Bootcamp

Christoph Thiele

Summer 2012

Contents

1 Analysis on N D, X 11.1 The natural numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Dyadic numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Dedekind cuts, real numbers . . . . . . . . . . . . . . . . . . . . . . . 81.4 Suprema and infima . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.5 Sequences, Bolzano Weierstrass . . . . . . . . . . . . . . . . . . . . . 121.6 Series, binary expansion . . . . . . . . . . . . . . . . . . . . . . . . . 151.7 Monotone functions from D . . . . . . . . . . . . . . . . . . . . . . . 161.8 The Riemann integral of monotone increasing functions . . . . . . . . 191.9 From monotone increasing to monotone increasing convex . . . . . . . 231.10 Continuous monotone functions from D . . . . . . . . . . . . . . . . . 261.11 The square function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.12 Linear functions, the product . . . . . . . . . . . . . . . . . . . . . . 301.13 Functions on product spaces, limsup, liminf . . . . . . . . . . . . . . 331.14 Theorems for series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351.15 Series theorems revisited with functions . . . . . . . . . . . . . . . . . 391.16 The power function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411.17 An integral equation for power function, the logarithm . . . . . . . . 431.18 The exponential series . . . . . . . . . . . . . . . . . . . . . . . . . . 451.19 Weierstrass approximation . . . . . . . . . . . . . . . . . . . . . . . . 471.20 Negative numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 511.21 Sequences of bounded variation . . . . . . . . . . . . . . . . . . . . . 54

1 Analysis on N D, X

1.1 The natural numbers

We discuss the natural numbers, a concept closely linked with the principle of mathe-matical induction. We also discuss the structure of addition and order on the naturalnumbers.

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We begin with the set N of natural numbers, which we define to include thenumber 0. The set of natural numbers comes with a successor functions ν : N → N.Properties of the set and the successor function are described by the Peano Axioms,which in particular manifest the principle of mathematical induction.

1. There is a unique element 0 ∈ N such that there does not exists any n ∈ N withν(n) = 0.

2. The function ν is injective, that is for all n,m ∈ N, ν(n) = ν(m) implies n = m.

3. Principle of mathematical induction: Let f : N→ {T, F} be any function (intothe set consisting of two elements T and F ), assume f(0) = T and for all n ∈ Nwith f(n) = T we als have f(ν(n)) = T . Then f(n) = T for all n ∈ N

Note: This version of the Peano axioms, that assumes we know what a set and afunction is. Alsonote that unqiueness of 0 can be concluded from the other axioms,but it helps citing uniqueness in the property to be able to refer to the element.

Lemma 1 1. We have that 0 6= ν(0)

2. We have that ν(0) 6= ν(ν(0))

Proof: To prove the first statement, not that ν(0) is the successor of 0, while 0 isnot the successor of any number. Hence the two are not equal. To prove the secondstatement, use the second property to conclude ν(0) 6= ν(ν(0)) from 0 6= ν(0). 2

Exercise 1 1. Prove that 0 6= ν(ν(ν(0).

2. Prove that ν(ν(0) 6= ν(ν(ν(ν(0)))).

The decimal scheme is a way to identify N with finite length words in the letters0, 1, 2, 3, 4, 5, 6, 7, 8, 9 that begin with a letter other than 0. We assume familiaritywith how the function ν is defined on these words.

Iterated application of the successor function leads to the concept of addition ofnatural numbers. For every natural number m we define

m+ 0 := m

and whenever m+ n is already defined, we define

m+ ν(n) := ν(m+ n) .

(such a way of defining a function n → m + n on natural number is called recursivedefinition. That it works requires a few observations.

Note that by the first and second properties of natural numbers, for each naturalnumber n, the value m+ n is defined at most once. Thus, if defined at all, m+ n iswell defined.

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This defines m+n for some elements of N, for example n = 0 and n = ν(0). To seethat it defines m+ n for all elements of N, we employ the principle of mathematicalinduction.

Define a function f : N → {T, F} by setting f(n) = T if m + n is defined andf(n) = F if m + n is not defined. Then clearly f(0) = T by the above definitions,and if f(n) = T then also f(ν(n)) = T . By the third property of naturla numbers,f(n) = T for all n ∈ N, and hence m+ n is defined for all n ∈ N.

One can prove that the following properties hold for any natural numbers n andm and k

1. Commutativity: n+m = m+ n

2. Associativity: n+ (m+ k) = (n+m) + k

3. Cancellation: n+m = n+ k implies m = k

Exercise 2 Prove these properties.

It is customary to write x+ y + z for (x+ y) + z and similarly for longer expressions((x+ y) + z) + a.

Exercise 3 Prove that the order in a triple sum of three elements x, y, z is irrelevantin the sense different orders produce the same sum.

Natural numbers have the structure of order, following the natural timeline of thecounting process. Formally, for any natural numbers n and m we define

n ≤ m

if there exists a natural number a such that m = n+a. One can show for any naturalnumbers n,m, k

1. Totality: We have n ≤ m or m ≤ n.

2. Antisymmetry: If n ≤ m and m ≤ n, then n = m.

3. Transitivity: n ≤ m and m ≤ k implies n ≤ k

4. Preservation under addition: If and only if n ≤ m then n+ k ≤ m+ k.

Exercise 4 Prove these properties.

One writes x < y if x ≤ y and x 6= y. One also write y ≥ x if x ≤ y and y > x ifx < y.

Lemma 2 Let N and N′ be two sets and ν : N → N and ν ′ : N′ → N′ two functionssuch that N satisfies the Peano axioms with ν and N′ satisfies the Peano axiomswith ν ′. Then there is a unique bijective map µ N → N′ such that µ(0) = 0′ andµ(ν(n)) = ν ′(µ(n)) for all n 6= n′.

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Proof: We may define µ(0) := 0′ and if µ(n) is already defined we define µ(ν(n)) =ν ′(ν(n)). This defines µ on all of N and µ obviously satisfies the desired properties.To see that it is bijective, we may define a similar map µ′ from N′ to N and proveby induction that µ′(µ(n) = n for all n ∈ N and symmetrically µ(µ(n′) = n for alln′ ∈ N′. To prove uniqueness, assume we are given another map µ̃ from N to N′, thenwe can prove by induction that m = m̃ 2

The following lemma elaborates the intuition that natural numbers have “gaps”.

Exercise 5 There is no natural number between n and ν(n), more precisely, if n ≤ mand m ≤ ν(n then m = n or m = ν(n.

Exercise 6 Assume there is a set X with a map + : X ×X → X satisfying commu-tativity, associativity, and cancellation. Assume further that if we define the relationx ≤ y if there exists a such that x+ a = y, then this relation is a total order. Finallyassume there is an element x which has opnly one element in X strictly less than x.then this set is isomorphic to N.

1.2 Dyadic numbers

Intuitively, imagine a yardstick with equidistant markings indicating natural numbers.When trying to measure the length of some object, i.e. holding the zero marking ofthe yardstick near one end of the object and comparing the other end of the objectwith markings on the yardstick, most likely the other end of the object will not fallexactly on one marking but inbetween two markings. This draw-back is tosome extentrelieved by introducing further markings. Most economically, one uses the concept ofmidpoints to introduce further markings, as the classical inch scale does.

The natural numbers have gaps: in particular two consequtive natural numbershave no midpoint between them. Thus we extend the natural numbers to the set ofdyadic numbers, which is closed under taking midpoints. It however still is lackingsome desireable featurs, for example there is no dyadic number x such that x + x =x = 1.

We begin by defining for each n ∈ N a new point which intuitively we think of asa point ont he yardstick in the middle between n and ν(n). We call this point ν ′(n).We assume ν ′(n) is not a natural number and ν ′(n) 6= ν ′(m) for n 6= m. Thus ν ′ is abijection from N to a set M of midpoints.

Now we like to obtain some structure on the new set of elements N ∪M. At aglance at a nicely drawn yardstick with these new markings will reveal, the new setof markings looks like a copy of the old one, just with shorter gaps. This suggests tobuild up structures on the new set analogously to the old set.

To obtain a successor function on all of N ∪M, we extend the map ν to all ofN ∪M by setting ν ′(ν ′(n) = ν(n) for all natural numbers n. Since everey element ofM is of the form ν ′(n) for some unique n, this properly defines ν ′ on M. (Wer alsouse that ν ′(n) 6∈ N, or esle we would have doubly defined ν ′(ν ′(n)). Thus we havedefined a map ν ′ : N ∪M→ N ∪M.

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Exercise 7 Prove that the function ν ′ on N ∪M satisfies the Peano axioms.

Thus we can define addition +′ and order ≤′ on N ∪M following the previousdiscussion for N.

What we need of this new addition and order is that restricted to the set N theycoincide with the old structure. This is the content of the following two lemmata.

Lemma 3 If n,m ∈ N, then n+′ m is again in N and equal to n+m.

Proof: We prove the lemma by induction onm. Ifm = 0, we have n+′0 = n = n+0by definition of either addition. Suppose we have

n+′ m = n+m

Then

n+′ν(m) = n+′ν ′(ν ′(m) = ν ′(n+′ν ′(m)) = ν ′(ν ′(n+′m) = ν ′(ν ′(n+m) = ν(n+m) = n+ν(m)

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Lemma 4 If n,m ∈ N, then n ≤′ m if and only if n ≤ m.

Proof: If n ≤ m, then m = n+a for some a ∈ N and by the above m = n+′ a andhence n ≤′ m. Now let n 6≤ m. Then n 6= m and m ≤ n. By the previous argument,m ≤′ n. Hence we cannot have n ≤′ m.2

Since + and +′ as well as ≤ and ≤′ coincide on all elements where they are bothdefined, we may drop the prime and just write + and ≤ for both without dangor ofmix-up.

Exercise 8 If x ∈M, y ∈M then x+ y ∈ N. If x ∈M and y ∈ N, then x+ y ∈M.

Now we iterate this process of adding midpoints inductively. To make the recursionplain, we introduce new notation: Let D0 : +N and ν0 := nu and write +0 for + and≤0 for ≤.

Assume Dn and νn and +n and ≤n have been defined. Then introduce the setMn of all formal expressions νν(n)(x) with x ∈ Dn. These expressions are assumed tonot be elements of Dn and different from each other for different x. Define Dν(n) =Dn ∪Mn and define νν(n)(νν(n)(x)) = νn(x) for all x ∈ Dn. By a reprise of the aboveobservations, νν(n) is a successor relation on Dν(n) satisfying the Peano axioms. Thecorresponding addition +ν(n) and order relation ≤ν(n), when restricted to the set Dn,coincide with +n and ≤n. These facts are all a reprise of the observations made abovein the case n = 0.

Thus we have a hierarchy of sets Dn. By induction, we can show the following

Exercise 9 Let n < m and let x, y ∈ Dn. Then x, y ∈ Dm as well and we have x+n yis an element of both Dn and Dm and equals x+m y. Moreover, x ≤n y if and only ifx ≤m y.

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To pass to one global set, we define D to be the union of all sets Dn with n anatural number.

As a set, this set is equal to the disjoint union of D0 and all the sets Mn withn ≥ 1. There is no canonical successor function on D, however, we can define acoherent addition and order thanks to Exercise 9.

For x, y ∈ D define there are n,m such that x ∈ Dn and y ∈ Dm. Let k be thelarger number of m and n, then both x and y are in Dk. We then define x+ y to bex+k y. Moreover, we define x ≤ y if and only if x ≤k y.

Exercise 10 Verify that this addition satisfies commutativity, associativity, and can-cellation and that the order satisfies totality, antisymmetry, transitivity, and preser-vation under addition.

Hint: for any of these statement, since only finitely many elements of D are involved,it suffices to work on some Dk with k large enough determined by the three elements,and use that addition and order in Dk coincide with addition and order on any otherDn used to define sum and order of respective elements.

Since all the pairs Dn, νn satisfy the Peano axioms, we have for n,m a canonicalisomorphism µn,m. We have the following extension proeprty:

Exercise 11 1. If n, n′,m ∈ N with n ≤ n′, then the map µn′,n′+m extends themap µn,n+m.

2. If n, n′,m ∈ N with n ≤ n′, then the map µn′+m,n′ extends the map µn+m,n.

Exercise 12 The map µ0,m has aunique extension to D. The map µm,0 has uniqueextension to D. The maps are inverses of each other.

We set µm := µ0,m and µ−1m := µm,0.The following exercise contrasts with Exercise 5; it says that there are no gaps in

D.

Exercise 13 Let x, y ∈ D with x ≤ y and x 6= y. then there is c ∈ D with x ≤ c ≤ ybut c 6= x and c 6= y.

A moderately stronger statement than the above exercise is the statement that Dis closed under taking midpoints.

Definition 1 A dyadic number m is a midpoint between the dyadic numbers x andy if m+m = x+ y.

Lemma 5 The midpoint of two dyadic numbers, if it exists, is unique.

Proof: Assume that there are two midpoints n and m. Then we have n+n = m+m.Assume without loss of generality that n ≤ m (Totality). Then, by preservationunder addition, m + n ≤ m + m = n + n. By preservation under addition again,m ≤ n. By antisymmetry, m = n. 2

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Lemma 6 The midpoint between x and y is the same as the midpoint between 0 andx+ y.

