Complex Analysis/Identity Theorem

The Identity Theorem is a statement about holomorphic functions, asserting that they are uniquely determined under relatively weak conditions.

Statement

Let $U\subseteq \mathbb {C}$ be a domain. For two holomorphic functions $f,g\colon U\to \mathbb {C}$, the following are equivalent:

(1) $f=g$ (i.e., $f(x)=g(x)$ for all $x\in U$)

(2) There exists a $z_{0}\in U$ such that $f^{(n)}(z_{0})=g^{(n)}(z_{0})$ for all $n\in \mathbb {N}$.

(3) The set ${z\in U:f(z)=g(z)}$ has a limit point in $U$.

Proof

By considering $f-g$, we may assume without loss of generality that $g=0$. Equivalently, the proof is reduced to showing the following three statements:

• (N1) $f=0$ (i.e., $f(x)=0$ for all $x\in U$)

• (N2) There exists a $z_{0}\in U$ such that $f^{(n)}(z_{0})=0$ for all $n\in \mathbb {N}$.

• (N3) The zero set $N_{f}:={z\in U:f(z)=0}$ has a limit point in $U$.

Proof Type

The equivalence is proven using a cyclic implication: $(1)\Rightarrow (2)\Rightarrow (3)\Rightarrow (1)$

Proof (N1 to N2)

(N1) $\Rightarrow$ (N2) is obvious, as all derivatives of the zero function $f$ are zero.

Proof (N2 to N3)

Assume (N2). Consider the power series expansion $f(z)=\sum _{n=0}^{\infty }a_{n}(z-z_{0})^{n}$ in $B_{r}(z_{0})$ with $r>0$. Here, $a_{n}={\frac {f^{(n)}(z_{0})}{n!}}=0$ for all $n\in \mathbb {N}$. Thus, $f|_{B_{r}(z_{0})}=0$, and (N3) follows.

Proof (N3 to N1) – Contradiction Proof

The step (N3) $\Rightarrow$ (N1) is proven by contradiction. Assume the zero set has a limit point and $f$ is not the zero function.

Proof 1 - (N3 to N1) - Power Series Expansion at Limit Point

Assume (N3), i.e., the set $N_{f}$ of zeros of $f$ has a limit point $z_{0}\in U$. Thus, there exists a sequence $(z_{n})\in U^{\mathbb {N} }$ with $z_{n}\to z_{0}$ and $f(z_{n})=0$ as well as $z_{n}\neq z_{0}$ for all $n\in \mathbb {N}$. Let $f(z)=\sum _{n=0}^{\infty }a_{n}(z-z_{0})^{n}$ be the power series expansion of $f$ around $z_{0}$.

Proof 2 - (N3 to N1) - Power Series Expansion

Suppose there exists $n\in \mathbb {N}$ with $a_{n}\neq 0$. Due to the well-ordering property of $\mathbb {N}$, there would also be a smallest such $n$. Then

$f(z)=(z-z_{0})^{n}\sum _{k=0}^{\infty }a_{n+k}(z-z_{0})^{k},\qquad |z-z_{0}|<r,a_{n}\neq 0$

Proof 3 - (N3 to N1) - Power Series Evaluation

For each $i\in \mathbb {N}$, we have

$0=f(z_{i})=(z_{i}-z_{0})^{n}\sum _{k=0}^{\infty }a_{n+k}(z_{i}-z_{0})^{k}$

Proof 4 - (N3 to N1) - Limit Process

Since $z_{i}\neq z_{0}$ and $\displaystyle \lim _{i\to \infty }z_{i}=z_{0}$, we get

$0=\sum _{k=0}^{\infty }a_{n+k}(z_{i}-z_{0})^{k}\to a_{n},\qquad i\to \infty$

As $a_{n+k}(z_{i}-z_{0})^{k}\to 0$ for all $k>0$ as $i\to \infty$. This contradicts $a_{n}\neq 0$. Therefore, $a_{n}=0$ for all $n\in \mathbb {N}$, and hence $f^{(n)}(z_{0})=0$ for all $n\in \mathbb {N}$, i.e., (N2) holds.

Proof 5 - (N3 to N1) - V is Closed

If (N2) holds, set $V:=\bigcap _{n\geq 0}{z\in U,|,f^{(n)}(z)=0}$. $V$ is closed in $U$ as the intersection of closed sets, because the $f^{(n)}$ are continuous, and preimages of closed sets (here ${0}$) are closed.

Proof 6 - (N3 to N1) - V is Open

$V$ is also open in $U$, as for every $w\in V$, the power series expansion of $f$ around $w$ vanishes. Thus, $f$ is locally zero around $w$. Since $z_{0}\in V$, $V$ is non-empty, and hence $V=U$ due to the connectedness of $U$.

Proof 7 - From (N1)-(N3) to (1)-(3)

The statement of the Identity Theorem (1)-(3) follows for arbitrary $g:U\to \mathbb {C}$ and $h:U\to \mathbb {C}$, by applying (N1)-(N3) to $f:=g-h$.

See Also

• Complex Analysis
• Casorati-Weierstrass Theorem

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• Source: Kurs:Funktionentheorie/Identitätssatz - URL: https://de.wikiversity.org/wiki/Kurs:Funktionentheorie/Identitätssatz
• Date: 12/17/2024