Identity theorem

In real analysis and complex analysis, branches of mathematics, the identity theorem for analytic functions states: given functions f and g analytic on a domain D (open and connected subset of $ℝ$ or $ℂ$), if f = g on some $S\subseteq D$, where $S$ has an accumulation point in D, then f = g on D.

Thus an analytic 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 together with its limit). This is not true in general for real-differentiable functions, even infinitely real-differentiable functions. In comparison, analytic functions are a much more rigid notion.

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 sets, $f$ 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 $n\ge 0$, since both functions are analytic.

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

$\begin{array}{cc}(f-g)(z)& =(z-c{)}^m\cdot [\frac{(f-g{)}^{(m)}(c)}{m!}+\frac{(z-c)\cdot (f-g{)}^{(m+1)}(c)}{(m+1)!}+\cdots ]\\ & =(z-c{)}^m\cdot h(z).\end{array}$

By continuity, $h$ is non-zero in some small open disk $B$ around $c$. But then $f-g\ne 0$ 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 $a\in ℂ$, the fiber $f^{-1}(a)$ is a discrete (and therefore countable) set, unless $f\equiv a$.

Proof

Define the set on which $f$ and $g$ have the same Taylor expansion: $$T=\{z\in D|f^{(k)}(z)=g^{(k)}(z)\text{ for all }k\ge 0\}=\bigcap _{k=0}^{\mathrm{\infty }}\{z\in D|(f^{(k)}-g^{(k)})(z)=0\}.$$

We'll show $T$ is nonempty, open, and closed. Then by connectedness of $D$, $T$ must be all of $D$, which implies $f=g$ on $T=D$.

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

As $f$ and $g$ are holomorphic on $D$, $\mathrm{\forall }w\in T$, 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 $T$ for some $r$. So $T$ is open.

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

Full characterisation

Since the identity theorem is concerned with the equality of two holomorphic functions, we can simply consider the difference (which remains holomorphic) and can simply characterise when a holomorphic function is identically $0$. The following result can be found in.

Claim

Let $G\subseteq ℂ$ denote a non-empty, connected open subset of the complex plane. For analytic $h:G\to ℂ$ the following are equivalent.

1. $h\equiv 0$ on $G$;
2. the set $G_0=\{z\in G\mid h(z)=0\}$ contains an accumulation point, $z_0$;
3. the set $G_{\ast }=\bigcap _{n\in {\mathbb{N}}_0}{G}_n$ is non-empty, where $G_n:=\{z\in G\mid h^{(n)}(z)=0\}$.

Proof

(1 $\Rightarrow$ 2) holds trivially.

(2 $\Rightarrow$ 3) is shown in section Lemma in part with Taylor expansion at accumulation point, just substitute g=0.

(3 $\Rightarrow$ 1) is shown in section Proof with set where all derivatives of f-g vanishes, just substitute g=0.

Q.E.D.

See also

• Analytic continuation
• Identity theorem for Riemann surfaces