CAUCHY INTEGRAL THEOREM

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1 CAUCHY INTEGRAL THEOREM XI CHEN 1. Differential Forms, Integration an Stokes Theorem Let X be an open set in R n an C (X) be the set of complex value C functions on X. A ifferential 1-form is (1.1) ω = f 1 (x 1, x 2,..., x n )x 1 + f 2 (x 1, x 2,..., x n )x f n (x 1, x 2,..., x n )x n = f i x i for some f i C (X). Let { } (1.2) Ω X = Ω 1 X = f i x i : f i C (X). We can consier Ω X as a free moule generate by x 1, x 2,..., x n over the ring C (X). Let Ω X = C (X) an { } (1.3) Ω k X = k Ω X = fi x I : f I C (X), I {1, 2,..., n}, I = k where we use the notation x I for (1.4) x I = x i1 x i2... x ik for I = {i 1, i 2,..., i k } with i 1 < i 2 <... < i k. So Ω k X is a free moule over C (X) generate by x I. Obviously, Ω k X = for k > n an Ω k X is a grae ring (noncommutative without multiplicative ientity) with multiplication efine by the wege prouct (1.5) : (ω 1, ω 2 ) ω 1 ω 2. Note that (1.6) ω 1 ω 2 = ( 1) k 1k 2 ω 2 ω 1 for ω i Ω k i X. In particular, Ω 2k X is a commutative ring without multiplicative ientity. The wege prouct also gives us a pairing (1.7) Ω k X Ω n k X ω 1 ω 2 ω 1 ω 2 Ω n X = C (X). Date: January 18,

2 2 XI CHEN We have the ifferential operator : C (X) Ω X efine by (1.8) f = f x 1 x 1 + f x 2 x f x n x n for f C an can extene to Ω k X Ωk+1 X by (1.9) (f I x I ) = f I x I. So we can think as an operator on n (1.1) It satisfies Leibniz rule k= Ω k X n Ω k X. k= (1.11) (ω 1 ω 2 ) = ω 1 ω 2 + ( 1) k ω 1 ω 2 for ω 1 Ω k X an (1.12) 2 =. An (1.12) shows that inuces a complex (1.13) Ω X on Ω X with (1.14) im(ω k 1 X Ω 1 X Ω 2 X Ω k X) ker(ω k X... Ω n X Ω k+1 X ). Definition 1.1. Let X an Y be open sets in R n an R m, respectively. For a C function F : Y X, we have a ring homomorphism (1.15) F : C (X) C (Y ) sening f to f F an corresponingly a homomorphism of moules (1.16) F : Ω k X Ω k Y efine by (1.17) for F = (F 1, F 2,..., F n ). F (f I x i1 x i2... x ik ) = (f I F )F i1 (y 1, y 2,..., y m ) Clearly, F satisfies (1.18) (F G) = F G (F ω) = F (ω) F i2 (y 1, y 2,..., y m )... F ik (y 1, y 2,...y m ) for all C functions F 1 : Y X an F 2 : Z Y an all ω Ω X. For a measurable subset M X an an n-form ω Ω n X, we efine the integral (1.19) ω = fx 1 x 2...x n for ω = fx 1 x 2... x n. M M

3 CAUCHY INTEGRAL THEOREM 3 For a C map F : Y X, a measurable subset M Y an an m-form ω Ω m X, we have a well-efine integral M F M. Theorem 1.2 (Stokes Theorem on Cubes). Let D n = { x i 1 : i = 1, 2,...n} R n be the n-cube, F : D X be a C map an ω be an (n 1)-form on X. Then (1.2) (F ω) = F ω = ( 1) i (Fi,ω Fi,1ω) D n D n D n 1 where F i,t : D n 1 X are the maps given by (1.21) F i,t (x 1, x 2,..., x n 1 ) = F (x 1, x 2,..., x i 1, t, x i,..., x n 1 ). Proof. Let (1.22) F ω = Then (1.23) (F ω) = an (1.24) (F ω) = D n = = f i x 1 x 2... x i... x n. D n ( 1) i+1 f i x i x 1 x 2... x n ( 1) i+1 f i x i x 1 x 2... x n ( 1) i+1 f i x 1 x 2...x n D n x i ( 1) i (f i (x 1, x 2,..., x i 1,, x i+1,...x n ) D n 1 f i (x 1, x 2,..., x i 1, 1, x i+1,...x n ))x 1 x 2... x i...x n = ( 1) i (Fi,ω Fi,1ω). D n 1 Although F an ω are C in the theorem, we just nee them to be C 1 in orer for (1.2) to hol. For a C 1 map F an a C 1 form ω, we can erive this from the theorem by approximating F an ω using a sequence of C functions F m an forms ω m uner C 1 -norm. For approximating continuous functions by smooth functions, we quote Stone-Weierstrass Theorem:

