Normal Convergence of Holomorphisms

At the moment, the only examples of holomorphic functions that we have are polynomials, exponentials, and functions that we can build out of these using sums, products, and composites. Having a small supply of examples can make it challenging to construct holomorphic functions with certain specific behaviors. One way to expand our supply of examples of holomorphic functions is to think about when the limit of a sequence of holomorphic functions is again holomorphic.

Normal Convergence

Let be a domain, and let be a sequence of functions on . We say that the sequence converges normally to or uniformly inside if for every compact subset , the sequence converges uniformly on to .

Theorem

normally on a domain is equivalent to for every , there is an open set such that uniformly on .

Proof First prove that local uniform convergence implies normal convergence. Let be compact. For each , there is an open set such that uniformly on . Then the collection forms an open cover for . By compactness, there is a finite subcover . Let . Then uniformly on , and hence on . Conversely, if normally, then for each , since is open, there is some . Notice that is compact, so uniformly on . Thus, locally uniformly on the open neighbourhood .

Proposition

Normal convergence implies pointwise convergence.

Proposition

Let be a sequence of continuous functions on a domain that converges normally to . Then is continuous.

Proof Fix some . Pick some such that and uniformly on . Since is continuous, is continuous at . Therefore is continuous on .

Proposition

Let be a domain and let be a sequence of functions on that converges normally to . Suppose is a piecewise path. Then

Proof We utilize ML estimate: since uniformly on the compact set .

Remark

The above arguments are just similar to the proposition in real analysis.

Theorem

Let be a domain, let for all , and let be a function. If normally on , then and we have normal convergence of the derivatives: normally on .

Proof Note that is continuous. Fix some , and pick some . Then each , by Cauchy integral formula, for any we have Claim that uniformly on the smaller disk . This can be proved by following arguments: where as , we have , so that Now because on the compact set , the right-hand side goes to zero as . Therefore the claim is proved. Consider the case , we obtain uniformly on . Since normally, pointwise on , by uniqueness of the limit, we conclude that on . It follows that is holonomic on . Since is arbitrary, . Then normal convergence of the derivatives follows from the above claim.