1. Summation. Let X be a set Finite summation. Suppose Y is a set and + : Y Y Y. is such that

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1 1. Summation. Let X be a set Finite summation. Suppose Y is a set and is such that + : Y Y Y (i) x + (y + z) (x + y) + z wheneve x, y, z Y ; (ii) x + y y + x wheneve x, y Y ; (iii) thee is 0 Y such that y + 0 y 0 + y wheneve y Y. Fo example, Y could be an belian goup o Y could be [0, ] whee + on [0, ) [0, ) is addition in the belian goup o R and whee y + + y wheneve y [0, ]. Deinition 1.1. Fo, g Y X we deine + g Y X by letting ( + g)(x) (x) + g(x) o x X and we note that appopiately eomulated vesions o (i),(ii) and (iii) hold. We let be such that 0(x) 0 o x X. Deinition 1.2. Fo Y X we let 0 : X Y spt {x X : (x) 0} and call this subset o X the suppot o. We let ( Y X ) 0 { Y X : spt is inite} and note that ( Y X) is closed unde addition. 0 Deinition 1.3. Wheneve X and Y X we let Y X be such that (x) { (x) i x, 0 i x X. Poposition 1.1. Suppose F is a inite subset o X. Thee is one and only one unction such that S F : Y X Y (i) S F (0) 0; (ii) S F () S( X {a} ) + (a) wheneve Y X and a ; (iii) S F ( + g) S F () + S F (g) wheneve, g Y X. 1

2 2 Poo. We deine S F by induction on F as ollows. We let S (0) 0. I F > 0 we let S F {(, S F {a} ( X {a} ) + (a)) : F F and a F }. It is obvious that S F is a unction i F 1. To veiy that S F is a unction in case F > 1 we suppose : F F, a, b F and a b and calculate S F {a} ( X {a} ) + (a) (S F {a,b} ( X {a,b} ) + (b)) + (a) S F {a,b} ( X {a,b} ) + ((b) + (a)) S F {a,b} ( X {a,b} ) + ((a) + (b)) (S F {a,b} ( X {a,b} + (a)) + (b) S F {b} ( X {b} ) + (b). We leave to the eade the staightowad veiication using induction on F that S F satisies (i)-(iii) Summing a unction with values in [0, ]. Fo each subset o X let 1 [0, ] X be such that { 1 i x, 1 (x) 0 i x X ; one calls 1 the indicato unction o. Note that p 1 p wheneve p [0, ] X. Deinition 1.4. Fo p [0, ] X we let p sup{sf (p) : F X and F is inite}. Theoem 1.1. We have (i) 0 0; (ii) 1 {a} 1 wheneve a X; (iii) cp c p wheneve 0 c and p [0, ] X ; (iv) (p + q) p + q wheneve p, q [0, ] X ; (v) p q wheneve p, q [0, ] X and p q. Poo. (i) and (ii) ae immediate. Let F be the amily o inite subsets o X. In what ollows we leave it to the eade to supply the simple poos o the popeties o S F, F F that we shall use. Suppose p, q [0, ] X and 0 c. Then cp sup S F (cp) sup cs F (p) c sup S F (p) c p F F F F F F so (ii) holds. Fo any F F we have S F (p + q) S F (p) + S F (q) p + q which implies that (p + q) p + q. Moeove, i F, G F we have S F (p) + S G (q) S F G (p) + S F G (q) S F G (p + q) (p + q). Thus (iv) holds.

3 3 Suppose p q. Fo any F F we have so (v) holds. S F (p) S F (q) q Suppose p [0, ] X. We will sometimes wite p o p(x) instead o x p. Coollay 1.1. Suppose p, q [0, ] X, p q and B X. Then p q. B Poo. Note that p q B. Example 1.1. Suppose 0 < 1. We deine p : N [0, 1) by letting Then (1) n p p(n) n o n N. sup{s F (p) : F is a inite subset o N} N sup{ n : N N} { 1 N+1 sup } : N N Poposition 1.2. Suppose is a patition o X and p [0, ] X. Then p p. Poo. One poves by induction on using (iv) o the peceding Theoem that the Poposition holds when is inite. Suppose F is a inite subamily o. Then p p p p F p. F F F Thus p p. Suppose F is a inite subset o X. Let F { : F }. Then pf p F pf p p. F F F Thus p p.

