Limits and Continuity

Transcription

1 Limits and Continuity Notes by Walter Noll, Real Sequences, Convergence Let an index set I be given. The set R I of all real families indexed on I has the natural structure of a commutative ring whose addition, multiplication, and opposition are defined by the termwise application of these operations, and whose zero and unity are the constant families whose terms are 0 R and 1 R, respectively. Thus we have (a + b) i := a i + b i (ab) i := a i b i for all a,b RI and i I. (1.1) ( a) i := a i The zero and unity of R I are also denoted by 0 and 1 when no confusion is likely. R I also acquires the natural structure of a real linear space when the scalar multiplication in R I is defined by (ξa) i := ξa i for all ξ R and i I. (1.2) We are here interested in the case when I := N or I := N, in which case the elements of R I are called real sequences. We deal explicitly only with the case I := N, but all the definitions and results below apply also to the case I := N. Definition 1: We say that the sequence b R N is a subsequence of a given sequence a R N if b = a σ for some strictly isotone σ : N N, so that b k = a σ(k) for all k N (1.3) Definition 2: Let a sequence a R N be given. We say that a is isotone if we say that a is strictly isotone if (i k) (a i a k ) for all i,k N j ; (1.4) (i < k) (a i < a k ) for all i,k N. (1.5) 1

2 We say that a is antitone if the second in (1.4) is replaced by, and we say that a is strictly anitone if the second < in (1.5) is replaced by >. We say that a is monotone if it is isotone or antitone, and we say that a is strictly monotone if it is strictly isotone or strictly antitone. Definition 3: We say that a sequence is bounded above, bounded below, or bounded if its range has the corresponding property. Definition 4: We say that a given sequence a R N converges to a given t R if for every ε P there is an n N such that or, equivalently, a > (n + N) ]t ε,t + ε[. (1.6) a k t < ε for all k n + N. (1.7) Proposition 1.1. Given a R N, there is at most one t R such that a converges to t. If this is the case, we write t = lim a = lim k a k. (1.8) The assertion that a given sequence a converges is often tacitly implied when lim a or lim a k is written down. Of course, these symbols make no k sense when it is not known that a converges. Proposition 1.2. For every ξ R, the constant sequence (ξ k N ) converges to ξ: lim(ξ k N ) = ξ. (1.9) We use the notation Conv R := { } a R N a converges (1.10) and we define the mapping lim : Conv R R in the obvious manner. Proposition 1.3. Conv R is a subring and a subspace of R N and lim : Conv R R is both a ring-homomorphism and a linear mapping. 2

3 Proof. It is not very difficult to show that Conv R is stable under (termwise) addition and that lim preserves sums. Now let a,b Conv R be given and put Note that and hence s : lim a, t := lim b. (ab) k st = a k b k st = (a k s)b k + s(b k t) (ab) k st a k s b k + s b k t for all k N (1.11) Now let η,ζ P be given. By (1.7) of Def. 4, we can determine m,r N such that a k s < η for all k m + N, and b k t < ζ for all k r + N. It follows that b k < t + ζ for all k r + N and, using (1.11), that (ab) k st η( t + ζ) + s ζ for all k n + N, (1.12) where n := max{m, r}. Now let ε P be given. We use (1.12) with the choices ζ := ε if s 0 2 s and ζ := 1 if s = 0 and ζ := 1 if s = 0 and η := ε, and then with m 2( t +ζ) and r and hence n := max{m, r} determined accordingly. We obtain (ab) k st < ε for all k n + N. Since ε P was arbitrary, it follows that ab converges to st according to (1.7) of Def. 4. Since a, b Conv R were arbitary, it follows that Conv R is stable under (termwise) multiplication and that lim preserves products. Let ξ R be given. By (1.1) and (1.2) we have ξa = (ξ k N )a for all a R N which shows that scalar multiplication in R N can be reduced to termwise multiplication with a constant sequence. Using Prop. 2 it follows that ξa Conv R and lim(ξa) = ξ lim a for all a Conv R. Since ξ R was arbitrary, it follows that Conv R is stable under scalar Multiplication and that lim preserves scalar multiplication. 3

4 Using (1.9) in the particular cases when ξ := 0 and ξ := 1, it follows that Conv R contains the zero and the unity of R N and that lim preserves zero and unity. Proposition 1.4. If a give sequence a is isotone and bounded above [antitone and bounded below] then a converges and we have lim a = sup Rng a [lim a = inf Rng a] (1.13) Proof. Assume that a R N is isotone and bounded above. Then Rng a is not empty and bounded above and hence has a supremum because R is a complete ordered field. Put t := sup Rng a. Let ε P be given. Since t ε < t we have t ε Ub Rng a. Hence, we may choose n N such that t ε < a n. Since a is isotone, we have a n a k and hence t ε < a k for all k n + N. On the other hand, since t Ub Rng a, we have a k t for all k N and hence, t ε < a k < t + ε for all k n + N. (1.14) Since ε P was arbitrary and since (1.14) is equivalent to a k t < ε for all k n + N, it follows by Def. 4 that a converges to g. Example: Consider the sequence a := 1 k! n N. (1.15) k n [ It is clear that a n+1 a n = n! > 0 for all n N and hence that a is isotone. It is easily seen that k! s k 1 for all k N and hence, that 4

