# Identity theorem

In complex analysis, a branch of mathematics, the identity theorem for holomorphic functions states: given functions f and g holomorphic on a domain D (open and connected subset), if f = g on some $S\subseteq D$ , $S$ having an accumulation point, then f = g on D.

Thus a holomorphic function is completely determined by its values on a single open neighborhood in D, or even a countable subset of D (provided this contains a converging sequence). This is not true for real-differentiable functions. In comparison, holomorphy, or complex-differentiability, is a much more rigid notion. Informally, one sometimes summarizes the theorem by saying holomorphic functions are "hard" (as opposed to, say, continuous functions which are "soft").

The underpinning fact from which the theorem is established is the expandability of a holomorphic function into its Taylor series.

The connectedness assumption on the domain D is necessary. For example, if D consists of two disjoint open set, $f$ can be $0$ on one open set, and $1$ on another, while $g$ is $0$ on one, and $2$ on another.

## Lemma

If two holomorphic functions f and g on a domain D agree on a set S which has an accumulation point c in D, then f = g on a disk in $D$ centered at $c$ .

To prove this, it is enough to show that $f^{(n)}(c)=g^{(n)}(c)$ for all $n\geq 0$ .

If this is not the case, let m be the smallest nonnegative integer with $f^{(m)}(c)\neq g^{(m)}(c)$ . By holomorphy, we have the following Taylor series representation in some open neighborhood U of c:

{\begin{aligned}(f-g)(z)&{}=(z-c)^{m}\cdot \left[{\frac {(f-g)^{(m)}(c)}{m!}}+{\frac {(z-c)\cdot (f-g)^{(m+1)}(c)}{(m+1)!}}+\cdots \right]\\[6pt]&{}=(z-c)^{m}\cdot h(z)\end{aligned}} By continuity, h is non-zero in some small open disk B around c. But then f − g ≠ 0 on the punctured set B − {c}. This contradicts the assumption that c is an accumulation point of {f = g}.

This lemma shows that for a complex number a, the fiber f−1(a) is a discrete (and therefore countable) set, unless f = a.

## Proof

Define the set on which $f$ and $g$ have the same Taylor expansion:

$S=\{z\in D\mid f^{(k)}(z)=g^{(k)}(z){\text{ for all }}k\geq 0\}=\bigcap _{k=0}^{\infty }\{z\in D\mid (f^{(k)}-g^{(k)})(z)=0\}.$ We'll show $S$ is nonempty, open, and closed. Then by connectedness of $D$ , $S$ must be all of $D$ , which implies $f=g$ on $S=D$ .

By the lemma, $f=g$ in a disk centered at $c$ in $D$ , they have the same Taylor series at $c$ , so $c\in S$ , $S$ is nonempty.

As $f$ and $g$ are holomorphic on $D$ , $\forall w\in S$ , the Taylor series of $f$ and $g$ at $w$ have non-zero radius of convergence. Therefore, the open disk $B_{r}(w)$ also lies in S for some r. So S is open.

By holomorphy of $f$ and $g$ , they have holomorphic derivatives, so all $f^{(n)},g^{(n)}$ are continuous. This means that $\{z\in D\mid (f^{(k)}-g^{(k)})(z)=0\}$ is closed for all $k$ . $S$ is an intersection of closed sets, so it's closed.