1. Inverse Function Theorem for Holomorphic Functions
The field of complex numbers C can be identified with R2 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 C∼= R2. 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)
F (z, z) = f z + z 2 ,z − z
2i
.
For convince, we denote f (x, y) by f (z, z). We define two partial differential operators
∂
∂z, ∂
∂z : C∞(U ) → C∞(U ) by
∂f
∂z = 1 2
∂f
∂x− i∂f
∂y
, ∂f
∂z = 1 2
∂f
∂x+ i∂f
∂y
. A smooth function f ∈ C∞(U ) is said to be holomorphic on U if
∂f
∂z = 0 on U .
In this case, we denote f (z, z) by f (z) and ∂f /∂z by f0(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 mapping f : U ⊂ R2 → R2 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 .
For each p ∈ U, the matrix representation of the derivative dfp : R2 → R2 with respect to the standard basis of R2 is given by
dfp =ux(p) uy(p) vx(p) vy(p)
. In this case, the Jacobian of f at p is given by
J (f )(p) = det dfp = ux(p)vy(p) − uy(p)vx(p) = u2x(p) + vx2(p) = |f0(p)|2.
Theorem 1.1. (Inverse Function Theorem for holomorphic Functions) Let f be a holomor- phic function on U and p ∈ U so that f0(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 f0(p) 6= 0, the Jacobian of f at p is J (f )(p) = |f0(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, 0 = ∂z
∂z = ∂
∂zg(f (z), f (z)) = ∂g
∂w
∂f
∂z + ∂g
∂w
∂f
∂z. Since f is holomorphic on U, ∂f /∂z = 0 on U. This implies
0 = ∂f
∂z
∂g
∂w.
Since f : V → W is a diffeomorphism, J (f )(z) = |f0(z)|2 > 0 for all z ∈ V. Then ∂f /∂z = f0(z) is nonzero on V, and hence
∂f
∂z = ∂f
∂z 6= 0, on V .
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 consisting of points a so that f0(a) 6= 0. Then U is open and f : U → C is locally univalent.
Theorem 1.2. Let U be an open set of C and f be a univalent function on U. Then f0 6= 0 on U and f : U → f (U ) is biholomorphic.
Since f is holomorphic on U , f0 is also holomorphic on U. Since f is a nonconstant function, f (U ) is open and f0(z) is not the zero function on U. The zeros of f0 forms a discrete subset of U. Hence if f0(z0) = 0, there exists an open neighborhood V = D(z0, δ) of z0 so that f0(z) 6= 0 for all z 6= z0 on V. Define a new function F (z) = f (z) − f (z0) on U.
Then F (z0) = F0(z0) = 0 and F0(z) = f0(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
m = 1
2min{|F (z)| : z ∈ C}.
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 <
|z| < ρ for some ρ < R. Assume that f (0) = 0 and f0(0) 6= 0. Let m0 = min{|f (z)| : |z| = ρ}
and |ω| < m0. Then the solution of f (z) = ω is given by z = 1
2πi I
C
ζ f0(ζ) f (ζ) − ωdζ.
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
z∗ = 1 2πi
I
C
ζF0(ζ)
F (ζ)dζ = 1 2πi
I
C
ζ f0(ζ) f (ζ) − ωdζ
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
1
f (ζ) − ω = 1
(f (ζ) − ω0) − (ω − ω0) =
∞
X
n=0
(ω − ω0)n (f (ζ) − ω0)n+1. By the uniform convergence, we obtain
g(ω) = 1 2πi
I
C
ζ f0(ζ) f (ζ) − ωdζ =
∞
X
n=0
1 2πi
I
C
ζ f0(ζ)
(f (ζ) − ω0)n+1dζ
(ω − ω0)n.
This shows that g is holomorphic at ω0. Since ω0 is arbitrary in U, g is holomorphic on U.
