Mathematica call Analysis T. T. M. Apostol
Chapter 3
Mathematica aticall Analys ysiis by Tom TomM. Apostol Chapter 3: Elements of Poi Point nt Set Theory: ory: Note Notes Let E1 denote the set of all real numbers (the real line). Let E 2 denote the set of all complex numbers complex numbers (the complex plane). An open interval (a, b) is defined as (a, b) = {x | a < x < b}. Closed intervals, half-open intervals and infinite intervals can be similarly defined. The real line is sometimes referred to as the open interval (- ∞, ∞). A single point is sometimes considered as a “degenerate” closed interval. Let h > 0, and let x be a given point. The open interval (x-h, x+h) is called a neighbourhood with x as centre and of radius h. We denote the neighbourhood by N(x; h), or simply by N(x) if the radius is unimportant . Let S be a set in E 1 and assume that x ∈ S. Then x is called an interior point of S if there is some neighbourhood N(x) all of whose points belong to S. Let S be a set in E 1. S is called an open set if every point of S is an interior point of S. Thus an open set is such that each of its points points can be enclosed in a neighbourhood which is completely contained in the set. The simplest kind of open set is an open interval. The empty set is also open, as is the real line E 1. The union union of any collection of open sets is an open set. The intersection intersection of a finite collection of open sets is open. Note that arbitrary intersections will intersections will not always lead to open sets. The union of a countable collection of disjoint open intervals is an open set and, remarkably enough, every open set on the real line can be obtained in this way. The structure of open sets in E1. A set of points in E 1 is said to be bounded if it is a subset of some finite interval. Let S be an open set in E1 and let (a, b) be an open interval which is contained in S, but whose endpoints are not in S. Then (a, b) is called a component interval of S. If S is a bounded open set, then (i) each point of S belongs to a uniquely determined component interval of S; and (ii) the component intervals of S form a countable collection of disjoint sets sets whose union union is S. From this, it follows that an open interval in E1 cannot be expressed as the union of two disjoint open sets when sets when neither set is empty. Every open set in E1 is the union of a countable collection of disjoint open intervals. intervals .
Accumulation points and the Bolzano-Weierstrass theorem in E1. Let S be a set in E1 and x a point in E1, x not necessarily in of S, provided necessarily in S. Then x is called an accumulation point of every neighbourhood of x contains at least one point of S distinct from x. If x is an accumulation point of S, then every neighbourhood every neighbourhood N(x) contains infinitely many points of S.
Theorem 3-13 ( Bolzano-Weierstrass): If a bounded set S in E1 contains infinitely many points, then there is at least one point in E1 which is an accumulation point of S. Examples: (1) The set of numbers of the form 1/n, n = 1, 2, 3, ... has 0 as an accumulation point . (2) The set of rational numbers has every real number as an accumulation point. (3) Every point of the closed interval [a, b] is an accumulation point of the set of numbers in the open interval (a, b). Closed sets in E1. A set is called closed if it contains all its accumulation points. Thus a closed interval is a closed set. An open interval, however, is not a closed set because it does not contain its endpoints, both of which are accumulation points of the set. A set which has no accumulation points is automatically closed and, in particular, every finite set is closed. The empty set is also closed, as is the whole real line E 1. (These last two sets are also open!) A set which is not closed need not be open, as, for example, the half-open interval (a, b], which is neither open nor closed. If S is open, then the complement E1 —S is closed. Conversely, if S is closed, then E1 —S is open. It follows from the above that the union of a finite collection of closed sets is closed and that the intersection of an arbitrary collection of closed sets is closed. Generalisation: If S is closed, then the complement of S (relative to any open set containing S) is open. If S is open, then the complement of S (relative to any closed set containing S) is closed. Extensions to higher dimensions. The definition of a neighbourhood , |x-x0| < h, still makes sense if x and x0 belong to higher dimensional spaces. Let n > 0 be an integer. An ordered set of n real numbers (x1, x2, ..., x n) is called an n-dimensional point or a vector with n components. Points or vectors will be denoted by single boldfaced letters, e.g. x = (x 1, x 2, ..., x n), where xk is the k th co-ordinate of the point or vector x. The set of all n-dimensional points is called the n-dimensional Euclidean space and is denoted by En. The usual vector rules apply. Note: Unit coordinate vectors: uk = ( δk,1, δk,2, ..., δk,n) (k = 1, 2, ..., n), where δk,j is the “ Kroneker delta”, defined by δk,j = 0 if k ≠ j, and δk,k = 1. Let x = (x1, ..., x n) and y = (y1, ..., yn) be in En. The absolute value, or length, or norm of x is given by |x| = x 21 + ... + x 2n . The distance between x and y is given by |x-y| = S in=1 ( x i − y i ) 2. |x| > 0, and |x| = 0 iff x = 0. |x-y| = |y-x|. |x+y| ≤ |x| + |y|. By an open sphere of radius r > 0 and centre at the point x0 in E n, we mean the set of all x in En such that |x-x0| < r. The set of x such that |x-x0| ≤ r form a closed sphere, the boundary of the sphere being the set of x with |x-x0| = r. An open sphere with centre at x0 is called a neighbourhood of x0 and is denoted by N( x0) — or by N(x0; r) if r is the radius. The corresponding closed sphere is denoted by N(x0). The open sphere with its centre removed is called a deleted neighbourhood of x0 and is denoted by N’( x0). Let a = (a1, ..., a n) and b = (b1, ..., b n) be two distinct points in En, such that a k ≤ bk for each k = 1, 2, ..., n. The n-dimensional closed interval [a, b] is defined to be the set [ a, b] = {(x1, ..., x n) | a k ≤ x k ≤ b k , k = 1, 2, ..., n}. If a k < b k for every k, the n-dimensional open interval (a, b) is the set (a, b) = {(x1, ..., x n) | ak < xk < bk , k = 1, 2, ..., n}.
Thus, for example, (a, b) can be considered as the “cartesian product ” (a, b) = (a1, b 1) × (a2, b2) × ... × (a n, bn) of the n one-dimensional open intervals (ak , bk ). An open interval in En could then be called a neighbourhood of any of its points, and in what follows, it makes no difference whether a neighbourhood is taken to be a “ sphere” or an “interval ” (we will use spheres). Let S be a set of points in E n, and assume that x ∈ S. Then x is called an interior point of S if there exists a neighbourhood N( x) ⊂ S. The set S is said to be open if each of its points is an interior point. The interior of S is the collection of its interior points. Examples of open sets in the plane are: the interior of a disk; the first quadrant ; the whole space. Any n-dimensional open sphere and any n-dimensional open interval is an open set in En. Caution: an open interval (a, b) in E 1 is no longer an open set when it is considered as a subset of the plane. In fact, no subset of E1 (except the empty set) can be open in E 2, because such a set can contain no two-dimensional neighbourhoods. The union of an arbitrary collection of open sets in E n is an open set in En, and the intersection of a finite collection of open sets in E n is an open set in En. Assume that S ⊂ En, x ∈ En. Then x is called an accumulation point of S if every neighbourhood N(x) contains at least one point of S distinct from x, that is, if N’( x) ∩ S is not empty. If x is an accumulation point of S, then every neighbourhood N(x) contains infinitely many points of S. A set S in E n is said to be bounded if S lies entirely within some sphere; that is, for some r > 0, we have S ⊂ N(0; r). Theorem 3-29 (Bolzano-Weierstrass): If a bounded set S in E n contains infinitely many points, then there exists at least one point in E n which is an accumulation point of S. A set S in E n is said to be closed if it contains all its accumulation points. Examples: A closed sphere in En is a closed set. An n-dimensional closed interval is a closed set. A set which is closed in E1 is also closed in E n for n > 1. A set S in E n is closed iff E n —S is open. The union of a finite collection of closed sets in En is closed and the intersection of an arbitrary collection of closed sets in E n is closed. Assume that S ⊂ A ⊂ En. If A is open and if S is closed , then A—S is open. If A is closed and if S is open, then A—S is closed . The Heine-Borel covering theorem. We begin by defining a covering of a set. A collection F of sets is said to be a covering of a given set S if S ⊂ ∪A∈F A. The collection F is also said to cover S. If F is a collection of open sets, then F is called an open covering of S.
Examples: (1) The collection of all intervals of the form 1/n < x < 2/n (n = 2, 3, 4, ...) is an open covering of the interval 0 < x < 1. This is an example of a countable covering. (2) The real axis E1 is covered by the collection of all open intervals (a, b). This covering is not countable. However, it contains a countable covering of E 1, namely all intervals of the form (n, n+2), where n runs through the integers. Let G = {A1, A2, ...} denote the countable collection of neighbourhoods in En having rational radii and centres at points with rational co-ordinates. Assume that x ∈ En, and let S be an open set in En which contains x. Then at least one of the neighbourhoods in G contains x and is contained in S. That is, we have x ∈ Ak ⊂ S for some Ak in G. Assume that A ⊂ E n, and let F be an open covering of A. Then there is a countable sub collection of F which also covers A. Let {Q1, Q 2, ...} be a countable collection of non empty sets in E n such that (1) Qk+1 ⊂ Qk (k = 1, 2, 3, ...); (ii) Each set Q k is closed and Q1 is bounded. Then the intersection ∩k=1∞ Q k is closed and non empty. Theorem 3-38 (Heine-Borel) Let F be an open covering of a closed and bounded set A in En. Then a finite sub collection of F also covers A. A set S in E n is said to be compact iff every open covering of S contains a finite sub collection which also covers S. The Heine-Borel theorem states that every closed and bounded set in En is compact. The converse result is as follows: Let S be a compact set in E n. We have (i) every infinite subset of S has an accumulation point in S; (ii) the set S is closed and bounded . By the extended real number system E1* we shall mean the union of the set of real numbers E1 with two symbols + ∞ and -∞ which satisfy the following properties: (1) If x ∈ E 1, then we have x+(+∞) = +∞, x+(-∞) = -∞, x-(+∞) = -∞, x-(-∞) = + ∞, and x/(+∞) = x/(-∞) = 0. (2) If x > 0, then we have x(+ ∞) = +∞ and x(-∞) = -∞. (3) If x < 0, then we have x(+ ∞) = -∞ and x(-∞) = +∞. (4) (+∞)+(+∞) = (+∞)(+∞) = (-∞)(-∞) = +∞, (-∞)+(-∞) = (+∞)(-∞) = -∞. (5) If x ∈ E1, then we have -∞ < x < + ∞. Every open interval (a, +∞) is called a neighbourhood of +∞, and every open interval (-∞, a) is called a neighbourhood of -∞. E 1 = (-∞, ∞); E1* = [-∞, ∞]. Every set in E1* has a sup — it is finite if the set is bounded above, and it is +∞ if the set is not bounded above. Note that E1* does not satisfy all the axioms for the real number system. By the extended complex number system E2* we shall mean the union of the complex plane E2 with a symbol ∞ which satisfies the following properties: (1) If z ∈ E2, then z+∞ = z-∞ = ∞, z/∞ = 0. (2) If z ∈ E2, but z ≠ 0, then z(∞) = ∞, and z/0 = ∞. (3) ∞+∞ = (∞)(∞) = ∞. Note that -∞ is not needed here because no ordering is involved with complex numbers. Every open set in E2 of the form {z| |z| > r > 0} is called a neighbourhood of ∞.
Chapter 3: Selected Exercises 3-1. Prove that an open interval in E 1 is an open set and that a closed interval is a closed
set . Answer: Let us first look at some definitions. Definition 1: An open interval (a, b) is defined by (a, b) = {x | a < x < b}. Definition 2: Let S be a set in E 1 and assume that x ∈ S. Then x is called an interior point of S if there is some neighbourhood N(x) all of whose points belong to S. Definition 3: If S is a set in E1, S is an open set if every point of S is an interior point of S.
To prove that an open interval in E1 is an open set , we need to prove that every point in an open interval is an interior point. To do this, we must prove that every point in an open interval has an associated neighbourhood, all of whose points belong to the open interval. Consider an arbitrary open interval (a, b). Let x be any point in this interval, x ∈ (a, b). A neighbourhood of x is given by the interval (x-h, x+h), where h > 0. What we have to show is that one of these neighbourhoods is entirely contained within the interval (a, b), i.e. we have to show that (x-h, x+h) ⊂ (a, b) for every x and for some h > 0. To do this, we have to show that for every y ∈ (x-h, x+h), we also have y ∈ (a, b). Let us choose h to be given by half the value of the minimum distance from x to each of the two endpoints, so ( that h = ½min(|x-a|, |x-b|). Because we know that x > a a and that x < b, we can rewrite h as h = ½min(x-a, b-x). From this definition, we see that (i) h ≤ ½(x-a), and that (ii) h ≤ ½(x-b).
x-h x x+h ( )
) b
Let us now analyse what we know about the point y. First of all, we know that y ( by definition) sits in the interval (x-h, x+h), so that x-h < y < x+h. In order to show that a < y < b, which is what we want to prove, all we have to show is that (1) a < x-h, and that (2) x+h < b. From (i) above, we can do the following manipulation: h ≤ ½(x-a) x-h ≥ x-½(x-a) ⇒ x-h ≥ ½(x+a) ⇒ x-h > ½(a+a) (because a < x) ⇒ proving part (1) x-h > a ⇒ From (ii) above, we can do the following manipulation: h ≤ ½(x-b) x+h ≤ x+½(x-b) ⇒ x+h ≤ ½(3x-b) ⇒ x+h < ½(3b-b) (because x < b) ⇒ proving part (2) x+h < b ⇒
Because we have shown that y ∈ (x-h, x+h) ⇒ y ∈ (a, b), then we have proved that an open interval is an open set. To show that a closed interval is a closed set , we use the following trick: the complement of an open set is a closed set. Let us consider an arbitrary closed interval [a, b]. The complement of this interval is given by E1 —[a, b] = (-∞, a)∪(b, ∞), the union of two open intervals, which we know are open sets by the above. Further, the union of two open sets is also an open set (Theorem 3.5), so that E1 —[a, b] is an open set. So, as the complement of a closed interval is an open set, then a closed interval must therefore be a closed set. QED. 3-2. Determine all the accumulation points of the following sets in E 1 and decide whether the sets are open or closed (or neither ). (a) All integers. (b) The interval (a, b]. (c) All numbers of the form 1/n, (n = 1, 2, 3, ...). (d) All rational numbers. (e) All numbers of the form 2-n + 5-m, (m, n = 1, 2, ...). (f) All numbers of the form (-1) n + (1/m), (m, n = 1, 2, ...). (g) All numbers of the form (1/n) + (1/m), (m, n = 1, 2, ...). (h) All numbers of the form (-1) n/[1+(1/n)], (n = 1, 2, ...). Answer: Let us first remind ourselves of the definition of an accumulation point: Let S be a set in E 1 and x a point in E 1, x not necessarily in S. Then x is called an accumulation point of S provided every neighbourhood of x contains at least one point of S distinct from x.
(a) Let S = {x | x ∈ Z}. S contains no accumulation points because every point x has a neighbourhood (x-½, x+½) which contains no other integers, so that x is not an accumulation point. Because S contains all its accumulation points (of which there are none!), it follows that S is a closed set, and hence not an open set (because of exercise 3-5). (b) Let S = (a, b]. Consider an arbitrary point x ∈ S. Taking an arbitrary neighbourhood N(x) of x, N(x) = (x-h, x+h), with h > 0, then it is clear that the point max{ x-a/2, x-h/2} will always be in S and in N(x) so that every neighbourhood of a point in S will always contain at least one other point from S. We therefore conclude that all the points in S are accumulation points. Further, a is an accumulation point of S so that the set of accumulation points of S is given by the set {x | a ≤ x ≤ b}. S is not closed because it doesn’t contain the accumulation point a. Further, S is not open because b is not an interior point (in the neighbourhood (b-h, b+h), with any h > 0, the points in (b, b+h) ⊂ (b-h, b+h) do not belong to S — but do belong to the neighbourhood. It follows that b is not an interior point). (e) Let S = 2 -n+5-m (n, m = 1, 2, ...). Claim: all numbers of the form x = 2 -n (n = 1, 2, ...) and of the form y = 5-m (m = 1, 2, ...) are accumulation points of S. Justification: to show that each point x = 2 -n is an accumulation point of S (n = 1, 2, ...), consider an arbitrary neighbourhood N(x) of x, N(x) = (x-h, x+h), with h > 0. If we now choose an integer l so that 5-l ≤ h/2 (and there will always be such an l , given by l = ceiling(-log(n/2)/log(5))), then the point y = 2-n+5-l will be in N(x) and in S.
It therefore follows that all points of the form x = 2 -n (n = 1, 2, ...) are accumulation points of S. A similar argument can be used to show that all points of the form y = 5 -m (m = 1, 2, ...) are accumulation points of S. There is one other accumulation point of S: zero (every neighbourhood of zero will contain a point from S if n and m are large enough), and because 0 ∉ S, then S is not a closed set. The question remains as to whether S is an open set. To prove that it is not, it is sufficient to find a single point in S which does not have a neighbourhood all of whose points belong to S. Take the point in S given by setting n = 1 and m = 1, namely the point 7/10. Clearly this is the largest element of the set S, and so every neighbourhood of this point (i.e. the interval (7/10-h, 7/10+h) for some h > 0) will contain elements which are larger than 7/10, elements which cannot be expressed in the form 2 -n+5-m, and so the point 7/10 ∈ S is not an interior point — so that S is not an open set. (h) Let S =
(−1) n 1+ 1n
(n = 1, 2, ...). Writing S out as a sequence, S = {- 1/2, 2/3, - 3/4, 4/5, - 5/6, ...}.
From this series representation, we see that S is an alternating series, one half of the series tending to -1, and the other half tending to 1. Therefore, -1 and 1 are accumulation points of the set S, because the set has elements which are arbitrarily close to -1 and 1. There are no other accumulation points in the set. Because S does not contain its two accumulation points (-1 ∉ S and 1 ∉ S), then S is not a closed set. By the same argument as in part (e), no neighbourhood of the largest element of S, 2 /3, will contain elements just from S, so that 2/3 is not an interior point of S, and so S is not an open set. 3-6. Show that every closed set in E 1 is the intersection of a countable collection of open
sets. Answer: Let us first consider some examples of closed sets and the intersection of open sets that is equivalent to the closed set in question. To start with, consider the set S consisting of a single point x ∈ E 1. This set is closed because it has no accumulation points, and it can be expressed as the intersection of a countable collection of open sets as follows:
S = ∩n=1∞ (x-1/n, x+1/n). Now consider a closed interval [a, b], which we have shown in a previous exercise to be a closed set, which we shall denote by T. This closed set can be expressed as the intersection of a countable collection of open sets as follows: T = ∩n=1∞ (a-1/n, b+1/n). We therefore have some evidence that the statement that we are trying to prove is correct. We now want to show that every closed set C can be expressed as the intersection of some countable collection of open sets, i.e. we want to show that C can be written as C = ∩n=1∞ On, where every On is an open set.
To start with, let us define D to be given by D i, j = (di-1/ j, di+1/ j), where di ∈ C, and j ∈ R >0. In other words, D is the ‘ 1/ j interval’ of a point from C. We see immediately that D i, j is an open set because it consists of an open interval, which we know ( from exercise 3-1) to be an open set. Now let us define E j to be the following expression: E j = ∪i Di, j , the union over all points of C of ‘ 1/ j intervals’ of all those points. Each E j is an open set because it is an arbitrary union of open sets, which (by theorem 3-5) is an open set. Finally, let us define F to be the set given by F = ∩ j=1∞ E j. It follows that F is the intersection of a countable collection of open sets. The question now arises as to whether we have the conclusion F = C. We have already shown that F is the intersection of an arbitrary collection of open sets, and so it is in the right form to be considered as the solution to this exercise. We will in fact show that F = C , i.e. show that F is the closure of the set C. If we can do this, then we will also show that F = C because for a closed set C, we have (looking at part (f) of exercise 3-12) C = C. Therefore, if we can show that F = C, then we reach the required conclusion. To do this, as C= C∪C’ (see exercise 3-12 for this definition), then we first need to show that F contains all of C’s accumulation points and all of the points from C. Claim 1: F contains all of C’s accumulation points, whether they are in C or not . Proof of Claim: Consider an arbitrary accumulation point of the set C, say the point x. If x ∈ C, then we refer to the proof of claim 2 below to show that x ∈ F. If the accumulation point x is not in C, then we must take a little more care. x Following the definition of an accumulation point, every ( neighbourhood N(x) of x will contain a point y ∈ C. But for every E 1 such y, there exists a corresponding interval D y,j which in turn is also E2 in every E j. When we take the intersection of all the E j’s, we “close E3 etc. in” on the point x (see the diagram on the right), the crucial feature being that every interval E j will contain the point x. Therefore, F will also contain the point x. End of Proof.
Claim 2: F contains all of the points from C, i.e. C ⊂ F. Proof of Claim. Consider a point y ∈ C. If y ∈ C, then there is a set D y, j = (y-1/ j, y+1/ j), where j ∈ R >0. It follows that y belongs to each E j, where E j = ∪i D i, j, and so y belongs to F = ∩ j=1∞ E j, so that y ∈ F. Because y ∈ C ⇒ y ∈ F, then C ⊂ F. End of Proof. Conclusion: Because we have shown that C ⊂ F and that C’ ⊂ F, then C ∪C’ ⊂ F. To complete the proof, we need to show that F ⊂ C∪C’. If z ∈ F, then z belongs to all of the sets E j, where E j = ∪i D i, j. If z belongs to all of the sets E j, then there must be a set D x, j = (x-1/ j, x+ 1/ j), where j ∈ R >0. It follows (by the definition of Di, j and by the proof of claims 1 and 2 above) that z ∈ C∪C’, and so z ∈ F ⇒ z ∈ C∪C’, so that F ⊂ C∪C’. QED. Conclusion: we have expressed an arbitrary closed set C as the intersection of a countable collection of open sets, namely the set F. QED.
3-9. Show that the interior of a set S in En is an open set. Answer: Let I be the set of all interior points from S, i.e. (by definition) I is the interior of S. For a set A to be an open set, all the points in A must be interior points. Because all the points in our set I are interior points of S, it follows that I is an open set, and therefore the interior of a set S in E n is an open set. QED. 3-12. If S is a set in E n, let S’ denote the set of accumulation points of S, and let S = S∪S’. (The set S’ is called the derived set of S and S is called the closure of S.) Show that (a) S’ is a closed set, that is, (S’)’ ⊂ S’. (b) If S ⊂ T, then S’ ⊂ T’. (c) (S∪T)’ = S’∪T’. (d) (S)’ = S’. (e) S is a closed set. (f) S is closed if, and only if, S = S. Answer: (a) If x ∈ E n and x ∈ (S’)’, then x is an accumulation point of the set S’. If x is an accumulation point of the set S’, then every neighbourhood N ε(x) of x contains at least one other point y ∈ S’ (y ≠ x), i.e. another accumulation point of S. It follows that all neighbourhoods N(y) of y will contain a point z ∈ S. Pick a N Ε (y) y neighbourhood NΕ (y) such that NΕ (y) ⊂ N ε(x) and x ∉ NΕ (y). Note that in order to do this, we must have Ε < max(dist(x,y), ε-dist(x,y)). So as x N ε(x) every neighbourhood N(x) of x contains a point z ∈ S (with z ≠ x), then it follows that x must be an accumulation point of the set S, so that x ∈ (S’)’ ⇒ x ∈ S’, and so (S’)’ ⊂ S’ as required. QED.
