Identity theorem

From Wikipedia, the free encyclopedia
Jump to navigation Jump to search

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, 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). 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[edit]

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[edit]

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[edit]

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.[1]

Claim[edit]

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

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

Proof[edit]

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 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 -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

 

 

 

 

(1)

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 (1) yields

 

 

 

 

(2)

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

See also[edit]

References[edit]

  1. ^ Guido Walz, ed. (2017). Lexikon der Mathematik (in German). 2. Mannheim: Springer Spektrum Verlag. p. 476. ISBN 978-3-662-53503-5.
  • Ablowitz, Mark J.; Fokas A. S. (1997). Complex variables: Introduction and applications. Cambridge, UK: Cambridge University Press. p. 122. ISBN 0-521-48058-2.