6.2 Mergelyan’s Theorem
Although many mathematicians have since tried after the efforts of Weierstrass and Runge to approximation continuous functions holomorphic on the interior restriction, it was only 67 years later when Armenian mathematician provided the first widely accepted proof. The proof of Runge’s Theorem (specifically in Proposition 6.1.1) relied heavily on the assumption of holomorphy on a neighborhood, a rational function was created by placing poles in prescribed points of a contour that lay outside of but within its domain of holomorphy. Obviously, these assumptions are null under the context of this new formulation.
The proof proposed by Mergelyan is almost trivial when compared with the results of many other mathematicians at the time. It even uses the concepts previously proposed by Runge. This begs the question: why was there such a prolonged time gap between the two similar formulations? Many mathematicians felt that the conclusion was “too good to be true”; during this elapsed time period there were many efforts of mathematicians that resulted in many technical partial results. Mergelyan’s Theorem came as a surprise as it encapsulated many of those results with simplicity.
As we have previously seen, there is a prevalent notion in complex analysis that regards intrinsically as essentially any other point of . An appertaining question relates to the complex derivative at . Although
may seem to be a natural object to consider, it is quite impractical; there exist functions which decay quickly to , while is unbounded as (take as an example). Even the assumption that does not imply that has a removable singularity at (consider ).
Definition 6.2.1.
Let , be holomorphic such that has a removable singularity at . Then we define the derivatives of at to be
In the case that , we have
This is precisely the first singular term of the Laurent expansion of at .
Remark.
This definition may feel unsatisfactory, but the underlying logic here is similar to the method used to generalize residues to .
If is bijective and meromorphic on some neighborhood of a point such that , then we informally define the derivative at the pole to be
Let . Then we have
Proposition 6.2.1.
For any connected compact set containing at least two distinct points such that is connected, let be an arbitrary biholomorphism mapping to such that . It follows that
where .
Proof.
Denote the derivative of at the infinity to be . By (6.2.2), we have
Fix and let , univalent on . By direct calculation, we have . Additionally,
By the Koebe Quarter Theorem (@ thm:koebequarter), whose proof is independent of results of this section, in accordance, . Let . Obviously, .
Let be injective on . For the sake of contradiction, assume that . Then such that . By injectivity, , which contradicts , and accordingly, .
Hence,
By taking the supremum for , the proof is complete.
Remark.
Such a biholomorphism will always exist; for arbitrary , the map maps to a simply connected, proper subset of , which is biholomorphic to by the Riemann Mapping Theorem (Theorem 5.3.1).
Proposition 6.2.2.
Let , , and suppose is compact such that is connected and . Then there is a family of holomorphic functions , where ,
and
- for any .
- for any .
- The function is jointly continuous in and .
Proof.
For brevity, assume .
Let be a conformal mapping from to , such that and . Let . It follows that , . By Proposition 6.2.1,
Consequently, we have the crucial estimate of . For fixed , define
where . The application of Cauchy’s Estimate (Theorem 3.2.2) on gives:
Hence,
This is Part 1. Suppose that . It follows from that (from the reverse triangle inequality) and hence disjoint from and . On this infinite annulus, we have the Laurent expansion (from Theorem 4.1.2) that
where because as . Since , we have the global Laurent expansion
Hence,
by letting . Since
from the definition of , we have
Hence, there exists some such that
for all satisfying . By Theorem 3.2.6, has a removable singularity at . On the other hand, for such that , we have
from Part 1. The Maximum Modulus Principle (Theorem 3.4.1) implies that
and thus Part 2 follows. The joint continuity of is immediate from the definition.
Lastly, if , we may define where is the family constructed above for the set .
Proposition 6.2.3.
Suppose
For fixed , the function satisfies:
- , where .
- and is compactly supported.
- .
- .
- for all , where denotes the vector differential operator.
-
For any such that is a holomorphic function on , we have the integral formula.
Proof.
Let . Then we have
which confirms Part 1. Let be arbitrary. The integral in (6.2.5) is equal to
by the mean-value property (Lemma 3.4.1), proving Part 6. For , we have
Hence,
which confirms Part 5. Since , we have
since which verifies the inequality in Part 4.
The Part 3 is also true since
Trivially, is continuous on and . Thus, we only need to prove the joint continuity of (the continuity of implies that of ) on an open neighborhood of .
Let . By simple calculation, we have
At , both partial derivatives vanish, and hence, they match the vanishing derivative on the complement of , completing the proof of Part 2.
Theorem 6.2.1 (Tietze–Urysohn–Brouwer).