(b) If S ⊂ T, then x ∈ S ⇒ x ∈ T. Now if y ⊂ S’, then y is an accumulation point of the set S. But if y is an accumulation point of S, then either y ∈ S, which implies that y ∈ T, and so y is also an accumulation point of the set T; or y ∉ S, but every neighbourhood of y contains a point from z from S, which is also (because S ⊂ T) a point from T, so that every neighbourhood of y contains a point z from T, and so y is an accumulation point of the set T. Bringing the two cases together, we see that if y ∈ S’, then y ∈ T’, and so S’ ⊂ T’. QED. (c) To prove that (S∪T)’ = S’∪T’, we must prove that (i) (S ∪T)’ ⊂ S’ ∪T’, and that (ii) S’∪T’ ⊂ (S∪T)’. (i) If x ∈ (S∪T)’, then x is an accumulation point of the set S ∪T, which implies that either (1) x is an accumulation point of S, or (2) an accumulation point of T, or (3) an accumulation point of both S and T; which implies that either (1) x ∈ S’ ⊂ S’ ∪T’, or (2) x ∈ T’ ⊂ S’∪T’, or (3) x ∈ S’∩T’ ⊂ S’∪T’. In all cases, x ∈ (S∪T)’ ⇒ x ∈ S’∪T’, so that (S∪T)’ ⊂ S’ ∪T’. (ii) If x ∈ S’ ∪T’, then either x ∈ S’ ⊂ (S∪T)’, or x ∈ T’ ⊂ (S ∪T)’, so that x ∈ S’∪T’ ⇒ x ∈ (S∪T)’, and so S’∪T’ ⊂ (S∪T)’. QED. (d) (S)’ = (S∪S’)’ = (from (c)) = S’∪(S’)’ = S’∪T (where T ⊂ S’) = S’. QED. (e) To be a closed set, S must contain all its accumulation points. From (d), we know that the set of the accumulation points of S is the set S’. But S’ ⊂ S as S = (S∪S’). So as S contains all its accumulation points, we conclude that Sis a closed set. QED.
(f) If. If S = S, then either S’ = φ, i.e. S contains no accumulation points, and so S is closed by definition; or S’ consists of points from S, i.e. S’ ⊂ S, so that again S contains all its accumulation points, and is again a closed set by definition. Only If. If S is closed, then it contains all its accumulation points. Therefore, S’ will consist of elements from S, so that S’ ⊂ S, and so S = (S∪S’) = (using S’ ⊂ S) = S in this case. QED. 3-15. The collection F of intervals of the form 1/n < x < 2/n, (n = 2, 3, 4, ...), is an open covering of the interval 0 < x < 1. Show (without using Theorem 3-40) that no finite subcollection of F covers the interval 0 < x < 1. Answer: Let us first consider some elements from F. n = 2: we have the interval (½, 1). n = 3: we have the interval ( 1/3, 2/3), etc. Let us denote by G any finite subcollection of F. Because G is a finite subcollection, it will have a finite number of elements, say g elements. If we order the elements in G with respect to the number n, then it follows that we can write G in the following form:
G = {(1/n1, 2/n1), (1/n2, 2/n2), ..., (1/n g, 2/ng)}, where n1 < n2 < ... < n g. We must now show that G does not cover the interval 0 < x < 1. To do this, it is sufficient to show that G does not cover the interval (0, 1/n g], where (0, 1/ng] ⊂ (0, 1), and ng ∈ R (n g < ∞). To cover at least part of this interval, the set G must contain elements of the form (1/α, 2/α), where α > ng. But G (by construction) has no elements of this type, so that the set G covers no part of the interval (0, 1/n g]. We therefore conclude that the set G does not fully cover the interval (0, 1). QED. (Note: a quicker proof is to show that 1/n g ∈ (0, 1) but 1/n g ∉ ∪G.) 3-19. Assume that S ⊂ En. A point x in E n is said to be a condensation point of S if every neighbourhood N(x) has the property that N( x)∩S is not countable. Show that if S is not countable, then there exists a point x in S such that x is a condensation point of S. S S1 S2 S3
Sn
Answer : Divide the set S up into n (possibly partially overlapping) sets S i, where n is a finite number; i = 1, 2, ..., n; and ∪i Si = S. If the set S is not countable, then at least one of these ‘regions’ will have to be uncountable as well, or else we will have a finite collection of countable sets and thus S will be countable — and we will have a contradiction.
Choose one of the sets Si which is uncountable, and repeat the above with this set (i.e. divide it up into a finite collection of smaller sets, at least one of which will be uncountable). If we continue to do this, then this process will converge onto a point z ∈ S which will be the condensation point we are looking for. Justification: with every subdivision, we find a smaller and smaller “region” which is uncountable. In the limit of this process, every neighbourhood of the point z we converge onto will contain one of the regions which we know to be uncountable (i.e. some region S j ⊂ N(z)). Therefore, every neighbourhood of this point z has the property that N(z)∩S is not countable, i.e. z is a condensation point . Conclusion: given an uncountable set S, we can always find a condensation point z ∈ S. QED.
3-20. Assume that S ⊂ En and assume that S is not countable. Let T denote the set of condensation points of S. Show that
(a) S—T is countable, (c) T is a closed set,
(b) S∩T is not countable, (d) T contains no isolated points.
Note that Exercise 3-19 is a very special case of (b). Answer: (a) If S—T was not countable, then by 3-19 it would contain a condensation point. But if T is the set of condensation points of S, then S—T contains no condensation points. It follows (by contradiction) that S—T is countable. QED
(b) The set S∩T is the set of condensation points which are in S. We must therefore prove that the set of condensation points in S is not countable. To do this, we use three pieces of information: (1) From set theory, we know that S ∩T = S—Tc; (2) We know that S is not countable; and (3) We know that S—T is countable. If S—T is countable, and if S is not countable, then S—Tc must be non countable. Using (1), it follows that S ∩T must also be non countable. QED. (c) By definition, T is closed if it contains all its accumulation points. We know that every point x ∈ T is a condensation point, so that every neighbourhood N(x) of a point x ∈ T is not countable (more precisely, N(x)∩S is not countable). This implies (by Exercise 3-19) that the neighbourhood contains another condensation point y ∈ T (N(x)∩S uncountable ⇒ ∃ a condensation point y ∈ N(x)∩S), so that the point x is an accumulation point of the set T. This applies for all condensation points, so that every point x ∈ T is an accumulation point. It remains to show that T has no accumulation points that are not in T. Assume that z is an accumulation point of the set T — with z ∉ T. Because z is not a condensation point, it follows that not all neighbourhoods N( z) are uncountable. Pick one such neighbourhood, say N1(z). By the definition of an accumulation point, N 1(z) contains an element z1 ∈ T. But the neighbourhoods of z1 are uncountable (by the definition of a condensation point). Pick a neighbourhood of z1 which is contained in N 1(z), say the neighbourhood N2(z1) (We can always do this — see the answer to N 2(z 1) z1 exercise 3-12, part (a)). Because N2(z1) is uncountable, N1(z) cannot z N 1(z) possibly be countable, contradicting our assumption that N 1(z) was countable. It follows that all the neighbourhoods of the point z must be uncountable, and so we must have z ∈ T, and so T has no accumulation points that are not in T . QED (d) Because we concluded in (c) that every point in T is an accumulation point, it follows that no point in T is an isolated point, i.e. not an accumulation point. QED.
3-21. A set S is said to be perfect if S = S’, that is, if S is a closed set which contains no isolated points. (Note: S’ is the set of accumulation points of S). Show that if F is an uncountable closed set in En, then F can be expressed in the form F = A ∪B, where A is a perfect set and B is a countable set (Cantor-Bendixon theorem). [Hint: Use Exercise 3-20.]. Answer: Let T be the set of condensation points of S. It follows that T ∩S is the set of condensation points in S, and that S—T is the set of points in S which are not condensation points. We can therefore write S = (T∩S)∪(S—T), i.e. S is the union of all the condensation points of S together with the points of S which are not condensation points. From 3-20, we know that all the points in T ∩S are accumulation points, i.e. we have T ∩S = (T∩S)’. It follows that T∩S is a perfect set. Also from 3-20, we know that S—T is countable. Therefore, S = (T∩S)∪(S—T) is the required form, where A = T ∩S is a perfect set, and B = S—T is countable. QED.
Possible Further Work / Evaluation Exercise 3-19: I could make a slight adjustment in the answer due to the comment in Draft 2 — to divide the set S up into a countable collection of sets instead of into a finite collection of sets. Adjusted answer: Answer: Divide the set S up into a countable collection of (possibly partially overlapping) sets Si, so that ∪i Si = S. If the set S is not countable, then at least one of these ‘regions’ will have to be uncountable as well, or else we will have a countable collection of countable sets and thus S will be countable — and we will have a contradiction.
Choose one of the sets Si which is uncountable, and repeat the above with this set (i.e. divide it up into a countable collection of smaller sets, at least one of which will be uncountable). If we continue to do this, then this process will converge onto a point z ∈ S which will be the condensation point we are looking for. Justification: with every subdivision, we find a smaller and smaller “region” which is uncountable. In the limit of this process, every neighbourhood of the point z we converge onto will contain one of the regions which we know to be uncountable (i.e. some region S j ⊂ N( z)). Therefore, every neighbourhood of this point z has the property that N( z)∩S is not countable, i.e. z is a condensation point . Conclusion: given an uncountable set S, we can always find a condensation point z ∈ S. QED.
Chapter 4
Chapter 4: The Limit Concept and Continuity: Notes Let us begin with the equation lim x x 0xn = A, which is meant to convey the idea that when x is sufficiently near to x0, then f(x) will be as near to A as desired . The statements “x sufficiently near to x0” and “f(x) will be as near to A as desired” are made mathematically precise by the following type of definition: the symbolism xlim x0 f ( x) = Ameans that for every number ε > 0, there is another number δ > 0 such that whenever 0 < |x-x0| < δ, then |f(x)-A| < ε. d
d
lim
We write n xn = A to mean that for every ε > 0, there is an integer N such that whenever n > N, then |x n-A| < ε. dº
If we use neighbourhood terminology, we see that both above definitions involve the same principle. To say that 0 < |x-x0| < δ means that x is in a neighbourhood of x 0, but x ≠ x 0. That is, x is in a deleted neighbourhood N’(x0). Also, to say that n > N is the same as saying that n is in a deleted neighbourhood of +∞. The parts of the definitions concerned with ε can also be rephrased in neighbourhood terminology. For example, the inequality |f(x)-A| < ε states that f(x) ∈ N(A; ε) and, similarly, |xn-A| < ε means that x n ∈ N(A; ε). The basic principle involved in both definitions of limit is that for every neighbourhood N(A) there must exist a neighbourhood N(x 0) such that x ∈ N’(x0) implies f(x) ∈ N(A). If f is a real-valued function defined at a point x = (x 1, ..., x m) in Em, we use either f(x 1, ..., xm) or f(x) to denote the value of f at that point. If we have several real-valued functions, say f 1, ..., f k defined on a common subset S of Em, it is extremely convenient to introduce a vector-valued function f , defined by the equation f (x) = (f 1(x), ..., f k( x)), if x ∈ S. Our definition of limit now assumes the following form: Let f be a function defined on a set S in E m and let the range of f be a subset T of Ek . If a is an accumulation point of S and if b ∈ Ek , then the symbolism lim f (x) = b is defined to mean the following : For every neighbourhood N(b) ⊂ Ek , there exists a neighbourhood N(a) ⊂ Em such that x ∈ N’(a)∩S implies f (x) ∈ N(b). x da
Note 1: we write x ∈ N’(a)∩S rather than x ∈ N’( a) in order to make certain that x is in the domain of f . Also, we require that a be an accumulation point of S in order to make certain that the intersection N’(a)∩S will never be the empty set. Note 2: If lim x→a f (x) exists ( finite or infinite), its value is uniquely determined. Strictly speaking, we should somehow indicate the fact that the limit just defined depends on the set S through which x is allowed to range. This will usually be clear from the lim f (x) = b to emphasise the fact more explicitly. An context but, if necessary, we will write x da
S
x `
important special case of this occurs when S is an interval in E 1 having a as its left endpoint. We then write lim f (x) = lim+ f (x) = b, and b is called the right-hand limit at a. (left-hand limits x da
S
x `
are similarly defined).
x da
Theorem 4-4. Let a be an accumulation point of a set S in E m. Then there exists an infinite sequence {xn} whose terms are distinct points of S, such that limn→∞ xn = a. Note: Suppose a sequence {xn} has a limit, say a = limn→∞ xn, and let S = {x1, x2, ...} denote the range of the sequence. If S is infinite, it follows at once from the definition of limit that a is an accumulation point of S. The above theorem tells us that S can have no further accumulation points. Therefore, a sequence {xn} whose range S is infinite has a limit if, and only if, S has exactly one accumulation point, in which case the accumulation point is also the limit of the sequence.
Theorem 4-5. Let f be defined on a set S in Em with function values in E k , and let a be an accumulation point of S. Let { xn} be an infinite sequence whose terms are points of S, such lim that each term xn ≠ a but such that ndº xn = a. Then we have (i) If limx→a f (x) = b, then limn→∞ f (xn) = b; (ii) Conversely, if for each such sequence we know that limn→∞ f (xn) exists, then all these sequences have the same limit (call it y) and also limx→a f (x) exists and equals y. In the definition of the statement limx→a f (x) = b, it was assumed that the limiting value b was given. The following theorem gives us a condition (called the Cauchy condition) which enables us to determine, without knowing its value in advance, whether such a point b exists. Theorem 4-6 (Cauchy condition for sequences). Let {xn} be an infinite sequence whose terms are points in E k . There exists a point y in Ek such that limn→∞ xn = y if, and only if, the following condition is satisfied: For every ε > 0, there exists an integer N such that n > N and m > N implies |xn-xm| < ε. Theorem 4-7 (Cauchy condition for functions). Let f be defined on a set S in E m, the function values being in Ek . Let a be an accumulation point of S. There exists a point b in Ek such that limx→a f (x) = b if, and only if, the following condition holds: For every ε > 0, there is a neighbourhood N(a) such that x and y in N’(a)∩S implies |f(x)-f(y)| < ε. Let f and g be two functions, each defined on a set S in E n, with function values in E 1 or lim lim in E2. Let a be an accumulation point of S and assume we have d f(x) = A, d g(x) = B. Then lim lim lim we also have (i) d [f(x)±g(x)] = A±B, (ii) d f(x)g(x) = AB, (iii) d f(x)/g(x) = A/B if B ≠ 0. x
x
a
x
a
a
x
x
a
a
Continuity. Definition 4-9. Let f be defined on a set S in E n with function values in E m, and let a be an accumulation point of S. We say that f is continuous at the point a provided that (i) f is defined at a, (ii) limx→a f (x) = f (a). If a is not an accumulation point of S, we say f is continuous at a provided only (i) holds. If f is continuous at every point of S, we say f is continuous on the set S.
It is convenient to note that whenever f is continuous at a, we can write part (ii) of the lim lim above definition as follows: d f(x) = f( d x). Thus, when we deal with continuous functions, the limit symbol may be interchanged with the function symbol. Observe also that continuity at a means that for every neighbourhood N(f (a)), there exists a neighbourhood N(a) such that f[N(a)∩S] ⊂ N[f (a)]. x
a
x
a
Theorem 4-10. Let f and g be continuous at the point a in En and assume that f and g have function values in E1 or E2. Then f+g, f-g, and f•g are each continuous at a. The quotient f/g is also continuous at a, provided that g(a) ≠ 0. Note that the product f•g should not be confused with the composition fg, defined by fg(x) = f[g(x)]. Theorem 4-11. Let f be a function defined on a set S in En and assume f (S) ⊂ Em. Let g be defined on f (S) with values in E k , and let gf be the composite function defined on S by the equation gf (x) = g[f (x)]. Suppose that we have (i) The point a is an accumulation point of S and f is continuous at a; (ii) The point f (a) is an accumulation point of f (S) and g is continuous at f (a). Then the composite function gf is also continuous at a; that is, lim (x)] = g[f (a)]. d g[ f x
a
Examples of continuous functions. In E2, constant functions (f(x) = c), the identity function (f(x) = x) and hence all polynomial functions are continuous. A rational function g/f is continuous whenever the denominator does not vanish. The familiar real-valued functions of elementary calculus such as exponential, trigonometric and logarithmic functions, are continuous whenever they are defined. Functions continuous on open or closed sets . Definition 4-12. Let f be a function whose domain is S and whose range is T, and let Y be a subset of T. By the inverse image of Y under f , denoted by f -1(Y), we shall mean the largest subset of S which f maps onto Y, that is to say, f -1(Y) = {x | x ∈ S, f(x) ∈ Y}.
Note: If f has an inverse function f -1, the inverse image of Y under f is the same as the image of Y under f -1, and in the case there is no ambiguity in the notation f -1(Y). Theorem 4-13. Let f be a function with domain S and range T. Assume X ⊂ S and Y ⊂ T. Then we have (i) If X = f -1(Y), then Y = f (X), (ii) If Y = f (X), then X ⊂ f -1(Y). It should be observed that, in general, we cannot conclude that Y = f (X) implies X = f -1(Y). Note that the statements in this theorem can also be expressed as follows: Y = f [f -1(Y)], X ⊂ f -1[f (X)]. Theorem 4-14. Let f be a function which is continuous on a closed set S in E m, and let the range of f be a set T in E k . Then if Y is a closed subset of T, the inverse image f -1(Y) will be a closed subset of S. Theorem 4-15. Let f be a function which is continuous on an open set S in E m, and let the range of f be a set T in E k . Then if Y is an open subset of T, the inverse image f -1(Y) will be an open subset of S. We have already seen that the image of an open set under a continuous mapping is not always open. The example in E 1, defined by the equation f(x) = tan -1(x), shows that the image of a closed set under a continuous mapping need not be closed, since f maps E 1 onto the open interval (-π/2, π/2). However, the image of a compact set under a continuous mapping is always compact. This will be proved in the next theorem.
Functions continuous on compact sets. Theorem 4-16. Let f be a function which is continuous on a compact set S in E m, and assume that f (S) ⊂ Ek . Then f (S) is a compact set.
Theorem 4-17. Let f be a function which is continuous on a compact set S in E m, and assume that f (S) ⊂ E k . Suppose further that f is one-to-one on S so that the inverse function f -1 exists. Then f -1 is continuous on f(S). Topological mappings. Definition 4-18. Let f be continuous on S and assume that f is one-to-one on S so that the inverse function f -1 exists. Then f is called a topological mapping or a homeomorphism if, in addition, f -1 is continuous on f (S). Note: A property of a set which remains invariant under every topological mapping is called a topological property. Properties of real-valued continuous functions. Definition 4-19. Let f be a real-valued function defined on a set S in E n. Then f is said to have an absolute maximum on the set S if there exists a point a in S such that f( x) ≤ f(a), for all x in S. If a ∈ S and if there is a neighbourhood N(a) such that f(x) ≤ f( a), for all x in N(a)∩S, then f is said to have a relative maximum at the point a. Absolute minimum and relative minimum are similarly defined, using f(x) > f(a).
Theorem 4-20. Let f be a real-valued function which is continuous on a compact set S in En. Then f has an absolute maximum and an absolute minimum on S. Theorem 4-21 (Bolzano). Let f be real-valued and continuous on a closed interval [a, b] in E1, and suppose that f(a) and f(b) have different signs; that is, assume f(a)f(b) < 0. Then there is at least one point x in the open interval (a, b) such that f(x) = 0. An immediate consequence of the above theorem is the intermediate value theorem for continuous functions (Theorem 4-22): If f is real-valued and continuous on a closed interval S in E1, then f assumes every value between its maximum, sup f(S), and its minimum, inf f(S). Uniform Continuity. Suppose we have a function f continuous on a set S, so that for each accumulation point a of S, we have limx→a f (x) = f (a). In the ε and δ terminology, this means that, given the accumulation point a and given ε, there is a number δ > 0 (depending on a and on ε) such that 0 < |x-a| < δ implies |f (x)-f (a)| < ε. In general, we cannot expect that for a fixed ε the same value of δ will serve equally well for every such point a. This might happen, however; when it does, the function is called uniformly continuous on the set S.
Definition 4-23. Let f be defined on a set S in E n with function values in Em. Then f is said to be uniformly continuous on S if the following statement holds: For every ε > 0, there exists a δ > 0 (depending only on ε) such that if x ∈ S and y ∈ S and |x-y| < δ, then |f (x)-f (y)| < ε. Observe that uniform continuity is a property of the whole set S, that is, a global property. It follows from the definition that uniform continuity on a set implies continuity on the set. The converse of this statement is also true when the set is compact.
Theorem 4-24 ( Heine). Let S be a compact set in En. If f is continuous on S, then f is also uniformly continuous on S. Discontinuities of real-valued functions. Definition 4-25. If f is a real-valued function defined on an open interval (a, b) in E 1 and if c is a point in [a, b), we write f(c+) to denote the right hand limit, limx→c+ f(x), whenever the limit exists. If c ∈ (a, b], we define f(c-) = lim x→cf(x).
It is clear that we have f(x+) = f(x-) = f(x) if, and only if, f is continuous at x. If f is defined at x, then x can be a discontinuity of f only under one of the following conditions: (a) If either f(x+) or f(x-) does not exist; (b) If both f(x+) and f(x-) exist but have different values; (c) If f(x+) = f(x-) ≠ f(x). Definition 4-26. Let f be a real-valued function defined on an interval [a, b]. If f(x+) and f(x-) both exist at some interior point x, then the difference f(x)-f(x-) is called the left-hand jump of f at x, the difference f(x+)-f(x) is called the right-hand jump of f at x, and their sum, f(x+)-f(x-), is called the jump at x. If any one of these three numbers is different from 0, then x is called a jump discontinuity of f. (For the endpoints a and b, only one-sided jumps are considered.) We say that f is continuous from the right at x if the right-hand jump is equal to 0, whereas f is discontinuous from the right if either f(x+) does not exist or the right-hand jump is ≠ 0. (Similar terminology applies from the left.) In the case that f(x+) = f(x-) ≠ f(x), the point x is also said to be a removable discontinuity, since the discontinuity can be removed by redefining f at x to have the value f(x+) = f(x-). If either limx→c+ f(x) or limx→c- f(x) is infinite (+∞ or - ∞), then f is sometimes said to have an infinite discontinuity at c. Example: The function f defined by f(x) = sin(1/x) if x ≠ 0, f(0) = A, has a discontinuity at x = 0, regardless of the value of A. In this case, neither f(0+) nor f(0-) exists. Monotonic Functions. Definition 4-27. Let f be a real-valued function defined on a set S in E1. If, for every pair of points x and y in S, x < y implies f(x) ≤ f(y), then f is said to be increasing (or nondecreasing) on S. If x < y implies f(x) < f(y), then f is said to be strictly increasing on S. (Decreasing functions are similarly defined.) A function is called monotonic on S if it is increasing on S or decreasing on S. If f is an increasing function, then -f is a decreasing function.
Theorem 4-28. If f is increasing on [a, b], then f(x0+) and f(x0-) both exist for each x0 in (a, b) and we have f(x 0-) ≤ f(x0) ≤ f(x0+). At the endpoints, we have f(a) ≤ f(a+) and f(b-) ≤ f(b). Theorem 4-29. Let f be strictly increasing on [a, b]. Then we have: (i) The inverse function f -1 exists and is strictly increasing on its domain; (ii) If f is continuous on [a, b], then f -1 is continuous on [f(a), f(b)]. This theorem tells us that a continuous, strictly increasing function is a topological mapping. Conversely, every topological mapping of an interval [a, b] onto an interval [c, d] must be a strictly monotonic function.
Necessary and sufficient conditions for continuity . Continuous functions can be characterised in a very elegant fashion if we introduce a slight generalisation of the notions of open and closed sets.
Definition 4-30. Let A and B be two sets in E n, with A ⊂ B. We say that A is closed relative to B if those accumulation points of A which lie in B are also in A. We say that A is open relative to B if the complement B—A is closed relative to B. It is clear that every closed set in En is closed relative to E n, and every open set in En is open relative to En. It is also clear that every set is both open and closed relative to itself . A set which is open relative to B need not, of course, be an open set open set S in E 2 (relative to E n). However, such a set must be the intersection of a B with an open set. This statement, which will form the basis of b B the following theorem, is illustrated as shown. The shaded c region (which includes the arc abc, except for the endpoints a B∩S open relative to B and c) is open relative to B. Theorem 4-31. Assume that A ⊂ B ⊂ En. Then A is open relative to B if, and only if, A is the intersection of B with a set which is open relative to E n. Note that this theorem also holds with “open” replaced by “closed ” throughout. Using these concepts, we can give the following characterisation of continuous functions: Theorem 4-32. Let f be a function defined on a set S in E n, and let T = f (S) be a subset of Em. Then the following three statements are equivalent: (a) The function f is continuous on S; (b) If Y is closed relative to T, then f -1(Y) is closed relative to S; (c) If Y is open relative to T, then f -1(Y) is open relative to S.
