# 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|

Because $D$ is open, it follows that $0 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, 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 ProofOfIdentityTheoremOfHolomorphicFunctions 2013-03-22 14:40:41 2013-03-22 14:40:41 rspuzio (6075) rspuzio (6075) 12 rspuzio (6075) Proof msc 30A99