Mathematical AnalysisT. M. Apostol

Chapter 3

Mathematical Analysis by Tom M. Apostol

Chapter 3: Elements of Point Set Theory: Notes

Let E1 denote the set of all real numbers (the real line). Let E2 denote the set of allcomplex 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. Thereal line is sometimes referred to as the open interval (-∞, ∞). A single point is sometimesconsidered as a “degenerate” closed interval.

Let h > 0, and let x be a given point. The open interval (x-h, x+h) is called aneighbourhood with x as centre and of radius h. We denote the neighbourhood by N(x; h), orsimply by N(x) if the radius is unimportant.

Let S be a set in E1 and assume that x ∈ S. Then x is called an interior point of S if thereis some neighbourhood N(x) all of whose points belong to S.

Let S be a set in E1. 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 can be enclosed in a neighbourhood which iscompletely contained in the set. The simplest kind of open set is an open interval. The emptyset is also open, as is the real line E1.

The union of any collection of open sets is an open set. The intersection of a finitecollection of open sets is open. Note that arbitrary intersections will not always lead to opensets. 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 E1 is said to be bounded if it is asubset of some finite interval. Let S be an open set in E1 and let (a, b) be an open intervalwhich is contained in S, but whose endpoints are not in S. Then (a, b) is called a componentinterval of S.

If S is a bounded open set, then (i) each point of S belongs to a uniquely determinedcomponent interval of S; and (ii) the component intervals of S form a countable collection ofdisjoint sets whose union is S. From this, it follows that an open interval in E1 cannot beexpressed as the union of two disjoint open sets when neither set is empty.

Every open set in E1 is the union of a countable collection of disjoint open intervals.

Accumulation points and the Bolzano-Weierstrass theorem in E1. Let S be a set in E1 andx a point in E1, x not necessarily in S. Then x is called an accumulation point of S, providedevery neighbourhood of x contains at least one point of S distinct from x. If x is anaccumulation point of S, then every neighbourhood N(x) contains infinitely many points of S.

Theorem 3-13 (Bolzano-Weierstrass): If a bounded set S in E1 contains infinitely manypoints, 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 ofrational numbers has every real number as an accumulation point. (3) Every point of the closedinterval [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 aclosed interval is a closed set. An open interval, however, is not a closed set because it doesnot contain its endpoints, both of which are accumulation points of the set. A set which has noaccumulation points is automatically closed and, in particular, every finite set is closed. Theempty set is also closed, as is the whole real line E1. (These last two sets are also open!) A setwhich is not closed need not be open, as, for example, the half-open interval (a, b], which isneither open nor closed.

If S is open, then the complement E1—S is closed. Conversely, if S is closed, thenE1—S is open. It follows from the above that the union of a finite collection of closed sets isclosed 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) isclosed.

Extensions to higher dimensions. The definition of a neighbourhood, |x-x0| < h, stillmakes 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, ..., xn) is called ann-dimensional point or a vector with n components. Points or vectors will be denoted by singleboldfaced letters, e.g. x = (x1, x2, ..., xn), where xk is the kth co-ordinate of the point or vector x.The set of all n-dimensional points is called the n-dimensional Euclidean space and is denotedby 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, ..., xn) and y = (y1, ..., yn) be in En. The absolute value, or length, or norm of xis given by |x| = . The distance between x and y is given by |x-y| = .x1

2 + ... + xn2 Si=1

n (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 En, we mean the set of allx in En such that |x-x0| < r. The set of x such that |x-x0| ≤ r form a closed sphere, the boundaryof the sphere being the set of x with |x-x0| = r. An open sphere with centre at x0 is called aneighbourhood of x0 and is denoted by N(x0) — or by N(x0; r) if r is the radius. Thecorresponding closed sphere is denoted by (x0). The open sphere with its centre removed isNcalled a deleted neighbourhood of x0 and is denoted by N’(x0).

Let a = (a1, ..., an) and b = (b1, ..., bn) be two distinct points in En, such that ak ≤ bk foreach k = 1, 2, ..., n. The n-dimensional closed interval [a, b] is defined to be the set [a, b] =(x1, ..., xn) | ak ≤ xk ≤ bk, k = 1, 2, ..., n. If ak < bk for every k, the n-dimensional open interval(a, b) is the set (a, b) = (x1, ..., xn) | ak < xk < bk, k = 1, 2, ..., n.

Thus, for example, (a, b) can be considered as the “cartesian product” (a, b) = (a1, b1) ×(a2, b2) × ... × (an, 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 nodifference whether a neighbourhood is taken to be a “sphere” or an “interval” (we will usespheres).

Let S be a set of points in En, and assume that x ∈ S. Then x is called an interior point ofS if there exists a neighbourhood N(x) ⊂ S. The set S is said to be open if each of its points isan 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; thewhole space. Any n-dimensional open sphere and any n-dimensional open interval is an openset in En. Caution: an open interval (a, b) in E1 is no longer an open set when it is considered asa subset of the plane. In fact, no subset of E1 (except the empty set) can be open in E2, becausesuch a set can contain no two-dimensional neighbourhoods.

The union of an arbitrary collection of open sets in En is an open set in En, and theintersection of a finite collection of open sets in En is an open set in En.

Assume that S ⊂ En, x ∈ En. Then x is called an accumulation point of S if everyneighbourhood N(x) contains at least one point of S distinct from x, that is, if N’(x) ∩ S is notempty. If x is an accumulation point of S, then every neighbourhood N(x) contains infinitelymany points of S.

A set S in En is said to be bounded if S lies entirely within some sphere; that is, for somer > 0, we have S ⊂ N(0; r). Theorem 3-29 (Bolzano-Weierstrass): If a bounded set S in En

contains infinitely many points, then there exists at least one point in En which is anaccumulation point of S.

A set S in En is said to be closed if it contains all its accumulation points. Examples: Aclosed sphere in En is a closed set. An n-dimensional closed interval is a closed set. A setwhich is closed in E1 is also closed in En for n > 1.

A set S in En is closed iff En—S is open. The union of a finite collection of closed sets inEn is closed and the intersection of an arbitrary collection of closed sets in En is closed.

Assume that S ⊂ A ⊂ En. If A is open and if S is closed, then A—S is open. If A isclosed and if S is open, then A—S is closed.

The Heine-Borel covering theorem. We begin by defining a covering of a set. Acollection F of sets is said to be a covering of a given set S if S ⊂ ∪A∈F A. The collection F isalso 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 anopen covering of the interval 0 < x < 1. This is an example of a countable covering. (2) Thereal axis E1 is covered by the collection of all open intervals (a, b). This covering is notcountable. However, it contains a countable covering of E1, 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 havingrational radii and centres at points with rational co-ordinates. Assume that x ∈ En, and let Sbe an open set in En which contains x. Then at least one of the neighbourhoods in G containsx and is contained in S. That is, we have x ∈ Ak ⊂ S for some Ak in G.

Assume that A ⊂ En, and let F be an open covering of A. Then there is a countable subcollection of F which also covers A.

Let Q1, Q2, ... be a countable collection of non empty sets in En such that (1) Qk+1 ⊂ Qk

(k = 1, 2, 3, ...); (ii) Each set Qk is closed and Q1 is bounded. Then the intersection ∩k=1∞ Qk is

closed and non empty.

Theorem 3-38 (Heine-Borel) Let F be an open covering of a closed and bounded set A inEn. Then a finite sub collection of F also covers A.

A set S in En is said to be compact iff every open covering of S contains a finite subcollection which also covers S.

The Heine-Borel theorem states that every closed and bounded set in En is compact. Theconverse result is as follows: Let S be a compact set in En. We have (i) every infinite subset ofS 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 realnumbers E1 with two symbols +∞ and -∞ which satisfy the following properties: (1) If x ∈ E1,then we have x+(+∞) = +∞, x+(-∞) = -∞, x-(+∞) = -∞, x-(-∞) = +∞, and x/(+∞) = x/(-∞) = 0. (2) Ifx > 0, then we have x(+∞) = +∞ and x(-∞) = -∞. (3) If x < 0, then we have x(+∞) = -∞ andx(-∞) = +∞. (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 -∞. E1 = (-∞, ∞); 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 thatE1* 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 complexplane 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. Everyopen 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 E1 is an open set and that a closed interval is a closedset.

Answer: Let us first look at some definitions. Definition 1: An open interval (a, b) isdefined by (a, b) = x | a < x < b. Definition 2: Let S be a set in E1 and assume that x ∈ S.Then x is called an interior point of S if there is some neighbourhood N(x) all of whose pointsbelong to S. Definition 3: If S is a set in E1, S is an open set if every point of S is an interiorpoint of S.

To prove that an open interval in E1 is an open set, we need to prove that every point inan open interval is an interior point. To do this, we must prove that every point in an openinterval 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 showis that one of these neighbourhoods is entirely contained within the interval (a, b), i.e. we haveto show that (x-h, x+h) ⊂ (a, b) for every x and for some h > 0. To do this, we have to showthat 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 theminimum distance from x to each of the two endpoints, sothat h = ½min(|x-a|, |x-b|). Because we know that x > aand 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).

Let us now analyse what we know about the point y. First of all, we know that y (bydefinition) 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)⇒ x-h > a proving part (1)

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)⇒ x+h < b proving part (2)

( )

a b

( )x-h x x+h

Because we have shown that y ∈ (x-h, x+h) ⇒ y ∈ (a, b), then we have proved that anopen interval is an open set. To show that a closed interval is a closed set, we use the followingtrick: the complement of an open set is a closed set.

Let us consider an arbitrary closed interval [a, b]. The complement of this interval isgiven by E1—[a, b] = (-∞, a)∪(b, ∞), the union of two open intervals, which we know are opensets by the above. Further, the union of two open sets is also an open set (Theorem 3.5), so thatE1—[a, b] is an open set. So, as the complement of a closed interval is an open set, then aclosed interval must therefore be a closed set. QED.

3-2. Determine all the accumulation points of the following sets in E1 and decidewhether the sets are open or closed (or neither). (a) All integers. (b) The interval (a, b]. (c) Allnumbers of the form 1/n, (n = 1, 2, 3, ...). (d) All rational numbers. (e) All numbers of the form2-n + 5-m, (m, n = 1, 2, ...). (f) All numbers of the form (-1)n + (1/m), (m, n = 1, 2, ...). (g) Allnumbers 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 Sbe a set in E1 and x a point in E1, x not necessarily in S. Then x is called an accumulation pointof 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 aneighbourhood (x-½, x+½) which contains no other integers, so that x is not an accumulationpoint. Because S contains all its accumulation points (of which there are none!), it follows thatS 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 neighbourhoodN(x) of x, N(x) = (x-h, x+h), with h > 0, then it is clear that the point maxx-a/2, x-h/2 willalways be in S and in N(x) so that every neighbourhood of a point in S will always contain atleast one other point from S. We therefore conclude that all the points in S are accumulationpoints. Further, a is an accumulation point of S so that the set of accumulation points of S isgiven by the set x | a ≤ x ≤ b.

S is not closed because it doesn’t contain the accumulation point a. Further, S is notopen because b is not an interior point (in the neighbourhood (b-h, b+h), with any h > 0, thepoints in (b, b+h) ⊂ (b-h, b+h) do not belong to S — but do belong to the neighbourhood. Itfollows 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 thateach point x = 2-n is an accumulation point of S (n = 1, 2, ...), consider an arbitraryneighbourhood N(x) of x, N(x) = (x-h, x+h), with h > 0. If we now choose an integer l so that5-l ≤ h/2 (and there will always be such an l, given by l = ceiling(-log(n/2)/log(5))), then the pointy = 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 accumulationpoints 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 (everyneighbourhood 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 issufficient to find a single point in S which does not have a neighbourhood all of whose pointsbelong to S. Take the point in S given by setting n = 1 and m = 1, namely the point 7/10. Clearlythis is the largest element of the set S, and so every neighbourhood of this point (i.e. theinterval (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 aninterior point — so that S is not an open set.

(h) Let S = (n = 1, 2, ...). Writing S out as a sequence, S = -1/2, 2/3, -3/4, 4/5, -5/6, ....(−1)n

1+ 1n

From this series representation, we see that S is an alternating series, one half of the seriestending to -1, and the other half tending to 1. Therefore, -1 and 1 are accumulation points ofthe set S, because the set has elements which are arbitrarily close to -1 and 1. There are noother accumulation points in the set.

Because S does not contain its two accumulation points (-1 ∉ S and 1 ∉ S), then S is nota 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 anopen set.

3-6. Show that every closed set in E1 is the intersection of a countable collection of opensets.

Answer: Let us first consider some examples of closed sets and the intersection of opensets that is equivalent to the closed set in question. To start with, consider the set S consistingof a single point x ∈ E1. This set is closed because it has no accumulation points, and it can beexpressed 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 bea closed set, which we shall denote by T. This closed set can be expressed as the intersection ofa 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 iscorrect. We now want to show that every closed set C can be expressed as the intersection ofsome 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 Di, 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 Di, j is anopen set because it consists of an open interval, which we know (from exercise 3-1) to be anopen set. Now let us define Ej to be the following expression: Ej = ∪i Di, j, the union over allpoints of C of ‘1/j intervals’ of all those points. Each Ej is an open set because it is an arbitraryunion of open sets, which (by theorem 3-5) is an open set. Finally, let us define F to be the setgiven by F = ∩j=1

∞ Ej. 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 alreadyshown that F is the intersection of an arbitrary collection of open sets, and so it is in the rightform to be considered as the solution to this exercise. We will in fact show that F = , i.e. showCthat F is the closure of the set C. If we can do this, then we will also show that F = C becausefor a closed set C, we have (looking at part (f) of exercise 3-12) C = . Therefore, if we canCshow that F = , then we reach the required conclusion. To do this, as = C∪C’ (see exerciseC C3-12 for this definition), then we first need to show that F contains all of C’s accumulationpoints and all of the points from C.

Claim 1: F contains all of C’s accumulation points, whether they are in C or not. Proofof Claim: Consider an arbitrary accumulation point of the set C, say the point x. If x ∈ C, thenwe refer to the proof of claim 2 below to show that x ∈ F. If the accumulation point x is not inC, then we must take a little more care.

Following the definition of an accumulation point, everyneighbourhood N(x) of x will contain a point y ∈ C. But for everysuch y, there exists a corresponding interval Dy,j which in turn is alsoin every Ej. When we take the intersection of all the Ej’s, we “closein” on the point x (see the diagram on the right), the crucial featurebeing that every interval Ej 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 pointy ∈ C. If y ∈ C, then there is a set Dy, j = (y-1/j, y+1/j), where j ∈ R>0. It follows that y belongs toeach Ej, where Ej = ∪i Di, j, and so y belongs to F = ∩j=1

∞ Ej, 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. Tocomplete the proof, we need to show that F ⊂ C∪C’.

If z ∈ F, then z belongs to all of the sets Ej, where Ej = ∪i Di, j. If z belongs to all of thesets Ej, then there must be a set Dx, j = (x-1/j, x+1/j), where j ∈ R>0. It follows (by the definitionof Di, j and by the proof of claims 1 and 2 above) that z ∈ C∪C’, and so z ∈ F ⇒ z ∈ C∪C’, sothat F ⊂ C∪C’. QED. Conclusion: we have expressed an arbitrary closed set C as theintersection of a countable collection of open sets, namely the set F. QED.

(x

E1

E2

E3

etc.

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 interiorof S. For a set A to be an open set, all the points in A must be interior points. Because all thepoints in our set I are interior points of S, it follows that I is an open set, and therefore theinterior of a set S in En is an open set. QED.

3-12. If S is a set in En, let S’ denote the set of accumulation points of S, and let =SS∪S’. (The set S’ is called the derived set of S and is called the closure of S.) Show that (a)SS’ is a closed set, that is, (S’)’ ⊂ S’. (b) If S ⊂ T, then S’ ⊂ T’. (c) (S∪T)’ = S’∪T’. (d) ( )’ =SS’. (e) is a closed set. (f) S is closed if, and only if, S = .S S

Answer: (a) If x ∈ En and x ∈ (S’)’, then x is an accumulation point of the set S’. If x isan accumulation point of the set S’, then every neighbourhood Nε(x) of x contains at least oneother point y ∈ S’ (y ≠ x), i.e. another accumulation point of S. It follows that allneighbourhoods N(y) of y will contain a point z ∈ S. Pick aneighbourhood NΕ(y) such that NΕ(y) ⊂ Nε(x) and x ∉ NΕ(y). Note thatin order to do this, we must have Ε < max(dist(x,y), ε-dist(x,y)). So asevery neighbourhood N(x) of x contains a point z ∈ S (with z ≠ x), thenit 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 theset S. But if y is an accumulation point of S, then either y ∈ S, which implies that y ∈ T, and soy is also an accumulation point of the set T; or y ∉ S, but every neighbourhood of y contains apoint from z from S, which is also (because S ⊂ T) a point from T, so that everyneighbourhood 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, whichimplies 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’)’ = (from (c)) = S’∪(S’)’ = S’∪T (where T ⊂ S’) = S’. QED.S

(e) To be a closed set, must contain all its accumulation points. From (d), we knowSthat the set of the accumulation points of is the set S’. But S’ ⊂ as = (S∪S’). So as S S S Scontains all its accumulation points, we conclude that is a closed set. QED.S

x Nε(x)

yNΕ(y)

(f) If. If S = , then either S’ = φ, i.e. S contains no accumulation points, and so S isSclosed by definition; or S’ consists of points from S, i.e. S’ ⊂ S, so that again S contains all itsaccumulation points, and is again a closed set by definition. Only If. If S is closed, then itcontains all its accumulation points. Therefore, S’ will consist of elements from S, so that S’ ⊂S, and so = (S∪S’) = (using S’ ⊂ S) = S in this case. QED.S

3-15. The collection F of intervals of the form 1/n < x < 2/n, (n = 2, 3, 4, ...), is an opencovering of the interval 0 < x < 1. Show (without using Theorem 3-40) that no finitesubcollection 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. Ifwe order the elements in G with respect to the number n, then it follows that we can write G inthe following form:

G = (1/n1, 2/n1), (1/n2, 2/n2), ..., (1/ng, 2/ng), where n1 < n2 < ... < ng.

