Sheng Li (gmachine1729) wrote,
Sheng Li
gmachine1729

An unpacking of Hurwitz’s theorem in complex analysis

Originally published at 狗和留美者不得入内. You can comment here or there.

Let’s first state it.

Theorem (Hurwitz’s theorem). Suppose \{f_k(z)\} is a sequence of analytic functions on a domain D that converges normally on D to f(z), and suppose that f(z) has a zero of order N at z_0. Then for every small enough \rho > 0, there is k large such that f_k(z) has exactly N zeros in the disk \{|z - z_0| < \rho\}, counting multiplicity, and these zeros converge to z_0 as k \to \infty.

As a refresher, normal convergence on D is convergence uniformly on every closed disk contained by it. We know that the argument principle comes in handy for counting zeros within a domain. That means

The number of zeros in |z - z_0| < \rho, \rho arbitrarily small, goes to the number of zeros inside the same circle of f, provided that

\frac{1}{2\pi i}\int_{|z - z_0| = \rho} \frac{f'_k(z)}{f_k(z)}dz \longrightarrow \frac{1}{2\pi i}\int_{|z - z_0| = \rho} \frac{f'(z)}{f(z)}dz.

To show that boils down to a few technicalities. First of all, let \rho > 0 be sufficiently small that the closed disk \{|z - z_0| \leq \rho\} is contained in D, with f(z) \neq 0 inside it everywhere except for z_0. Since f_k(z) converges to f(z) uniformly inside that closed disk, f_k(z) is not zero on its boundary, the domain integrated over, for sufficiently large k. Further, since f_k \to f uniformly, so does f'_k / f_k \to f' / f, so we have condition such that convergence is preserved on application of integral to the elements of the sequence and to its convergent value. With \rho arbitrarily small, the zeros of f_k(z) must accumulate at z_0.

Tags: complex analysis
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