5 a n = k n [ k! 1 + k (n 1) ] 2 k 1 = 1 + k (n 1) [ 2 k = 1 + (1 2 )n = 3 for all n N 2 Hence, a is bounded and sup Rng a 3. By Prop. 3, it follows that a converges. Moreover, since a 3 = = 2.5 sup Rng a, it follows that 2 (1.13) that (Actually, e ) 2.5 e := lim a 3. (1.16) Proposition 1.5. Let a,b Conv R be given. If there is n N such that then lim a lim b. a k b k for all k n + N (1.17) Proof. Assume that n N is such that (1.17) holds. Let ε P be given. Since a and b converge, we may choose m,r N such that a k lim a < ε for all k m + N (1.18) b k lim b < ε for all k r + N. It follows from (1.18) and (1.17) that lim a = (lim a) a k + a k ε + a k ε + b k = ε + (b k lim b) + lim b 2ε + lim b k (1.19) for all k max{n,m,r}. Since ε P was arbitrary, we conclude that lim a lim b. The following is an immediate consequence of Prop. 4. 5

6 Proposition 1.6. Let ξ,η R and a Conv R be given. If there is n N such that then ξ lim a η. ξ a k η for all k n + N (1.20) Proposition 1.7. Let a Conv R be given and let b be a subsequence of a. Then b Conv R and lim b = lim a. Proof. By Def. 1, we may choose a strictly isotone mapping σ : N N such that b = a σ. Put t := lim a. Let ε P be given. By Def. 4 we may choose n N such that a > (n+n) [t ε,t+ε[. Since σ is strictly isotone, it is easily seen that σ(n) n and σ > (n + N) n + N. It follows that b > (n + N) = a > (σ > (n + N)) a > (n + N) ]t ε,t + ε[. Since ε P was arbitrary, we conclude that b converges to t. 2 Cluster points, fundamental sequences Definition 1: We say that a given sequence a R N clusters at a given t R, and that t is a cluster point of a, if for every ε P and every n N we have or, equivalently, a > (n + N) ]t ε, t + ε[ (2.1) a k t < ε for some k n + N. (2.2) Proposition 2.1. Let a R N and t R be given. Then a clusters at t if and only if some subsequence of a converges to t. Proof. Assume that a clusters at t. We define the mapping σ : N N by recursive choice as follows: We put σ(1) := 1. Let n N be given and assume that σ(n) has been determined. Applying (2.2) to the case when ε := and when n is replaced by σ(n)+1, we may choose σ(n+1) σ(n)+1+n n 6

7 such that a σ(n) t <. It is easily seen that the mapping σ obtained in n this way is strictly isotone and that a σ converges to t. Now assume that a σ converges to t when σ : N N is a suitable strictly isotone mapping. Let ε P and n N be given. By Def. 4 of Sect. 1, we may determine m N such that (a σ) > (m+n) ]t ε,t+ε[. Since σ is strictly isotone, σ > (m + N) cannot be bounded and hence σ > (m + N) (n + N). It follows that a > (n + N) a > (σ > (m + N)) a > (n + N) ]t ε,t + ε[. Since ε P and n N were arbitrary, it follows that from (2.1) of Def. 1 that a clusters at t. Proposition 2.2. If a is a convergent sequence then a has exactly 1 cluster point, namely lim a. Pitfall: A sequence may have exactly one cluster point without being convergent. For example, 0 is the only cluster point of the sequence. ((1 + ( 1) n )n n N ), yet this sequence fails to converge. Proposition 2.3. Monotone unbounded sequences do not have cluster points. Cluster Point Theorem Every bounded sequence has at least one cluster point. Lemma 2.1. Every sequence in R has a monotone subsequence. Proof. Let a R N be given and put There are two cases: S := {k N a j a k for all j k + N} (2.3) (i) S is an infinite set. Since every non-empty subset of N has a minimum, we then can define σ : N N recursively by σ(1) := min S and σ(n + 1) := min(s σ(n) + N ) for all n N. It is easily seen that σ is strictly isotone and that a σ is isotone. 7

8 (ii) S is a finite set. Then S has a maximum and {j k + N a j < a k } for all k maxs + N We define σ : N N recursively by σ(1) := maxs + 1 and σ(n + 1) := min{j σ(n) + N a j < a σ(n) } for all n N It is easily seen that σ is strictly isotone and that a σ is (strictly) antitone. Proof of Theorem: Let a bounded sequence a be given. By the Lemma, we may choose a monotone subsequence b of a. It is clear from Def. 1 of Sect. 1 that Rng b Rng a and hence from Def. 3 of Sect. 1 that b is bounded. By Prop. 4 of Section 1 b converges and by Prop. 1 lim b is a cluster point of a. Definition 2: We say that a given sequence a R N for every ε P there is an n N such that is fundamental if, or, equivalently, a > (n + N) a > (N) ] ε,ε[, (2.4) a k a j < ε for all k,j n + N. (2.5) Basic Convergence Criterion: A sequence converges if and only if it is fundamental. Proof. Let a R be given. First assume that a converges and put t := lim a. Let ε P be given. By Def. 4 of Sect. 1 we may determine n N such that a k t < ε for all 2 k n + N. It follows that a k a j = (a k t) (a j t) a k t + a j t < ε 2 + ε 2 = ε. for all j,k n + N. Since ε P was arbitrary, it follows that a is fundamental. Now assume that a is fundamental. Using Def. 2 with ε := 1 we may choose m N such that, by (2.4), 8