4 4 XI CHEN Theorem 1.3 (Stone-Weierstrass Theorem). Let X be a compact Hausorff space, C(X, C) be the complex-value continuous functions on X an S be a subset of C(X, C) which separates points. Then the complex unital -algebra generate by S is ense in C(X, C). 2. Homotopy Version of Cauchy Integral Theorem Let X an Y be two topological spaces. We say two continuous maps f : X Y an g : X Y are homotopy if there exists a continuous function F : X [, 1] Y such that F (x, ) f(x) an F (x, 1) g(x). We consier a close curve : S 1 X, which is a continuous function from the circle R/Z to X. Alternatively, we can regar it as a continuous function : [, 1] X satisfying () = (1). Two close curves 1 : S 1 X an 2 : S 1 X are homotopic if there exists a continuous function F : S 1 [, 1] X such that F (t, ) 1 (t) an F (t, 1) 2 (t). Alternatively, we can regar F as a continuous function F : [, 1] [, 1] X satisfying (2.1) F (t, ) 1 (t), F (t, 1) 2 (t) an F (, s) F (1, s). For a continuous complex function f(z) on an open set G C an a C 1 curve : [, 1] G, we efine 1 (2.2) f(z)z = (f(z)z). We nee to be piece-wise C 1 (more generally rectifiable) in orer for the integral to converge. However, we can efine the integral for continuous curves by 1 (2.3) f(z)z = lim f(z)z = lim α (f(z)z) α α α if the limit exists for α C ([, 1], G), where α is the C norm. As it turns out, when f(z) is holomorphic, f(z)z only epens on the homotopy type of. So the right han sie (RHS) of (2.3) always converges for f(z) holomorphic. Theorem 2.1 (Cauchy Integral Theorem, Homotopy Version). Let f(z) be an analytic function on an open set G C with f (z) continuous. If 1 : S 1 G an 2 : S 1 G are two homotopic close curves on G, then (2.4) f(z)z = 1 f(z)z. 2

5 CAUCHY INTEGRAL THEOREM 5 Proof. Since f(z) is analytic, we have f/ z = an hence ω = (f(z)z) = (f(z)) z ( ) f f (2.5) = z + z z z z = f z z = z on G for ω = f(z)z. Let F : D 2 = [, 1] [, 1] G be a continuous function satisfying (2.1). If F is C, then by Stokes Theorem 1.2, = F (ω) = F (ω) = F (ω) D 2 D 2 D 2 (2.6) 1 1 = (F1,ω F1,1ω) + (F2,ω F2,1ω) ( 1 1 ) = f(f (, s))f (, s) f(f (1, s))f (1, s) ( 1 + f(f (t, ))F (t, ) = ω 1 ω. 2 1 ) f(f (t, 1))F (t, 1) By Stone-Weierstrass theorem 1.3, we can fin a sequence of C functions F n : S 1 [, 1] G such that (2.7) lim n F F n =. Let α 1,n (t) = F n (t, ) an α 2,n = F n (t, 1). Then (2.8) lim 1 α 1,n = lim 2 α 2,n = n n an (2.9) ω = α 1,n ω α 2,n for all n by (2.6). We claim that for every continuous close curve : S 1 G, there exists r > such that (2.1) f(z)z const for all C close curves β : S 1 G satisfying β < r. By the compactness of [, 1], we can fin r > such that (2.11) {z : z (t) < r} G β for all t [, 1]. For two C close curves β 1 an β 2 satisfying (2.12) β 1 < r an β 2 < r,