Chapter 4: Selected Exercises 4-1. Prove statements (i) and (iii) of Theorem 4-8. Answer: Let us first remind ourselves of the statements that we are trying to prove.
Theorem 4-8. Let f and g be two functions, each defined on a set S in En, with function values in E 1 or in E2. Let a be an accumulation point of S and assume that we have lim d f(x) = A lim lim lim and d g(x) = B. Then we also have (i) d [f( x) ± g(x)] = A ± B, (ii) d f(x)g(x) = AB, and lim (iii) d f(x)/g(x) = A/B if B ≠ 0. x
x
a
x
a
x
a
x
a
a
Proof. (i) Let an arbitrary ε > 0 be given and let ε’ be a second number (ε’ > 0) which will be made to depend on ε in a way to be described later. Choose a neighbourhood N(a) such that if x ∈ N’(a)∩S, then we have both |f(x)-A| < ε’ and |g(x)-B| < ε’. Now |f(x)+g(x) - (A+B)| = |f(x)-A + g(x)-B| ≤ |f(x)-A| + |g(x)-B| (by the triangle rule) < ε’ + ε’ = 2ε’. And |f(x)-g(x) - (A-B)| = |f(x)-A - (g(x)-B)| ≤ |f(x)-A| + |g(x)-B| (using |α-β| ≤ |α|+|β|) < ε’ + ε’ = 2ε’. Conclusion: |f(x)±g(x) - (A±B)| < 2ε’, and choosing ε = ε’/2, we see that |f(x)±g(x) - (A±B)| < ε as required . QED. (iii) Let an arbitrary ε > 0 be given and let ε’ be a second number (ε’ > 0) which will be made to depend on ε in a way to be described later. Choose a neighbourhood N( a) such that if x ∈ N’(a)∩S, then we have both |f(x)-A| < ε’ and |g(x)-B| < ε’. In this question, the conclusion we want is to show that | f(x)/g(x) - A/B| < ε whenever x ∈ N’(a)∩S. f ( x )
f ( x ) B− AB+ AB− Ag ( x )+ AB− AB | Bg ( x ) − A( g ( x )− B) − Ae  Be  | Bg ( x ) | | Bg ( x ) | | Bg ( x ) |.
A
Now | g ( ) − B | = | x
[
B( f ( x )− A) Bg ( x ) |
|
+
B( f ( x )− A)+ AB− A( g ( x )− B)− AB | Bg ( x )
= |
<
B( f ( x )− A)− A( g ( x )− B) | Bg ( x )
= |
+
Now |g(x)| = |B+g(x)-B| < |B|+ε’ < |B|+1, so that Be Â
− Ae Â
| Bg ( ) | + | Bg ( ) | = x
x
|Be  | |B|| g ( x )|
+
|− Ae  | |B|| g ( x )|
<
|e  | |B|+1
+
|− Ae  | |B|( |B|+1)
=
e
Â
B| | +1 +
|A|e  |B|( |B|+1)
=
Â
( B| | + |A|) |B|( |B|+1) .
e
Choosing ε’ = (ε(|B|(|B|+1)))/(|B|+|A|), we see that we have |f(x)/g(x) - A/B| < ε whenever x ∈ N’(a)∩S, and this proves (iii). QED.
4-3. Let f be defined on the interval (a, b) and assume x ∈ (a, b). Consider the two statements lim (i) hd0 |f(x+h)-f(x)| = 0. lim (ii) hd0 |f(x+h)-f(x-h)| = 0.
(a) (b)
Show that (i) always implies (ii). Give an example in which (ii) holds but (i) does not hold.
Answer: (a) Consider an arbitrary point x ∈ (a, b). If (i) holds, then hlim |f(x+h)-f(x)| = 0. d0 In other words, given an arbitrary ε > 0, there exists a δ > 0 such that if |h| < δ, then |f(x+h)-f(x)| < ε/2 (---(1)). Similarly, if we replace h by -h (which holds since |-h| = |h|), then if |-h| < δ (which is equivalent to the condition |h| < δ), we have |f(x-h)-f(x)| < ε/2 (---(2)).
Adding equations (1) and (2) together, we get |f(x+h)-f(x)| + |f(x-h)-f(x)| < ε, |f(x+h)-f(x)+f(x-h)-f(x)| < ε, (using the triangle inequality, |A+B| ≤ |A|+|B|) |f(x+h)+f(x-h)| < ε. Therefore, given an arbitrary ε > 0, there exists a δ > 0 such that if |h| < δ, then |f(x+h)+f(x-h)| < ε, i.e. hlim |f(x+h)-f(x-h)| = 0 as required . QED. d0 (b) Let us define a function f to be given by the following : f(x) = x² if x ≠ 0, and f(x) = 1 if x = 0 (with x ∈ E1). At x = 0, hlim |f(x+h)-f(x-h)| = |0-0| = 0, but hlim |f(x+h)-f(x)| = |0-1| = 1. d0 d0 4-5. If x is any real number in [0, 1], show that the following limit exists: lim lim 2n mdº [ ndº cos (m!o x)]. and that its value is 0 or 1, according to whether x is irrational or rational . Answer: When n → ∞ in the inner limit, we will have an ‘infinite power’ of cos(m!πx). For any m > 2, m!π will always be an even number of π’s, so that it is of the form 2k π for some integer k > 0. Recall that cos(2k π) = 1 for all k > 0. If x is rational, say x = a/ b, where a, b ∈ R , then m!πx = m!πa/b. If m is large enough, then b will be a factor of m! (i.e. we will have m! = 1×2×...×b×...×m), and thus the b’s cancel in the numerator and the denominator, leaving us with A = απa, where α = m!/b. A will still consist of an even number of π’s (a large m! will still have an even factor — lots of them in fact!), so that cos(A) = 1 for large m.
Therefore, when x is rational and when m is large enough, our expression consists of an lim ‘infinite power’ of 1’s which will still be 1, so that we have ndº cos 2n (m!o x)= 1 provided that m m m is large enough. In other words, mdº cos 2n (m!o x)] = 1 when x is rational. Conclusion: [ ndº the limit in question exists when x is rational, and it has the value 1. If x is irrational, then m!x cannot possibly be an even or an odd number (no irrational number is ever even or odd). It follows that we must have |cos(m!πx)| ≠ 1 in this case (for any m), and so it follows that cos(m! πx) ∈ (-1, 1).
lim
Claim: if y ∈ (-1, 1), then ndº yn = 0. Proof of Claim. We need to find an N such that if n > N, then for an arbitrary ε > 0 and a particular y ∈ (-1, 1) we have |y|n < ε (We need the modulus signs in because if we don’t put them in, then for any ε > 0 we will always have y n < ε for any y ∈ (-1, 0) and any odd n. We will therefore prove a stronger condition than the one we need). But |y|n < ε happens exactly when nlog|y| < log(ε), or when n > log( ε)/log|y|. Therefore, if we choose N = log(ε)/log|y|, then we will always have |y| n < ε for all n > N. End of Proof. n Therefore, because for any y ∈ (-1, 1) we have nlim dºy = 0, it follows that for any lim cos(m!πx) ∈ (-1, 1) we have ndº cos2n(m!πx) = 0. This will hold for any m and so we have lim lim 2n mdº [ ndº cos (m!o x)] = 0 when x is irrational. Conclusion: the limit in question exists when x is irrational, and it is given by zero. QED.
4-7. Let f be continuous on [a, b] and let f(x) = 0 when x is rational . Show that f(x) = 0 for every x in [a, b]. Answer: If f is continuous on [a, b], then f is continuous at every point in [a, b]. We know that f(x) = 0 when x is rational . But every irrational number y is an accumulation point of the set of rational numbers, so by definition 4-9, and by knowing that f is continuous at the point y, it follows that (i) f is defined at y, and (ii) lim x→y f(x) = f(y).
We now know that f(y) = limx→y f(x) for any irrational point y, but what is this limit? It must be zero, as in every neighbourhood N(y), because y is an accumulation point, there is a rational number z such that f(z) = 0 and so f(x) ∈ N(0), thus satisfying the definition for a limit. Conclusion: f(x) = 0 for all x ∈ [a, b]. 4-8. Let f be continuous at the point a = (a1, a 2, ..., a n) in En. Keep a2, a 3, ..., a n fixed and define a new function g of one real variable by the equation g(x) = f(x, a2, ..., a n). Show that g is continuous at the point x = a 1. (This is sometimes stated as follows: “ A continuous function of n variables is continuous in each variable separately.”) Answer: Let us define a new function s: E 1 → En by the following definition: s(x) = (x, a2, a3, ..., a n) for an arbitrary x ∈ E1 and fixed a2, a3, ..., a n ∈ E1.
Looking at the above definition and at the question itself, we see that we can now write g(x) as follows: g(x) = f(s(x)) for an arbitrary x ∈ E1. Looking at Theorem 4-11, we see that in order to prove that g(x) is continuous at the point a = (a 1, a 2, ..., a n) in En, we need to show that (i) the arbitrary point x ∈ E1 is an accumulation point of E1, (ii) the function s(x) is continuous at x, (iii) The point s(x) is an accumulation point of En, and (iv) f is continuous at the point s(x). If we can show that the four conditions above hold, then we can say that g is continuous at the point x = a. (i) We know that any real number is an accumulation point of the real line E 1, so that condition (i) is satisfied for an arbitrary real number x = a 1.
(ii) To prove that the function s(x) is continuous at an arbitrary point x ∈ E 1, we need to lim show that the function is defined at x and that ad x s(a) = s( x). The first part follows immediately because the function s(x) is defined for all real numbers x ∈ E1 by its definition on the previous page. It remains to show that we have alim x). d x s(a) = s( lim
lim
lim
lim
lim
But ad x s(a) = ad x (a, a 2 , a 3 , ..., a n ) = ( ad x (a), ad x (a 2 ),..., ad x (a n )) = ( x, a 2 ,..., a n ) = s(x). We can therefore say that condition (ii) holds. For condition (iii) to hold, we must show that the point s(x) is an accumulation point of the set E n. But by the same argument as in condition (i), as s(x) is a point in E n, and as all points in E n are accumulation points, then s(x) = (x, a 2, ..., an) is an accumulation point of the set E n, satisfying condition (iii). Finally, we must show that f is continuous at the point s(x). But if we let x = a 1, then s(a1) = (a1, a2, ..., a n) = a, and we know by the assumption made in the question that f is continuous at the point a in En. Therefore, condition (iv) is satisfied, and thus we have shown that all four conditions hold, and can subsequently say that the function g is continuous at the point x = a1. QED. 4-11. For each x in [0, 1], let f(x) = x if x is rational , and let f(x) = 1-x if x is irrational . Show that (a) f is continuous only at the point x = ½; (b) f assumes every value between 0 and 1. Answer: Let us first prove that f is continuous at the point x = ½. If x = ½, then x is rational and so f(x) = ½. To show that f is continuous at this point, then for every arbitrary ε > 0, we need to find a δ > 0 such that if |x-y| < δ, with y ∈ [0, 1], then |f(x)-f(y)| < ε. Now for any arbitrary δ > 0, the deleted neighbourhood N’(½) ∩[0, 1] will always contain irrational and rational numbers from [0, 1]. If y is a rational number in N’(½)∩[0, 1], then |f(x)-f(y)| = |½-y|. If we choose δ = ε, then we want to show that if |x-y| < δ = ε, then |f(x)-f(y)| < ε. But in this case, |x-y| = |½-y| = |f(x)-f(y)|, so if |x-y| < δ = ε, then we will always have |f(x)-f(y)| < ε.
Now if y is an irrational number, then |f(x)-f(y)| = |½-(1-y)| = |y-½|. But if we choose δ = ε again, as before we need to show that if |x-y| < δ = ε, then |f(x)-f(y)| < ε. But in this case, |x-y| = |½-y| = |y-½| = |f(x)-f(y)|, so that if |x-y| < δ = ε, then we will always have |f(x)-f(y)| < ε. (In the above we used the fact that |A-B| = |B-A|). So we have shown that if x = ½, then f(x) is continuous. Now consider the case when x ≠ ½. If f(x) is continuous for x ≠ ½, then given an arbitrary ε > 0, we need to find a δ > 0 such that if |x-y| < δ, with y ∈ [0, 1] ≠ x, then |f(x)-f(y)| < ε. In order to prove that the above definition cannot be satisfied, it is sufficient to find a single ε1 > 0 such that there is no neighbourhood N’(x)∩[0, 1] in which all the points within δ of x in [0, 1] satisfy |f(x)-f(y)| < ε1. Consider that we fix x to be a number in [0, 1] which is not a half. As stated above, every neighbourhood of this point x will contain both rational and irrational numbers from [0, 1]. Consider the case when x is rational and we therefore have f(x) = x.
Consider also that we define a number ε1 = |2x-1| > 0. Given this ε1 (which will be greater than zero if x is not a half), we need to find a δ > 0 such that if |x-y| < δ for y ∈ [0, 1], then we will always have |f(x)-f(y)| < ε1. Assume that such a δ exists, and let us pick out an element y from the neighbourhood N’(x)∩[0, 1] which is an irrational number. In this case, we have f(x) = x and f(y) = 1-y, with |x-y| < δ. It follows that |f(x)-f(y)| = |x-(1-y)| = |x+y-1|. Now if |x-y| < δ, then y is bounded from above by x+ δ, i.e. y < x+ δ. Therefore, we will have |x+y-1| < |x+(x+δ)-1| = |2x-1+δ|. But if f is continuous at x, then we need |2x-1+ δ| < ε2, i.e. |2x-1+δ| < |2x-1|. There is no possible δ > 0 that we can pick that will satisfy the above inequation. Therefore, we conclude that when x is rational and not equal to a half, if we pick ε1 = |2x-1| > 0, then there is no δ > 0 that will suffice so that if |x-y| < δ, then |f(x)-f(y)| < ε1 for ALL irrational y in this neighbourhood. Conclusion: f(x) is not continuous when x is a rational number in [0, 1] and x ≠ ½. The other case to consider is when x is a fixed irrational number in [0, 1]. Consider now that for this fixed irrational number x, we define a number ε2 = 1. Given this ε2, we need to find a δ > 0 such that if |x-y| < δ for y ∈ [0, 1] ≠ x, then we will always have |f(x)-f(y)| < ε2. Assume that such a δ exists, and let us pick out an element y from the neighbourhood N’(x) ∩[0, 1] which is a rational number. In this case, we have f(x) = 1-x and f(y) = y, with |x-y| < δ. It follows that |f(x)-f(y)| = |1-x+y| = |-x+y+1|. Now if |x-y| < δ, then y is again bounded from above by x+ δ, i.e. y < x+ δ. Therefore, we will have |-x+y+1| < |-x+(x+δ)+1| = |δ+1|. But if f is continuous at x, then we need |δ+1| < ε2, i.e. |δ+1| < 1. But there is no possible δ > 0 that we can pick that will satisfy the above inequation. Therefore, we conclude that when x is irrational, if we pick ε2 = 1, then there is no δ > 0 that will suffice so that if |x-y| < δ, then |f(x)-f(y)| < ε1 for ALL rational y in this neighbourhood. Conclusion: f(x) is not continuous when x is an irrational number in [0, 1]. We have now considered all cases for x in [0, 1]: x = ½, so that f(x) is continuous; and x ≠ ½, so that f(x) is not continuous. QED.
(b) It is a trivial matter to show that f assumes the value of every rational number in the interval [0, 1], as f(x) = x for all rational numbers in [0, 1]. Our only problem occurs in showing that f assumes the value of every irrational number in the interval [0, 1], i.e. if we have an arbitrary irrational number y ∈ [0, 1], then we need to find a number z ∈ [0, 1] such that f(z) = y. If y is our arbitrary irrational number in the interval [0, 1], then 1-y will also be an arbitrary irrational number in this interval. Thus f(1-y) = 1-(1-y) = y (because 1-y is irrational), and thus z = 1-y is the number we are looking for so that f(z) = y for all irrational numbers y ∈ [0, 1]. Conclusion: f assumes every value between 0 and 1. QED. 4-14. Let f be defined and bounded on the closed interval S = [a, b]. (That is, assume that f(S) is a bounded set.) If T is a subset of S, the number Ωf (T) = sup{f(x)-f(y) | x ∈ T, y ∈ T} is called the oscillation of f on T. If x ∈ S, the oscillation of f at x is defined to be the number ωf (x) = hlim Ωf (N(x; h)∩S). Show that this limit always exists and that ωf (x) = 0 if, and only if, d0+ f is continuous at x. Answer: By definition, Ωf (T) is the biggest value of f(x)-f(y) in T (x, y ∈ T). It follows that ωf (x) is the biggest value of f(x)-f(y) in N(x; h) ∩S (x, y ∈ N(x; h)∩S), i.e. in the neighbourhood of x in S. The value ωf (x) is thus determined by two points (say x1 and x2) in N(x; h) ∩S. Because f(S) is bounded (i.e. f(S) is a subset of some finite interval), then the two points in question will have finite values, and so their difference (i.e. f(x 1)-f(x2)) will also be finite. Therefore, for a particular h, Ωf (N(x; h)∩S) will exist and is a finite value.
Does the limit hlim Ωf (N(x; h)∩S) exist? To show that it does exist, all we need do is to d0+ show that if h 2 > h1, then Ωf (N(x; h1)∩S) ≤ Ωf (N(x; h2)∩S). If this is true, then the sequence of numbers Ωf (N(x; h)∩S) given by decreasing h will be a bounded monotonic sequence of finite positive numbers (bounded above by, say, the entry where h = 1), and thus we can apply Theorem 12-6 to say that this sequence converges to a value, say A, and so we can write A=
lim hd0+
Ωf (N(x; h)∩S), where 0 ≤ A ≤ Ωf (N(x; 1)∩S), say.
Claim: If h2 > h1, then Ωf (N(x; h1)∩S) ≤ Ωf (N(x; h2)∩S). Proof of Claim. To start with, let us look at what exactly is sup{f(x1)-f(x2) | x1 ∈ (N(x; h)∩S), x2 ∈ ((x; h)∩S)}? Well, it is the maximum value of f(x1)-f(x2) in the neighbourhood N(x; h)∩S. But if this is so, then f(x 1) must be the highest value of f(x) in the neighbourhood N(x; h) ∩S, and f(x2) must be the smallest value of f(x) in the neighbourhood N(x; h)∩S. Therefore, we have sup{f(x1)-f(x2) | x1 ∈ (N(x; h)∩S), x2 ∈ ((x; h)∩S)} = sup{f(x1) | x1 ∈ ((x; h)∩S)} - inf{f(x2) | x2 ∈ ((x; h)∩S}. The proof centres on the fact that if h 2 > h 1, then sup{f(x) | x ∈ N(x; h1)∩S} ≤ sup{f(x) | x ∈ N(x; h2)∩S}, and inf{f(x) | x ∈ N(x; h1)∩S} > inf{f(x) | x ∈ N(x; h2)∩S}
I will not prove these statements (intuitively, if we are considering a smaller neighbourhood, there are less points to consider, and so the maximum value in a smaller neighbourhood will be smaller and the smallest value in a smaller neighbourhood will be bigger). Assuming that these statements are correct, we have sup{f(x1) | x1 ∈ ((x; h1)∩S)} - inf{f(x2) | x2 ∈ ((x; h1)∩S} ≤ sup{f(x1) | x1 ∈ ((x; h2)∩S)} - inf{f(x2) | x2 ∈ ((x; h2)∩S}, or sup{f(x1)-f(x2) | x1 ∈ (N(x; h1)∩S), x2 ∈ ((x; h1)∩S)} ≤ sup{f(x1)-f(x2) | x1 ∈ (N(x; h2)∩S), x2 ∈ ((x; h2)∩S)}, or Ωf (N(x; h1)∩S) ≤ Ωf (N(x; h2)∩S) as required. QED. We now need to show that ωf (x) = 0 ⇔ f is continuous at x. If. If f is continuous at x, then given an arbitrary ε > 0, there exists a h > 0 such that if |x-y| < h, then |f(x)-f(y)| < ε, where y ∈ N(x; h)∩S. Now if x1, x 2 ∈ N(x; h)∩S, then the distance between x1 and x2 is at most 2h, so that for a given ε > 0, |x1-x2| < 2h ⇒ |f(x1)-f(x2)| < 2ε. But if h → 0+, then |f(x1)-f(x2)| → 0+ so that sup{f(x1)-f(x2) | x1 ∈ (N(x; h)∩S), x2 ∈ ((x; h)∩S)} → 0 when h → 0+, or, in other words, lim hd0+ sup{f(x1)-f(x2) | x1 ∈ (N(x; h)∩S), x2 ∈ ((x; h)∩S)} = 0 as required. QED. Only If. If ωf (x) = 0, then we need to show that f is continuous at x. To do this, we need to show that f is defined at x and that limy→x f(y) = f(x). The first bit follows immediately from the fact that f is defined and bounded on the interval [a, b], so that f is defined for all x ∈ [a, b]. For the second bit, we note that limy→x f(y) = f(x) can be equivalently written as the two lim lim following equations: hd0+ ( x + h) = f ( x) and hd0+ ( x − h) = f ( x); or hlim f ( x) − ( x − h) = 0. ( x + h) − f ( x) = 0 and hlim d0+ d0+ For a given h, the neighbourhood given by N(x; h)∩S will contain the point x and the points (x+h) and (x-h). Now if ωf (x) = 0, then lim hd0+ sup{f(x1)-f(x2) | x1 ∈ (N(x; h)∩S), x2 ∈ ((x; h)∩S)} = 0. From our previous analysis of sup{f(x1)-f(x2) | x1 ∈ (N(x; h)∩S), x2 ∈ ((x; h)∩S)}, we know that f(x1) must be the highest value of f(x) in the neighbourhood N(x; h)∩S, and that f(x2) must be the smallest value of f(x) in the neighbourhood N(x; h) ∩S. Knowing this, then for a point x3 in the neighbourhood N(x; h)∩S, then we must have f(x 3)-f(x) ≤ sup{f(x1)-f(x2) | x1 ∈ (N(x; h)∩S), x2 ∈ ((x; h)∩S)}. Note also that we must have 0 ≤ f(y)-f(z) ≤ sup{f(x1)-f(x2) | x1 ∈ (N(x; h)∩S), x2 ∈ ((x; h)∩S)} if y, z ∈ N(x; h)∩S, and if y > z.
If we let x3 = x ± h, then we see that we have (i)
f(x + h) - f(x) ≤ sup{f(x1)-f(x2) | x1 ∈ (N(x; h)∩S), x2 ∈ (N(x; h)∩S)}. Therefore, lim lim hd0+ f(x + h) - f(x) ≤ hd0+ sup{f(x1)-f(x2) | x1 ∈ (N(x; h)∩S), x2 ∈ (N(x; h)∩S)}.
But we have assumed that hlim sup{f(x1)-f(x2) | x 1 ∈ (N(x; h)∩S), x2 ∈ ((x; h)∩S)} = 0, d0+ so that hlim f(x + h) - f(x) ≤ 0, i.e. we have hlim f(x + h) - f(x) = 0 as required (because we must d0+ d0+ have f(x+h)-f(x) > 0). (ii)
f(x) - f(x-h) ≤ sup{f(x1)-f(x2) | x1 ∈ (N(x; h)∩S), x2 ∈ ((x; h)∩S)}. Therefore, lim lim f(x) f(x-h) ≤ hd0+ hd0+ sup{f(x1)-f(x2) | x1 ∈ (N(x; h) ∩S), x2 ∈ ((x; h)∩S)}. lim
But again we have assumed that hd0+sup{f(x1)-f(x2) | x1 ∈ (N(x; h)∩S), x2 ∈ (N(x; lim lim h)∩S)} = 0, so that hd0+ f(x) - f(x-h) ≤ 0, i.e. we have hd0+ f(x) - f(x-h) = 0 as required (because we must have f(x)-f(x-h) > 0). QED. Technical Note: There is a slight mistake in the above in that f(x ± h) ∉ N(x; h)∩S, but I have assumed that in taking the limit this is of no consequence to the proof. 4-17. Let f be defined and continuous on a closed set S in E n. Let A = { x | x ∈ S, f(x) = 0}. Show that A is a closed set. Answer: Assume that the range of f is a set T in E 1 (i.e. T = f(S)). Let Y be the subset of T consisting only of the zero point, i.e. Y = {0}. Because Y consists of just a single point, then Y contains all its accumulation points (of which there are none of), and thus Y is a closed subset of T.
