax(k) + b, k = 0, 1, . . . ,
and we can choose any point x(0) ∈ Rn as the initial point. Each of the conditions 17.4–17.6 is sufﬁcient for applying this method. None is necessary; it is easy to give examples of A where one condition holds but not the others.
18
Integral Equations
The most interesting applications of ﬁxed point theorems arise when the underlying metric space is a function space. Here we discuss the existence and uniquencess of solutions to Fredholm and Volterra equations.
Fredholm Equation
A Fredholm equation (of the second kind) is an integral equation of the form
b
18.1
f (x) = φ(x) + λ
a
K(x, y)f (y)dy.
Here, the functions K : [a, b] × [a, b] → R and φ : [a, b] → R are given, λ ∈ R is an arbitrary parameter, and f : [a, b] → R is the unknown function. The function K is called the kernel of the equation. The equation is said to be homogeneous if φ = 0 and nonhomogeneous otherwise. The Fredholm equation is the continuous version of the system of linear equations 17.1. To see this, suppose that the interval is discretized and is replaced by n + 1 equidistant points a = x0 < x1 < · · · < xn = b. Then, writing yi = f (xi ) and bi = φ(xi ) and aij = λK(xi , xj )/n, we see that 18.1 becomes y i = bi +
j
aij yj .
Whether this discretization is appropriate is a different matter.
72
DIFFERENTIAL AND INTEGRAL EQUATIONS
Let C be the collection of all continuous functions f from [a, b] into R, and let the metric on C be deﬁned through the maximum norm: 18.2 d(f, g) = f − g = sup f (x) − g(x).
a≤x≤b
With this metric, C is a complete metric space (see Theorem 15.9 in Chapter ). Suppose that K is continuous on the square [a, b] × [a, b] and that φ is continuous on [a, b]. Then, the function T f deﬁned by
b
18.3
T f (x) = φ(x) + λ
a
K(x, y)f (y)dy
is continuous on [a, b] for each continuous function f on [a, b]. In other words, the mapping f → T f is a transformation on C. Now, the Fredholm equation 18.1 becomes 18.4 f = T f,
and thus, solving 18.1 is equivalent to ﬁnding the ﬁxed points of the transformation T on C. To this end, in order to apply the ﬁxed point theorem 16.3, all we need to show is that T is a contraction (recall that C is complete). The following shows that T is indeed so if the parameter λ is small enough. 18.5 THEOREM. Suppose that φ and K are continuous. Then there exists λ0 > 0 such that the equation 18.1 has a unique solution f for each λ in (−λ0 , λ0 ). Moreover, the solution f is continuous. PROOF. Since K is continuous on the square [a, b] × [a, b], it is bounded there (continuous functions on compact spaces are bounded). So, there is a constant c > 0 such that K(x, y) ≤ c for all x, y. Thus,
b
Tf − Tg
= ≤
max λ
x a
K(x, y)(f (y) − g(y))
y
λ · c · (b − a) max f (y) − g(y)
= λ · c · (b − a) · f − g . Choose λ0 = 1/c · (b − a). Then, for each λ ∈ (−λ0 , λ0 ), the preceding shows that T is a contraction on C. By Theorem 16.3, consequently, there is a unique ﬁxed point f in C of the transformation T . 2
18.6 EXAMPLE. Suppose that K(x, y) = xy on [0, 1] × [0, 1]. Let φ ∈ C be arbitrary and consider the Fredholm equation
1
18.7
f (x) = φ(x) + λ
0
xyf (y)dy.
18. INTEGRAL EQUATIONS The proof of 18.5 shows that, for λ < 1, there is a unique solution f . And the solution is the limit of the sequence f0 = φ, where, in general,
1
73
f1 = T f0 ,
f2 = T f1 ,
f3 = T f2 , . . .
T f (x) = φ(x) + λx
0
yf (y)dy.
Now, we start computing. Deﬁning a = f0 (x) = f1 (x) = φ(x) T f0 (x) =
1 0
yφ(y)dy, we have
φ(x) + λx
1 0
yφ(y)dy
= φ(x) + aλx f2 (x) = T f1 (x) = φ(x) + λx
1 0
y(φ(y) + aλy)dy
2
= φ(x) + aλx + a λ x 3 f3 (x) = T f2 (x) . . . fn (x) = T fn−1 (x) = φ(x) + aλx 1 +
λ 3
=
φ(x) + λx
1 0
y(φ(y) + aλy + a λ y)dy 3
2 3
2
= φ(x) + aλx + a λ x + a λ x 3 9
λ2 3 λ n−1 3
+
+ ··· +
.
In fact, it becomes clear from this that a ﬁxed point f exists for all λ ∈ (−3, 3) and the solution to 18.7 is 3aλ 18.8 f (x) = lim fn (x) = x + φ(x) n 3−λ with a = 0 φ(y)dy. Going back to 18.7, the special form of the kernel K suggests a quicker method. Indeed, let
1 1
c=
0
yf (y)dy.
Then, using 18.7 in the form f (x) = φ(x) + λxc, we get
1 1 1
c=
0
xf (x)dx =
0
xφ(x)dx +
0
xλxcdx = a +
λ c. 3
Solving this for c, we see that f (x) = φ(x) + λxc = φ(x) + 3aλ x 3−λ
74
DIFFERENTIAL AND INTEGRAL EQUATIONS
as before provided that λ = 3. Note that this is the solution for arbitrary λ = 3. But the method of successive approximations works for λ < 3 only. Studying the iterative method in the preceding example, we can get a theoretical understanding of the nature of solutions. To this end, we redo the computations of f0 = φ, f1 = T f0 , f2 = T f1 , . . . once more, now with an arbitrary kernel K, and omitting the limits of integration we get f0 (x) = f1 (x) f2 (x) = = φ(x) T f0 (x) T f1 (x) = = φ(x) + λ K(x, y)φ(y)dy φ(x) + λ K(x, y)f1 (y)dy K(x, y)[φ(y) + λ K(y, z)φ(z)dz]dy K2 (x, z)φ(z)dz K(x, y)φ(y)dy + λ2
= φ(x) + λ = where K2 (x, z) = Continuing, f3 (x) = T f2 (x) = φ(x) + λ +λ2 φ(x) + λ
K(x, y)K(y, z)dy.
K(x, y)[φ(y) + λ
K(y, z)φ(z)dz
K2 (y, z)φ(z)dz] K(x, z)φ(z)dz K3 (x, z)φ(z)dz
= φ(x) + λ +λ2 where K3 (x, z) = The pattern is now clear. We have
K2 (x, z)φ(z)dz + λ3
K(x, y)K2 (y, z)dz.
n
b
18.9
fn (x) = φ(x) +
i=1
λi
a
Ki (x, y)φ(y)dy
with K1 = K, and K2 , K3 , . . . deﬁned recursively via
b
18.10
Ki+1 (x, y) =
a
K(x, z)Ki (z, y)dy.
Theorem 18.5 shows that when λ < λ0 , the sequence fn converges to the ﬁxed point f , where
∞ b
18.11
f (x) = φ(x) +
i=1
λi
a
Ki (x, y)φ(y)dy.
18. INTEGRAL EQUATIONS Since this is true for arbitrary φ, we can change the order of summation and integration. Thus, with
∞
75
18.12 we have 18.13
Rλ (x, y) =
i=1 b
λi Ki (x, y),
f (x) = φ(x) +
a
Rλ (x, y)φ(y)dy.
Although 18.10, 18.12, 18.13 together give an “explicit” solution to the Fredholm equation, this explicitness is only theoretical. For, computing Rλ is of the same order of difﬁculty as solving 18.1 (in fact, even harder). On the other hand, if the kernel K is simple enough, analytic solutions might be possible. The following illustrates the computations for such a special case. 18.14 EXAMPLE. Suppose that
n
K(x, y) =
j=1
pj (x)qj (y) x, y ∈ [a, b]
for some continuous functions p1 , . . . , pn and q1 , . . . , qn on [a, b]. For φ continuous on [a, b], consider the Fredholm equations 18.1. Now, if f ∈ C satisﬁes 18.1, then
n
18.15 where 18.16 In view of 18.15, then
b
f (x) = φ(x) + λ
j=1
zj pj (x)
b
zj =
a
qj (y)f (y)dy,
j = 1, . . . , n.
zi
=
a b
qi (x)f (x)dx
n b
=
a
qi (x)φ(x)dx + λ
j=1 a
qi (x)pj (x)dx zj .
Thus, letting
b b
18.17 we obtain 18.18
ci =
a
qi (x)φ(x)dx,
n
aij =
a
qi (x)pj (x)dx,
zi = ci + λ
j=1
aij zj ,
i = 1, 2, . . . , n.
