Hurwitz's theorem (complex analysis)

In mathematics and in particular the field of complex analysis, Hurwitz's theorem is a theorem associating the zeroes of a sequence of holomorphic, compact locally uniformly convergent functions with that of their corresponding limit. The theorem is named after Adolf Hurwitz.

Statement
Let {fk} be a sequence of holomorphic functions on a connected open set G that converge uniformly on compact subsets of G to a holomorphic function f which is not constantly zero on G. If f has a zero of order m at z0 then for every small enough ρ > 0 and for sufficiently large k ∈ N (depending on ρ), fk has precisely m zeroes in the disk defined by |z − z0| < ρ, including multiplicity. Furthermore, these zeroes converge to z0 as k → ∞.

Remarks
The theorem does not guarantee that the result will hold for arbitrary disks. Indeed, if one chooses a disk such that f has zeroes on its boundary, the theorem fails. An explicit example is to consider the unit disk D and the sequence defined by


 * $$f_n(z) = z-1+\frac 1 n, \qquad z \in \mathbb C$$

which converges uniformly to f(z) = z − 1. The function f(z) contains no zeroes in D; however, each fn has exactly one zero in the disk corresponding to the real value 1 − (1/n).

Applications
Hurwitz's theorem is used in the proof of the Riemann mapping theorem, and also has the following two corollaries as an immediate consequence:
 * Let G be a connected, open set and {fn} a sequence of holomorphic functions which converge uniformly on compact subsets of G to a holomorphic function f. If each fn is nonzero everywhere in G, then f is either identically zero or also is nowhere zero.
 * If {fn} is a sequence of univalent functions on a connected open set G that converge uniformly on compact subsets of G to a holomorphic function f, then either f is univalent or constant.

Proof
Let f be an analytic function on an open subset of the complex plane with a zero of order m at z0, and suppose that {fn} is a sequence of functions converging uniformly on compact subsets to f. Fix some ρ > 0 such that f(z) ≠ 0 in 0 < |z − z0| ≤ ρ. Choose δ such that |f(z)| > δ for z on the circle |z − z0| = ρ. Since fk(z) converges uniformly on the disc we have chosen, we can find N such that |fk(z)| ≥ δ/2 for every k ≥ N and every z on the circle, ensuring that the quotient fk′(z)/fk(z) is well defined for all z on the circle |z − z0| = ρ. By Weierstrass's theorem we have $$f_k' \to f'$$ uniformly on the disc, and hence we have another uniform convergence:


 * $$\frac{f_k'(z)}{f_k(z)} \to \frac{f'(z)}{f(z)}.$$

Denoting the number of zeros of fk(z) in the disk by Nk, we may apply the argument principle to find


 * $$ m = \frac 1 {2\pi i}\int_{\vert z -z_0\vert = \rho} \frac{f'(z)}{f(z)} \,dz = \lim_{k\to\infty} \frac 1 {2\pi i} \int_{\vert z -z_0\vert = \rho} \frac{f'_k(z)}{f_k(z)} \, dz = \lim_{k\to\infty} N_k$$

In the above step, we were able to interchange the integral and the limit because of the uniform convergence of the integrand. We have shown that Nk → m as k → ∞. Since the Nk are integer valued, Nk must equal m for large enough k.