Proof: x+ y = 0 + (x+ y). 2

Exercise 14 If x ≤ y and x 6= y, and if m is the midpoint of x and y, then x ≤ m ≤ yand m 6= x and m 6= y

Given two natural numbers, there may not be a natural number which is themidpoint of the two natural numbers. However, dyadic numbers always allow formidpoints.

Lemma 7 For any two dyadic numbers x and y there is a unique dyadic numberwhich is their midpoint.

Proof: We only need to prove existence. We may assume that x and y are in Dk forsome k. We may also assume that x ≤ y. Unless x = y or y = νk(x), we may replacex by its successor and y by its predecessor in Dk without changing the sum x + yand without changing the property x ≤ y. This process can be iterated. Howeversince y has only finitely many iterated predecessors, this process needs to terminate.It can only terminate with a pair x′ y′ with x′ ≤ y′ and x′ = y′ or y′ = νk(x

′) andx + y = x′ + y′. In the first case x′ is the midpoint of x and y. In the second case,νν(k)(x

′) is the midpoint of x and y. 2

However, despite this density, one quickly encounters some obvious insufficienciesof the dyadic numbers. For example

Lemma 8 There is no dyadic number x such that x+ x+ x = 1.

Proof: We first prove that there is no natural number x satisfying x + x + x = 1.Clearly 0 + 0 + 0 6= 1, and for x ∈ N, x 6= 0 we have x+ x+ x ≥ 1 + 1 + 1 ≥ 3, whichshows x+ x+ x 6= 1.

It remains to show that for every n there is no x ∈ Mn satisfying x + x + x =1. Assume to get a contradiction that there is an n ≥ 1 and x in Mn such thatx+ x+ x = 1. Then

x+ x+ x = x+ (x+ x)

is the sum of an element x in Mn and an element (x + x) in Dn, and hence is itselfin Mn (See Exercise 8) for these statements.But this contradicts x+ x+ x = 1 since1 6∈Mn. 2

On the other hand, there is certainly a dyadic number x such that x+ x+ x = 3,namely x = 1. In view of the natural scaling of physics (take a yardstick made ofrubber and stretch it), it is unintuitive that some distances should be divisible intothree equal distances and others should not. Hence one is lead to introducing yetfurther numbers beyond dyadic numbers.

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1.3 Dedekind cuts, real numbers

The main purpose of this section is to construct the nonnegative real numbers fromthe dyadic numbers using Dedekind cuts. We also discuss the extended real numbers.

We will extend our set D, i.e., we will have a set X of nonnegative real numberswhich is the union D∪P of dyadic numbers and a certain set of non-dyadic numbersP. The extension X should extend the structures addition and order, in particularwe will insist on totality of the extended order.

Except at some exposed locations, we will for simplicity we will say “real numbers”in the discussion below when we mean “nonnegative real number”, there is no casuefor nix-up until such time when we have started discussing negative real numbers aswell.

Before defining P precisely, note that every element x of P will be comparablewith every dyadic number. Thus D = A ∪ B where A is the set of elements y ∈ Dwith y < x and B is the set of all elements y ∈ D with x ≤ y. Note that we havesplit the symmetry by defining the set A with a strict inequality and the set B witha non-strict inequality. Doing it conversely would lead to a similar theory. Choosingthe same type of inequality for both sets would just push the asymmetry issue furtherdown the road with unpleasant consequences.

By transitivity, we have a < b for all a ∈ A and b ∈ B and in particular A and Bare disjoint.

The idea then is that we can identify a real number by this partition of dyadicsets. In particular we emphasize that there will be at most one real number whichproduces a given partition. In non-standard analysis, one introduces further nonstan-dard real numbers, and there are different nonstandard real numbers producing thesame partition. We will not delve further into nonstandard analysis.

The second important point about real numbers is that we demand that everypartition of the above type comes with a real number, with the sole exception ofthe partition (D, ∅) . This is the assumption that guarantees an abundance of realnumbers.

We formalize the above ideas. The subtlety of introducing an ε in the definitionbelow continues the split of symmetry discussed above and will help in dealing withDedkind cuts produced by dyadic numbers.

Definition 2 A Dedekind cut is a partition D = A ∪ \A into two disjoint sets suchthat for every a ∈ A there is ε > 0 such that for all a′ ∈ D \ A we have a + ε ≤ a′.The Dedekind cut is called finite if A 6= D.

Note that a Dedekind cut is determined by the set A and we will also refer to Ais the Dedekind cuts.

We call a Dedekind cut A dyadic if there is a dyadic number d such that A ={a ∈ D : a < d}. Note that given a dyadic number d, the sets A defined in thisfashion forms indeed a Dedekind cut. Namely, for every a < d there exists ε 6= 0with a + ε = d and we have a + ε ≤ d ≤ a′ for every a′ ∈ D \ A. For different d, d′

these Dedekind cuts are different since the midpoint of d and d′ belongs to only one

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of A and A′. Hence the dyadic Dedekind cuts are in bijective correspondence withthe dyadic numbers. Note that in particular the number 0 corresponds to the dyadicDedekind cut ∅.

Definition 3 1. A finite Dedekind cut is called a (nonnegative) real number andthe set of such Dedekind cuts is denoted by X.

2. The Dedekind cut D is denoted by ∞ and the set X∪{∞} is called the extendedset of real numbers.

We identify the dyadic numbers with the dyadic Dedekind cuts.As an example for a non-dyadic Dedekind cut, we have

Lemma 9 The setA = {x ∈ D : x+ x+ x < 1}

is a finite dyadic Dedekind cut. it is not dyadic.

Proof: Given a ∈ A, find c such that a + a + a + c = 1. Let c′ be the midpointof 0 and c and let c′′ be the midpoint of 0 and c′, hence c = c′′ + c′′ + c′′ + c′′. Thena+ a+ a+ c′′ + c′′ + c′′ < 1. Hence Then

a+ a+ a+ c′′ + c′′ + c′′ < 1 ≤ a′ + a′′ + a′′

for every a′ ∈ D \A. This implies a+ c′′ ≤ a′, since the asusmption a+ c′′ > a′ wouldlead to a contradiction by the order preservation by addition. Hence A is a Dedekindcut.

It is a finite Dedekind cut since 1 + 1 + 1 > 1 and hence 1 ∈ D \ A.It is not a dyadic Dedekind cut, for assume it was the dyadic number d. We

do not have d + d + d < 1, since we would similarly to above produce a d′ > dwith d′ + d′ + d′ < 1, a contradiction to A being the dedekind cut for d. Then1 ≤ d+d+d. Assume to get a contradiction we have the strict inequality 1 < d+d+d,then 1 + c ≤ d + d + d for some c > 0. Arghuing as above we find c′ such that1 + c′ + c′ + c′ < d + d + d. Replacing c′ by a smaller number if necessary we mayassume c′ + c′ < d. Then there exists b with c′ + b = d and we have b < d < b + c′.Then b+ b+ b > 1, a contradiction to b < d. Hence 1 = d+ d+ d, a contradiction toLemma 8. 2

We now proceed to define structure on the real numbers.

Definition 4 If are two extended real numbers with Dedekind cuts A and B, wedefine x+ y to be the extended real number with Dedekind cut the set C of al dyadicnumber c such that there exists ε > 0 such that for all a′ ∈ D \ A and b′ ∈ D \ B wehave c+ ε ≤ a′ + b′.

This definition requires checking that C is indeed a Dedekind cut.Suppose c ∈ C, and let ε > 0 be as in the definition of C. Now let c′ ∈ D \ c′

and assume to get a contradiction that c′ < c + ε. Then there exists ε′ such thatc′ + ε′ = c + ε and we have for for all a′ ∈ D \ A and b′ ∈ D \ B the inequalityc′ + ε′ ≤ a′ + b′. Hence c′ ∈ C, a contradiction. Hence C is a Dedekind cut.

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Lemma 10 If x and y are finite (not ∞), then so is x+ y.

Proof: With the notation as above: if x and y are finite there exist a′ ∈ D \ A andb′ ∈ D \B. Then it is easy to check that a′ + b′ 6∈ C. 2

Exercise 15 Prove that the addition of Dedekind cuts, when restricted to dyadicdeedekind cuts, coincides with the addition of dyadic numbers.

Exercise 16 1. Prove that for every nonnegative extended real number x

x+∞ =∞+ x =∞

2. Prove that x+∞ = y +∞ does not imply x+ y.

Exercise 17 Prove commutative and associative law for addition in the extended realnumbers.

Exercise 18 Prove Cancellation property for addition when restricted to X.

Exercise 19 Consider the real number x with Dedekind cut A = {a ∈ D : a + a =a < 1} Prove that x+ x+ x = 1.

Now we extend the order:

Definition 5 For two nonnegative extended real numbers x, y ∈ X ∪ {∞} we definex ≤ y if there exists a ∈ X ∪ {∞} with x+ a = y.

There is a very nice alternative description of the order:

Exercise 20 If x, y ∈ X ∪ {∞} with Dedekind cuts A, B, then x ≤ y if and only ifA ⊂ B.

Exercise 21 Prove totality, antisymmetry, transitivity for this order. Prove thatrestricted to the set of dyadic numbers, this order coincides with the previously definedorder. Prove that when restricted to X, addition preserves order.

Exercise 22 For every real number x > 0 there is a dyadic number d such that0 < d < x.

The following is a useful observation, reiterating some remark.

Exercise 23 Two real numbers x, y are equal if for all dyadic numbers d ≤ x wehave d ≤ x′ and for all dyadic numbers x ≤ d we have x′ ≤ d

In practice this criterion is often used in the following form:

Exercise 24 Let x, y ∈ X. Then x = y if and only if for every real ε > 0 we havey ≤ x+ ε and x ≤ y + ε.

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1.4 Suprema and infima

We discuss supremum and infimum, whose existence is one of the incarnations of thecompleteness principle of the real numbers.

We turn to least upper bounds and greatest lower bounds. The ease of handlingthese in the world of X is one of the major features of real numbers.

Definition 6 Assume we are given a set X ⊂ X ∪ {∞}.An upper bound for x is a number s ∈ X ∪ {∞} such that x ≤ s for all x ∈ X.

1.2. A least upper bound of X is an upper bound s such s ≤ s′ for every upper bounds′ of X.

Least upper bounds are unique:

Lemma 11 There is at most one least upper bound for any set X ⊂ X ∪ {∞}Proof: If s and s′ are least upper bounds, then by the “least” part of the least upperbound property of s we have s ≤ s′ and by the “least” part of the least upper boundproperty of s′ we have s′ ≤ s. By antisymemtry we have s = s′ 2

We also have existence of least upper bounds.

Lemma 12 Any subset X ⊂ X ∪ {∞} has a least upper bound.

Proof: Consider the set of all Dedekind cuts A in X. Take the union B =⋃A∈X A.

We claim B is a Dedekind cut and a least upper bound for X. Let b ∈ B, then b ∈ Afor some A ∈ X . Find ε > 0 such that b+ ε ≤ a′ for all a′ ∈ D\A. Hence a ∈ A ⊂ Bfor all a < b+ ε and hence b+ ε ≤ b′ for all b ∈ D \B. Hence B is a Dedekind cut.

Clearly A ⊂ B for every A ∈ X, hence B is an upper bound for A.Let B′ be any other upper bound of X. then for every A ∈ X we have A ⊂ B′.

But then we have⋃A∈X A ⊂ B′ and hence B ⊂ B′. Hence B ≤ B′ and B is the least

upper bound. 2

Note that if X is the empty set, then the above proof gives ∅ as the least upperbound of X. This is reasonable in the setting of X∪{∞}, but when we later introducenegative numbers, this will no longer be the least upper bound. Thus one may wantto exercise some care when taking suprema of empty sets.

Exercise 25 Define a greatest lower bound of a set X ⊂ X∪{∞} and prove existenceand uniqueness for the greatest lower bound.

Note: due to the asymmetry in the use of strict and nonstrict inequality in thedefinition of Dedekind cuts, the proof of existence of greatest lower bound is slightlydifferent. Given a set X of Dedekind cuts A, one will have to define B to be the setof all b ∈ D such that there exists ε > 0 with b+ ε ∈

⋂A∈X A.

Definition 7 1. We write sup(X) or supx∈Xx for the least upper bound of X andcall it the supremum of X.

2. We write inf(X) or infx∈X x for the greatest lower bound of X and call it theinfimum of X.

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1.5 Sequences, Bolzano Weierstrass

We turn to functions with domain or range the sets we have constructed.We first consider functions f : N→ X.Any such function with domain N is also called a sequences. One often writes fn

for f(n) to save some space and brackets.It is instructive to consider classes of sequences respecting some of the structures

we have introduced, additon and order. As for addition, we have:

Exercise 26 Suppose we have a sequence f : N → X that respects addition in thesense

f(n+m) = f(n) + f(m)

for all n,m ∈ N. Then f is determined by the value x := f(1). Conversely, for eachx ∈ X there exists such a sequence with x = f(1).

If f is a sequence as in the exercise and f(1) = x we also write nx for f(x). Thisintroduces multiplication by a natural number and one can check the standard mul-tiplication rules for this. In particular, respecting addition becomes the distributivelaw.

This more or less fully describes the sequences which respect addition.We then turn to sequences which only respect order.

Definition 8 A sequence is monotone increasing if for any natural numbers n,mwith n ≤ m we have f(n) ≤ f(m).

Note that multiplication y a natural number is a monotone incresing sequence.One can reduce the condition of monotonicity to a local one:

Lemma 13 Prove: A sequence is montone increasing if and only if for every n

f(n) ≤ f(n+ 1) .