By Theorem 4-14, the inverse image f -1(Y) is a closed subset of S. But f -1(Y) is the set A, i.e. the elements of S which map onto 0, and thus A must be a closed subset of S. QED. Note: in the case that 0 ∉ f(S), then A = {φ}, and is a closed set by definition. 4-19. Let f be continuous on a closed interval [a, b]. Suppose that f has a local maximum at x1 and a local maximum at x 2. Show that there must be a third point between x1 and x2 where f has a local minimum. Answer: Consider the interval [x1, x2] ⊂ [a, b]. Because f is continuous on the closed interval [a, b], then it will be continuous on the subinterval [x 1, x2] as well. Further, the interval [x1, x2] is a closed interval. By Exercise 3-1, this interval is a closed set, and so by the note following Definition 3-39 (and by knowing that [x1, x2] is a bounded set (it is a subset of [a, b]), we can say that [x1, x2] is a compact set. Therefore, we can apply Theorem 4-20 to say that f has an absolute maximum and an absolute minimum on [x1, x2].
If f has an absolute minimum on [x1, x2], then there exists a point c ∈ [x1, x2] such that f(x) > f(c) for all x ∈ [x1, x2]. If we now take any neighbourhood around the point c such that N(c) ⊂ [x1, x2], then we can say that f has a local minimum at the point c in question. The only detail left to mop up is to make sure that c is not either x 1 or x2 (we need a third point between x 1 and x2). But, using the above, if we find that we arrive at the conclusion c = x1 or c = x2, then we proceed as follows: for the case c = x1, because x1 is a local maximum, then there is a neighbourhood N(x1) such that for all x ∈ N(x1)∩[a, b], we have f(x) ≤ f(x1). But because c is the absolute minimum of f on [x1, x2], then we must have f(x) = f(x1) for x ∈ [x1, x1+h), where h is the radius of the neighbourhood N(x1), or else c would have been incorrectly found. Therefore, f(x1) = f(x1+h), and so we can equally well define c to be given by c = x1+h to represent our absolute/local minimum on [x1, x2]. Note: if x 1+h > x2, then just take a point half way between x1 and x2 to represent c — in this situation f must be a constant function between x1 and x2. The case c = x2 goes through in much the same way as above (i.e. we redefine c to be given by c = x2-h or by c = (x1+x2)/2 as appropriate), so that we have found a point c between x1 and x2 which is a local minimum in the interval [x 1, x2]. QED. 4-21. Show that a function which is uniformly continuous on a set S is also continuous
on S. Answer: Let f be defined on a set S in E n with function values in Em. f is said to be uniformly continuous on S if the following statement holds: For every ε > 0, there exists a δ > 0 (depending only on ε) such that if x ∈ S and y ∈ S and |x-y| < δ, then |f (x)-f (y)| < ε. f is said to be continuous on S provided that at each accumulation point a, and given ε, there is a number δ (depending on a and on ε) such that 0 < | x-a| < δ ⇒ |f (x)-f (a)| < ε.
Let a denote any arbitrary accumulation point in S, and let ε > 0 be an arbitrary number. If f is uniformly continuous, then for every ε there exists a δ > 0 (depending only on ε) such that if x ∈ S, and |x-a| < δ, then |f (x)-f (a)| < ε. But this is exactly the condition we want so that the function f is continuous on S (it doesn’t matter that δ does not depend on a), and therefore we have reached the conclusion we want, in that uniform continuity on S does imply continuity on S. QED. 4-22. Show that the function f defined by f(x) = x² is not uniformly continuous on E1. Answer: Consider that we are given two arbitrary points x and y in E 1. We need to show that if the distance between x and y is less than δ, then the distance between f(x) and f(y) is less than ε, where δ depends only on ε. In other words, given ε, if |x-y| < δ, then we need to show that |f(x)-f(y)| < ε.
Now |f(x)-f(y)| = |x²-y²| = |(x-y)(x+y)| = |(x-y)||(x+y)|. If |x-y| < δ, then |f(x)-f(y)| < δ|x+y|. In the example in the book where we considered points in the interval (0, 1], the value |x+y| was bounded above by 2, and so as |x+y| ≤ 2, then |f(x)-f(y)| < 2δ, so that if we choose δ = ε/2 for a given arbitrary ε > 0, then we get the required result: |f(x)-f(y)| < ε if |x-y| < δ. However, if x, y ∈ E1, then |x+y| is not bounded above, so that we cannot choose a value n so that |f(x)-f(y)| < δn, and therefore cannot define δ = ε/n for a given arbitrary ε > 0 so that |f(x)-f(y)| < ε. Proof by Contradiction: Suppose that there was a positive number ‘n’ so that if we defined δ = ε/n for a given arbitrary ε > 0, then we would get |f(x)-f(y)| < ε for every |x-y| < δ (i.e. the definition of uniform continuity). Note: with our definition of δ, δ = ε/n, we get ε = δn. But we can always choose two numbers c and d which are greater in magnitude than n (because c, d ∈ E1) and which are at a distance of δ/2 apart, so that we have |c-d| < δ, but |f(c)-f(d)| = |(c-d)||(c+d)| > |(c-d)||(n+n)| = 2n(δ/2) = nδ = ε. So we have a contradiction, and thus cannot define δ in terms of ε only so that the function f(x) = x² in uniformly continuous on E1. End of Proof , and QED. 4-25. Let f be a function defined on a set S in E n and assume that f (S) ⊂ Em. Let g be defined on f (S) with values in Ek , and let gf denote the composite function defined by gf (x) = g[f (x)], if x ∈ S. If f is uniformly continuous on S and if g is uniformly continuous on f (S), show that gf is uniformly continuous on S. Answer: (1) If f is uniformly continuous on S, then given an ε1 > 0, we can always find a δ1 > 0 (dependent only on ε1) such that |x-y| < δ1 ⇒ |f (x)-f (y)| < ε1. (x, y ∈ En). (2) If g is uniformly continuous on f (S), then given an ε2 > 0, we can always find a δ2 > 0 (dependent only on ε2) such that |p-q| < δ2 ⇒ |g(p)-g(q)| < ε2. (p, q ∈ Em).
Now let a = f (x) and let b = f (y). We know that if | x-y| < δ1, then |a-b| < ε1. Assume for the moment that δ2 = ε1, so that |a-b| < ε1 ⇒ |a-b| < δ2. But if | a-b| < δ2, then |g(a)-g(b)| < ε2, i.e. |g(f (x))-g(f (y))| < ε2. Therefore, given an ε2 > 0, by (2) we can always find a δ2 > 0 (dependent only on ε2) such that |f (x)-f (y)| < δ2 ⇒ |g(f (x))-g(f (y))| < ε2. If we let δ2 = ε1, then given that particular ε1 (which is dependent only on ε2), we can (by (1)) always find a δ1 > 0 (which now is dependent only on ε2) such that |x-y| < δ1 ⇒ |f (x)-f (y)| < ε1. Putting the above all together, we conclude that given an ε2 > 0, we can always find a δ1 > 0 (dependent only on ε2) such that |x-y| < δ1 ⇒ |f (x)-f (y)| < ε1 = δ2 ⇒ |g(f (x))-g(f (y))| < ε2, i.e. |x-y| < δ1 ⇒ |g(f (x))-g(f (y))| < ε2 (x, y ∈ En); and so gf is uniformly continuous on S. QED.
4-27. Locate and classify the discontinuities of the function f defined on E 1 by the following equations: (a) f(x) = (sin x)/x if x ≠ 0, f(0) = 0. 1/x (b) f(x) = e if x ≠ 0, f(0) = 0. (c) f(x) = e1/x + sin(1/x) if x ≠ 0, f(0) = 0. (d) f(x) = 1/(1-e1/x) if x ≠ 0, f(0) = 0. Answer: (a) f(x) is f(x) f(x) (b) continuous at every point in E 1 (a) 1 except at x = 0, where f(x) has a x x removable discontinuity. As limx→0- = 1 and lim x→0+ = 1, we can make the function continuous by f(x) f(x) (d) redefining f(0) to be f(0) = 1. (b) (c) Again the function is continuous at 1 x x every point in E1 except at x = 0, where f(x) has an infinite jump discontinuity: f(0-) = 0, f(0) = 0, and f(0+) = ∞. As the function has an infinite right-hand jump, then it is discontinuous at x = 0 (and cannot be made continuous at this point).
(c) In this case, as before, f(x) is continuous everywhere except at x = 0. This time, f(x) has an infinite right-hand jump discontinuity as in part (b), and f(0-) does not exist. We therefore conclude that f(x) is discontinuous at x = 0. (d) In this final example, we have f(0-) = 1, f(0+) = 0, and f(0) = 0. As f(x) has a left-hand jump at x = 0, then f(x) has a left-hand jump discontinuity at x = 0. Apart from this, f(x) is continuous at every other point in E 1. 4-29. Let f be defined in the open interval (a, b) and assume that for each interior point x of (a, b) there exists a neighbourhood N(x) in which f is increasing. Show that f is an increasing function throughout (a, b). Answer: We saw in exercise 3-1 that an open interval (a, b) is an open set — so therefore all points x ∈ (a, b) are interior points. This implies that for all points x ∈ (a, b), there exists a neighbourhood N(x) in which f is increasing. Let us now prove that f is increasing by contradiction.
If f is not increasing on (a, b), then there is a subinterval (say (c, d)) of (a, b) in which f is not increasing, i.e. f is strictly decreasing in this subinterval. Consider a point z ∈ (c, d). If z ∈ (c, d), then because (c, d) ⊂ (a, b), we have z ∈ (a, b), and thus there is some neighbourhood of z in which f is increasing. Because of this, then we must throw away this neighbourhood from our subinterval (c, d) in which we are claiming that f is strictly decreasing. But we can apply the above argument with any point in the subinterval (c, d), and thus there cannot possibly be any such subinterval (c, d) in which f is strictly decreasing because for any point in such an interval, f will be increasing in a neighbourhood of that point. We therefore conclude that f must be increasing on the interval (a, b). QED.
Here is a second proof for the above question. Suppose that we pick two arbitrary numbers x1 and x2 from the interval (a, b) so that a < x 1 < x2 < b. To show that f is an increasing function throughout (a, b), all we must show is that f(x 1) ≤ f(x2) for our arbitrary numbers x1 and x2. The interval [x1, x2] is a closed interval. By Exercise 3-1, this interval is a closed set, and so by the note following Definition 3-39 (and by knowing that [x 1, x 2] is a bounded set (it is a subset of (a, b)) we can say that [x 1, x2] is a compact set. Now for every point x ∈ [x1, x2], there will be a corresponding neighbourhood N(x) in which f is increasing . Consider that we define a covering of the compact set [x1, x 2] given by the union of all such neighbourhoods of points in [x1, x2] in which f is increasing. In other words, we can write [x1, x2] =
4
( ).
all x c [ x 1 , x 2 ] N x
Now because [x1, x2] is a compact set, a finite number of these neighbourhoods will cover [x1, x2], say a collection of n neighbourhoods. Therefore, we can write [x1, x2] = 4 ni=1 N i ( x) , with x ∈ [x1, x2]. Because f is increasing throughout all of these neighbourhoods, and because this finite collection of neighbourhoods covers [x1, x2], then f will be increasing throughout the interval [x1, x2], enabling us to say that f(x1) ≤ f(x2) as required. QED. 4-31. If f is one-to-one and continuous on [a, b], show that f must be strictly monotonic on [a, b]. That is, prove that every topological mapping of [a, b] onto an interval [c, d] must be strictly monotonic. Answer: If f is one-to-one on [a, b], then for all points y in the range, there exists one and only one point x in the domain such that f(x) = y. If f is continuous on [a, b], then we know that f is defined for all points in the interval [a, b]. Because of this, we know that f(a) will have a value: say f(a) = y1; and that f(b) will have a value: say f(b) = y 2.
Because f is one-to-one, then we cannot have y1 = y2, and thus we must have either y 1 > y2 or y1 < y2. Consider the case where y1 < y2 (we will only consider this case — the other case can be derived from the following by replacing all ‘<’ by ‘>’ and vice-versa). Because of the continuity of f on the closed interval [a, b], then the intermediate value theorem says that f assumes all values between y1 and y2.
Consider an arbitrary point c ∈ (a, b). What is f(c)? First of all, f(c) cannot be equal to either y1 or y2 because f is a one-to-one function. Further (as seen in Diagram 1), we cannot f(p) have f(c) > y2 as then (according to the IVT) there will be a point p in the interval (a, c) where f(p) = y2, and this cannot be the case if f is one-to-one (because then we will have f(p) = f(b) = f(a) y2). Similarly (as seen in Diagram 2), we cannot have f(c) < y1 as then (again according to the IVT) there will be a point q in the f(a) interval (c, b) where f(q) = y1, and this cannot be the case if f is one-to-one (because then we will have f(q) = f(a) = y 1). Therefore, we conclude that we must have y1 < f(c) < y2.
Diagram 1
f(c)
f(b) f(b) f(q)
f(c) Diagram 2
If we apply the same sort of argument for the point d ∈ (c, b), we would reach the following conclusion: y1 < f(c) < f(d) < y2. In other words, given any pair of points c and d in [0, 1], with c < d, then we must have f(c) < f(d), and thus f is strictly increasing on [a, b]. With the other case, where y1 > y2, we would reach the conclusion that f would be strictly decreasing on [a, b]. Putting all of this together, we conclude that f is either a strictly monotonically increasing function or a strictly monotonically decreasing function, i.e. f is a monotonic function. QED.
Possible Further Work / Evaluation Exercise 4-7: Clarification in the wording , or possibly an alternative solution, e.g. if we consider an arbitrary irrational number in [a, b], and because we know that there exists a sequence of rational numbers xn converging to y, the result follows because of the continuity of f? Exercise 4-11: A ‘small gap’ left unfilled in the first part of the answer. In the second part, I made a slip up with a negative sign. New version:
...The other case to consider is when x is a fixed irrational number in [0, 1]. Consider now that for this fixed irrational number x, we define a number ε2 = 1. Given this ε2, we need to find a δ > 0 such that if |x-y| < δ for y ∈ [0, 1] ≠ x, then we will always have |f(x)-f(y)| < ε2. Assume that such a δ exists, and let us pick out an element y from the neighbourhood N’(x)∩[0, 1] which is a rational number. In this case, we have f(x) = 1-x and f(y) = y, with |x-y| < δ. It follows that |f(x)-f(y)| = |1-x-y| = |-x-y+1| = |x+y-1| (as |A-B| = |B-A|). But this is what we had in the first section of the answer so the result follows by the validity of this first section. QED. Exercise 4-14: I was mistaken in thinking that “ sup” was the same thing as “maximum”. I therefore looked at the following definitions in Chapter 1: Definition 1.6: Let A be a set of real numbers. If there is a real number x such that a ∈ A implies a ≤ x, then x is called an upper bound for the set A and we say that A is bounded above. Definition 1.7: Let A be a set of real numbers bounded above. Suppose there is a real number x satisfying the following two conditions:
(i) (ii)
x is an upper bound for A, and if y is any upper bound for A, then x ≤ y.
Such a number x is called a least upper bound, or a supremum, of the set A. Exercise 4-22: Need to prove the general case — not just for a particular formula for δ in terms of ε. Exercise 4-29: Need to correct the indexing in the penultimate paragraph:
....Now because [x1, x2] is a compact set, a finite number of these neighbourhoods will cover [x1, x2], say a collection of n neighbourhoods N(y1), ...., N(yn). Therefore, we can write [x1, x2] ⊂
n
4 i=1 N ( y i ) ,
with yi ∈ [x1, x2].....
Chapter 13
Chapter 13: Sequences of Functions In this chapter, we will be dealing with complex-valued functions defined on certain subsets of E2 (but if results hold only for real-valued functions defined on subsets of E 1, this will be explicitly stated). Given a sequence {f n}, each term of which is a function defined on a set S, for each x in S we can form another sequence {f n(x)} whose terms are the corresponding function values. Let T denote the set of these points x in S for which the second sequence converges. The lim function f defined by f(x) = ndº f n(x), if x ∈ T, will be referred to as the limit function of the sequence {f n} and we will say that {f n} converges pointwise on the set T. If each term of the sequence {f n} has a certain property, then to what extent does the function f also possess this property? In general, we need a study of “ stronger ” methods that do preserve these properties. The most important of these is the notion of uniform convergence. When we ask whether continuity at each f n at x 0 implies continuity of the limit function f lim at x0, we are really asking whether the equation xlim d x 0 f n(x) = f n(x0) implies the equation xd x 0 f(x) lim lim lim = f(x0). The last equation can also be written as follows: xlim . d x 0 ndº f n ( x) = ndº xd x 0 f n ( x) Therefore, our question about continuity amounts to this: Can we change the limit symbols in the above? In general, we shall see that we cannot interchange the symbols. First of all, the limit in lim the equation ( xd x0 fn (x) = f n(x0)) may not exist; or, even if it does exist, it may not be equal to f(x0). In chapter 12, it is stated that Σ m=1∞ Σn=1∞ f(m, n) is not necessarily equal to Σn=1∞ Σm=1∞ f(m, n). For example, consider the following function: f(m, n) = 1 if m = n+1, n = 1, 2, ...; f(m, n) = -1 if m = n-1, n = 1, 2, ...; and f(m, n) = 0 otherwise. Then Σm=1∞Σn=1∞ f(m, n) = -1, but
Σn=1∞Σm=1∞ f(m, n) = 1.
The question arises frequently as to whether we can change the order of the limit processes. We shall find that uniform convergence is a far-reaching sufficient condition for the validity of interchanging certain limits, but it does not provide the complete answer to the question. Examples can be found in which the order of the two limits can be interchanged although the sequence is not uniformly convergent. An example of a sequence of real-valued functions: Let f n(x) = n²x(1-x)n if x ∈ E1, n = 1, 2, ... Here, lim n→∞ f n(x) exists if 0 ≤ x ≤ 1, and the limit function has the value 0 at each point in [0, 1]. f n has a local maximum at x = 1/(n+1), but f n(1/(n+1)) → +∞ as n → ∞.
Let {f n} be a sequence of functions which converges pointwise on a set T to a limit function f. Going back to the basic definition of limit, this means that for each point x in T and for each ε > 0, there exists an N (depending on at most both x and ε) such that n > N implies |f n(x)-f(x)| < ε. If the same N works equally well for every point in T, the convergence is said to be uniform on T. That is, we have the following definition: A sequence of functions {f n} is said to converge uniformly to f on a set T if, for every ε > 0, there exists an N (depending only on ε) such that n > N implies |f n(x)-f(x)| < ε, for every x in T. We denote this symbolically by writing f n → f uniformly on T. When each term of the sequence {f n} is real-valued, there is a useful geometric interpretation of uniform convergence. The inequality |f n(x)-f(x)| < ε is then equivalent to the two inequalities f(x)-ε < f n(x) < f(x)+ε. If this is to hold for y = f(x)+ε y = f n(x) y = f(x) all n > N and for all x in T, this means that the entire graph of f n (that is, the set {(x, y) | y = f n(x), x ∈ T}) lies within a “band ” of height 2ε situated symmetrically about the graph y = f(x)-ε of f (see the diagram on the right). A sequence {f n} is said to be uniformly bounded on T if there exists a constant M > 0 such that |f n(x)| ≤ M for all x in T and for all n = 1, 2, ... The number M is called a uniform bound for {f n}. If each individual function is bounded and if f n → f uniformly on T, then it is easy to prove that {f n} is uniformly bounded on T. This observation often enables us to conclude that a sequence is not uniformly convergent. Theorem 13-2. Let f be a double sequence and let P denote the set of positive integers. For each n = 1, 2, ..., define a function g n on P as follows: g n(m) = f(m, n), if m ∈ P. Assume that gn → g uniformly on P, where g(m) = lim n→∞ f(m, n). If the iterated limit lim m→∞ (limn→∞ f(m, n)) exists, then the double limit lim m,n→∞ f(m, n) also exists and has the same value. Uniform Convergence and Continuity. Theorem 13-3: Assume that f n → f uniformly on T. If each f n is continuous at a point x0 of T, then the limit function f is also continuous at x0. Note 1: If x0 is an accumulation point of T, the conclusion implies that m m m m xd x 0 ndº f n ( x) = ndº xd x 0 f n ( x) . Note 2: Uniform convergence of {f n} is sufficient but not necessary to transmit continuity from the individual terms to the limit function. The Cauchy condition for uniform convergence. Theorem 13-4: Let {f n} be a sequence of functions defined on a set T. There exists a function f such that f n → f uniformly on T if, and only if, the following condition (called the Cauchy condition) is satisfied: For every ε > 0 there exists an N such that m > N and n > N implies |f m(x)-f n(x)| < ε, for every x in T. Uniform Convergence on infinite series. Definition 13-5: Given a sequence {f n} of functions defined on a set T. For each x in T, let s n(x) = Σk=1n f k( x) (n = 1, 2, ...). If there exists a function f such that sn → f uniformly on T, we say that the series Σ f n(x) converges uniformly on T and we write Σn=1∞ f n(x) = f(x) (uniformly on T).
Theorem 13-6: (Cauchy condition for uniform convergence of series). The series Σf n(x) converges uniformly on T if, and only if, for every ε > 0 there is an N such that n > N implies n+ p | S k =n+1 f k( x)| < ε, for each p = 1, 2, ..., and every x in T. Theorem 13-7: (Weierstrass’ M-test). Let {M n} be a sequence of non-negative numbers such that 0 ≤ |f n(x)| ≤ Mn, for n = 1, 2, ..., and for every x in T. Then Σ f n(x) converges uniformly on T if ΣMn converges. Theorem 13-8: Assume that Σf n(x) = f(x) (uniformly on T). If each f n is continuous at a point x0 of T, then f is also continuous at x0. Note: If x0 is an accumulation point of T, this lim º º theorem permits us to write xlim d x 0 S n=1f n(x) = S n=1 xd x 0f n(x). A space-filling curve. We can apply the above theorem to construct an example of what is known as a space-filling curve. This is a continuous curve in E2 that passes through every point of the unit square [0, 1]×[0, 1]. Consider the following example: Let φ be defined on the interval [0, 2] by the following formulas:
φ(t) = 0, if 0 ≤ t ≤ 1/3, or if 5/3 ≤ t ≤ 2; φ(t) = 3t-1, if 1/3 ≤ t ≤ 2/3; φ(t) = 1, if 2/3 ≤ t ≤ 4/3; and φ(t) = -3t+5, if 4/3 ≤ t ≤ 5/3. Extend the definition of φ to all of E 1 by the equation φ(t+2) = φ(t). This makes φ periodic with period 2 (as shown below). Now define two functions α1 and α2 by the following equations:
α1(t) = S ºn=1
v(3 2n−2 t ) , 2n
α2(t) = S ºn=1
v(3 2n−1 t ) . 2n
-2
-1
0
1
2
3
4
Both series converge absolutely for each real t and they converge uniformly on E1. In fact, since |φ(t)| ≤ 1 for all t, the Weierstrass M-test is applicable with M n = 2-n. Since f is continuous on E 1, Theorem 13-8 tells us that α1 and α2 are also continuous on E1. Let α = (α1, α2), and let Γ denote the image of the unit interval [0, 1] under α. We will show that Γ ‘ fills’ the unit square, i.e. that Γ = [0, 1]×[0, 1]. First, it is clear that 0 ≤ α1(t) ≤ 1 and that 0 ≤ α2(t) ≤ 1 for each t, since Σn=1∞ 2-n = 1. Hence, Γ is a subset of the unit square. Next, we must show that (a, b) ∈ Γ whenever (a, b) ∈ [0, 1]×[0, 1]. For this purpose, we write a and b in the binary system. That is, we write a
bn
n º a = Sº n=1 2 n , b = S n=1 2 n , c where each an and each bn is either 0 or 1. Now let c = 2 S nº=1 3nn, where c2n-1 = an and c2n = bn, n = 1, 2, ... Clearly, 0 ≤ c ≤ 1 since 2Σ n=1∞ 3-n = 1. We will show that α1(c) = a and that α2(c) = b. If we can prove that φ(3k c) = ck+1, for each k = 0, 1, 2, ... (---(1)), then we will have φ(32n-2c) = c2n-1 = an, and φ(32n-1c) = c2n = bn, and this will give us α1(c) = a, α2(c) = b.
c
c
n To prove equation (1), we write 3k c = 2 S k n=1 3 nn−k + 2 S º n=k +1 3 n−k = (an even integer) + dk , where dk = Σ n=1∞ c n+k /2n. Since φ has period 2, it follows that φ(3k c) = φ(dk ). If ck+1 = 0, then we have 0 ≤ dk ≤ 2 Σn=2∞ 3-n = 1/3, and hence φ(dk ) = 0. Therefore, φ(3k c) = ck+1 in this case. The only other case to consider is the case c k+1 = 1. But then we get 2/3 ≤ dk ≤ 1, and hence φ(dk ) = 1. Therefore, φ(3k c) = ck+1 in all cases and this proves that α1(c) = a, α2(c) = b. Hence, Γ fills the unit square.