Note that the ci and aij are known. If we can solve 18.18 for the zi ’s, then 18.15 gives the solution f .
76
DIFFERENTIAL AND INTEGRAL EQUATIONS In vectormatrix notation, 18.18 becomes z = c + λAz,
whose solution is easy to discern. We can solve it for z (for arbitrary c) as long as I − λA is invertible, that is, as long as 1/λ is not an eigenvalue for A. Thus, we have a solution z for arbitrary b provided that λ ∈ (−1/λ0 , 1/λ0 ) where λ0 is the modulus of the largest eigenvalue of A.
Volterra Equation
Let K be a continuous function on [a, b] × [a, b] and let φ be a continuous function on [a, b]. Consider the equation
x
18.19
f (x) = φ(x) + λ
a
K(x, y)f (y)dy,
x ∈ [a, b].
It is called the Volterra equation. It differs from the Fredholm equation only slightly, and in form only. If we deﬁne ˆ K(x, y) = K(x, y) 0 if y ≤ x, if y > x,
ˆ then 18.19 becomes the Fredholm equation 18.1 with kernel K. However, it is easier to attack 18.19 directly. 18.20 THEOREM. For each λ ∈ R, the Volterra equation 18.19 has a unique solution f that is continuous on [a, b].
PROOF. Let C = C([a, b], R), the set of all continuous functions from [a, b] into R, with the usual uniform metric f − g . Let c be the maximum of K(x, y) over all x, y ∈ [a, b]; this number is ﬁnite since K is continuous. Deﬁne the transformation T : f → T f on C by
x
T f (x) = φ(x) + λ
a
K(x, y)f (y)dy.
Now, for f and g in C,
x
T f (x) − T g(x) = λ
a
K(x, y)[f (y) − g(y)]dy
≤
λc(x − a) f − g , x ∈ [a, b].
18. INTEGRAL EQUATIONS We use this, next, to bound T 2 f − T 2 g = T (T f − T g):
x
77
T 2 f (x) − T 2 g(x) = λ
a
K(x, y)[T f (y) − T g(y)]dy
x
≤
λ
a
K(x, y)λc(y − a) f − g dy
x
≤ λ2 c2
a
(y − a)dy f − g
≤ Iterating in this manner, we see that
λ2 c2 (x − a)2 f −g . 2
T k f (x) − T k g(x) ≤ for all x ∈ [a, b]. Hence, T kf − T kg ≤
λk ck (x − a)k f −g k!
[λc(b − a)]k f −g . k!
Recalling that rn /n! tends to 0 as n → ∞ for any r ∈ R, we conclude that there exists k such that T k is a contraction: simply take k large enough to have [λc(b − a)]k /k! < 1. Finally, the existence and uniqueness of f ∈ C satisfying f = T f follows from the next theorem. Obviously, if f = T f , then f solves 18.19. 2
Generalization of the Fixed Point Theorem
18.21 THEOREM. Let E be a complete metric space and let T be a continuous transformation on E. If T k is a contraction for some k ≥ 1, then T has a unique ﬁxed point.
PROOF. Fix k such that U = T k is a contraction. By Theorem 16.3, then, U has a unique ﬁxed point x, and limn U n x0 = x for every point x0 in E. Now, by the continuity of T , Tx = = lim T U n x0
n n
lim T T kn x0
= lim T kn T x0 = lim U n T x0 n = x,
78
DIFFERENTIAL AND INTEGRAL EQUATIONS
x(t)
1
v(t,x(t))
x0 t0
t
Figure 14: A moving particle. that is, x is a ﬁxed point of T . To show that it is the only ﬁxed point of T we note that every ﬁxed point of T is a ﬁxed point of T k = U , whereas U has only one ﬁxed point, namely x. 2
Exercises:
18.1 Solve the Fredholm equation 18.1 for arbitrary φ, on [a, b] = [0, 2π], with the kernel K(x, y) = sin(x + y). 18.2 Do the same with [a, b] = [0, 1] and K(x, y) = (x − y)2 . 18.3 Let p be a continuous function of [0, b]. Show that
x
f (x) = φ(x) +
0
p(y)f (x − y)dy,
x ∈ [0, b],
has a unique solution f for each continuous function φ.
19
Differential Equations
We continue with applications of the ﬁxed point theorem by discussing Picard’s method of successive approximations for solving systems of differential equations. We start with the simplest case where the differential equation describes the position of a particle moving on R. The picture of the motion is given in Figure 14. The motion is described by the initial data t0 and x0 and by a continuous function v : R × R → R as follows. The particle starts from x0 at time t0 ; its velocity at time t is v(t, x) if its position then is x. Thus, letting x(t) denote the position of the particle at time t, we have
t
19.1
x(t) = x0 +
t0
v(s, x(s))ds,
t ≥ t0 .
19. DIFFERENTIAL EQUATIONS The points t0 and x0 and the velocity function v are given. We are interested in the existence and uniqueness of the function x. In the classical formulation of this problem, it is usual to express 19.1 as a differential equation: dx 19.2 = v(t, x), x(t0 ) = x0 . dt The following is Picard’s Theorem: 19.3 THEOREM. Let v be deﬁned and continuous on [t0 , ∞) × [a, b], and x0 be in (a, b), and suppose that v satisﬁes a Lipschitz condition in its spatial argument: 19.4 v(t, x) − v(t, y) ≤ Kx − y, x, y ∈ [a, b].
79
Then, there is a t1 > t0 such that 19.1 has a unique solution {x(t) : t0 ≤ t ≤ t1 }. PROOF. By the continuity of v, we have 19.5 v(t, x) ≤ c, t 0 ≤ t ≤ t1 , a≤x≤b
for some constant c. Choose δ > 0 so that 19.6 Kδ < 1 and a ≤ x0 − cδ < x0 < x0 + cδ ≤ b.
Let t1 = min{t1 , t0 + δ}. Let C ∗ be the space of all continuous functions x : [t0 , t1 ] → [x0 −cδ, x0 +cδ] with the usual supremum metric; that is, x−y = supt0 ≤t≤t1 x(t)− y(t). The set C ∗ is a closed subset of the space C([t0 , t1 ], R). Since the latter is complete, ∗ C is complete. Consider the transformation T deﬁned by
t
19.7
T x(t) = x0 +
t0
v(s, x(s))ds,
t ∈ [t0 , t1 ].
