Identity theorem

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In complex analysis, the identity theorem for holomorphic functions states: given functions f and g holomorphic on a connected open set D, if f = g on some neighborhood of z that is in D, then f = g on D. Thus a holomorphic function is completely determined by its values on a (possibly quite small) neighborhood in D. 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 developability of a holomorphic function into its Taylor series.

[edit] Proof

The connectedness assumption on the domain D is necessary and is in fact key to a short proof given here. (Obviously if D consisted of two disjoint open sets, the result does not hold.) Under this assumption, since we are given that the set is not empty, topologically the claim amounts to that f and g coincide on a set that is both open and closed. Closedness is immediate from the continuity of f and g.

Therefore the main issue is to show that the set on which f = g coincide on an open set. Because a holomorphic function can be represented by its Taylor series everywhere on its domain, it is sufficient to consider the set

$S = \{ z \in D | f^{(k)}(z) = g^{(k)}(z) \quad \mbox{for all} \; k \geq 0\}.$

Suppose w lies in S. Then, because the Taylor series of f and g at w have non-zero radius of convergence, the open disk B_r(w) also lies in S for some r. (In fact, r can be anything less than the distance from w to the boundary of D). This shows S is open and proves the theorem.

[edit] An improvement

The hypotheses on this theorem can be relaxed slightly while still producing the same conclusion. Specifically, if two holomorphic functions f and g on a domain D agree on a set which has an accumulation point c in D then f = g on all of D.

To prove this, it is enough to show that f^(k)(c) = g^(k)(c) for all k ≥ 0. If this is not the case, let m be the smallest nonnegative integer with f^(m)(c) ≠ g^(m)(c). By holomorphy, we have the following Taylor series representation in some open neighborhood of c:

$\begin{align} (f - g)(z) &{}=(z - c)^m \cdot [ \, (f - g)^{(m)} + (f - g)^{(m+1)} (z-c) + \cdots \,] \\ &{}=(z - c)^m \cdot h(z) \\ \\ \end{align}$

Evidently 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} and therefore the claim is proved.

This formulation of the theorem shows that for a complex number a, the fiber f ^-1(a) is a discrete (in fact countable) set unless f = a.