9 a > (m + N) a m a > (m + N) a > (m + N) ] 1, 1[ and hence a > (m + N) a m +] 1, 1[. It follows that Rng a = a > ((m 1) ] ) a > (m + N) a > ((m 1) ] ) (a m +] 1, 1[) and hence that Rng a is bounded. By the Cluster Point Theorem, we may choose a cluster point t of a. Let ε P be given. Using Def. 2 with ε replaced by ε we can determine n 2 N such that a k a j < ε for all k,j n + N. 2 Since t is a cluster point of a, by Def. 1 we can fix j n + N such that a j t < ε. We conclude that 2 a k t = a k a j + a j t a k a j + a j t < ε 2 + ε 2 = ε for all k n + N. Since ε P was arbitrary, it follows that a converges to t. 3 Nullsequences, Sum-sequences A sequence is called a nullsequence if it converges to zero. Thus, the set of all nullsequences is Null lim, the nullspace of the linear mapping lim : Conv R R (see Prop.3 of Sect. 1). Hence Null lim is a subspace of Conv R, i.e, termwise sums and scalar multiples of nullsequences are again nullsequences. A given sequence a R N converges to a given t R if and only if ((a k t) k N ) is a nullsequence. Since lim : Conv R R is also a ring-homomorphism, it follows that Null lim is an ideal in Conv R, i.e., that the termwise product of a nullsequence and a convergent sequence is again a nullsequence. A stronger result than this is the following: Proposition 3.1. The termwise product of a nullsequence and a bounded sequence is a nullsequence. Proposition 3.2. Let a R N and a nullsequence b with terms in P be given. If a k b k for all k N, then a is also nullsequence. Proposition( 3.3. If a given ) sequence a (P ) N is isotone and not bounded 1 above, then k N is a nullsequence. The sequence a itself has no a k cluster points and hence fails to converge. 9

10 Since R is a archimedean field, the sequence (n n N ) is not bounded above. Hence ( 1 ( n N ) is a nullsequence by Prop. 3. More generally, n 1 n N ) is a nullsequence for every p P. n p Definition 1: The mappings ssq : R N R N and dsq : R N R N are defined by (ssq a) n := k n ] a k for all n N (3.1) and a 1 if n = 1 (dsq a) n := for all n N (3.2) a n a n 1 if n > 1 for all a R N. Given a R N we call ssq a the sum-sequence of a and dsq a the difference-sequence of a. Proposition 3.4. The mappings ssq and dsq are linear isomorphisms and we have dsq = ssq and ssq = dsq. The following result is an easy consequence of the Basic Convergence Criterion of Sect. 2. Proposition 3.5. Let a R N be given. The sum-sequence of a converges, i.e., ssq a Conv R, if and only if for every ε P there is an n N such that a k < ε for all p n + N and q p + N. (3.3) k p...q Proposition 3.6. Let a R N and b P N be given such that ssq b converges. If there is an n N and a c P such that a k cb k for all k n + N then ssq a converges also. Proof. This is an easy consequence of Prop

11 Proposition 3.7. Given a R N, if ssq a converges then a is a nullsequence. Equivalently, dsq > (Conv R) = ssq < (Conv R) Null lim. (3.4) Proof. Assume that ssq a converges. Let ε P be given and determine n N. Since ε P was arbitrary, a converges to zero according to Def. 4 of Sect. 1. Pitfall: The inclusion (3.4) is strict, i.e., there are nullsequences ( whose sum-sequences do not converge. An example is the nullsequence a := k k N. Indeed, we have k (n+1)...2n k n 2n = 2 for all n N. Hence there is ε P, namely ε := 2, such that for all n N there is p n + N, namely p := n + 1, and there is q p + N, namely q := 2n, such that the inequality (3.3) fails to be valid. By Prop. 5, ssq a fails to converge. Proposition 3.8. Let a P N be given. If a is an antitone nullsequence, then ssq (( 1) k+1 a k k N ) converges. Proof. For each r N, consider the statement P(r) given by 0 j r [ ( 1) j a k+j a k for all k N. (3.5) The statement P(0) is trivially valid. Now let r N be given and assume that P(r) is valid. Let k N be given. Using (3.5) with k replaced by (k + 1) we get 0 j r [ ( 1) j a k+1+j = j r ] ( 1) j 1 a k+j a k+1 11

12 and hence a k a k + j r ] ( 1) j 1 a k+j = Therefore we have a k a k+1 j (r+1) [ ( 1) j a k+j a k+1 a k. j (r+1) [ ( 1) j a k+j a k. (3.6) Since a is antitone we have a k a k+1 0 and, since k N was arbitrary, it follows from (3.6) that the statement P(r + 1) is valid. Since r N was arbitrary we can conclude, by induction, that (3.5) holds for all r N. Now let ε P be given. Since a is a nullsequence we can determine n N such that a k < ε for all k n + N. By (3.5), we conclude that 0 j r [ ( 1) j a k+j ε for all k n + N and r N. The assertion now follows from Prop Summability Let an index set I be given. Definition 1: Given a P I and given J Sub I, we put a := sup { a K Fin J} P = P { } (4.1) J K and call it the sum of a over J. In particular, we call a = a the I sum of a and we say that a is summable if a <. It is clear that for all a P I and all J,J Sub I, we have a = a J (4.2) and 12 J