For x ∈ C ∗ , we have from 19.5 that
t
T x(t) − x0  ≤
t0
v(s, x(s))ds ≤ c(t − t0 ) ≤ cδ,
which shows that T x ∈ C ∗ . Moreover, for x, y ∈ C ∗ ,
t
T x(t) − T y(t) ≤
t0 t
v(s, x(s)) − v(s, y(s))ds Kx(s) − y(s)ds
t0
≤
≤ Kδ x − y
80
DIFFERENTIAL AND INTEGRAL EQUATIONS
in view of 19.4. Thus, T x − T y ≤ Kδ x − y and Kδ < 1 by the way δ was chosen. So, T is a contraction on C ∗ . Since C ∗ is complete, Theorem 16.3 applies to show that T has a unique ﬁxed point x. But, x = T x means that x solves 19.1. This completes the proof. 2 The preceding can be easily generalized to the case of systems of differential equations dxi 19.8 = vi (t, x1 , . . . , xn ), i = 1, 2, . . . , n. dt Before listing it, we mention that the term “domain” means “an open and connected subset of a Euclidean space”, and we note that 19.1 can be interpreted for t < t0 by the convention that integrals from t0 to t are the nega − x)T (x − x) > 0. ¯ ¯
While z might be further from x than x, we shall show that some points on the line ¯ segment connecting x to z are closer than x (see Figure 19). To this end, put ¯ ¯ z(t) = tz + (1 − t)¯ x and f (t) = z(t) − x 2 . It is easy to check that f (0) = 2(z − x)T (¯ − x), which is strictly negative. Therefore, ¯x ¯ ¯ ¯ there exists a 0 < t < 1 such that f (t) < f (0). But z(t) ∈ C and so x cannot be the ¯ projection of x on C. This contradiction implies that the strict inequality (21.3) must be wrong. 2 When the set C is a linear subspace of Rn , an explicit formula can be given for the projection onto C: 21.4 THEOREM. Suppose that C = {z : z = AT y for some y ∈ Rm } where A is an m × n matrix of rank m. Then the following are equivalent:
21. PROJECTION
_ x x
89
}
z
these points are closer
C
Figure 19: Clearly some points on the line segment connecting x to z lie closer to x ¯ than x when the angle is acute as shown here. ¯ 1. x is the projection of x on C. ¯ 2. x = AT (AAT )−1 Ax. ¯ 3. x ∈ C and xT z = xT z for all z ∈ C. ¯ ¯ Note: The set C is the span of the set of nvectors given by the rows of A. The rank assumption simply means that these vectors are linearly independent. It is easy to check that A has rank m if and only if AAT is nonsingular. PROOF. (1) implies (2): By deﬁnition, x solves miny∈Rn f (y) where f (y) = x − ¯ AT y 2 = xT x − 2(Ax)T y + y T AAT y. Let y denote a point at which the gradient of ¯ f vanishes: f (¯) = −2Ax + 2AAT y = 0. y ¯ Since AAT is nonsingular, y is uniquely given by ¯ y = (AAT )−1 Ax. ¯ Hence, x = AT y = AT (AAT )−1 Ax. ¯ ¯ (2) implies (3): Suppose that x = AT (AAT )−1 Ax. Then, x = AT y , where ¯ ¯ ¯ y = (AAT )−1 Ax. Hence, x belongs to C. Suppose that z also belongs to C. That is, ¯ ¯ z = AT y for some y ∈ Rm . Then, z T x = y T AAT (AAT )−1 Ax = y T Ax = z T x. ¯ (3) implies (1): Suppose that x ∈ C and xT z = xT z for all z ∈ C. Picking z = x, ¯ ¯ ¯ we see that xT x = xT x. That is, ¯ ¯¯ xT (x − x) = 0. ¯ ¯ Yet, for any z in C we have z T (x − x) = 0. ¯
90
CONVEX ANALYSIS
H
x _ x
C
Figure 20: The separating hyperplane theorem. Combining these two equations, we see that (z − x)T (x − x) = 0. ¯ ¯ Therefore, Theorem 21.2 implies that x is the projection of x on C. ¯ 2
22
Supporting Hyperplane Theorem
22.1 DEFINITION. A halfspace H is a set of the form {z : aT z ≤ b}, where a = 0. The boundary ∂H is the hyperplane {z : aT z = b}. The projection theorems of the previous section provide the key tool to proving the important supporting hyperplane theorem: 22.2 THEOREM. Suppose that C is a nonempty closed convex set in Rn and that x is a point not in C. Then there exists a halfspace H such that C ⊂ H, C ∩ ∂H = ∅, and x ∈ H. PROOF. Let x denote the projection of x on C. Let a = x − x. Since x ∈ C ¯ ¯ x ∈ C, we see that a = 0. Put H = {z : aT z ≤ aT x}. By Theorem 21.2, C ¯ ¯ subset of H. Since aT x − aT x = a 2 > 0, it follows that x ∈ H. Since x ∈ C ¯ ¯ x ∈ ∂H, we get that C ∩ ∂H = ∅. ¯ and is a and 2
Measure and Integration
This chapter is devoted to integration on abstract spaces. As special cases, it covers the Riemann integral, line and surface integrals, and Stieltjes integrals.
23
Motivation
The integral introduced in elementary calculus courses is called the Riemann integral. Let’s brieﬂy review the deﬁnition of the integral from a to b of a realvalued function f . Let P denote a partition of the interval [a, b]: a = x0 < x1 < x2 < · · · < xn−1 < xn = b. Associated with this partition, is an upper estimate of the integral
n
U (f, P) = and a lower estimate
n
sup
i=1 xi−1 ≤x≤xi
f (x)(xi − xi−1 )
L(f, P) =
i=1
xi−1 ≤x≤xi
inf
f (x)(xi − xi−1 ).
Clearly, L(f, P) ≤ U (f, P). The function f is said to be Riemann integrable over the interval [a, b] if sup L(f, P) = inf U (f, P).
P P
The basic result regarding Riemann integration is that if f is continuous, then the Riemann integral exists. There are at least three problems with the Riemann integral. The ﬁrst problem is that highly discontinuous functions aren’t integrable. For example, consider the function f that is one at every irrational point and is zero at every rational. Then, for every partition P, U (f, P) = b − a 91
92 and L(f, P) = 0.
MEASURE AND INTEGRATION
The second problem is that one would like to be able to integrate functions whose domain is more general than simply the reals. Of course, Riemann integrals are extended to functions deﬁned on Rn , but even that is not as general as one would prefer. The third problem is that one would often like to interchange a limit with an integral. Although it is not apparent from the deﬁnition given above, it turns out that justifying such an interchange for Riemann integrals is difﬁcult. To circumvent these difﬁculties, the idea is to partition the range instead of the domain (after all, the range is always the reals). Suppose ﬁrst that f is a positive function deﬁned on an arbitrary set E and partition [0, n) using dyadic intervals [(k − 1)/2n , k/2n ). Let Bk,n = {x ∈ E : f (x) ∈ [(k − 1)/2n , k/2n )} denote the set of points in the domain that map into [(k − 1)/2n , k/2n ). The following sum is a lower estimate of the area under f :
n2n
k=1
k µ(Bk,n ), 2n
where µ(Bk,n ) denotes the length or, more generally, the measure of Bk,n . As n increases, this sum increases. Therefore, it has a limit (possibly inﬁnite) which is called the Lebesgue integral of f over E:
n2n
f (x)µ(dx) = lim
E n k=1
k µ(Bk,n ). 2n
Note that µ is a function from subsets of E into R+ . To capture the notion of being a “measure” of the subsets, µ should possess the following properties: 1. if A1 , A2 , . . ., are disjoint subsets of E, then µ(∪n An ) = 2. µ(∅) = 0. A function on subsets of E with these two properties is called a measure on E. At this point the picture seems pretty clear. All that remains is to construct the measure µ in the cases of interest (such as the usual notion of length on R). However, the following theorem due to Ulam shows that there aren’t many measures that can be constructed this way. 23.1 THEOREM. If µ is a ﬁnite measure deﬁned on all subsets of [0, 1], then there exists a countable collection of points x1 , x2 , . . . in [0, 1] such that µ({x1 , x2 , . . .}c ) = 0.
n
µ(An );
24. ALGEBRAS Hence, there does not exist a measure deﬁned on all subsets of [0, 1] for which µ([a, b]) = b − a. That is, there does not exist a measure which corresponds to our idea of length. The problem is that we have asked for too much. It is not necessary (and evidently not possible) to deﬁne our measures on all subsets of E. The collections of sets on which we will deﬁne our measures will be called algebras. This is the subject of the next section.
93
24
Algebras
Let E be a set (generally this set will be uncountably inﬁnite although we by no means require this). We wish to assign “measures” to the sizes of various subsets of E. It would be nice to assign a measure to arbitrary subsets, but as we shall see this is impossible to do in such a way that certain natural additivity properties hold. Hence, we must restrict our attention only to certain subsets of E. We will call such subsets measurable. If a set A is measurable, it stands to reason that its complement should also be measurable (and its measure should be the total measure of E minus the measure of A). Given a ﬁnite disjoint collection of measurable sets, it makes sense that their union should be measurable since the measure of the union should be the sum of the measures of each set. Using the fact that complements of measurable sets are measurable, it is easy to see that ﬁnite nondisjoint unions of measurable set should also be measurable since they can be pieced together from disjoint measurable sets. Finally, it is reasonable to assume that countable unions of measurable sets should also be measurable, since the sums involved in the appropriate deﬁnition involves only positive numbers and so must either converge to a ﬁnite number or to inﬁnity. A collection of measurable sets will be called a σalgebra on E. To summarize the foregoing, a σalgebra is a nonempty collection E of subsets of E with the following two properties: A ∈ E ⇒ E \ A ∈ E,
A1 , A2 , . . . ∈ E ⇒ ∪∞ An ∈ E. 1 In other words, a σalgebra is a collection of subsets of E that is closed under the operations of complementation and countable unions. It follows that a σalgebra is closed under ﬁnite unions, ﬁnite intersections, and countable intersections as well. In particular, the sets ∅ and E belong to every σalgebra on E. The simplest σalgebra on E is E = {∅, E}; it is called the trivial σalgebra. The largest is the collection of all subsets; it is called the discrete σalgebra. The intersection of an arbitrary family (countable or uncountable) of σalgebra on E is again a σalgebra. If C is a collection of subsets of E, the intersection of all σalgebras containing C is the smallest σalgebra that contains C; it is called the σalgebra generated by C and is denoted by σ(C). If E is a metric space, then the σalgebra generated by the collection of all open subsets is called the Borel σalgebra on E; it is denoted by B(E), and its elements are
94
MEASURE AND INTEGRATION
called Borel sets. Thus, every open set, every closed set, every set obtained from open and closed sets through various set operations are all Borel sets.