We must now show that G does not cover the interval 0 < x < 1. To do this, it issufficient to show that G does not cover the interval (0, 1/ng], where (0, 1/ng] ⊂ (0, 1), and ng ∈R (ng < ∞). 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 Gcovers no part of the interval (0, 1/ng]. We therefore conclude that the set G does not fullycover the interval (0, 1). QED. (Note: a quicker proof is to show that 1/ng ∈ (0, 1) but 1/ng ∉∪G.)

3-19. Assume that S ⊂ En. A point x in En is said to be a condensation point of S if everyneighbourhood N(x) has the property that N(x)∩S is not countable. Show that if S is notcountable, then there exists a point x in S such that x is a condensation point of S.

Answer: Divide the set S up into n (possibly partially overlapping) sets Si,where n is a finite number; i = 1, 2, ..., n; and ∪i Si = S. If the set S is notcountable, then at least one of these ‘regions’ will have to be uncountable aswell, or else we will have a finite collection of countable sets and thus Swill 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). Ifwe continue to do this, then this process will converge onto a point z ∈ S which will be thecondensation point we are looking for. Justification: with every subdivision, we find a smallerand smaller “region” which is uncountable. In the limit of this process, every neighbourhood ofthe point z we converge onto will contain one of the regions which we know to be uncountable(i.e. some region Sj ⊂ N(z)). Therefore, every neighbourhood of this point z has the propertythat N(z)∩S is not countable, i.e. z is a condensation point. Conclusion: given an uncountableset S, we can always find a condensation point z ∈ S. QED.

S1 S2 S3 Sn

S

3-20. Assume that S ⊂ En and assume that S is not countable. Let T denote the set ofcondensation points of S. Show that

(a) S—T is countable, (b) S∩T is not countable, (c) T is a closed set, (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 condensationpoint. But if T is the set of condensation points of S, then S—T contains no condensationpoints. 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 thereforeprove that the set of condensation points in S is not countable. To do this, we use three piecesof information: (1) From set theory, we know that S∩T = S—Tc; (2) We know that S is notcountable; and (3) We know that S—T is countable. If S—T is countable, and if S is notcountable, then S—Tc must be non countable. Using (1), it follows that S∩T must also be noncountable. QED.

(c) By definition, T is closed if it contains all its accumulation points. We know thatevery point x ∈ T is a condensation point, so that every neighbourhood N(x) of a point x ∈ T isnot countable (more precisely, N(x)∩S is not countable). This implies (by Exercise 3-19) thatthe neighbourhood contains another condensation point y ∈ T (N(x)∩S uncountable ⇒ ∃ acondensation point y ∈ N(x)∩S), so that the point x is an accumulation point of the set T. Thisapplies 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 isan accumulation point of the set T — with z ∉ T. Because z is not a condensation point, itfollows that not all neighbourhoods N(z) are uncountable. Pick one such neighbourhood, sayN1(z). By the definition of an accumulation point, N1(z) contains an element z1 ∈ T. But theneighbourhoods of z1 are uncountable (by the definition of a condensation point). Pick aneighbourhood of z1 which is contained in N1(z), say theneighbourhood N2(z1) (We can always do this — see the answer toexercise 3-12, part (a)). Because N2(z1) is uncountable, N1(z) cannotpossibly be countable, contradicting our assumption that N1(z) wascountable. It follows that all the neighbourhoods of the point z must beuncountable, and so we must have z ∈ T, and so T has no accumulationpoints that are not in T. QED

(d) Because we concluded in (c) that every point in T is an accumulation point, itfollows that no point in T is an isolated point, i.e. not an accumulation point. QED.

z N 1(z)

z1N2(z1)

3-21. A set S is said to be perfect if S = S’, that is, if S is a closed set which contains noisolated points. (Note: S’ is the set of accumulation points of S). Show that if F is anuncountable closed set in En, then F can be expressed in the form F = A∪B, where A is aperfect 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 ofcondensation points in S, and that S—T is the set of points in S which are not condensationpoints. We can therefore write S = (T∩S)∪(S—T), i.e. S is the union of all the condensationpoints of S together with the points of S which are not condensation points. From 3-20, weknow that all the points in T∩S are accumulation points, i.e. we have T∩S = (T∩S)’. It followsthat 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 iscountable. QED.

Possible Further Work / Evaluation

Exercise 3-19: I could make a slight adjustment in the answer due to the comment inDraft 2 — to divide the set S up into a countable collection of sets instead of into a finitecollection of sets. Adjusted answer:

Answer: Divide the set S up into a countable collection of (possibly partiallyoverlapping) 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 ofcountable 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 beuncountable). If we continue to do this, then this process will converge onto a point z ∈ Swhich 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, everyneighbourhood of the point z we converge onto will contain one of the regions which we knowto be uncountable (i.e. some region Sj ⊂ N(z)). Therefore, every neighbourhood of this point zhas the property that N(z)∩S is not countable, i.e. z is a condensation point. Conclusion: givenan 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 xn = A, which is meant to convey the idea thatlimxdx0

when x is sufficiently near to x0, then f(x) will be as near to A as desired. The statements “xsufficiently near to x0” and “f(x) will be as near to A as desired” are made mathematicallyprecise by the following type of definition: the symbolism means that for everyxdx0

lim f(x) = Anumber ε > 0, there is another number δ > 0 such that whenever 0 < |x-x0| < δ, then |f(x)-A| < ε.

We write xn = A to mean that for every ε > 0, there is an integer N such thatlimnd∞

whenever n > N, then |xn-A| < ε.

If we use neighbourhood terminology, we see that both above definitions involve thesame principle. To say that 0 < |x-x0| < δ means that x is in a neighbourhood of x0, but x ≠ x0.That is, x is in a deleted neighbourhood N’(x0). Also, to say that n > N is the same as sayingthat n is in a deleted neighbourhood of +∞.

The parts of the definitions concerned with ε can also be rephrased in neighbourhoodterminology. For example, the inequality |f(x)-A| < ε states that f(x) ∈ N(A; ε) and, similarly,|xn-A| < ε means that xn ∈ N(A; ε). The basic principle involved in both definitions of limit isthat for every neighbourhood N(A) there must exist a neighbourhood N(x0) such that x ∈N’(x0) implies f(x) ∈ N(A).

If f is a real-valued function defined at a point x = (x1, ..., xm) in Em, we use either f(x1,..., xm) or f(x) to denote the value of f at that point. If we have several real-valued functions,say f1, ..., fk defined on a common subset S of Em, it is extremely convenient to introduce avector-valued function f, defined by the equation f(x) = (f1(x), ..., fk(x)), if x ∈ S.

Our definition of limit now assumes the following form: Let f be a function defined on aset S in Em 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 f(x) = b is defined to mean the following: For everylim

xda

neighbourhood N(b) ⊂ Ek, there exists a neighbourhood N(a) ⊂ Em such that x ∈ N’(a)∩Simplies f(x) ∈ N(b).

Note 1: we write x ∈ N’(a)∩S rather than x ∈ N’(a) in order to make certain that x is inthe domain of f. Also, we require that a be an accumulation point of S in order to make certainthat the intersection N’(a)∩S will never be the empty set. Note 2: If limx→a f(x) exists (finite orinfinite), its value is uniquely determined.

Strictly speaking, we should somehow indicate the fact that the limit just defineddepends on the set S through which x is allowed to range. This will usually be clear from thecontext but, if necessary, we will write f(x) = b to emphasise the fact more explicitly. Anlim

xdax`S

important special case of this occurs when S is an interval in E1 having a as its left endpoint.We then write f(x) = f(x) = b, and b is called the right-hand limit at a. (left-hand limitslim

xdax`S

limxda+

are similarly defined).

Theorem 4-4. Let a be an accumulation point of a set S in Em. Then there exists aninfinite 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 limitthat a is an accumulation point of S. The above theorem tells us that S can have no furtheraccumulation points. Therefore, a sequence xn whose range S is infinite has a limit if, andonly if, S has exactly one accumulation point, in which case the accumulation point is also thelimit of the sequence.

Theorem 4-5. Let f be defined on a set S in Em with function values in Ek, and let a be anaccumulation point of S. Let xn be an infinite sequence whose terms are points of S, suchthat each term xn ≠ a but such that xn = a. Then we have (i) If limx→a f(x) = b, then limn→∞

limnd∞

f(xn) = b; (ii) Conversely, if for each such sequence we know that limn→∞ f(xn) exists, then allthese 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 bwas given. The following theorem gives us a condition (called the Cauchy condition) whichenables 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 whoseterms are points in Ek. There exists a point y in Ek such that limn→∞ xn = y if, and only if, thefollowing condition is satisfied: For every ε > 0, there exists an integer N such that n > N andm > N implies |xn-xm| < ε.

Theorem 4-7 (Cauchy condition for functions). Let f be defined on a set S in Em, thefunction 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 isa 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 En, with function values in E1 orin E2. Let a be an accumulation point of S and assume we have f(x) = A, g(x) = B. Thenlim

xdalimxda

we also have (i) [f(x)±g(x)] = A±B, (ii) f(x)g(x) = AB, (iii) f(x)/g(x) = A/B if B ≠ 0.limxda

limxda

limxda

Continuity. Definition 4-9. Let f be defined on a set S in En with function values in Em,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 iscontinuous at a provided only (i) holds. If f is continuous at every point of S, we say f iscontinuous on the set S.

It is convenient to note that whenever f is continuous at a, we can write part (ii) of theabove definition as follows: f(x) = f( x). Thus, when we deal with continuous functions,lim

xdalimxda

the limit symbol may be interchanged with the function symbol. Observe also that continuityat a means that for every neighbourhood N(f(a)), there exists a neighbourhood N(a) such thatf[N(a)∩S] ⊂ N[f(a)].

Theorem 4-10. Let f and g be continuous at the point a in En and assume that f and ghave function values in E1 or E2. Then f+g, f-g, and f•g are each continuous at a. The quotientf/g is also continuous at a, provided that g(a) ≠ 0. Note that the product f•g should not beconfused 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 gbe defined on f(S) with values in Ek, and let gf be the composite function defined on S by theequation gf(x) = g[f(x)]. Suppose that we have (i) The point a is an accumulation point of Sand f is continuous at a; (ii) The point f(a) is an accumulation point of f(S) and g is continuousat f(a). Then the composite function gf is also continuous at a; that is, g[f(x)] = g[f(a)].lim

xda

Examples of continuous functions. In E2, constant functions (f(x) = c), the identityfunction (f(x) = x) and hence all polynomial functions are continuous. A rational function g/fis continuous whenever the denominator does not vanish. The familiar real-valued functionsof elementary calculus such as exponential, trigonometric and logarithmic functions, arecontinuous whenever they are defined.

Functions continuous on open or closed sets. Definition 4-12. Let f be a functionwhose domain is S and whose range is T, and let Y be a subset of T. By the inverse image of Yunder f, denoted by f-1(Y), we shall mean the largest subset of S which f maps onto Y, that isto 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 theimage 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 beobserved that, in general, we cannot conclude that Y = f(X) implies X = f-1(Y). Note that thestatements 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 Em, and letthe range of f be a set T in Ek. Then if Y is a closed subset of T, the inverse image f-1(Y) will bea closed subset of S.

Theorem 4-15. Let f be a function which is continuous on an open set S in Em, and let therange of f be a set T in Ek. Then if Y is an open subset of T, the inverse image f-1(Y) will be anopen subset of S.

We have already seen that the image of an open set under a continuous mapping is notalways open. The example in E1, defined by the equation f(x) = tan-1(x), shows that the imageof a closed set under a continuous mapping need not be closed, since f maps E1 onto the openinterval (-π/2, π/2). However, the image of a compact set under a continuous mapping is alwayscompact. This will be proved in the next theorem.

Functions continuous on compact sets. Theorem 4-16. Let f be a function which iscontinuous on a compact set S in Em, 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 Em, andassume that f(S) ⊂ Ek. 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 isone-to-one on S so that the inverse function f-1 exists. Then f is called a topological mapping ora homeomorphism if, in addition, f-1 is continuous on f(S). Note: A property of a set whichremains invariant under every topological mapping is called a topological property.

Properties of real-valued continuous functions. Definition 4-19. Let f be a real-valuedfunction defined on a set S in En. Then f is said to have an absolute maximum on the set S ifthere exists a point a in S such that f(x) ≤ f(a), for all x in S. If a ∈ S and if there is aneighbourhood N(a) such that f(x) ≤ f(a), for all x in N(a)∩S, then f is said to have a relativemaximum at the point a. Absolute minimum and relative minimum are similarly defined, usingf(x) > f(a).

Theorem 4-20. Let f be a real-valued function which is continuous on a compact set S inEn. 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. Thenthere 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 forcontinuous functions (Theorem 4-22): If f is real-valued and continuous on a closed intervalS 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 foreach accumulation point a of S, we have limx→a f(x) = f(a). In the ε and δ terminology, thismeans that, given the accumulation point a and given ε, there is a number δ > 0 (depending ona and on ε) such that 0 < |x-a| < δ implies |f(x)-f(a)| < ε. In general, we cannot expect that for afixed 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 En with function values in Em. Then f issaid to be uniformly continuous on S if the following statement holds: For every ε > 0, thereexists 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. Itfollows from the definition that uniform continuity on a set implies continuity on the set. Theconverse 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 isalso uniformly continuous on S.

Discontinuities of real-valued functions. Definition 4-25. If f is a real-valued functiondefined on an open interval (a, b) in E1 and if c is a point in [a, b), we write f(c+) to denote theright hand limit, limx→c+ f(x), whenever the limit exists. If c ∈ (a, b], we define f(c-) = limx→c-

f(x).

It is clear that we have f(x+) = f(x-) = f(x) if, and only if, f is continuous at x. If f isdefined 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+) andf(x-) both exist at some interior point x, then the difference f(x)-f(x-) is called the left-handjump 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 xis called a jump discontinuity of f. (For the endpoints a and b, only one-sided jumps areconsidered.) 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 removablediscontinuity, since the discontinuity can be removed by redefining f at x to have the valuef(x+) = f(x-). If either limx→c+ f(x) or limx→c- f(x) is infinite (+∞ or -∞), then f is sometimes saidto 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 setS 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 beincreasing (or nondecreasing) on S. If x < y implies f(x) < f(y), then f is said to be strictlyincreasing on S. (Decreasing functions are similarly defined.) A function is called monotonicon S if it is increasing on S or decreasing on S. If f is an increasing function, then -f is adecreasing 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(x0-) ≤ 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 inversefunction f-1 exists and is strictly increasing on its domain; (ii) If f is continuous on [a, b], thenf-1 is continuous on [f(a), f(b)]. This theorem tells us that a continuous, strictly increasingfunction 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 becharacterised in a very elegant fashion if we introduce a slight generalisation of the notions ofopen and closed sets.

Definition 4-30. Let A and B be two sets in En, with A ⊂ B. We say that A is closedrelative to B if those accumulation points of A which lie in B are also in A. We say that A isopen 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 En, and every open set in En isopen relative to En. It is also clear that every set is both open and closed relative to itself. A setwhich is open relative to B need not, of course, be an open set(relative to En). However, such a set must be the intersection ofB with an open set. This statement, which will form the basis ofthe following theorem, is illustrated as shown. The shadedregion (which includes the arc abc, except for the endpoints aand 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, Ais the intersection of B with a set which is open relative to En. Note that this theorem alsoholds with “open” replaced by “closed” throughout. Using these concepts, we can give thefollowing characterisation of continuous functions:

Theorem 4-32. Let f be a function defined on a set S in En, and let T = f(S) be a subset ofEm. 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.

B

a

b

c

open set S in E2

B∩S open relative to B

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 functionvalues in E1 or in E2. Let a be an accumulation point of S and assume that we have f(x) = Alim

xda

and g(x) = B. Then we also have (i) [f(x) ± g(x)] = A ± B, (ii) f(x)g(x) = AB, andlimxda

limxda

limxda

(iii) f(x)/g(x) = A/B if B ≠ 0.limxda

Proof. (i) Let an arbitrary ε > 0 be given and let ε’ be a second number (ε’ > 0) whichwill be made to depend on ε in a way to be described later. Choose a neighbourhood N(a) suchthat 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 bemade to depend on ε in a way to be described later. Choose a neighbourhood N(a) such that ifx ∈ N’(a)∩S, then we have both |f(x)-A| < ε’ and |g(x)-B| < ε’. In this question, the conclusionwe want is to show that |f(x)/g(x) - A/B| < ε whenever x ∈ N’(a)∩S.