Proof: If f is monotone and n ∈ N, then n ≤ n + 1 and we conclude by mono-tonicity f(n) ≤ f(n+1). Conversely, assume f satisfies f(n) ≤ f(n+1) for all n ∈ N.Fix n, we shall prove that f(n) ≤ f(n + m) for all m,∈ N. For m = 0 this is clear.Assume f(n) ≤ f(n+m) is true for spome m. Then f(n) ≤ f(n+m) ≤ f(n+m+1)by assumption. By induction we have f(n) ≤ f(n + m) for all m. Now if n ≤ k,then there exists ,m such that n + m = k and hence f(n) ≤ f(k). This provesmonotonicity.

Given a monotone increasing sequence, an interesting quantity is supn∈N f(n). Wehave the following observation

Lemma 14 Let f be a monotone increasing sequence. Then for every x < supn∈N f(n)there is an n0 such that for all m ≥ n0 we have

x ≤ f(m) ≤ supn∈N

f(n)

12

Proof: The second inequality is simply by definition of the supremum. For thefirst inequality note that since x < supn∈N f(n), x is not an upper bound for the setof values f(n) and therefore there is an n0 such that x < f(n0). By monotonicityof f we have for all n0 ≤ m the inequality x < f(n0) ≤ f(m). This proves the firstinequality. 2

This lemma is most interesting in the case of finite supn∈N f(n) and then in stan-dard language the conclusion is tantamount to saying that the sequence converges tothe limit supn∈N f(n). In case of infinite supn∈N f(n), some standard language saysf(n) diverges to ∞.

Definition 9 A sequence f : N→ X∪{∞} is called bounded if there exists a x <∞such that f(n) ≤ x for all n ∈ N.

Exercise 27 Prove that supn∈N f(n) <∞ if and only if the sequence f is bounded.

The class of sequences N→ X form a class with addition and order.

Definition 10 The sum f + g of two sequences is defined by

(f + g)(n) = f(n) + g(n)

We define f ≤ g if there exists an h : N → X that f + h = g for all n ∈ N we havef(n) ≤ g(n).

Exercise 28 Prove that addition satisfies associative and commuttive law, and can-cellation.

Note that canclelation establishes that there is a unique sequence f such that g+f = gfor all sequences g.

Exercise 29 Prove that f ≤ g if and only iof f(n) ≤ g(n) for all n. Prove that theorder on sequences is a partial order, i.e., it satisfies relflexivity, antisymmetry andtransitivity. Prove that addition rpeserves the order.

Even when restricted to the class of monotone functions, the order is not total.

Lemma 15 There exist two monotone increasing sequences f and g such that neitherf ≤ g not g ≤ f .

Proof: Define f(n) = 1 for all n and g(0) = 0 and g(n) = 2 for all n ≥ 1. 2

The supremum behaves as follows with respect to these structures:

Exercise 30 Let f, g : N→ X be sequences

1. If (f(n) ≤ g(n)) for all n ∈ N, then then also

supm∈N

f(m) ≤ supm∈N

g(m)

13

2. We havesupn∈N

(f(n) + g(n)) ≤ supn∈N

f(n) + supn∈N

g(n)

3. if in addition gf and g are monotone increasing, then

supn∈N

(f(n) + g(n)) = supn∈N

f(n) + supn∈N

g(n)

Similar statements can be proven by induction for finite sums of sequences.

Lemma 16 Let f, g : N → X be two sequences and x a number such that f(n) +g(n) = x for all n ∈ N. Then

supn∈N

f(n) + infn∈N

g(n) = x

Proof: For every n ∈ N

f(n) + infk∈N

g(k) ≤ f(n) + g(n) = x

Hencesupn∈N

(f(n) + infk∈N

g(k)) ≤ x

and hencesupn∈N

f(n) + infk∈N

g(k) ≤ x

The other inequality is proved similarly. 2

One also has the functions reversing the order structure:

Definition 11 A sequence is called anti-monotone if for any natural numbers n,mwith n ≤ m we have f(n) ≥ f(m).

Exercise 31 Formulate and prove a corresponding statements as above for anti-monotone sequences.

Lemma 17 Every sequence has monotone subsequecne or an antimonotone subse-quence.

Here a subsequence g of a sequence f is one satisfying g(k) = f(n(k)) for somemonotone increasing injective sequence n : N→ N.

Proof: We shall distinguish two cases.Case 1: For every n there exists m ≥ n such that f(m) = infk≥n f(k). In this case

we shall construct a monotone increasing subsequence. Pick n0 such that

f(n0) = infn≥0

f(n)

14

Assume we have already picked fnk, Then pick nk+1 > nk + 1 such that

f(nk+1) = infn≥nk+1

f(n)

This defines a sequence g(k) = f(nk) which is a subsequence of f . It is clear that wehave fnk+1

≥ fnkfor every n and hence g is monotone increasing.

Case 2: There exists n such that there does not exist m ≥ n with

f(m) = infk≥n

f(k)

. In this case we shall construct a monotone decreasing subsequence. Set i =infk≥nf(k) for this value of n. Pick n0 = n. Assume nk is already chosen withf(m) ≥ f(nk) for all m with n ≤ m ≤ nk. Since f(nk) 6= i, there exists nk+1 ≥ nsuch that f(nk+1) < f(nk). By assumpiton on nk, we have nk+1 > nk. We mayassume nk+1 is the minimal value with this property, then f(m) > f(nk+1) for allnk ≤ m < nk+1 and hence for all n ≤ m ≤ nk+1. Then the resulting subsequenceg(k) = f(nk) is monotone decreasing. 2

Theorem 1 (Bolzano Weierstrass on X) Given any sequence f and finite realnumbers a and b with a ≤ f(n) ≤ b for every n, then there exists a subsequence whichhas a limit x with a ≤ x ≤ b

Proof: Chose a monotone subsequence.2One advantage of finding an actual monotone subsequence rather than passing to

a hull sequence is that this can be easily iterated. Assume we have a sequence ofpairs of real numbers, f : N → X × X, f(n) = (x(n), y(n)) Then we can first choosea subsequence nk such that x(nk) is monotone, and then we can choose a furthersubsequence thereof such that y(nkm) is monotone. Clearly x(nkm) is also monotone.Hecne both components x and y of the subsubsequence have a limit. This kind ofiteration is not possible for the hulls.

1.6 Series, binary expansion

Given a monotone increasing sequence f : N→ X, we have f(n) ≤ f(n+ 1) for everyn, hence there exists a(n) ∈ X such that f(n)+a(n) = f(n+1) and we have a derivedsequence a : N→ X.

Exercise 32 The above described derivation of sequecnes provides a bijection betweenthe class of monotone sequecnes and the class of sequences.

Exercise 33 Define the sum symbol on the right hand side of the following displayrecursively and then show the identity by induction. For any m < n

f(n) = f(m) +n−1∑k=m

a(k)

15

This exercise describes some discrete version of the fundamental theorem of cal-culus. This validates the naming of a as the derived sequence of f .

One then writes for a monotone increasing sequecne and its derived sequence

supn∈N

f(n) =:∞∑k=1

a(k) .

The right hand side is called a series. Note that we have not made any assumptionabout boundedness of the sequence f , so series may be ∞.

Exercise 34 Define a series by an(0) = νn+1(0) for all n ∈ N. Prove

∞∑n=0

= 1

Exercise 35 Define a series by an(0) = νn+n+2(0) for all n ∈ N. Prove that for

x =∞∑n=0

= 1

we have x+ x+ x = 1.

Exercise 36 Prove that for every x ∈ X with 0 ≤ x ≤ 1 there exists a subset A of Nsuch that

x =∞∑k=0

a(k)

where a(k) = 1 if k ∈ A and a(k) = 0 if k 6∈ A

Exercise 37 Prove that for x as in the previosu exercise,

1. if x 6∈ D, there eists exactly one such sequence. The sets A and N \ A areinfinite.

2. If x ∈ D there exist two such sequences. For one sequence the set A is finite,for the other sequence the set N \ A is finite.

3. Find properties that relate the two sequences in case x ∈ D to each other.

1.7 Monotone functions from DWe turn to studying functions mapping D to X. As for sequences, much that can besaid extends to functions f : D → X ∪ {∞}, but the added value of considering thismore generalization is so small that we shall simply retrict attention to f : D→ X.

We again restrict attention mostly to functions which preserve (monotone) orreverse (anti monotone) the order. Both types of functions follow a similar theory, sowe will in fact mostly focus on monootne functions.

16

Definition 12 A function f : D→ R+0 is is monotone or monotone increasing if for

all x, y ∈ D with x ≤ y we have f(x) ≤ f(y). It is called anti monotone (monotonedecreasing), if for every x, y ∈ D with x ≤ y we have f(x) ≥ f(y).

For a monotone increasing function and given y we define left and right limits asfollows

limy↗x

f(x) = supy∈D:y<x

f(y)

limy↘x

= infy∈D:x>y

f(y)

Lemma 18 For a monotone increasing functions f : D→ X and x ∈ D we have

limx↗y

f(x) ≤ f(y) ≤ limx↘y

f(x)

Proof: The two inequalities are proved similarly, we focus on the first inequality.Let limx↗y f(x) = h and assume h > f(y). Then there is x < y with f(x) > f(y) bydefinition of the supremum. This however contradicts monotonicity of f .

Exercise 38 Make reasonable respective definitions and statements for monotone de-creasing functions.

A monotone function is called left semicontinuous if for all f we have

limx↗y

f(x) = f(y)

it is called right semicontinuous if

f(y) = limx↘y

f(x)

and it is called continuous if it is both left and right semicontinuous.Given a monotone function, one can define new functions

fl(y) = limx↗y

f(x)

fr(y) = limx↘y

f(x)

These are called the semicontinuous functions associated to f .

Lemma 19 The functions fl and fr are monotone again.

Proof: This follows form the fact that subsets have smaller suprema than thesurrounding set. 2

Define f ≤ g for two functions f, g if f(x) ≤ g(x) in every point of the domain.Note that this order of functions does not satisfy totality.

Since fl and fr are monotone again, we may again take their associated left con-tinuous functions. The following lemma states among other things that for monotoneincreasing functions (fl)l = fl,

17

Lemma 20 For a monotone increasing function f , the function fl = limx↗y f(x) isleft semicontinuous.

Proof: We need to showlimx↗y

f(x) = limx↗y

fl(x)

Since fl ≤ f , one of the inequalities is trivial and it suffices to show

limx↗y

f(x) ≤ limx↗y

fl(x) + ε

for every ε > 0. Let x < y such that f(x) + ε ≥ fl(y). Let x < z < y, thenfl(z) + ε ≥ fl(y). Hence limx↗y fl(x) + ε ≥ fl(y). This is the inequality we needed toprove. 2

Exercise 39 Formulate and prove a corresponding statement for fr.

Note that we have an equivalence class of functions by defining f ∼ g if fl = gl.(Equivalence classes are a notion in set theory. An account on equivalence classesoccurs later in the section on negative numbers.)

The equivalence properties follow from the same properties for =. The aboveLemma states that for each f we have f ∼ fl, and hence each equivalence classcontains a unique special element h wich satisfies h = fl for every element f in theequivalence class. It is also the unique left semicontinuous element in the class, andit is the minimal element in the class with respect to the order of functions.

Similarly we may define equivalence classes using the associated right semicontin-uous function. The following lemma together with its symmetric counterpart statesthat these two possibly different equivalence structures are actually the same. Thuseach equivalence class contains two unique elements h and g with h = fl and g = frfor every f in the equivalence class.

Lemma 21 For a monotone increasing function f we have (fr)l = fl.

Proof: We have fr ≥ f , hence (fr)l ≥ fl. It suffices then to show for every ε > 0 andx

(fr)l(x) ≤ fl(x) + ε .

Find y < x such that fr(y) + ε ≥ (fr)l(x). Find y < z < x, then f(z) ≥ fr(y). Butthen fl(x) ≥ f(z) Putting the last three inequalities together we have fl(x) + ε ≥(fr)l(y), which proves the desired inequality. 2

Exercise 40 Prove the symmetric counterpart with left and right interchanged.

Exercise 41 Prove that fl is the unique left semicontinuous functiuon in the equiv-alence class of f . Prove that fl is the minimal function in the equivalence class of f ,i.e. fl ≤ g for all g equivalent to f .

18

The theory of monotone functions discussed above suggests that one should oftenconsider monotone functions not as such, but as equivalence classs of functions. Thisprecedes lebesgue theory, which also builds on equivalence classes of fucntions. Unlikefor monotone functions however, in lebesgue theory these equivalence classes do notalways have distinguished representatives.

Exercise 42 Let f and g are monotone increasing. If for all x < y we have gr(x) ≤fl(y) and fr(x) ≤ gl(y), then f and g are in the same equivalence class of monotoneincreasing functions.

1.8 The Riemann integral of monotone increasing functions

We continue to work with monotone increasing functions f : D→ X.Let f : D → X and let a < b be two dyadic numbers, and assume a, b ∈ Dm. We

recall the isomorphism µn which is an extension of the unique isomorphism sendingN to Dn Then for every n > m define the upper and lower Riemann sum

L(n, f) =∑

x∈Dn:a≤x<b

µn(fr(x))

L(n, f) =∑

x∈Dn:a≤x<b

µn(fl(νn(x)))

Definition 13 A function is called Riemmann integrable over a < b if

supn≥m

L(n, f) = infn≥m

U(n, f)

This common value is then said to be the Riemann integral∫ b

a

f(t) dt

Note that the lower and upper Riemann sum depend only on the equivalence classof the monotone increasing function f .

