# proof of identity theorem of holomorphic functions

Since ${z}_{0}\in D$, there exists an ${\u03f5}_{0}>0$ the closed disk of radius $\u03f5$ 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 $x_{k}{}_{k=1}{}^{\mathrm{\infty}}$ of distinct points of $s$ which converges to ${z}_{0}$. We may further assume that $$ 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 ${\u03f5}_{0}$. 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 $$.

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

$${f}^{(n)}({z}_{0})=\underset{{y}_{1},\mathrm{\dots}{y}_{n}\to {z}_{0}}{lim}\frac{{\mathrm{\Delta}}^{n}f({y}_{1},\mathrm{\dots},{y}_{n})}{{\mathrm{\Delta}}^{n}({y}_{1},\mathrm{\dots},{y}_{n})}$$ |

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

$${f}^{(n)}({z}_{0})=\underset{m\to \mathrm{\infty}}{lim}\frac{{\mathrm{\Delta}}^{n}f({x}_{m+1},\mathrm{\dots},{x}_{m+n})}{{\mathrm{\Delta}}^{n}({x}_{m+1},\mathrm{\dots},{x}_{m+n})}$$ |

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 $$.

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 $|{z}_{1}-{z}_{0}|\ge \u03f5$. 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 $M:D\to \mathbb{R}$ as follows

$$ |

Because $D$ is open, it follows that $$ 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 $M({w}_{2})\ge M({w}_{1})-|{w}_{1}-{w}_{2}|$. Therefore, for any two points ${w}_{1}$ and ${w}_{2}$, it is the case that $|M({w}_{1})-M({w}_{2})|\le |{w}_{1}-{w}_{2}|$, 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 ${\u03f5}_{0}$. 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 ${y}_{1},{y}_{2},\mathrm{\dots}{y}_{n}$ such that, if $z\in C$, then $$ for some $j\in \{1,2,\mathrm{\dots},n\}$. We may assume that the ${y}_{j}$’s are ordered so that, as one traverses $C$ from ${z}_{0}$ to ${z}_{1}$, one encounters ${y}_{j}$ before one encounters ${y}_{j+1}$. This imples that $$. 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 $$ for all $j$ by induction^{}. From our definitions it follows that ${f}_{1}(z)={f}_{2}(z)$ when $$. Next, we shall now show that if ${f}_{1}(z)={f}_{2}(z)$ when $|z-{y}_{j}|\le m/2$, then ${f}_{1}(z)={f}_{2}(z)$ when $|z-{y}_{j+i}|\le m/2$. Since $$, there exists a point $w\in C$ and a constant $\u03f5>0$ such that $$ implies $|z-{y}_{j}|\le \mu $ and $|z-{y}_{j+i}|\le \mu $. By the induction hypothesis, ${f}_{1}(z)={f}_{2}(z)$ when $$. Consider a disk of radius $m$ about $w$. By the definition of $m$, this disk lies inside $D$ and, by what we have already shown, ${f}_{1}(z)={f}_{2}(z)$ when $|z-w|\le m$. Since $$, it follows from the triangle inequality^{} that ${f}_{1}(z)={f}_{2}(z)$ when $$.

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 |