Complex Analysis/Identity theorem, Liouville-type theorems, Riemann's theorem

After now having established the main tools of complex analysis, we may deduce the first corollaries from them, which are theorems about general holomorphic functions; for, it is the case that holomorphic functions follow a wide range of rules just by their holomorphy, and this and the following chapter cover some of them.

Connectedness and the identity theoremEdit

For the famous identity theorem, we first need to frame the intuitively plausible notion of a connected set in mathematically precise terms.

Definition 7.1:

A topological space   is called connected if and only if the only nonempty set which is simultaneously open and closed is  .

It seems somewhat implausible that this is actually the "right" definition of connectedness. The following theorem shows that in fact, if the topological space   equals   (and thus in particular if  ) and the subset is open, then there is a very intuitive characterisation of connectedness.

Theorem 7.2:

Let   with the standard Euclidean topology. Then any nonempty open subset   is connected (w.r.t. to the subspace topology induced by the standard Euclidean topology) if and only if for all   there exists a continuous function   such that   and   (such sets are also called path-connected, and the continuous map   is called a path).

Proof:

Assume first that   is path-connected and not connected. We first prove the following claim, which is actually an important alternative characterisation of connectedness:

Claim: A topological space   is connected if and only if there do not exist open   such that  ,  ,   and   (that is, we may not split up   into two nonempty open disjoint subsets).

Proof of claim: Suppose first that   is connected, and assume for a contradiction that we can write it as   with   nonempty, open, disjoint. Then   is open, closed, nonempty and not the whole space, contradiction. Let now   not be connected. We pick a nontrivial open and closed set  , and   contradicts the condition of the claim. 

Further, we have the following:

Claim: If   is connected and   is a continuous function between topological spaces, then   is connected.

Proof of claim: Assume otherwise. Then there exists   open, closed, nontrivial. By the continuity of  ,   is open, closed, nontrivial, contradiction. 

Claim: The set   is connected.

Proof:

Let now   be connected, and let   be an arbitrary point within   (  was assumed to be nonempty). We define

 .

This set is certainly open, because if there exists a path from   to   and   is any ball around   contained in  , then in fact   since we may connect   to   using a straight line and   to   using the path we have. But it is also closed, since the complement is open for the same reason: Pick   and   such that  ; if any point in   is in  , then, since   is convex, the previous argument applies and  , contradiction. Since   is connected and   nonempty ( ),   and thus, by connecting any two points   by first connecting   and   and then   and  ,   is path-connected.


Now we can state and prove the identity theorem.

Theorem 7.3:

An accumulation point of a sequence   of complex numbers is a point   such that for all  ,   contains infinitely many distinct points

Let   be a connected set, and let   be holomorphic. If there exists a sequence   that has an accumulation point such that  , then  .

Proof:

Liouville-type theoremsEdit

Riemann's theoremEdit

Theorem 7.6:

Let   be a holomorphic function, where   is an interior point of   and assume that   is bounded in a neighbourhood of  . Then there exists a holomorphic function   which extends  , that is,

 .

Thus,   is an extension of   to  .

Proof:

Consider the function

 .

We claim that   is holomorphic in  . Indeed, for   it is differentiable by the product rule, and for   we have

 , i.e.  .

Hence, we may write   as a Taylor series

 

where   due to   and  . Thus,

 ,

and hence for  ,   in the domain of convergence of the power series, we have

 ,

where the latter expression makes sense also at   and is complex differentiable there. This means that

 

is the desired continuation of   to  . 

ExercisesEdit

  1. Prove that if   is an entire function such that  , then   is constant.
  2. Prove that every uncountable subset of   as an accumulation point. Conclude that if for uncountably many   we have  , then  .