Now |f(x)g(x) − A

B | = |f(x)B−AB+AB−Ag(x)+AB−AB

Bg(x) | = |B(f(x)−A)+AB−A(g(x)−B)−AB

Bg(x) | = |B(f(x)−A)−A(g(x)−B)

Bg(x) |

.[ |B(f(x)−A)

Bg(x) | + |−A(g(x)−B)

Bg(x) | < | Be∏

Bg(x) | + | −Ae∏

Bg(x) |

Now |g(x)| = |B+g(x)-B| < |B|+ε’ < |B|+1, so that

.| Be∏

Bg(x) | + | −Ae∏

Bg(x) | = |Be∏||B||g(x)| + |−Ae∏|

|B||g(x)| < |e∏||B|+1 + |−Ae∏|

|B|(|B|+1) = e ∏

|B|+1 + |A|e∏

|B|(|B|+1) = e∏(|B|+|A|)|B|(|B|+1)

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 twostatements(i) |f(x+h)-f(x)| = 0.lim

hd0

(ii) |f(x+h)-f(x-h)| = 0.limhd0

(a) Show that (i) always implies (ii).(b) 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 |f(x+h)-f(x)| = 0.limhd0

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. |f(x+h)-f(x-h)| = 0 as required. QED.lim

hd0

(b) Let us define a function f to be given by the following: f(x) = x² if x ≠ 0, and f(x) = 1if x = 0 (with x ∈ E1). At x = 0, |f(x+h)-f(x-h)| = |0-0| = 0, but |f(x+h)-f(x)| = |0-1| = 1.lim

hd0limhd0

4-5. If x is any real number in [0, 1], show that the following limit exists: .lim

md∞ [ limnd∞ cos2n(m!ox)]

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’ ofcos(m!πx). For any m > 2, m!π will always be an even number of π’s, so that it is of the form2kπ for some integer k > 0. Recall that cos(2kπ) = 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. wewill 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 largem! 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‘infinite power’ of 1’s which will still be 1, so that we have = 1 provided thatlim

nd∞ cos2n(m!ox)m is large enough. In other words, = 1 when x is rational. Conclusion:lim

md∞ [ limnd∞ cos2n(m!ox)]

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 irrationalnumber is ever even or odd). It follows that we must have |cos(m!πx)| ≠ 1 in this case (for anym), and so it follows that cos(m!πx) ∈ (-1, 1).

Claim: if y ∈ (-1, 1), then yn = 0. Proof of Claim. We need to find an N such that if nlimnd∞

> N, then for an arbitrary ε > 0 and a particular y ∈ (-1, 1) we have |y|n < ε (We need themodulus signs in because if we don’t put them in, then for any ε > 0 we will always have yn < εfor any y ∈ (-1, 0) and any odd n. We will therefore prove a stronger condition than the one weneed). 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.

Therefore, because for any y ∈ (-1, 1) we have yn = 0, it follows that for anylimnd∞

cos(m!πx) ∈ (-1, 1) we have cos2n(m!πx) = 0. This will hold for any m and so we have limnd∞

= 0 when x is irrational. Conclusion: the limit in question exists when xlimmd∞ [ lim

nd∞ cos2n(m!ox)]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) = 0for every x in [a, b].

Answer: If f is continuous on [a, b], then f is continuous at every point in [a, b]. Weknow that f(x) = 0 when x is rational. But every irrational number y is an accumulation pointof the set of rational numbers, so by definition 4-9, and by knowing that f is continuous at thepoint y, it follows that (i) f is defined at y, and (ii) limx→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? Itmust be zero, as in every neighbourhood N(y), because y is an accumulation point, there is arational number z such that f(z) = 0 and so f(x) ∈ N(0), thus satisfying the definition for alimit. Conclusion: f(x) = 0 for all x ∈ [a, b].

4-8. Let f be continuous at the point a = (a1, a2, ..., an) in En. Keep a2, a3, ..., an fixed anddefine a new function g of one real variable by the equation g(x) = f(x, a2, ..., an). Show that gis continuous at the point x = a1. (This is sometimes stated as follows: “A continuous functionof n variables is continuous in each variable separately.”)

Answer: Let us define a new function s: E1 → En by the following definition: s(x) = (x, a2, a3, ..., an) for an arbitrary x ∈ E1 and fixed a2, a3, ..., an ∈ E1.

Looking at the above definition and at the question itself, we see that we can now writeg(x) as follows: g(x) = f(s(x)) for an arbitrary x ∈ E1. Looking at Theorem 4-11, we see that inorder to prove that g(x) is continuous at the point a = (a1, a2, ..., an) 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 continuousat x, (iii) The point s(x) is an accumulation point of En, and (iv) f is continuous at the points(x). If we can show that the four conditions above hold, then we can say that g is continuousat the point x = a.

(i) We know that any real number is an accumulation point of the real line E1, so thatcondition (i) is satisfied for an arbitrary real number x = a1.

(ii) To prove that the function s(x) is continuous at an arbitrary point x ∈ E1, we need toshow that the function is defined at x and that . The first part followslim

adx s(a) = s(x)immediately because the function s(x) is defined for all real numbers x ∈ E1 by its definitionon the previous page. It remains to show that we have .lim

adx s(a) = s(x)

But = = s(x).limadx s(a) lim

adx (a, a2, a3, ..., an) = ( limadx (a), lim

adx (a2), ..., limadx (an)) = (x, a2, ..., an)

We can therefore say that condition (ii) holds. For condition (iii) to hold, we must show thatthe point s(x) is an accumulation point of the set En. But by the same argument as in condition(i), as s(x) is a point in En, and as all points in En are accumulation points, then s(x) = (x, a2, ...,an) is an accumulation point of the set En, satisfying condition (iii).

Finally, we must show that f is continuous at the point s(x). But if we let x = a1, thens(a1) = (a1, a2, ..., an) = a, and we know by the assumption made in the question that f iscontinuous at the point a in En. Therefore, condition (iv) is satisfied, and thus we have shownthat all four conditions hold, and can subsequently say that the function g is continuous at thepoint 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 and1.

Answer: Let us first prove that f is continuous at the point x = ½. If x = ½, then x isrational 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 anyarbitrary δ > 0, the deleted neighbourhood N’(½)∩[0, 1] will always contain irrational andrational 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 thiscase, |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. (Inthe above we used the fact that |A-B| = |B-A|). So we have shown that if x = ½, then f(x) iscontinuous.

Now consider the case when x ≠ ½. If f(x) is continuous for x ≠ ½, then given anarbitrary 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 asingle ε1 > 0 such that there is no neighbourhood N’(x)∩[0, 1] in which all the points within δof 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 begreater 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 anelement 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 ALLirrational y in this neighbourhood. Conclusion: f(x) is not continuous when x is a rationalnumber in [0, 1] and x ≠ ½.

The other case to consider is when x is a fixed irrational number in [0, 1]. Consider nowthat for this fixed irrational number x, we define a number ε2 = 1. Given this ε2, we need to finda δ > 0 such that if |x-y| < δ for y ∈ [0, 1] ≠ x, then we will always have |f(x)-f(y)| < ε2. Assumethat 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 thatwill 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 theinterval [0, 1], as f(x) = x for all rational numbers in [0, 1]. Our only problem occurs inshowing that f assumes the value of every irrational number in the interval [0, 1], i.e. if wehave an arbitrary irrational number y ∈ [0, 1], then we need to find a number z ∈ [0, 1] suchthat f(z) = y.

If y is our arbitrary irrational number in the interval [0, 1], then 1-y will also be anarbitrary 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 thatf(S) is a bounded set.) If T is a subset of S, the number Ωf(T) = supf(x)-f(y) | x ∈ T, y ∈ T iscalled the oscillation of f on T. If x ∈ S, the oscillation of f at x is defined to be the numberωf(x) = Ωf(N(x; h)∩S). Show that this limit always exists and that ωf(x) = 0 if, and only if,lim

hd0+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 followsthat 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 theneighbourhood of x in S. The value ωf(x) is thus determined by two points (say x1 and x2) inN(x; h)∩S. Because f(S) is bounded (i.e. f(S) is a subset of some finite interval), then the twopoints in question will have finite values, and so their difference (i.e. f(x1)-f(x2)) will also befinite. Therefore, for a particular h, Ωf(N(x; h)∩S) will exist and is a finite value.

Does the limit Ωf(N(x; h)∩S) exist? To show that it does exist, all we need do is tolimhd0+

show that if h2 > h1, then Ωf(N(x; h1)∩S) ≤ Ωf(N(x; h2)∩S). If this is true, then the sequence ofnumbers Ωf(N(x; h)∩S) given by decreasing h will be a bounded monotonic sequence of finitepositive numbers (bounded above by, say, the entry where h = 1), and thus we can applyTheorem 12-6 to say that this sequence converges to a value, say A, and so we can write

A = Ωf(N(x; h)∩S), where 0 ≤ A ≤ Ωf(N(x; 1)∩S), say. limhd0+

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 supf(x1)-f(x2) | x1 ∈ (N(x; h)∩S), x2 ∈ ((x; h)∩S)? Well, it is themaximum value of f(x1)-f(x2) in the neighbourhood N(x; h)∩S. But if this is so, then f(x1)must be the highest value of f(x) in the neighbourhood N(x; h)∩S, and f(x2) must be thesmallest value of f(x) in the neighbourhood N(x; h)∩S. Therefore, we have

supf(x1)-f(x2) | x1 ∈ (N(x; h)∩S), x2 ∈ ((x; h)∩S) = supf(x1) | x1 ∈ ((x; h)∩S) - inff(x2) | x2 ∈ ((x; h)∩S.

The proof centres on the fact that if h2 > h1, then supf(x) | x ∈ N(x; h1)∩S ≤ supf(x) |x ∈ N(x; h2)∩S, and inff(x) | x ∈ N(x; h1)∩S > inff(x) | x ∈ N(x; h2)∩S

I will not prove these statements (intuitively, if we are considering a smallerneighbourhood, there are less points to consider, and so the maximum value in a smallerneighbourhood will be smaller and the smallest value in a smaller neighbourhood will bebigger). Assuming that these statements are correct, we have

supf(x1) | x1 ∈ ((x; h1)∩S) - inff(x2) | x2 ∈ ((x; h1)∩S≤ supf(x1) | x1 ∈ ((x; h2)∩S) - inff(x2) | x2 ∈ ((x; h2)∩S,or supf(x1)-f(x2) | x1 ∈ (N(x; h1)∩S), x2 ∈ ((x; h1)∩S)≤ supf(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, wherey ∈ N(x; h)∩S. Now if x1, x2 ∈ 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 supf(x1)-f(x2) | x1 ∈ (N(x; h)∩S), x2 ∈ ((x; h)∩S) → 0 when h → 0+, or, in other words,

supf(x1)-f(x2) | x1 ∈ (N(x; h)∩S), x2 ∈ ((x; h)∩S) = 0 as required. QED.limhd0+

Only If. If ωf(x) = 0, then we need to show that f is continuous at x. To do this, we needto show that f is defined at x and that limy→x f(y) = f(x). The first bit follows immediately fromthe 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 twofollowing equations: and ;lim

hd0+ (x + h) = f(x) limhd0+ (x − h) = f(x)

or and .limhd0+ (x + h) − f(x) = 0 lim

hd0+ f(x) − (x − h) = 0

For a given h, the neighbourhood given by (x; h)∩S will contain the point x and theNpoints (x+h) and (x-h). Now if ωf(x) = 0, then

supf(x1)-f(x2) | x1 ∈ (N(x; h)∩S), x2 ∈ ((x; h)∩S) = 0. limhd0+

From our previous analysis of supf(x1)-f(x2) | x1 ∈ (N(x; h)∩S), x2 ∈ ((x; h)∩S), weknow that f(x1) must be the highest value of f(x) in the neighbourhood N(x; h)∩S, and thatf(x2) must be the smallest value of f(x) in the neighbourhood N(x; h)∩S. Knowing this, thenfor a point x3 in the neighbourhood N(x; h)∩S, then we must have f(x3)-f(x) ≤ supf(x1)-f(x2) |x1 ∈ (N(x; h)∩S), x2 ∈ ((x; h)∩S). Note also that we must have 0 ≤ f(y)-f(z) ≤ supf(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) ≤ supf(x1)-f(x2) | x1 ∈ (N(x; h)∩S), x2 ∈ (N(x; h)∩S).Therefore,

f(x + h) - f(x) ≤ supf(x1)-f(x2) | x1 ∈ (N(x; h)∩S), x2 ∈ (N(x; h)∩S).limhd0+

limhd0+

But we have assumed that supf(x1)-f(x2) | x1 ∈ (N(x; h)∩S), x2 ∈ ((x; h)∩S) = 0,limhd0+

so that f(x + h) - f(x) ≤ 0, i.e. we have f(x + h) - f(x) = 0 as required (because we mustlimhd0+

limhd0+

have f(x+h)-f(x) > 0).

(ii) f(x) - f(x-h) ≤ supf(x1)-f(x2) | x1 ∈ (N(x; h)∩S), x2 ∈ ((x; h)∩S).Therefore,

f(x) - f(x-h) ≤ supf(x1)-f(x2) | x1 ∈ (N(x; h)∩S), x2 ∈ ((x; h)∩S).limhd0+

limhd0+

But again we have assumed that supf(x1)-f(x2) | x1 ∈ (N(x; h)∩S), x2 ∈ (N(x;limhd0+

h)∩S) = 0, so that f(x) - f(x-h) ≤ 0, i.e. we have f(x) - f(x-h) = 0 as required (becauselimhd0+

limhd0+

we must have f(x)-f(x-h) > 0). QED. Technical Note: There is a slight mistake in the above inthat f(x ± h) ∉ N(x; h)∩S, but I have assumed that in taking the limit this is of no consequenceto the proof.

4-17. Let f be defined and continuous on a closed set S in En. 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 E1 (i.e. T = f(S)). Let Y be the subset ofT consisting only of the zero point, i.e. Y = 0. Because Y consists of just a single point, thenY contains all its accumulation points (of which there are none of), and thus Y is a closedsubset 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 maximumat x1 and a local maximum at x2. Show that there must be a third point between x1 and x2 wheref has a local minimum.

Answer: Consider the interval [x1, x2] ⊂ [a, b]. Because f is continuous on the closedinterval [a, b], then it will be continuous on the subinterval [x1, 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 notefollowing 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 thatf 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 thatf(x) > f(c) for all x ∈ [x1, x2]. If we now take any neighbourhood around the point c such thatN(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 x1 or x2 (we need a thirdpoint between x1 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 beenincorrectly found. Therefore, f(x1) = f(x1+h), and so we can equally well define c to be given byc = x1+h to represent our absolute/local minimum on [x1, x2]. Note: if x1+h > x2, then just take apoint half way between x1 and x2 to represent c — in this situation f must be a constantfunction between x1 and x2.

The case c = x2 goes through in much the same way as above (i.e. we redefine c to begiven by c = x2-h or by c = (x1+x2)/2 as appropriate), so that we have found a point c betweenx1 and x2 which is a local minimum in the interval [x1, x2]. QED.

4-21. Show that a function which is uniformly continuous on a set S is also continuouson S.

Answer: Let f be defined on a set S in En with function values in Em. f is said to beuniformly 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 saidto be continuous on S provided that at each accumulation point a, and given ε, there is anumber 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) suchthat if x ∈ S, and |x-a| < δ, then |f(x)-f(a)| < ε. But this is exactly the condition we want so thatthe function f is continuous on S (it doesn’t matter that δ does not depend on a), and thereforewe have reached the conclusion we want, in that uniform continuity on S does imply continuityon 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 E1. We need to showthat if the distance between x and y is less than δ, then the distance between f(x) and f(y) isless than ε, where δ depends only on ε. In other words, given ε, if |x-y| < δ, then we need toshow 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/2for 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 valuen 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 ifwe 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 thuscannot 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 En and assume that f(S) ⊂ Em. Let g bedefined 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 aδ1 > 0 (dependent only on ε1) such that |x-y| < δ1 ⇒ |f(x)-f(y)| < ε1. (x, y ∈ En). (2) If g isuniformly continuous on f(S), then given an ε2 > 0, we can always find a δ2 > 0 (dependent onlyon 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 forthe 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 (dependentonly on ε2) such that |f(x)-f(y)| < δ2 ⇒ |g(f(x))-g(f(y))| < ε2. If we let δ2 = ε1, then given thatparticular ε1 (which is dependent only on ε2), we can (by (1)) always find a δ1 > 0 (which nowis 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 E1 by thefollowing equations:

(a) f(x) = (sin x)/x if x ≠ 0, f(0) = 0.(b) f(x) = e1/x 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) iscontinuous at every point in E1

except at x = 0, where f(x) has aremovable discontinuity. Aslimx→0- = 1 and limx→0+ = 1, we canmake the function continuous byredefining f(0) to be f(0) = 1. (b)Again the function is continuous atevery point in E1 except at x = 0,where f(x) has an infinite jumpdiscontinuity: 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. Wetherefore 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 jumpdiscontinuity at x = 0. Apart from this, f(x) is continuous at every other point in E1.

4-29. Let f be defined in the open interval (a, b) and assume that for each interior point xof (a, b) there exists a neighbourhood N(x) in which f is increasing. Show that f is anincreasing function throughout (a, b).