13 J J = J a J a. (4.3) Proposition 4.1. The usual summation rules for sums over finite families in a monoid extent to sums over infinite families in P as defined in Def. 1, with sums in the additive monoid P. For example, given a P I, we have a = a j + a for all J Sub I,j J, (4.4) J J\{j} a = a + J K J\Ka for all J,K Sub I, (4.5) a π = a for all π Perm I. (4.6) The set P I of all families indexed on I with values in P has the structure of an additive monoid, whose addition is defined by termwise addition and whose zero is the family (0 i I). Also, P I has a natural order defined termwise, i.e., by a b : (a i b i for all i I) (4.7) for all a,b P I. Of course, P also has the structure of an additive monoid and it is naturally totally ordered. Proposition 4.2. The mapping (a a) : P I P (4.8) is a monoid-homomorphism and it is isotone. In particular, given a,b P I such that a b, if b is summable so is a, and if a fails to be summable, so does b. Proposition 4.3. If a given a P I is summable, then Supt a is countable. Proof. Put s := a. Assume that a is summable, i.e., that s P. By Def. 1 we may choose, for each n N, a finite subset K n of I such that s n K n a 13

14 and hence, since s = I\K n a + K n a by (4.3), I\K n a n. (4.9) Put H := n N K n. Since I\K n I\H for all n N, it follows from (4.9) and (4.3) that a 1 n for all n N and hence that = 0. We conclude I\H I\Ha that a I\H = 0 and hence that Supt a H, being the union of a countable family of finite sets, is countable and hence Supt a is countable. As we pointed out in Sect. 1, R I has the natural structure of a linear space. It also has a natural order defined termwise, i.e., by (4.7) for all a,b R I. This order in R I is a lattice order and we have (sup {a,b}) i = max{a i,b i } for all i I (4.10) (inf {a,b}) i = min{a i,b i } for all a,b R I. For every a R I, we put a + := sup{a, 0} P I, a := inf{a, 0} P I, (4.11) and we define a P I termwise, i.e., by We then have a i = a j for all i I. (4.12) a = a + a, a = a + + a. (4.13) Definition 2: We say that a given a R I is summable if both a + P I and a P I are summable in the sense of Def. 1. If this is the case, we put a := a + a R (4.14) and call it the sum of a. We denote the set of all summable families indexed on I with values in R by Sbl (I). The following result is evident from (4.13) and Defs. 1 and 2: 14

15 Proposition 4.4. A given a R I is summable if and only if a P I, as defined in (4.12), is summable. Proposition 4.5. The set Sbl (I) is a subspace of the linear space R I and the mapping is linear and isotone. (a a) : R I R (4.15) Proof. It is easily seen that 0 Sbl (I), that Sbl (I) is stable under scalar multiplication, and that the mapping (4.15) preserves scalar multiplication. It is also easily seen that the mapping (4.15) is isotone. The fact that Sbl (I) is stable under addition and that the mapping (4.15) preserves sums is an easy consequence of the following Lemma 4.1. Let a,b P I be given. If a and b are summable, so is a b R I and we have (a b) = a b. Proof. Assume that a and b are summable and put K := {i I a i > b i } (4.16) We then have, for all i I, a i b i if i K (a b) + i = a i (4.17) 0 if i I\K and b i a i if i I\K (a b) i = b i (4.18) 0 if i K Using Prop. 2, we conclude that (a b) + and (a b) are summable and using (4.5), we have (a b) + = (a b) K, (a b) = (b a) I\K and hence, by Def. 2 15

16 (a b) = (a b) K (b a) I\K (4.19) Since (a b) K P K and (b a) I\K P I\K we can apply Prop. 2 to conclude that (a b) K + b K = a K (4.20) and (b a) I\K + a I\K = b I\K. (4.21) If follows from (4.19) - (4.21) and (4.2) that (a b) = a + K I\Ka K b + I\Kb and hence, by (4.5), that the assertion is valid. Proposition 4.6. A given a R I is summable with sum s if and only if, for every ε P, there is K FinI such that a s < ε for all J Fin I with K J. (4.22) J Proof. Assume that a is summable and let ε P be given. Bu Def. 2 and Def. 1, we can then determine K +,K Fin I such that a + ε < K + a +, a ε < Put K := K + K. Let J Fin I with K J be given. Then and hence, by (4.23) a + a + a +, K + J K a. (4.23) a a a K J a + ε < a + a +, J a J 16 a < ( a ε). (4.24)

17 Since s = a + a, and J a + J a = J a, we conclude, by adding the inequalities (4.24), that s ε < J a < s + ε, which shows that (4.22) holds. Now assume that a satisfies the condition of Prop. 6. Let ε P be given and determine K Fin I such that (4.22) holds. Let H Fin I be given. We put J := (H Supt a + ) K. Since Supt a + Supt a =, we have a = J K a and J a + H a +. Hence, using (4.22) and the fact that a + = a J J J a + a a < ε + s a. H J K K a we obtain Since H Fin I was arbitrary, we conclude by Def. 1 that a + is summable. Applying the argument just given to the case when a is replaced by a and s by s, we conclude that a = ( a) + is also summable. Hence, by Def. 2, a is summable. It is easily seen that there can be at most one s R such that condition of Prop. 6 is satisfied and hence that s = a. We now deal with the special case when I := N, i.e., with sequences. Proposition 4.7. A given sequence a R N is summable if and only if the sum-sequence ssq a of a P N converges. If this is the case, then ssqa converges and a = lim ssq a (4.25) Proof. In view of Prop. 4 and (4.14), the desired result is an easy consequence of the following Lemma 4.2. A sequence a P N is summable if and only if ssq a converges. If this is the case, then (4.25) holds. 17