Let be compact and be continuous. Then such that on .
Proof.
For any two disjoint closed , consider the continuous separation function
so that and .
For simplicity, by the boundedness of , we may assume that (by a scaling and shift). We now aim to construct a sequence inductively such that
In the case that , define the disjoint closed sets
Let . It is clear that on . If , then , , and hence . If , then , , and thus . If , then and . Thus, ,
This proves the base case. For the inductive step, assume the claim holds for each . Define
for . By the inductive hypothesis, we have on . Define the disjoint closed sets
and
Let , so that on , and
for all by the same argument as in the base case. Hence,
for all , completing the induction. Because
the Weierstrass –Test (Theorem 2.3.2) implies that the series converges uniformly on to . Since each is continuous, Theorem 2.3.5 gives the continuity of on . Finally, for any , we have
Corollary 6.2.1.
If is compact and is continuous, then such that on and has compact support.
Proof.
Let where are continuous. By Tietze–Urysohn–Brouwer (Theorem 6.2.1), such that and on . Let be such that , provided by compactness. Define the piecewise-linear function
such that and is compactly supported. Let , and the assertion follows.
Let be holomorphic on . Then has a continuous extension to all of by virtue of Corollary 6.2.1. Define the modulus of continuity of to be the function with
Because has compact support, it must be uniformly continuous; hence we have .
For , define
where employs the same definition as in (6.2.4).
Proposition 6.2.4.
The function as in (6.2.6) satisfies:
- and has compact support.
- on .
- for all .
- For all , .
- for , where .
Proof.
Because and is compact, for sufficiently large , the two supports will be disjoint and hence the integrand vanishes for all . We can explicitly find that
Because is continuous and vanishes on a compact set, it is bounded. Similarly, Part 2 of Proposition 6.2.3 implies that is bounded. Hence, by Lebesgue’s Dominated Convergence Theorem, we have
and similarly,
Hence, and this is Part 1. Because
by Part 1 of Proposition 6.2.3, we have
which implies Part 3. For , now implies that and hence is holomorphic in on . By Part 6 of Proposition 6.2.3, (6.2.7) becomes
which proves Part 2. Because ,
by Part 3 of Proposition 6.2.3. Hence,
by Part 5 of Proposition 6.2.3, confirming Part 4. Finally, Part 5 follows from Corollary 3.1.1 (since outside the support the integral trivially vanishes and within , vanishes as a consequence of holomorphy).
Theorem 6.2.2 (Mergelyan).
Let be compact such that has finitely many connected components. Let contain exactly one point from each of the connected components of . Suppose is holomorphic on . Then , there exists a rational function with poles in such that
Proof.
Let contain precisely one point from each connected component of (such that each is finite). Suppose that is chosen such that so that for each not equal to ,
Define the extension of , , , and as in the previous results (see Figure 15). Hence, (see Figure 16)
covers a (compact) -neighborhood of (so that each , and is labeled so that each is in the same connected component as ) (in the case that , let the disk inside be the empty set). Thus, the collection also covers . A finite subcover covering exists by the Heine–Borel Theorem (Theorem 1.1.3).
By the connectivity of each component of , there exists a piecewise-linear simple curve for all , , joining and , which lies entirely within . The compact disks are all disjoint from their corresponding since each by definition.
Hence, the intersection consists of at least one connected component joining to a point on . Denote the connected component of this intersection by , satisfying and (see Figure 17).
Now for each and , Proposition 6.2.2 now provides the existence of a family of holomorphic functions given with such that
Let , for each and construct the disjoint sets
as done in Figure 18. Thus the union
since the set of all covers . Let
where . Because
and both and the integrand is continuous on a set (we only need to consider the factors involving since is independent from ) by Cauchy’s Estimates and the first bound of (6.2.8), Lebesgue’s Dominated Convergence gives that
and in analogous fashion,
Hence, is holomorphic on , a neighborhood of . Since , by Part 2 of Proposition 6.2.2,
The estimates in Part 4 of Proposition 6.2.4, in tandem with those from (6.2.8) now give that
through a linear change of variables. Now evaluation via polar coordinates (with , ) yields a revised upper bound of
Runge’s Theorem (Theorem 6.1.1) provides the existence of some rational function with poles in such that
since is holomorphic on a neighborhood of (to assure this bound is positive, we assume is not identically zero, otherwise the assertion is trivial). Therefore, for all , we have (the third supremum term coming from Part 3 of Proposition 6.2.4)
Because , for any , there exists a such that
Hence for any such , we now construct in accordance with an satisfying .