Introduction and Preliminaries

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1 Chapter 1 Introduction and Preliminaries This chapter serves two purposes. The first purpose is to prepare the readers for the more systematic development in later chapters of methods of real analysis through some introductory accounts of a few specific topics. The second purpose is, in view of the possible situation where some of the readers might not be well conversant with basic operations of elementary analysis, to acquaint general readers with fundamental background of analysis. 1.1 Summability of Systems of Real Numbers For a set S, the family of all nonempty finite subsets of S will be denoted by FS). Consider now a system {c α } α I, of real numbers indexed by an index set I. The system {c α } α I will be simply denoted by {c α } if the index set I is assumed either explicitly or implicitly. The system is called summable if there is l Ê such that for any ǫ > 0 there is A FI) with the property that whenever B FI) and B A, then c α l < ǫ 1.1) α B Exercise Show that if l in the preceding definition exists, then it is unique. Exercise Show that if {c α } is summable then {α I : c α 0} is finite or countably infinite. 1

2 2 CHAPTER 1. INTRODUCTION AND PRELIMINARIES If {c α } is summable, the uniquely determined l in the definition above is called the sum of {c α } and is denoted by α I c α. Before we go further we note that if I = {1, 2, 3, }, n=1 c n depends on the order 1 < 2 < 3 < and n I c n, if it exists, does not depend on how I is ordered, hence n I c n may not exist while n=1 c n exists. We will come back to this remark in a later exercise. Theorem If {c 1) α } α I and {c 2) α } α I are summable, then so is {ac 1) bc 2) α } α I for fixed real numbers a and b, and ) ac 1) α + bc2) α = a c 1) α + b c 2) α. α I α I α I Proof. Obvious. α + Theorem If c α 0 α I, then {c α } is summable if and only if { } c α : A FI) α A 1.2) is bounded. Proof. That 1) is necessary for {c α } to be summable is left as an exercise. Now we show that 1) is sufficient for {c α } to be summable. Let l be the least upper bound of { α A c α : A FI)}, for any ǫ > 0 there is A FI) such that 0 l c α < ǫ 1.3) α A Let now B FI) and B A, then c α l = l c α l c α < ǫ α B α B α A Exercise Show that 1.2) is necessary for {c α } to be summable. Because of Theorem if {c α } is a system of nonnegative real numbers and is not summable, then we write α I c α = +. Hence α I c α always has a meaning. Theorem {c α } is summable if and only if for any ǫ > 0 there is A FI) such that α B c α < ǫ whenever B FI) and A B =. Sufficiency: Choose A FI) such that α B c α < 1 for B FI) and A B =, then obviously the following claim holds:

3 1.2. DOUBLE SERIES 3 Claim: For B as above, we have α B c+ α < 1, where c+ α = c α or 0 according as c α 0 or < 0. Now for B FI) we have c + α = c + α + c + α < c + α + 1 α B α B A α B\A i.e. { α B c+ α : B FI)} is bounded, hence by Theorem {c+ α } is summable. Similarly {c α } is summable, where c α = c α or 0 according as c α 0 or > 0. Now c α = c + α c α, hence {c α} is summable by Theorem 1.1). Necessity part is left as an exercise. α A Exercise Prove the necessity part of Theorem Exercise Show that {c α } is summable if and only if { c α } is summable; show also that {c α } is summable if and only if { } c α : A FI) α A is bounded. Exercise Let Æ = {1, 2, 3,...}. Show that α Æ c α is summable if and only if α=1 c α is absolutely convergent. Show also that if α Æ c α = n=1 c α if {c α } α Æ is summable. Exercise Suppose that for each n = 1, 2, 3,, there is A n FI) with the property that for each A FI), there is a positive integer N such that A A n for all n N. Show that if {c α } α I is summable, then c α = lim c α. α I n α A n Give an example to show that it is possible that lim n α A n c α exists and finite, but {c α } is not summable. We shall recognize in Example that summability considered in this section is the integrability with respect to counting measure. 1.2 Double Series Let I = Æ Æ = {i, j) : i, j = 1, 2, } and write c ij for c i,j). The double series c ij is called summable if {c ij } = {c i,j) } is summable and i,j) I c ij is called the sum of the double series c ij. For a double sequence {a mn }, we say that lim m,n a mn = l if for any ǫ > 0 there is positive integer N such that a mn l < ǫ whenever m, n N.

