proof of identity theorem of holomorphic functions

Since $z_{0}\in D$ , there exists an $\epsilon_{0}>0$ the closed disk of radius $\epsilon$ about $z_{0}$ is contained in $D$ . Furthermore, both $f_{1}$ and $f_{2}$ are analytic inside this disk. Since $z_{0}$ is a limit point, there must exist a sequence ${x_{k}}_{k=1}^{\infty}$ of distinct points of $s$ which converges to $z_{0}$ . We may further assume that $|x_{k}-z_{0}|<\epsilon_{0}$ for every $k$ .

By the theorem on the radius of convergence of a complex function, the Taylor series of $f_{1}$ and $f_{2}$ about $z_{0}$ have radii of convergence greater than or equal to $\epsilon_{0}$ . Hence, if we can show that the Taylor series of the two functions at $z_{0}$ ar equal, we will have shown that $f_{1}(z)=g_{1}(z)$ whenever $z<\epsilon$ .

The $n$ -th coefficient of the Taylor series of a function is constructed from the $n$ -th derivative of the function. The $n$ -th derivative may be expressed as a limit of $n$ -th divided differences

$f^{(n)}(z_{0})=\lim_{y_{1},\ldots y_{n}\to z_{0}}{\Delta^{n}f(y_{1},\ldots,y_{% n})\over\Delta^{n}(y_{1},\ldots,y_{n})}$

Suppose we choose the points at which to compute the divided differences as points of the sequence $x_{i}$ . Then we have

$f^{(n)}(z_{0})=\lim_{m\to\infty}{\Delta^{n}f(x_{m+1},\ldots,x_{m+n})\over% \Delta^{n}(x_{m+1},\ldots,x_{m+n})}$

Since $f_{1}(x_{i})=f_{2}(x_{i})$ , it follows that $f_{1}^{(n)}(z_{0})=f_{2}^{(n)}(z_{0})$ for all $n$ and hence $f_{1}(z)=f_{2}(z)$ when $|z-z-0|<\epsilon_{0}$ .

If $D$ happens to be a circle centred about $z_{0}$ , we are done. If not, let $z_{1}$ be any point of $D$ such that $|z_{1}-z_{0}|\geq\epsilon$ . Since every connected open subset of the plane is arcwise connected, there exists an arc $C$ with endpoints $z_{1}$ and $z_{0}$ .

Define the function $M\colon D\to\mathbb{R}$ as follows

$M(z)=\sup\{r\mid|z-w|<r\Rightarrow w\in D\}\cap[0,1]$

Because $D$ is open, it follows that $0<M(z)\leq 1$ for all $z\in D$ .

We will now show that $M$ is continuous. Let $w_{1}$ and $w_{2}$ be any two distinct points of $D$ . If $M(w_{1})>|w_{1}-w_{2}|$ , then a disk of radius $M(w_{1})-|w_{1}-w_{2}|$ about $w_{2}$ will be contained in the disk of radius $M(w_{1})$ about $w_{1}$ . Hence, by the definition of $M$ , it will follow that $M(w_{2})\geq M(w_{1})-|w_{1}-w_{2}|$ . Therefore, for any two points $w_{1}$ and $w_{2}$ , it is the case that $|M(w_{1})-M(w_{2})|\leq|w_{1}-w_{2}|$ , which implies that $M$ is continuous.

Since $M$ is continuous and the arc $C$ is compact, $M$ attains a minimum value $m$ on $C$ . Let $\mu>0$ be chosen smaller strictly less than both $m/2$ and $\epsilon_{0}$ . Consider the set of all open disks of radius $\mu$ centred about ponts of $C$ . By the way $\mu$ was selected, each of these disks lies inside $D$ . Since $C$ is compact a finite subset of these disks will serve to cover $D$ . In other words, there exsits a finite set of points $y_{1},y_{2},\ldots y_{n}$ such that, if $z\in C$ , then $|z-y_{j}|<\mu$ for some $j\in\{1,2,\ldots,n\}$ . We may assume that the $y_{j}$ ’s are ordered so that, as one traverses $C$ from $z_{0}$ to $z_{1}$ , one encounters $y_{j}$ before one encounters $y_{j+1}$ . This imples that $|y_{j}-y_{j+1}|<\mu$ . Without loss of generality, we may assume that $y_{1}=z_{0}$ and $y_{n}=z_{1}$ .

We shall now show that $f_{1}(z)=f_{2}(z)$ when $|z-y_{j}|<\mu$ for all $j$ by induction. From our definitions it follows that $f_{1}(z)=f_{2}(z)$ when $|z-y_{1}|<\mu$ . Next, we shall now show that if $f_{1}(z)=f_{2}(z)$ when $|z-y_{j}|\leq m/2$ , then $f_{1}(z)=f_{2}(z)$ when $|z-y_{j+i}|\leq m/2$ . Since $|y_{j}-y_{j+1}|<\mu$ , there exists a point $w\in C$ and a constant $\epsilon>0$ such that $|w-z|<\epsilon$ implies $|z-y_{j}|\leq\mu$ and $|z-y_{j+i}|\leq\mu$ . By the induction hypothesis, $f_{1}(z)=f_{2}(z)$ when $|z-y_{j}|<\mu$ . Consider a disk of radius $m$ about $w$ . By the definition of $m$ , this disk lies inside $D$ and, by what we have already shown, $f_{1}(z)=f_{2}(z)$ when $|z-w|\leq m$ . Since $|w-y_{j+1}|<\mu<m/2$ , it follows from the triangle inequality that $f_{1}(z)=f_{2}(z)$ when $|z-y_{j+1}|<\mu$ .

In particular, the proposition just proven implies that $f_{1}(z_{1})=f_{1}(z_{1})$ since $z_{1}=y_{n}$ . This means that we have shown that $f_{1}(z)=f_{2}(z)$ for all $z\in D$ .

Title	proof of identity theorem of holomorphic functions
Canonical name	ProofOfIdentityTheoremOfHolomorphicFunctions
Date of creation	2013-03-22 14:40:41
Last modified on	2013-03-22 14:40:41
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	12
Author	rspuzio (6075)
Entry type	Proof
Classification	msc 30A99