1. Inverse Function Theorem for Holomorphic Functions the Field Of

Home , Inverse function theorem

1. Inverse Function Theorem for Holomorphic Functions 2 The field of complex numbers C can be identified with R as a two dimensional real vector space via x + iy 7→ (x, y). On C, we define an inner product hz, wi = Re(zw). With respect to the the norm induced from the inner product, C becomes a two dimensional real Hilbert space. Let C∞(U) be the space of all complex valued smooth functions on an open subset U of ∼ 2 C = R . Since x = (z + z)/2 and y = (z − z)/2i, a smooth complex valued function f(x, y) on U can be considered as a function F (z, z) z + z z − z F (z, z) = f , . 2 2i For convince, we denote f(x, y) by f(z, z). We define two partial differential operators ∂ ∂ , : C∞(U) → C∞(U) ∂z ∂z by ∂f 1 ∂f ∂f ∂f 1 ∂f ∂f = − i , = + i . ∂z 2 ∂x ∂y ∂z 2 ∂x ∂y A smooth function f ∈ C∞(U) is said to be holomorphic on U if ∂f = 0 on U. ∂z In this case, we denote f(z, z) by f(z) and ∂f/∂z by f 0(z). A function f is said to be holomorphic at a point p ∈ C if f is holomorphic defined in an open neighborhood of p. For open subsets U and V in C, a function f : U → V is biholomorphic if f is a bijection from U onto V and both f and f −1 are holomorphic. A holomorphic function f on an open subset U of C can be identified with a smooth 2 2 mapping f : U ⊂ R → R via f(x, y) = (u(x, y), v(x, y)) where u, v are real valued smooth functions on U obeying the Cauchy-Riemann equation