Answer: We saw in exercise 3-1 that an open interval (a, b) is an open set — sotherefore all points x ∈ (a, b) are interior points. This implies that for all points x ∈ (a, b), thereexists a neighbourhood N(x) in which f is increasing. Let us now prove that f is increasing bycontradiction.

If f is not increasing on (a, b), then there is a subinterval (say (c, d)) of (a, b) in which fis 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 neighbourhoodof z in which f is increasing. Because of this, then we must throw away this neighbourhoodfrom our subinterval (c, d) in which we are claiming that f is strictly decreasing. But we canapply the above argument with any point in the subinterval (c, d), and thus there cannotpossibly be any such subinterval (c, d) in which f is strictly decreasing because for any point insuch an interval, f will be increasing in a neighbourhood of that point. We therefore concludethat f must be increasing on the interval (a, b). QED.

x

f(x)(a)

x

f(x)(b)

x

f(x)(c)

x

f(x)(d)

1

1

Here is a second proof for the above question. Suppose that we pick two arbitrarynumbers x1 and x2 from the interval (a, b) so that a < x1 < x2 < b. To show that f is an increasingfunction throughout (a, b), all we must show is that f(x1) ≤ 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, andso by the note following Definition 3-39 (and by knowing that [x1, x2] is a bounded set (it is asubset of (a, b)) we can say that [x1, 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 wedefine a covering of the compact set [x1, x2] given by the union of all such neighbourhoods ofpoints in [x1, x2] in which f is increasing. In other words, we can write

[x1, x2] = .4all x c [x1,x2] N(x)

Now because [x1, x2] is a compact set, a finite number of these neighbourhoods willcover [x1, x2], say a collection of n neighbourhoods. Therefore, we can write

[x1, x2] = , with x ∈ [x1, x2].4 i=1n N i(x)

Because f is increasing throughout all of these neighbourhoods, and because this finitecollection 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 monotonicon [a, b]. That is, prove that every topological mapping of [a, b] onto an interval [c, d] must bestrictly monotonic.

Answer: If f is one-to-one on [a, b], then for all points y in the range, there exists oneand only one point x in the domain such that f(x) = y. If f is continuous on [a, b], then weknow 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) = y2.

Because f is one-to-one, then we cannot have y1 = y2, and thus we must have either y1 >y2 or y1 < y2. Consider the case where y1 < y2 (we will only consider this case — the other casecan be derived from the following by replacing all ‘<’ by ‘>’ and vice-versa). Because of thecontinuity of f on the closed interval [a, b], then the intermediate value theorem says that fassumes all values between y1 and y2.

Consider an arbitrary point c ∈ (a, b). What is f(c)? First ofall, f(c) cannot be equal to either y1 or y2 because f is aone-to-one function. Further (as seen in Diagram 1), we cannothave f(c) > y2 as then (according to the IVT) there will be a pointp in the interval (a, c) where f(p) = y2, and this cannot be thecase if f is one-to-one (because then we will have f(p) = f(b) =y2). Similarly (as seen in Diagram 2), we cannot have f(c) < y1 asthen (again according to the IVT) there will be a point q in theinterval (c, b) where f(q) = y1, and this cannot be the case if f isone-to-one (because then we will have f(q) = f(a) = y1).Therefore, we conclude that we must have y1 < f(c) < y2.

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 anypair of points c and d in [0, 1], with c < d, then we must have f(c) < f(d), and thus f is strictlyincreasing on [a, b].

With the other case, where y1 > y2, we would reach the conclusion that f would bestrictly decreasing on [a, b]. Putting all of this together, we conclude that f is either a strictlymonotonically increasing function or a strictly monotonically decreasing function, i.e. f is amonotonic function. QED.

f(a)f(b)

f(q)

f(c)

f(a)f(b)

f(c)

f(p)

Diagram 1

Diagram 2

Possible Further Work / Evaluation

Exercise 4-7: Clarification in the wording, or possibly an alternative solution, e.g. if weconsider an arbitrary irrational number in [a, b], and because we know that there exists asequence of rational numbers xn converging to y, the result follows because of the continuity off?

Exercise 4-11: A ‘small gap’ left unfilled in the first part of the answer. In the secondpart, 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]. Considernow that for this fixed irrational number x, we define a number ε2 = 1. Given this ε2, we needto 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 neighbourhoodN’(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 firstsection. 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 ∈ Aimplies a ≤ x, then x is called an upper bound for the set A and we say that A is boundedabove.

Definition 1.7: Let A be a set of real numbers bounded above. Suppose there is a realnumber x satisfying the following two conditions:

(i) x is an upper bound for A, and(ii) 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 δin 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 willcover [x1, x2], say a collection of n neighbourhoods N(y1), ...., N(yn). Therefore, we can write

[x1, x2] ⊂ , with yi ∈ [x1, x2].....4 i=1n N(y i)

Chapter 13

Chapter 13: Sequences of Functions

In this chapter, we will be dealing with complex-valued functions defined on certainsubsets of E2 (but if results hold only for real-valued functions defined on subsets of E1, thiswill be explicitly stated).

Given a sequence fn, each term of which is a function defined on a set S, for each x inS we can form another sequence fn(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. Thefunction f defined by f(x) = fn(x), if x ∈ T, will be referred to as the limit function of thelim

nd∞

sequence fn and we will say that fn converges pointwise on the set T.

If each term of the sequence fn has a certain property, then to what extent does thefunction f also possess this property? In general, we need a study of “stronger” methods thatdo preserve these properties. The most important of these is the notion of uniformconvergence.

When we ask whether continuity at each fn at x0 implies continuity of the limit function fat x0, we are really asking whether the equation fn(x) = fn(x0) implies the equation f(x)lim

xdx0lim

xdx0

= f(x0). The last equation can also be written as follows: .limxdx0

limnd∞ fn(x) = lim

nd∞lim

xdx0 fn(x)Therefore, our question about continuity amounts to this: Can we change the limit symbols inthe above?

In general, we shall see that we cannot interchange the symbols. First of all, the limit inthe equation ( fn(x) = fn(x0)) may not exist; or, even if it does exist, it may not be equal tolim

xdx0

f(x0). In chapter 12, it is stated that Σm=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 limitprocesses. We shall find that uniform convergence is a far-reaching sufficient condition forthe validity of interchanging certain limits, but it does not provide the complete answer to thequestion. Examples can be found in which the order of the two limits can be interchangedalthough the sequence is not uniformly convergent.

An example of a sequence of real-valued functions: Let fn(x) = n²x(1-x)n if x ∈ E1, n = 1,2, ... Here, limn→∞ fn(x) exists if 0 ≤ x ≤ 1, and the limit function has the value 0 at each point in[0, 1]. fn has a local maximum at x = 1/(n+1), but fn(1/(n+1)) → +∞ as n → ∞.

Let fn be a sequence of functions which converges pointwise on a set T to a limitfunction f. Going back to the basic definition of limit, this means that for each point x in T andfor each ε > 0, there exists an N (depending on at most both x and ε) such that n > N implies|fn(x)-f(x)| < ε. If the same N works equally well for every point in T, the convergence is saidto be uniform on T. That is, we have the following definition:

A sequence of functions fn 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 |fn(x)-f(x)| < ε, for every xin T. We denote this symbolically by writing fn → f uniformly on T.

When each term of the sequence fn is real-valued, there is a useful geometricinterpretation of uniform convergence. The inequality |fn(x)-f(x)| < ε is then equivalent to thetwo inequalities f(x)-ε < fn(x) < f(x)+ε. If this is to hold forall n > N and for all x in T, this means that the entire graphof fn (that is, the set (x, y) | y = fn(x), x ∈ T) lies within a“band” of height 2ε situated symmetrically about the graphof f (see the diagram on the right).

A sequence fn is said to be uniformly bounded on T if there exists a constant M > 0such that |fn(x)| ≤ M for all x in T and for all n = 1, 2, ... The number M is called a uniformbound for fn. If each individual function is bounded and if fn → f uniformly on T, then it iseasy to prove that fn is uniformly bounded on T. This observation often enables us toconclude 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 gn on P as follows: gn(m) = f(m, n), if m ∈ P. Assumethat gn → g uniformly on P, where g(m) = limn→∞ f(m, n). If the iterated limit limm→∞ (limn→∞

f(m, n)) exists, then the double limit limm,n→∞ f(m, n) also exists and has the same value.

Uniform Convergence and Continuity. Theorem 13-3: Assume that fn → f uniformlyon T. If each fn is continuous at a point x0 of T, then the limit function f is also continuous atx0. Note 1: If x0 is an accumulation point of T, the conclusion implies that

. Note 2: Uniform convergence of fn is sufficient but notlimxdx0

limnd∞ fn(x) = lim

nd∞lim

xdx0 fn(x)necessary to transmit continuity from the individual terms to the limit function.

The Cauchy condition for uniform convergence. Theorem 13-4: Let fn be asequence of functions defined on a set T. There exists a function f such that fn → f uniformlyon T if, and only if, the following condition (called the Cauchy condition) is satisfied: Forevery ε > 0 there exists an N such that m > N and n > N implies |fm(x)-fn(x)| < ε, for every x inT.

Uniform Convergence on infinite series. Definition 13-5: Given a sequence fn offunctions defined on a set T. For each x in T, let sn(x) = Σk=1

n fk(x) (n = 1, 2, ...). If there exists afunction f such that sn → f uniformly on T, we say that the series Σfn(x) converges uniformly onT and we write Σn=1

∞ fn(x) = f(x) (uniformly on T).

y = fn(x) y = f(x)+ε

y = f(x)-ε

y = f(x)

Theorem 13-6: (Cauchy condition for uniform convergence of series). The series Σfn(x)converges uniformly on T if, and only if, for every ε > 0 there is an N such that n > N implies

fk(x)| < ε, for each p = 1, 2, ..., and every x in T.| Sk=n+1n+p

Theorem 13-7: (Weierstrass’ M-test). Let Mn be a sequence of non-negative numberssuch that 0 ≤ |fn(x)| ≤ Mn, for n = 1, 2, ..., and for every x in T. Then Σfn(x) converges uniformlyon T if ΣMn converges.

Theorem 13-8: Assume that Σfn(x) = f(x) (uniformly on T). If each fn is continuous at apoint x0 of T, then f is also continuous at x0. Note: If x0 is an accumulation point of T, thistheorem permits us to write fn(x) = fn(x).lim

xdx0 Sn=1∞ Sn=1

∞ limxdx0

A space-filling curve. We can apply the above theorem to construct an example of whatis known as a space-filling curve. This is a continuous curve in E2 that passes through everypoint of the unit square [0, 1]×[0, 1]. Consider the following example: Let φ be defined on theinterval [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; andφ(t) = -3t+5, if 4/3 ≤ t ≤ 5/3. Extend the definition of φ to all of E1 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 bythe following equations:

α1(t) = , α2(t) = .Sn=1∞ v(32n−2t)

2n Sn=1∞ v(32n−1t)

2n

Both series converge absolutely for each real t and they converge uniformly on E1. Infact, since |φ(t)| ≤ 1 for all t, the Weierstrass M-test is applicable with Mn = 2-n. Since f iscontinuous on E1, Theorem 13-8 tells us that α1 and α2 are also continuous on E1. Let αα = (α1,α2), and let Γ denote the image of the unit interval [0, 1] under αα. 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 = , b = ,Sn=1∞ an

2n Sn=1∞ bn

2n

where each an and each bn is either 0 or 1. Now let c = , where c2n-1 = an and c2n =2 Sn=1∞ cn

3n

bn, n = 1, 2, ... Clearly, 0 ≤ c ≤ 1 since 2Σn=1∞ 3-n = 1. We will show that α1(c) = a and that α2(c)

= b. If we can prove that φ(3kc) = 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.

To prove equation (1), we write 3kc = 2 + 2 = (an even integer) + dk,Sn=1k cn

3n−k Sn=k+1∞ cn

3n−k

where dk = Σn=1∞ cn+k/2n. Since φ has period 2, it follows that φ(3kc) = φ(dk). If ck+1 = 0, then we

have 0 ≤ dk ≤ 2Σn=2∞ 3-n = 1/3, and hence φ(dk) = 0. Therefore, φ(3kc) = ck+1 in this case. The only

other case to consider is the case ck+1 = 1. But then we get 2/3 ≤ dk ≤ 1, and hence φ(dk) = 1.Therefore, φ(3kc) = ck+1 in all cases and this proves that α1(c) = a, α2(c) = b. Hence, Γ fills theunit square.

-2 -1 0 1 2 3 4

An application to repeated series. As a second application of Theorem 13-8, we derivethe following generalisation of Theorem 12-43 (which went as follows: Theorem 12-43:Suppose f(m, n) > 0 for all m, n. Assume that Σn=1

∞ f(m, n) converges for each fixed m = 1, 2,..., and that Σm=1

∞Σn=1∞ f(m, n) converges. Then: (a) Σm=1

∞ f(m, n) converges for each n = 1, 2, ...;and (b) Σn=1

∞Σm=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. Assume that Σn=1∞ f(m, n)

converges absolutely for each fixed m and that Σm=1∞Σn=1

∞ |f(m, n)| converges. Then (a) Σm=1∞

f(m, n) converges absolutely for each n, and (b) Σn=1∞Σm=1

∞ f(m, n) converges absolutely andequals Σm=1

∞Σn=1∞ f(m, n).

Uniform convergence and differentiation. Theorem 13-13: Assume that each term offn is a real-valued function having a finite derivative at each point of an open interval (a, b).Assume that for at least one point x0 in (a, b) the sequence fn(x0) converges. Assume furtherthat there exists a function g such that f’n → g uniformly on (a, b). Then (a) There exists afunction f such that fn → f uniformly on (a, b); (b) For each x in (a, b), the derivative f’(x)exists and equals g(x).

When we reformulate the above theorem in terms of series, we obtain the following:Theorem 13-14: Assume that each fn is a real-valued function defined on (a, b) such that thederivative f’n(x) exists for each x in (a, b). Assume that, for at least one point x0 in (a, b), theseries Σfn(x0) converges. Assume further that there exists a function g such that Σf’n(x) = g(x)(uniformly on (a, b)). Then (a) There exists a function f such that Σfn(x) = f(x) (uniformly on (a,b)); (b) If x ∈ (a, b), the derivative f’(x) exists and equals Σf’n(x).

Sufficient conditions for uniform convergence of a series. In some of the precedingtheorems, the importance of uniformly convergent series has been illustrated. It seems naturalto seek some simple ways of testing a series for uniform convergence without resorting to thedefinition in each case. We have seen one such test already (Weierstrass’ M-test), but when theM-test is not applicable, the following may be useful:

Theorem 13-15: (Dirichlet’s test for uniform convergence). Let Fn(x) denote the nth

partial sum of the series Σfn(x), where each fn is a complex-valued function defined on a set T.Assume that Fn is uniformly bounded on T. Let gn be a sequence of real-valued functionssuch that gn+1(x) ≤ gn(x) for each x in T and for every n = 1, 2, ..., and assume that gn → 0uniformly on T. Then the series Σfn(x)gn(x) converges uniformly on T.

Example: Let Fn(x) = Σk=1n eikx. In a previous chapter, the inequality |Fn(x)| ≤ 1/|sin(x/2)|

was derived, valid for every real x ≠ 2mπ (m is an integer). Therefore, if 0 < δ < π, we have theestimate |Fn(x)| ≤ 1/sin(δ/2) if δ ≤ x ≤ 2π-δ. Hence, Fn is uniformly bounded on the interval [δ,2π-δ]. If gn satisfies the conditions of the above theorem, we can conclude that the seriesΣgn(x)einx converges uniformly on [δ, 2π-δ]. In particular, if we take gn(x) = 1/n, this establishesthe uniform convergence of the series Σn=1

∞ (einx/n) on [δ, 2π-δ] if 0 < δ < π. Note that theWeierstrass M-test cannot be used to establish uniform convergence in this case, since |einx| =1.

Bounded convergence. Arzelà’s theorem. Definition 13-16. A sequence of functionsfn is said to be boundedly convergent on T if fn is pointwise convergent and uniformlybounded on T. Theorem 13-17 (Arzelà). Assume that fn is boundedly convergent on [a, b]and suppose each fn is Riemann-integrable on [a, b]. Assume also that the limit function f isRiemann-integrable on [a, b]. Then lim

n→∞ ∫ba fn(x)dx = ∫b

a limn→∞ fn(x)dx = ∫b

a f(x)dx.

Note: it is easy to give an example of a boundedly convergent sequence fn ofRiemann-integrable functions whose limit f is not Riemann-integrable. If r1, r2, ... denotesthe set of rational numbers in [0, 1], define fn(x) to have the value 1 if x = rk for some k = 1, 2,..., n, and put fn(x) = 0 otherwise. Then the integral ∫0

1 fn(x)dx = 0 for each n, but the limitfunction f is not Riemann-integrable on [0, 1].

Mean convergence. Definition 13-18. Let fn be a sequence of Riemann-integrablefunctions defined on [a, b]. Assume that f ∈ R on [a, b]. The sequence fn is said to convergein the mean to f on [a, b], and we write l.i.m.n→∞ fn = f on [a, b], if limn→∞ ∫b

a |fn(x)-f(x)|²dx = 0.If the inequality |f(x)-fn(x)| < ε holds for every x in [a, b], then we have ∫b

a |f(x)-fn(x)|²dx ≤ε²(b-a). Therefore, uniform convergence of fn to f on [a, b] implies mean convergence,provided that each fn is Riemann-integrable on [a, b].

