1550251515-Classical_Complex_Analysis__Gonzalez_

448 Chapter 7

Now any two admissible sequences s 1 and s 2 have a common refinement

sa = s1 U s 2. Since the result agrees for s1 and sa as well as for s2 and sa,

we conclude that the sum in (7.10-10) is the same for any two admissible

sequences s1 and s2.

Definitions 7 .8 Following Eberlein, consider the lattice

Lmn = {(j,k): j = 0, 1, ... ,m;k = 0,1, ... ,n}

m ~ 1, n ~ 1, and a complex function F: Lmn --+ C. Then define

(fj_xF)(j, k) = F(j + 1, k) - F(j, k) for (j, k) E Lm-1,n

and

(fj_yF)(j, k) = F(j, k + 1) - F(j, k) for (j, k) E Lm,n-1

Also, when F: Lm-1,n-1 --+ C define

n-1 m-1

ff Ftlx/).y = L ~ F(j,k)

L mn h=O J=O

For a pair of complex functions P, Q such that P: Lm-l,n --+ C and

Q: Lm,n-1 --+ C set

m-1 n-1 m-1 n-1

f P~1x+Q/).y = ~ P(j,O)+ LQ(m,k)-~ P(j,n)-LQ(O,k)

8Lmn J=O k=O J=O k=O

Lemma 7.3 (Discrete Green Theorem). We have

f P ~x + Q /).y = j f ( /).x Q - /).yP) /).x /).y

8Lmn Lmn

Proof A simple computation gives

m-1 n-1

• L: L[P(j, k + 1) - P(j, k)]
  j=O k=O
  n-1 m-1

= L[Q(m,k)-Q(O,k)]-L[P(j,n)-P(j,O)]

k=O j=O

m-1 n-1 m-1 n-1

= LP(j,0)+ LQ(m,k)-LP(j,n)-LQ(O,k)

j=O k=O j=O k=O