18 Proof. Let a P N be given. Assume that a is summable and put s = a. Let ε P be given. By Def. 1 we may determine K Fin N such that s ε < J a s for all J Fin I with K J. (4.26) Put n := maxk. Then K k ] for all k n + N and hence, by (3.1), we have s ε < (ssq a) k s for all k n + N. Since ε P was arbitrary, ssq a converges to s according to Def. 4 of Sect. 1 and (4.25) holds. Now assume that ssq a converges and put s := lim ssq a. Since ssq a is an isotone sequence it follows from (1.13) that Since, for every K Fin N, s = sup { (ssq a) k k Fin N }. (4.27) a (ssq a) max K s, we conclude that a summable according to Def. 1. We add here two results without proof. K Proposition 4.8. A sequence a R N is summable if and only if, for every σ Perm N, ssq (a σ) converges. If this is the case, then a = lim ssq (a σ) for all σ Perm N. (4.28) Proposition 4.9. Let a R N be a sequence which fails to be summable but has a convergent sum-sequence ssq a. For every s R we can then find σ Perm N such that ssq (a σ) converges and s = lim ssq (a σ). (4.29) 18

19 5 Applications and examples Proposition 5.1. Let x R be given. The sequence g x = (x k k N) has the following properties: (i) If x < 1 then g x is a nullsequence. (ii) If x > 1 then g x has no cluster points and hence fails to converge. (iii) g 1 is the constant sequence with term 1 and hence converges to 1. (iv) g 1 has the cluster points 1 and 1 and hence fails to converge. (v) If x < 1, then g x is summable and ssq g x = g x = k N x k = 1 1 x (5.1) (vi) If x 1, then g x fails to be summable and ssq g x has no cluster points and hence ssq g x fails to converge. Proof of (i): Assume first that 0 < x < 1. Then g x is easily seen to be antitone. Since it is also bounded below by 0 it must converge by Prop. 4 of Sect. 1. Denote the limit by s. By (1.13), we have 0 s x k for all k N. Suppose that s 0. Then s x 1 = x and hence s x xk for all k N. Thus s x inf Rng g x = s by (1.13), which is inconsistent with 0 < x < 1. We conclude that s = 0. If 1 < x < 0, then g x = g x and the assertion follows by appklication of Prop. 2 of Sect. 3. The case when x = 0 is trivial. Proof of (v): Using simple algebra, we find that (1 x) k n [ x k = (1 x n ) for all n N. (5.2) Now assume that x < 1. Then (ssq g x ) n = k n [ x k = 1 xn 1 x for all n N 19

20 Since (x n n N) is a nullsequence by (i), we can use Prop. 3 of Sect. 1 to conclude that ssq g x converges to (5.1). The remaining assertions follow from Prop. 7 of Sect. 4. The proofs of (ii), (iii), (iv) and (vi) are easy. Proposition 5.2. Let a sequence a R N or a R N there is n N and cion]0, 1[ such that either be given. Assume that or a k 0 and k ak c for all k n = N. (5.3) Then a is summable and henced ssq a is convergent. a k+1 a k c for all k n + N. (5.4) Proof. Let n N and c ]0, 1[ be given. Assume that (5.3) holds. Then a k c k for all k n + N. Since ssq(c k k N) converges by Prop.1, we can use Prop. 6 of Sect. 3 to conclude that ssq a converges. By Prop. 7 of Sect. 4 a is summable and ssq a is convergent. Assume now that (5.4) holds. It then follows immediately, by induction, a n+j c j a j for all j N and hence a k a n c n ck for all k n+n. Using the same reasons as in the case when (5.3) holds, we conclude that a is summable and hence ssq a convergent. Proposition 5.3. The sequence e x := x R. We define exp : R R by ( ) x k k! k N is summable for all exp(x) := k N x k k! = e x for all x R. (5.5) The sequence ((1 + x n )n n N ) converges for all x R and we have exp(x) := lim n (1 + x n )n for all x R. (5.6) 20

21 Proof. Let x R be given. It is easily seen that (e x ) k+1 (e x ) k = x for all k N. k + 1 Since R is archimedean, we can determine n N x such that x n. Then (e x ) k+1 (e x ) k n k + 1 n for all k n + N. n + 1 Therefore, the condition (5.4) is satisfied for a := e x and c := n ]0, 1[, n + 1 and we conclude from Prop. 2 that e x is summable. To prove the second part of Prop.2, we assume that x R and ε P are given. By the Binomial Theorem, we have ( 1 + x n) n = = = k (n+1) [ k (n+1) [ k (n+1) [ ( n k) ( x n ( (x n ( x k ) k ) ) k 1 k! Π (n j) j k [ k! Π j k [ ( 1 j )) n for all n N. (5.7) By (5.5), we can determine m N such that xk k! exp(x) < ε for all p m + N. (5.8) 3 k (p+1) [ ( ) x k Since ssq k N converges by the first part of Prop. 2, we can use k! Prop. 5 of Sect. 3 to determine m N such that x k k! < ε for all p m + N and n p + N. (5.9) 3 k (p+1)...n 21