Immediately from the definitions we have L(n, f) ≤ U(n, f) for every n ≥ m,because by monotonicity fr(x) ≤ fl(νn(x)) for all x ∈ Dn.

Lemma 22 With notation as above, the sequence n → L(n, f) defined for n ≥ mis monotone increasing. The sequence n → U(n, f) defined for n ≥ m is monotonedecreasing.

Proof: The statements are similar, we only prove the one for the lower Riemann sum.It suffices to prove L(n, f) ≤ L(n+ 1, f) for every n ≥ m. We have

L(n, f) =∑

x∈Dn:a≤x<b

µn(fr(x))

19

=∑

x∈Dn:a≤x<b

µn+1(fr(x) + fr(x))

≤∑

x∈Dn:a≤x<b

µn+1(fr(x) + fr(νm+1(x))

=∑

x∈Dn+1:a≤x<b

µn+1(fr(x))

= L(n+ 1, f)

2

Lemma 23 For any two numbers n, n′ ≥ m we have

L(n, f) ≤ U(n′, f)

Proof: Use monotonicity of L and U to compare at the maximum of n and n′. 2

As a corollary, we have

supn≥m

L(n, f) ≤ infn≥m

U(n, f)

The essence of Riemann integrability of monotone functions lies in the opposite in-equality.

Theorem 2 (Newton’s integrability theorem) For every monotone increasingfunction f : D→ X and any a < b with a, b ∈ Dm we have

supn≥m

L(n, f) = infn≥m

U(n, f)

Proof: We have for n ≥ m

L(n, f) + µn(fl(b)) = µn(fr(a)) +

( ∑x∈Dn:a<x<b

µn(fr(x))

)+ µn(fl(b))

µn(fr(a)) +

( ∑x∈Dn:a<x<b

µn(fl(x))

)+ µn(fl(b)) = µn(fr(a)) + U(n, f) ≥ inf

k≥mU(k, f)

But since µn(fl(b)) can be made smaller than any given ε > 0, we conclude

supk≥m

L(k, f) ≥ infk≥m

U(k, f)

This proves the theorem 2

If we are given a monotone increasing function f : D→ X, we define the primitivefunction F : D→ X of f by

F (x) =

∫ x

0

f(t) dt

for every x ∈ D.

20

Exercise 43 The primitive F of a monotone increasing function f is monotone in-creasing.

Exercise 44 With the definitions as above and x < y: F (x) +∫ yxf(t) dt = F (x+ y)

Exercise 45 If f and g are monotone increasing, then∫ b

a

(f(t) + g(t))dt =

∫ b

a

f(t)dt+

∫ b

a

g(t)dt

Exercise 46 If f and g are monotone increasing and f(x) ≤ g(x) for all a < x < b,then ∫ b

a

f(t)dt ≤∫ b

a

g(t)dt

Lemma 24 (Translation invariance of the Riemann integral) Let x, y, z ∈ Dwith x ≤ y and let f : D→ X be a monotone increasing function. Define g : D→ Xby g(t) = f(t+ z) for all t ∈D, then g is also monotone increasing, and we have∫ y

x

g(t) dt =

∫ y+z

x+z

f(t) dt

Note that we also write ∫ y

x

f(z + t) dt

for the left hand side of the concluded identity of this lemma.Proof: Monotonicity of g is follows since for all s ≤ t we have

g(s) = f(s+ z) ≤ f(t+ z) = g(t)

Not also that fr(t+ z) = gr(t) for all t ∈ D.Pick m such that x, y, z ∈ Dm. Then there is a bijection

{t ∈ Dm+n : x ≤ t < y}

to{t ∈ Dm+n : x+ z ≤ t < y + z}

given by t→ t+ z. hence w emay frite for the lower Riemann sum∑t∈Dm+n,x≤t<y

µm+ngr(t) =∑

t∈Dm+n,x+z≤t<y+z

µm+nfr(t)

Taking supremum in n gives the desired identity. 2

21

Convex functions

Definition 14 Assume we are given a monotone increasing function f : D → XThen for m ∈ N the derived sequence am : Dm → X is defined by

f(x) + am(x) = f(νm(x))

for every x ∈ Dm.

Note that the derived sequence is well defined since for every x ∈ Dm we havef(x) ≤ f(νm(x)).

Definition 15 A monotone increasing function f : D → X is called convex, if forevery m the derived sequecne am is monotone increasing.

Lemma 25 A function f : D→ R is convex if and only if it is monotone increasingand for every x, y in D we have sub-midpoint inequality

f(x+ y) + f(x+ y) ≤ f(x) + f(x+ y + y)

Proof:Now assume the sequence am is monotone increasing for every m. Let x, y ∈ D

and pick m such that x, y ∈ Dm. Then we can write∑z∈Dm:x≤z<x+y

am(z) ≤∑

z∈Dm:x+y≤z<x+y+y

am(z)

Namely, the map z → z + y is a bijection from the set {z ∈ Dm : x ≤ z < x + y} tothe set {z ∈ Dm : x + y ≤ z < x + y + y} and for each corresponding pair of pointsunder this map the difference on the left hand side of the equation is less than thedifference on the right hand side of the equation. Adding f(x) + f(x + y) on bothsides gives

f(x+ y) + f(x+ y) ≤ f(x) + f(x+ y + y)

Now assume that f is monotone increasing and satisfies the sub-midpoint inequal-ity. For every m and x, y ∈ Dm

am(x)+f(x)+f(x+y) = f(x+y)+f(x+y) ≤ f(x)+f(x+y+y) = f(x)+f(x+y)+am(x+y)

Cancelling f(x) + f(x+ y) on both sides give

am(x) ≤ am(x+ y)

Applying this with y = νm(0) proves that am is monotone increasing.2

Exercise 47 Let f and g be convex functions. Then f + g is a convex function.

22

Lemma 26 Let m,n ∈ D. Assume x, y ∈ Dm+n such that x, νm+n(y) ∈ Dm. Then

µmam+n(x) ≤ µm+nam(x) ≤ µmam+n(y)

Proof: We prove the first inequality by induction on n. If n = 0, we have evidently

µmam(x)µmam(x)

Now assume we have proven the desired inequality for some n.. Then

µman+m+1(x) = µm+1(an+m+1(x) + an+m+1(x))

≤ µm+1(an+m+1(x) + an+m+1(x+ νm+n=1(0))

= µm+1an+m(x) ≤ µmn+1an(x)

where in the last inequality we have used the induction hypothesis.The second inequality is proved similarly. 2

Lemma 27 A monotone increasing convex function F : D→ X is continuous.

Proof: To see right continuity, ot suffices to prove that for x ∈ Dm we haveinfn am+n(x) = 0 since then, by adding f(x) on both sides

infnf(x+ νm+n(0)) = f(x)

from which we see right continuity. However, it then suffices to show

infnµman+m(x) = 0

which follows frominfnµm+nam(x) = 0

by the previous lemma.Similarly we show right continuity. 2

1.9 From monotone increasing to monotone increasing con-vex

Lemma 28 The primitive F of a monotone increasing function f : D→ X is convex.

Proof: We have already seen that the primitive is monotone increasing.Let x, y ∈ D and pick m such that x, y ∈ Dm Then, by monotonicoty of the

integral,

F (x+ y) + F (x+ y) = F (x) +

∫ x+y

x

f(x) + F (x+ y)

23

≤ F (x) +

∫ x+y

x

f(x+ y) + F (x+ y)

By translation invariance of the Riemann integral we have for the last display

= F (x) +

∫ x+y+y

x+y

f(x) + F (x+ y)

= F (x) + F (x+ y + y)

This shows convexity of F . 2

The following is an injectivity statement for the primitive.

Lemma 29 (Injectivity of the primitive operation) Assume that f and g aremonotone increasing functions form D to X. Assume that for all x ∈ D we have∫ x

0

f(t) dt =

∫ x

0

g(t) dt

Then f and g are in the sema equivalence class of monotone increasing functions.

Proof It suffices to show for every x, y ∈ D with x < y that fr(x) ≤ gl(y). Namely,then the reverse inequality follows by symmetry and then f and g are in the sameequivalence class by the previously shown Criterion 42.

Let x, y ∈ Dn and pick z ∈ Dn+m such that νn+m(z) = y. Let F denote thecommon primitive of f and g and let an+m denote the derived sequence of F onDn+m. Then, by looking at lower and upper Riemann sums at level n+m we obtain

µn+mfr(x) ≤∫ νn+m(x)

x

f(t) dt = an+m(x)

and

an+m(z) =

∫ νm+n(z)

x

g(t) dt ≤ µn(gr(y)

By monotonicity of the sequecne an+m we have

µn+mfr(x) ≤ µn+m(gr(y)

Since µn preserves inequalities, we conclude the desired inequality. 2

We now work on surjectivity of the primitive onto convex monotone increasingfunctions. We begin with

Lemma 30 Assume the function F : D → X is convex and monotone increasing.Given n ∈ N denote by an the derived sequecne of F on Dn. Then for every x ∈ Dm

we havef(x) = sup

nµ−1m+nan+m(x)

is a finite number, and we have f(x) ≤ f(y) whenever x ≤ y.

24

The function f in this lemma will be called the right derivative of F .Proof: Finiteness of f(x) follows since F is convex and hence the sequence µ−1m+nan+m(x)

is monotone decreasing by a previous observation.If x ≤ y and x, y ∈ Dm, then

µ−1m+nan+m(x) ≤ µ−1m+nan+m(y)

for every m, and hence taking supremum of these monootne sequences proves f(x) ≤f(y). 2

Exercise 48 Assume the function F : D → X is convex and monotone increasing.Then the right derivative f of F decsribed in the previous lemma is right continuous.

Exercise 49 Formulate and prove the corresponding statement for the left derivativeg.

Exercise 50 Assume the function F : D → X is convex. Prove that left and rightderivative are equivalent monotone functions. (Recall the criterion of Exercise 42).

Exercise 51 Assume the function F : D → X is convex. Prove that left and rightderivative are equivalent monotone functions. (Recall the criterion of Exercise 42).

If F is convex monotone increasing, then we may call the common equivalenceclass of f and g the derivative of F .

Theorem 3 (Surjectivity of the primitive operation) Let F : D→ X be a con-vex function and let f denote its right derivative. Then for every x ∈ D we have

F (0) +

∫ x

0

f(t) t = F (x)

Proof: Let an denote the derived sequecne of F on Dn. Let x ∈ Dm. We haveseen before that for every n and every t ∈ Dn+m we have

µn+mfr(t) ≤ an+m(t)

by considering Riemann sums containign a single term. Adding over all t ∈ Dn+m

with 0 ≤ t < x givesF (0) + L(n+m, f) ≤ F (x)

Taking sup in m gives

F (0) +

∫ x

0

f(t) dt ≤ F (x)

To see the reverse inequality, Note that

µn+mfr(νn+m(t)) ≥ an+m(t)

Adding this ienquality similarly to before we obtain

F (0) + L(n+m, f) + µn+mfr(x) ≥ F (x) + µn+mfr(0)

Observing that the two terms involving µn+m tend to zero as n tends to ∞ gives

F (0) +

∫ x

0

f(t) dt ≥ F (x)

This proves the desired inequality. 2

25

1.10 Continuous monotone functions from DIn the past discussion, we only had the need to discuss limy↗x f(x) and limy↘x f(y)of monotone functions. However, these are well defined for any x ∈ X. Define for amonotone function fD→ X and for x ∈ X

limy↗x

f := supy∈D:y<x

f(y)

limy↘x

f := infy∈D:y>x

f(y)

TheIf f is left continous, we define the left continuous extension f̃l as the subclass of

the class of pairs 〈x, f〈 such that

limy↗x

f := supy∈D:y<x

f(y)

If fr is right continous, we define the right continuous extension f̃l as the subclass ofthe class of pairs 〈x, f〈 such that

limy↗x

f := infy∈D:y>x

f(y)

If the two extensions f̃l and f̃r coincide at all points x ∈ X, we call f and f̃ continuous.

Exercise 52 Each of the functions f̃l ,f̃r, and f̃ , whenever defined, are monotoneincreasing, that is for all x, y ∈ R with x ≤ y we have

f̃l(x) ≤ f̃l(y), f̃r(x) ≤ f̃r(y), f̃(x) ≤ f̃(y)

for whichever of these functions is defined.

For convex functions we also prove essentially as before:

Exercise 53 If f : D→ X is monotone increasing and convex, then f has a contin-uous extension to X.

Theorem 4 (Intermediate value theorem for monotone functions) Assume amonotone function is continuous at all points y. Assume f(x1) = a < c < b = f(x2).then there exists x such that f(x) = c.

Proof: Consider the set X of all points x such that f(x) ≤ c. By monotonicity, thisset contains no points larger than x2 and thus is bounded above. it is also nonemptysince x1 ∈ X. Henxe it has a finite supremum s with x1 < s < x2. Right limit ats has to be at least c, while the left limit has to be at most c. By continuity, bothlimits are c and we have f(s) = c. 2

We describe the bisection algorithm for finding the point that is guaranteed toexist by the above theorem. Assume for simplicity that c = 0, this situation can be

26

achgieved by subtacting c from the function. Thus our problem is to find a zeo of thefunction f . Note that f(x1) < 0 and f(x2) > 0.

