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 $U\subseteq ℂ$ be a domain. For two holomorphic functions $f,g:U\to ℂ$, 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 $f(z)=\sum _{n=0}^{\mathrm{\infty }}{a}_n(z-z_0{)}^n$ 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 $z_n\ne z_0$ for all $n\in \mathbb{N}$. Let $f(z)=\sum _{n=0}^{\mathrm{\infty }}{a}_n(z-z_0{)}^n$ 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 $a_n\ne 0$. Due to the well-ordering property of $\mathbb{N}$, there would also be a smallest such $n$. Then

$f(z)=(z-z_0{)}^n\sum _{k=0}^{\mathrm{\infty }}{a}_{n+k}(z-z_0{)}^k,|z-z_0|<r,a_n\ne 0$

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

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

$0=f(z_i)=(z_i-z_0{)}^n\sum _{k=0}^{\mathrm{\infty }}{a}_{n+k}(z_i-z_0{)}^k$

Proof 4 - (N3 to N1) - Limit Process

Since $z_i\ne z_0$ and $\underset{i\to \mathrm{\infty }}{\lim }{z}_i=z_0$, we get

$0=\sum _{k=0}^{\mathrm{\infty }}{a}_{n+k}(z_i-z_0{)}^k\to a_n,i\to \mathrm{\infty }$

As $a_{n+k}(z_i-z_0{)}^k\to 0$ for all $k>0$ as $i\to \mathrm{\infty }$. This contradicts $a_n\ne 0$. 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 $V:=\bigcap _{n\ge 0}z\in U,|,f^{(n)}(z)=0$. $V$ 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 $g:U\to ℂ$ and $h:U\to ℂ$, by applying (N1)-(N3) to $f:=g-h$.

See Also

• Complex Analysis
• Casorati-Weierstrass Theorem

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• Date: 12/17/2024

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