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Lecture

# Chapter 10

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School
Department
Economics
Course
ECON 2P30
Professor
Lester Kwong
Semester
Winter

Description
Chapter 10 Theory of Functions ∗ Lester M.K. Kwong Department of Economics Brock University Version: 3.1.19.o Revised: November 8, 2012 ∗ Copyright ' 2004-2012 Lester M.K. KwonDepartment of Economics, Brock University, 500 Glenridge AveCatharines, Ontario, L2S 3A1, CanEmail: [email protected] Tel: +1 (905) 688-5550, Ext. 5137. This document was typeset with the LE X Documentation System usingYthe X -piE, the LT XDraw, and the MnSymbol packages. 1 Introduction 2 Sequences Before we devote time into the study of generalized functions, we begin our analysis of functions by ▯rst considering a very specialized one. In particular, consider a function f ∶ N → R . That is, the domain of f is the set of natural numbers N and the codomain the Euclidean n-space R . Such a function is called a real-valued sequence, or in the case where in context, the codomain R is obvious, f will be referred simply to as a sequence. As can be seen in its de▯nition, the concept of a sequence can be rather arbitrary. For example, a random assignment of real numbers, if n = 1, to the set of natural numbers such as: N 1 2 3 4 5 6 7 8 ⋯ f ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ” ” R 2 2 1 2 3 100:32 −3 ln(6) 0 ⋯ is, by de▯nition, a sequence even though there appears to be no pattern attached to the image of f. Conversely, sequences could take on a precisely de▯ned form such as f(x) = x~2 in which case the mapping may be illustrated as: N 1 2 3 4 5 6 7 8 ⋯ f ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ R 2 1 2 2 2 3 2 4 ⋯ or simply written as { ;1; ;2; ;3;:::}. In general, however, since sequences 2 2 2 map from N, it is sometimes of convenience and expositional simplicity to index the sequence and hence write f ∶ N → R as f . Therefore, no confusion i should arise when we refer to, for example, {x } as being a sequence meaning x ∶ N → R where each x ∈ R for all i ∈ N. A sequence {x } ‘ R is said to converge to x ∈ R if for every ▯ > 0, there exists some s ∈ N so that for all t ≥ s, x ∈▯B (x). In words, convergence of a sequence states that for every positive ▯ > 0, there must exist some point in the sequence after which the sequence remains within the open ball with radius ▯ around the point x. In the even that {x } converges to x, we write i i {x } → x or limx = x. A sequence that does not converge is said to diverge. For example, the sequence {x }, as illustrated in Figure 1, converges to the point x ∈ R. 1 B (ǫ) ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ x − ǫ x + ǫ b b b b b b b b bbbbb b b b 1 2 3 ⋯ xs x R x x x Figure 1: A convergent sequence {x } to x ∈ R. i It should be noted that while the above de▯nition of convergence makes use of open balls in R , we could alternatively de▯ne it using the Euclidean metric. In particular, the requirement that x ∈ B (x) could be restated ▯s d(x;x ) < ▯. Hence, a sequence {x } is said to converge to x if for every ▯ > 0, t there exists some s ∈ N so that for all t ≥ s, d(x;x ) < ▯. i i For example, consider the sequence {x } where x = 1~i. This sequence converges to 0 since as i gets larger and larger, 1~i gets smaller and smaller. However, this explanation was rather \hand-waving" in that it o▯ers no real correlation to the de▯nition of convergence. Hence, let us check, using the de▯nition, whether it does, in fact, converge to 0. We begin by picking any ▯ > 0 and construct B (0) = (−▯;▯). Note tha▯ x > 0 for all i ∈ N, hence, only the positive part of B (0) is relevant in the ▯ sense that the negative portion of B (0) will not play any role here. For ▯ a given ▯, when will the sequence {x } fall into B (0)? Naturally, it thi▯ occurs at t ∈ N, it follows that 1~t ≤ ▯. We must now verify that for all s ≥ t, s 1 s t x ∈ B (0)▯ But this is readily veri▯ed since x ≤ x if and only if 1~s ≤ 1~t s if and only if t ≤ s. Furthermore, x > 0 for all s ≥ t, or for all s ∈ N for that matter, so x ∈ (0;▯) ‘ B (x). It, therefore, follows from de▯nition that ▯ limx = 0. Theorem 2.1. If limx = x and limx = y then x = y. i Proof. Suppose limx = x and limx = y and x ≠ y. It follows that d(x;y) ≠ 0. Hence, by the triangle inequality: d(x ;y) ≥ d(x;y) − d(x;x ) i (1) 1It should be noted that even in the event that even if ∃s ≥ t so that x ∉ B (0) does s ▯ i not automatically imply that {x } does not converge to 0. Recall that convergence simply requires the existence of some point along the sequence where by the sequence remains in the proximity of 0. However, there is no requirement that this point must come from its initial entrance into the proximity. 2 i i i Naturally, by our assumption that limx = x and limx = y, d(x ;y) → 0 and d(x;x ) → 0 provides the necessary contradiction in that 0 ≥ d(x;y) > 0. ∎ The result in Theorem 2.1 gives us uniqueness in convergence points for i convergent sequences. We now decompose convergent sequences {x } into each of its n dimensions in hopes to simplify the problem slightly going for- ward. Not surprisingly, for a sequence {x } in R , if limx = x = (x ;:::1x ) ∈ n n i i R , then for each dimension j along the sequence in {x }, limx = x . We j j state and prove this formally in the following theorem. Theorem 2.2. Suppose {x } is a sequence in R . Then limx = x ∈ R if i n i and only if limx =jx foj every j ∈ {1;:::;n}. i Proof. Suppose limx = x. It follows from the de▯nition of convergence that for every ▯ > 0, there exists some s ∈ N so that for all t ≥ s, x ∈ B (x). ▯ Therefore, ▯x ▯ and consider the corresponding s. Note that by the Euclidean metric: n 1~2 t t 2 t 2 1~2 t d(x;x ) =  Q (xm− x m )  ≥ (x −jx ) j  = d(xj;x j (2) m=1 for every j ∈ {1;:::;n}. Therefore, from the assumption of convergence of the sequence {x }, we have ▯ > d(x;x ) ≥ d(x ;x ) foj every t ≥ s and for i j every j ∈ {1;:::;n}. Hence, the sequences {x } converje to x for every j j ∈ {1;:::;n}. Now suppose every sequence {x } converges to x for j ∈j{1;:::;n}. By j t 2 de▯nition, for every ▯ > 0 there exists some s for which d(x ;x j 0 the sequence 5 ′ must eventually be in the set B (x)▯ Since for ▯ < ▯ , B (x) ▯ B (x), ▯hat is, the set B ▯x) is decreasing in ▯, one can relate convergence to the sequence being eventually in a set this way; every convergent sequence must eventually be in a decreasing set of sets. Lastly, consider the sequence {x } in Eq. (4) again. While the sequence is bounded by [−1;1] and hence, eventually in [−1;1], what about the set [0;1]? It is certainly not bounded by this set, nor is it eventually in this set since every other point along the sequence exits, or enters, the set. In instances such as this, we say that the sequence {x } isnfrequently in the set [0;1]. More formally, we say that a sequence {x } is frequently in a set X if n for every s ∈ N, there exists some t ≥ s so that x ∈ Xt It naturally follows that a sequence that is eventually in a set X is frequently in a set X but that a sequence frequently in a set X may not be eventually in the set X. The set X = [0;−1] for the sequence in Eq. (4) and the set Y = {1~5} for the sequence in Eq. (5) are two prime examples of this. Theorem 2.5. Let {x } and {y } be two convergent sequences so that limx = n n n x and limy =ny. Then: 1. for all ▯ ∈ R, lim▯x = nx. 2. lim(x + y ) = x + y. n n 3. lim(x n n = xy. 4. if y ≠ 0, then lim(x ~n )n= x~y. 4 We leave the proof of Theorem 2.5 as an exercise for the reader. Theorem 2.6. Let {x } and {y } be two convergent sequences so that limx = n n n x and limy =ny. Then: 1. if xn≥ 0 for all n ∈ N, then x ≥ 0. 2. if x ≥ y for all n ∈ N, then x ≥ y. n n 3. if for some ▯ ∈ R, ▯ ≥ (≤)y , nor all n ∈ N, then ▯ ≥ (≤)y. Proof. For Part 1, suppose limx = x and that x ≥ 0 for aln n ∈ N but x < 0. Then by assumption, for ▯ = SxS, there exists some s ∈ N so that for all 3Note that the correct interpretation of the concept of a sequence being frequently in a set X implies that it belongs to the set X for an in▯nite number of points in the sequence. That is, a convergent sequence to x may not frequently be in the set {x} whereas it is frequently in the set ▯ (x) for some ▯ > 0. 