4 4 CHAPTER 1. INTRODUCTION AND PRELIMINARIES Theorem If the double series c ij is summable, then i,j) I c ij = lim n m,n j=1i=1 m c ij = c ij = c ij. j=1i=1 i=1j=1 Proof. We show first that i,j) I c ij = lim n m n,m j=1 i=1 c ij. Let l = i,j) I c ij. Given ε > 0, there is A FI) such that c ij l < ε i,j) B whenever B FI) and B A. Let N = max{i j : i, j) A}, where i j is the larger of i and j. For n, m N, let B mn = {i, j) I : 1 i m, 1 j n}, then B mn FI) and B mn A, hence n m c ij l = c ij l < ε. j=1i=1 i,j) B mn n m This means that l = lim m,n j=1 i=1 c ij. Since i,j) I c ij = i,j) I c+ ij i,j) I c ij, in the remaining part of the proof, we may assume c ij 0 for all i, j) I. Observe then that Hence for each n and consequently On the other hand, l = l = sup n l lim m lim n,m j=1i=1 = c ij. j=1i=1 n m c ij. n,m 1 j=1i=1 j=1 j=1 ) m c ij = n l c ij. j=1i=1 n m c ij lim lim n We have shown that l = j=1 i=1 c ij; similarly, l = c ij. i=1j=1 c ij j=1i=1 ) n m c ij m j=1i=1 Exercise Let be the set of all integers and let I =. Suppose that {c ij } i,j) I is summable. State and prove a theorem corresponding to Theorem

5 1.3. COIN TOSSING Coin Tossing Coin tossing of a fair coin is modeled by the Bernoulli trail with two outcomes H and T with PH) = PT) = 1 2, where H and T stand for head and tail respectively, while PH) and PT) are the probalities of the outcomes H and T respectively. For convenience we replace H and T by 1 and 0 in this order. Then an infinite sequence ω = ω 1, ω 2,..., ω n,...) where each ω n is either 1 or 0 represent a realization of a sequence of coin tossings. We consider the case that tossings are carried out independently at this stage independence is understood intuitively). Let now Ω = {1, 0} {1, 0} = {ω = ω 1, ω 2,...) : ω n = 1 or 0 for each n}, and for each n let E n) 1 = {ω Ω : ω n = 1}; E n) 0 = {ω Ω : ω n = 0}, i.e. E n) 1 E n) 0 ) is the event that at the nth toss head tail) appears. Since the tossings are independent for any finite sequence 1 n 1 < < n k of integers we assign 1 2 )k as the probability of the event E ǫ n1) 1 E n k) ǫ k for any choice ǫ 1,...,ǫ k of 1 or 0, i.e. ) P E ǫ n1) 1 E n k) ǫ k = k j=1 P E nj) ǫ j ) = ) k 1. 2 Sets of the form E ǫ n1) 1 E n k) ǫ k = {ω Ω : ω n1 = ǫ 1,..., ω nk = ǫ k } are called cylinder sets is Ω. Let us denote by QΩ) = the family of all finite union of cylinder sets in Ω. Obviously Ω is in QΩ); is also in QΩ) because it is the union of empty family of cylinder sets. Exercise Show that QΩ) is an algebra of subsets of Ω. i.e. i) Ω QΩ), ii) if A QΩ), then A c = Ω\A is in QΩ); and iii) if A, B are in QΩ), then A B is in QΩ). Observe now that each element Z of QΩ) can be expressed as disjoint union of cylinder sets: Z = l C j, define PZ) = l j=1 PC j). j=1 Exercise Show that P is well-defined and that P is additive on QΩ) i.e. PZ 1 Z 2 ) = PZ 1 ) + PZ 2 ) if Z 1, Z 2 are disjoint elements of QΩ) and PΩ) = PE n1) 1 ) + PE n) 2 ) = 1 for each n.