Monotone Class Theorem
This is a very useful theorem which simpliﬁes the task of showing that a given colleciton is a σalgebra. Throughout this subsection, E is an arbitrary set. A collection C of subsets of E is called a πsystem if it is closed under ﬁnite intersections, that is, if 24.1 A, B ∈ C ⇒ A ∩ B ∈ C. A collection D of subsets of E is called a dsystem on E if (i) (ii) (iii) E ∈ D, A, B ∈ D and B ⊂ A ⇒ A \ B ∈ D, (An ) ⊂ D and An A ⇒ A ∈ D.
24.2
On the last line, we wrote (An ) ⊂ D to mean that (An ) is a sequence of elements of D, and we wrote An A to mean that A1 ⊂ A2 ⊂ · · · and ∪n An = A. 24.3 PROPOSITION. Let E be a collection of subsets of E. Then, E is a σalgebra on E if and only if E is both a πsystme and a dsystem on E. PROOF. If E is σalgebra then it is obviously a πsystem and a dsystem. To show the converse, suppose that E is both a πsystem and a dsystem. Now, 24.2i and 24.2ii show that E is closed under complements. Since A ∪ B = (Ac ∩ B c )c , this implies that E is closed under unions (if A, B ∈ E then Ac , B c ∈ E, and thus Ac ∩ B c ∈ E since E is a πsystem, and hence (Ac ∩ B c )c ∈ E). This implies that E is closed under countable unios as well: if A1 , A2 , . . . ∈ E, put B1 = A1 , B2 = A2 , B3 = A3 , . . . .
Each Bn belongs to E by what we have just shown. Obviously, B1 ⊂ B2 ⊂ · · · and ∪n Bn = cupn An . Thus, using property 24.2iii of athe dsystem E, we see that ∪n An ∈ E. 2 The following lemma is needed in the proof of the main theorem. Its proof is obtained by checking the conditions of 24.2 one by one; we leave it as an exercise. 24.4 LEMMA. Let D be a dsystem on E. Fix D ∈ D and let ˆ d = {A ∈ D : A ∩ D ∈ D}. Then, ∩D is again a dsystem.
24. ALGEBRAS The following the main result of this section. It is called Dynkin’s monotome class theorem. 24.5 THEOREM. If a dsystem contains a πsystem, then it contains also the σalgebra generated by that πsystem. PROOF. Let C be a πsystem. Let D be the smallest dsystem on E that contains C. We need to show that D ⊃ σ(C). To that end, since σ(C) is the smallest σalgebra ˙ containing C, it is sufﬁcient to show that D is a σalgebraFor this, it is in turn sufﬁcient to show that D is a πsystem (and then Proposition 24.3 implies that the dsystem D is a σalgebra). Fix B ∈ C and let D1 = {A ∈ D : A ∩ B ∈ D}. Since B ∈ C ⊂ D, Lemma 24.4 shows that D1 is a dsystem. Moreover, D1 ⊃ C since A ∩ B ∈ C ⊂ D for every A ∈ C by the fact that C is a πsystem. So D1 must contain the smallest dsystem containing C, that is, D1 ⊃ D. In other words, A ∩ B ∈ D for every A ∈ D and B ∈ C. Next, ﬁx A ∈ D and let D2 = {B ∈ D : A ∩ B ∈ D}. We have just shown that D2 ⊃ C. Moreover, by Lemma 24.4 again, D2 is a dsystem. THus, D2 ⊃ D. In other words, A ∩ B ∈ D for every A ∈ D and B ∈ D, that is, D is a πsystem. This completes the proof. 2
95
Exercises:
24.1 Partitions. A partition of E is a countable disjointed collection of subsets whose union is E. It is called a ﬁnite partition if it has only ﬁnitely many elements. 1. Let {A, B, C} be a partition of E. Describe the σalgebra generated by this partition. 2. Let C be a partition of E. Let E be the collection of all countable unions of elements of C. Show that E is a σalgebra. Show that, in fact, E = σ(C). Generally, if C is not a partition, the elements of σ(C) cannot be obtained through such explicit constructions. 24.2 Let B and C be two collections of subsets of E. If B ⊂ C, then σ(B) ⊂ σ(C). If B ⊂ σ(C) ⊂ σ(B), then σ(B) = σ(C). Show these. 24.3 Borel σalgebra on R. Show that B(R) is generated by the collection of all open intervals. Hint: recall that every open subset of R is a countable union of open intervals. 24.4 Continuation. Show that every interval of R is a Borel set. In particular, (−∞, x), (−∞, x], (x, y], [x, y] are all Borel sets. Every singleton {x} is a Borel set.
96
MEASURE AND INTEGRATION
24.5 Show that B(R) is also generated by any one of the following: 1. the collection of all intervals of the form (x, ∞), 2. the collection of all intervals of the form (x, y], 3. the collection of all intervals of the form [x, y], 4. the collection of all intervals of the form (−∞, x], 5. the collection of all intervals of the form (x, ∞) with x rational.
25
Measurable Spaces and Functions
A measurable space is a pair (E, E) where E is a set and E is a σalgebra on E. Then, the elements of E are called measurable sets. When E is a metric space and E = B(E), the Borel σalgebra on E, the measurable sets are also called Borel sets. Let (E, E) and F, F) be measurable spaces and let f be a mapping from E into F . Then, f is said to be measurable relative to E and F if f −1 (B) ∈ E for every B ∈ F (these are the functions we wish to be able to integrate). If E and F are metric spaces and E = B(E) and F = B(F ) and f : E → F is measurable relative to E and F, tthen f is also called a Borel function.
Measurable Functions
The following proposition reduces the checks for measurability: 25.1 PROPOSITION. Let (E, E) and (F, F) be measurable spaces. In order for f : E → F to be measurable relative to E and F, it is necessary and sufﬁcient that f −1 (B) ∈ E for every B ∈ F0 for some collection F0 that generates F.
PROOF. Necessity part is trivial. To prove the sufﬁciency, let F0 ⊂ F be such that σ(F0 ) = F and suppose that f −1 (B) ∈ E for every B ∈ F0 . We need to show that, then, F1 = {B ∈ F : f −1 (B) ∈ E} is equal to F. For this, it is sufﬁcient to show that F1 is a σalgebra, since F1 ⊃ F0 by hypothesis and F is the smallest σalgebra containing F0 . But checking that F1 is a σalgebra is easy in view of the relations given in Exercise 2.1. 2
25. MEASURABLE SPACES AND FUNCTIONS
97
Borel Functions
Let E and F be metric spaces and let E and F be their respective Borel σalgebras. Let f : E → F . Since F is generated by the open subsets of F , in order for f to be a Borel function, it is necessary and sufﬁcient that f −1 (B) ∈ E for every open subset B of F ; this is an immediate corollary of the preceding proposition. In particular, if f is continuous, then f −1 (B) is open in E for every open B ⊂ F . Thus, every continuous function f : E → F is Borel measurable. The converse is generally false.
Compositions of Functions
Let (E, E), (F, F), and (G, G) be measurable spaces. Let f : E → F and g : F → G. Then, their composition g ◦ f : x → g(f (x)) is a mapping from E into G. The following proposition will be recalled by the phrase “measurable functions of measurable functions are measurable”. 25.2 PROPOSITION. If f is measurable relative to E and F, and if g is measurable relative to F and G, then g ◦ f is measurable relative to E and G. PROOF. Recall that (g ◦ f )−1 (C) = f −1 (g −1 (C)) for every C ⊂ G. If C ∈ G and g is measurable, then B = g −1 (C) is in F. Therefore, if f is measurable, f −1 (B) = f −1 (g −1 (C)) is in E for every C ∈ G. 2
Numerical Functions
¯ By a numerical function on E, we mean a mapping from E into R or some subset ¯ thereof. Such a function is said to be positive if all its values are in R+ and is said to be realvalued if all its values are in R. If (E, E) is a measurable space and f is a numerical function on E, then f is said to be Emeasurable if it is measurable with ¯ respect to E and B(R). Let (E, E) be a measurable space and let f be a numerical function on E. Using ¯ ¯ Proposition 25.1 with F = R and F = B(R) and recalling Exercise 24.5, we see that the following holds. 25.3 PROPOSITION. The numerical function f is Emeasurable if and only if any one of the following is true: 1. {x : f (x) ≤ r} ∈ E for every r ∈ R, 2. {x : f (x) > r} ∈ E for every r ∈ R, 3. {x : f (x) < r} ∈ E for every r ∈ R, etc.
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MEASURE AND INTEGRATION
¯ 25.4 COROLLARY. Suppose that f : E → F where F is a countable subset of R. Then, f is Emeasurable if and only if {x : f (x) = a} ∈ E for every a ∈ F . PROOF. Necessity is trivial since each singleton {a} is a Borel set. For the sufﬁciency, ﬁx r ∈ R and note that {x : f (x) ≤ r} is the union of {x : f (x) = a} over all a ≤ r, a ∈ F , and therefore belongs to E since it is a countable union of the sets {x : f (x) = a} ∈ E. Thus, f is Emeasurable by the preceeding proposition. 2
Positive and Negative Parts of a Function
Let (E, E) be a measurable space. Let f be a numerical function on E. Then,3 f + = f ∨ 0, f − = −(f ∧ 0)
are called the positive part of f and negative part of f , respectively. Note that both f + and f − are positive functions and f = f + − f −.