A rather surprising fact is that convergence in the mean need not imply pointwiseconvergence at any point of the interval. This can be seen as follows: For each integer n > 0,subdivide the interval [0, 1] into 2n equal subintervals and let I denote that subinterval2n+k

whose right endpoint is (k+1)/2n, where k = 1, 2, ..., 2n-1. This yields a collection I1, I2, ... ofsubintervals of [0, 1] of which the first few are: I1 = [0, 1], I2 = [0, 1/2], I3 = [1/2, 1], I4 = [0, 1/4],I5 = [1/4, 1/2], I6 = [1/2, 3/4], and so forth.

Define fn on [0, 1] as follows: fn(x) = 1 if x ∈ In, and fn(x) = 0 if x ∈ [0, 1]—In. Then fnconverges in the mean to 0, since ∫1

0|fn(x)|²dx is the length of In, and this approaches 0 as n→∞.On the other hand, for each x in [0, 1], we have fn(x) = 1 and fn(x) = 0. Hence,

lim supnd∞

lim infnd∞

fn(x) does not converge for any x in [0, 1].

Theorem 13-19. Assume that l.i.m.n→∞ fn = f on [a, b]. If g ∈ R on [a, b], define h(x) = ∫ba

f(t)g(t)dt, hn(x) = ∫xa fn(t)g(t)dt, if x ∈ [a, b]. Then hn→h uniformly on [a, b].

Theorem 13-20. Assume that l.i.m.n→∞ fn = f and l.i.m.n→∞ gn = g on [a, b]. Define h(x) =∫x

a f(t)g(t)dt, hn(x) = ∫xa fn(t)gn(t)dt, if x ∈ [a, b]. Then hn→h uniformly on [a, b].

Power Series. An infinite series of the form a0+Σn=1∞an(z-z0)n, written more briefly as

Σn=0∞an(z-z0)n, is called a power series in z-z0. Here, z, z0 and an (n = 0, 1, 2, ...) are complex

numbers. With every power series, there is associated a circle, called the circle ofconvergence, such that the series converges absolutely for every z interior to this circle anddiverges for every z outside this circle. The centre of the circle is at z0 and its radius is calledthe radius of convergence of the power series. (The radius may be 0 or +∞ in extreme cases.)The next theorem establishes the existence of the circle of convergence and provides us with away for calculating its radius.

Theorem 13-21. Given a power series Σn=0∞ an(z-z0)n, let λ = , r = 1/λ,

lim supnd∞

n|an |

(where r = 0 if λ = +∞ and r = +∞ if λ = 0). Then the series converges absolutely if |z-z0| < rand diverges if |z-z0| > r. Furthermore, the series converges uniformly on every compact subsetinterior to the circle of convergence.

Note: If the limit limn→∞ |an/an+1| exists (or if this limit is +∞), its value is also equal to theradius 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, butthe second converges everywhere. This illustrates why the theorem at the top of the pagemakes no assertion about the behaviour of a power series on the boundary of the circle ofconvergence.

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) = Σn=0∞an(z-z0)n, if z ∈ N(z0;r), is

continuous on N(z0;r). Proof. Since each point in N(z0;r) belongs to some compact subset ofN(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(z0;r), is said to represent f in N(z0;r). It is

also called a power series expansion of f about z0. Functions having power series expansionsare continuous inside the circle of convergence. Much more than this is true, however. Wewill later prove that such functions have derivatives of every order inside the circle ofconvergence. The proof will make use of the following theorem:

Theorem 13-23. Assume that Σan(z-z0)n converges if z ∈ N(z0;r). Suppose that theequation f(z) = Σn=0

∞an(z-z0)n is known to be valid for each z in some open subset of N(z0;r).Then, for each point z1 in S, there exists a neighbourhood N(z1;R) ⊂ S in which f has a powerseries expansion of the form f(z) = Σk=0

∞ bk(z-z1)k (equation (1)), where bk = Σn=k∞

(k = 0, 1, 2, ...) (equation (2)). Note that the proof of this theorem asserts that( nk )an(z1 − z0)n−k

we may use any R > 0 that satisfies the condition N(z1;R) ⊂ S.

Theorem 13-24. Assume that Σan(z-z0)n converges for each z in N(z0;r). Then thefunction f defined by the equation f(z) = Σn=0

∞an(z-z0)n, if z ∈ N(z0;r), has a derivative f’(z) foreach z in N(z0;r), given by f’(z) = Σn=1

∞nan(z-z0)n-1.

Note 1: The two series above have the same radius of convergence. Note 2: By repeatedapplication of f’(z) = Σn=1

∞nan(z-z0)n-1, we find that for each k = 1, 2, ..., the derivative f(k)(z)exists in N(z0;r) and is given by the series f(k)(z) = Σn=k

∞ n!/(n-k)!an(z-z0)n-k. If we put z = z0 in thisseries, we obtain the important formula f(k)(z0) = k!ak (k = 1, 2, ...).

This equation tells us that if two power series Σan(z-z0)n and Σbn(z-z0)n both represent thesame function in a neighbourhood N(z0;r), then an = bn for every n. That is, the power seriesexpansion of a function f about a given point z0 is uniquely determined (if it exists al all), andis given by the formula f(z) = Σn=0

∞ (f(n)(z0)/n!)(z-z0)n, valid for each z in the circle ofconvergence.

Theorem 13-25. Assume that Σan(z-z0)n converges for each z in N(z0;r). Let Γ be apiece-wise smooth curve in N(z0;r) joining z0 to z1. Then Σn=0

∞an(z-z0)ndx = Σn=0∞¶G(z0,z1)

(an/n+1)(z1-z0)n+1.

Multiplication of power series. Theorem 13-26. Given two power series expansionsabout 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=0nakbn-k (n = 0, 1, 2, ..).

Note: If the two series are identical, we get f(z)² = Σn=0∞cnzn, where cn = Σk=0

nakan-k = . The symbol indicates that the summation is to be extended over allSm1+m2=n am1am2 Sm1+m2=n

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

m1+...+mp=n am1 ...amp

The substitution theorem. Theorem 13-27. Consider that we are given two powerseries 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∞ckzk, where the coefficients ck are obtained as follows: Define the numbers bk(n) by the

equation g(z)n = (Σk=0∞bkzk)n = Σk=0

∞bk(n)zk. Then ck = Σn=0∞anbk(n) for k = 0, 1, 2, ... Note: The

series Σk=0∞ckzk 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 apower 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. Thenthere exists a neighbourhood N(0;δ) in which the reciprocal of p has a power series expansionof the form 1/p(z) = Σn=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 intervalcalled the interval of convergence. If neighbourhoods are taken to mean one-dimensionalneighbourhoods, all the results of the last three sections can be interpreted in an obvious wayas 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 areal-valued function f defined in some neighbourhood of a point x0 in E1, and suppose that fhas derivatives of every order in this neighbourhood. Then we can certainly form the powerseries Σn=0

∞ (x-x0)n.f(n)(x0)

n!

Does this series converge for any x besides x = x0? If so, is its sum equal to f(x)? Ingeneral, the answer to both questions is “No”. A necessary and sufficient condition foranswering both question in the affirmative can be given by using Taylor’s formula (f(x) = f(x0)+ Σk=1

n-1 (x-x0)k + (x-x0)n), but first we introduce further terminology and notation.f(k)(x0)k!

f(n)(x1)n!

Definition 13-29. Let f be a real-valued function defined on an interval I in E1. If f hasderivatives of every order at each point of I, we write f ∈ C∞ on I. If f ∈ C∞ on someneighbourhood of a point x0, the power series Σn=0

∞(f(n)(x0)/n!)(x-x0)n is called the Taylor’sseries 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 thesymbol =? Taylor’s formula states that if f ∈ C∞ on the closed interval [a, b] and if x0 ∈ [a, b],then, for every x in [a, b], and for every n, we have f(x) = Σk=0

n-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, andon n. Hence a necessary and sufficient condition for the Taylor’s series to converge to f(x) isthat lim

n→∞(f(n)(x1)/n!)(x-x0)n = 0.

In practise it may be quite difficult to deal with this limit because of the unknownposition of x1. In some cases, however, a suitable upper bound can be obtained for f(n)(x1) andthe limit can be shown to be zero. Since (x-x0)n/n! → 0 as n → ∞, lim

n→∞(f(n)(x1)/n!)(x-x0)n = 0will certainly hold if the sequence f(n) is uniformly bounded on [a, b]. Thus we can state thefollowing sufficient condition for representing a function by a Taylor’s series.

Theorem 13-30. Assume that f ∈ C∞ on [a, b] and let x0 ∈ [a, b]. Assume that there is aneighbourhood N(x0) and a constant M (which might depend on x0) such that |f(n)(x)| ≤ M forevery x in N(x0)∩[a, b] and every n = 1, 2, .... Then, for each x in N(x0)∩[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 theform (a-δ, b), where δ > 0, and suppose that f and all its derivatives are non-negative in thehalf-open interval [a, b). Then, for every x0 in [a, b), we have f(x) = Σk=0

∞ (f(k)(x0)/k!)(x-x0)k, ifx0 ≤ x < b.

Note that Bernstein’s Theorem gives us a power series expansion about x0 valid in ahalf-open interval of the form [x0, b). By Theorem 13-21, this series also converges in aneighbourhood of radius b-x0 about x0, 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 ofthe series f(x) = Σk=0

∞ (f(k)(x0)/k!)(x-x0)k, if x0 ≤ x < b, so that it extends to the left of x0.

Theorem 13-32. Assume that f satisfies the hypotheses of the previous theorem andsuppose that x0 is an interior point of [a, b). Let R be such that N(x0;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(x0;R).

The binomial series. As an example illustrating the use of Bernstein’s Theorem, we willobtain the following expansion, known as the binomial series: (1+x)a = Σn=0

∞ xn, if -1 < x <( an )

1, where a is an arbitrary real number, and . Bernstein’s theorem is not( an ) = a(a−1)...(a−n+1)

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, providedthat 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∞ (-1)kxk, if -1 < x < 1. Replacing c by -a and x by -x in the above, we find that the−c

k( )

formula (1+x)a = Σn=0∞ xn, if -1 < x < 1, is valid for each a < 0. But now the validity of this( a

n )formula can be extended to all real a by successive integration. Of course, if a is a positiveinteger, say a = m, then = 0 for n > m, and the formula (1+x)a = Σn=0

∞ xn, if -1 < x < 1,( mn ) ( a

n )reduces to a finite sum (the Binomial Theorem).

Abel’s limit theorem. If -1 < x < 1, integration of the geometric series 1/1-x = Σn=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 thisquestion in the affirmative.

Theorem 13-33 (Abel’s limit theorem). Assume that we have f(x) = Σn=0∞ anxn, if -r < x

< r. If the series also converges at x = r, then the limit limx→r-f(x) exists and we have limx→r-f(x)= Σn=0

∞ anrn. Example: we may put x = -1 in log(1-x) = -Σn=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 onmultiplication of series:

Theorem 13-34. Let Σn=0∞an and Σn=0

∞bn be two convergent series and let Σn=0∞cn denote

their Cauchy product. If Σn=0∞cn converges, we have Σn=0

∞cn = (Σn=0∞an)(Σn=0

∞bn).

Tauber’s theorem. The converse of Abel’s limit theorem is false in general. That is, if fis given by f(x) = Σn=0

∞ anxn, and if -r < x < r, then the limit f(r-) may exist but yet the seriesΣanrn 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 furtherrestrictions on the coefficients an, one can obtain a converse to Abel’s theorem. A largenumber of such results are now known and they are referred to as Tauberian theorems. Thesimplest 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→∞ nan

= 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 fn → f uniformly on S and that each fn is bounded on S. Show that fnis uniformly bounded on S.

Answer: A sequence fn is said to be uniformly bounded on S if there exists a constantM > 0 such that |fn(x)| ≤ M for all x in S and all n = 1, 2, ... The number M is called a uniformbound for fn.

If each fn is bounded on S, then there exists a constant Mn such that |fn(x)| ≤ Mn for all xin S.

If fn → f uniformly on S, then for every ε > 0, there exists an N (depending only on ε)such that n > N implies |fn(x)-f(x)| < ε for every x in S.

If fn → f uniformly on S, then given an ε > 0 and an associated (finite) N, we know thatfor all values of n > N, all function values fn(x) are within ε of f(x). In other words, every valueof fn(x) for n > N is bounded above by f(x)+ε. But f(x) itself is bounded above as it is within εof every function value fn(x) (using |a-b| = |b-a|). Knowing that fn(x) is bounded above by thevalue Mn, it follows that f(x) is bounded above by Mn+ε, and so every value of fn(x) for n > Nis bounded above by Mn+2ε. Choosing n = N+1 for simplicity, we can say that fn(x) is boundedabove for n > N by the value MN+1+2ε.

We now know that the sequence fn is uniformly bounded on S if we have n > N, whereN is associated with a particular ε > 0. But what about the function values f1, f2, ..., fN? Well,each of these is bounded above by a particular M, so if we define O = Mi, then O is anmax

i=1,2,...,N

upper bound for each the fi from i = 1 to i = N.

Summary: we now have a bound (O) for f1, f2, ..., fN; and another bound (MN+1+2ε) forfN+1, fN+2, .... Defining P = max(O, MN+1+2ε), it follows that P is an upper bound such that |fn(x)|≤ P for all x in S and all n = 1, 2, ... In other words, we have shown that the sequence fn isuniformly bounded, and that P is an uniform bound for fn. QED.

13-2. Define two sequences fn and gn as follows:fn(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) = fn(x)gn(x).

(a) Show that both fn and gn converge uniformly on every finite interval.(b) 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 fn(x),we want to show that fn(x) → f(x) on [c, d] for some function f(x). Claim: f(x) = x. To provethis claim, we need to show that for every ε > 0, there exists an N (depending only on ε) suchthat n > N implies |fn(x)-f(x)| < ε for all x ∈ S.

Consider an arbitrary ε > 0. We need to find an N such that n > N implies|fn(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 anarbitrary ε > 0, we have found a suitable N (N = ceiling(max|c|, |d|/ε) such that n > N implies|fn(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 toshow that gn(x) → g(x) on [c, d] for some function g(x). Claim: g(x) = 0 if x = 0 or if x isirrational; and g(x) = b if x is rational, say x = a/b, with b > 0. To prove this claim, we need toshow 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 |gn(x)-g(x)| <ε. In all cases (with our definition of g(x)), |gn(x)-g(x)| = |1/n| = 1/n. So we need to find an N suchthat 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 fn → f uniformly on S and suppose there is a constant M > 0 such that|fn(x)| ≤ M for all x in S and all n. Let g be continuous on the closed disk (0; M) and defineNhn(x) = g[fn(x)], h(x) = g[f(x)], if x ∈ S. Show that hn → h uniformly on S.

Answer: Because g is continuous on (0; M), and because |fn(x)| ≤ M for all x in S andNall n, then g is uniformly continuous on (0; M). It follows that given an ε1 > 0, there exists aNδ1 > 0 (dependent only on ε1) such that if |x-y| < δ1, then |g(x)-g(y)| < ε1. Because we know thatfn → f uniformly on S, then given a δ1 > 0 (the same one as above), then there exists an N suchthat for all n > N and all x ∈ S, we have |fn(x)-f(x)| < δ1.

But if |fn(x)-f(x)| < δ1, then we have |g(fn(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 turnwas 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(fn(x))-g(f(x))| < ε1, i.e. hn → h uniformly on S. QED.

13-6. Let fn be a sequence of continuous functions defined on a compact set S andassume that fn converges pointwise on S to a limit function f. Show that fn → f uniformly onS if, and only if, the following two conditions hold:

(i) The limit function f is continuous on S.(ii) For every ε > 0, there exists an m > 0 and a δ > 0 such that n > m and |fk(x)-f(x)| <

δ implies |fk+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 aneighbourhood N(x0) and an integer k (depending on x0) such that |fk(x)-f(x)| < δ if x ∈ N(x0).By compactness, a finite set of integers, say A = k1, ..., kr, has the property that, for each x inS, some k in A satisfies |fk(x)-f(x)| < δ. Uniform convergence is an easy consequence of thisfact.

Answer: If. Assume that fn → f uniformly on S. We know that every function fn iscontinuous at every point x0 in S. It therefore follows by Theorem 13-3 that the limit function fis continuous on S. This proves part (i).

If fn → f uniformly on S, then given an ε > 0, there always exists an m > 0 (dependentonly on ε) such that if n > m, then |fn(x)-f(x)| < ε. If we now choose any arbitrary δ > 0, then ifwe have |fk(x)-f(x)| < δ, then we automatically have |fk+n(x)-f(x)| < ε precisely because|fn(x)-f(x)| < ε. Conclusion: For every ε > 0, there exists an m > 0 and a δ > 0 such that n > mand |fk(x)-f(x)| < δ implies |fk+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 fn → f uniformlyon S.