6 6 CHAPTER 1. INTRODUCTION AND PRELIMINARIES From now on we write d j ω) = ω j, j = 1, 2,..., if ω = ω 1, ω 2,...) Ω; and for each n define a function S n on Ω by S n ω) = n d j ω) Exercise Show that, for each k = 0, 1,..., n, the set {ω Ω : S n ω) = k} is in QΩ) and ) n 1 P{S n = k}) = k 2 n, where ) n k = n! k!n k)!. For a given realization ω of an independent sequence of coin tossings S n ω) is the number of heads that appear in the first n tosses and Snω) n measures the frequency of appearance of heads in the first n tosses. Let E = j=1 { S n ω) ω Ω : lim = 1 }, n n 2 E is easily seen to be not in QΩ). Nevertheless, we expect that P can be extended to be defined on a larger family of sets than QΩ) so that PE) is defined and PE) = 1. Eventually we shall answer positively to this expectation. 1.4 Metric Spaces Let M be a nonempty set and let ρ : M M [0, + ) satisfy i) ρx, y) = ρy, x) 0 for all x, y M and ρx, y) = 0 if and only if x = y; ii) ρx, z) ρx, y) + ρy, z) for all x, y, and z in M. ρ is then called a metric on M and M, ρ) is called a metric space. Usually we say that M is a metric space with metric ρ, or simply that M is a metric space when a certain metric ρ is implicitly implied. Example Let M = Ê n and for x, y Ê n let ρx, y) = x y, where x = n i=1 x2 i )1 2 if x = x 1,..., x n ) Ê n. To show that ρ is a metric on Ê n we first establish the well-known Schwarz inequality: x y x y if x, y Ê n, where, for x = x 1,..., x n ) and y = y 1,..., y n ) in Ê n, x y = n i=1 x iy i is called the inner product of x and y. For this purpose we may assume that x 0 and y 0, hence x > 0 and y > 0. For t Ê we have 0 x + ty 2 = x + ty) x + ty) = x 2 + 2tx y) + t 2 y 2 = x + t y ) 2 + 2t x y x y ),

7 1.4. METRIC SPACES 7 from which by taking t = x / y we obtain x y x y. Then x y x y follows, because x y) x y = x y. Now for x, y, and z in Ê n we have ρx, z) 2 = x z 2 = x y + y z 2 = x y 2 + 2x y, y z) + y z 2 x y x y y z + y z 2 = x y + y z ) 2 = [ ρx, y) + ρy, z) ] 2, i.e. ρx, z) ρx, y) + ρy, z). Hence Ê n is a metric space with metric ρ defined above. This metric is called the Euclidean metric on Ê n. Unless stated otherwise, Ê n is considered as a metric space with this metric, then Ê n is called n-dimensional Euclidean space. Example For a closed finite interval [a, b] in Ê let C[a, b] denote the space of all real-valued continuous functions defined on [a, b]. For f, g C[a, b] let ρf, g) = max a t b ft) gt). It is easily verified that C[a, b] is a metric space with metric ρ so defined. Unless stated otherwise C[a, b] is equiped with this metric which is often referred to as the uniform metric on C[a, b]. C[a, b] is also used to denote the space of all complex-valued continuous functions on [a, b] with metric defined similarly. Exercise Show that Ê n is also a metric space with metric ρ defined by ρx, y) = max 1 i n x i y i if x = x 1,...,x n ) and y = y 1,...,y n ). A sequence {x n } M is said to converge to x M if for any ε > 0 there is n 0 Æ such that ρx n, x) < ε whenever n n 0. Since x is uniquely determined, x is denoted by lim n x n. If lim n x n exists, then we say {x n } converges in M. Example {f n } C[0, 1] converges if and only if f n x) converges uniformly for x [0, 1]. A sequence {x n } M is called a Cauchy sequence if for any ε > 0, there is n 0 Æ such that ρx n, x m ) < ε whenever n, m n 0. Exercise Show that if {x n } M converges, then {x n } is a Cauchy sequence. Exercise let {x n } be a Cauchy sequence. Show that if {x n } has a convergent subsequence, then {x n } converges. A metric space M is called complete if every Cauchy sequence in M converges in M. Exercise Show that C[a, b] is complete.