25.5 PROPOSITION. The function f is Emeasurable if and only if both f + and f − are Emeasurable. The proof is left as an exercise. The decomposition f = f + − f − enables us to state most results for positive functions only, since it is easy to obtain the corresponding result for arbitrary f .
Indicators and Simple Functions
Let A ⊂ E. Its indicator, denoted by 1A , is deﬁned by 1A (x) = 1 0 if x ∈ A, if x ∈ A.
Obviously, 1A is Emeasurable if and only if A ∈ E. A function f on E is said to be simple if it has the form
n
25.6
f=
1
ai 1Ai
3 For a, b ∈ R we write a ∨ b for the maximum of a and b, and a ∧ b for the minimum. The notation ¯ extends to functions: f ∨ g is the function whose value at x is f (x) ∨ g(x); similarly for f ∧ g.
25. MEASURABLE SPACES AND FUNCTIONS for some integer n, real numbers a1 , . . . , an , and measurable sets A1 , . . . , An . It is clear that, then, there exist an integer m ≥ 1, distinct real numbers b1 , . . . , bm , and a m measurable partition {B1 , . . . , Bm } of E such that f = 1 bi 1Bi , this latter representation is called the canonical form of the simple function f . Every simple function of E is Emeasurable; this is immediate from Corollary 25.4 applied to the canonical form of f . Conversely, if f is Emeasurable, takes only ﬁnitely many values, and all those values are real, then f is simple. In particular, every constant is a simple function. Moreover, if f and g are simple, then so are f + g, f − g, f g, f /g, f ∨ g, f ∧ g, except that, in the case of f /g one must make sure that g is never 0.
99
Approximations by Simple Functions
We start by constructing a sequence of simple functions that approximate the identity ¯ ¯ function d from R+ into R+ . For each n ∈ N, let 25.7 dn (x) = k/2n n if 2k ≤ x < n if x ≥ n.
k+1 2n ,
k ∈ {0, 1, . . . , n2n − 1},
The ﬁgure below is for d2 . The following lemma should be selfevident. ¯ 25.8 LEMMA. Each dn is a simple Borel function on R+ . Each dn is rightcontinuous and increasing. The sequence (dn ) is increasing pointwise to the function d : x → x. The following theorem characterizes all Emeasurable positive functions, and via Proposition 25.5, all Emeasurable functions. 25.9 THEOREM. A positive function on E is Emeasruable if and only if it is the limit of an increasing sequence of simple positive functions. ¯ PROOF. Necessity. Let f : E → R+ be Emeasurable. Let the dn be deﬁned by 25.7. ¯ ¯ Since each dn is a measurable function from R+ into R+ , and since measurable functions of measurable functions are measurable, the function fn = dn ◦f is Emeasurable for each n. Since dn is simple, so is fn . Finally, lim fn (x) = lim dn (f (x)) = f (x) ¯ since lim dn (y) = y for all R+ . Thus, f is the limit of the sequence (fn ) of simple positive functions and f1 ≤ f2 ≤ · · · since d1 ≤ d2 ≤ · · ·. Sufﬁciency. Let f1 ≤ f2 ≤ · · · be simple and positive and let f = lim fn . Now, for each x ∈ E and r ∈ R, we have f (x) ≤ r if and only if fn (x) ≤ r for all n; thus, {x ∈ E : f (x) ≤ r} = ∩∞ {x ∈ E : fn (x) ≤ r} n=1 for each r ∈ R. Since the fn are simple (and therefore measurable), each factor on the right side belongs to E and, therefore, so does the intersection. Hence, f is Emeasurable by Proposition 25.3. 2
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MEASURE AND INTEGRATION
Limits of Sequences of Functions
Let (E, E) be a measurable space and let (fn ) be a sequence of numerical functions on E. 25.10 THEOREM. Suppose that each fn is Emeasurable. The, each one of inf fn , sup fn , lim inf fn , lim sup fn
is again Emeasurable. Moreover, if lim fn exists, then it is Emeasurable. PROOF. For x ∈ E and r ∈ R, we have inf fn (x) ≥ r if and only if fn (x) ≥ r for all n. Thus, for each r ∈ R, {x ∈ E : inf fn (x) ≥ r} = ∩n {x ∈ E : fn (x) ≥ r}. Now, {x : fn (x) ≥ r} ∈ E for each n by the measurability of fn , and therefore the intersection on the right side belongs to E since E is closed under countable intersections. Thus, inf fn is Emeasurable by Proposition 25.3. The proof that sup fn is Emeasurable follows via similar reasoning upon noting that {x ∈ E : sup fn (x) ≤ r} = ∩n {x ∈ E : fn (x) ≤ r}. It follows from these that lim inf fn = sup inf fn ,
m n≥m
lim sup fn = inf sup fn
m n≥m
are both Emeasurable. Finally, lim fn exists if and only if lim inf fn = lim sup fn , and then lim fn is the common limit; so, it must be Emeasurable. 2
Monotone Classes of Functions
Often we are interested in showing that a certain property holds for all measurable functions. The following are useful in such quests. Let M be a collection of positive functions on E. Then, M is called a monotone class of functions provided that 25.11 (i) (ii) (iii) 1 ∈ M, f, g ∈ M, and a, b ∈ R+ ⇒ af + bg ∈ M, (fn ) ⊂ M, and fn f ⇒ f ∈ M.
The following is called the monotone class theorem for functions. 25.12 THEOREM. Let M be a monotone class of functions on E. Suppose that 1A ∈ M for every A ∈ C for some πsystem C that generates the σalgebra E. Then, M
25. MEASURABLE SPACES AND FUNCTIONS includes all positive Emeasurable functions and all bounded Emeasurable functions.
101
PROOF. We start by showing that 1A ∈ M for every A ∈ E. To this end, let D = {A ∈ E : 1A ∈ M}. Using the properties 25.11 of M, it is easy to check that D is a dsystem. Moreover, D ⊃ C by hypothesis. Thus, by Dynkin’s monotone class theorem, D ⊃ σ(C) = E. In other words, 1A ∈ M for every A ∈ E. Consequently, in view of property 25.11(ii), M includes all simple Emeasurable functions. Let f be a positive Emeasurable function. By Theorem 25.9, there exists a sequence of positive simple functions fn f . Since each fn in in M by the preceeding step, 25.11(iii) implies that f is in M. 2
Notation
We shall write f ∈ E to mean that f is an Emeasurable function. Thus, E stands both for a σalgebra and for the collection of all numerical functions measurable with respect to it. Furthermore, we shall use the notation F+ = {f ∈ F : f ≥ 0} for any collection of F of numerical functions. Thus, in particular, E+ is the collection of all positive Emeasurable functions.
Exercises:
25.1 Trace spaces. Let (E, E) be a measurable space and let D ⊂ E be ﬁxed. Show that D = {A ∩ D : A ∈ E} is a σalgebra on D. Then, D is called the trace of E on D, and (D, D) is called the trace of (E, E) on D. 25.2 σalgebra generated by a function. Let E be a set and let (F, F) be a measurable space, Let f be a mapping from E into F and set f −1 (F) = {f −1 (B) : B ∈ F}. Use Exercise 2.1 to show that f −1 (F) is a σalgebra on E; it is called the σalgebra on E generated by f . It is the smallest σalgebra on E such that f is measurable relative to it and F.