Now because we know that fn converges pointwise on the set S, then for every x0 ∈ S,we know that given an ε > 0, there exists an N > 0 (depending only on x0 and on ε) such that n> N implies |fn(x0)-f(x0)| < ε/3. Our task is to find an N (depending only on a given ε > 0) suchthat n > N implies |fn(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 isuniformly continuous on S. Therefore, given an ε > 0, there is a δ1 > 0 (dependent only on ε)such that if we have |x-x0| < δ1, then |f(x)-f(x0)| < ε/3 (x, x0 ∈ S). In other words, if x ∈ Nδ1(x0),then |f(x)-f(x0)| < ε/3 (---(1)).

Similarly, the functions fn are all uniformly continuous on the set S. Therefore, for everygiven integer k > 0, and given an ε > 0, there is a δ2 > 0 such that if |x-x0| < δ2, then |fk(x)-fk(x0)|< ε/3 (x, x0 ∈ S). In other words, if x ∈ Nδ2(x0), then |fk(x)-fk(x0)| < ε/3 (---(2)). Now what canwe say about |fk(x)-f(x)|? If we add equations (1) and (2) together, then for a given x ∈ Nδ3(x0),where δ3 = minδ1, δ2, we have (see over)

|f(x)-f(x0)| + |fk(x)-fk(x0)| < ε/3 + ε/3|f(x0)-f(x)| + |fk(x)-fk(x0)| < 2ε/3 (using |A-B| = |B-A|)|f(x0)-f(x)+fk(x)-fk(x0)| < 2ε/3 (using |A+B| ≤ |A| + |B|)|fk(x)-f(x) + (f(x0)-fk(x0))| < 2ε/3.

Now as |fk(x0)-f(x0)| < ε/3 for all k > N, then |f(x0)-fk(x0)| < ε/3 for all k > N, and so|fk(x)-f(x)| < ε/3 + 2ε/3 (for all x ∈ Nδ3(x0)),|fk(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 aninteger k > N (where N corresponds to a particular ε > 0 and is dependent on x0) such that|fk(x)-f(x)| < ε for x ∈ Nδ3(x0).

Let us now define an open covering of S given by the union of all neighbourhoodsNδ3(x), where x ∈ S. Therefore, we have defined an open covering P = Nδ3(x) ⊃ S. Because4

xcS

S is a compact set, this open covering of S reduces to a finite covering of S, so that a finitenumber of such neighbourhoods cover the set S. Let this collection of neighbourhoods bedenoted as Q = Nδ3(x1), Nδ3(x2), ..., Nδ3(xq), where each xi ∈ S.

For each of the above neighbourhoods, we have an integer ki so that for any x in aparticular neighbourhood, we have |fn(x)-f(x)| < ε for n > ki. If we define K = ki, where i =max

i

1, ..., q, then we see that for any x ∈ S, we have |fn(x)-f(x)| < ε for any n > K, where K dependssolely on ε. Therefore, we conclude that fn → f uniformly on S. QED.

13-7. (a) Use Exercise 13-6 to prove the following theorem of Dini: If fn is asequence of real-valued continuous functions converging pointwise to acontinuous limit function f on a compact set S, and if fn(x) > fn+1(x) for eachx in S and every n = 1, 2, ..., then fn → f uniformly on S.

(b) Use the sequence in Exercise 13-5(a) to show that compactness of S isessential in Dini’s theorem.

Answer: (a) Exercise 13-6 says the following:

Let fn be a sequence of continuous functions defined on a compact set S and assumethat fn converges pointwise on S to a limit function f. To show that fn → f uniformly on S, weonly need to show that the following two conditions hold:

(i) The limit function f is continuous on S.(ii) For every ε > 0, there exists an m > 0 and a δ > 0 such that n > m and |fk(x)-f(x)| <

δ implies |fk+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 fn → f uniformly on S, all we need show is that condition (ii) holds.

Because we know that the sequence of functions fn converges pointwise to acontinuous limit function f, then given any δ > 0, we can always find an integer k such that|fk(x)-f(x)| < δ. But if we are then given another number ε > 0, then (a) if ε > δ, then it followsimmediately that |fk+n(x)-f(x)| < ε for any integer n because we know (from the question) thatfn(x) > fn+1(x) for all x ∈ S. (b) If ε < δ, then again because fn(x) > fn+1(x) for all x ∈ S, andbecause fn converges pointwise to f, there will always be an integer n such that |fk+n(x)-f(x)| <ε. Therefore, for every ε > 0, there exists an m > 0 and a δ > 0 such that n > m and |fk(x)-f(x)| <δ implies |fk+n(x)-f(x)| < ε for all x in S and all k = 1, 2, ... This shows that condition (ii) aboveholds, and therefore we conclude that fn → f uniformly on S. QED.

(b) Recall that the sequence in Exercise 13-5(a) was the following sequence: fn(x) =1/(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 fn(x) is continuous because fn(x) is a rational function andthe range of x does not include the single discontinuity of fn(x) which occurs at x = -1/n for eachn = 1, 2, ... Secondly, the functions fn(x) converge pointwise to a function f(x) (Exercise 13-5asks you to show that this is true). The function f(x) to which the functions fn(x) convergepointwise to is f(x) = 0 (as n → ∞, the denominator of fn(x) tends to ∞ and thus fn(x) tends to0). It follows immediately that f(x) is a continuous function because it is a constant function.

Thirdly, we see that we have fn(x) > fn+1(x) for all x ∈ (0, 1): if we fix x, then we see thatthe denominator of fn(x) is smaller than the denominator of fn+1(x), and thus fn(x) is larger thanfn+1(x). Therefore, as fn(x) > fn+1(x), then we definitely have fn(x) > fn+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 ofDini’s theorem are satisfied but for the compactness of the set S on which the functions fn andthe function f are defined upon. But Exercise 13-5(a) shows that fn does not convergeuniformly on (0, 1). Therefore, we conclude that the compactness of S is essential in Dini’stheorem. QED.

13-10. Assume that gn+1(x) ≤ gn(x) for each x in T and each n = 1, 2, ..., and suppose thatgn → 0 uniformly on T. Show that Σ(-1)n+1gn(x) converges uniformly on T.

Answer: Let us define the function fn(x) = (-1)n+1 defined for all x ∈ T. Let Fn(x) denotethe nth partial sum of the series Σfn(x). Now

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 Fn is uniformly bounded above by the number 1. Knowingthis, knowing that gn is a sequence of real-valued functions such that gn+1(x) ≤ gn(x) for eachx in T and for all n = 1, 2, 3, ...; and knowing that gn → 0 uniformly on T, then we can applyDirichlet’s test for uniform convergence (Theorem 13-15) to say that the series Σfn(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-valuedfunctions such that gn+1(x) ≤ gn(x) for each x in T and for every n = 1, 2, ... If gn is uniformlybounded on T and if Σfn(x) converges uniformly on T, then Σfn(x)gn(x) also convergesuniformly on T.

Answer: If gn is uniformly bounded on T, then there exists a positive real number Msuch that |gn(x)| < M for all x ∈ T and for all n = 1, 2, ... To prove that Σfn(x)gn(x) convergesuniformly on T, we need to show that for every ε > 0, there is an integer N such that n > Nimplies that we have fk(x)gk(x)| < ε for each p = 1, 2, ... and for every x in T.| Sk=n+1

n+p

Now because gn is uniformly bounded above by M, we have (for any integers n and p) fk(x)gk(x)| ≤ fk(x)|×|M|.| Sk=n+1

n+p | Sk=n+1n+p

Knowing that Σfn(x) converges uniformly on T, then we know that given an ε2 > 0, thereexists an integer N2 such that if n2 > N2, then we have fk(x)| < ε2 for each p = 1, 2, ...| Sk=n2+1

n2+p

and for every x in T. Therefore, fk(x)gk(x)| ≤ fk(x)|×|M| < ε2×|M|.| Sk=n2+1

n2+p | Sk=n2+1n2+p

So, given an ε > 0, pick ε2 so that ε2 < ε/|M|, or ε2×|M| < ε. By the above discussion, thereexists an integer N2 (associated with ε2 and thus now with ε) such that if n > N2, then we have

fk(x)gk(x)| < ε2×|M| < ε for each p = 1, 2, ... and for every x in T. This proves that| Sk=n+1n+p

Σfn(x)gn(x) converges uniformly on T. QED.

13-14. Let fn be a sequence of real-valued continuous functions defined on [0, 1] andassume that fn → f uniformly on [0, 1]. Prove or disprove

.limnd∞ ¶0

1−1/nfn(x)dx = ¶0

1f(x)dx

Answer: In this question, we shall apply Theorem 13-17:

Theorem (Arzelà): Assume that fn is boundedly convergent on [a, b] and suppose eachfn is Riemann-integrable on [a, b]. Assume also that the limit function f is Riemann-integrableon [a, b]. Then lim

n→∞ ∫ba fn(x)dx = ∫b

a limn→∞ fn(x)dx = ∫b

a f(x)dx.

We know that our sequence fn is a sequence of real-valued continuous functionsdefined on [0, 1]; and we know that fn → f uniformly on [0, 1]. Is fn boundedly convergenton [0, 1]? To be boundedly convergent, fn must be pointwise convergent and uniformlybounded on [0, 1]. Because we know that fn → f uniformly on [0, 1], then we certainly knowthat fn is pointwise convergent. Is fn uniformly bounded on [0, 1]?

To be uniformly bounded, all we must show is that each individual function is boundedbecause we know that fn → f uniformly on [0, 1] (see Exercise 13-1). Because each function fn

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 fn is uniformlycontinuous on [0, 1] (see Theorem 4-24).

Because each function fn is uniformly continuous on the bounded set [0, 1], then f mustbe bounded on the set [0, 1] — see Exercise 4-23. Therefore, knowing that each function fn isbounded on [0, 1], we know that fn is uniformly bounded on [0, 1], and so fn isboundedly convergent on [0, 1].

The next question to ask ascertains as to whether each function fn is Riemann integrableor not on [0, 1]. Looking at Theorem 9-27, and knowing that each function fn is continuous on[0, 1], then we can say with certainty that each fn is Riemann integrable on [0, 1]. Similarly,because f is continuous on [0, 1] (we know that fn is a sequence of continuous functionsdefined on a compact set [0, 1], and that fn → 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 satisfiedwhen applied to this exercise, we can say the following:

limn→∞ ∫1

0 fn(x)dx = ∫10 lim

n→∞ fn(x)dx = ∫10 f(x)dx.

But as limn→∞ ∫1

0 fn(x)dx = limn→∞ fn(x)dx (as lim

n→∞ 1-1/n = 1), then we can say that¶01−1/n

, and so we have proved the hypothesis asked in the question. QED.limnd∞ ¶0

1−1/nfn(x)dx = ¶0

1f(x)dx

13-17. Show that the series Σn=1∞ ((-1)n/√(n))sin(1+(x/n)) converges uniformly on E1.

Answer: Let us define fn(x) =((-1)n/√(n))sin(1+(x/n)). Taking the derivative, we getf’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 ∈ E1 as cos(1+x/n)∈ [-1, 1] for all x ∈ E1. If we consider the point x = 0, then the series Σfn(0) is given by

Σfn(0) = Σn=1∞ ((-1)n/√(n))sin(1+(0/n)) = sin(1)Σn=1

∞ ((-1)n/√(n)).

Let us now define the terms an = 1/√n. It is obvious that an 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 |an-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 an converges to zero has been proven.

Theorem 12-16 says that if an is a decreasing sequence converging to zero, which ourseries an is, then the alternating series Σ(-1)nan converges. Therefore, using this theorem, wecan say that the series Σfn(0) converges (Σn=1

∞ ((-1)n/√(n)) converges so sin(1)Σn=1∞ ((-1)n/√(n))

converges as well), thus proving that the series Σfn(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, weshall use Weierstrass’ M-test (Theorem 13-7). Looking at Σf’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] forany 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, theneach 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) convergesuniformly 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 andThomas & Finney, Chapter 8, Section 8.4, Example 3: The p-series Σn=1

∞ 1/np converges if p >1 and diverges if p ≤ 1). Therefore, we have shown that because ΣMn converges, we have theconclusion that Σf’n(x) converges uniformly on E1.

Therefore, knowing that each fn 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 E1,knowing that the series Σfn(x0) converges for x0 = 0, and knowing that Σf’n(x) convergesuniformly on E1, we can apply Theorem 13-14 to say that the series Σfn(x) converges uniformlyon E1. QED.

13-19. Show that Σn=1∞ ansin(nx) and Σn=1

∞ ancos(nx) are uniformly convergent on E1 ifΣn=1

∞ |an| converges.

Answer: Given any numbers n and x, recall that |sin(nx)| ≤ 1 and |cos(nx)| ≤ 1. Let usdefine fn(x) = ansin(nx) and define gn(x) = ancos(nx). Because of the above, we see that |fn(x)| ≤|an|×1 = |an|, and that |gn(x)| ≤ |an|×1 = |an|. In other words, 0 ≤ |fn(x)| ≤ |an| and 0 ≤ |gn(x)| ≤ |an|.Applying Weierstrass’ M-test, we see that Σn=1

∞ fn(x) = Σn=1∞ ansin(nx) converges uniformly on

E1 if Σn=1∞ |an| converges; and that Σn=1

∞ gn(x) = Σn=1∞ ancos(nx) converges uniformly on E1 if

Σn=1∞ |an| converges. QED.

13-25. Let fn(x) = cosnx if 0 ≤ x ≤ π.

(a) Show that l.i.m.n→∞ fn = 0 on [0, π] but that fn(π) does not converge.(b) 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 ofthe notes: Definition 13-18. Let fn be a sequence of Riemann-integrable functions defined on[a, b]. Assume that f ∈ R on [a, b]. The sequence fn is said to converge in the mean to f on [a,b], and we write l.i.m.n→∞ fn = f on [a, b], if limn→∞ ∫b

a |fn(x)-f(x)|²dx = 0.

(a) Looking at the above definition, we need to show that limn→∞ ∫0

π |(cosnx-0)|²dx = 0. Wecan remove the modulus signs because we are squaring one term, leaving us to prove that

limn→∞ ∫0

π (cosnx)²dx = 0, or that limn→∞ ∫0

π cos2nxdx = 0.

Looking at the back of Thomas & Finney in the Table of Integrals, we find that∫cosnaxdx = .cosn−1ax sin ax

na + n−1n ¶ cosn−2axdx

Therefore, limn→∞ ∫0

π cos2nxdx= lim

n→∞ ([(cos2n-1(x)sin(x))/n]π0 + (2n-1/2n)∫0

π cos2n-2(x)dx).Now as sin(0) = 0 and as sin(π) = 0, the above reduces to

limn→∞ (2n-1/2n)∫0

π cos2n-2xdx.

We can apply the result from Thomas & Finney iteratively, so we obtainlim

n→∞ (2n-1/2n)∫0π cos2nxdx

= limn→∞ (2n-1/2n)(2n-3/2n-2)(2n-5/2n-4)....(7/8)(5/6)(3/4)∫0

π 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]π

0 = [(π/2+sin(2π)/4) - (0/2+sin(0)/4)] = π/2.

It follows that limn→∞ (2n-1/2n)(2n-3/2n-2)(2n-5/2n-4)....(7/8)(5/6)(3/4)∫0

π cos2xdx= lim

n→∞ (2n-1/2n)(2n-3/2n-2)(2n-5/2n-4)....(7/8)(5/6)(3/4)(π/2).= (π/2)lim

n→∞ (2n-1/2n)(2n-3/2n-2)(2n-5/2n-4)....(7/8)(5/6)(3/4).= (π/2)lim

n→∞ Πk=1n (1+2k/2+2k)

= (π/2)Πk=1∞ (1+2k/2+2k)

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 sequencebounded 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 pn converges. But what does itconverge to?

It must converge to zero because (intuitively) we are constantly multiplying togethernumbers in the range (0, 1). I will not prove that it converges to zero and assume that this is thecase, but just note that a proof can be obtained by modifying the proof in Exercise 4-5 or theproof 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 lim

n→∞ ∫0π |(cosnx-0)|²dx = (π/2)Πk=1

∞ (1+2k/2+2k) = 0, we can say that l.i.m.n→∞fn = 0 on [0, π].

Our next task is to show that fn(π) does not converge. We can do this simply byobserving that fn(π) = (-1)n for all n, a series that alternates between -1 and 1 and so does notconverge. QED.

(b) We now want to show that fn converges pointwise on [0, π/2]. There are two casesto deal with here: when x = 0, and when x ∈ (0, π/2]. If x = 0, then because cos(0) = 1, we havecosn(0) = 1n = 1 for all n. It follows that cosn(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 = logε/logy, and so if n > ceiling(logε/logy), then yn < εas required. Note that this will only work if y ∈ [0, 1) (which is all right!) and if ε ∈ (0, 1). Butif ε > 1, then yn < ε holds automatically for y ∈ [0, 1), so we don’t have to worry about thiscase.

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

fn(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 fn does not converge uniformly on [0, π/2]. I will skip this!

13-26. Let fn(x) = 0 if 0 ≤ x ≤ 1/n or if 2/n ≤ x ≤ 1, and let fn(x) = n if 1/n < x < 2/n. Showthat fn converges pointwise to 0 on [0, 1] but that l.i.m.n→∞ fn ≠ 0 on [0, 1].

Answer: To show that fn converges pointwise to 0 on [0, 1], we must show that limn→∞

fn(x) = 0 for all x ∈ [0, 1]. To show this, for every x ∈ [0, 1], and for a given ε > 0, we mustfind an N so that if n > N, we have fn(x) < ε.