We will define a monotone increasing sequence sn and a monotone decreasingsequence tn with sn ≤ tn for all n, f(sn) ≤ 0 for all n and f(tn) ≥ 0 for all n.Define s0 = x1 and t0 = x2. Assume we have already defined sn and tn. Considerthe midpoint m of sn and tn. If f(m) = 0 we have found a zero and may stop. Iff(m) < 0 we define sn+1 = m and tn+1 = tn. If f(m) > 0 we define sn+1 = sn andtn+1 = m. Then clearly all properties claimed for tn+1 and sn+1 are satisfied. Weclaim sup(sn) = inf(tn). Call this point c. Then left limit of f at c is ≤ 0, for assumenot then there exists x < c with f(c) > 0, but there also exists n with x < sn ≤ cwith f(sn) < 0, a contradiction to monotonicity. Similarly the right limit of f at c is≥ 0. By continuity of f we have f(c) = 0.

Proof of claim: exercise.2

Exercise 54 Let f : D → X be a convex monotone function with f(0) = 0 andf(x) 6= 0 for all x ≥ 0 Prove that the unique continuous extension f̃ of f to X hasan inverse, i.e., for every y ∈ X there exists x ∈ X such that f̃(x) = y.

An observation related to the intermediate value theorem that does not needcontinuity is the following fixed point theorem for monotone increasing functions:

Exercise 55 Assume f is a monotone increasing function, a ≤ b and with f(a) ≥ aand f(b) ≤ b. Then there exists a c with f(c) = c

We conclude with a remark on functions that are not necessarily monotone. Givenf : X → X and a point y ∈ X, we can restrict the function to the interval [0, y) anddefine a lower and upper monotone hull for the function analoguously to sequences.We can then define lim infx↗y f(x) and lim supx↗y f(x). Similarly we can definelim supx↘y f(x) and lim supx↘y f(x). We say that f is continuosu at y, if all these fourquantities agree wit f(y). Recall the classical definition of continuity at y, namely forall ε > 0 there exists δ > 0 such that for all x with |x−y| ≤ δ we have |f(x)−f(y)| ≤ ε.Note that this statement has two absolute value signs involved. Resolving each ofthese into two inequalities results roughly 4 conditions, corresponding roughly to thefour conditions in our setup.

1.11 The square function

The following will establish the half square function.

Lemma 31 There is a unique function s : D→ X such that

1. s(0) = 0

2. s(1) = ν1(0)

27

3. (Parallelogram law) For all x, y ∈ D:

s(x) + s(x+ y + y) = s(x+ y) + s(x+ y) + s(y) + s(y)

The function is monotone, convex, injective, and the unique extension to Xsatisfies the parallelogram law for all x, y ∈ X.

The third property is called the parallelogram law. To explain this terminology, weshall discours to some elementary facts about Euclidean space. Setting x+ y = z wemay write it for some function f as

f(z − y) + f(z + y) = f(z) + f(z) + f(y) + f(y)

If y and z are elements in Rn and f(x) = ‖x‖2, then this identity says that the sum ofsquare of the length of the diagonals of a parallelogram spanned by the vectors y andz is equal to the sum of squares of the length of the four sides. This also holds in onedimension, where the length squared is simply the absolut value squared. If restrictedto X, the function is simply the square function. Since we build it is primitive of theidentity function, it is convenient here to consider the half of the suqare function.

We proceed to prove the lemma.Proof:We first prove uniqueness. Note that s(ν0(0)) is unique. Assume s(νn(0)) is

unique. Applying the third property with x = 0 and y = νn+1(0) gives

s(νn(0)) = s(νn+1(0)) + s(νn+1(0)) + s(νn+1(0)) + s(νn+1(0))

Hence s(νn+1(0)) is the unique midpoint of 0 and zn where zn is the unique midpointof 0 and s(νn(0)). This proves by induction that s(νn(0)) is unique.

Fix n. Now we prove by induction on n that for every x ∈ Dn we have that s(x)and s(νn(x)) are unique. For x = 0 this has been established. Assume this is true forsome n. We only need to show that s(νn(ν(x))) is unique. However, we have by thesecond property applied with x and y = νn(0)

s(x) + s(νn(ν(x))) = s(νn(x)) + s(νn(x)) + s(νn(0)) + s(νn(0))

By cancellation, s(νn(ν(x))) is unique. This establishes uniqueness of s.To prove existence, we let s be the primitive of the monotone function g(x) = x:

s(x) :=

∫ x

0

t dt

To show the desired properties for s we consider the auxiliary integral∫ x

0

x dt

28

Assume x ∈ Dm and let n > m. Then the n-th lower Riemann sum for this integralis ∑

u∈Dn:0≤u<y

µn(y)

For each u ∈ Dn with 0 ≤ u < y there exists exactly on v ∈ Dn with 0 < v ≤ y andu + v = y and this is a bijective correspondence. Hence we may write for the abovesum, reshuffling the terms, ∑

u∈Dn:0≤u<y

µn(u) +∑

u∈Dn:0<v≤y

µn(v)

= µn(y) +∑

u∈Dn:0≤u<y

µn(u) +∑

u∈Dn:0≤v<y

µn(v)

Now taking lim sup in n gives∫ x

0

x dt =

∫ x

0

t dt+

∫ x

0

t dt

To check the first property, it then suffices to prove∫ 1

0

1 dt = 1

It suffices to prove that every Riemann sum for the left hand side is 1. We have

L(0, 1) =∑

n∈N:0≤n<1

1 = 1

and, assuming L(n, 1) = 1, we have

L(n+ 1, f) =∑

x∈Dn+1:0≤x<1

µn+1(1)

Since

{x : x ∈ Dn+1, 0 ≤ x < 1} = {x : x ∈ Dn, 0 ≤ x < 1} ∪ {νn+1(x) : x ∈ Dn, 0 ≤ x < 1}

we haveL(n+ 1, f) =

∑x∈Dn:0≤x<1

µn+1(1) + µn+1(1) = L(n, 1)

This verifies the first property for s.To verify the second property, we first observe that for x, y ∈ Dm and n ≥ m there

is a bijection between {z ∈ Dn : 0 ≤ z ≤ y} and {z ∈ Dn : x ≤ z ≤ x + y} given byz → z + x. hence we have ∑

z∈Dn:0≤z≤y

(x+ z) =∑

z∈Dn:x≤z≤x+y

z

29

Taking supremum in n gives ∫ y

0

(x+ t) dt =

∫ x+y

x

t dt

Applying this one more time with x replaced by x+ y and adding the two identitiesgives ∫ y

0

(x+ t) dt+

∫ x+y+y

x+y

t dt =

∫ x+y

x

t dt+

∫ y

0

(x+ y + t) dt

Using additivity in the integral and cancelling the integral in x gives∫ y

0

t dt+

∫ x+y+y

x+y

t dt =

∫ x+y

x

t dt+

∫ y

0

t dt+

∫ y

0

y dt

Cancelling one integral from 0 to y and using the previously shown identity for theintegral of y gives ∫ x+y+y

x+y

t dt ==

∫ x+y

x

t dt+

∫ y

0

t dt+

∫ y

0

t dt

Adding s(x+ y) and s(x) on both sides gives

s(x) + s(x+ y + y) = s(x+ y) + s(x+ y) + s(y) + s(y)

This is the parallelogram law for s.Being the primitive of a monotone function, s is convex and monotone. Since

the derivative does not vanish except at 0, the function s is strict monotone andthus injective. The function s has a unique continuous extension to X. That theextension also satisfies the parallelogram law follows by the general fact that for anytwo monotone functions f and g we have (f + g)l = fl + gl. 2

Definition 16 (Square function) Define the square function S(x) = s(x) + s(x)where s is the function of the previous lemma.

Exercise 56 Note that S satisfies all properties of the previous lemma except thevalue at 1 is not ν1(0) but 1.

1.12 Linear functions, the product

We now look at functions preserving both addition and order. We have

Exercise 57 For every number z ∈ R+0 there is a unique function fz : D→ R+

0 whichsatisfies

1. x ≤ y implies fz(x) ≤ fz(y)

2. fz(1) = z

30

3. fz(x) + fz(y) = fz(x+ y)

The function fz is monotone and convex and its unique continuous extension to Xalso satisfies all these properties with x, y ∈ X.

The second property says that the function preserves addition. The functions inthe lemma are called linear functions. We will derive these functions using the squarefunction: this process is called polarization. (this process is called polarization).

Proof: We fix z and write f for fz. We first show uniqueness. Since f(0) =f(0 + 0) = f(0) + f(0) we conclude f(0) = 0. Next f(ν0(0)) = f(1) = 1 is uniquelydetermined. Assume that f(νn(0)) is uniquely determined, then f(νn+1(0)) is themidpoint of 0 and f(νn(0)) and hence uniquely determined. By induction, f(νn(0))is uniquely determined. Now assume f(x) is uniquely determined for x ∈ Dn. Thenf(νm(x)) = f(x) + f(νm(0)) is uniquely determined. By induction, f(x) is uniquelydetermined for x ∈ Dn. Hence f is unique.

We turn to existence. Let s denote the half square function and let f(x) be theunique solution to

s(x) + s(z) + f(x) = s(x+ z)

Note f(0) = 0. Adding this equation to itself and adding the parallelogram law inthe form

s(z) + s(z) + s(x+ z) + s(x+ z) = s(x+ z + z) + s(x)

gives the alternative definition of f(x):

s(x) + s(z) + s(z) + s(z) + s(z) + f(x) + f(x) = s(x+ z + z)

Assume x ≥ y, then he parallelogram law for x− y and z + y gives

s(x− y) + s(z + z + x+ y) = s(z + x) + s(z + x) + s(z + y) + s(z + y)

Applying the definition of f(x) and f(y) and the alternative definiiton for f(x + y)and cancelling four copies of s(z) on both sides gives

s(x−y)+f(x+y)+f(x+y)s(x+y) = f(x)+f(x)+f(y)+f(y)+s(x)+s(x)+s(y)+s(y)

Cancelling the parallelogram law for x− y and y gives

f(x+ y) + f(x+ y) = f(x) + f(x) + f(y) + f(y)

which implies f(x + y) = f(x) + f(y). If y ≥ x, a symmetric argument yields thesame conclusion.

We conclude in particular that f is convex and monotone increasing.We calculate the right derivative of f at 0. The expressions

µ−1n (f(νn(0))− f(0)) = µ−1n (f(νn(0)))

31

are constant in n as one proves by induction:

µ−1n+1(f(νn+1(0))) = µ−1n (f(νn+1(0))+f(νn+1(0)))µ−1n (f(νn+1(0)+νn+1(0))) = µ−1n (f(νn(0))−0)

Hence the right derivative of f at 0 is

f(1)− f(0) = f(1)

On the other hand, taking the right derivative on the equation

s(x+ z) = f(x) + s(x) + s(z)

givesz + 0 = f(1) + 0 + 0

Hence f(1) = z as desired.That the continuous extension of f to X also satisfies the second property follows

from the general property(f + g)l = fl + gl

of left limits. 2

Definition 17 We define the product zx to be f(x) where f is the unique linearfunction with f(1) = z.

The symmetric definition of f in the above proof immediately gives

Exercise 58 The product is commutative:

zx = xz

Also, the above definition gives that the product is convex, monotone, and con-tinuous in each argument. If we retsrict to RP \ {0} ×RP \ {0} then it is bijective ineach argument.

The linearity of linear functions translates into the distributive law for the product.

Lemma 32 Prove the associative law of the product.

Proof: We need to showfx(fy(z)) = ffx(y)(z)

We will identify show that the composite functiuon on the left hand side equals thefunction ffx(y) by identifying ffx(y) by its properties.

We first havefx(fy(1)) = fx(y) = ffx(y)(1)

which shows the second property for all. We furthermore have

fx(fy(z + z′)) = fx(fy(z) + fy(z′)) = fx(fy(z)) + fx(fy(z

′))

This proves the third property for all x, y ∈ X.Monotonicity follows similarly. This identifies the composite function as ffx(y). 2

Exercise 59 Prove S(x) = fx(x) for all x ∈ R where S is the square function andfx is the linear function associated with x.

32

1.13 Functions on product spaces, limsup, liminf

Given two sets and a function on a product space X × Y , i.e. a function in twovariables (x ∈ X and y ∈ Y ), one can take supremum and infimum separately ineach variable. The two operations commute if they are of the same type, i.e. bothsupremum or both infimum, but they do not necessarily commute if they are ofdifferent type. There is however an inequality in the latter case. We will lim sup andlim inf and they satisfy an inequality which is a special case of the above.

Functions f in two variable can also be viewed as functions g in one variable whosevalues are functions in the other variable, with the natural identity:

f(n,m) = (g(n))(m)

We are mainly interested in the domains N,D, leading to four different combinationsof pairs of domains. All these cases have their separate interest and go under separatenames.

In case the domain is N×N, this leads to the study of sequences of sequences. Ifsequences are monotone, then we can form derived sequences. The consequences ofthe previous observation about commuting supremum or infimum in the two variableshas consequenes in the setting of series that are useful to know due to their frequentencounter. We will discuss these in detail.

