Hurwitz's theorem (complex analysis)

This article is about a theorem in complex analysis. For other uses, see Hurwitz's theorem.

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.

Theorem Statement

Suppose that $\{f_{k}(z)\}$ is a sequence of holomorphic functions on an connected open set $G$ that converge uniformly on compact subsets of $G$ to a holomorphic function $f$. If $f(z)$ has a zero of order $m$ at $z_{0}$ then there exists $\rho >0$ such that for sufficiently large $k\in \mathbb {N} ,f_{k}(z)$ has precisely $m$ zeroes in the disk $\{|z-z_{0}|<\rho \}$, including multiplicity. Furthermore, these zeroes converge to $z_{0}$ as $k\to \infty$.

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 $\mathbb {D}$ 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 $\mathbb {D}$; however, each $f_{n}(z)$ has exactly one zero in the disk corresponding to the real value $1-1/n.$

The result holds more generally for any bounded convex sets but it is most useful to state for disks.

Applications

Hurwitz 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 an connected, open set and $\{f_{n}(z)\}$ a sequence of holomorphic functions which converge uniformly on compact subsets of $G$ to a holomorphic function $f.$ If $f_{n}$ is not zero at any point in $G$, then $f$ is either identically zero or also is never zero.
• If $\{f_{n}(z)\}$ is a sequence of univalent functions on a connected open set $G$ that converge uniformly on compact subsets of $G$ to a holomorphic function $f(z)$, then either $f(z)$ is univalent or constant.

Proof

Let $f(z)$ be an analytic function on an open subset of the complex plane with a zero of order $m$ at $z_{0}$, and suppose that $\{f_{n}(z)\}$ is a sequence of functions converging uniformly on compact subsets to $f(z)$. Fix some $\rho >0$ such that $f(z)\neq 0$ for any $0<\vert z-z_{0}\vert <\rho$. Choose $\delta$ such that $\vert f(z)\vert >\delta$ for $z$ on the circle $\vert z-z_{0}\vert =\rho$. Since $f_{k}(z)$ converges uniformly on the disc we have chosen, we can find an $N$ such that $\vert f_{k}(z)\vert \geq \delta /2$ for every $k\geq N$, ensuring that the quotient $f_{k}'(z)/f_{k}(z)$ is well defined for all $z$ on the circle $\vert z-z_{0}\vert =\rho$. By Morera's theorem, $f_{k}'(z)/f_{k}(z)$ converges uniformly to $f'(z)/f(z)$. Denoting the number of zeros of $f_{k}(z)$ in the disk by $N_{k}$, we may apply the argument principle to find

$m={\frac {1}{2\pi i}}\int _{\vert z-z_{0}\vert =\rho }{\frac {f'(z)}{f(z)}}\,dz=\lim _{k\rightarrow \infty }{\frac {1}{2\pi i}}\int _{\vert z-z_{0}\vert =\rho }{\frac {f'_{k}(z)}{f_{k}(z)}}\,dz=\lim _{k\rightarrow \infty }N_{k}$

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 $N_{k}\rightarrow m$ as $k\rightarrow \infty$. Since $N_{k}$ are integer valued, $N_{k}$ must equal $m$ for large enough $k$.

See also

• Rouché's theorem

References

• 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.
• Solomentsev, E.D. (2001) [1994], "Hurwitz theorem", Encyclopedia of Mathematics, EMS Press

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