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Identity theorem for holomorphic functions

Statement

If two holomorphic functions on a complex domain agree on a set with an accumulation point in the domain, then they agree everywhere on the domain.

Precisely, let $Ω \subseteq C$ be a complex domain (A complex domain is a nonempty connected open subset of $C$ ), let $f, g : Ω \to C$ be holomorphic, and suppose that some $a \in Ω$ is an accumulation point of ${z \in Ω : f (z) = g (z)}$ . Then $f = g$ on $Ω$ . The requirement $a \in Ω$ is essential.

Facts & Assumptions

Given: A complex domain $Ω$ , holomorphic functions $f, g : Ω \to C$ , an accumulation point $a \in Ω$ of their agreement set, and the holomorphic difference $h : = f - g$ supplied by Linearity, product, reciprocal, and quotient rules for complex derivatives. A nonempty subset of a connected space that is both open and closed is the whole space (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets).

[L1]

For a holomorphic function $h$ on an open set $U$ , the set of points having a neighbourhood on which $h$ vanishes is both open and closed in $U$ (The locally zero locus of a holomorphic function is clopen).

[L2]

A holomorphic function has finite order $m$ at $a$ exactly when it factors near $a$ as $(z - a)^{m} q (z)$ with $q (a) \neq 0$ ; its order is $+ \infty$ exactly when it vanishes on a neighbourhood of $a$ (The order of a zero is the exponent in its local holomorphic factorization).

[L3]

If $f : U \to C$ is complex differentiable at $a \in U$ , then $f$ is continuous at $a$ (Complex differentiability at a point implies continuity there).

Proof

technique · direct

1.1

The function $h$ vanishes at points arbitrarily close to $a$ . If it had finite order there, [L2] would give $h (z) = (z - a)^{m} q (z)$ with $q (a) \neq 0$ , and [L3] would make $q$ nonzero on a smaller neighbourhood; then $h$ would have no zeros there other than possibly $a$ , contrary to accumulation. Hence $h$ has infinite order at $a$ , so [L2] makes it vanish on a neighbourhood of $a$ .

givenL2L3

2.1

By [L1], the locally zero locus of $h$ is open and closed in $Ω$ ; it is nonempty by step 1.1. Since $Ω$ is connected, that locus is all of $Ω$ .

step 1.1L1given

3.1

Therefore $h (z) = 0$ for every $z \in Ω$ , which means $f (z) = g (z)$ throughout $Ω$ .

step 2.1∎

Remarks

The accumulation point must belong to the domain. Accumulation only at a boundary point does not force identity, as the companion counterexample shows.

Depends on

Used by

The ring of holomorphic functions on a complex domain is an integral domain Corollary
Agreement accumulating only at the boundary does not force a holomorphic identity Counterexample
Local degree of a nonconstant holomorphic map Definition
Open mapping theorem for holomorphic functions Theorem
The complex Pythagorean identity by the identity theorem Theorem
Zeros of a nonzero holomorphic function are isolated Theorem

Dependency tree · two levels

28 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Guide to Cultivating Complex Analysis, Theorem 2.4.7 (standard reference, not scraped)
B. V. Shabat, Introduction to Complex Analysis, Theorem 2.28 (standard reference, not scraped)