Lemma 33 Let X and Y be sets and let f X × Y → X be a function. then

supx∈X

(supy∈Y

f(x, y) = supy′∈Y

( supx′∈X

f(u, v)

Proof: Since the supremum is an upper bound, we have for all x, y

f(x, y) ≤ supy′∈Y

f(x, y′)

By the same token we have for all x and y

f(x, y) ≤ supx′∈X

(supy′∈Y

f(x′, y′))

Hence, for fixed y, the right hand side is an upper bound for all f(x, y). Since thesupremum is the least upper bound, we have

supx∈X

f(x, y) ≤ supx′∈X

(supy′∈Y

f(x′, y′))

By the same token,

supy∈Y

(supx∈X

f(x, y)) ≤ supx′∈X

(supy′∈Y

f(x′, y′))

By symmetry, we also have the reverse inequality. Hence we have equality. 2

33

Lemma 34 Let X and Y be sets and let f : X × Y → X be a function. Then

supx∈X

( infy′∈Y

f(x, y′)) ≤ infy∈Y

( supx′∈X

f(x′, y))

Proof: Note first that we have for all x, y

infy′∈Y

f(x, y′) ≤ f(x, y) ≤ supx′∈U

f(x′, y)

since inf is a lower bound and sup is an upper bound. Since sup is a least upperbound, we have for all y

supx∈X

( infy′∈Y

f(x, y′)) ≤ f(x, y) ≤ supx′∈X

f(x′, y)

Since inf is the greatest lower bound, we have

supx∈X

( infy′∈Y

f(x, y′)) ≤ f(x, y) ≤ infy∈Y

( supx′∈X

f(x′, y))

2

To see that this inequality cannot be improved to equality in general, Pick X =Y = N and define f(n,m) = 0 if n ≤ m and f(n,m) = 1 if n > m. Then

supn∈N

( infm∈N

f(n,m)) = 0

infm∈N

(supn∈N

f(n,m)) = 1

Note that if for some sequence g : N→ X we define a double sequence

f(n,m) = g(n+m)

The partial suprema and infima of this function lead to sequenes natuarlly associatedsequec nes for a given sequence.

Definition 18 If f is any sequence, then the lower hull sequence is the sequence gdefined by

g(n) = infm∈N

f(n+m)

The upper hull sequence is the sequence g defined by

g(n) = supm∈N

f(n+m)

Lemma 35 The lower hull sequence of a sequence is monotone increasing.

Proof: This hinges on the simple fact that for any two subsets X, Y of X withX ⊂ Y we have

inf(X) ≥ inf(Y ) .

2

A natural concept then is

34

Definition 19 Definelim supn→∞

f(n) = infn∈N

g

where g is the upper hull sequence of f , and

lim infn→∞

f(n) = supn∈N

g

where g is the lower hull sequence of f .

Exercise 60 1. We have

lim infn→∞

f(n) ≤ lim supn→∞

f(n) .

2. If f is monotone, then one has actually equality in this inequality.

3. If a ≤ f(n) ≤ b for every n ∈ N, then

a ≤ lim infn→∞

f(n) ≤ lim supn→∞

f(n) ≤ b .

If lim inf and lim sup of f are equal, one says that f converges and calls thecommon value of lim sup and lim inf the limit of f . However, even if the limit exists, inmany instances the real interest lies in either lim inf or lim sup rather than the limit.One is well advised to use the concepts of lim inf and lim sup where appropriate.Unlike the limit, the definitions of lim inf and lim sup are not conditioned on anidentity (between lim sup and lim inf), and in case one merely is interested in one ofthem, the information as to which of these one is interested in is helpful.

Exercise 61 Formulate and prove further theorems about lim sup and lim inf relativeto addition and order structure of seqences.

1.14 Theorems for series

When we start passing to the derived sequence in one of the variables of a doublesequence, then of the various ways one can turn the above facts into statements inthis setting there is a threesome of fundamental statements that turn out particularlyconvenient in applications: the monotone convergence theorem, Fatou’s lemma, andthe dominated convergence theorem.

Lemma 36 (Monotone convergence of series) Let an : N → X be a monotoneincreasing sequence of sequences, that is an(m) ≤ an+1(m) for every n,m ∈ N. Then∑∞

m=0 an is monotone increasing and

limn→∞

∞∑m=0

an(m) =∞∑m=0

( limn→∞

an(m))

35

Proof: Note thatM∑m=0

an(m)

is monotone increasing, by order preservation of (finite) addition. Since sup respectsinequality of sequences,

∞∑m=0

an(m) = supM

∞∑m=0

an(m)

is monotone increasing in n.We then have by definition for the left hand side

supn∈N

supM∈N

m∑n=0

an(m)

Commuting the suprema by Lemma 33 and then a supremum with the finite sumequates the last display with

supM∈N

m∑n=0

supn∈N

an(m)

This by definiton is the right hand side of the identity to be shown. 2

Lemma 37 (Fatou for sequences) Let an : N→ X be a sequence of sequences,

∞∑m=0

(lim infn∈N

an(m)) ≤ lim infn∈N

∞∑m=0

an(m)

Proof: The left hand side is by definition

supM∈N

M∑m=0

supn∈N

infk∈N

an+k(m)

Commuting the finite sum to the right of the infimum (gainign an inequality sign)and commuting the suprema by Lemma 33 and then a supremum with the infimumby Lemma 34 gives

supn∈N

infk∈N

supM∈N

M∑m=0

an+k(m)

This by definition is the right hand side of the inequality to be shown. 2

Consider the following example closely related to the example after Lemma 34.Let nn(m) = 1 if n = m, and an(m) = 0 if n 6= m. Then

∞∑m=0

(lim infn∈N

an(m)) = 0

36

lim infn∈N

∞∑m=0

an(m) = 1

The intuition here is as follows. While every sequence an has total integral 1, theposition which carries the value 1 wanders off to infinity as time (the parameter n)increases, and hence the limit sequence has total integral 0.

Lemma 38 (Dominated convergence) Let an : N → X be a convergent sequenceof sequences, i.e.

lim infn→∞

an(m) = lim supn→∞

an(m)

for every m. Assume that for every n we have an < b for some sequence b with∑∞m=0 b(m) <∞. Then

∞∑m=0

limn→∞

an(m) = limn→∞

∞∑m=0

an(m)

and both sides are finite.

Proof: By Fatou we have

∞∑m=0

lim infn→∞

an(m) ≤ lim infn→∞

∞∑m=0

an(m)

Now define sequences cn : N→ X such that an + cn = b for every n, this can be doneby domination of an by b. Note that by the previous lemma for each k

supn≥k

an(m) + infn≥k

cn = b

and hence, by the same lemma, for all m,

lim supn→∞

an(m) + lim infn→∞

cn(m) = b(m)

Furthermore,M∑m=0

an(m) +M∑m=0

cn(m) =M∑m=0

b(m)

Since supremum of a sum of monotone sequences is the sum of the suprema of themonotone sequecnes, we have

∞∑m=0

an(m) +∞∑n=0

cn(m) =∞∑m=0

b(m)

By the same argument as above,

lim supn→∞

∞∑m=0

an(m) + lim infn→∞

∞∑n=0

cn(m) =∞∑m=0

b(m)

37

Now we conclude with Fatou again

∞∑m=0

b(m) + lim supn→∞

∞∑m=0

an(m)

=∞∑m=0

lim supn→∞

an +∞∑m=0

lim supn→∞

cn + lim supn→∞

∞∑m=0

an(m)

≤∞∑m=0

lim supn→∞

an + lim supn→∞

∞∑m=0

cn + lim supn→∞

∞∑m=0

an(m)

=∞∑m=0

lim supn→∞

an + lim supn→∞

∞∑m=0

b(m)

Since the series for b has a finite value, we can cancel it and obtain

lim supn→∞

∞∑m=0

an(m) ≤∞∑m=0

lim supn→∞

an

Since the sequences an converges in n,

lim supn→∞

∞∑m=0

an(m) ≤∞∑m=0

limn→∞

an ≤ lim infn→∞

∞∑m=0

an(m)

By the natural inequality between lim inf and lim sup, All terms in the chain ofinequalities have to coincide. 2

2

The threesome of the above theorems for series is joined by one for double series,a Fubini type theorem.

Lemma 39 (Fubini) Let a : N× N→ X be a double sequence. Then

∞∑n=0

(∞∑m=0

a(n,m)) =∞∑m=0

(∞∑n=0

a(n,m))

Proof: The left hand side is∞∑m=0

limn→∞

n∑k=0

ak(m)

Sincen∑k=0

ak(m)

is a monotone increasing sequence in n, we can apply the monotone convergencetheorem to equate the last display with

limn→∞

∞∑m=0

n∑k=0

ak(m)

38

= limn→∞

n∑k=0

∞∑m=0

ak(m)

=∞∑k=0

∞∑m=0

ak(m)

2

We now turn to the cases when the product space is N × D or D × N or D × D.Each of these cases has the pendant of the four theorems. We list them as exercises.

1.15 Series theorems revisited with functions

Exercise 62 (Monotone convergence) Let fn : D→ X be a monotone increasingsequence of monotone increasing functions, that is fn(x) ≤ fn+1(x) for every n ∈ Nand x ∈ D. Then for every x ∈ D we have that

∫ x0fn(t) dt is monotone increasing in

n and

limn→∞

∫ x

0

fn(t) dt =

∫ x

0

limn→∞

fn(t) dt

Exercise 63 (Fatou) Let fn : D → X be a sequence of monotone increasing func-tions, then lim infn→∞ fn is also monotone increasing (see a former exercise)and forevery x ∈ D ∫ x

0

lim infn∈N

fn(t) dt ≤ lim infn∈N

∫ x

0

fn(t) dt

Alternatively, passing to the primitive Fn of fn, we obtain for every seuqence of convexfunctions Fn that lim infn→∞ Fn is convex and

lim infn→∞

F ′n = (lim infn→∞

Fn)′

Exercise 64 (Dominated convergence) Let fn : N→ X be a convergent sequenceof monotone increasing functions, i.e.

lim infn→∞

fn(x) = lim supn→∞

fn(x)

for every x ∈ D. Let x ∈ D and assume that for every n we have an < b for somemonotone increasing function b with

∫ x0b(t) dt <∞. Then∫ x

0

limn→∞

fn(t) dt = limn→∞

∫ x

0

fn(t) dt

and both sides are finite.

Exercise 65 (Fubini) Let fn : D → X be a sequence of monotone increasing func-tions. Then for x ∈ D

∞∑n=0

∫ x

0

fn(t) dt =

∫ x

0

∞∑n=0

fn(t) dt

39

Alternatively, if we have for a sequence Fn of convex functions for all x

(∞∑n=0

F (x))′ =∞∑n=0

F ′(x)

In the caes of functions mapping D to the space of sequences, monotone conver-gence and Fubini are identical to the above by symmetry.

Exercise 66 Formulate and proof the pendants of Fatou and Dominated convergence.

We turn to functions whose values are functions.

Exercise 67 (Monotone convergence) Let f : D×D→ X be a doubly monotoneincreasing function, i.e., for every x, x′, y, y′ in D with x ≤ x′ and y ≤ y′ we havef(x, y) ≤ f(x′, y) and f(x, y) ≤ f(x, y′).

Then for every x, y ∈ D we have that∫ x0f(t, y′) dt is monotone increasing in y′

and

limy′↗y

∫ x

0

f(t, y′) dt =

∫ x

0

limy′↗y

fn(t, y′) dt

and similarly for right limits.

Exercise 68 (Fatou) Let f : D×D→ X be monotone increasing in the first variable.Then for every x, y ∈ D∫ x

0

lim infy′↗y

f(t, y) dt ≤ lim infy′↗y

∫ x

0

f(t, y) dt

and similarly for right limits.

Exercise 69 (Dominated convergence) Let f : D×D→ X be monotone increas-ing in the first variable. Let y ∈ D and assume

lim infy′↗y

f(x, y) = lim supy′↗y

f(x, y)

Let x ∈ D and assume that for every y′ and t ∈ D we have f(t, y′) < b(t) for somemonotone increasing function b with

∫ x0b(t) dt <∞. Then∫ x

0

limy′↗y

f(t, y′) dt = limy′↗y

∫ x

0

f(t, y′) dt

and both sides are finite. A similar statement holds for right limits.

Exercise 70 (Fubini) Let f : D×D→ X be a doubly monotone increasing function.Then for x, y ∈ D ∫ x

0

(

∫ y

0

f(s, t) dt) ds) =

∫ y

0

(

∫ x

0

f(s, t) ds) dt)

Alternatively, let F be convex in x and monotone in y, then

(

∫ y

0

F dt)′ =

∫ y

0

F ′(t) dt

40

1.16 The power function

The following lemma establishes integer powers of a number a.

Lemma 40 For every a ∈ X there exists a unqiue sequence f : N→ X satisfying forall n,m ∈ N

1. f(0) = 1

2. f(1) = a

3. f(n+m) = f(n)f(m)

where on the right hand side of the last property we have the product of f(n) and f(m).If a ≥ 1, then the sequence is monotone increasing. If a ≤ 1, then the sequence ismonotone decreasing

Proof: We first prove uniqueness. By assumption the function is unique at 0.Assume we have shown that the function is unqiue at some n ∈ N. Then

f(n+ 1) = f(n)f(1)

and the assumed uniqueness at f(n) and the given value at 1 proves uniqueness atn+ 1. This completes the induction.