8 8 CHAPTER 1. INTRODUCTION AND PRELIMINARIES Let à = Ê or and let E be a vector space over Ã. Suppose that for each x E, there is a nonnegative number x associated with it so that i) x = 0 if and only if x is the zero element of E; ii) αx = α x for all α à and x E; iii) x + y x + y for all x, y in E. Then E is called a normed vector space n.v.s.) with norm, and is called a norm on E. If E is a n.v.s., for x, y in E let ρx, y) = x y, then ρ is a metric on E and is called the metric associated with norm. Unless stated otherwise, for a n.v.s., we always consider this metric. A normed vector space is called a Banach space if it is a complete metric space. Both Ê n and C[a, b] are Banach spaces. Let M 1, M 2 be metric spaces with metrics ρ 1 and ρ 2 respectively. A mapping T : M 1 M 2 is said to be continuous at x M 1 if for any ε > 0, there is δ > 0 such that ρ 2 Tx), Ty)) < ε whenever ρ 1 x, y) < δ. A set G in a metric space M is called open if for every x G, there is ε > 0 such that y G whenever ρx, y) < ε. Complement of an open set is called a closed set. For x M and r > 0, let B r x) = {y M : ρy, x) < r} and C r x) = {y M : ρy, x) 0}. It is easily verified that B r x) is an open set and C r x) is a closed set. B r x) C r x)) is usually referred to as the open closed) ball centered at x and with radius r. A point x M is called isolated if B r x) = {x} for some r > 0. Exercise Let M 1, M 2 be metric spaces with metrics ρ 1 and ρ 2 respectively and let T : M 1 M 2. i) Show that T is continuous at x M 1 if and only if for any open set G 2 M 2 with Tx) G 2, the set T 1 G 2 = {y M 1 : Ty) G 2 } contains an open set which contains x. ii) Show that T is continuous on M 1 i.e. T is continuous at every point of M 1 ) if and only if for any open set G 2 M 2, T 1 G 2 is an open subset of M 1. Exercise Let T be the family of all open subsets of a metric space. Show that i) A, B T A B T ; ii) If {A i } i I T, then i I A i T, where I is any index set.