4 4 Deinition 1.5. Suppose B is a set and p : B [0, ] X. (Some would say p b, b B, is an indexed amily o [0, ] valued unctions with domain X.) We let ( p b X x ) p b (x) [0, ] [0, ] X. b B b B Poposition 1.3. Suppose p b, b B is an indexed amily o [0, ] valued unctions with a common domain X and X. Then p b. b B p b b B Poo. Let P (b, x) p b (x) o (b, x) B X. pply the pevious Poposition twice to P with thee equal {{b} X : b B} and {B {x} : x X}. Poposition 1.4. Suppose p [0, ] X and p <. Fo each ɛ > 0 thee is a inite subset F o X such that p < ɛ wheneve F X. X Poo. Suppose ɛ > 0. Let F be a inite subset o X such that p < p + ɛ. Since p p F + p X F we have p + p p F + p X F p < F X F F I F X we have p X p X F so F X p X F p < ɛ. p + ɛ. Poposition 1.5. Suppose p [0, ] X and p <. Then spt p is countable. Poo. Suppose n is positive intege and n {x X : p(x) 1/n}. Then 1 n np which implies n 1 n np n p < so n is inite. Thus spt p n1 n is countable.

5 Vecto valued summation. We now assume that V is a Banach space which, by deinition, means that V is a complete nomed linea space and, unde this assumption, extend the notion o summation when the suppot o is ininite. Deinition 1.6. Suppose V X and X. We say is summable ove i <. We say is summable i is summable ove X. I X and V X is summable then, as we ind that is summable ove. Evidently, { V X : i summable} is a linea subspace o V X. Theoem 1.2. Thee is one and only one linea unction : { V X : is summable} V such that F S F () i V X and F is a inite subset o X and wheneve V X and is summable. Poo. Let ρ() o V X. Then and ρ(c) c ρ() wheneve c R and V X ρ( + g) ρ() + ρ(g) wheneve, g V X. By induction on spt one shows that S F () S F ( ) ρ() wheneve V X and F is a inite subset o F. The Theoem now ollows by applying the bstact Closue Pinciple to ( V X ) 0 S spt (). Remak 1.1. n altenative appoach to deining when is summable is as ollows. Fo each positive intege ν let F ν {x X : (x) 1/ν}, note that F ν is inite and set y ν x F ν (x). Given ɛ > 0 thee is a positive intege N such that X F N < ɛ. Thus i µ, ν ae positive integes and µ, ν N we have y µ y ν x X F N (x) < ɛ whee we have used that act that F N F µ F ν. Thus y is a Cauchy sequence whose limit is.

6 6 Deinition 1.7. Wheneve V X, is a subset o X and is summable we let. Remak 1.2. Note that i is summable and ɛ > 0 thee is a inite subset F o X such that < ɛ wheneve F X. Theoem 1.3. Suppose V X, is summable and is a patition o X. Then and (1) (i) < o each ; (ii) < ;. Poo. We have < so (i) holds. We have < and (ii) holds. Suppose ɛ > 0. Let F be a subset o such that () F < ɛ 2. Let F { : F }; note that F is inite, F F and ( ) ( F) () F < ɛ 2. Using the act that ( m i1 g i) m i1 gi wheneve g 1,..., g m ae summable we ind that sup F F ( ) ( ). F F F

7 7 Thus + F F ( ) ( F) + F + 2 < ɛ. ( ) ( F) ( ) ( F) ; F hee we have used again that summation is initely additive and the vesion o the cuent Theoem in the case o nonnegative unctions The complex exponential unction. Wheneve 0 < < we let { } n M() sup : n N. Suppose 0 < <. Let N() be that intege such that N() < N() + 1. Wheneve n N and n > N() we have n n N() m N()! < N() N()! om which it ollows that mn()+1 M() N() N()! <. Suppose z C. Let be such that z < <. Then z n z n n so, by Example (1), z n M() Thus we may deine ( z ) n M() ( ) n z M() e z (One also wites exp(z) o e z.) I claim that z n. e z+w e z e w o z, w C. We pove this as ollows. Fix z, w C. Let and let T {(m, n) N N : m n} (m, n) zm w n m (n m)! ( z 1 1 z ) n o (m, n) T. <.

8 8 Fo each n N let n {m N : m n} and o each m N let B m {n N : m n}. Note that We have { n : n N} and {B m : m N} ae patitions o T. (m,n) T (m, n) so that, using moe taditional notation, (m,n) T z m w n m (n m)! m n (m, n) n m0 z m ( z + w ) n e z + w <. pplying the pevious theoem twice we ine that (m, n) (m, n) m n so that Thus n m0 (m,n) T w n m (n m)! e z+w m0 nm (z + w) n m0 n m0 nm m0 e z e w. ew m0 w n m (n m)! w n m (n m)! w n m (n m)! n B m (m, n) w n m (n m)!. Finally, let us show that e w e z (3) lim w z w z ez. Suppose z, w C and 0 < w z < 1. Then e w z 1 + (w z) + (w z) 2 n2 (w z) n 2

9 9 so Since (3) holds. e w z 1 w z 1 (w z) n 2 w z w z w z n 2 n2 1 w z 1 w z. e w e z w z ( e ez e z w z ) 1 1 w z n2

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