22 We now put p := max{m,m ( }. Then(5.8) and (5.9) are both valid for this value of p. Since 0 1 j ) 1 for all n N and j (n + 1) [ we have, n by (5.9), ( ( n)) x k k! Π 1 j j k [ k (p+1)...n x k k! k (p+1)...n < ε for all n p + N. 3 ( ) (5.10) j Noting that n n N is a nullsequence for every j N, we can use Prop. 3 of Sect. 1 to conclude that ( ( x k lim n k! Π 1 j )) = xk j k [ n k!. k (p+1) [ k (p+1) [ Hence we can determine p N such that k (p+1) [ ( x k k! Π j k [ ( 1 j )) n xk k! k (p+1) [ < ε 3 for all n p + N. and therefore, by (5.8), such that k (p+1) [ ( x k k! Π j k [ ( 1 j )) n exp(x) < 2ε 3 for all n p + N. (5.11) Since a sum over (n+1) [ can be split into a sum over (p+1) [ and a sum over (p + 1)...n when n p + N, it follows from (5.7), (5.11), and (5.10) that ( 1 + n) x n exp(x) < ε for all n max{p,p } + N. Since ε P was arbitrary, we conclude that (5.6) holds. 22

23 Proposition 5.4. Let a P be given. Then the sequence (n a n N ) is an antitone nullsequence and we have if a 1 n a = (5.12) n N < if a > 1 The sequence (( 1) n 1 n a n N ) has a convergent sum-sequence for all a P but is summable only if a > 1. Proof. The first part of Prop. 4 will be proved later. The second part is an immediate consequence of the first part and Prop. 8 of Sect. 3. We define ζ : 1 + P R (the Riemann Zeta-function ) by ζ(a) = n N n a for all a 1 + P. (5.13) Proposition 5.5. Put F := Q ]0, 1] and define, for each r F, den(r) to be the denominator of the reduced fraction representing r, so that r = n den(r) for some n r ] relative prime to den(r). (5.14) Let r F and a P be given. Then σ(a) := 1 (den(r)) = a r F Proof. We consider the sets if a 2 ζ(a 1) ζ(a) if a > 2 (5.15) and, for each k N, P := { (n,d) (N ) 2 n d } (5.16) Q k := {(n,d) P gcd(d,n) = k}. (5.17) ( It is clear that (n,d) n ) : Q 1 F is invertible and that d 23

24 Also, we have kq 1 = Q k and hence, by (5.18), σ(a) = 1 da. (5.18) (n,d) Q 1 (n,d) Q k 1 d a = 1 k aσ(a) for all k N. (5.19) On the other hand, {Q k k N } is a partition of P and hence, by (5.19), 1 d = 1 = σ(a) 1 a d a k. (5.20) a (n,d) P k N (n,d) Q k k N On the other hand, since #{n N (n,d) P } = d for every d N, we have (n,d) P 1 d = d 1 a d = 1 a da 1. (5.21) d N d N If a > 2, we infer from (5.20), (5.21), and Prop. 4 that σ(a)ζ(a) = ζ(a 1). If a = 2 and hence a 1 = 1, we( find that σ(2)ζ(2) ) = and hence σ(2) =. 1 Since the terms of the family r F increase when a decreases, it den(r) a follows that σ(a) = if a 2. 6 Continuity Definition 1: Let D,C Sub R and a function f : D C be given. We say that f is continuous at a given t D if for every ε P there is δ P such that or, equivalently, f > (]t δ,t + δ[ D) ]f(t) ε,f(t) + ε[ (6.1) ( s t < δ = f(s) f(t) < ε) for all s D. (6.2) We say that f is continuous if f is continuous at every t D. Clearly, all constant functions are continuous. 24

25 Adjustments preserve continuity. In particular, if a given f : D C is continuous at t D and if A,B Sub R satisfy A D, t A, and f > (A) B, then f B A is continuous at t. However, if f B A is continuous at t, it does not necessarily follow that f is continuous at t. Proposition 6.1. Let D,C,E Sub R, functions f : D C and g : C E, and t D be given. If f is continuous at t and if g is continuous at f(t) then g f is continuous at t. Proof. Let ε P be given. Since g is continuous at f(t) we may determine ε P such that g > (](f(t) ε, f(t) + ε [ C) ]g(f(t)) ε, g(f(t)) + ε[. (6.3) Since f is continuous at t we may determine δ P such that f > (]t δ, t + δ[ D) ]f(t) ε, f(t) + ε [ C. (6.4) Using the fact that (g f) > = g > f > and the fact that g > : Sub C Sub E is isotone, we conclude from (6.4) and (6.3) that (g f) > (]t δ, t + δ[ D) ](g f)(t) ε, (g f)(t) + ε[. Since ε P was arbitrary, g f is continuous at t. Corollary 6.1. Composites of continuous functions are continuous. It is easily seen that the absolute value function : R R is continuous. Let f : D R be a continuous function. By the Corollary, the value-wise absolute value f := f is then also continuous. Let D Sub R be given. Recall that Fun D := Map (D, R) has the natural structure of a commutative ring whose addition, opposition, zero, unity, and multiplication are defined value-wise. Similarly, Fun D has the natural structure of a linear space. Given t D, we use the notation Cont t D := {f Fun D f is continuous at t}, (6.5) 25