To prove existence, we define f(0) := 1 and recursively

f(n+ 1) := f(n)a

To establish the claimed properties, note that f(1) = f(0)a = a. We fix n and proveby induction on m that

f(n+m) = f(n)f(m)

If m = 0, we havef(n+ 0) = f(n) = f(n)1 = f(n)f(0)

Assume we have proven the desired identity for some m. Then we have, using asso-ciativity of the product,

f(n+(m+1)) = f((n+m)+1) = f(n+m)a = (f(n)f(m))a = f(n)(f(m)a) = f(n)f(m+1)

This proves the desired property.To see the claimed monotonicity of the sequence, note that f(n + 1) = f(n)a ≥

f(n) if a ≥ 1 andf(n+ 1) = f(n)a ≤ f(n)

if a ≤ 1. 2

Note in particular that if a = 0 then f(0) = 1 and f(n) = 0 for n ≥ 1.Buy the isomorphism between N and Dm for any m, we have

41

Exercise 71 For every m ∈ N and a ∈ X there exists a unqiue sequence f : Dm → Xsatisfying for all x, y ∈ Dm

1. f(0) = 1

2. f(νm(0)) = a

3. f(x+ y) = f(x)f(y)

If a ≥ 1, then the sequence is monotone increasing. If a ≤ 1, then the sequence ismonotone decreasing

Now we extend this function to D by virtue of the following lemma. We assumethe freedom to normalize the function at any point νm(0).

Lemma 41 Let a ∈ X. Then there is a unique function f : D→ X such that for allx, y ∈ D

1. f(0) = 1

2. f(1) = a

3. f(x+ y) = f(x)f(y), where the right hand side is the product of the two valuesof the function at x and y.

Prove:We first establish uniqueness. First we prove by induction that for all m ∈ N the

function is unique at νm(0). For m = 0 this is part of the hypothesis as f(1) = a.Assume we have proven that f is unique at νm(0) for some m. Then the assumedidentity f(νm+1(0) + νm+1(0)) = f(νm+1(0))f(νm+1(0)) shows

f(νm(0)) = S(f(νm+1(0)))

and uniqueness of the inverse of S implies that f is unique at f(νm(0)). This completesthe induction. If f is unique at νm(0), then uniqueness at all points of Dm follows byan application of the previous lemma. Since D =

⋃m≥k Dm, the function is unique at

all points of D.Now we prove existence. First we define a sequence of consecutive square roots as

follows: Define s0 = a Assume sm is already defined for some m ∈ N. Then let sm+1

be the unique number such that S(sm+1) = sm.On D0, define f to be the unique sequence given by the previous lemma with

f(1) = s0. Assume we have already defined f on Dm, satisfying the assumptions of theprevious lemma with f(νm(0)) = sm. Let g be the unique sequence given by the previ-ous lemma which satisfies g(νm+1(0)) = sm+1. We claim that on Dm, this sequence co-incides with f . Namely, g(νm(0)) = g(νm+1(0) + νm+1(0)) = g(νm+1(0))g(νm+1(0)) =sm+1sm+1 = sm = f(νm(0)). Since both f and g restricted to Dm satisfy the as-sumptions of the previous lemma, they coincide. Hence we may extned f to Dm+1 by

42

setting f(x) = g(x) for all x ∈ Dm. Since D =⋃m∈N Dm, this defines f on all of D.

To see the identityf(x+ y) = f(x)f(y)

for any x, y ∈ D, note that for any such x, y there is a m ∈ N such that x, y ∈ Dm

and then the identity holds since it holds in Dm. This proves existence.

1.17 An integral equation for power function, the logarithm

Lemma 42 Assume a function f : D → X satisfies f(x + y) = f(x)f(y) for allx, y ∈ D. and it is monotone increasing. Then the function

g(x) := 1 + c

∫ x

0

f(t) dt

satisfies g(x+ y) + f(y)g(0) = f(y)g(x) + g(y) for all x, y ∈ D.

Proof: We need to establish the identity, since upon multiplying by c and addingf(y)g(0) + g(y) on both sides we will obtain the desired identity.∫ x+y

x

f(t) dt = f(x)

∫ y

0

f(t) dt

Assume x, y ∈ Dm. It suffices to show the corrseponding identity for the lower riemnansums for n ≥ m: ∑

z∈Dm:x≤z<x+y

νm(f(z)) = f(x)∑

z∈Dm:0≤z<y

νm(f(z))

But using the bijection x → x + y which maps {z ∈ Dm : 0 ≤ z < y} to the set{z ∈ Dm : x ≤ z < x + y}, we may conclude for the right hand side (using linearityof multiplication by f(z) and that it commutes with νm,

f(x)∑

z∈Dm:0≤z<y

νm(f(z))

∑z∈Dm:0≤z<y

νm(f(x)f(z))

=∑

z∈Dm:0≤z<y

νm(f(x+ z))

=∑

z∈Dm:x≤z<x+y

νm(f(z))

Hence the lower Riemann sums are identical. 2

43

Lemma 43 There is a unique c such that for the power function f(x) = ax we have

f(x) = 1 + c

∫ x

0

f(t) dt

Proof: Uniqueness follows by evaluating at x = 1. Existence: Choose ck so that

f(νk(0)) = 1 + ck

∫ νk(0)

0

f(t) dt

and define

gk(x) = 1 + ck

∫ x

0

f(t) dt

Then by an application of the previous lemma we obtain for any y ∈ D

gk(νk(0) + y) + f(y)g(0) = f(y)gk(νk(0)) + gk(y)

This simplifes togk(νk(0) + y) = f(y)gk(νk(0))

No by induction we may easily verify gk(y) = f(y) for every y ∈ Dk. Then gk andgk+1 coincide on Dk and hence ck = ck+1. By induction all values of ck are equal tosdome c. Defining

g(x) = 1 + c

∫ x

0

f(t) dt

we see that g coincides with f on all x ∈ D. This proves existence of c. 2

The c established in the previous lemma is called the logarithm of a.Note that we can conclude that the power function si convex fro a ≥ 1, and the

fundamental theorem gives that the derivative of the power function ax is cax.The next lemma shows that one can characterize the power function by the integral

equation.

Lemma 44 Let c ∈ X. Let f, g be two monotone functions such that f(x) = f(0) +c∫ x0f(t) dt. g(x) = g(0) + c

∫ x0g(t) dt. and assume that f(0) < g(0). Then f(x) ≤

g(x) for all x ∈ D

Proof:Pick δ > 0 such that f(0) + δ = g(0). Assume to get a contradiction that

there exists y ∈ D with f(y) > g(y). Then let x = infy∈D f(y) > g(y). Thenwe have x ∈ X and x ≥ ε by the above. Then limy↘x f(y) ≥ limy↘x g(y) andlimy↗x f(y) ≤ limy↗x g(y). Since f and g are continuous, limy↗x f(y) = limy↗x g(y).Now let y ≤ x with y ∈ D, then we have

f(x) + δ = f(0) + δ + c

∫ x

0

f(t) dt

≤ g(0) + c

∫ x

0

g(t) = g(x)

But this contradicts limy↗x f(y) = limy↗x g(y). 2

44

Lemma 45 Let c ∈ X. Let f, g be two monotone functions such that f(x) = f(0) +c∫ x0f(t) dt. g(x) = g(0) + c

∫ x0g(t) dt. and assume that f(0) = g(0). Then f(x) =

g(x) for all x ∈ D

Proof: Note that if f satisfies the inetgral equation, then so does df for everydyadic d. By the previous lemma, we have g(x) ≤ df(x) for all x whenever d ≤ 1 andwe have g(x) ≥ df(x) for all x whenever d ≥ 1. By continuity of multiplication by dwe have f(x) = dg(x) for d = 1 2.

1.18 The exponential series

The exponential function is the function ex where e is the positive real number whoselogarithm is 1. We also write exp(x) = ex.

From the previous discussions, the primitive and derivative of exp are exp again.We provide to representations of exp.Consider the sequence satisfying

f(x+ y) = f(x)f(y)

on Dn which takes value1 + νn(0)

at the point νn(0). By Previous lemmata it is unique and has unique extension to Dof the form an

x.Note that the sequence of functions axn is monotone increasing, and thus has a

limit.By convexity, the derivative at 0 of this function is bounded above by 1 + νn(0)

Thus all these functions are bounded above by any power function with derivative 2at 0

Hence the monotone increasing sequence has a limit.

Exercise 72 The limit of this function is exp

Define the n-th monomial f : D→ X recursively as follows. First set

f0(x) = 1

and note that f0 is monotone increasing and convex. Assume fn is defined andmonotone increasing and convex, then define

fn+1(x) = xf(x)

and note that fn+1 is monotone increasing (easy) and convex:

fn+1(x+ y) + fn+1(x+ y) =

(x+ y)[fn(x+ y) + fn(x+ y)]

45

≤ (x+ y)[fn(x) + fn(x) + fn(x+ y + y) + fn(x+ y + y)]

≤ x[fn(x)+fn(x)]+y[fn(x+y+y)+fn(x+y+y)]+(x+y)[fn(x+y+y)+fn(x+y+y)]

= fn+1(x) + fn+1(x+ y + y)

Note that(x− y)(

∑(k,m):k+m=n

xkym)

∑(k,m):k+m=n

xk + 1ym)−∑

(k,m):k+m=n

xkym+1)

= xn+1 − yn+1

Hence

limy→x

xn+1 − yn+1

x− y=

limy→x

(∑

(k,m):k+m=n

xkym) = (∑

(k,m):k+m=n

xn) = (n+ 1)xn

Definition 20 Define 0! = 1 and if n! is defined define (n+ 1)! = (n+ 1)n!

Lemma 46 For given n ≥ 1 and m ≥ n we have

m!nn ≥ nm

Proof: First we prove n! ≤ n. Both sides are 1 if n = 1. Assume the desiredinequality is proven for some n, then

(n+ 1)! = (n+ 1)n! ≤ (n+ 1)nn ≤ (n+ 1)(n+ 1)n = (n+ 1)n+1

Now we prove by the inequality of the lemma by induction on m ≥ n. The casem = n has just been established. Then we have

(m+ 1)!nn = (m+ 1)m!nn ≥ (m+ 1)nm ≥ nm+1

2

Note thatn∑

m=1

xm

m!

is a monotone increasing sequence of monotone increasing functions. Let n be aninteger n ≥ 2x and m > n, then

xm

m!=xmnn

nnm!≤ xm

nnnm ≤ nn

1

2m

The latter is a convergent series, hence the series

46

exp(x) =∞∑m=1

xm

m!

Converges for every x to some finite number.By monotone convergence for monotoen functions we have∫ x

0

exp(t) dt = 1 +

∫ x

0

limn→∞

n∑m=0

xm

m!

= 1 + limn→∞

∫ x

0

n∑m=0

xm

m!

= 1 + limn→∞

n+1∑m=1

xm

m!

limn→∞

n+1∑m=0

xm

m!= exp(x)

Note that by the fundamental theorem we also have exp′ = exp.As acorollary, we have exp(x) = ex for the function ex defined before.

Exercise 73 Give an alternative proof of exp(x) = limn→∞(1 + xn)n By expanding

the function (1 + xn)n show it is a monotone increasing sequence of convex monotone

functions that converges to the exponential series.

Lemma 47 For all n we haven! ≤ (

n

e)n

Proof: Write the exponential series for ex and observe that nn

n!is one term in the

series. 2

1.19 Weierstrass approximation

Denote by f the primitive of e−x2

and denote by g the primitive of f .

Lemma 48 There are a constants a, b such that

af(x) ≤ 1 ≤ af(x) + be−x

Proof: First note thate−t

2 ≤ e−t

for t > 1. Since ∫ x

0

e−t dt = 1− e−x

47

is bounded, so is f . Hence we can pick a so that

a supx≥

f(x) = 1

This proves the first inequality.The second inequality we prove separately for t ≤ 1 and t ≥ 1. For t ≤ 1 this

is trivial by choosing b appropriately. For t ≥ 1 pick ε > 0 and choose t1 > 1 largeenough so that

a

∫ t1

0

e−s2

ds+ ε ≥ 1

a

∫ t

0

e−s2

ds+

∫ t1

t

e−s2

ds+ ε ≥ 1

a

∫ t

0

e−s2

ds+

∫ t1

t

e−s ds+ ε ≥ 1

a

∫ t

0

e−s2

ds+ e−t + ε ≥ 1

Since ε was arbitrary, this proves the desired inequality.2

Lemma 49 There are constanst a, b such that

ag(x) ≤ x ≤ ag(x) + b

Proof: Lat a as in the previous lemma. The first inequality follows by integrating

af(x) ≤ 1

To see the second inequality, integrate the second inequality of the previous lemma

x ≤ ag(x) + b

∫ x

0

e−t dt

= ag(x) + b(1− e−x) ≤ ag(x) + b

2

Lemma 50 For each N ≥ 0

a

Ng(Nx) ≤ x ≤ a

Ng(Nx) +

b

N

Proof: Just note that (Nx)/N = x . 2

Note that we have a convergent series for the function in the last lemma:

a

Ng(Nx) =

∞∑k=0

a1

(2k + 1)(2k + 2)N

(Nx)2k+2

k!

48

Lemma 51 If we truncate the series for

a

Ng(Nx)

at the 40N2 th term, then the error for 0 ≤ x ≤ 1 is less than

C2−N2

for some constant C.