An application to repeated series. As a second a second application application of Theorem 13-8, we derive the following generalisation generalisation of Theorem 12-43 (which went as follows: follows: Theorem 12-43: Suppose f(m, n) > 0 for all m, n. Assume that Σn=1∞ f(m, n) converges for each fixed each fixed m = 1, 2, ∞ ∞ ∞ ..., and that Σm=1 Σn=1 f(m, n) converges. converges. Then: (a) Σm=1 f(m, n) converges for each n each n = 1, 2, ...; and (b) Σn=1∞Σ m=1∞ f(m, n) converges and equals Σm=1∞Σn=1∞ f(m, n).):
Theorem 13-9. Let f be a complex-valued double sequence. sequence. Assume that Σn=1∞ f(m, n) converges absolutely absolutely for each fixed m and that Σm=1∞Σn=1∞ |f(m, n)| converges. converges. Then (a) Σm=1∞ f(m, n) converges absolutely for each n, and (b) Σ n=1∞Σm=1∞ f(m, n) converges absolutely and equals Σm=1∞Σn=1∞ f(m, n). Uniform convergence and differentiation. Theorem 13-13: Assume that each term of {f n} is a real-valued function having a finite derivative at each point of an open interval open interval (a, b). Assume that for at least one point x0 in (a, b) the sequence {f n(x0)} converges. Assume further that there exists a function g such that f’ n → g uniformly uniformly on (a, b). Then (a) There exists a function f such that f n → f uniformly on (a, b); (b) For each x in (a, b), the derivative f’(x) exists and equals g(x). equals g(x).
When we reformulate reformulate the above theorem in terms of series, we obtain the following: Theorem 13-14: Assume that each f n is a real-valued function defined function defined on (a, b) such that the derivative f’n(x) exists for each x each x in (a, b). Assume that, for at least one point x0 in (a, b), the series Σf n(x0) converges. Assume further that there exists a function g such that Σf’n(x) = g(x) (uniformly on uniformly on (a, b)). Then (a) There exists a function a function f f such that Σ f n(x) = f(x) (uniformly (uniformly on on (a, b)); (b) If x ∈ (a, b), the derivative f’(x) derivative f’(x) exists and equals Σf’n(x). Sufficient conditions for uniform convergence of a series. In some of the preceding theorems, the importance of importance of uniformly convergent series has been illustrated. It seems natural to seek some simple ways of testing a series for uniform convergence without resorting to the definition in definition in each case. We have seen one such test already (Weierstrass’ ( Weierstrass’ M-test ), ), but when the M-test is not applicable, the following may be useful:
Theorem 13-15: ( Dirichlet’s Dirichlet’s test for uniform convergence). convergence). Let Fn(x) denote the nth partial sum of the series Σf n(x), where each f n is a complex-valued function defined function defined on a set T. Assume that {Fn} is uniformly bounded uniformly bounded on T. Let {g n} be a sequence of real-valued functions functions such that gn+1(x) ≤ gn(x) for each x in T and for every n = 1, 2, ..., and assume assume that gn → 0 uniformly on T. Then the series Σf n(x)gn(x) converges uniformly on uniformly on T. Example: Let Fn(x) = Σk=1n eikx. In a previous chapter, previous chapter, the inequality |Fn(x)| ≤ 1/|sin(x/2)| was derived, valid for every real x x ≠ 2mπ (m is an integer ). ). Therefore, if 0 < δ < π, we have the estimate |Fn(x)| ≤ 1/sin(δ/2) if δ ≤ x ≤ 2π-δ. Hence, {Fn} is uniformly bounded on on the interval [ δ, 2π-δ]. If {gn} satisfies the conditions conditions of the above theorem, we can conclude that the series In particular , if we take g n(x) = 1/n, this establishes Σgn(x)einx converges uniformly on uniformly on [ δ, 2π-δ]. In particular the uniform convergence convergence of the series Σn=1∞ (einx/n) on [δ, 2π-δ] if 0 < δ < π. Note that the Weierstrass M-test cannot cannot be used to establish uniform convergence in this case, since |e inx| = 1.
Bounded convergence. Arzelà’s theorem. Definition 13-16. A sequence sequence of functions {f n} is said to be boundedly convergent on T if {f n} is pointwise convergent and uniformly bounded on T. Theorem 13-17 (Arzelà). Assume that {f n} is boundedly convergent on [a, b] and suppose each f n is Riemann-integrable is Riemann-integrable on on [a, b]. Assume also that the limit function f is lim b Riemann-integrable on [a, b]. Then n→∞ ∫ a f n(x)dx = ∫ ba limn→∞ f n(x)dx = ∫ ba f(x)dx.
Note: it is easy to give an example example of a boundedly convergent sequence {f n} of Riemann-integrable functions whose limit f is not Riemann-integrable. If {r 1, r 2, ...} denotes the set of rational numbers numbers in [0, 1], define f n(x) to have the value 1 if x = r k k for for some k = 1, 2, ..., n, and put f n(x) = 0 otherwise. Then the integral ∫ 01 f n(x)dx = 0 for each n, but the limit function f function f is not Riemann-integrable on [0, 1]. Mean convergence. Definition 13-18. Let {f n} be a sequence of Riemann-integrable functions defined functions defined on [a, b]. Assume that f ∈ R on [a, b]. The sequence {f n} is said to converge in the mean to mean to f on [a, b], and we write l.i.m. n→∞ f n = f on [a, b], if lim n→∞ ∫ ba |f n(x)-f(x)|²dx = 0. If the inequality inequality |f(x)-f n(x)| < ε holds for every x in [a, b], then we have ∫ ba |f(x)-f n(x)|²dx ≤ ε²(b-a). Therefore, uniform uniform convergence of {f n} to f on [a, b] implies mean mean convergence, provided that each f n is Riemann-integrable on [a, b].
A rather surprising fact is that convergence in the mean mean need not imply pointwise convergence at convergence at any point of the interval. This can be seen as follows: For each integer each integer n > 0, subdivide the interval [0, 1] into 2 n equal equal subi subint nter erva vals ls and and let let I 2 n +k deno denote te tha thatt subi subint nter erva vall whose right endpoint endpoint is (k+1)/2n, where k = 1, 2, ..., 2 n-1. This yields a collection {I collection {I 1, I2, ...} of 1 subintervals of subintervals of [0, 1] of which the first few are: I1 = [0, 1], I2 = [0, /2], I 3 = [1/2, 1], I4 = [0, 1/4], I5 = [1/4, 1/2], I6 = [1/2, 3/4], and so forth. Define f n on [0, 1] as follows: f n(x) = 1 if x ∈ In, and f n(x) = 0 if x ∈ [0, 1]—In. Then {f n} converges in the mean to mean to 0, since ∫ 10|f n(x)|²dx is the length of length of I n, and this approaches 0 as n→∞. lim lim sup sup lim inf inf On the other hand, for each x in [0, 1], we have ndº f n(x) = 1 and lim ndº f n(x) = 0. Hence, {f n(x)} does not converge for any x in [0, 1]. Theorem 13-19. Assume that l.i.m. n→∞ f n = f on [a, b]. If g ∈ R on [a, b], define h(x) = f(t)g(t)dt, hn(x) = ∫ xa f n(t)g(t)dt, if x ∈ [a, b]. Then h n→h uniformly on uniformly on [a, b].
∫ ba
Theorem 13-20. Assume that l.i.m.n→∞ f n = f and l.i.m. n→∞ g n = g on [a, b]. Define h(x) = ∫ xa f(t)g(t)dt, hn(x) = ∫ xa f n(t)gn(t)dt, if x ∈ [a, b]. Then hn→h uniformly on uniformly on [a, b]. Power Series. An infinite series of the form a 0+Σn=1∞an(z-z0)n, written more briefly as Σn=0∞an(z-z0)n, is called a power series series in z-z0. Here, z, z 0 and an (n = 0, 1, 2, ...) are complex numbers. numbers. With every power series, there is associated a circle, circle, called the circle of convergence, such that the series converges absolutely absolutely for every z interior to this circle and diverges for every z outside this outside this circle. The centre of the circle is at z 0 and its radius is called the radius of convergence of convergence of the power series. (The radius may be 0 or +∞ in extreme cases.) extreme cases.) The next theorem establishes the existence of existence of the circle of convergence and provides us with a way for calculating its radius.
Theorem 13-21. Given a power series
Σn=0∞ an(z-z0)n, let λ =
lim lim sup sup n ndº
|a n |, r = 1/λ,
(where r = 0 if λ = +∞ and r = + ∞ if λ = 0). Then the series converges absolutely if absolutely if |z-z0| < r and diverges if diverges if |z-z0| > r. Furthermore, the series converges uniformly on every compact subset subset interior to the circle of convergence. Note: If the the limit lim limn→∞ |an/an+1| exists (or if this limit is + ∞), its value is also equal to the radius of convergence of convergence of Σn=0∞an(z-z0)n. Example: The two series Σn=0∞zn and Σn=1∞an/n² have the same radius of convergence, namely r = 1. On the boundary of the circle of convergence, the first converges nowhere, nowhere, but the second converges everywhere. everywhere. This illustrates why the theorem at the top of the page makes no assertion assertion about the behaviour of a power series on the boundary of the circle of convergence. Theorem 13-22. Assume that the power series Σn=0∞ an(z-z0)n converges for each z in N(z0;r). Then the function f defined by the equation f(z) = Σ n=0∞an(z-z0)n, if z ∈ N(z0;r), is continuous on continuous on N(z0;r). Proof . Since each point in N(z0;r) belongs to some compact subset compact subset of of N(z0;r), the conclusion follows at once from Theorem 13-8. Note: The series f(z) = Σn=0∞an(z-z0)n, if z ∈ N(z 0;r), is said to represent f f in N(z0;r). It is also called a power a power series expansion of expansion of f about z 0. Functions having power series expansions are continuous inside the circle of convergence. Much more than this is true, however. We will later prove that such functions have derivatives of every order inside the circle of convergence. The proof will make use of the following the following theorem: theorem: Theorem 13-23. Assume that Σan(z-z0)n converges if z ∈ N(z0;r). Suppose that the equation f(z) equation f(z) = Σn=0∞an(z-z0)n is known to be valid for each z in some open subset of N(z 0;r). Then, for each point each point z1 in S, there exists a neighbourhood N(z N(z 1;R) ⊂ S in which f has a power (z-z1)k (equation (1)), series expansion of the form f(z) = Σk=0∞ bk (z-z (1)), where bk = Σn=k ∞ ( equation (2)). (2)). Note that the proof the proof of of this theorem asserts that z 1 − z 0 ) n−k (k = 0, 1, 2, ...) (equation ( nk )a n ( z we may use any R > 0 that satisfies the condition N(z condition N(z1;R) ⊂ S. Theorem 13-24. Assume that Σan(z-z0)n converges for each z in N(z0;r). Then the function f defined by the equation f(z) = Σ n=0∞an(z-z0)n, if z ∈ N(z0;r), has a derivative f’(z) for each z each z in N(z0;r), given by f’(z) = Σ n=1∞nan(z-z0)n-1. Note 1: The two series above above have the same radius of convergence. Note 2: By repeated ∞ application of f’(z) = Σn=1 nan(z-z0)n-1, we find that for each each k = 1, 2, ..., the derivative f (k)(z) exists in N(z0;r) and is given by the series f series f (k)(z) = Σn=k ∞ n!/(n-k)!an(z-z0)n-k . If we put z = z 0 in this series, we obtain the important formula f formula f (k)(z0) = k!ak (k (k = 1, 2, ...). This equation tells us that if two power series Σan(z-z0)n and Σ bn(z-z0)n both represent the same function in a neighbourhood N(z0;r), then an = bn for every n. That is, the power series expansion of expansion of a function f about a given point z0 is uniquely determined (if it exists al all), and is given by the formula f(z) = Σ n=0∞ (f (n)(z0)/n!)(z-z0)n, valid for each z in the circle of convergence. convergence.
Theorem 13-25. Assume that Σan(z-z0)n converges for each z in N(z0;r). Let Γ be a piece-wise smooth curve in N(z0;r) joining z0 to z1. Then °G( z 0 , z 1 ) Σn=0∞an(z-z0)ndx = Σn=0∞ (an/n+1)(z1-z0)n+1. Multiplication of power series. Theorem 13-26. Given two power series expansions about the origin, say f(z) = Σn=0∞anzn, if z ∈ N(0;r), and g(z) = Σn=0∞ bnzn, if z ∈ N(0;R), then the product f(z)g(z) is given by the power series f(z)g(z) = Σn=0∞cnzn, if z ∈ N(0;r)∩ N(0;R), where cn = Σk=0nak bn-k (n = 0, 1, 2, ..).
Note: If the two series are identical, we get f(z)² = Σn=0∞cnzn, where cn = Σk=0nak an-k = S m 1+m 2 =n a m 1 a m 2 . The symbol S m 1+m 2 =n indicates that the summation is to be extended over all non-negative integers m1 and m2 whose sum is n. Similarly, for any integer p > 0, we have f(z) p = Σn=0∞cn(p)zn, where cn(p) =
S
m 1 +...+m p =n
a m 1 ...a m p .
The substitution theorem. Theorem 13-27. Consider that we are given two power series expansions about the origin, say f(z) = Σn=0∞anzn, if z ∈ N(0;r), and g(z) = Σn=0∞ bnzn, if z ∈ N(0;R). If, for a fixed z in N(0;R), we have Σn=0∞|bnzn| < r, then for this z we can write f[g(z)] = Σk=0∞ck zk , where the coefficients ck are obtained as follows: Define the numbers bk (n) by the equation g(z)n = (Σk=0∞ bk zk )n = Σk=0∞ bk (n)zk . Then ck = Σn=0∞an bk (n) for k = 0, 1, 2, ... Note: The series Σk=0∞ck zk is the power series which arises formally by substituting the series for g(z) in place of z in the expansion of f and then rearranging terms in increasing powers of z.
As an application of the substitution theorem, we will show that the reciprocal of a power series in z is again a power series in z, provided that the constant term is not 0. Theorem 13-28. Assume that we have p(z) = Σn=0∞ pnzn, if z ∈ N(0;h), where p(0) ≠ 0. Then there exists a neighbourhood N(0;δ) in which the reciprocal of p has a power series expansion of the form 1/ p(z) = Σ n=0∞qnzn. Furthermore, q0 = 1/p0. Real Power Series. If x, x0 and an (n = 0, 1, 2, ...) are real numbers, the series Σan(x-x0)n is called a real power series. Its circle of convergence intersects the real axis in an interval called the interval of convergence. If neighbourhoods are taken to mean one-dimensional neighbourhoods, all the results of the last three sections can be interpreted in an obvious way as theorems on the real line.
We shall restrict our considerations in the remainder of this chapter to real power series. In this section, we shall deal with the following type of problem: Suppose that we are given a real-valued function f defined in some neighbourhood of a point x0 in E1, and suppose that f has derivatives of every order in this neighbourhood. Then we can certainly form the power (n) ∞ f ( x0 )
series Σn=0
n!
(x-x0)n.
Does this series converge for any x besides x = x0? If so, is its sum equal to f(x)? In general, the answer to both questions is “No”. A necessary and sufficient condition for answering both question in the affirmative can be given by using Taylor’s formula (f(x) = f(x0) +Σ
(k ) n-1 f ( x 0 ) k=1 k !
k
(x-x0) +
f (n) ( x 1 ) n!
(x-x0)n), but first we introduce further terminology and notation.
Definition 13-29. Let f be a real-valued function defined on an interval I in E 1. If f has derivatives of every order at each point of I, we write f ∈ C∞ on I. If f ∈ C∞ on some neighbourhood of a point x0, the power series Σn=0∞(f (n)(x0)/n!)(x-x0)n is called the Taylor’s series about x0 generated by f. To indicate that f generates this series, we write f(x) ~ Σn=0∞ (f (n)(x0)/n!)(x-x0)n. The question we are interested in is this: When can we replace the symbol ~ by the symbol =? Taylor’s formula states that if f ∈ C ∞ on the closed interval [a, b] and if x 0 ∈ [a, b], then, for every x in [a, b], and for every n, we have f(x) = Σk=0n-1 (f (k)(x0)/k!)(x-x0)k + (f (n)(x1)/n!)(x-x0)n, where x1 is some point between x and x0. The point x1 depends on x, x0, and on n. Hence a necessary and sufficient condition for the Taylor’s series to converge to f(x) is that limn→∞(f (n)(x1)/n!)(x-x0)n = 0. In practise it may be quite difficult to deal with this limit because of the unknown position of x 1. In some cases, however, a suitable upper bound can be obtained for f (n)(x1) and the limit can be shown to be zero. Since (x-x0)n/n! → 0 as n → ∞, limn→∞(f (n)(x1)/n!)(x-x0)n = 0 will certainly hold if the sequence {f (n)} is uniformly bounded on [a, b]. Thus we can state the following sufficient condition for representing a function by a Taylor’s series. Theorem 13-30. Assume that f ∈ C∞ on [a, b] and let x 0 ∈ [a, b]. Assume that there is a neighbourhood N(x0) and a constant M (which might depend on x0) such that |f (n)(x)| ≤ M for every x in N(x0)∩[a, b] and every n = 1, 2, .... Then, for each x in N(x 0)∩[a, b], we have f(x) = Σn=0∞ (f (n)(x0)/n!)(x-x0)n. Theorem 13-31 (Bernstein’s Theorem). Assume that f ∈ C∞ on an open interval of the form (a-δ, b), where δ > 0, and suppose that f and all its derivatives are non-negative in the half-open interval [a, b). Then, for every x0 in [a, b), we have f(x) = Σk=0∞ (f (k)(x0)/k!)(x-x0)k , if x0 ≤ x < b. Note that Bernstein’s Theorem gives us a power series expansion about x 0 valid in a half-open interval of the form [x0, b). By Theorem 13-21, this series also converges in a neighbourhood of radius b-x0 about x 0, but we have no guarantee that it represents f(x) if x < x0. However , if x0 is an interior point of [a, b), we can always increase the range of validity of the series f(x) = Σk=0∞ (f (k)(x0)/k!)(x-x0)k , if x0 ≤ x < b, so that it extends to the left of x 0. Theorem 13-32. Assume that f satisfies the hypotheses of the previous theorem and suppose that x0 is an interior point of [a, b). Let R be such that N(x 0;R) ⊂ (a, b). Then f(x) = Σk=0∞ (f (k)(x0)/k!)(x-x0)k , if x0 ≤ x < b, is valid for x in N(x 0;R). The binomial series. As an example illustrating the use of Bernstein’s Theorem, we will obtain the following expansion, known as the binomial series: (1+x)a = Σn=0∞ ( an x) n, if -1 < x < a(a−1)...(a−n+1) a 1, where a is an arbitrary real number, and ( n ) = . Bernstein’s theorem is not n! directly applicable in this case. However, we can argue as follows: Let f(x) = (1-x) -c, where c > 0 and x < 1. Then f (n)(x) = c(c+1)...(c+n-1)(1-x)-c-n, and hence f (n)(x) > 0 for each n, provided that x < 1.
Therefore, we may apply Theorem 13-32, with x0 = 0, a = -1 and b = 1. We find (1−1 x) c = Σk=0∞( −k c )(-1)k xk , if -1 < x < 1. Replacing c by -a and x by -x in the above, we find that the formula (1+x)a = Σn=0∞ ( an )xn, if -1 < x < 1, is valid for each a < 0. But now the validity of this formula can be extended to all real a by successive integration. Of course, if a is a positive integer, say a = m, then ( mn ) = 0 for n > m, and the formula (1+x) a = Σn=0∞ ( an )xn, if -1 < x < 1, reduces to a finite sum (the Binomial Theorem). Abel’s limit theorem. If -1 < x < 1, integration of the geometric series 1/1-x = Σ n=0∞xn gives us the series expansion log(1-x) = -Σn=1∞ (xn/n), also valid for -1 < x < 1. If we put x = -1 in the right-hand side of log(1-x) = -Σn=1∞ (xn/n), we obtain a convergent alternating series, namely Σ(-1)n+1/n. Can we also put x = -1 in the left-hand side? The next theorem answers this question in the affirmative.
Theorem 13-33 (Abel’s limit theorem). Assume that we have f(x) = Σn=0∞ a nxn, if -r < x < r. If the series also converges at x = r, then the limit lim x→r-f(x) exists and we have limx→r-f(x) = Σn=0∞ anr n. Example: we may put x = -1 in log(1-x) = - Σ n=1∞ (xn/n) to obtain log 2 = Σn=1∞ ((-1)n+1/n). As an application of Abel’s theorem, we can derive the following result on multiplication of series: Theorem 13-34. Let Σn=0∞an and Σn=0∞ bn be two convergent series and let their Cauchy product . If Σn=0∞cn converges, we have Σn=0∞cn = (Σn=0∞an)(Σn=0∞ bn).
Σn=0∞cn denote
Tauber’s theorem. The converse of Abel’s limit theorem is false in general. That is, if f is given by f(x) = Σn=0∞ anxn, and if -r < x < r, then the limit f(r-) may exist but yet the series Σanr n may fail to converge. For example, take an = (-1)n. Then f(x) = 1/(1+x) if -1 < x < 1 and f(x) → ½ as x → 1-. However, Σ(-1)n diverges. A. Tauber (1897) discovered that by placing further restrictions on the coefficients an, one can obtain a converse to Abel’s theorem. A large number of such results are now known and they are referred to as Tauberian theorems. The simplest of these, sometimes called Tauber’s first theorem, is the following:
Theorem 13-35 (Tauber). Let f(x) = Σn=0∞anxn for -1 < x < 1, and assume that limn→∞ na n = 0. If f(x) → S as x → 1-, then Σn=0∞an converges and has sum S.
Chapter 13: Selected Exercises 13-1. Assume that f n → f uniformly on S and that each f n is bounded on S. Show that {f n} is uniformly bounded on S. Answer: A sequence {f n} is said to be uniformly bounded on S if there exists a constant M > 0 such that |f n(x)| ≤ M for all x in S and all n = 1, 2, ... The number M is called a uniform bound for {f n}.