26 and we put Cont D := t D Cont t D. (6.6) so that Cont D = {f Fun D f is continuous} (6.7) Proposition 6.2. Let t D be given. Then Cont t D is a subring and a subspace of Fun D. Also, Cont D is a subring and a subspace of Cont t D and hence of Fun D. The identity function ι := 1 R (6.8) and its adjustments are easily seen to be continuous. Put Pol R := Lsp {ι n n N} (6.9) It follows that from Prop.2 that Pol R is a subring and a subspace of Cont R. The members of Pol R are called polynomial functions. The sequence (ι n n N) is a basis of Pol R, i.e., is an invertible linear mapping. (c k Nc i ι k ) : R (N) Pol R (6.10) Proposition 6.3. Let D,C Sub R, a function f : D C, and t D be given. Then f is continuous at t if and only if, for every sequence a D N that converges to t, the sequence f a C N converges to f(t). Proof. Assume that f is continuous at t D. Let a D N be given and assume that lim a = t. Let ε P be such that (6.1) holds. By Def. 4 of Sect. 1, we may determine n N such that a > (n+n) ]t δ,t+δ[. Since a > (n + N) D and (f a) > = f > a >, we conclude that (f a) > (n + N) f > (]t δ,t + δ[ D) ]f(t) ε,f(t) + ε[. Since ε P was arbitrary, it follows that f a converges to t. 26

27 Now assume that f fails to be continuous at t. By Def. 1, we may choose ε P such that, for every δ P, there is s D such that s t < δ but f(s) f(t) > ε. In particular, for every n N, we may choose a n D such that a n t < 1 n but f(a n) f(t) > ε. It is clear that the sequence (a n n N ) obtained in this way converges to t but that f a = (f(a n ) n N) fails to converge to f(t). Corollary 6.2. A given function f : D C, with D,C continuous if and only if Sub R, is f(lim a) = lim(f a) (6.11) for every convergent sequence a D N with lim a D. Proposition 6.4. Let a function f : D C, with D,C Sub R, be given and assume that f is continuous at a given t D. Then, for every c R satisfying f(t) > c, there is δ P such that f ]t δ,t+δ[ > c (valuewise). (6.12) Intermediate Value Theorem: Let a,b, R with a < b and f Cont [a,b] be given. Then for every c [f(a),f(b)] there is t [a,b] such that f(t) = c. Proof. We may assume, without loss, that f(a) f(b). Let c [f(a),f(b)] be given. We put S := {s [a,b] f(s) c}. (6.13) S is bounded by b and S is not empty because a S. Hence S has a supremum t := sup S [a,b]. Suppose that t S and hence, by (6.13), that f(t) > c f(a). Since t a and hence t > a, we may apply Prop. 4 and choose δ ]0,t a] such that f ]t δ,t] > c. In view of (6.13), we conclude that S ]t δ,t] =. 27

28 Since t Ub S, it follows that ]t δ,t] Ub S, which is inconsistent with t := min Ub S. We conclude that t S and hence, by (6.13), that f(t) c and t = maxs. (6.14) Now suppose that f(t) < c f(b). Since t b and hence t < b, we may apply Prop. 4 to f and c instead of f and c and choose δ ]0,b t] such that f [t,t+δ[ < c. Hence, by (6.13), we [t,t + δ[ S, which is inconsistent with (6.14) 2. We conclude that f(t) = c. Corollary 6.3. The image of an interval under a continuous function is again an interval. Theorem on Attainment of Extrema: Let a,b R with a < b and f Cont [a,b] be given. Then f attains both a maximum and a minimum, i.e., Rng f has a maximum and a minimum. In other words, there are c,d [a,b] such that Moreover, we have f(d) f(t) f(c) for all t [a,b]. (6.15) Rng f = [f(d),f(c)]. (6.16) Proof. We construct a sequence s [a,b] N as follows: Case (i): Rng f fails to be bounded above. For each n N, we then may choose s n [a,b] such that f(s n ) n. The sequence f s = (f(s n ) n N) then has no cluster points. Case (ii): Rng f is bounded above. We then put σ := sup Rng f and, for each n N, we choose s n [a,b] such that σ 1 n f(s n) σ. The sequence f s = (f(s n ) n N) then converges to σ. We now apply the Cluster Point Theorem of Sect. 2 and choose a cluster point c of s. Since Rng s [a,b], it is easily seen that c [a,b]. By Prop. 1 of Sect. 2 we may choose a subsequence r of s that converges to c. By Prop. 3 above, f r converges to f(c). Now, f r is a subsequence of f s. Thus, f(c) is a cluster point of f s. This is inconsistent with case (i) above. Hence case (ii) must obtain and we have f(c) = lim(f r) = σ = max Rng f. 28

29 Applying the argument above to f instead of f, we can determine d [a,b] such that f(d) = min Rng f. The result (6.16) is an immediate consequence of the Corollary to the Intermediate Value Theorem. 7 Limits of functions: Definition 1: Let S Sub R and t R be given. We say that t is an accumulation point of S if, for every ε P, we have or equivalently (]t ε, t + ε[\{t}) S 0 (7.1) 0 < s t < ε for some s S. (7.2) The set of all accumulation points of S will be denoted by Acc S. The memebers of S\Acc S will be called isolated points of S. If S is finite, then Acc S =. We have { } 1 Acc N = Acc Z =, Acc Q = R, Acc n n N = {0}. The following result is an easy consequence of the Cluster Point Theorem. Proposition 7.1. Every bounded infinite subset of R has at least one accumulation point. Definition 2: Let D,C R, a function f : D C and t Acc D be given. We say that f converges to a given c R at t if the function defined by is continuous at t. f : D {t} C {c} (7.3) { } f(s) if s D\{t} f(s) := c if s = t (7.4) We assume now that D,C Sub R, f : D C, and t Acc D are given. 29

