Let {fk} be a sequence of holomorphic functions on a connected open setG 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 → ∞.[1]
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 diskD and the sequence defined by
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).
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.[2]
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 uniformly on the disc, and hence we have another uniform convergence:
Denoting the number of zeros of fk(z) in the disk by Nk, we may apply the argument principle to find
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.[1]
Ahlfors, Lars V. (1966), Complex analysis. An introduction to the theory of analytic functions of one complex variable, International Series in Pure and Applied Mathematics (2nd ed.), McGraw-Hill
Ahlfors, Lars V. (1978), Complex analysis. An introduction to the theory of analytic functions of one complex variable, International Series in Pure and Applied Mathematics (3rd ed.), McGraw-Hill, ISBN0070006571
John B. Conway. Functions of One Complex Variable I. Springer-Verlag, New York, New York, 1978.
E. C. Titchmarsh, The Theory of Functions, second edition (Oxford University Press, 1939; reprinted 1985), p. 119.