Hurwitz's theorem (complex analysis)

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

In complex analysis, a field within mathematics, Hurwitz's theorem, named after Adolf Hurwitz, roughly states that, under certain conditions, if a sequence of holomorphic functions converges uniformly to a holomorphic function on compact sets, then after a while those functions and the limit function have the same number of zeros in any open disk.

More precisely, let $G$ be an open set in the complex plane, and consider a sequence of holomorphic functions $(f_{n})$ which converges uniformly on compact subsets of $G$ to a holomorphic function $f.$ Let $D(z_{0},r)$ be an open disk of center $z_{0}$ and radius $r$ which is contained in $G$ together with its boundary. Assume that $f(z)$ has no zeros on the disk boundary. Then, there exists a natural number $N$ such that for all $n$ greater than $N$ the functions $f_{n}$ and $f$ have the same number of zeros in $D(z_{0},r).$

The requirement that $f$ have no zeros on the disk boundary is necessary. For example, consider the unit disk, and the sequence

$f_{n}(z)=z-1+{\frac {1}{n}}$

for all $z.$ It converges uniformly to $f(z)=z-1$ which has no zeros inside of this disk, but each $f_{n}(z)$ has exactly one zero in the disk, which is $1-1/n.$

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

An immediate consequence of this theorem is the following corollary. If $G$ is an open set and a sequence of holomorphic functions $(f_{n})$ converges uniformly on compact subsets of $G$ to a holomorphic function $f,$ and furthermore if $f_{n}$ is not zero at any point in $G$, then $f$ is either identically zero or also is never zero.

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 $\vert z-z_{0}\vert =\rho$, then 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 $k\geq N$. In this case, 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 $f'(z)/f(z)$. Now, denote the number of zeros of $f_{k}(z)$ in the disk by $N_{k}$ and apply the argument principle: $m={\frac {1}{2\pi i}}\int _{\vert z-z_{0}\vert =\rho }{\frac {f'(z)}{f(z)}}\,dz=\lim _{k\rightarrow \infty }{\frac {1}{2\pi i}}\int _{\vert z-z_{0}\vert =\rho }{\frac {f'_{k}(z)}{f_{k}(z)}}\,dz=\lim _{k\rightarrow \infty }N_{k}$ Where 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.
• Theodore W. Gamelin. Complex Analysis. Springer, New York, New York, 2001.
• 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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