§6.2 Mergelyan’s Theorem

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 $Proposition 6.1.1 Proposition 6.1.1$ ) 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 $𝐾$ 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 $\infty$ intrinsically as essentially any other point of $\hat{ℂ}$ . An appertaining question relates to the complex derivative at $\infty$ . Although

𝑓^{'} (\infty) \overset{?}{=} lim_{𝑧 \to \infty} 𝑓^{'} (𝑧)

may seem to be a natural object to consider, it is quite impractical; there exist functions which decay quickly to $0$ , while $𝑓^{'} (𝑧)$ is unbounded as $𝑧 \to \infty$ (take $𝑧 \mapsto \frac{sin (𝑧^{2})}{\sqrt{𝑧}}$ as an example). Even the assumption that ${lim}_{𝑧 \to \infty} 𝑓^{'} (𝑧) = 0$ does not imply that $𝑓 (𝑧)$ has a removable singularity at $\infty$ (consider $𝑧 \mapsto \sqrt{𝑧}$ ).

Definition 6.2.1.

Let $𝑅 > 0$ , $𝑓 : ℂ ∖ \overline{𝐷 (0, 𝑅)} \to ℂ$ be holomorphic such that $𝑓$ has a removable singularity at $\infty$ . Then we define the derivatives of $𝑓$ at $\infty$ to be

𝑓^{(𝑛)} (\infty) = {\frac{d^{𝑛}}{d 𝑧^{𝑛}} (𝑓 (\frac{1}{𝑧})) |}_{𝑧 = 0} .

In the case that $𝑛 = 1$ , we have

𝑓^{'} (\infty) = - lim_{𝑧 \to \infty} 𝑧^{2} 𝑓^{'} (𝑧) . (6.2 . 1)

This is precisely the first holomorphic coefficient of the Laurent expansion of $𝑓$ at $\infty$ (coefficient of $𝑧^{- 1}$ ).

Remark.

This definition may feel unsatisfactory, but the underlying logic here is similar to the method used to generalize residues to

\infty

If $𝑓$ is bijective and meromorphic on some neighborhood of a point $𝑎 \in ℂ$ such that $𝑓 (𝑎) = \infty$ , then we informally define the derivative at the pole $𝑎$ to be

\begin{aligned} 𝑓^{'} (𝑎) = \frac{1}{{(𝑓^{- 1})}^{'} (\infty)} & = - lim_{𝑤 \to \infty} \frac{1}{𝑤^{2} {(𝑓^{- 1})}^{'} (𝑤)} \\ = - lim_{𝑤 \to \infty} \frac{𝑓^{'} (𝑓^{- 1} (𝑤))}{𝑤^{2}} . (6.2 . 2) \end{aligned}

Let $𝑧 = (𝑓^{- 1}) (𝑤)$ . Then we have

𝑓^{'} (𝑎) = - lim_{𝑧 \to 𝑎} \frac{𝑓^{'} (𝑧)}{{𝑓 (𝑧)}^{2}} = {\frac{d}{d 𝑧} (\frac{1}{𝑓 (𝑧)}) |}_{𝑧 = 𝑎} . (6.2 . 3)

Proposition 6.2.2.

For any connected compact set $𝐾 \subseteq ℂ$ containing at least two distinct points such that $\hat{ℂ} ∖ 𝐾$ is connected, let $𝜙$ be an arbitrary biholomorphism mapping $\hat{ℂ} ∖ 𝐾$ to $𝔻$ such that $𝜙 (\infty) = {lim}_{𝑧 \to \infty} 𝜙 (𝑧) = 0$ . It follows that

| 𝜙^{'} (\infty) | \geq \frac{1}{4} diam (𝐾),

where

diam (𝐾) = {sup}_{𝑧, 𝜁 \in 𝐾} | 𝜁 - 𝑧 |

Proof.

Denote the derivative of $𝜙$ at the infinity to be $𝛼$ . By (6.2.2) $(6.2.2) (6.2.2)$ , we have

{(𝜙^{- 1})}^{'} (0) = \frac{1}{𝜙^{'} (\infty)} = \frac{1}{𝛼} = - lim_{𝑧 \to 0} \frac{{(𝜙^{- 1})}^{'} (𝑧)}{𝜙^{- 1} {(𝑧)}^{2}} ⟺ - lim_{𝑧 \to 0} \frac{𝜙^{- 1} {(𝑧)}^{2}}{𝛼 {(𝜙^{- 1})}^{'} (𝑧)} = 1 .

Fix $𝜏 \in 𝐾$ and let $𝜓 (𝑧) = \frac{𝛼}{𝜙^{- 1} (𝑧) - 𝜏}$ , univalent on $𝔻$ . By direct calculation, we have $𝜓 (0) = 0$ . Additionally,

𝜓^{'} (0) = - lim_{𝑧 \to 0} \frac{𝛼 {(𝜙^{- 1})}^{'} (𝑧)}{{(𝜙^{- 1} (𝑧) - 𝜏)}^{2}} = lim_{𝑧 \to 0} \frac{𝛼 {(𝜙^{- 1})}^{'} (𝑧)}{{(𝜙^{- 1} (𝑧) - 𝜏)}^{2}} \cdot \frac{𝜙^{- 1} {(𝑧)}^{2}}{𝛼 {(𝜙^{- 1})}^{'} (𝑧)} = 1 .

