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, if f = g on some, where has 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. 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".
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, can be on one open set, and on another, while is on one, and 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 centered at.
To prove this, it is enough to show that for all.
If this is not the case, let m be the smallest nonnegative integer with. By holomorphy, we have the following Taylor series representation in some open neighborhood U of c:
By continuity, h is non-zero in some small open disk B around c. But then f − g ≠ 0 on the punctured set B − . This contradicts the assumption that c is an accumulation point of.
This lemma shows that for a complex number a, the fiber f−1 is a discrete set, unless f ≡ a.

Proof

Define the set on which and have the same Taylor expansion:
We'll show is nonempty, open, and closed. Then by connectedness of, must be all of, which implies on.
By the lemma, in a disk centered at in, they have the same Taylor series at, so, is nonempty.
As and are holomorphic on,, the Taylor series of and at have non-zero radius of convergence. Therefore, the open disk also lies in S for some r. So S is open.
By holomorphy of and, they have holomorphic derivatives, so all are continuous. This means that is closed for all. is an intersection of closed sets, so it's closed.