6.1 Runge’s Theorem
In the earliest formulation by Carl Runge in 1885, he provided the sufficiency of the holomorphy of on (in effect, a neighborhood of ).
The proof can be well-organized through the use of the results that we will now introduce. In essence, it involves applying Cauchy–Goursat to and the subsequent use of Riemann sums to approximate the complex line integral.
Proposition 6.1.1.
Let be compact and suppose is a neighborhood of that is relatively compact in . Let be an arbitrary holomorphic function. Then for fixed , there exists a rational function with only simple poles (all of which lie in ) such that
Proof.
By assumption of relative compactness, , or the distance (infimum) between and , is positive and finite. More concretely, let
The longest distance between two points in any square is the length of the diagonal. Hence, any square that intersects with a side length less than will lie completely within .
Choose to satisfy and consider the grid generated by compact squares in the form of
(where and are integers) and let be the collection of all such squares in this grid that intersect , and it follows that (refer to Figure 14).
As a consequence of Cauchy–Goursat (Theorem 3.1.8), we have
in the case that . The boundary may be written as the union of lines parameterized by ; more concretely, we have . Hence we have in equivalent formulation,
The distance between and is strictly positive. Suppose instead that the distance were zero. Then some point of would lie on the boundary of a square that intersects . If this point lies on an edge of (but not at a vertex), then the square adjacent along that edge must also intersect , and hence belong to , contradicting the assumption that the point lies on . If the point lies at a vertex of , then all three adjacent squares also intersect , so they too belong to , leading to the same contradiction. Thus, the distance must be positive.
Hence, each integrand as defined in (6.1.1) is jointly continuous for and . By compactness of the product, it is in fact uniformly continuous by Heine–Cantor (Theorem 1.2.15).
Hence, , such that , (uniform in as we can take the minimum of each ), and satisfying ,
Partition by such that , . It follows that
uniformly in . The summation
defines a rational function with simple poles at each , which is disjoint from .
In its full generality, we will now apply a technique to push a pole to a prescribed point, while ensuring that the resulting function remains uniformly approximated outside of a given connected compact set that contains both the original and target pole locations.
Lemma 6.1.1 (Pole-Pushing Lemma).
Let and let be a rational function with a single singularity, a pole at , whose Laurent expansion consists solely of its principal part. Then , , there exists a rational function whose only singularity is a pole at such that
Proof.
By assumption, can be expressed as a polynomial of
This series locally uniformly converges on
and uniformly converges on . Hence, for , we have
For fixed (where ), we aim to prove the existence of an such that , we have at least
where . Since , we can restrict (6.1.2) further with
Additionally, the difference on the left-hand side is also equal to
For any , we have
Since the dominating sequence of partial sums are monotonically increasing, it follows that the sequence of partial sums is uniformly bounded by
on . Thus, (6.1.3) is bounded by , and we apply further restriction by setting this to be bounded by . By uniform convergence, for any , such that ,
For , (6.1.2) is satisfied, and , , we have
which completes the proof as
is rational with a pole at .
Lemma 6.1.2 (Generalized Pole-Pushing Lemma).
Let be compact and choose . Let be the connected component of containing . Then , , there exists a rational function with a pole only at such that
Proof.
Define the set
Since satisfies the condition with , it follows that , ensuring that is nonempty.
Consider , where lies in the complement of . The distance from to , denoted , is positive, and the open disk is disjoint from . Let be an arbitrary point in this disk. By the definition of , for every , there exists a rational function with a pole only at such that
By Lemma 6.1.1, there exists a rational function with a pole only at such that
which implies
Thus,
and by definition, . Hence, is relatively open in .
Now, consider . Suppose there exists . By repeated application of the preceding argument, this would imply , contradicting the assumption that . Therefore, no such exists, and is relatively closed in .
Since is connected and is both relatively open and closed in , it follows from Theorem 3.2.9 that , completing the proof under the assumption that .
Now suppose . In essence, we pole push to a point outside a disk on which we can make approximations by Taylor polynomials. Let satisfy and let be an arbitrary point. By Lemma 6.1.1, there exists some rational function with a pole at such that
Since is holomorphic on some neighborhood of , we have
and it uniformly converges on . Hence, such that
Since polynomials have poles at , we have
Theorem 6.1.1 (Runge).
Let be compact such that has finitely many connected components and suppose is holomorphic on a neighborhood of . Let be a subset of containing one point from each of its connected components. Then , there is a rational function whose poles lie in such that
Proof.
By Proposition 6.1.1, there is a rational function with simple poles in satisfying such that
Let the poles of be , and as a consequence, we have where is entire. Since , we have by Liouville’s Theorem (Theorem 3.2.3). By Lemma 6.1.2, there exist rational functions whose only poles lie in such that ,
and it follows that
Let . From (6.1.4), we have