9 1.5. SEMI-CONTINUITIES Semi-continuities For real-valued functions, the fact that the real field Ê is ordered plays an important role in the analysis of functions. In particular, for real-valued functions defined on a metric space, lower semi-continuity and upper semi-continuity are useful concepts that owe their existence to Ê being ordered. Semi-continuities are our concern in this section. For a sequence x n, n = 1, 2,..., of real numbers let lim inf x n = lim n n lim sup x n = lim n n ) inf x k, 1.4) k n x k ). 1.5) sup k n Notice that inf k n x k is increasing as n increases, hence both limits on the right-hand sides of 1.4) and 1.5) exist, although may not be finite. Thus liminf n x n and limsup n x n always exist. Exercise i) Show that liminf n x n limsup n x n ; ii) show that lim n x n exists if and only if liminf n x n = limsup n x n, and lim n x n is the common value liminf n x n = limsup n x n if it exists; iii) show that liminf n x n +y n ) liminf n x n +liminf n y n limsup n x n +y n ) limsup n x n +limsup n y n ), if liminf n x n +liminf n y n limsup n x n + limsup n y n ) is meaningful. Let now f be a real-valued function defined on a metric space M with metric ρ. f is said to be lower semi-continuous upper semi-continuous) at x M if, for every sequence x n, n = 1, 2,..., in M with x = lim n x n, fx) liminf n fx n ) fx) limsup n fx n )) holds. Exercise i) Show that f is lower semi-continuous upper semi-continuous) at x if and only if [ ] [ fx) = lim inf fy) fx) = lim sup fy)] ) ; δց0 y M, ρx,y)<δ δց0 y M, ρx,y)<δ ii) show that f is continuous at x if and only if f is both lower semi-continuous and upper semi-continuous at x. Because of assertions of Exercise 1.5.2, if x is not an isolated point of M we define liminf y x fy) and limsup y x fy) by [ ] lim inf fy) = lim inf fy) ; y x δց0 y M, 0<ρx,y)<δ [ ] lim sup fy) = lim y x δց0 sup fy) y M, 0<ρx,y)<δ,

10 10 CHAPTER 1. INTRODUCTION AND PRELIMINARIES since inf y M, 0<ρx,y)<δ fy) increases as δ decreases and sup y M, 0<ρx,y)<δ fy) decreases as δ decreases, both liminf y x fy) and limsup y x fy) exist, although may not be finite. If liminf y x fy) = limsup y x fy), the common value is called the limit of fy) as y x and is denoted by lim y x fy). Exercise Show that liminf y x fy) limsup y x fy) and that f is continuous at x if and only if lim y x fy) = fx). If f is lower semi-continuous upper semi-continuous) at every point of M, then f is called a lower semi-continuous upper semi-continuous) function on M. Exercise f is lower semi-continuous upper semi-continuous) on M if and only if {x M : fx) > α} {x M : fx) < α}) is open for every α Ê. Exercise Let f α, α I, be a family of real-valued continuous functions defined on M and assume that sup α I f α x) inf α I f α x)) is finite for each x M show that sup α I f α x) inf α I fx)) is lower upper) semi-continuous on M. Exercise A metric space M is called compact if every sequence in M has a subsequence which converges in M. Show that if f is lower semi-continuous upper semi-continuous) on a compact metric space M, then f assumes its minimum maximum) on M. 1.6 The Space l p ) Let be the set of all integers and consider the space L of all real-valued functions defined on. L is a real vector space. For f L and j if we denote fj) by f j, then f can be identified with the two-way sequence f j ) j of real numbers and L is the space of all sequences a j ) j of real numbers. For f L and 1 p let ) 1 fj) p p if p < ; f p = j sup j fj) if p =. Now consider the space l p ), 1 p, defined by l p ) = {f L : f p < }. Presently we shall prove that p is a norm on l p ), but for this purpose we show first an inequality which is a generalization of Schwarz inequality and

11 1.6. THE SPACE l P ) 11 is called Hölder s inequality. Two extended real numbers p, q 1 are called conjugate exponents if 1 p + 1 q = 1; while two nonnegative numbers α and β will be called a convex pair if α + β = 1. Lemma If α and β is a convex pair, then for any 0 ζ, η < the following inequality holds: ζ α η β αζ + βη. 1.6) Proof. We may assume that 0 < α, β < 1 and ζ, η > 0. Since 1 + x) α αx + 1, for x 0, we have y α αy + β, y ) Now either ζη 1 1 or ζ 1 η 1; if ζη 1 1 take y = ζη 1 in 1.4), while if ζ 1 η 1 take y = ζ 1 η in 1.5) with α and β interchanged, then we lead to 1.6). Lemma Hölder s Inequality) If x = x 1,...,x n ) and y = y 1,..., y n ) are in Ê n, then for conjugate exponents p and q we have n x j y j x p y q. j=1 Remark. Since an element x of Ê n can be identified with an element f of L by f1) = x 1,..., fn) = x n, and fj) = 0 for other j, x p is defined. Proof of Lemma It is clear that if one of p and q is, the Lemma is trivial, hence we suppose that 1 < p, q <. Since x p = 0 if and only if x = 0, we ) may assume that x p > 0 and y p > 0. For 1 j n choose p ) q xj yj ζ = x p and η = y q in Lemma With α = 1 p and β = 1 q, then x j y j 1 x j p x p y q p x p + 1 y j q p q y q, q and consequently n 1 j y j x p y q j=1 x p + 1 ) = x p y q. q We are now in position to prove that p is a norm on l p ). That f p = 0 if and only if f = 0, and λf p = λ f p for f l p ) and λ Ê is obvious. It remains only to show that f + g p f p + g p. For this purpose, we

