15 - Identity Theorem

Identity Theorem

Let $U$ be a domain
Suppose $f_1,f_2$ are holomorphic functions defined on $U$

If $S=\{z\in U:f_1(z)=f_2(z)\}$ has a limit point in $U$ then

$$S=U$$

In other words

$$f_1(z)=f_2(z)\text{ for all }z\in U$$

Proof

Let $g=f_1-f_2$ so $S=g^{-1}(\{0\})$

Since $g$ is clearly holomorphic in $U$ then by Zeros of holomorphic functions then
If $z_0\in S$ then either $z_0$ is an isolated point of $S$ or it lies in an open ball contained in $S$

So

$$S=V\cup T$$

where $T=\{z\in S:z\text{ is isolated}\}$ and $V=\text{int}(S)$

As $g$ is continuous then $S=g^{-1}(\{0\})$ is closed in $U$ thus $V\cup T$ is closed
So ${\mathrm{Cl}}_U(V)$ which is the closure of $V$ in $U$ lies in $V\cup T$

However by definition, no limit point of $V$ can lie in $T$ so ${\mathrm{Cl}}_U(V)=V$
Hence $V$ is open and closed in $U$ so as $U$ is connected then $V=\varnothing$ or $V=U$

If $V=\varnothing$ then all zeros of $g$ are isolated so $S$ has no limit points
If $V=U$ then $V=S=U$