30 Proposition 7.2. f converges to a given c R at t if and only if for every ε P there is δ P such that or, equivalently, f > ((]t δ, t + δ[\{t}) D) ]c ε, c + ε[ (7.5) 0 < s t < δ = f(s) c < ε for all s D. (7.6) Proposition 7.3. There is at most one c R such that f converges to c at t. If this is the case, we write c = lim t f = lim s t f(s). (7.7) Proof. Suppose that f converges to both c R and c R. Let ε P be given. By Prop. 2, we can choose δ,δ P such that 0 < s t < min(δ,δ ) = f(s) c < ε and f(s) c < ε for all s D. Since t Acc D, we can use (7.2) with ε replaced by min(δ,δ ) to fix s D such that 0 < s t < min(δ,δ ). We then have f(s) c < ε and c f(s) < ε and hence c c < 2ε. Since ε P was arbitrary, we conclude that c = c. The assertion that f converges at t is often tacitly implied when lim t f or limf(s) is written down. Of course, these symbols have no meaning when s t it is not known that f converges at t. Pitfall: The assumption that t Acc D is essential when deciding whether or not f converges at t. If we had left it out in Def. 2, then the conclusion of Prop. 3 would no longer hold. In fact, if t Acc D but Def. 2 would stay in force, one easily sees that f converges to every c R at t. The following result is an easy consequence of Def.2 and Prop.3 of Sect.6. Proposition 7.4. f converges to a given c at t if and only if, for every sequence a (D\{t}) N that converges to t, the sequence f a C N converges to c. 30

31 Proposition 7.5. f is continuous at a given r D if and only if either r is isolated, i.e. r Acc d, or r Acc D and lim r f = f(r). The following result is an easy consequence of Def.2 and Prop.1 of Sect.6. Proposition 7.6. Let D,C,E Sub R and functions f : D C and g : C E be given. Let t Acc D be given and assume that f converges at t. Then: (a) If lim t f C and if g is continuous at lim t f then g f converges at t and lim(g f) = g(lim f). (7.8) t t (b) If lim t f C then lim t f Acc C. If, in this case, g converges at lim t f then g f converges at t and lim(g f) = lim g. (7.9) t limt f Now let D Sub R and t Acc D be given. We use the notation Conv t D := {f Fun D f converges at t} (7.10) and define the mapping lim t : Conv t D R in the obvious manner suggested by Prop.3. Proposition 7.7. Conv t D is a subring and a subspace of Fun D and lim t is a ring-homomorphism as well as a linear form. Fun D has also a natural order defined value-wise, i.e. by f g := (f(t) g(t) for all t D). (7.11) Proposition 7.8. The mapping lim t : Conv t D R is isotone. Proposition 7.9. Let f,g,h Fun D be given such that f g h. If f and h converge at t and if lim t f = lim t h, then g also converges at t and lim t g = lim t f = lim t h. 31

32 8 Uniform Continuity Definition 1:We say that a given function f : D C, with D,C Sub R is uniformly continuous if, for every ε P, there is a δ P such that s t < δ = f(s) f(t) < ε for all s,t D. (8.1) It is evident from Def. 1 of Sect. 6 that uniformly continuous functions are continuous. But ι 2 : R R is an example of a function that is continuous but not uniformly continuous. All constants are uniformly continuous and so is ι. Adjustment preserve uniform continuity. Proposition 8.1. Composites of uniformly continuous functions are uniformly continuous. Let D Sub R be given. We use the notation UconD := {f Fun D f is uniformly continuous} (8.2) Proposition 8.2. Ucon D is a subspace of the linear space Fun D. Uniform Continuity Theorem: If the domain of a continuous function is a closed and bounded interval then the function is uniformly continuous. Proof. Let a closed and bounded interval [a,b] with a,b R and a b, and a function f : [a,b] R be given. Assume that f fails to be uniformly continuous. Then we may choose ε P such that, for every δ P we can determine s,t [a,b] such that s t < δ but f(s) f(t) ε. In particular, for every n N, we can determine c n,d n [a,b] such that c n d n < 1 n but f(c n) f(d n ) ε (8.3) The sequence c := (c n n N ) has terms in [a,b] and hence is bounded. By the Cluster Point Theorem and Prop. 1 of Sect. 2, a subsequence of c must converge, i.e. we can choose a strictly isotone mapping σ : N N such that c σ converges. Now, the subsequence d c of d is also bounded and hence must have a convergent subsequence, i.e. we can choose a strictly isotone mapping τ : N N such that d := (d c) τ converges. The sequence c := (c σ) τ, being a subsequence of c σ, converges by Prop. 7 of Sect. 1. It is easily seen that limc, lim d, [a,b]. It follows from (8.3) that 32

33 c d d n < 1 σ(τ(n)) and (f c) n (f d) n ε (8.4) for all n N. We conclude from (8.4) 1 that lim c = lim d and from (8.4) 2 that f c and f d, if they converge at all, cannot have the same limit. It follows from Prop. 4 of Sect. 7 that f cannot be continuous. 33