If each f n is bounded on S, then there exists a constant M n such that |f n(x)| ≤ M n for all x in S. If f n → f uniformly on S, then for every ε > 0, there exists an N (depending only on ε) such that n > N implies |f n(x)-f(x)| < ε for every x in S. If f n → f uniformly on S, then given an ε > 0 and an associated ( finite) N, we know that for all values of n > N, all function values f n(x) are within ε of f(x). In other words, every value of f n(x) for n > N is bounded above by f(x)+ε. But f(x) itself is bounded above as it is within ε of every function value f n(x) (using |a-b| = |b-a|). Knowing that f n(x) is bounded above by the value Mn, it follows that f(x) is bounded above by M n+ε, and so every value of f n(x) for n > N is bounded above by Mn+2ε. Choosing n = N+1 for simplicity, we can say that f n(x) is bounded above for n > N by the value M N+1+2ε. We now know that the sequence {f n} is uniformly bounded on S if we have n > N, where N is associated with a particular ε > 0. But what about the function values f 1, f 2, ..., f N? Well, max each of these is bounded above by a particular M, so if we define O = i=1,2,..., N Mi, then O is an upper bound for each the f i from i = 1 to i = N. Summary: we now have a bound (O) for f 1, f 2, ..., f N; and another bound (M N+1+2ε) for f N+1, f N+2, .... Defining P = max(O, M N+1+2ε), it follows that P is an upper bound such that |f n(x)| ≤ P for all x in S and all n = 1, 2, ... In other words, we have shown that the sequence {f n} is uniformly bounded, and that P is an uniform bound for {f n}. QED. 13-2. Define two sequences {f n} and {gn} as follows: f n(x) = x(1+1/n), if x ∈ E1, n = 1, 2, ..., gn(x) = 1/n if x = 0 or if x is irrational , and gn(x) = b+1/n is x is rational , say x = a/ b, b > 0. Let hn(x) = f n(x)gn(x).
(a) (b)
Show that both {f n} and {gn} converge uniformly on every finite interval. Show that {hn} does not converge uniformly on any finite interval.
Answer: (a) Consider that our arbitrary finite interval is given by S = [c, d]. For f n(x), we want to show that f n(x) → f(x) on [c, d] for some function f(x). Claim: f(x) = x. To prove this claim, we need to show that for every ε > 0, there exists an N (depending only on ε) such that n > N implies |f n(x)-f(x)| < ε for all x ∈ S.
Consider an arbitrary ε > 0. We need to find an N such that n > N implies |f n(x)-f(x)| < ε, |x(1+1/n)-x| < ε, |x/n| < ε, |x|/n < ε If we choose n > |x|/ε, then |x|/n < ε certainly holds. But we know that |x| ≤ max{|c|, |d|}, so that if we choose n > max{|c|, |d|}/ε, then |x|/n < ε certainly holds. Therefore, given an arbitrary ε > 0, we have found a suitable N (N = ceiling(max{|c|, |d|}/ ε) such that n > N implies |f n(x)-f(x)| < ε for all x ∈ S, i.e. |x|/n < ε for all x ∈ S. This proves our initial claim. For gn(x), consider that our arbitrary interval is again given by S = [c, d]. We want to show that gn(x) → g(x) on [c, d] for some function g(x). Claim: g(x) = 0 if x = 0 or if x is irrational; and g(x) = b if x is rational, say x = a/ b, with b > 0. To prove this claim, we need to show that for every ε > 0, there exists an N (depending only on ε) such that n > N implies |gn(x)-g(x)| < ε for all x ∈ S. Consider an arbitrary ε > 0. We need to find an N such that n > N implies |g n(x)-g(x)| < ε. In all cases (with our definition of g(x)), |g n(x)-g(x)| = |1/n| = 1/n. So we need to find an N such that n > N implies 1/n < ε. This happens exactly when n > 1/ε, so that N = ceiling( 1/ε) will do. Therefore, given an arbitrary ε > 0, we have found a suitable N (N = ceiling( 1/ε)) such that n > N implies |gn(x)-g(x)| < ε for all x ∈ S. This proves our second claim. 13-4. Assume that f n → f uniformly on S and suppose there is a constant M > 0 such that |f n(x)| ≤ M for all x in S and all n. Let g be continuous on the closed disk N(0; M) and define hn(x) = g[f n(x)], h(x) = g[f(x)], if x ∈ S. Show that h n → h uniformly on S. Answer: Because g is continuous on N(0; M), and because |f n(x)| ≤ M for all x in S and all n, then g is uniformly continuous on N(0; M). It follows that given an ε1 > 0, there exists a δ1 > 0 (dependent only on ε1) such that if |x-y| < δ1, then |g(x)-g(y)| < ε1. Because we know that f n → f uniformly on S, then given a δ1 > 0 (the same one as above), then there exists an N such that for all n > N and all x ∈ S, we have |f n(x)-f(x)| < δ1.
But if |f n(x)-f(x)| < δ1, then we have |g(f n(x))-g(f(x))| < ε1. In other words, given an ε1 > 0, there exists an N (the same one as above — which was dependent only on δ1, which in turn was only dependent on ε1 — so the N is only dependent on ε1) such that if n > N and if x ∈ S, then we have |g(f n(x))-g(f(x))| < ε1, i.e. h n → h uniformly on S. QED.
13-6. Let {f n} be a sequence of continuous functions defined on a compact set S and assume that {f n} converges pointwise on S to a limit function f. Show that f n → f uniformly on S if, and only if, the following two conditions hold:
(i) (ii)
The limit function f is continuous on S. For every ε > 0, there exists an m > 0 and a δ > 0 such that n > m and |f k( x)-f(x)| < δ implies |f k+n(x)-f(x)| < ε for all x in S and all k = 1, 2, ...
Hint: To prove the sufficiency of (i) and (ii), show that for each x0 in S there is a neighbourhood N(x 0) and an integer k (depending on x 0) such that |f k( x)-f(x)| < δ if x ∈ N(x 0). By compactness, a finite set of integers, say A = {k 1, ..., k r }, has the property that, for each x in S, some k in A satisfies |f k( x)-f(x)| < δ. Uniform convergence is an easy consequence of this fact. Answer: If . Assume that f n → f uniformly on S. We know that every function f n is continuous at every point x0 in S. It therefore follows by Theorem 13-3 that the limit function f is continuous on S. This proves part (i).
If f n → f uniformly on S, then given an ε > 0, there always exists an m > 0 (dependent only on ε) such that if n > m, then |f n(x)-f(x)| < ε. If we now choose any arbitrary δ > 0, then if we have |f k( x)-f(x)| < δ, then we automatically have |f k+n(x)-f(x)| < ε precisely because |f n(x)-f(x)| < ε. Conclusion: For every ε > 0, there exists an m > 0 and a δ > 0 such that n > m and |f k( x)-f(x)| < δ implies |f k+n(x)-f(x)| < ε for all x in S and all k = 1, 2, .... This proves part (ii). Only If. Assume that conditions (i) and (ii) hold. We need to show that f n → f uniformly
on S. Now because we know that {f n} converges pointwise on the set S, then for every x 0 ∈ S, we know that given an ε > 0, there exists an N > 0 (depending only on x 0 and on ε) such that n > N implies |f n(x0)-f(x0)| < ε/3. Our task is to find an N (depending only on a given ε > 0) such that n > N implies |f n(x)-f(x)| < ε for all x ∈ S. Because the limit function f is continuous on S, and because S is a compact set, then f is uniformly continuous on S. Therefore, given an ε > 0, there is a δ1 > 0 (dependent only on ε) such that if we have |x-x 0| < δ1, then |f(x)-f(x0)| < ε/3 (x, x0 ∈ S). In other words, if x ∈ N δ1(x0), then |f(x)-f(x0)| < ε/3 (---(1)). Similarly, the functions f n are all uniformly continuous on the set S. Therefore, for every given integer k > 0, and given an ε > 0, there is a δ2 > 0 such that if |x-x 0| < δ2, then |f k( x)-f k( x0)| < ε/3 (x, x0 ∈ S). In other words, if x ∈ Nδ2(x0), then |f k( x)-f k( x0)| < ε/3 (---(2)). Now what can we say about |f k( x)-f(x)|? If we add equations (1) and (2) together, then for a given x ∈ Nδ3(x0), where δ3 = min{δ1, δ2}, we have (see over)
|f(x)-f(x0)| + |f k( x)-f k( x0)| < ε/3 + ε/3 |f(x0)-f(x)| + |f k( x)-f k( x0)| < 2ε/3 |f(x0)-f(x)+f k( x)-f k( x0)| < 2ε/3 |f k( x)-f(x) + (f(x0)-f k( x0))| < 2ε/3.
(using |A-B| = |B-A|) (using |A+B| ≤ |A| + |B|)
Now as |f k( x0)-f(x0)| < ε/3 for all k > N, then |f(x0)-f k( x0)| < ε/3 for all k > N, and so |f k( x)-f(x)| < ε/3 + 2ε/3 (for all x ∈ Nδ3(x0)), |f k( x)-f(x)| < ε (for all x ∈ Nδ3(x0)). Therefore, for each x0 ∈ S, and given an ε > 0, there is a neighbourhood Nδ3(x0) and an integer k > N (where N corresponds to a particular ε > 0 and is dependent on x 0) such that |f k( x)-f(x)| < ε for x ∈ Nδ3(x0). Let us now define an open covering of S given by the union of all neighbourhoods 4 Nδ3(x), where x ∈ S. Therefore, we have defined an open covering P = xcS Nδ3(x) ⊃ S. Because S is a compact set, this open covering of S reduces to a finite covering of S, so that a finite number of such neighbourhoods cover the set S. Let this collection of neighbourhoods be denoted as Q = {N δ3(x1), Nδ3(x2), ..., Nδ3(xq)}, where each xi ∈ S. For each of the above neighbourhoods, we have an integer k i so that for any x in a particular neighbourhood, we have |f n(x)-f(x)| < ε for n > k i. If we define K = max i k i, where i = 1, ..., q, then we see that for any x ∈ S, we have |f n(x)-f(x)| < ε for any n > K, where K depends solely on ε. Therefore, we conclude that f n → f uniformly on S. QED. 13-7. (a)
Use Exercise 13-6 to prove the following theorem of Dini: If {f n} is a sequence of real-valued continuous functions converging pointwise to a continuous limit function f on a compact set S, and if f n(x) > f n+1(x) for each x in S and every n = 1, 2, ..., then f n → f uniformly on S.
(b)
Use the sequence in Exercise 13-5(a) to show that compactness of S is essential in Dini’s theorem.
Answer: (a) Exercise 13-6 says the following:
Let {f n} be a sequence of continuous functions defined on a compact set S and assume that {f n} converges pointwise on S to a limit function f. To show that f n → f uniformly on S, we only need to show that the following two conditions hold: (i) (ii)
The limit function f is continuous on S. For every ε > 0, there exists an m > 0 and a δ > 0 such that n > m and |f k( x)-f(x)| < δ implies |f k+n(x)-f(x)| < ε for all x in S and all k = 1, 2, ...
But we already know that condition (i) holds because it is assumed in the question. Therefore, to show that f n → f uniformly on S, all we need show is that condition (ii) holds.
Because we know that the sequence of functions {f n} converges pointwise to a continuous limit function f, then given any δ > 0, we can always find an integer k such that |f k( x)-f(x)| < δ. But if we are then given another number ε > 0, then (a) if ε > δ, then it follows immediately that |f k+n(x)-f(x)| < ε for any integer n because we know (from the question) that f n(x) > f n+1(x) for all x ∈ S. (b) If ε < δ, then again because f n(x) > f n+1(x) for all x ∈ S, and because {f n} converges pointwise to f, there will always be an integer n such that |f k+n(x)-f(x)| < ε. Therefore, for every ε > 0, there exists an m > 0 and a δ > 0 such that n > m and |f k( x)-f(x)| < δ implies |f k+n(x)-f(x)| < ε for all x in S and all k = 1, 2, ... This shows that condition (ii) above holds, and therefore we conclude that f n → f uniformly on S. QED. 1
(b) Recall that the sequence in Exercise 13-5(a) was the following sequence: f n(x) = /(nx+1) for 0 < x < 1 and n = 1, 2, ... Let us analyse the properties of this sequence of functions.
First of all, every function f n(x) is continuous because f n(x) is a rational function and the range of x does not include the single discontinuity of f n(x) which occurs at x = - 1/n for each n = 1, 2, ... Secondly, the functions f n(x) converge pointwise to a function f(x) (Exercise 13-5 asks you to show that this is true). The function f(x) to which the functions f n(x) converge pointwise to is f(x) = 0 (as n → ∞, the denominator of f n(x) tends to ∞ and thus f n(x) tends to 0). It follows immediately that f(x) is a continuous function because it is a constant function. Thirdly, we see that we have f n(x) > f n+1(x) for all x ∈ (0, 1): if we fix x, then we see that the denominator of f n(x) is smaller than the denominator of f n+1(x), and thus f n(x) is larger than f n+1(x). Therefore, as f n(x) > f n+1(x), then we definitely have f n(x) > f n+1(x) for all x in (0, 1). Finally, we note that the set S = (0, 1) is not compact (it is an open set). Looking at the information on the previous page, we note that all the conditions of Dini’s theorem are satisfied but for the compactness of the set S on which the functions f n and the function f are defined upon. But Exercise 13-5(a) shows that {f n} does not converge uniformly on (0, 1). Therefore, we conclude that the compactness of S is essential in Dini’s theorem. QED. 13-10. Assume that gn+1(x) ≤ gn(x) for each x in T and each n = 1, 2, ..., and suppose that gn → 0 uniformly on T. Show that Σ(-1)n+1gn(x) converges uniformly on T. Answer: Let us define the function f n(x) = (-1)n+1 defined for all x the nth partial sum of the series Σ f n(x). Now
∈ T. Let Fn(x) denote
F1(x) = (-1)² = 1, F2(x) = (-1)²+(-1)³ = 1-1 = 0, F3(x) = (-1)²+(-1)³+(-1)4 = 1-1+1 = 1, F4(x) = (-1)²+(-1)³+(-1)4+(-1)5 = 1-1+1-1 = 0, etc. From the above, we see that {F n} is uniformly bounded above by the number 1. Knowing this, knowing that {gn} is a sequence of real-valued functions such that g n+1(x) ≤ gn(x) for each x in T and for all n = 1, 2, 3, ...; and knowing that gn → 0 uniformly on T, then we can apply Dirichlet’s test for uniform convergence (Theorem 13-15) to say that the series Σf n(x)gn(x) = Σ(-1)n+1gn(x) converges uniformly on T. QED.
13-11. Prove Abel’s test for uniform convergence: Let {gn} be a sequence of real-valued functions such that gn+1(x) ≤ gn(x) for each x in T and for every n = 1, 2, ... If {g n} is uniformly bounded on T and if Σf n(x) converges uniformly on T, then Σ f n(x)gn(x) also converges uniformly on T. Answer: If {gn} is uniformly bounded on T, then there exists a positive real number M such that |gn(x)| < M for all x ∈ T and for all n = 1, 2, ... To prove that Σf n(x)gn(x) converges uniformly on T, we need to show that for every ε > 0, there is an integer N such that n > N n+ p implies that we have | S k =n+1 f k( x)gk (x)| < ε for each p = 1, 2, ... and for every x in T.
Now because {gn} is uniformly bounded above by M, we have (for any integers n and p) n+ p n+ p | S k =n+1 f k( x)gk (x)| ≤ | S k =n+1 f k( x)|×|M|. Knowing that Σf n(x) converges uniformly on T, then we know that given an ε2 > 0, there n + p exists an integer N2 such that if n2 > N2, then we have | S k =2n 2 +1 f k( x)| < ε2 for each p = 1, 2, ... and for every x in T. Therefore, n + p n + p | S k =2n 2 +1 f k( x)gk (x)| ≤ | S k =2n 2 +1 f k( x)|×|M| < ε2×|M|. So, given an ε > 0, pick ε2 so that ε2 < ε/|M|, or ε2×|M| < ε. By the above discussion, there exists an integer N2 (associated with ε2 and thus now with ε) such that if n > N2, then we have n+ p | S k =n+1 f k( x)gk (x)| < ε2×|M| < ε for each p = 1, 2, ... and for every x in T . This proves that Σf n(x)gn(x) converges uniformly on T. QED. 13-14. Let {f n} be a sequence of real-valued continuous functions defined on [0, 1] and assume that f n → f uniformly on [0, 1]. Prove or disprove 1−1/n 1 lim f n ( x)dx = ° 0 f ( x)dx. ndº ° 0 Answer: In this question, we shall apply Theorem 13-17:
Theorem (Arzelà): Assume that {f n} is boundedly convergent on [a, b] and suppose each f n is Riemann-integrable on [a, b]. Assume also that the limit function f is Riemann-integrable on [a, b]. Then limn→∞ ∫ ba f n(x)dx = ∫ ba limn→∞ f n(x)dx = ∫ ba f(x)dx. We know that our sequence {f n} is a sequence of real-valued continuous functions defined on [0, 1]; and we know that f n → f uniformly on [0, 1]. Is {f n} boundedly convergent on [0, 1]? To be boundedly convergent, {f n} must be pointwise convergent and uniformly bounded on [0, 1]. Because we know that f n → f uniformly on [0, 1], then we certainly know that {f n} is pointwise convergent. Is {f n} uniformly bounded on [0, 1]? To be uniformly bounded, all we must show is that each individual function is bounded because we know that f n → f uniformly on [0, 1] (see Exercise 13-1). Because each function f n is continuous and because [0, 1] is a compact set (every closed and bounded set such as [0, 1] is a compact set — see the note following definition 3-39), then every function f n is uniformly continuous on [0, 1] (see Theorem 4-24).
Because each function f n is uniformly continuous on the bounded set [0, 1], then f must be bounded on the set [0, 1] — see Exercise 4-23. Therefore, knowing that each function f n is bounded on [0, 1], we know that {f n} is uniformly bounded on [0, 1], and so {f n} is boundedly convergent on [0, 1]. The next question to ask ascertains as to whether each function f n is Riemann integrable or not on [0, 1]. Looking at Theorem 9-27 , and knowing that each function f n is continuous on [0, 1], then we can say with certainty that each f n is Riemann integrable on [0, 1]. Similarly, because f is continuous on [0, 1] (we know that {f n} is a sequence of continuous functions defined on a compact set [0, 1], and that f n → f uniformly on [0, 1] — so that we can apply part (i) of Theorem 13-6 to say that f is continuous on [0, 1]), then f is Riemann integrable on [0, 1] as well. Now that we have shown that all the conditions of Theorem 13-17 have been satisfied when applied to this exercise, we can say the following: lim 1 1 lim 1 n→∞ ∫ 0 f n(x)dx = ∫ 0 n→∞ f n(x)dx = ∫ 0 f(x)dx. 1−1/n
lim ndº
But as limn→∞ ∫ 10 f n(x)dx = limn→∞ ° 0 f n(x)dx (as limn→∞ 1-1/n = 1), then we can say that ° 01−1/n f n ( x)dx = ° 01 f ( x)dx, and so we have proved the hypothesis asked in the question. QED. 13-17. Show that the series Σn=1∞ ((-1)n/√(n))sin(1+(x/n)) converges uniformly on E1. Answer: Let us define
f n(x) =((-1)n/√(n))sin(1+(x/n)). Taking the derivative, we get f’n(x) = ((-1)n/n×√(n))cos(1+(x/n)).
If n = 1, 2, 3, ...., then we see that the derivative f’ n(x) exists for all x ∈ E 1 as cos(1+x/n) ∈ [-1, 1] for all x ∈ E1. If we consider the point x = 0, then the series Σf n(0) is given by
Σf n(0) = Σn=1∞ ((-1)n/√(n))sin(1+(0/n)) = sin(1)Σn=1∞ ((-1)n/√(n)). Let us now define the terms a n = 1/√n. It is obvious that a n is a decreasing sequence, i.e. that an+1 ≤ an, and we can prove that an → 0 as n → ∞: Claim: limn→∞ an = 0. To prove the above claim, we need to find an N such that if n > N, then |a n-0| < ε, or |an| < ε, |1/√n| < ε, 1/√n < ε, 1/n < ε², n > 1/ε². Therefore, if we choose N = ceiling(1/ε²), then given an arbitrary ε > 0, and if we have n > N, then we get |an-0| < ε. Thus the claim that a n converges to zero has been proven.
Theorem 12-16 says that if {an} is a decreasing sequence converging to zero, which our series {an} is, then the alternating series Σ(-1)nan converges. Therefore, using this theorem, we can say that the series Σf n(0) converges (Σn=1∞ ((-1)n/√(n)) converges so sin(1)Σn=1∞ ((-1)n/√(n)) converges as well), thus proving that the series Σf n(x) converges for at least one point x0 in E1. Our next task is to show that the series Σf’n(x) converges uniformly on E1. To do this, we shall use Weierstrass’ M-test (Theorem 13-7). Looking at Σ f’n(x) = Σn=1∞ ((-1)n/n×√(n))cos(1+(x/n)), we can split it up into two parts:
Σf’n(x) = Σn=1∞((-1)n/n×√(n))cos(1+(x/n)) = Σn=1∞((-1)n/n×√(n))×cos(1+(x/n)) Now the blue bit is bounded above by 1 for all possible x and n as cos(z) ∈ [-1, 1] for any real number z. The red bit is also bounded above by 1 for all possible n because 1/(n× √n) = 1/n3/2 is another decreasing sequence bounded above by 1 (for n = 1, 2, ...). Therefore, we can say that because cos(1+(x/n)) ∈ [-1, 1] for all x and for all n, then each term f’n(x) is bounded above by |((-1)n/n3/2)|×1 or simply by (1/n3/2) (to repeat, f’n(x) cannot possibly be larger than this if cos(1+(x/n)) ∈ [-1, 1]). In other words, we have 0 ≤ |f’n(x)| ≤ Mn, where Mn = 1/n3/2. By Weierstrass’ M-test , to show that Σf’n(x) converges uniformly on E1, all we need do is to show that ΣMn converges. To show that ΣMn = 1/n3/2 converges, we use the Integral Test (see Theorem 12-23 and Thomas & Finney, Chapter 8, Section 8.4, Example 3: The p-series Σn=1∞ 1/n p converges if p > 1 and diverges if p ≤ 1). Therefore, we have shown that because ΣMn converges, we have the conclusion that Σf’n(x) converges uniformly on E1. Therefore, knowing that each f n is a real-valued function defined on E1 (if n = 1, 2, 3, ..., and because sin(z) ∈ [-1, 1] for all z), knowing that the derivative f’n(x) exists for each x in E 1, knowing that the series Σf n(x0) converges for x0 = 0, and knowing that Σf’n(x) converges uniformly on E1, we can apply Theorem 13-14 to say that the series Σf n(x) converges uniformly on E1. QED. 13-19. Show that Σn=1∞ |an| converges.
Σn=1∞ ansin(nx) and Σn=1∞ ancos(nx) are uniformly convergent on E1 if
Answer: Given any numbers n and x, recall that |sin(nx)| ≤ 1 and |cos(nx)| ≤ 1. Let us define f n(x) = ansin(nx) and define gn(x) = ancos(nx). Because of the above, we see that |f n(x)| ≤ |an|×1 = |an|, and that |gn(x)| ≤ |an|×1 = |an|. In other words, 0 ≤ |f n(x)| ≤ |an| and 0 ≤ |gn(x)| ≤ |a n|. Applying Weierstrass’ M-test , we see that Σn=1∞ f n(x) = Σ n=1∞ ansin(nx) converges uniformly on E1 if Σn=1∞ |an| converges; and that Σ n=1∞ gn(x) = Σn=1∞ ancos(nx) converges uniformly on E1 if Σn=1∞ |an| converges. QED.
13-25. Let f n(x) = cosnx if 0 ≤ x ≤ π.
(a) (b)
Show that l.i.m.n→∞ f n = 0 on [0, π] but that {f n(π)} does not converge. Show that {fn } converges pointwise but not uniformly on [0, π/2].
Answer: Before we start on the answer, recall the following definition from page 5 of the notes: Definition 13-18. Let {f n} be a sequence of Riemann-integrable functions defined on [a, b]. Assume that f ∈ R on [a, b]. The sequence {f n} is said to converge in the mean to f on [a, b], and we write l.i.m. n→∞ f n = f on [a, b], if limn→∞ ∫ ba |f n(x)-f(x)|²dx = 0.
(a) Looking at the above definition, we need to show that limn→∞ ∫ 0π |(cosnx-0)|²dx = 0. We can remove the modulus signs because we are squaring one term, leaving us to prove that π lim n→∞ 0
∫ (cosnx)²dx = 0, or that limn→∞ ∫ 0π cos2nxdx = 0.