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MEASURE AND INTEGRATION
25.3 Product spaces. Let (E, E) and (F, F) be measurable spaces. A rectangle A × B is said to be measurable if A ∈ E and B ∈ F. Show that the collection of all measurable rectangles form a πsystem. The σalgebra on E × F generated by that πsystem is denoted by E ⊗ F and is called the product σalgebra. Further, (E × F, E ⊗ F) is called the product of (E, E) and (F, F), and is denoted by (E, E) × (F, F) also. If (E, E) = (F, F), then it is usual to write E 2 for E × F and E 2 = E ⊗ F. In particular, (R2 , B(R2 )) = (R, B(R)) × (R, B(R)), and by an obvious extension, (Rn , B(Rn )) = (R, B(R)) × · · · × (R, B(R)), n times. 25.4 Continuation. Let (E, E), (F, F), (G, G) be measurable spaces. Let f : E → F be measurable relative to E and F, and let g : E → G be measurable relative to E and G. Then, h(x) = (f (x), g(x)), x ∈ E,
deﬁnes a mapping from E into F × G. Show that h is measurable relative to E and F ⊗ G. In particular, a function f : E → Rn is measurable relative to E and B(Rn ) if and only if its coordinates are measurable relative to E and B(R); recall that the coordinates of f are the functions f1 , . . . , fn such that f (x) = (f1 (x), . . . , fn (x)), x ∈ E. 25.5 Discrete spaces. A measurable space (E, E) is said to be discrete if E is countable and E is the σalgebra of all subsets of E. Then, show that every numerical function of E is Emeasurable. 25.6 Suppose that E is generated by a countable partition of E. Show that, then, a numerical function on E is Emeasurable if and only if it is constant over each member of that partition. 25.7 Approximation by simple functions. Show that a numerical function of E is Emeasruable if and only if it is the limit of a sequence of simple functions. 25.8 Arithmetic operations. Let f and g be Emeasurable. Show that, then, each one of f + g, f − g, f · g, f /g, f ∨ g, f ∧g
is Emeasurable provided that it be welldeﬁned. 25.9 Functions on R. Let f : R → R+ be increasing. Show that it is a Borel function. 25.10 Step functions. A function f : R → R is called a step function if it has the form
∞
f=
1
ai 1Ai
26. MEASURES where each Ai is an interval. Show that every such f is a Borel function. 25.11 Rightcontinuous functions. Show that every rightcontinuous function f : R → R is Borel measurable. SImilarly, every leftcontinuous function is Borel. Hint for rightcontinuous f : deﬁne dn (x) = (k + 1)/2n if k/2n ≤ x < (k + 1)/2n for some k = 0, 1, 2, . . . for n = 1, 2, . . .. Show that dn is Borel. Let fn (x) = f (dn (x)). Show that each fn is a step function, and show that fn → f as n → ∞.
103
26
Measures
¯ Let E, E) be a measurable space. A measure on (E, E) is a mapping µ : E → R+ such that 1. µ(∅) = 0 , 2. µ(∪n An ) =
n
µ(An ) for every disjointed sequence (An ) ⊂ E.
The latter condition is called countable additivity. A measure space is a triplet (E, E, µ) where E is a set, E is a σalgebra on E, and µ is a measure on (E, E). 26.1 PROPOSITION. Let µ be a measure on (E, E). Then, the following hold for all measurable sets A, B, and An , n ≥ 1: Finite additivity: A ∩ B = ∅ implies that µ(A ∪ B) = µ(A) + µ(B). Monotonicity: A ⊂ B implies that µ(A) ≤ µ(B). Sequential continuity: An A implies that µ(An )
n
µ(A).
Boole’s inequality: µ(∪n An ) ≤
µ(An ).
PROOF. Finite additivity is a particular instance of the countable additivity of µ: take A1 = A, A2 = B, A3 = A4 = · · · = ∅. Monotonicity follows from it and the positivity of µ: if A ⊂ B, µ(B) = µ(A) + µ(B \ A) ≥ µ(A) since µ(B\A) ≥ 0. Sequential continuity follows from (and is equivalent to) countable additivity: suppose that An A; then, B1 = A1 , B2 = A2 \ A1 , B3 = A3 \ A2 , · · ·
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are disjoint, their union is A, and the union of the ﬁrst n is An ; hence, the sequence of numbers µ(An ) increases by the monotonicity of µ, and
n ∞
lim µ(An ) = lim µ(∪n Bi ) = lim 1
n 1
µ(Bi ) =
1
µ(Bi ) = µ(∪∞ Bi ) = µ(A). 1
Finally, Boole’s inequality follows from the observation that µ(A ∪ B) = µ(A) + µ(B \ A) ≤ µ(A) + µ(B). 2
Arithmetic of Measures
Let (E, E) be a measurable space. If µ is a measure on it and if c ≥ 0 is a constant, then cµ is again a measure. If µ and ν are measures on (E, E), so is µ+ν. If µ1 , µ2 , . . . are measures, then so is µ = µm : it is obvious that µ(∅) = 0, and if A1 , A2 , . . . are disjoint then µ(∪n An ) =
m
µm (∪n An ) µm (An )
m n
= =
n m
µm (An ) µ(An ),
n
=
where the crucial step (where the order of summation is changed) is justiﬁed by the elementary fact that amn =
m n n m
amn
if amn ≥ 0 for all m, n.
Finite, σﬁnite, Σﬁnite measures
Let µ be a measure on (E, E). It is said to be ﬁnite if µ(E) < ∞. It is called a probability measure if µ(E) = 1. It is said to be σﬁnite if there exists a measurable partition (En ) of E such that µ(En ) < ∞ for each n. It is said to be Σﬁnite if there exist ﬁnite measures µ1 , µ2 , . . . such that µ = µn . Note that every ﬁnite measure is trivially σﬁnite, every σﬁnite measure is Σﬁnite. The converses are false (see Exercise 26.4).
26. MEASURES
105
Speciﬁcation of Measures
It is generally difﬁcult to specify µ(A) for each A, simply because there are too many A in a σalgebra. The following proposition is helpful in reducing the task to specifying µ(A) for those A belonging to a πsystem that generates the given σalgebra. 26.2 PROPOSITION. Let µ and ν be measures on (E, E). Suppose that µ(E) = ν(E) < ∞, and that µ and ν agree on a πsystem generating E. Then, µ = ν. PROOF. Let C be a πsystem with σ(C) = E. Suppose that µ(A) = ν(A) for every A ∈ C. We need to show that, then, µ(A) = ν(A) for every A ∈ E. This amounts to showing that D = {A ∈ E : µ(A) = ν(A)} contains E. Now, D ⊃ C by hypothesis, and it is straightforward to check that D is a dsystem. Thus, by Dynkin’s monotone class theorem, D ⊃ σ(C) = E. 2
26.3 COROLLARY. LEt µ and ν be probability measures on R, B(R)). Then, µ = ν if and only if, for every x ∈ R, µ((−∞, x]) = ν((−∞, x]).
PROOF. The collection C of all intervals of the form (−∞, x] is a πsystem generating B(R). THus, the preceding proposition applies to prove sufﬁciency. Necessity is trivial. 2 The following proposition extends 26.2 to σﬁnite measures. 26.4 PROPOSITION. Let µ and ν be σﬁnite measures on (E, E). Suppose that they agree on a πsystem C generating E. Suppose further that there is a partition (En ) of E such that En ∈ C and µ(En ) = ν(En ) < ∞ for every n. Then, µ = ν. PROOF. For each n, deﬁne the measures µn and νn on (E, E) by µn (A) = µ(A ∩ En ), νn (A) = ν(A ∩ En ), A ∈ E.
SInce En ∈ C, and since A ∩ En ∈ C for every A ∈ C, we have µn (A) = µ(A ∩ En ) = ν(A ∩ En ) = νn (A) for A ∈ C. And, by hypothesis, µn (E) = µ(E) = ν(E) = νn (E) < ∞. Thus, the last proposition applies to show that µn = νn for each n. This completes the proof since µ = µn and ν = νn . 2
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MEASURE AND INTEGRATION
Image of Measure
Let (E, E) and (F, F) be measurable spaces. Let µ be a measure on (E, E) and let f : E → F be measurable relative to E and F. Then, 26.5 µ ◦ f −1 (B) = µ(f −1 (B)), B ∈ F,
is welldeﬁned since f −1 (B) ∈ E for each B ∈ F. It is easy to check that ν = µ ◦ f −1 is a measure on (F, F). It is called the image of µ under f .
Almost Everywhere
Often we face situations where a certain statement is true for every x ∈ E0 and E0 is almost the same as E in the sense that E0 ∈ E and µ(E \ E0 ) = 0. In that case, we say that the statement is true for almost every x in E or that the statement is true almost everywhere. Incidentally, a set N ⊂ E is said to be neglibible if there is an A ∈ E such that N ⊂ A and µ(A) = 0. So, a statement holds almost everywhere if and only if it fails only over a neglibible set. EXAMPLES. 26.6 Dirac measure. Let (E, E) be a measurable space. Fix x ∈ E. Deﬁne δx (A) = 1 0 if x ∈ A if x ∈ A
for each A ∈ E. Then, δx is a measure on (E, E). It is called the Dirac measure sitting at x. 26.7 Counting measures. Let (E, E) be a measurable space and let D be a countable subset of E. Deﬁne a measure ν on (E, E) by ν=
x∈D
δx .
Note that ν(A) is the number of points in A ∩ D. Such measures are called counting measures. 26.8 Discrete measure spaces. Let E be countable and E be the collection of all subsets of E. Specifying a measure on (E, E) is equivalent to assigning a number ¯ m(x) in R+ to each point x in E and then letting µ(A) =
x∈A
m(x),
A ∈ E.