Proof: We estimate the k-th term of the series by

a1

(2k + 1)(2k + 2)N

(Nx)2k+2

k!

≤ CN

k2(2N)2k

ek

kk

= CN

k2(4N2e

k)k

Now we pick k = 40N2 +m, then we can estimate the previosu display by

= C(e

10)N+m ≤ C(

1

2)N+m

2

Lemma 52 There is a polynomial of degree 40N2 with even powers only which ap-proximates the function x in [0, 2] within C/N

Proof: This is a consequence of the previous, summing a geometric series, esti-mating the tail by C2−N . 2

The elementary linear spline of width 1/N is the function h which satisfies

1. h(0) = 1

2. h(x) = 0 fro |x| ≥ 1/N

3. h restricted to [−1/N, 0] is affine linear

4. h restricted to [0, 1/N ] is affine linear

Lemma 53 There is a polynomial of degree 40N6 which estimates the elementarylienar spline h within 1/N2 on the interval [−1, 1]

49

Proof: We write for the elemntary linear spline

h(x) =N

2(|x+ 1/N | − 2|x− 1/N |)

Since we can approximate |x| on [−2, 2] by the previous lemma, we can estimate|x± 1/N | on [−1, 1]. Note that we need an accuracy of 1/N3 for the approximationof |x|, which needs a polynomial of degree 40N6.

2

Exercise 74 Show that one can estimate the coefficients of the approximating poly-nomial by C2100N6

(This estimate is not very good)

Definition 21 A function f : D→ R is called Lipschitz with constant c, if for everyx, y ∈ D we have

f(x+ y) ≤ f(x) + cy

andf(x) ≤ f(x+ y) + cy

Exercise 75 Prove that if f : D → X is Lipshitz with constant 1, then there is aspline function g which is continuous and affine linear on every interval [x, νm(x)]with x ∈ Dm and which satisfies

f(x) ≤ g(x) ≤ f(x) + 21−m

Moreover, we can choose g so that its slope is either 1 or −1 on each such interval.

Lemma 54 (Weierstrass approximation of Lipschitz functions) Assume f :[0, 1] → X is Lipschitz with constant 1, then for each large N there exists a poly-nomial of degree at most CN6 such that we can approximate f within 1/N by thepolynomial.

Proof: First subtract a constant so that the Lipschitz function, whose range hasdiameter at most 1, is bounded by 2. Approximate teh lipshitz function by a splineas in the above exercise, then the spline is still bounded by 3, say. This spline isthe linear combination of N translates of the elementary spline. Approxmating eachelementary spline within 1/N2 and adding the approximating polynomials gives thedesired polynomial.

2

Remark: The bound CN6 is not very good. Using better proofs, one can lowerthe power. The power N1 would be best possible, as the follwoing exercise shows.

Exercise 76 Let m ≥ 2 and consider a Lipshitz function on [0, 1) which is affinelinear on each interval [x, νm(x)) with x ∈ Dm with slope ±1 and which oscillatesabove and below 0 as often as possible. Show that a polynomial approximating thisfunction with accuracy νm+1(0) has to have degree at least cm for some constant c.

50

Lemma 55 (Weierstrass approximation for continuous functions) Assume f :[0, 1]→ X is continuous. Then we can approximate f within 1/N by the polynomial.

Proof: We construct a sequence xn of points. Let x0 = 1. Assume xn is alreadydefined. Pick xn+1 > xn such that

1. |f(x)− f(xn)| ≤ 1/2N

2. xn+1 = 1 or there exists x ≤ f(xn)+2(f(xn+1−f(xn)) with f(x)−f(xn) ≥ 1/4N

If xn+1 = 1 we stop. Assuem we never stop. Then supn(xn) = y ≤ 1. By continuitythere exists δ such that |f(y) − f(x)| ≤ 1/8N for all |x − y| ≤ δ. Pick xn such that|xn − y| ≤ δ/4. Then |xn=1 − xn| < δ. This can be seen to contradict the choice ofxn+1.

Hence we stop. Now we cand efien a linear continuous spline with points xn andprove that this approximates f within 1/2N . Since this spline consists of finitelymany affine linear pieces, it is lipschitz. We may then approximate it within 1/2Nby some polynomial.

2

Note that no bound on the degree of the polynomial is provided. Indeed, providingsuch a bound is impossible, it is not just an artifact of our proof.

Lemma 56 Given a Lipshitz function f : [0, x)→ R and ε > 0 with Lipshitz constant1, There is a polynomial P of degree at most CeC/ε suich that

P (x) ≤ f(x) ≤ P (x) + ε

for all x ∈ [0, 1)

1.20 Negative numbers

One purpose of negative numbers is to solve the equation s+x = t in x unrestrictedlyfor any given s, t. Another purpose is to model all kind of physical phenomenaproperly, usually the world does not stop at 0.

First we note that for the set X we have analogously to the dyadic numbers:

Exercise 77 For s, t ∈ X there exists x ∈ X with s+ x = t if and only if s ≤ t.

One might introduce the real numbers by taking two copies of X and identifyingthe 0 of both copies as suggested by the standard picture of the real line, and thenwork out the structure of addition and order. We choose another method that ismore general and analoguous to the standard introduction of rational numbers asequivalence classes of fractions. This more general approach will then also appliedto monotone sequences and monotone functions etc. As a rule of thumb, to get anelegant theory one should first work out new concepts (sup, seqeunces, functions) onX and in the end apply this general approach.

51

Consider all pairs (x, y) of elements of X (think ′′x− y′′). We write

(x, y) ∼ (x′y′) ifx+ y′ = x′ + y

A symbol such as ∼ or ≤ or = that connects two elements and thereby forms astatement that may be true or false is called a relation (e.g. ≤ is an order relation).

Lemma 57 The relation ∼ is an equivalence relation, which means that it satisfiesthe following three properties for any pairs (x, y), (x′, y′), (x′′, y′′)

1. Reflexivity: (x, y) ∼ (x, y)

2. Symmetry: (x, y) ∼ (x′, y′) implies (x′, y′) ∼ (x, y)

3. Transitivity: (x, y) ∼ (x′, y′) and (x′, y′) ∼ (x′′, y′′) imply (x, y) ∼ (x′′, y′′)

Note that one can deduce the first proeprty from the other two for any (x, y) forwhich there exists some (x′, y′) , may it be equal or different from (x, y), which isrelated to (x, y). Since such an (x′, y′) is not guaranteed to exists without the fristproperty, this property is not entirely superfluous.

Proof: The first two proeprties are immediate from the definition. We prove thethird property, which is the most interesting to prove. Assume that

x+ y′ = x′ + y

x′ + y′′ = x′′ + y′

Add the two identities and use associative and commutative law to write

x+ y′′ + (x′ + y′) = x′′ + y + (x′ + y′)

Using the cancellation property (note that at this point we have to do subtleties andmodifications when we try to include ∞ in the scheme) we obtain

x+ y′′ = x′′ + y

. 2

Note that = is also an equivalence realtion. Indeed, the properties of equivalencerelation are soemthing like the defining properties of equality. One may call to pairs“equal” if they are equivalent under ∼.

Formally, one observes that the set of pairs is partitioned into a disjoint sets called“equivalence classes” of pairwise equivalent elements, such that elements from differ-ent equivalence classes are not equivalent. That this partition exists is a consequenceof the three proeprties of equivalence relation.

The set of real numbers is then the set of equivalence classes of pairs.We now define addition and order on pairs:

1. (x, y) + (x′, y′) := (x+ x′, y + y′)

52

2. (x, y) ≤ (x′, y′) if x+ y′ ≤ x′ + y

If this is to define additiona and order on the set of equivalence classes themselves,these structures need to respect the equivalence relation in the sense of the followinglemma:

Lemma 58 Let (x, y) ∼ (u, v) and (x′, y′) ∼ (u′, v′). Then

(x, y) + (x′, y′) ∼ (u, v) + (u′, v′)

and(x, y) ≤ (x′, y′)

if and only if(u, v) + (u′, v′) .

Proof: The assumptions give

x+ v = u+ y

x′ + v′ = u′ + y′

Adding the two identities and using associative and commutative law gives

x+ x′ + v + v′ = u+ u′ + y + y′

Hence (x + x′, y + y′) and (u + u′, v = v′) are equivalent again. If (x, y) ≤ (x′, y′),that is

x+ y′ ≤ x′ + y

then we can add u+ v+u′+ v′ on both sides, and using associative and commutativelaw we obtain

u+ v′ + (x+ v) + (u′ + y′) ≤ u′ + v + (u+ y) + (x′ + v′)

Now using preservation of order and cancelling equal terms according to the knownequivalences we obtain

u+ v′ ≤ u′ + v

and hence (u, v) ≤ (u′, v′). This proves the only if part of the last statement, and theif part follows symnmetrically. 2

Hence one has addition and order on equivalence classes of pairs, by simply eval-uating on individual representatives of the equiovalence classes and having indepen-dence of which representative we choose.

Exercise 78 Prove commutative, associative law and cancellation for addition ofreals.. Prove totality, antisymmetry, and transitivity for the order of reals prove thataddition respects order.

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The following lemma identifies distinguished representatives in each equivalenceclass.

Lemma 59 Each equivalence class of pairs has exactly one representative which hasat least one entry being equal to 0.

Proof: We first prove uniqueness. Suppose we have two such members. There arethree possibilities

1. The representatives are (x, 0) and (u, 0). Then x+ 0 = u+ 0 and hence x = u.

2. The representatives are (0, y) and (0, v). Then 0 + v = 0 + y and hence y = v.

3. The representatives are (x, 0) and (0, v). Then x+ v = 0 + 0 = 0. Since v ∈ Xawe have x + v ≤ v and by order preserving of addition we have x ≤ 0. Hencex = 0 and by the previous case v = 0 as well.

This proves uniqueness of the representative. To prove existence, we pick an equiva-lence class and some representative (x, y). There are two cases

1. x ≤ y. Then there exists a ∈ X such that x+a = y. This means (x, y) ∼ (0, a).

2. y ≤ x. Then there exists a ∈ X such that x = y+a. This means (x, y) ∼ (a, 0).

2The uniqueness part of the last lemma is quite universal in this setup but uninter-esting without the existence part. The existence part however relies on the speciallink between addition and order of X and takes different form in different settigns, e.g.in the case of constructing the rationals as equivalence classes of fractions one findsdifferent types of specific representatives, liekwise in our discussions further below.in general, existence of some canonical representative is not guaranteed.

Exercise 79 Identify the reals of the form (x, 0) with the set X by showing that thebijective map x→ (x, 0) preserves the structure of addition and order.

Note that we have (x, y) + (y, x) = (0, 0) so each element in X has an additiveinverse. From there it is immediate that we can always solve the equation s + x = tfor a real number x.

1.21 Sequences of bounded variation

Consider the class M of bounded monotone increasing sequences, where boundednessmeans supn∈N f(n) < ∞. We mimick the passage from nonnegative real numbers toreal numbers by considering all pairs (f, g) of elements of M . Again, we intuitivelythink of f − g for the pair.

We introduce the same equivalence relation (f, g) ∼ (f ′, g′) if f+g′ = f ′+g. Thatthis is an equivalence class follows similarly as above, one uses cancellation propertyfor every index n.

A canonical distinguished member in each equivalence class is described as follows:

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Exercise 80 Prove that in each equivalence class there is exactly one representative(f, g) such that for every n we have f(n) = f(n+ 1) or g(n) = g(n+ 1).

Note that while in the setting of R it is feasable to build the theory from the specialrepresentatives, in the current setting such endeavour would be quite cumbersome.Hecne the virtue of our general approach.

The equivalence classes of pairs in M ×M are called sequences of bounded vari-ation. One may think of them as sequences from N to the set R of real numbers,though not every sequence N→ R comes from a sequence of bounded variation.

Note we can define a limit of a sequecne of bounded variation by

limn→∞

(f, g) = (supn∈N

f(n), supn∈N

g(n))

where the right hand side is a real number. That this real number is well defined isthe reason we restricted attention to bounded sequences f and g. More precisely, westill need to verify that the limit thus defined depends only on the equivalence classof (f, g). Thus let (f, g) ∼ (f ′, g′), i.e. f + g′ = f ′ + g. Then

supn∈N

(f + g′)(n) = supn∈N

(f + g′)(n)

If we can conclude

supn∈N

f(n) + supn∈N

g′(n) = supn∈N

f(n) + supn∈N

g′(n)

then we have shown that the two have the same limit by definition of the equivalenceclassses of the real numbers. But this follows from the following lemma.

Exercise 81 If f and g are two monotone increasing sequences, then

supn∈N

(f(n) + g(n)) = supn∈N

f(n) + supn∈N

g(n)

We remark that one can show that given a sequence of bounded variation, the limitwe defined above coincides with the limit of the corresponding sequence N → R inthe classical sense, in particular the latter exists whenever the sequence is of boundedvariation.

Now given a sequence of bounded variation (f, g), one can formally write downthe derived sequence (a, b) where

f(n) + a(n) = f(n+ 1)

g(n) + b(n) = f(n+ 1)

This is a sequence of real numbers.It turns out that these are exactly the sequences for which

∞∑k=0

(a(n), b(n))

and the sum in the classical sense coincides with the limit

limn→∞

(f, g)

that we defined above.

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