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 + 1 n , z ∈ C {\displaystyle 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 ′ → f ′ {\displaystyle f_{k}'\to f'} uniformly on the disc, and hence we have another uniform convergence:
f k ′ ( z ) f k ( z ) → f ′ ( z ) f ( z ) . {\displaystyle {\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 = 1 2 π i ∫ | z − z 0 | = ρ f ′ ( z ) f ( z ) d z = lim k → ∞ 1 2 π i ∫ | z − z 0 | = ρ f k ′ ( z ) f k ( z ) d z = lim k → ∞ N k {\displaystyle 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.
See also Rouché's theorem
References
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