26. MEASURES
107
Then, m is called the mass function corresponding to µ. In particular, if E = {1, 2, . . . , n}, every measure µ on (E, E) can be regarded a a vector in Rn . 26.9 Purely atomic measures. Let (E, E) be a measurable space, let D be a countable subset of E, and let m(x) be a positive number for each x ∈ D. Deﬁne µ(A) =
x∈D
m(x)δx (A),
A ∈ E.
Then, µ is a measure on (E, E). It puts the mass m(x) at the point x, and there are only countable many points x like that. Such µ are said to be purely atomic, the points x with µ({x}) > 0 are called the atoms of µ. 26.10 Lebesgue measures. A measure µ on (R, B(R)) is called the Lebesgue measure on R if µ(A) is the length of A for every interval A. The collection C of all intervals form a πsystem that generates B(R) and thus, by Proposition 26.4, there can be at most one such measure. The whole point of all measure theory is the following theorem which, unfortunately, we don’t prove. 26.11 THEOREM. There exists a measure on (R, B(R)) which assigns to each interval A its length. It is impossible to display µ(A) explicity for each Borel set A, but countable additivity and various properties list in Proposition 26.1 enable us to ﬁgure µ(A) out for most reasonable sets A. For instance, µ({x}) = 0 for every x ∈ R, µ(A) = 0 for every countable set A ⊂ R, µ(A) = 0 for the cantor set A, and so on. Of course, there are many sets with strictly positive measure. Similarly, Lebesgue measure on R2 is the “area” measure, Lebesgue measure on 3 R is the “volume” measure, and so on. All Lebesgue measures on R, R2 , R3 , etc. are σﬁnite. More generally, given an interval E ⊂ R, it makes sense to talk of Lebesgue measure on (E, B(E)); this is the restriction of Lebesgue measure on R to the trace space (E, B(E)). Similarly, one can talk of Lebesgue measure on a domain in R2 or on a domain in Rn . In all cases we shall use λn to denote the Lebesgue measure on a domain in Rn .
Exercises:
26.1 Show that D in the proof of 26.2 is a dsystem. 26.2 Restrictions. Let (E, E, µ) be a measure space. Let D ∈ E and let D = {A ∈ E : A ⊂ D}. Then, (D, D) is the trace of (E, E) on D. Deﬁne ν(A) = µ(A) for A ∈ D. Then, ν is a measure on (D, D); it is called the restriction of µ to D.
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26.3 Uniform distribution. Let D ⊂ R be an interval of ﬁnite length. Let µ(B) = λ1 (B)/λ1 (D) for Borel subsets B of D. Show that µ is a probability measure on (D, D) where D = B(D). It is called the uniform distribution on D. 26.4 Σﬁniteness. Let E = {a, b} with the discrete σalgebra, and deﬁne µ({a}) = 0, µ({b}) = +∞. Show that this deﬁnes a Σﬁnite measure µ that is not σﬁnite. 26.5 Atoms, atomic measures, diffuse measures. Let (E, E) be such the {x} ∈ E for every x ∈ E. A point x is said to be an atom for the measure µ if µ({x}) > 0. If µ has no atoms, then it is said to be diffuse. If µ puts no mass outside the set of its atoms, then it is purely atomic. In general, µ will have some atomic part and some diffuse part. This is to show this decomposition. 1. Let µ be ﬁnite. Show that it has at most countably many atoms. Hint: let D be the set of atoms, note that D = ∪n Dn where Dn = {x : µ({x}) ∈ [1/n, 1/(n − 1)), n = 1, 2, . . .. Use the ﬁniteness of µ to conclude that each Dn is a ﬁnite set, and therefore, that D must be countable. 2. Let µ be Σﬁnite. Show that it has at most countably many atoms. 3. Let D be the set of atoms of a Σﬁnite measure µ. Deﬁne ν(A) = µ(A ∩ D), λ(A) = µ(A ∩ Dc ), A ∈ E.
Then, ν is purely atomic, λ is diffuse, and µ = ν + λ.
27
Integration
Let (E, E) be a measurable space. Recall that E stands also for the collection of all Emeasurable functions and that E+ is the subcollection consisting of positive Emeasurable functions. Given a measure µ on (E, E), our aim is to deﬁne the “integral of f with respect to µ” for all reasonable functions f in E. We shall denote it by any of the following: µf =
E
µ(dx)f (x) =
E
f dµ.
When E is an interval of R and f is continuous and µ is the Lebesgue measure, the integral will coincide with the usual Riemann integral of f on E. When E = {1, . . . , n} and E is the discrete σalgebra, every measure µ is speciﬁed by a row vector (µ1 , . . . , µn ) with µi denoting µ({i}), and every function f ∈ E corresponds to a column vector (f1 , . . . , fn ) with fi = f (i); in this case the integral µf will coincide
27. INTEGRATION with the product of the row vector (µ1 , . . . , µn ) with the column vector with entires f1 , . . . , fn . As this last case illustrates, it is best to think of the integral as a product. After we deﬁne it, we shall show that it has the properties of products.
109
Deﬁnition of the Integral
We deﬁne the integral µf in three steps: ﬁrst for simple positive f , then for f ∈ E+ , ﬁnally for reasonable f ∈ E. Step 1. Let f be a nonnegative simple function. If its cannonical form is f = n 1 ai 1Ai , then we deﬁne
n
27.1
µf =
1
ai µ(Ai ).
Step 2. Let f ∈ E+ . Let (dn ) be deﬁned by 25.7 and recall from the proof of Theorem 25.9 that lim dn ◦ f = f . Now, for each n, the function dn ◦ f is simple and positive, and the integral µ(dn ◦ f ) is deﬁned by the preceding step. We shall show in the remarks below that the numbers µ(dn ◦ f ) form an increasing sequence, and hence, lim µ(dn ◦ f ) exists (it may be +∞). Since f = lim dn ◦ f , we deﬁne 27.2 µf = lim µ(dn ◦ f ). Step 3. Let f ∈ E be arbitrary. Then, f + and f − belong to E+ , and their integrals are deﬁned by the preceding step. Noting that f = f + − f − , we deﬁne 27.3 µf = µf + − µf − provided that at least one term on the right is ﬁnite. Otherwise, if µf + = µf − = +∞, then µf does not exist. REMARKS: (a) Formula 27.1 holds for nonnegative simple functions even when n 1 ai 1Ai is not the canonical representation for f :
n m n m
f=
1
ai 1Ai =
1
bj 1Bj
⇒
µf =
1
ai µ(Ai ) =
1
bj µ(Bj ).
This is easy to check using the ﬁnite additivity of µ. (b) If f and g are nonnegative simple functions and a, b ∈ R+ , then af + bg is again a nonnegative simple function, and µ(af + bg) = a µf + b µg. This can be checked using the preceding remark. (c) If f is a nonnegative simple function, then 27.1 shows that µf ≥ 0 (it can be +∞). (d) If f and g are nonnegative simple functions and f ≤ g, then the preceding two remarks applied to f and g − f show that µf ≤ µg. (e) In Step 2 of the deﬁnition, we have d1 ◦ f ≤ d2 ◦ f ≤ · · · and the preceding remark shows that µ(d1 ◦ f ) ≤ µ(d2 ◦ f ) ≤ · · · as claimed.
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MEASURE AND INTEGRATION
Integral over a Set
Let f be a measurable function and A a measurable set. Then, f 1A ∈ E. The integral of f over A is deﬁned to be the integral of f 1A ; it exists if and only if µ(f 1A ) exists. The following notations are used for it: 27.4 µ(f 1A ) =
A
µ(dx)f (x) =
A
f dµ.
Integrability
A function f ∈ E is said to be integrable if µf exists and is a ﬁnite number. Thus, f ∈ E is integrable if and only if µf + < ∞ and µf − < ∞, or equivalently, if and only if µf  < ∞ (note that f  = f + + f − ).
Elementary Properties
Here are some familiar properties of the integrals. A few others are put into the exercises. 27.5 PROPOSITION. (a) Positivity. If f ∈ E+ , then µf ≥ 0. (b) For f ∈ E+ , µf = 0 if and only if f = 0 almost everywhere. (c) Monotonicity. If f, g ∈ E+ and f ≤ g, then µf ≤ µg. If f, g ∈ E and f, g are integrable, and f ≤ g, then µf ≤ µg. (d) Finite additivity over sets. Let f ∈ E+ . If {A1 , . . . , Am } is a measurable partition of A ∈ E, then
m
27.6
A
f dµ =
i=1 Ai
f dµ.