Looking at the back of Thomas & Finney in the Table of Integrals, we find that n−1 ∫ cosnaxdx = cos naax sin ax + n−n1 ° cosn−2 axdx. Therefore,
π lim n→∞ 0 lim n→∞
∫ cos2nxdx
= ([(cos2n-1(x)sin(x))/n]π0 + (2n-1/2n)∫ 0π cos2n-2(x)dx). Now as sin(0) = 0 and as sin(π) = 0, the above reduces to lim 2n-1 /2n)∫ 0π cos2n-2xdx. n→∞ ( We can apply the result from Thomas & Finney iteratively, so we obtain lim 2n-1 /2n)∫ 0π cos2nxdx n→∞ ( = limn→∞ (2n-1/2n)(2n-3/2n-2)(2n-5/2n-4)....(7/8)(5/6)(3/4)∫ 0π cos2xdx. Looking at the Table of Integrals again, we find that ∫ cos²ax = x/2 + sin(2ax)/4a + C. Therefore, ∫ 0πcos²xdx = [x/2+sin(2x)/4]π0 = [(π/2+sin(2π)/4) - (0/2+sin(0)/4)] = π/2. It follows that
π lim 2n-1 2n-3 2n-5 7 5 3 2 n→∞ 2n 2n-2 2n-4 8 6 4 0 π lim 2n-1 2n-3 2n-5 7 5 3 n→∞ 2n 2n-2 2n-4 8 6 4 2 π lim 2n-1 2n-3 2n-5 7 5 3 2 n→∞ 2n 2n-2 2n-4 8 6 4 π lim n 1+2k 2 n→∞ k=1 2+2k π ∞ 1+2k 2 k=1 2+2k
/ )( / )( / )....( / )( / )( / )∫ cos xdx = ( / )( / )( / )....( / )( / )( / )( / ). =(/) ( / )( / )( / )....( / )( / )( / ). =(/) Π ( / ) = ( / )Π ( / ) (
Now let us define pn = Πk=1n(1+2k /2+2k ). Examples: p1 = ¾, p2 = 3/4×5/6, p3 = p2(7/8), etc.
From the above, we see that {pn} is a strictly monotonically decreasing sequence bounded above by ¾. Therefore, applying Theorem 12-6 (A monotonic sequence converges if, and only if, it is bounded ), we can say that the sequence {p n} converges. But what does it converge to? It must converge to zero because (intuitively) we are constantly multiplying together numbers in the range (0, 1). I will not prove that it converges to zero and assume that this is the case, but just note that a proof can be obtained by modifying the proof in Exercise 4-5 or the proof that appears below in part (b). Therefore, limn→∞ pn = 0, and so ( π/2)Πk=1∞ (1+2k /2+2k ) = (π/2)×0 = 0, and it follows as π π ∞ 1+2k lim n /2+2k ) = 0, we can say that l.i.m. n→∞f n = 0 on [0, π]. n→∞ ∫ 0 |(cos x-0)|²dx = ( /2)Πk=1 ( Our next task is to show that {f n(π)} does not converge. We can do this simply by observing that f n(π) = (-1)n for all n, a series that alternates between -1 and 1 and so does not converge. QED. (b) We now want to show that {f n} converges pointwise on [0, π/2]. There are two cases to deal with here: when x = 0, and when x ∈ (0, π/2]. If x = 0, then because cos(0) = 1, we have cosn(0) = 1n = 1 for all n. It follows that cos n(x) converges pointwise to 1 if x = 0. If x ∈ (0, π/2], then cos(x) ∈ [0, 1). Now consider an arbitrary y ∈ [0, 1). Claim: limn→∞ yn = 0. In order to prove this claim, given an arbitrary ε > 0, we need to find an N such that if n > N, we have yn < ε. But if yn = ε, then n = log ε/logy, and so if n > ceiling(log ε/logy), then yn < ε as required. Note that this will only work if y ∈ [0, 1) (which is all right!) and if ε ∈ (0, 1). But if ε > 1, then yn < ε holds automatically for y ∈ [0, 1), so we don’t have to worry about this case. Therefore, we have just proved that limn→∞ yn = 0. It follows that if we set y = cos(x) (which we can do because cos(x) ∈ [0, 1) for x ∈ (0, π/2]), then limn→∞ cosnx = 0. It follows that f n(x) converges pointwise for x ∈ [0, π/2]: to one if x = 0, and to zero if x ∈ (0, π/2]. It remains to show that {f n} does not converge uniformly on [0, π/2]. I will skip this! 13-26. Let f n(x) = 0 if 0 ≤ x ≤ 1/n or if 2/n ≤ x ≤ 1, and let f n(x) = n if 1/n < x < 2/n. Show that {f n} converges pointwise to 0 on [0, 1] but that l.i.m.n→∞ f n ≠ 0 on [0, 1]. Answer: To show that {f n} converges pointwise to 0 on [0, 1], we must show that limn→∞ f n(x) = 0 for all x ∈ [0, 1]. To show this, for every x ∈ [0, 1], and for a given ε > 0, we must find an N so that if n > N, we have f n(x) < ε.
Now if x = 0, then f n(x) = 0 (by definition) for all n. Therefore, f n(x) < ε for all ε > 0 and all n if x = 0, i.e. limn→∞ f n(x) = 0 if x = 0. If x ∈ (0, 1], then in order to prove that limn→∞ f n(x) = 0, all we need do is to find an N so that if n > N, then 2 /n < x. If this is the case, then for all n > N, we have x ∈ [2/n, 1] so that f n(x) = 0 < ε for all n > N, and so limn→∞ f n(x) = 0 for this particular x ∈ (0, 1]. Claim: For a particular x ∈ (0, 1], if N = ceiling( 2/x), then for all n > N, we have 2/n < x. Proof of claim: if n > 2/x, then we can manipulate to give 2/n < x as required (the inequality doesn’t change because we have n > 0 and x > 0). It automatically follows that if n > ceiling(2/x), then 2/n < x as required. Therefore, we have found an N (dependent on x) so that if n > N, then we have x ∈ [ 2/n, 1], and so f n(x) = 0 < ε for all n > N, and so limn→∞ f n(x) = 0 for all x ∈ (0, 1]. It remains to show that l.i.m. n→∞ f n ≠ 0 on [0, 1]. To do this, we must show that lim 1 n→∞∫ 0 |f n(x)-0|²dx ≠ 0. Manipulating the left hand side of this equation, we get
∫ |f n(x)-0|²dx ∫ 10 |f n(x)|²dx [∫ 01/n|0|²dx + ∫ 1/n2/n |n|²dx + ∫ 2/n1 |0|²dx] (because n = 1, 2, ...) ∫ 1/n2/n n²dx
lim 1 n→∞ 0 lim n→∞ lim n→∞ lim n→∞ lim n→∞ lim n→∞ lim n→∞ lim n→∞
= = = = [n²x]1/n2/n = [n²(2/n-1/n)] = [n(2-1)] = n = ∞ ≠ 0. QED. 13-27.
If r is the radius of convergence of Σan(z-z0)n, show that lim sup a n lim inf a n ndº | a n+1 | [ r [ ndº | a n+1 |.
Answer: The important factor about this question is to show that the radius of convergence can be defined in another way (under certain conditions), different to the definition that forms part of Theorem 13-21 (that derives from the root test, Theorem 12-26).
Consider that we derive a definition for r by considering the ratio test. Consider the power series Σan(z-z0) . For each n > 1, define r n = n
Now
lim ndº r n
|z − z 0 | r .
=
lim a n+1 ndº | a n
|| z − z 0 | =
|z − z 0 | an lim ndº | a n+1
|
a n+1 ( z − z 0 ) n+1 | a n ( z − z 0 ) n
| = |
a n+1 an
. If we define r =
|| z − z 0 |.
an lim ndº | a n+1
|, then we see that
Claim: r is the radius of convergence of the series Σ an(z-z0)n. Proof of Claim. By the ratio test (not the one shown in Theorem 12-25 — see the one for example in Thomas & |z − z 0 | Finney, Section 8.6), the series converges absolutely when r < 1, that is, |z-z0| < r; and z − z 0 diverges when r > 1, that is, |z-z0| > r. If r = 0, the test shows divergence except for z = z 0 (I a lim will not prove this). If r = ∞, so that ndº | a n+n1 | = 0, the test shows that the series converges absolutely for all z. Comparing the above to the definition of radius of convergence in an m Theorem 13-21, it follows that the r defined above as r = ndº | a n+1 |is the radius of convergence of the series Σ an(z-z0)n. End of Proof. lim ndº r n
=
Note: Portions of the above proof were taken from “ Advanced Calculus with Applications” by Nicholas J. DeLillo, 1982, QA303.D4: see section 7.8, Theorem 7-30. Therefore, we have found an alternative definition for the radius of convergence: an r = nlim dº | a n+1 |. If this limit exists, then there is no problem and we have lim sup a n an lim inf a n lim ndº | a n+1 | = ndº | a n+1 | = ndº | a n+1 |.
However, if the limit does not exist, then we have (using Theorem 12-3) lim sup a n an lim inf a n lim ndº | a n+1 | [ ndº | a n+1 | [ ndº | a n+1 |, lim sup a n inf a n or lim | | r [ [ ndº a n+1 ndº | a n+1 | as required . QED. 13-30. Let f(x) = e-1/x² if x ≠ 0, f(0) = 0.
(a) (b)
Show that f(n) (0) exists for all n > 1. Show that the Taylor’s series about x0 = 0 generated by f converges everywhere on E1 but that it represents f only at the origin.
Answer: In this question, I shall use f’(c) = lim f ( x)− f (0) xd0 x−0
lim f ( x)− f (c) xdc x−c .
2 lim e −1/ x −0 xd0 x−0
1 −1/ x Case n = 1. f’(0) = . = = xlim d0 x e Let t = 1/x, then t → ∞ as x → 0+, and t → -∞ as x → 0-. Further, lim 1 −1/ x 2 lim 1 −1/ x 2 lim −t 2 = 0. e e te = = x − x d º t xd0 xd0 Hence f’(0) = 0. If x ≠ 0, then f’(x) = ( 2/x³)e-1/x².
lim xd0
2 −1/ x 2 −0 e x 3
Case n = 2. f’’(0) = x−0 1 If we use t = /x again, we have lim xd0
2 −1/ x 2 e x 3
x
=
lim 4 −t 2 t dº 2t e
Hence f’’(0) = 0. If x
=
lim xd0
2 −1/ x 2 e x 3
x
2
.
= 0.
≠ 0, then f’’(x) = ( x46 − x64 )e −1/ x2 .
We may continue calculating f (n)(0) and f (n)(x) for every positive integer in this manner. However, no matter which n is chosen, we always get (for x ≠ 0) f (n)(x) = g(1/x)e-1/x², where g(1/x) is some polynomial in 1/x. Therefore, if we continue to apply t = 1/x, then we shall always have 2 lim f (n)(0) = t dº tf (t )e −t = 0, since e-t² approaches 0 more quickly than any polynomial in t approaches infinity. Thus we conclude that f ∈ C ∞ everywhere in E1, and, in particular, f (n)(0) exists for every n — and we have f (n)(0) = 0 for every n. QED. Note: The above solution was taken from “ Advanced Calculus with Applications” by Nicholas J. DeLillo, 1982, QA303.D4: see section 7.9, Exercise 7-38.
(b) The Taylor’s series about x0 = 0 generated by f is given by f (k ) (0)
0
º k k f(x) = S º k =0 k ! ( x − 0) = S k =0 k ! x = 0 for all x. But looking at the definition of f, we see that f is zero only at x = 0 (e -1/x² > 0 for all x ≠ 0), so that the Taylor series only represents f at the origin i.e. at x = 0. QED.
Possible Further Work / Evaluation − n
Exercise 13-14: Further explanation of the result limn→∞ ∫ 10 f n(x)dx = limn→∞ ° 0 f n(x)dx. Possible method: define gn(x) = f n(x) for 0 ≤ x ≤ 1-1/n, and gn(x) = 0 for 1-1/n < x ≤ 1. Then
°
1
1− n 0 f n x
( ) = ° 01 g n ( x) and we need to show that gn(x) → f(x) uniformly.
Chapter 15
Chapter 15: Fourier Series and Fourier Integrals 15-1.
Derive the Minkowski inequality ||f+g||
≤ ||f|| + ||g|| from the Cauchy-Schwarz
inequality. Answer:
The Cauchy-Schwarz inequality assumes the form |(f, g)| ≤ ||f|| ||g||.
Now ||f+g|| = = =
≤ = = = 15-2.
(f+g, f+g)½ [(f, f) + (f, g) + (g, f) + (g, g)]½ [(f, f) + 2(f, g) + (g, g)]½ [(f, f) + 2(f, f)½(g, g)½ + (g, g)] ½ [ {(f, f)½ + (g, g)½}² ]½ (f, f)½ + (g, g)½ ||f|| + ||g|| as required . QED.
(by definition) (using elementary property 3) (using elementary property 2) (using Cauchy-Schwarz )
Verify that the trigonometric system in equation (6) is orthonormal on [0, 2π].
Answer:
The trigonometric system in question is as follows:
φ0(x) =
1 2o
, φ2n-1(x) =
cos nx o
, φ2n(x) =
sin nx
(n = 1, 2, ...).
o
Now in order to prove that this system is orthonormal on [0, 2π], we must show that the following holds: (1) (φn, φm) = 0 whenever m ≠ n, and (2) ||φn|| = 1 for all n. Looking at the system in question, we see that in order to prove that it is orthonormal, we must prove the following: (a) (d)
(φ0, φ2n-1) = 0, (φ0, φ0)½ = 1,
(a) (φ0, φ2n-1) =
° 02o
= (b) (φ0, φ2n) =
(d) (φ0, φ0) =
[
° 02o cos(nx)dx =
]=
1 o 2
(φ2n-1, φ2n) = 0, (φ2n, φ2n) = 1. 1 o 2
[ sinnnx ] 20o
[0 − 0] = 0.
= o 12 ° 20o sin(nx)dx = o 12 [ −cosn nx ] 20o cos(0) + n ] = o 12 [−1 + 1] = 0.
[
cos nx sin nx
1 2o
o
[
o
dx =
−cos(2nx) 2o 2n
[
1
o
]0 =
1/2 2o 1 2 dx 0 2o [ 21o 2o 0 ] 1/2
° (
=
(c) (e)
1 sin nx dx o 2o −cos(2on) 1 n o 2
° 02o
= ½
= −
1 cos nx 1 dx o o 2 2o sin(2on) sin(0) 1 n n o 2
° 02o
= (c) (φ2n-1, φ2n) =
(φ0, φ2n) = 0, (φ2n-1, φ2n-1)½ = 1,
(b) (d)
) − ]
° 02o cos(nx) sin(nx)dx =
1 4no
1 2o
[− cos(4on) + cos(0)] =
= 21o ° 02o dx = 1 = 1.
1/2
° 02o sin(2nx)dx 1 4no
= [ 21o [ x] 20o ] 1/2
[−1 + 1] = 0.
(e) (φ2n-1, φ2n-1)½ =
° 02o(
= (f) (φ2n, φ2n) = ½
1
o
1
o
o
[ 22o +
° 02o(
=
cos(nx) 2
) dx
sin(4on) 4n
sin(nx) 2 o
[ 22o −
1/2
) dx
1/2
sin(2nx) 2o 1/2 0 4n
= [ + ] 1/2 sin(0) − 02 − 4n ] = [ o1 [o + 0 − 0 − 0] ] 1/2 = 1 = 1.
1/2
sin(4on) 4n
= o ° 02o cos2 (nx)dx 1
=
° 02o sin 2 (nx)dx
1/2
1 x o 2
o
=
sin(2nx) 2o 1/2 0 4n
[ − ] 1/2 sin(0) − 02 + 4n ] = [ o1 [o − 0 − 0 + 0] ] 1/2 = 1 = 1. 1
1 x o 2
QED. Note: I have used some entries from the Table of Integrals at the back of Thomas & Finney to do some of the calculations above. 15-9.
(a) (b)
Show that the expansions below are valid in the ranges indicated. (−1) n cos nx
x = 2 S n=1 , if -π < x < π. n n cos nx 2 (− ) 1 o º x² = 3 + 4 S n=1 , if -π ≤ x ≤ π. n2 º
Answer:
(a) The function f(x) = x is an odd function, so that we can apply the result from Exercise 15-6 which said that if f ∈ R on [-π, π], if f has period 2 π, and if f(-x) = -f(x) when 0 < x < π, then f(x) ~ Σn=1∞ bnsin(nx), where bn = 2/π∫ 0π f(t)sin(nt)dt. For f(x) = x on -π < x < π, we have x ~ now calculate bn for every n > 1.
Σn=1∞ bnsin(nx), where bn = 2/π∫ 0π t sin(nt)dt. Let us
bn = 2/π∫ 0π xsin(nx)dx = 2/π[1/n²sin(nx) - x/ncos(nx)]π0 = 2/π[1/n²sin(nπ) - π/ncos(nπ) - 1/n²sin(0) + 0] = 2/π[0 - π/2(-1)n - 0 + 0] = 2/π(-π/2(-1)n) = -2/n(-1)n = 2/n(-1)n-1. So for f(x) = x on - π < x < π, we have x = (−1) n cos nx º = 2 S n=1 as required. QED. n
(Result taken from Thomas & Finney)
Σn=1∞ bnsin(nx) = Σn=1∞ 2/n(-1)n-1sin(nx)
(b) The function f(x) = x² is an even function, so that we can apply the result from Exercise 15-6 which said that if f ∈ R on [-π, π], if f has period 2 π, and if f(-x) = f(x) when 0 < a x < π, then f(x) ~ 20 +Σn=1∞ ancos(nx), where an = 2/π∫ 0π f(t)cos(nt)dt. For f(x) = x² on -π < x < π, which can be extended to -π ≤ x ≤ π because f(-π) = f(π) for a an even function, we have x² ~ 20 +Σn=1∞ancos(nx), where an = 2/π∫ 0π t² cos(nt)dt. Now a0 = 2/π∫ 0π x²cos(0)dt = 2/π[x³/3]π0 = 2/π[π³/3-0³/3] = 2π³/3π = 2π²/3.
Let us now calculate an for every n > 1. We want to calculate a n = 2/πè02 x²cos(nx)dx. Integrating by parts, we let u = x² so that du/dx = 2x, and let dv/dx = cos(nx) so that v = 1/nsin(nx). It follows that /π∫ 0π x²cos(nx)dx = 2/π{[x²/nsin(nx)]π0 - ∫ 0π 2x/n sin(nx)dx} = 2/π{[π²/nsin(πn)-0] - ∫ π0 2x/nsin(nx)dx} = -2/π∫ π0 2x/nsin(nx)dx = -4/nπ∫ 0π xsin(nx)dx =-4/nπ[1/n²sin(nx) - x/ncos(nx)]π0 =-4/nπ[1/n²sin(πn) - π/ncos(nπ) - 1/n²sin(0) + 0] =-4/nπ[0 - π/n(-1)n - 0 + 0] = 4π/n²π(-1)n = 4/n²(-1)n. 2
Using x² ~
a0 2
(Result taken from Thomas & Finney)
+Σn=1∞ ancos(nx), where a0 = 2π²/3 and an = 4/n²(-1)n, we find that
x² = (2π²/3)½ + Σn=1∞ 4/n²(-1)ncos(nx) (−1) n cos nx o2 º = 3 + 4 S n=1 as required. QED. n2 15-10.
Show that the expansion x² =
x < 2π. Answer:
4 2 3o
+ 4 S nº=1 ( cosn 2nx − o sinn nx ) is valid in the range 0 <
Recall that the Fourier Series generated by f is given by
a
f(x) ~ 20 + S nº=1 (a n cos nx + b n sin nx), where an = 1/π∫ 02π f(t)cos(nt)dt, and bn = 1/π∫ 02π f(t)sin(nt)dt. For f(x) = x², we have a0 = 1/π∫ 02π x²cos(0)dt = 1/π[x³/3]2π0 = 1/π[(2π)³/3-(0)³/3] = 8π³/3π = 8π²/3. Further , a n = 1/π∫ 02π x²cos(nx)dx. Integrating by parts, we let u = x² so that let dv/dx = cos(nx) so that v = 1/nsin(nx). It follows that /π∫ 02π x²cos(nx)dx = 1/π{[x²/nsin(nx)]2π0 - ∫ 02π 2x/n sin(nx)dx} = 1/π{[4π²/nsin(2πn)-0] - ∫ 2π0 2x/nsin(nx)dx} = -1/π∫ 2π0 2x/nsin(nx)dx = -2/nπ∫ 02π xsin(nx)dx =-2/nπ[1/n²sin(nx) - x/ncos(nx)]2π0 =-2/nπ[1/n²sin(2πn) - 2π/ncos(2nπ) - 1/n²sin(0) + 0] =-2/nπ[0 - 2π/n - 0 + 0] = 4π/n²π = 4/n².
du
/dx = 2x, and
1
(Result taken from Thomas & Finney)
Similarly, bn = 1/π∫ 02π x²sin(nx)dx. Integrating by parts again, we let u = x² so that 2x, and let dv/dx = sin(nx) so that v = -1/ncos(nx). It follows that
du
/dx =
/π∫ 02π x²sin(nx)dx = 1/π{[-x²/ncos(nx)]2π0 + ∫ 02π 2x/n cos(nx)dx} = 1/π{[-4π²/ncos(2πn)-0] - ∫ 2π0 2x/ncos(nx)dx} = 1/π(-4π²/n) - 1/π∫ 2π0 2x/ncos(nx)dx = -4π/n - 2/nπ∫ 2π0 xcos(nx)dx = -4π/n - 2/nπ[1/n²cos(nx)+x/nsin(nx)]2π0 (Result taken from Thomas & Finney) 4π 2 1 2π 1 0 = - /n - /nπ[ /n²cos(2πn) + /nsin(2πn) - /n²cos(0) - /nsin(0)] = -4π/n - 2/nπ[1/n² + 0 - 1/n² - 0] = -4π/n. 1
Now that we know that a 0 = 8π²/3, that an = 4/n² for n > 1, and that b n = -4π/n for n > 1, we a can apply the formula f(x) ~ 20 + S nº=1 (a n cos nx + b n sin nx) to say that for 0 < x < 2π, we have x² = (8π²/3)(½) + Σn=1∞ (4/n²cos(nx) - 4π/nsin(nx)) o sin nx cos nx = 43 o 2 + 4 S nº=1 ( n 2 − n ) . QED. 15-22:
(a)
(b)
If f satisfies the hypothesis of the Fourier integral theorem, show that
If f is even, that is, if f(-t) = f(t) for every t, then = o2 ° 0º cos(vx)[° º0 f (u) cos(vu)du ]dv.
f ( x+)+ f ( x−) 2
If f is odd, that is, f(-t) = -f(t) for every t, then = o2 ° 0º sin(vx)[° 0º f (u) sin(vu)du ]dv.
f ( x+)+ f ( x−) 2
Answer:
Let us first recall the Fourier Integral Theorem:
Assume that f ∈ R*(-∞, + ∞). Suppose there is a point x in E 1 and an interval [x-δ, x+ δ] about x such that either (a)
f is of bounded variation on [x-δ, x+δ], or else
(b)
both limits f(x+) and f(x-) exist and both improper integrals f ( x+t )− f ( x+) f ( x−t )− f ( x−) d ° d0+ dt and ° 0+ dt are absolutely convergent . t t
Then we have the formula = o1 ° 0º [° º−º f (u) cos(v(u − x))du]dv.
f ( x+)+ f ( x−) 2
If we manipulate the formula on the previous page, we see that f ( x+)+ f ( x−) 1 º º [ −º f u cos v u x du ]dv o 0 2 º º 1 = o 0 [ −º f u cos vu vx du ]dv º º = o1 0 [ −º f u cos vu cos vx sin vu sin vx du ]dv º 1 º º = o 0 [ −º f u cos vu cos vx du −º f u sin vu sin vx º º 1 º = o 0 [cos vx −º f u cos vu du sin vx −º f u sin vu
° ° ° °
= ° ° ° ( ) ° ( )( ° ( ) ( )°
( ) ( ( − )) ( − )) ( ) ( ) + ( ) ( )) ( ) ( ) +° ( ) ( ) ( ) ( ) + ( )° ( )
( )du ]dv ( )du ]dv (---(1)).