Now if x = 0, then fn(x) = 0 (by definition) for all n. Therefore, fn(x) < ε for all ε > 0 andall n if x = 0, i.e. lim

n→∞ fn(x) = 0 if x = 0. If x ∈ (0, 1], then in order to prove that limn→∞ fn(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 fn(x) = 0 < ε for all n > N, and so lim

n→∞ fn(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 inequalitydoesn’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 ifn > N, then we have x ∈ [2/n, 1], and so fn(x) = 0 < ε for all n > N, and so lim

n→∞ fn(x) = 0 for allx ∈ (0, 1].

It remains to show that l.i.m.n→∞ fn ≠ 0 on [0, 1]. To do this, we must show that lim

n→∞∫10 |fn(x)-0|²dx ≠ 0. Manipulating the left hand side of this equation, we get

limn→∞∫1

0 |fn(x)-0|²dx = lim

n→∞∫10 |fn(x)|²dx

= limn→∞[∫0

1/n|0|²dx + ∫1/n2/n |n|²dx + ∫2/n

1 |0|²dx]= lim

n→∞ ∫1/n2/n n²dx (because n = 1, 2, ...)

= limn→∞[n²x]1/n

2/n = lim

n→∞[n²(2/n-1/n)] = lim

n→∞[n(2-1)] = lim

n→∞ n= ∞ ≠ 0. QED.

13-27. If r is the radius of convergence of Σan(z-z0)n, show that .lim inf

nd∞ | anan+1 | [ r [

lim supnd∞ | an

an+1 |

Answer: The important factor about this question is to show that the radius ofconvergence can be defined in another way (under certain conditions), different to thedefinition 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)n. For each n > 1, define rn = .|an+1(z−z0)n+1

an(z−z0)n | = |an+1an ||z − z0 |

Now . If we define r = , then we see that limnd∞ rn = lim

nd∞ |an+1an ||z − z0 | = |z−z0 |

limnd∞ | an

an+1 |

limnd∞ | an

an+1 |

. Claim: r is the radius of convergence of the series Σan(z-z0)n. Proof of Claim. Bylimnd∞ rn = |z−z0|

r

the ratio test (not the one shown in Theorem 12-25 — see the one for example in Thomas &Finney, Section 8.6), the series converges absolutely when < 1, that is, |z-z0| < r; and

|z−z0|r

diverges when > 1, that is, |z-z0| > r. If r = 0, the test shows divergence except for z = z0 (I|z−z0|

r

will not prove this). If r = ∞, so that = 0, the test shows that the series convergeslimnd∞ | an

an+1 |absolutely for all z. Comparing the above to the definition of radius of convergence inTheorem 13-21, it follows that the r defined above as r = is the radius of convergencelim

nd∞ | anan+1 |

of the series Σan(z-z0)n. End of Proof.

Note: Portions of the above proof were taken from “Advanced Calculus withApplications” 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:r = .lim

nd∞ | anan+1 |

If this limit exists, then there is no problem and we have .lim inf

nd∞ | anan+1 | = lim

nd∞ | anan+1 | = lim sup

nd∞ | anan+1 |

However, if the limit does not exist, then we have (using Theorem 12-3),lim inf

nd∞ | anan+1 | [ lim

nd∞ | anan+1 | [

lim supnd∞ | an

an+1 |or as required. QED.lim inf

nd∞ | anan+1 | [ r [

lim supnd∞ | an

an+1 |

13-30. Let f(x) = e-1/x² if x ≠ 0, f(0) = 0.

(a) Show that f(n)(0) exists for all n > 1.(b) 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) = .limxdc

f(x)−f(c)x−c

Case n = 1. f’(0) = .limxd0

f(x)−f(0)x−0 = lim

xd0e−1/x2

−0x−0 = lim

xd01x e−1/x2

Let t = 1/x, then t → ∞ as x → 0+, and t → -∞ as x → 0-. Further,.lim

xd01x e−1/x2 = lim

xd0−1x e−1/x2 = lim

td∞ te−t2 = 0Hence f’(0) = 0. If x ≠ 0, then f’(x) = (2/x³)e-1/x².

Case n = 2. f’’(0) = .limxd0

2x3 e−1/x2

−0

x−0 = limxd0

2x3 e−1/x2

x

If we use t = 1/x again, we have

.limxd0

2x3 e−1/x2

x = limtd∞ 2t4e−t2 = 0

Hence f’’(0) = 0. If x ≠ 0, then f’’(x) = .( 4x6 − 6

x4 )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 havef(n)(0) = = 0, since e-t² approaches 0 more quickly than any polynomial in tlim

td∞ tf(t)e−t2

approaches infinity.

Thus we conclude that f ∈ C∞ everywhere in E1, and, in particular, f(n)(0) exists for everyn — and we have f(n)(0) = 0 for every n. QED.

Note: The above solution was taken from “Advanced Calculus with Applications” byNicholas 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(x) = for all x. But looking at the definition of f, we seeSk=0∞ f(k)(0)

k! (x − 0)k = Sk=0∞ 0

k! xk = 0

that f is zero only at x = 0 (e-1/x² > 0 for all x ≠ 0), so that the Taylor series only represents f atthe origin i.e. at x = 0. QED.

Possible Further Work / Evaluation

Exercise 13-14: Further explanation of the result limn→∞ ∫1

0 fn(x)dx = limn→∞ fn(x)dx.¶0

1−1/n

Possible method: define gn(x) = fn(x) for 0 ≤ x ≤ 1-1/n, and gn(x) = 0 for 1-1/n < x ≤ 1. Then

and we need to show that gn(x) → f(x) uniformly.¶01− 1

n fn(x) = ¶01

gn(x)

Chapter 15

Chapter 15: Fourier Series and Fourier Integrals

15-1. Derive the Minkowski inequality ||f+g|| ≤ ||f|| + ||g|| from the Cauchy-Schwarzinequality.

Answer: The Cauchy-Schwarz inequality assumes the form |(f, g)| ≤ ||f|| ||g||.

Now ||f+g|| = (f+g, f+g)½ (by definition)= [(f, f) + (f, g) + (g, f) + (g, g)]½ (using elementary property 3)= [(f, f) + 2(f, g) + (g, g)]½ (using elementary property 2)≤ [(f, f) + 2(f, f)½(g, g)½ + (g, g)]½ (using Cauchy-Schwarz)= [ (f, f)½ + (g, g)½² ]½

= (f, f)½ + (g, g)½

= ||f|| + ||g|| as required. QED.

15-2. Verify that the trigonometric system in equation (6) is orthonormal on [0, 2π].

Answer: The trigonometric system in question is as follows:

φ0(x) = , φ2n-1(x) = , φ2n(x) = (n = 1, 2, ...).12o

cos nxo

sin nxo

Now in order to prove that this system is orthonormal on [0, 2π], we must show that thefollowing holds: (1) (φn, φm) = 0 whenever m ≠ n, and (2) ||φn|| = 1 for all n. Looking at thesystem in question, we see that in order to prove that it is orthonormal, we must prove thefollowing:

(a) (φ0, φ2n-1) = 0, (b) (φ0, φ2n) = 0, (c) (φ2n-1, φ2n) = 0,(d) (φ0, φ0)½ = 1, (d) (φ2n-1, φ2n-1)½ = 1, (e) (φ2n, φ2n) = 1.

(a) (φ0, φ2n-1) = ¶02o 1

2ocos nx

o dx = 1o 2

¶02o cos(nx)dx = 1

o 2[ sin nx

n ]02o

= 1o 2

[ sin(2on)n − sin(0)

n ] = 1o 2

[0 − 0] = 0.

(b) (φ0, φ2n) = ¶02o 1

2osin nx

o dx = 1o 2

¶02o sin(nx)dx = 1

o 2[ −cos nx

n ]02o

= 1o 2

[ −cos(2on)n + cos(0)

n ] = 1o 2

[−1 + 1] = 0.

(c) (φ2n-1, φ2n) = ¶02o cos nx

osin nx

o dx = 1o ¶0

2o cos(nx) sin(nx)dx = 12o ¶0

2o sin(2nx)dx

= 12o [ −cos(2nx)

2n ]02o = 1

4no [− cos(4on) + cos(0)] = 14no [−1 + 1] = 0.

(d) (φ0, φ0)½ = ¶02o( 1

2o)2dx

1/2

= 12o ¶0

2odx

1/2= [ 1

2o [x]02o ]1/2

= [ 12o [2o − 0]]1/2 = 1 = 1.

(e) (φ2n-1, φ2n-1)½ = ¶02o( cos(nx)

o )2dx1/2

= 1o ¶0

2o cos2(nx)dx1/2

= 1o [ x

2 + sin(2nx)4n ]0

2o1/2

.= 1o [ 2o

2 + sin(4on)4n − 0

2 − sin(0)4n ]

1/2= [ 1

o [o + 0 − 0 − 0]]1/2 = 1 = 1

(f) (φ2n, φ2n)½ = ¶02o( sin(nx)

o )2dx1/2

= 1o ¶0

2o sin2(nx)dx1/2

= 1o [ x

2 − sin(2nx)4n ]0

2o1/2

.= 1o [ 2o

2 − sin(4on)4n − 0

2 + sin(0)4n ]

1/2= [ 1

o [o − 0 − 0 + 0]]1/2 = 1 = 1

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. Show that the expansions below are valid in the ranges indicated.

(a) x = , if -π < x < π.2 Sn=1∞ (−1)n cos nx

n

(b) x² = , if -π ≤ x ≤ π.o2

3 + 4 Sn=1∞ (−1)n cos nx

n2

Answer: (a) The function f(x) = x is an odd function, so that we can apply the resultfrom Exercise 15-6 which said that if f ∈ R on [-π, π], if f has period 2π, 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 ~ Σn=1∞bnsin(nx), where bn = 2/π∫0

π tsin(nt)dt. Let usnow calculate bn for every n > 1.

bn = 2/π∫0π xsin(nx)dx

= 2/π[1/n²sin(nx) - x/ncos(nx)]π0 (Result taken from Thomas & Finney)

= 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 = Σn=1∞bnsin(nx) = Σn=1

∞ 2/n(-1)n-1sin(nx)= as required. QED.2 Sn=1

∞ (−1)n cos nxn

(b) The function f(x) = x² is an even function, so that we can apply the result fromExercise 15-6 which said that if f ∈ R on [-π, π], if f has period 2π, and if f(-x) = f(x) when 0 <x < π, then f(x) ~ +Σn=1

∞ ancos(nx), where an = 2/π∫0π f(t)cos(nt)dt. a0

2

For f(x) = x² on -π < x < π, which can be extended to -π ≤ x ≤ π because f(-π) = f(π) foran even function, we have x² ~ +Σn=1

∞ancos(nx), where an = 2/π∫0π t²cos(nt)dt. a0

2

Now a0 = 2/π∫0π x²cos(0)dt = 2/π[x³/3]π

0 = 2/π[π³/3-0³/3] = 2π³/3π = 2π²/3.

Let us now calculate an for every n > 1. We want to calculate an = 2/πfi02 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

2/π∫0π x²cos(nx)dx

= 2/π[x²/nsin(nx)]π0 - ∫0

π 2x/n sin(nx)dx = 2/π[π²/nsin(πn)-0] - ∫π

0 2x/nsin(nx)dx = -2/π∫π

0 2x/nsin(nx)dx= -4/nπ∫0

π xsin(nx)dx=-4/nπ[1/n²sin(nx) - x/ncos(nx)]π

0 (Result taken from Thomas & Finney)=-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.

Using x² ~ +Σn=1∞ ancos(nx), where a0 = 2π²/3 and an = 4/n²(-1)n, we find thata0

2

x² = (2π²/3)½ + Σn=1∞ 4/n²(-1)ncos(nx)

= as required. QED.o2

3 + 4 Sn=1∞ (−1)n cos nx

n2

15-10. Show that the expansion x² = is valid in the range 0 <43 o2 + 4 Sn=1

∞ ( cos nxn2 − o sin nx

n )x < 2π.

Answer: Recall that the Fourier Series generated by f is given by

f(x) ~ ,a0

2 + Sn=1∞ (an cos nx + bn sin nx)

where an = 1/π∫02π f(t)cos(nt)dt, and bn = 1/π∫0

2π f(t)sin(nt)dt.

For f(x) = x², we have a0 = 1/π∫02π x²cos(0)dt = 1/π[x³/3]2π

0 = 1/π[(2π)³/3-(0)³/3] = 8π³/3π = 8π²/3.

Further, an = 1/π∫02π 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

1/π∫02π x²cos(nx)dx

= 1/π[x²/nsin(nx)]2π0 - ∫0

2π 2x/n sin(nx)dx = 1/π[4π²/nsin(2πn)-0] - ∫2π

0 2x/nsin(nx)dx = -1/π∫2π

0 2x/nsin(nx)dx= -2/nπ∫0

2π xsin(nx)dx=-2/nπ[1/n²sin(nx) - x/ncos(nx)]2π

0 (Result taken from Thomas & Finney)=-2/nπ[1/n²sin(2πn) - 2π/ncos(2nπ) - 1/n²sin(0) + 0]=-2/nπ[0 - 2π/n - 0 + 0]= 4π/n²π = 4/n².

Similarly, bn = 1/π∫02π x²sin(nx)dx. Integrating by parts again, we let u = x² so that du/dx =

2x, and let dv/dx = sin(nx) so that v = -1/ncos(nx). It follows that

1/π∫02π x²sin(nx)dx

= 1/π[-x²/ncos(nx)]2π0 + ∫0

2π 2x/n cos(nx)dx = 1/π[-4π²/ncos(2πn)-0] - ∫2π

0 2x/ncos(nx)dx = 1/π(-4π²/n) - 1/π∫2π

0 2x/ncos(nx)dx= -4π/n - 2/nπ∫2π

0 xcos(nx)dx= -4π/n - 2/nπ[1/n²cos(nx)+x/nsin(nx)]2π

0 (Result taken from Thomas & Finney)= -4π/n - 2/nπ[1/n²cos(2πn) + 2π/nsin(2πn) - 1/n²cos(0) - 0/nsin(0)]= -4π/n - 2/nπ[1/n² + 0 - 1/n² - 0]= -4π/n.

Now that we know that a0 = 8π²/3, that an = 4/n² for n > 1, and that bn = -4π/n for n > 1, wecan apply the formula f(x) ~ to say that for 0 < x < 2π, we havea0

2 + Sn=1∞ (an cos nx + bn sin nx)

x² = (8π²/3)(½) + Σn=1∞ (4/n²cos(nx) - 4π/nsin(nx))

= . QED.43 o2 + 4 Sn=1

∞ ( cos nxn2 − o sin nx

n )

15-22: If f satisfies the hypothesis of the Fourier integral theorem, show that

(a) If f is even, that is, if f(-t) = f(t) for every t, then.

f(x+)+f(x−)2 = 2

o ¶0∞ cos(vx)[¶0

∞f(u) cos(vu)du]dv

(b) If f is odd, that is, f(-t) = -f(t) for every t, then.

f(x+)+f(x−)2 = 2

o ¶0∞ sin(vx)[¶0

∞f(u) sin(vu)du]dv

Answer: Let us first recall the Fourier Integral Theorem:

Assume that f ∈ R*(-∞, +∞). Suppose there is a point x in E1 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 and are absolutely convergent.¶0+

d f(x+t)−f(x+)t dt ¶0+

d f(x−t)−f(x−)t dt

Then we have the formula.

f(x+)+f(x−)2 = 1

o ¶0∞[¶−∞

∞f(u) cos(v(u − x))du]dv

If we manipulate the formula on the previous page, we see that

f(x+)+f(x−)2 = 1

o ¶0∞[¶−∞

∞f(u) cos(v(u − x))du]dv

= 1o ¶0

∞[¶−∞∞

f(u) cos(vu − vx))du]dv= 1

o ¶0∞[¶−∞

∞f(u)(cos(vu) cos(vx) + sin(vu) sin(vx))du]dv

= 1o ¶0

∞[¶−∞∞

f(u) cos(vu) cos(vx)du + ¶−∞∞

f(u) sin(vu) sin(vx)du]dv= (---(1)).1

o ¶0∞[cos(vx) ¶−∞

∞f(u) cos(vu)du + sin(vx) ¶−∞

∞f(u) sin(vu)du]dv

Now if f(x) is an odd function, then ;¶−∞∞

f(x)dx = 0and if f(x) is an even function, then .¶−∞

∞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 oddor where f is even.