ux = vy and uy = −vx on U. 2 2 For each p ∈ U, the matrix representation of the derivative dfp : R → R with respect to 2 the standard basis of R is given by ux(p) uy(p) dfp = . vx(p) vy(p) In this case, the Jacobian of f at p is given by 2 2 0 2 J(f)(p) = det dfp = ux(p)vy(p) − uy(p)vx(p) = ux(p) + vx(p) = |f (p)| . Theorem 1.1. (Inverse Function Theorem for holomorphic Functions) Let f be a holomorphic function on U and p ∈ U so that f 0(p) 6= 0. Then there exists an open neighborhood V of p so that f : V → f(V ) is biholomorphic. Proof. Since f is holomorphic on U, we can represent f by f = f(z) on U. Since f 0(p) 6= 0, the Jacobian of f at p is J(f)(p) = |f 0(p)|2 > 0. By inverse function theorem, there exists an open neighborhood V of p so that f : V → W = f(V ) is a C∞-diffeomorphism and J(f)(q) 6= 0 for all q ∈ U. Let g : W → V be the inverse map of f. Then g is smooth on V and hence we may represent g by g(w, w). Then g(f(z), f(z)) = z for all z ∈ V and 1 2 f(g(w, w)) = w for all w ∈ W. To show that g is holomorphic on W, we only need to show ∂g/∂w = 0. In fact, by chain rule, ∂z ∂ ∂g ∂f ∂g ∂f 0 = = g(f(z), f(z)) = + . ∂z ∂z ∂w ∂z ∂w ∂z Since f is holomorphic on U, ∂f/∂z = 0 on U. This implies ∂f ∂g 0 = . ∂z ∂w Since f : V → W is a diffeomorphism, J(f)(z) = |f 0(z)|2 > 0 for all z ∈ V. Then ∂f/∂z = f 0(z) is nonzero on V, and hence ∂f ∂f = 6= 0, on V. ∂z ∂z We obtain ∂g/∂w = 0 on W. This shows that g is holomorphic. Let Ω be a region in C. A holomorphic function f :Ω ⊂→ C is univalent if it is an injection. A locally univalent function on Ω is a holomorphic function f :Ω → C so that every point a of Ω has an open neighborhood U so that f : U → C is univalent. Corollary 1.1. Let f :Ω → C be a holomorphic function. Let U be the subset of Ω 0 consisting of points a so that f (a) 6= 0. Then U is open and f : U → C is locally univalent. 0 Theorem 1.2. Let U be an open set of C and f be a univalent function on U. Then f 6= 0 on U and f : U → f(U) is biholomorphic. Since f is holomorphic on U, f 0 is also holomorphic on U. Since f is a nonconstant function, f(U) is open and f 0(z) is not the zero function on U. The zeros of f 0 forms a 0 discrete subset of U. Hence if f (z0) = 0, there exists an open neighborhood V = D(z0, δ) 0 of z0 so that f (z) 6= 0 for all z 6= z0 on V. Define a new function F (z) = f(z) − f(z0) on U. 0 0 0 Then F (z0) = F (z0) = 0 and F (z) = f (z) 6= 0 for z 6= z0 on V. Since F is holomorphic and nonconstant, we can choose δ small enough so that F (z) 6= 0 for 0 < |z − z0| < δ. Let C be the circle |z − z0| = δ/2. Then F is nonzero on C. Since C is compact, F is continuous on C, |F (z)| with z ∈ C has minimum. Let 1 m = min{|F (z)| : z ∈ C}. 2 Then m > 0. For any 0 < |w| < m, |w| < |F (z)| on C. Rouche’s theorem implies that F (z) − w and F (z) have the same number of zeros (counting multiplicities) in D(z0, δ/2). Hence #ZC (F (z) − w) = #ZC (F (z)) ≥ 2. This shows that F (z) = w has at least two solutions in D(z0, δ/2). In other words, F is not univalent on D(z0, δ/2) which leads to a contradiction to F being univalent. Now, let us prove that f −1 : f(U) → U is holomorphic. To do this, we need the following lemma. Lemma 1.1. Let f be a holomorphic function on |z| < R. Suppose f is nonzero on 0 < 0 |z| < ρ for some ρ < R. Assume that f(0) = 0 and f (0) 6= 0. Let m0 = min{|f(z)| : |z| = ρ} and |ω| < m0. Then the solution of f(z) = ω is given by 1 I f 0(ζ) z = ζ dζ. 2πi C f(ζ) − ω Here C is the circle |z| = ρ 3

Proof. Let F (z) = f(z) − ω on |z| < R. Then F is holomorphic on |z| < R. Since |ω| < m0, F (z) and f(z) has the same number of zero in |z| < ρ by Rouche’s theorem. We know f(z) = 0 has a unique zero 0 in |z| < ρ. Hence F (z) = 0 has a unique solution. Denote this solution by z∗. Then 1 I F 0(ζ) 1 I f 0(ζ) z∗ = ζ dζ = ζ dζ 2πi C F (ζ) 2πi C f(ζ) − ω which proves our assertion. Now, assume that f : U → C is univalent. To show that the inverse map g : f(U) → U is holomorphic, we only need to show that given any point ω0 ∈ f(U), the function z = g(ω) has a Taylor expansion in a neighborhood of ω0. Let f(z0) = ω0. Choose ρ > 0 so that f(z) 6= ω0 for 0 < |z − z0| < ρ. Denote m0 = min{|f(z) − ω0| : |z − z0| = ρ} as above. For all |ω − ω0| < m0(1 − δ), the inﬁnite series converges uniformly ∞ n 1 1 X (ω − ω0) = = . f(ζ) − ω (f(ζ) − ω ) − (ω − ω ) (f(ζ) − ω )n+1 0 0 n=0 0 By the uniform convergence, we obtain ∞ 1 I f 0(ζ) X 1 I f 0(ζ) g(ω) = ζ dζ = ζ dζ (ω − ω )n. 2πi f(ζ) − ω 2πi (f(ζ) − ω )n+1 0 C n=0 C 0 This shows that g is holomorphic at ω0. Since ω0 is arbitrary in U, g is holomorphic on U.