4See Exercise ??. 6 t ≥ s, d(x;x t = Sx − xtS < SxS. Since x < 0, this implies that x t 0 which is a contradiction that x ≥n0 for all n ∈ N. For Part 2, from Theorem 2.5, Part 2, it follows that lim(x −y n = xny. Since x n y ≥n0, we may apply Theorem 2.6, Part 1. For Part 3, let x n ▯ for all n ∈ N and apply Theorem 2.6, Part 2. ∎ Consider the sequence {x } niven by: 1 x n (6) 2n − 1 If we computed the points of {x }, we would have {1; ; ; ;:::}. This1 n 3 5 7 collection of points is exactly the same collection of points of the sequence {y } from Eq. (5) restricted to odd values of n. In instances a sequence, n {x n, is obtained by considering an in▯nite subset of points from another sequence, {y n, the sequence {x } in said to be a subsequence of {y }. Mone formally, let ▯ ∶ N → N be a strictly increasing function so that ▯(x) > ▯(y) if and only if x > y, then x ▯(n) is a subsequence of x .nIn the context of the example above, the function ▯(n) = 2n − 1 so that y = y . ▯(n) n Heuristically, a subsequence is, itself, a sequence that is obtained by elim- inating some points from another sequence. So, for example, if we have a sequence: {1;2;3;4;5;6;7;8;9;10;:::} then: {1;3;5;7;9;:::}; {2;4;6;8;10;:::};and {1;2;3;5;7;8;10;:::} are all examples of subsequences. Note that the de▯nition only required ▯(n) to be a strictly increasing function. The explicit statement of the function ▯, such as in the above example where ▯(n) = 2n−1 may not always be possible. Theorem 2.7. Let {x } bena sequence. Then, limx = x if and only if every subsequence {x ▯(n)}, limx ▯(n) = x. Proof. For the if part, suppose limx = n. Then by de▯nition, for every ▯ > 0, there exists some s ∈ N so that for all t ≥ s, xt∈ B▯(x). It immediately follows that for all t ≥ s, x ∈ B▯(x) since ▯(n) ≥ for all n ∈ N since ▯(⋅) is a strictly ▯(t) increasing function. Therefore, limx ▯(n)= x. For the only if part, de▯ne X = NRange(▯) and let ▯ ∶ N → X be a strictly increasing surjection. By construction {x } =n{x ▯(n)} ∪ {x▯(n)}. By 7 assumption, {x ▯(n)} and {x ▯(n)} are convergent subsequences to x hence for all ▯ > 0, there exists some s ;s ▯ N ▯o that for all t ≥ s and ▯ ≥ s ▯ ▯ ▯ x▯(t▯) ∈ B ▯x) and x ▯(t▯) ∈ B ▯x). De▯ne s = s + s▯. It ▯ollows that for all t ≥ s, xt∈ B (▯). To see this, pick some t ≥ s. Then either t ∈ Range(▯) or t ∈ Range(▯). If t ∈ Range(▯), then ▯ (t) = t ≥ s ≥ s and hence x ∈ B (x) ▯ ▯ t ▯ since limx ▯(n) = x. Conversely, if t ∈ Range(▯), then ▯ −1 = t▯≥ s ≥ s a▯d hence, again, x ∈tB (x▯ since limx ▯(n) = x. In either case, for any given ▯, we have found some s ∈ N so that for all t ≥ s, x ∈ B tx) a▯d so limx = x. n ∎ A sequence {x } is said to be: n (i) increasing, nondecreasing, or weakly increasing if x ≤ x for all n ∈ N. n n+1 (ii) strictly increasing if x n x n+1 for all n ∈ N. (iii) decreasing, nonincreasing, or weakly decreasing if x ≥ xn n+1 for all n ∈ N. (iv) strictly decreasing if x n x n+1 for all n ∈ N. Increasing and decreasing sequences are also referred to as monotone se- quences whereas strictly increasing and strictly decreasing sequences are also referred to as strictly monotone sequences. Theorem 2.8. If {x } is n bounded monotone sequence, then {x } is con- n vergent. Proof. Suppose {x } in increasing and let M = sup{x }. Now cnnsider ▯ > 0. Since M = sup{x }, nt implies that M −▯ ≠ sup{x }. This nmplies that there exists some s ∈ N so that x ≥ M − ▯. By monotonicity of {x }, for all t ≥ s, s n xt≥ M − ▯ and so x ∈ B (t) wh▯ch is the de▯nition of convergence. For the case when {x } in decreasing, let M = inf{x } and consider the point M + ▯. ∎ Theorem 2.9. Every sequence {x } has a nonotone subsequence. Proof. Let {x } be a sequence and de▯ne S = {s ∈ N ∶ ƒt > s;x < x }. If the n t s set S is an in▯nite set (i.e., S = {s 1s 2:::}), then the sequence {x } fosms a n decreasing, hence monotone, subsequence. On the other hand, if S is a ▯nite set so that S = {s ;:1:;s } tnen let y = s +11. Bn assumption, y ∉ S so 1 ∃y ∈ N, with y > y