6 6 XI CHEN we have a C homotopy E : S 1 [, 1] G efine by (2.13) E(t, s) = sβ 1 (t) + (1 s)β 2 (t). So we have (2.14) ω = ω. β 1 β 2 Therefore, we have (2.1). Consequently, (2.15) ω = lim ω = lim ω. 1 β 1 β n α 1,n Similarly, (2.16) 2 ω = an (2.4) follows from (2.9). lim ω = lim ω β 2 β n α 2,n Corollary 2.2. Let f(z) be an analytic function on an open set D C. Then f(z)z exists for all continuous curves : [, 1] D. Proof. We have prove it for close. In case that () (1), we fin a C curve β : [, 1] D such that () = β() an (1) = β(1). We let { (2t) if t 1/2 (2.17) α = β(2 2t) if 1/2 < t 1 Then α is close an hence α f(z)z exists. Therefore, (2.18) f(z)z = f(z)z f(z)z exists. α We say that a topological space X is simply connecte if every close curve in X is homotopic to a fixe point p X. We say that two topological spaces X an Y are homotopic if there exists continuous function f : X Y an g : Y X such that g f is homotopic to 1 X an f g is homotopic to 1 Y. In other wors, f an g are homotopy inverse to each other. We say a topological space X is contractible if there is a point p X such that f : {p} X an g : X {p} are homotopy inverse to each other. Obviously, every contractible space is simply connecte. A set X R n is star-shape if there exists a point p X such that the line segment (2.19) pq = {(1 s)p + sq : s 1} X for all q X. A star-shape set X is contractible an hence simply connecte by the homotopy F : X [, 1] X (2.2) F (q, s) = (1 s)p + sq. β

7 CAUCHY INTEGRAL THEOREM 7 Every convex set is star-shape an hence simply connecte. We call 3. Homology Version of Cauchy Integral Theorem (3.1) n = {(t, t 1,..., t n ) : t k = 1, t k for k n} R n+1 k= the n-simplex. The k-th face α k : n 1 n is the map (3.2) α k (t, t 1,..., t n 1 ) = (t, t 1,..., t k 1,, t k,..., t n 1 ). For a topological space X, we let C n (X) be the free abelian groups generate by continuous maps f : n X, i.e., { a } (3.3) C n (X) = m i f i : m i Z, f i C( n, X). We have a group homomorphism : C n (X) C n 1 (X) inuce by (3.4) f = ( 1) k f α k k= for all f C( n, X). One can check that (3.5) 2 =. So efines a complex on C (X) an the n-th homology group of X is (3.6) H n (X) = ker(c n(x) C n 1 (X)). Im(C n+1 (X) C n (X)) Let G be an open set in C. For an analytic function f(z) on G an C 1 (X), we efine a (3.7) f(z)z = m i f(z)z i if = m i i with i : 1 = [, 1] G continuous. Theorem 3.1 (Cauchy Integral Theorem, Homology Version). Let f(z) be an analytic function on an open set G C with f (z) continuous. If C 1 (G) satisfies = an = in H 1 (G), then (3.8) f(z)z =. Proof. Since = an = in H 1 (G), by the efinition of H 1 (G), = σ for some σ C 2 (G). It suffices to show that (3.9) f(z)z = σ for all continuous maps σ : 2 G.

8 8 XI CHEN Let α : [, 1] 2 be the close curve (, 3t, 1 3t) for t 1/3 (3.1) α(t) = (3t 1, 2 3t, ) for 1/3 t 2/3 (3 3t,, 3t 2) for 2/3 t 1. Clearly, (3.11) σ f(z)z = σ α f(z)z. Since 2 is contractible, α is homotopic to a point in 2. Hence σ α is homotopic to a point in G. Then by the homotopy version of CIT, we have (3.12) f(z)z = f(z)z =. σ The avantage of the homology version of CIT is that it is very easy to verify two curves are homologous by triangularization. For example, let G = C\{z 1 z 2 }. It is easy to see that σ α (3.13) 1 = in H 1 (G) by Figure 1, where i are the circles (3.14) { z = R}, { z z 1 = r 1 } an { z z 2 = r 2 }, respectively, oriente counter-clockwise with r 1 +r 2 < z 1 z 2 << R. Then we can conclue (3.15) f(z)z = 1 f(z)z + 2 f(z)z 3 for all analytic functions f(z) on G. 632 Central Acaemic Builing, University of Alberta, Emonton, Alberta T6G 2G1, CANADA aress: xichen@math.ualberta.ca

9 CAUCHY INTEGRAL THEOREM Figure 1. The sum of the bounaries of all triangles, oriente counterclockwise, is 1 2 3