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  or ), if f = g on some , where  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. Informally, one sometimes summarizes the theorem by saying analytic 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 sets,  can be  on one open set, and  on another, while  is  on one, and  on another.

Lemma 

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

To prove this, it is enough to show that  for all . 

If this is not the case, let  be the smallest nonnegative integer with . By holomorphy, we have the following Taylor series representation in some open neighborhood U of :

By continuity,  is non-zero in some small open disk  around . But then  on the punctured set . This contradicts the assumption that  is an accumulation point of .

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

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  for some . So  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.

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 . The following result can be found in.

Claim 
Let  denote a non-empty, connected open subset of the complex plane.
For  the following are equivalent.

  on ;
 the set  contains an accumulation point, ;
 the set  is non-empty, where .

Proof 

The directions (1  2) and (1  3) hold trivially.

For (3  1), by connectedness of  it suffices to prove that the non-empty subset, , is clopen (since a topological space is connected if and only if it has no proper clopen subsets).
Since holomorphic functions are infinitely differentiable, i.e. , it is clear that  is closed. To show openness, consider some .
Consider an open ball  containing , in which  has a convergent Taylor-series expansion centered on .
By virtue of , all coefficients of this series are , whence  on .
It follows that all -th derivatives of  are  on , whence .
So each  lies in the interior of .

Towards (2  3), fix an accumulation point .
We now prove directly by induction that  for each .
To this end let  be strictly smaller than the convergence radius of the power series expansion of  around , given by . Fix now some  and assume that  for all . Then for  manipulation of the power series expansion yields

Note that, since  is smaller than radius of the power series, one can readily derive that the power series  is continuous and thus bounded on .

Now, since  is an accumulation point in , there is a sequence of points  convergent to .
Since  on  and since each , the expression in () yields

By the boundedness of  on ,
it follows that , whence .
Via induction the claim holds.
Q.E.D.

See also 
Analytic continuation
Identity theorem for Riemann surfaces

References

 

Theorems_in_real_analysis
Theorems in complex analysis
Articles containing proofs