and x ≥ x . Similarly, y ∉ S implies that ∃y ∈ N, 2 2 1 y2 y1 2 3 with y 3 y an2 x 3 ≥ x y2 ≥ x y1Continuing on with this argument, {x } yn forms an increasing, hence monotone, subsequence. ∎ 8 Theorems 2.8 and 2.9 immediately gives us the Bolzano-Weierstrass The- orem which we formally state. Theorem 2.10. Every bounded sequence has a convergent subsequence. A sequence {x } ns said to diverge to “, x → “, ifnfor every M > 0, ∃s ∈ N so that for all t ≥ s, x ≥tM. Similarly, the sequence {x } is sand to diverge to −“, x → −“, if for every M > 0, ∃s ∈ N so that for all t ≥ s, n xt≤ −M. The following theorem naturally follows. Theorem 2.11. Let {x } be an unbounded sequence. If {x } is: n 1. an increasing sequence, then x → “n 2. a decreasing sequence, then x → −n. Proof. For part 1, if {x } is unbounded, that means that for every M > 0, n ∃s ∈ N so that x ≥sM. But by the assumption that {x } is incrensing, this means for all t ≥ s, x ≥ t ≥ M.sHence, {x } divergesnto “. A mirror argument for part 2 may be easily established. ∎ Recall that the notion of a convergent sequence is that x → x if fon every ▯ > 0, there exists some s ∈ N so that for all t ≥ s, x ∈ B (x). Since t ▯ xt∈ B (▯), it follows that d(x ;t) < ▯. If, rather than thinking about the point of convergence x, we replaced it with some point su▯ciently far o▯ in the tail end along the sequence {x }, ne have what is known as the Cauchy criterion. More formally, a sequence {x } isnsaid to satisfy the Cauchy criterion if for all ▯ > 0, there exists some s ∈ N so that for all t;u ≥ s, d(x ;x ) < ▯. t u A sequence which satis▯es the Cauchy criterion resembles very much the concept of a convergent sequence since, by de▯nition, the Cauchy criterion needs to hold for every ▯ > 0. Hence, the smaller the ▯ chosen, the smaller the distance between points along the sequence in the tail of the sequence needs to be. A sequence which satis▯es the Cauchy criterion is referred to as a Cauchy sequence. Theorem 2.12. A sequence {x } is a Cnuchy sequence if and only if the sequence {x }nconverges. Proof. For the if part, let {x } ne a Cauchy sequence. We begin by ▯rst showing that {x } is bounded. It follows that for all ▯ > 0, there exists some n s ∈ N so that for all t;u ≥ s, d(x tx u < ▯. Hence, let ▯ = 1, then we have some s so that for all t ≥ s, d(x tx s = Sx t x Ss< 1. It follows that Sx St< Sx s + 1 for 9 all t ≥ s and hence, we only have the ▯rst s − 1 points remaining. However, we can, much like the proof of Theorem 2.3 construct: M = max{Sx S;1::;Sx s−1S;Sx2S + 1} and so {x } is bounded by M. Applying the Bolzano-Weierstrass Theorem, n Theorem 2.10, we may construct a convergent subsequence, given by {x ▯(n)} so that x ▯(n) → x. Now pick any ▯ > 0. It follows that there exists some s ∈ N so that for all t;u ≥ s, Sx −tx S u ▯~2. Since x ▯(n) → x, there exists some ▯(k) ≥ s so that Sx − x ▯(k) < ▯~2. It also follows that since ▯(k) ≥ s, Sx▯(k)− x t < ▯~2. Clearly: Sx − xtS = Sx −▯(k) + x▯(k)− x t ≤ Sx − ▯(k)S + S▯(k)− x t < ▯~2 + ▯~2 = ▯ Thus, the Cauchy sequence {x } converges to x. For the only if part, suppose {x } cnnverges to x. It follows that for every ▯ > 0, de▯ne ▯ = ▯~2, then there exists some s ∈ N so that for all t ≥ s, d(x tx) < ▯. Naturally, if u ≥ s, then d(x ;u) < ▯. Therefore, by the triangle inequality: d(x tx u ≤ d(x ;t) + d(x ;xu < ▯ + ▯ = ▯ Hence, the sequence {x } is a Cauchy sequence. ∎ n By Theorem 2.12 every Cauchy sequence is a convergent sequence and every convergent sequence is a Cauchy sequence. Hence, Theorems 2.1 and 2.3 are applicable. Therefore, Cauchy sequences have unique points of con- vergence and are also bounded. 3 Functional Limit With the concept of sequences established above, we now turn our attention to functional limits. In principle, our objective is to understand and be able to evaluate the statement: limf(x) (7) x→z where f ∶ X → R with X ⊆ R . The statement presented in Eq. (7) examines the image of the function f(⋅) when the value of x, from the domain, is in the proximity of z.
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