12 12 CHAPTER 1. INTRODUCTION AND PRELIMINARIES may assume f + g p > 0. Under this assumption, there is A F ) such that fj) + gj) p > 0. For such A we have 0 < fj) + gj) p fj) + gj) p 1 fj) + gj) ), from which by using Hölder s inequality see Lemma ) We have 0 < fj) + gj) p ) 1 { fj) + gj) p 1)q q ) 1 fj) p p + ) 1 fj) + gj) p q f p ) + q p, ) 1 } gj) p p and thus, on dividing the last sequence of inequalities by fj)+gj) p) 1 q, we obtain ) 1 fj) + gj) p p f p + g p. 1.8) Now observe that 1.8) holds for any A F ). Taking supremum on the left hand side of 1.8) over A F ), we see that f +g p f p + g p. Therefore p is a norm on l p ). We shall always refer to l p ) as a normed vector space with this norm. Exercise Let k 1 < < k n be a finite sequence in of length n, define a map T from l p ) to the n-dimensional Euclidean Ê n by Tf) = fk 1 ),..., fk n ) ), f l p ). Show that T is continuous from l p ) onto Ê n. Show also that image under T of any open set in l p ) is an open set in Ê n. Exercise Use Hölder s in equality to show that if 1 p <, then a + b p 2 p q a p + b p) 2 p a p + b p) for a, b in Ê. Exercise Let f 1, f 2,..., f n,... be a Cauchy sequence in l p ), show that for every j lim n f n j) exists and finite. Exercise Show that l ) is a Banach space. Theorem l p ) is a Banach space for 1 p.

13 1.6. THE SPACE l P ) 13 Proof. The case p = is relatively easy and is left as an exercise see Exercise 1.6.4). Consider now the case 1 p <. Let f 1, f 2,...,f n,... be a Cauchy sequence in l p ), then lim n f n j) exists and finite for each j see Exercise 1.6.3), say fj) = lim n f n j). We show first that f l p ). Since f 1, f 2,...,f n,... is a Cauchy sequence, it is necessarily bounded. Let f n p M for all n. There is n 0 Æ such that f n f m < 1, n, m n 0. Fix now m n 0 and let A F ), then fj) p = lim f n j) p = lim n limsup n n f n j) f m j) + f m j) p { fn j) f m j) + f m j) } p, from which by Exercise we have { fj) p limsup 2 p f n j) f m j) p + } f m j) p Thus n { } 2 p lim sup f n f m p p + f m p p n < 2 p {1 + M}. fj) p = sup j A F ) fj) p 2 p 1 + M p ) <, which shows f l p ). We claim now lim n f n = f in l p ). Actually given ε > 0, there is N Æ such that f n f m p < ε, n, m N. Now for n N and A F ), fj) f n j) p = lim which implies or f f n p p = sup m f m j) f n j) p liminf m f m f n p p εp, A F ) fj) f n j) p ε p, f f n p ε, n N. In other words lim n f n = f in l p ). This shows that l p ) is complete and hence is a Banach space.

14 14 CHAPTER 1. INTRODUCTION AND PRELIMINARIES

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