º
Now if f(x) is an odd function, then ° −º f ( x)dx = 0; º º and if f(x) is an even function, then ° −º f ( x)dx = 2 ° 0 f ( x)dx. Further, if f(x) is an odd function, then f(-t) = -f(t) ⇒ f(-t)² = f(-t)f(-t) = (-f(t))(-f(t)) = f(t)² so that f²(x) is an even function. If f(x) is odd and if g(x) is even, then f(-t) = -f(t) and g(-t) = g(t), ⇒ f(-t)g(-t) = -f(t)g(t) so that f(x)g(x) is an odd function. Finally, if g(x) is an even function, then g(-t) = g(t) ⇒ g(-t)² = g(-t)g(-t) = g(t)g(t) = g(t)² so that g²(x) is an
even function.
We can use the above information to simplify expression (1) in the cases where f is or where f is even.
odd
(a) If f is even, then f(u)cos(vu) will be even (as cos(vu) is even), and f(u)sin(vu) will be odd (as sin(vu) is odd). Therefore,
° 0º [cos(vx) ° º−º f (u) cos(vu)du + sin(vx) ° º−º f (u) sin(vu)du ]dv º º = o1 ° 0 [cos(vx) % 2 ° 0 f (u) cos(vu)du + sin(vx)(0) ]dv º º = o2 ° 0 [cos(vx) ° 0 f (u) cos(vu)du ]dv. QED. 1
o
(b) If f is odd, then f(u)cos(vu) will be odd (as cos(vu) is even), and f(u)sin(vu) will be even (as sin(vu) is odd ). Therefore,
° 0º [cos(vx) ° º−º f (u) cos(vu)du + sin(vx) ° º−º f (u) sin(vu)du ]dv º 1 º = o ° 0 [cos(vx)(0) + sin(vx) % 2 ° 0 f (u) sin(vu)du ]dv º 2 º = o ° 0 [sin(vx) ° 0 f (u) sin(vu)du ]dv. QED. 1
o
15-28:
Verify the entries in the following table of Laplace transforms: f(t)
(a) (b) (c) (d)
eαt cos(αt) sin(αt) t peαt
F(z) =
° 0º e-ztf(t) dt, z = x + iy (z-α)-1 z/(z²+α²) α/(z²+α²) Γ (p+1)/(z-α) p+1
(x > α) (x > 0) (x > 0) (x > α, p > 0)
Answer:
(a) F(z) = ∫ 0∞ e-zteαtdt = ∫ 0∞ e(α-z)tdt = lim N→∞[∫ 0 N e(α-z)tdt] 1 = lim N→∞[ a− z e (a− z )t ] N 0 1 lim ( a− z ) N = N→∞[ a− z e − a1− z e 0 ] = 1/z-α + lim N→∞[1/α-ze(α-z)N] = 1/(z-α) + 0 = (z-α)-1 for x > α. QED.
(z = x + iy)
(for x > α)
(b)
F(z) = ∫ 0∞ e-ztcos(αt)dt = lim N→∞[∫ 0 N e-ztcos(αt)] e − zt = lim N→∞[ z 2 +a 2 (− z cos at + a sin at )] N0 (Result taken from Thomas & Finney) − zN = (lim N→∞ z e2 +a 2 (− z cos a N + a sin a N ) - (e0/z²+α²)(-zcos(0)+αsin(0)) = 0 - (1/z²+α²)(-z+0) (for x > 0) z = /z²+α² for x > 0. QED.
(c)
F(z) = ∫ 0∞ e-ztsin(αt)dt = lim N→∞[∫ 0 N e-ztsin(αt)] e − zt = lim N→∞[ z 2 +a 2 (− z sin at − a cos at )] N0 (Result taken from Thomas & Finney) e − zN = (lim N→∞ z 2 +a 2 (− z sin a N − a cos a N ) - (e0/z²+α²)(-zsin(0)-αcos(0)) = 0 - (1/z²+α²)(-0-α) (for x > 0) α = /z²+α² for x > 0. QED.
(d)
F(z) = ∫ 0∞ e-ztt peαtdt = lim N→∞[∫ 0 N e(α-z)tt pdt]
Now let us integrate the above by parts, setting u = t p so that du/dx = pt p-1, and setting dv/dx = e(α-z)t so that v = 1/(α-z)e(α-z)t. It follows that lim N→∞[∫ 0 N e(α-z)tt pdt] = lim N→∞{[(t p/(α-z))e(α-z)t] N0 - ∫ 0 N (pt p-1/(α-z))e(α-z)tdt} = lim N→∞{[0-0] - ∫ 0 N (pt p-1/(α-z))e(α-z)tdt} = lim N→∞[ p/z-α∫ 0 N e(α-z)tt p-1dt] We can now integrate by parts repeatedly, and assuming that p following: lim p N (α-z)t p-1 N→∞ z-α 0 lim p(p-1) N (α-z)t p-2 N→∞ (z-α)² 0 lim p N (α-z)t N→∞ 0
[/ = [ = ... =
> 0, we arrive at the
∫ e
t dt] / ∫ e t dt] [(p!/(z-α) )∫ e dt]
(if p > 0).
We can now use the information gained by doing part (a) to say that [(p!/(z-α) p)∫ 0 N e(α-z)tdt]
lim N→∞
p!
= ( z −a) p ( z −a ) = p!/(z-α) p+1. 1
(if p > 0 and x > α)
Now looking at Exercise 14-31, we see that Γ (n+1) = n! for any integer n > 0 so that p!/(z-α) p+1 = Γ (p+1)/(z-α) p+1 if p > 0 and x > α as required . QED.
Chapter 16
Chapter 16: Cauchy’s Theoremand the ResidueCalculus Exercise 16-1: Assume that f is analytic on a neighbourhood N(z0; R). If 0 < r < R, let C(r) denote the circle of radius r with z 0 as centre and let M(r) denote the maximum of |f| on C(r). Deduce Cauchy’s inequalities:
|f (n) ( z 0 )| [
M (r )n! r n
(n = 0, 1, 2, ...).
Answer: Let us first remind ourselves of a few definitions. Definition. Let f = u + iv be a complex-valued function defined on an open set S in E 2. The function f is said to be analytic on S if the derivative f’ exists and is continuous at every point of S. Theorem (Cauchy’s integral formula). Assume that f is analytic on an open region S in E2. Let Γ be a rectifiable Jordan curve such that both Γ and its inner region lie within S. Then, for every point z0 inside Γ , we have
f ( z 0 ) =
1 2oi
f ( z )
°G[ z ] z − z 0 dz ,
provided that the path Γ [z] is positively orientated. Further, for every integer n > 1, the derivative f (n)(z0) exists and is given by the integral f (n) ( z 0 ) =
n! 2oi
°G[ z ]
f ( z )
( z − z 0 ) n+1 dz ,
again provided that the path Γ [z] is positively oriented. Now knowing that f (n) ( z 0 ) =
n! 2oi
°G[ z ]
f ( z )
( z − z 0 ) n+1 dz ,
if we define the path Γ [z] to be the circle
C(r) defined above, then we can write f (n) ( z 0 ) =
n! 2oi
°C (r )
f ( z )
( z − z 0 ) n+1 dz .
Because C(r) is a circle with radius r centred at z 0, then we can write (z-z0)n+1 = r n+1. f ( z ) f ( z ) Therefore, 2n!oi °C (r ) ( z − z 0 ) n+1 dz = 2n!oi °C (r ) r n+1 dz = (2oin!)r n+1 °C (r ) f ( z )dz . We now need to use Theorem 9-60: If Γ is a rectifiable curve of length Λ(Γ ) described by a complex-valued continuous function z, and if ∫ Γ [z] F(z)dz exists, then we have the inequality
°G[ z ] F ( z )dz [ M L(G), where M is such that |F(z)| ≤ M for all z on Γ .
Before applying the above theorem, let us manipulate our expression to get it into the desired form: f (n) ( z 0 ) =
n! (2oi)r n+1
°C (r ) f ( z )dz u |f (n) ( z 0 )| = | (2oin!)r n+1 °C (r ) f ( z )dz | =
n! (2o)r n+1 | C (r ) f z dz |.
°
()
Applying Theorem 9-60 to | °C (r ) f ( z )dz | (knowing that |f(z)| ≤ M(r) on C(r), where C(r) is a curve of length 2πr), we obtain the following: |f (n) ( z 0 )| =
()
n! n! (2o)r n+1 | C (r ) f z dz | [ 2or n+1
°
Thus |f (n) ( z 0 )| [
n! M (r ) r n
( M (r ) % 2or ) =
n! M (r ) r n+1
.
as required. QED.
Exercise 16-3: Let f = u + iv be analytic on a neighbourhood N(z0; R). If 0 < r < R, show that f’(z0) = 1/πr ∫0 2π u(z0+reiθ)e-iθdθ. Answer: Recall that in the previous exercise, we encountered the formula f ( z ) f ( z ) n! 1 f (n) ( z 0 ) = 2oi °C (r ) ( z − z 0 )n+1 dz . Putting in n = 1, we get the formula f’(z 0) = 2oi °C (r ) ( z − z 0 ) 2 dz . Now let
Γ be a circle with centre at z and radius r, described by the function z, where z(θ) = z0+reiθ, if 0 ≤ θ ≤ 2π. Using this information, we can write f’(z0) =
1 2oi
° 20o
f [ z (h)]
( z (h)− z 0 ) 2 dz (h) =
1 2oi
° 02o
f ( z 0 +re ih )
( z (h)− z 0 ) 2 z (h)d h. Â
Since z’(θ) = ireiθ = i(z(θ)-z0), we obtain 2o f ( z 0 +re ih )(i( z (h)− z 0 )) 1 d h 2oi 0 ( z (h)− z 0 ) 2 2o f ( z 0 +re ih ) 2o 1 d f z 0 h h i 0 0 o 2 r re
°
f’(z0) = =
1 2o
°
=
°
f ( z +re ih )
= 21o ° 02o z (0h)− z 0 d h ( + reih )e −ih d h .
But as f = u + iv, the above can be written as f’(z0) =
1 2or
° 20o u( z 0 + re ih )e −ih d h +
2o ih −ih d h. ° 2or 0 v( z 0 + re )e i
Now because we are integrating over a circle, then we have |u| = |v|, and so 1 2or
° 20o u( z 0 + re ih )e −ih d h =
2o i 2or 0
° v( z 0 + re ih )e −ih d h. (Note: I am not sure about the last bit,
but the conclusion reached is clearly the one we want). Therefore, f’(z0) = =
2o 1 2or 0 u z 0 2o 1 or 0 u z 0
°
°
2o
( + re ih )e −ih d h + 21or ° 0 u( z 0 + re ih )e −ih d h ( + re ih )e −ih d h as required. QED.
Exercise 16-9: Assume that f has the Taylor expansion f(z) = N(z0; R).
(a)
If 0 ≤ r < R, deduce Parseval’s identity: 1 2o
(b)
Σn=0∞ an(z-z0)n, valid in
° 02o |f ( z 0 + re ih )| 2 d h = S nº=0 |a n | 2 r 2n .
Use (a) to deduce the inequality Σn=0∞ |an|²r 2n ≤ M(r)², where M(r) is the maximum of |f| on the circle |z-z0| = r.
Answer: Note: Portions of the following solution have been taken from “ Problems in Complex Variable Theory” by Jan G. Krzyz, 1971, Exercise 4.2.6.
The vital key to answering this question is to notice that |f|² = f .f Now f(z0+reiθ) = Σn=0∞ an(z0+reiθ-z0)n = Σn=0∞ anr neinθ. Similarly, f ( z 0 + re ih ) = Σn=0∞ a n r ne-inθ. The product ff = |f|² is calculated as follows: ff = (Σn=0∞ anr neinθ)(Σn=0∞ a n r ne-inθ) = (a0+a1reiθ+a2r²e2iθ+...)( a 0+ a 1re-iθ+ a 2r²e-2iθ+...) = (|a0|² + a0a 1 re-iθ + a0 a 2 r²e-2ιθ + ... + a1reiθa 0 + a1reiθa 1 re-iθ + a1reiθ a 2r²e-2iθ + ... + a2r²e2iθa 0 + a2r²e2iθa 1 re-iθ + a2r²e2iθa 2 r²e-2iθ + ... + ....) = (|a0|² + (a0a 1 r)e-iθ + (a0 a 2r²)e-2iθ + ... + |a1|²r² + (a1a 0 r)eiθ + (a1a 2 r³)e-iθ + ... + |a2|r 4 + (a2a 0 r²)e2iθ + (a2a 1 r³)eiθ + ... + ....) = Σn=0∞ |an|²r 2n + A1(r)eiθ + B1(r)e-iθ + ..., where A1(r), B1(r), etc. are functions of r and the constants an, e.g. A1(r) = a1a 0 r + a2a 1 r³ + ... Because we are only dealing with values of r inside the circle of convergence of the Taylor series, then we can integrate the expression for |f|² term-by-term with respect to θ from θ = 0 to θ = 2π, i.e. we can find
∫ 02π |f|²dθ = ∫ 02π (Σn=0∞ |an|²r 2n + A1(r)eiθ + B1(r)e-iθ + ...)dθ. But knowing that ∫ 02π einθdθ = 0 for any integer n ≠ 0, the above integral reduces to
∫ 02π (Σn=0∞ |an|²r 2n)dθ = (Σn=0∞ |an|²r 2n)∫ 02π dθ = (Σn=0∞ |an|²r 2n)2π. Thus
1 2o
° 02o |f ( z 0 + re ih )| 2 d h =
1 2o
(Σn=0∞ |an|²r 2n)2π = Σn=0∞ |an|²r 2n as required. QED.
(b) Let us now define a new integral I = ∫ 02π g(θ)dθ, where g(θ) = |f(z0+reiθ)|². Because we know that |f| ≤ M(r) on the circle |z-z 0| = r, then it follows that |f|² ≤ M(r)² on the circle |z-z 0| = r. By definition, we see that g( θ) ≤ M(r)² on the circle |z-z0| = r. But knowing this, we can apply Theorem 9-60 (see the solution for Exercise 16-1 to see this) to say that
∫ 02π g(θ)dθ ≤ M(r)²×2π. 2o
1 2o
It follows that 21o ° 0 |f ( z 0 + re ih )| 2 d h ≤ 21o M (r ) 2 2o = M(r)². But we know from part (a) that ° 02o |f ( z 0 + re ih )| 2 d h = S nº=0 |a n | 2 r 2n , so it follows that º
S n=0 |a n | 2 r 2n
=
1 2o
° 20o |f ( z 0 + re ih )| 2 d h ≤ M(r)² as required . QED.
Exercise 16-27: Evaluate the following integral by means of residuals: 2 cos(2t )dt ° 02o 1−2a cos(t )+a 2 = 21o−aa 2 , if a² < 1. Answer: Let us first introduce some theory to be used in the solution. The Cauchy Residue Theorem. Let f be analytic at all points on an open region S in E 2, with the exception of a finite number of isolated singularities (“ poles”). Let Γ be a rectifiable Jordan curve such that both Γ and its interior lie within S. Assume that Γ contains a certain number of singularities of f in its interior, say z 1, ..., z n, but there are no singularities on Γ itself . Then we have
°G[ z ] f ( z )dz = 2oi S k n=1 z Res = z k f ( z ), provided that the path Γ [z] is positively orientated .
° 02o
In this question, we want to apply the above theorem to the integral given by , if a² < 1. The first thing to do is to reparametrise the integral to see the poles, 1−2a cos(t )+a 2 cos(2t )dt
and we do this by defining z = e iθ for 0 ≤ θ ≤ 2π. If z = eiθ, then dz = izdθ and (because e±iθ = cosθ ± isinθ) cosθ = ½(z+1/z) and cos2θ = ½(z²+z-2). Therefore, applying all these substitutions, we see that cos(2t )dt
h=2o
° h=0
=
1 2i
1
=°
1−2a cos(t )+a 2
° 01
2 −2 2 −2 z =1 dz 1 1 1 1 z 2 + z −2 2 ( z + z ) 2 ( z + z ) dz 1 z =0 1−2a( 1 ( z + 1 ))+a 2 iz i 0 (1−a( z + z 2i 0 z −az 2 −a+a 2 z dz )+a 2 ) z z 2 1 4 1 a ( z +1) 1 1 1 z 4 +1 z 4 +1 −1 1 dz dz z − 2 1 2 2 2 2i 0 z ( a − z −1+az ) 2ai 0 z (a− z )( z −a ) 2ai 0 z ( z −a)( z −a −1 ) dz .
1 2 −2 a ( z + z ) z 2 a − z −1+az
dz =
1
= ° =
°
°
=
°
=
Let us now find the poles of f(z), where f(z) is defined as f(z) =
°
z +1
z 2 ( z −a)( z −a −1 )
.
The first pole to consider is the pole of order 2 at z = 0. So b 1 = Res(f(z), 0) d m−1 1 m (m−1)! dz m−1 z z 0 f z 1 −1 3 4 lim ( z −a)( z −a )(4 z )−( z +1)(2 z − a −a) z d0 (( z −a)( z −a −1 )) 2 lim
= z d z 0 =
d z +1 2 ( − ) ( ) = (m = 2, z 0 = 0) = z lim d0 [ dz ( z − 0) ( z 2 ( z −a)( z −a −1 ) ) ] 4
=
0−1(− 1a −a)
(1) 2
= 1a + a.
The second pole to consider is the simple pole at z = a. So b 2 = Res(f(z), a) a 4 +1 lim z 4 +1 lim z 4 +1 = z lim = d z 0 [( z − z 0 ) f ( z ) ] = ( z 0 = a) = z da ( z − a)( z 2 ( z −a)( z −a −1 ) ) = z da z 2 ( z −a −1 ) = 2 1 a (a− ) a
a 4 +1 . a(a 2 −1)
The final pole to consider is the simple pole at z = 1/a. But because 1/a is outside the area of integration (a² < 1, and 0 < z < 1), we ignore it. Now that we have found all of the poles, we can apply Cauchy’s Residue Theorem to our problem: cos(2t )dt
−1 −1 −1 z +1 Res n ° ° dz f z 2 i % = = ( ) = ( ) o S z = z k f ( z ) 2 2 1 − k 1 = 0 0 2ai 2ai 2ai 1−2a cos(t )+a z ( z −a)( z −a ) 2 2 2 4 1 a 4 +1 a 2 −1+a 4 −a 2 +a 4 +1 −1 −o a −1+a (a −1)+a +1 = ( 2ai )(2oi)( a + a + a(a 2 −1) ) = ( a )( ) = − o ( )= a(a 2 −1) a 2 (a 2 −1)
° 02o
4
1
1
−2oa 2 . 1−a 2
QED.
Exercise 16-30: Evaluate the following integral by means of residuals:
° −ºº x 2+1 x+1 dx =
2o 3 3
.
Answer: To do this question, we need some further theory: R
n ° Theorem: Rlim d+º − R f ( x)dx = 2oi S k =1 z = z k f ( z ). This is automatically satisfied if f is the quotient of two polynomials, say f = P/Q, provided that the degree of Q exceeds the degree of P by at least 2. Res
º ° − For º x 2 + x+1 dx , let f(z) =
1
. Then P(z) = 1 and Q(z) = z²+z+1, and hence the properties of the above theorem are satisfied. The poles of f are given by the roots of the equation z²+z+1 = 0, i.e. z =
z 2 + z +1
−1! 1−4 2
−1!i 3
=
2
. Of these, only the ‘positive’ root lies in the −1+i 3
upper half-plane. The residue at this pole (at z = Res(f(z), lim
= z d
−1+i 3 2
−1+i 3
)=
2
( z − (
lim z d
−1+i 3
i 3 2
)( z + 12 +
2
i 3 2
)
=
1
−1+i 3 2
1
z + 2 +
i 3 2
z + 12 −
lim z d
lim z d
) is given by
)) f ( z ) =
2
i 3 z + 12 − 2
( z + 12 −
−1+i 3
2
−1+i 3
=
i 3 2
z 2 + z +1
2
1 −1+i 3 2
+ 12 +
i 3 2
=
Therefore, using the above theorem, we can say that
° −ºº x 2+1 x+1 dx = 2oi S k n=1 z Res = z k f ( z ) = 2oi
3 3i
=
2o 3 3
as required. QED.
1 i 3 2
+
i 3 2
=
1 i 3
=
3 3i
.
Exercise 16-38: If f and g are Möbius transformations, show that the composite function fg is also a Möbius transformation. Answer: Let us first remind ourselves of the definition for a Möbius transformation: they are functions f defined as follows: if a, b, c and d are complex numbers such that ad-bc ≠ az +b 0, we define f ( z ) = cz +d whenever cz+d ≠ 0. It is convenient to define f everywhere on the extended plane E2* by setting f(-d/c) = ∞ and f(∞) = a/c. (If c = 0, these last two equations are to be replaced by the single equation f(∞) = ∞.) pz +q
Consider that f is defined as above and we also have g defined as g ( z ) = rz + s whenever rz+s ≠ 0, where p, q, r and s are complex numbers such that ps-rq ≠ 0. We define g everywhere on the extended plane E2* as above. Let us now consider the composite function fg = f(g(z)). Manipulating, f(g(z))
pz +q = f ( rz + s
) =
pz +q
a( rz + s )+b
pz +q c( rz + s )+d
=
a( pz +q)+b(rz + s) rz + s c( pz +q)+d (rz + s) rz + s
=
a( pz +q)+b(rz + s) c( pz +q)+d (rz + s)
=
(ap+br ) z +(aq+bs) (cp+dr ) z +(cq+ds) .
We now need to show that if ad-bc ≠ 0, and if ps-rq ≠ 0, then we have (ap+br)(cq+ds) (cp+dr)(aq+bs) ≠ 0. But (ap+br)(cq+ds) - (cp+dr)(aq+bs) = apcq + apds + brcq + brds - cpaq - cpbs - draq - drbs = apds + brcq - cpbs - draq = ad(ps-rq) - bc(ps-rq) = (ad-bc)(ps-rq) ≠ 0 (as (ad-bc) ≠ 0 and as (ps-rq) ≠ 0.) So we have shown that fg is a Möbius transformation (defined whenever (cp+dr)z+(cq+ds) ≠ 0). We can also define fg everywhere on the extended plane E2* by setting cq+ds ap+br fg(− cp+dr ) = ∞ and fg(∞) = cp+dr . (If cp+dr = 0, these last two equations are to be replaced by the single equation fg(∞) = ∞.)
Exercise 16-39: Describe geometrically what happens to a point z in E 2 when it is carried into f(z) by the following special Möbius transformations:
(a) f(z) = z+b (b) f(z) = az, where a > 0 (c) f(z) = eiαz, where α is real (d) f(z) = 1/z y z+b b z x
(a)
(Translation). (Stretching). (Rotation). (Inversion).
Answer: (a) Assuming that b is a complex number, say b = p+qi, then if z = x+iy, then the Möbius transformation in this case moves the point z to the point (x+p)+(q+y)i, i.e. we translate the point z to the point z+b. This is useful if a change of origin is required, from the original origin O to the new origin b.
y 2z
z
½z
(b)
(b) Assuming that a is a real number, the Möbius transformation in this case in effect changes the scale of the axes in an Argand diagram, to x’ = x×a and to y’ = y×a. Geometrically, the transformation moves the point along the vector defined in an Argand diagram by z by an amount x specified by the number a.
y
az = re
i(θ+α)
i
α
r θ
z = re θ x
(c)
(c) Recalling that a complex number can be written in the form z = reiθ, then the Möbius transformation in this case rotates the point z by an angle of α around the origin (in the anticlockwise direction). In other words, f(z) = ei(α+θ)r.
y z x 1/z (d)
(d) Finally, recall that if z = p+iq, then 1/z = z-1 = p-iq/ p²+q². Geometrically, we reflect the point z in the x-axis of an Argand diagram, and then apply a “Stretching ” transformation (as in part (b)), where a in this case is given by a = 1/ p²+q².