(a) If f is even, then f(u)cos(vu) will be even (as cos(vu) is even), and f(u)sin(vu) will beodd (as sin(vu) is odd). Therefore,

1o ¶0

∞[cos(vx) ¶−∞∞

f(u) cos(vu)du + sin(vx) ¶−∞∞

f(u) sin(vu)du]dv= 1

o ¶0∞[cos(vx) % 2 ¶0

∞f(u) cos(vu)du + sin(vx)(0)]dv

= . QED.2o ¶0

∞[cos(vx) ¶0∞

f(u) cos(vu)du]dv

(b) If f is odd, then f(u)cos(vu) will be odd (as cos(vu) is even), and f(u)sin(vu) will beeven (as sin(vu) is odd). Therefore,

1o ¶0

∞[cos(vx) ¶−∞∞

f(u) cos(vu)du + sin(vx) ¶−∞∞

f(u) sin(vu)du]dv= 1

o ¶0∞[cos(vx)(0) + sin(vx) % 2 ¶0

∞f(u) sin(vu)du]dv

= . QED.2o ¶0

∞[sin(vx) ¶0∞

f(u) sin(vu)du]dv

15-28: Verify the entries in the following table of Laplace transforms:

(x > α, p > 0)Γ(p+1)/(z-α)p+1tpeαt(d)(x > 0)α/(z²+α²)sin(αt)(c)(x > 0)z/(z²+α²)cos(αt)(b)(x > α)(z-α)-1eαt(a)

F(z) = e-ztf(t) dt, z = x + iy¶0∞f(t)

Answer: (a) F(z) = ∫0∞ e-zteαtdt (z = x + iy)

= ∫0∞ e(α-z)tdt

= limN→∞[∫0

N e(α-z)tdt]= lim

N→∞[ 1a−z e(a−z)t]0

N

= limN→∞[ ]1

a−z e(a−z)N − 1a−z e0

= 1/z-α + limN→∞[1/α-ze(α-z)N]

= 1/(z-α) + 0 (for x > α)= (z-α)-1 for x > α. QED.

(b) F(z) = ∫0∞ e-ztcos(αt)dt

= limN→∞[∫0

N e-ztcos(αt)]= lim

N→∞[ ]N0 (Result taken from Thomas & Finney)e−zt

z2+a2 (−z cos at + a sin at)= (lim

N→∞ ) - (e0/z²+α²)(-zcos(0)+αsin(0))e−zN

z2+a2 (−z cos aN + a sin aN= 0 - (1/z²+α²)(-z+0) (for x > 0)= z/z²+α² for x > 0. QED.

(c) F(z) = ∫0∞ e-ztsin(αt)dt

= limN→∞[∫0

N e-ztsin(αt)]= lim

N→∞[ ]N0 (Result taken from Thomas & Finney)e−zt

z2+a2 (−z sin at − a cos at)= (lim

N→∞ ) - (e0/z²+α²)(-zsin(0)-αcos(0))e−zN

z2+a2 (−z sin aN − a cos aN= 0 - (1/z²+α²)(-0-α) (for x > 0)= α/z²+α² for x > 0. QED.

(d) F(z) = ∫0∞ e-zttpeαtdt

= limN→∞[∫0

N e(α-z)ttpdt]

Now let us integrate the above by parts, setting u = tp so that du/dx = ptp-1, and setting dv/dx

= e(α-z)t so that v = 1/(α-z)e(α-z)t. It follows that limN→∞[∫0

N e(α-z)ttpdt]

= limN→∞[(tp/(α-z))e(α-z)t]N

0 - ∫0N (ptp-1/(α-z))e(α-z)tdt

= limN→∞[0-0] - ∫0

N (ptp-1/(α-z))e(α-z)tdt= lim

N→∞[p/z-α∫0N e(α-z)ttp-1dt]

We can now integrate by parts repeatedly, and assuming that p > 0, we arrive at thefollowing:

limN→∞[p/z-α∫0

N e(α-z)ttp-1dt]= lim

N→∞[p(p-1)/(z-α)²∫0N e(α-z)ttp-2dt]

= ... = limN→∞[(p!/(z-α)p)∫0

N e(α-z)tdt] (if p > 0).

We can now use the information gained by doing part (a) to say that

limN→∞[(p!/(z-α)p)∫0

N e(α-z)tdt]= (if p > 0 and x > α)

p!(z−a)p ( 1

z−a )= p!/(z-α)p+1.

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 Theorem and the Residue Calculus

Exercise 16-1: Assume that f is analytic on a neighbourhood N(z0; R). If 0 < r < R, letC(r) denote the circle of radius r with z0 as centre and let M(r) denote the maximum of |f| onC(r). Deduce Cauchy’s inequalities:

(n = 0, 1, 2, ...).|f(n)(z0)| [M(r)n!

rn

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 E2.The function f is said to be analytic on S if the derivative f’ exists and is continuous at everypoint of S.

Theorem (Cauchy’s integral formula). Assume that f is analytic on an open region S inE2. 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(z0) = 12oi ¶G[z]

f(z)z−z0 dz,

provided that the path Γ[z] is positively orientated. Further, for every integer n > 1, thederivative f(n)(z0) exists and is given by the integral

f(n)(z0) = n!2oi ¶G[z]

f(z)(z−z0)n+1 dz,

again provided that the path Γ[z] is positively oriented.

Now knowing that , if we define the path Γ[z] to be the circlef(n)(z0) = n!2oi ¶G[z]

f(z)(z−z0)n+1 dz

C(r) defined above, then we can write

.f(n)(z0) = n!2oi ¶C(r)

f(z)(z−z0)n+1 dz

Because C(r) is a circle with radius r centred at z0, then we can write (z-z0)n+1 = rn+1.Therefore, . n!

2oi ¶C(r)f(z)

(z−z0)n+1 dz = n!2oi ¶C(r)

f(z)rn+1 dz = n!

(2oi)rn+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 continuousfunction z, and if ∫Γ[z] F(z)dz exists, then we have the inequality

,¶G[z] F(z)dz [ ML(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 thedesired form:

.f(n)(z0) = n!(2oi)rn+1 ¶C(r) f(z)dz u |f(n)(z0)| = | n!

(2oi)rn+1 ¶C(r) f(z)dz| = n!(2o)rn+1 | ¶C(r) f(z)dz|

Applying Theorem 9-60 to (knowing that |f(z)| ≤ M(r) on C(r), where C(r) is| ¶C(r) f(z)dz|a curve of length 2πr), we obtain the following:

.|f(n)(z0)| = n!(2o)rn+1 | ¶C(r) f(z)dz| [ n!

2orn+1 (M(r) % 2or) = n!M(r)rn+1

Thus as required. QED.|f(n)(z0)| [n!M(r)

rn

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 . Putting in n = 1, we get the formula f’(z0) = . Now letf(n)(z0) = n!

2oi ¶C(r)f(z)

(z−z0)n+1 dz 12oi ¶C(r)

f(z)(z−z0)2 dz

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) = .12oi ¶0

2o f[z(h)](z(h)−z0)2 dz(h) = 1

2oi ¶02o f(z0+re ih)

(z(h)−z0)2 z ∏(h)dh

Since z’(θ) = ireiθ = i(z(θ)-z0), we obtain

f’(z0) = 12oi ¶0

2o f(z0+reih)(i(z(h)−z0))(z(h)−z0)2 dh = 1

2o ¶02o f(z0+re ih)

z(h)−z0dh

= .12o ¶0

2o f(z0+re ih)re ih dh = 1

2or ¶02o f(z0 + reih)e−ihdh

But as f = u + iv, the above can be written as

f’(z0) = .12or ¶0

2ou(z0 + reih)e−ihdh + i

2or ¶02o

v(z0 + reih)e−ihdh

Now because we are integrating over a circle, then we have |u| = |v|, and so

. (Note: I am not sure about the last bit,12or ¶0

2ou(z0 + reih)e−ihdh = i

2or ¶02o

v(z0 + reih)e−ihdhbut the conclusion reached is clearly the one we want).

Therefore,

f’(z0) = 12or ¶0

2ou(z0 + reih)e−ihdh + 1

2or ¶02o

u(z0 + re ih)e−ihdh = as required. QED.1

or ¶02o

u(z0 + reih)e−ihdh

Exercise 16-9: Assume that f has the Taylor expansion f(z) = Σn=0∞ an(z-z0)n, valid in

N(z0; R).

(a) If 0 ≤ r < R, deduce Parseval’s identity:

.12o ¶0

2o |f(z0 + re ih)|2dh = Sn=0∞ |an |2r2n

(b) Use (a) to deduce the inequality Σn=0∞ |an|²r2n ≤ 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 inComplex 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

∞ anrneinθ.

Similarly, = Σn=0∞ rne-inθ.f(z0 + reih) an

The product f = |f|² is calculated as follows:ff = (Σn=0

∞ anrneinθ)(Σn=0∞ rne-inθ)f an

= (a0+a1reiθ+a2r²e2iθ+...)( + re-iθ+ r²e-2iθ+...)a0 a1 a2

= (|a0|² + a0 re-iθ + a0 r²e-2ιθ + ...a1 a2

+ a1reiθ + a1reiθ re-iθ + a1reiθ r²e-2iθ + ...a0 a1 a2

+ a2r²e2iθ + a2r²e2iθ re-iθ + a2r²e2iθ r²e-2iθ + ...a0 a1 a2

+ ....)= (|a0|² + (a0 r)e-iθ + (a0 r²)e-2iθ + ...a1 a2

+ |a1|²r² + (a1 r)eiθ + (a1 r³)e-iθ + ...a0 a2

+ |a2|r4 + (a2 r²)e2iθ + (a2 r³)eiθ + ...a0 a1

+ ....)= Σn=0

∞ |an|²r2n + 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) = a1 r + a2 r³ + ... a0 a1

Because we are only dealing with values of r inside the circle of convergence of theTaylor series, then we can integrate the expression for |f|² term-by-term with respect to θ fromθ = 0 to θ = 2π, i.e. we can find

∫02π |f|²dθ = ∫0

2π (Σn=0∞ |an|²r2n + 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|²r2n)dθ = (Σn=0∞ |an|²r2n)∫0

2π dθ = (Σn=0∞ |an|²r2n)2π.

Thus = (Σn=0∞ |an|²r2n)2π = Σn=0

∞ |an|²r2n as required. QED.12o ¶0

2o |f(z0 + re ih)|2dh 12o

(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-z0| = r, then it follows that |f|² ≤ M(r)² on the circle |z-z0| = r.By definition, we see that g(θ) ≤ M(r)² on the circle |z-z0| = r. But knowing this, we can applyTheorem 9-60 (see the solution for Exercise 16-1 to see this) to say that

∫02π g(θ)dθ ≤ M(r)²×2π.

It follows that ≤ = M(r)². But we know from part (a) that12o ¶0

2o |f(z0 + re ih)|2dh 12o M(r)22o

, so it follows that 12o ¶0

2o |f(z0 + re ih)|2dh = Sn=0∞ |an |2r2n

≤ M(r)² as required. QED.Sn=0∞ |an |2r2n = 1

2o ¶02o |f(z0 + reih)|2dh

Exercise 16-27: Evaluate the following integral by means of residuals:, if a² < 1.¶0

2o cos(2t)dt

1−2a cos(t)+a2 = 2oa2

1−a2

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 E2,with the exception of a finite number of isolated singularities (“poles”). Let Γ be a rectifiableJordan curve such that both Γ and its interior lie within S. Assume that Γ contains a certainnumber of singularities of f in its interior, say z1, ..., zn, but there are no singularities on Γitself. Then we have

,¶G[z] f(z)dz = 2oi Sk=1n Res

z=zk f(z)

provided that the path Γ[z] is positively orientated.

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,¶0

2o cos(2t)dt

1−2a cos(t)+a2

and we do this by defining z = eiθ 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

¶h=0h=2o cos(2t)dt

1−2a cos(t)+a2 = ¶z=0z=1

12 (z2+z−2)

1−2a( 12 (z+ 1

z ))+a2

dziz = 1

i ¶01

12 (z2+z−2)

(1−a(z+ 1z )+a2)z

dz = 12i ¶0

1 z2+z−2

z−az2−a+a2z dz

.= 12i ¶0

1 1a (z2+z−2)

za −z2−1+az dz = 1

2i ¶01 1

a (z4+1)z2( z

a −z2−1+az) dz = 12ai ¶0

1 z4+1z2(a−z)(z−a−1) dz = −1

2ai ¶01 z4+1

z2(z−a)(z−a−1) dz

Let us now find the poles of f(z), where f(z) is defined as f(z) = .z4+1z2(z−a)(z−a−1)

The first pole to consider is the pole of order 2 at z = 0. So b1 = Res(f(z), 0) = ]lim

zdz01

(m−1)!dm−1

dzm−1 (z − z0)mf(z) = (m = 2, z0 = 0) = limzd0 [ d

dz (z − 0)2( z4+1z2(z−a)(z−a−1) )

= .limzd0

(z−a)(z−a−1)(4z3)−(z4+1)(2z− 1a −a)

((z−a)(z−a−1))2 = 0−1(− 1a −a)

(1)2 = 1a + a

The second pole to consider is the simple pole at z = a. So b2 = Res(f(z), a)= .lim

zdz0[(z − z0)f(z)] = (z0 = a) = lim

zda (z − a)( z4+1z2(z−a)(z−a−1) ) = lim

zdaz4+1

z2(z−a−1) = a4+1

a2(a− 1a )

= a4+1a(a2−1)

The final pole to consider is the simple pole at z = 1/a. But because 1/a is outside the areaof integration (a² < 1, and 0 < z < 1), we ignore it. Now that we have found all of the poles, wecan apply Cauchy’s Residue Theorem to our problem:

¶02o cos(2t)dt

1−2a cos(t)+a2 = −12ai ¶0

1 z4+1z2(z−a)(z−a−1) dz = −1

2ai ¶01

f(z) = ( −12ai ) % 2oi Sk=1

n Resz=zk f(z)

= . QED.( −12ai )(2oi)( 1

a + a + a4+1a(a2−1) ) = ( −o

a )( a2−1+a2(a2−1)+a4+1

a(a2−1) ) = −o( a2−1+a4−a2+a4+1a2(a2−1) ) = −2oa2

1−a2

Exercise 16-30: Evaluate the following integral by means of residuals:

.¶−∞∞ 1

x2+x+1 dx = 2o 33

Answer: To do this question, we need some further theory:

Theorem: . This is automatically satisfied if f is thelimRd+∞ ¶−R

R f(x)dx = 2oi Sk=1n Res

z=zk f(z)quotient of two polynomials, say f = P/Q, provided that the degree of Q exceeds the degree ofP by at least 2.

For , let f(z) = . Then P(z) = 1 and Q(z) = z²+z+1, and hence the¶−∞∞ 1

x2+x+1 dx 1z2+z+1

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 = . Of these, only the ‘positive’ root lies in the−1! 1−4

2 = −1!i 32

upper half-plane. The residue at this pole (at z = ) is given by−1+i 3

2

Res(f(z), ) = −1+i 3

2lim

zd−1+i 3

2

(z − (−1+i 3

2 ))f(z) = lim

zd−1+i 3

2

z+ 12 −

i 32

z2+z+1

= .lim

zd−1+i 3

2

z+ 12 −

i 32

(z+ 12 −

i 32 )(z+ 1

2 +i 3

2 )= lim

zd−1+i 3

2

1

z+ 12 +

i 32

= 1−1+i 3

2 + 12 +

i 32

= 1i 3

2 +i 3

2

= 1i 3

=3

3i

Therefore, using the above theorem, we can say that

= = as required. QED.¶−∞∞ 1

x2+x+1 dx 2oi Sk=1n Res

z=zk f(z) 2oi3

3i = 2o 33

Exercise 16-38: If f and g are Möbius transformations, show that the compositefunction 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 ≠0, we define whenever cz+d ≠ 0. It is convenient to define f everywhere on thef(z) = az+b

cz+d

extended plane E2* by setting f(-d/c) = ∞ and f(∞) = a/c. (If c = 0, these last two equations are tobe replaced by the single equation f(∞) = ∞.)

Consider that f is defined as above and we also have g defined as wheneverg(z) = pz+qrz+s

rz+s ≠ 0, where p, q, r and s are complex numbers such that ps-rq ≠ 0. We define g everywhereon the extended plane E2* as above. Let us now consider the composite function fg = f(g(z)).Manipulating,

f(g(z)) = .f(pz+qrz+s ) = a(

pz+qrz+s )+b

c(pz+qrz+s )+d

=a(pz+q)+b(rz+s)

rz+sc(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 settingfg( ) = ∞ and fg(∞) = . (If cp+dr = 0, these last two equations are to be replaced by− cq+ds

cp+drap+brcp+dr

the single equation fg(∞) = ∞.)

Exercise 16-39: Describe geometrically what happens to a point z in E2 when it iscarried into f(z) by the following special Möbius transformations:

(a) f(z) = z+b (Translation).(b) f(z) = az, where a > 0 (Stretching).(c) f(z) = eiαz, where α is real (Rotation).(d) f(z) = 1/z (Inversion).

Answer: (a) Assuming that b is a complex number, say b = p+qi, thenif z = x+iy, then the Möbius transformation in this case moves the point zto 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 Oto the new origin b.

(b) Assuming that a is a real number, the Möbius transformation inthis case in effect changes the scale of the axes in an Argand diagram, tox’ = x×a and to y’ = y×a. Geometrically, the transformation moves thepoint along the vector defined in an Argand diagram by z by an amountspecified by the number a.

(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 anangle of α around the origin (in the anticlockwise direction). In otherwords, f(z) = ei(α+θ)r.

(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 inthis case is given by a = 1/p²+q².

x

y

z

b

z+b

(a)

x

y

z

2z

(b)

½z

x

y

z = re

(c)

x

y

(d)

iθ

θα

az = rei(θ+α)

z

1/z

r

Possible Further Work / Evaluation

Exercise 16-3: To get the result , we12or ¶0

2ou(z0 + reih)e−ihdh = i

2or ¶02o

v(z0 + reih)e−ihdhcould try using the analyticity of f (i.e. using the Cauchy Riemann equations in polar form).