PROOF. (a) If f ≥ 0, then the deﬁnition of µf yields µf ≥ 0. (c) If 0 ≤ f ≤ g, then dn ◦ f ≤ dn ◦ g and so µ(dn ◦ f ) ≤ µ(dn ◦ g) by the monotonicity of integration for simple functions. Now, the lefthand side converges to µf and the righthand side converges to µg. Hence µf ≤ µg. The general case is similar. (b) Linearity for simple functions and monotonicity imply the following chain of inequalities: 0≤ 1 1 1 1 1 1 1 µ({x : f (x) ≥ }) = µ(1f ≥ n ) = µ( 1f ≥ n ) ≤ µ(f 1f ≥ n ) ≤ µf = 0. n n n n
27. INTEGRATION Since the two ends of this chain of inequalities are equal, it follows that all the inequalities are in fact equalities. Hence, µ({x : f (x) ≥ 1/n}) = 0 and so {x : f (x) > 0} = ∪n {x : f (x) ≥ 1/n}. Taking the measure of both sides, we get 0 ≤ µ({x : f (x) > 0}) ≤
n
111
∀n
µ({x : f (x) ≥ 1/n}) = 0.
Again, equating this anchored chain of inequalities, we see that f = 0 a.e. (d) Fix f ∈ E+ . Let A1 , . . . , Am ∈ E be disjoint with union A. If f is simple, 27.6 is immediate from Remark b applied to the simple functions f 1A1 , . . . , f 1Am whose sum is f 1A . Applying this to simple functions dn ◦ f , we see that
m
µ(1Ai dn ◦ f ) = µ(1A dn ◦ f ).
1
Note that 1B (x)dn ◦ f (x) = dn (1B (x)f (x)) for each x by the way the function dn is deﬁned. Putting this observation into the preceding expression and letting n → ∞ we obtain
m m
µ(f 1Ai )
1
=
1
lim µ(dn ◦ (f 1Ai ))
n m
= = =
lim
n 1 n n
µ(dn ◦ (f 1Ai ))
lim µ(dn ◦ (f 1A )) lim µ(f 1A ), 2
where the interchange of the limit and the sum is justiﬁed by the ﬁniteness of m.
Monotone Convergence Theorem
This is the key result in the theory of integration. It allows interchanging the order of taking limits and integrals under reasonable conditions. 27.7 THEOREM. Let (fn ) ⊂ E+ be increasing. Then, µ(lim fn ) = lim µfn .
112
MEASURE AND INTEGRATION
PROOF. Let f = lim fn ; it is welldeﬁned since f1 ≤ f2 ≤ · · · and is positive and Emeasurable. So, µf is welldeﬁned. By the monotonicity of integration, µf1 ≤ µf2 ≤ · · · ≤ µf . Therefore lim µfn exists and lim µfn ≤ µf.
n
It remains to show that limn µfn ≥ µf . This is accomplished in steps. Step 1. If b ∈ R+ , B ∈ E, and f (x) > b for x ∈ B, then limn µ(fn 1B ) ≥ bµ(B). First, note that {f1 > b} ⊂ {f2 > b} ⊂ · · · and that ∪n {fn > b} = {x : fn (x) > b for some n} = {f > b}. Put Bn = {fn > b} ∩ B. Then, Bn 27.8
n
and ∪n Bn = {f > b} ∩ B = B. Thus,
lim µ(Bn ) = µ(B)
by the sequential continuity of µ under increasing limits. Now, note that fn 1B ≥ fn 1Bn ≥ b1Bn , and so the monotonicity of integration yields that µ(fn 1B ) ≥ µ(b1Bn ) = bµ(Bn ). Taking limits on both sides and using 27.8, we get 27.9 lim µ(fn 1B ) ≥ bµ(B).
Step 2. The same inequality holds even if f (x) ≥ b for x ∈ B. For b = 0, this is trivial. For b > 0, apply Step 1 with b− to see that limn µ(fn 1B ) ≥ (b− )µ(B). Since is arbitrary, we can let it go to zero to obtain the desired inequality. Step 3. If g is a simple function and g ≤ f , then limn µfn ≥ µg. m Let 1 bi 1Bi denote the canonical representation for g. Then, our assumptions imply that f (x) ≥ g(x) = bi for x ∈ Bi . Hence, we may apply the result of Step 2 to conclude that lim µ(fn 1Bi ) ≥ bi µ(Bi ) i = 1, . . . , m.
n
Hence, by Proposition 27.5d applied to the function fn , we see that
m
lim µfn
n
= =
lim
n 1 m
µ(fn 1Bi ) lim µ(fn 1Bi )
1 m
≥
1
bi µ(Bi ) = µg.
27. INTEGRATION Step 4. limn µfn ≥ µf . Put g = dm ◦ f . Step 3 applied with this g yields limn µfn ≥ µ(dm ◦ f ). Letting m → ∞ we get the desired result. 2 A particular consequence of the monotone convergence theorem is that, in deﬁnition 27.2, the special sequence (dn ◦ f ) can be replaced by any sequence (fn ) ⊂ E+ increasing to f .
113
Linearity of Integration
27.10 PROPOSITION. If f, g ∈ E+ and a, b ∈ R+ , then µ(af + bg) = aµf + bµg. The same holds for arbitrary f, g ∈ E and a, b ∈ R provided that both sides are welldeﬁned. It holds, in particular, if f and g are integrable.
PROOF. If f, g are simple, the result is established by direct checking as was remarked in b. For f, g ∈ E+ , and a, b ∈ R+ , choose (fn ) and (gn ) to be sequences of simple positive functions increasing to f and g, respectively. Then, µ(afn + bgn ) = aµfn + bµgn , and afn + bgn af + bg, fn f , gn f . Taking limits on both sides and using the monotone convergence theorem completes the proof. If f, g ∈ E are arbitrary, write f = f + − f − and g = f + − g − and go through the same steps. 2
Fatou’s Lemma
This gives a useful inequlaity for arbitrary sequences of positive measurable functions. 27.11 LEMMA. Let (fn ) ⊂ E+ . Then, µ(lim inf fn ) ≤ lim inf µfn .
PROOF. Deﬁne gm = inf n≥m fn . Then, lim inf fn is the limit of the increasing sequence (gm ) ⊂ E+ , and thus µ(lim inf fn ) = µ(lim gm ) = lim µgm by the monotone convergence theorem. On the other hand, gm ≤ fn for all n ≥ m, which yields µgm ≤ µfn for all n ≥ m, which in turn means that µgm ≤ inf n≥m µfn . Hence, as needed, lim µgm ≤ lim inf µfn = lim inf µfn .
m n≥m
114
MEASURE AND INTEGRATION 2
27.12 COROLLARY. (a) Let (fn ) ⊂ E. If fn ≥ g for all n for some integrable function g, then µ(lim inf fn ) ≤ lim inf µfn . (b) Let (fn ) ⊂ E. If fn ≤ g for all n for some integrable function g, then µ(lim sup fn ) ≥ lim sup µfn .
PROOF. Let g be an integrable function. Suppose that g is realvalued so that
2
Dominated Convergence Theorem
This is the second important tool for interchanging the order of taking limits and integrals. A function f is said to be dominated by a function g if f  ≤ g; note that g ≥ 0 necessarily. A sequence of functions (fn ) is said to be dominated by g if fn  ≤ g for each n. If g can be taken to be a ﬁnite constant, the (fn ) is said to be bounded. 27.13 THEOREM. Suppose that (fn ) ⊂ E is dominated by an integrable function g. If lim fn exists, then it is integrable and µ(lim fn ) = lim µfn .
n n
PROOF. By assumption, −g ≤ fn ≤ g for every n, and g and −g are both integrable. Thus, µfn exists and is sandwiched between the ﬁnite numbers −µg and µg. Now, both statements of the last corollary apply and we get µ(lim inf fn ) ≤ lim inf µfn ≤ lim sup µfn ≤ µ(lim sup fn ). If lim fn exists, then lim inf fn = lim sup fn = lim fn , and lim fn is integrable since it is dominated by g. Hence, the extreme members of the preceding expression are ﬁnite and equal, which means that equality holds throughout. 2
27. INTEGRATION If (fn ) ⊂ E is bounded, say by the constant b, and if the measure µ is ﬁnite, then we can take g = b in the preceding theorem. The resulting corollary is called the bounded convergence theorem: 27.14 THEOREM. Let (fn ) ⊂ E be bounded. Suppose that µ is ﬁnite. If lim fn exists, then µ(lim fn ) = lim µfn .
n n
115
27.15 EXAMPLE. Let (E, E) = (R+ , B(R+ )) and let fn be the sequence of functions shown in Figure ??. Note that the functions are not monotone and there is no integrable function that dominates them. Also, µfn = 1 for all n and so lim µfn = 1, whereas, lim fn = 0 and so µ lim fn = 0.
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