By the Koebe Quarter Theorem (Theorem 9.2.6 $Theorem 9.2.6 Theorem 9.2.6$ ), whose proof is independent of results of this section, in accordance, $𝐷 (0, \frac{1}{4}) \subseteq 𝜓 (𝔻)$ . Let $𝜇 \in 𝐾 ∖ {𝜏}$ . Obviously, $𝜇 \notin (𝜙^{- 1}) (𝔻) = \hat{ℂ} ∖ 𝐾$ .

Let $𝑧 \mapsto \frac{𝛼}{𝑧 - 𝜏}$ be injective on $\hat{ℂ}$ . For the sake of contradiction, assume that $(𝑧 \mapsto \frac{𝛼}{𝑧 - 𝜏}) (𝜇) \in 𝜓 (𝔻)$ . Then $\exists 𝜁 \in (𝜙^{- 1}) (𝔻)$ such that $\frac{𝛼}{𝜁 - 𝜏} = \frac{𝛼}{𝜇 - 𝜏}$ . By injectivity, $𝜁 = 𝜇$ , which contradicts $𝜇 \in 𝐾$ , and accordingly, $\frac{𝛼}{𝜇 - 𝜏} \notin 𝜓 (𝔻) \supseteq 𝐷 (0, \frac{1}{4})$ .

Hence,

| \frac{𝛼}{𝜇 - 𝜏} | \geq \frac{1}{4} ⟺ | 𝛼 | \geq \frac{| 𝜇 - 𝜏 |}{4} .

By taking the supremum for $𝜇, 𝜏 \in 𝐾$ , the proof is complete. ⁠ $□$

Remark.

Such a biholomorphism will always exist; for arbitrary

𝜁 \in 𝐾

, the map

𝑧 \mapsto \frac{1}{𝑧 - 𝜁}

maps

\hat{ℂ} ∖ 𝐾

to a simply connected, proper subset of

ℂ

, which is biholomorphic to

𝔻

by the Riemann Mapping Theorem (Theorem 5.3.2

Theorem 5.3.2 Theorem 5.3.2

Proposition 6.2.3.

Let $𝑎 \in ℂ$ , $𝑟 > 0$ , and suppose $𝐾 ⋐ 𝐷 (𝑎, 𝑟)$ is compact such that $\hat{ℂ} ∖ 𝐾$ is connected and $diam (𝐾) \geq \frac{𝑟}{2}$ . Then there is a family of holomorphic functions $ℱ︀ = {𝜑_{𝜁}}_{𝜁 \in 𝐷 (𝑎, 𝑟)}$ , where $\forall 𝜁 \in 𝐷 (𝑎, 𝑟)$ ,

𝜑_{𝜁} : \hat{ℂ} ∖ 𝐾 \to ℂ,

and

$| 𝜑_{𝜁} (𝑧) | \leq \frac{584}{𝑟}$ for any $𝑧 \in \hat{ℂ} ∖ 𝐾$ .
$| 𝜑_{𝜁} (𝑧) - \frac{1}{𝑧 - 𝜁} | \leq \frac{4676 𝑟^{2}}{{| 𝜁 - 𝑧 |}^{3}}$ for any $𝑧 \in \hat{ℂ} ∖ (𝐾 \cup {𝜁})$ .
The function $𝜑 (𝜁, 𝑧) \equiv 𝜑_{𝜁} (𝑧)$ is jointly continuous in $𝜁$ and $𝑧$ .

Proof.

For brevity, assume $𝑎 = 0$ .

Let $\tilde{𝜑}$ be a conformal mapping from $\hat{ℂ} ∖ 𝐾$ to $𝔻$ , such that $\tilde{𝜑} (\infty) = 0$ and $𝛼 = {\tilde{𝜑}}^{'} (\infty) \in ℝ_{> 0}$ . Let $𝜑 (𝑧) = \frac{1}{𝛼} \tilde{𝜑} (𝑧)$ . It follows that $𝜑^{'} (\infty) = 1$ , $𝜑 (\infty) = 0$ . By Proposition 6.2.2 $Proposition 6.2.2 Proposition 6.2.2$ ,

| 𝛼 | \geq \frac{1}{4} diam (𝐾) ⟺ | 𝜑 (𝑧) | \leq \frac{4 | \tilde{𝜑} (𝑧) |}{diam (𝐾)} .

Consequently, we have the crucial estimate of $𝜑 (\hat{ℂ} ∖ 𝐾) \subseteq 𝐷 (0, \frac{4}{diam (𝐾)}) \subseteq 𝐷 (0, \frac{8}{𝑟})$ . For fixed $𝜁 \in 𝐷 (0, 𝑟)$ , define

𝜑_{𝜁} (𝑧) = 𝜑 (𝑧) + (𝜁 - 𝛽) 𝜑^{2} (𝑧), 𝑧 \in \hat{ℂ} ∖ 𝐾

where $𝛽 = \frac{𝜑^{″} (\infty)}{2}$ . The application of Cauchy’s Estimate (Theorem 3.2.3 $Theorem 3.2.3 Theorem 3.2.3$ ) on $(𝑧 \mapsto \frac{1}{𝑧}) (\hat{ℂ} ∖ 𝐷 (0, 𝑟)) = 𝐷 (0, \frac{1}{𝑟})$ gives:

| 𝛽 | = \frac{1}{2} | {\frac{d^{2}}{d 𝑧^{2}} 𝜑 (\frac{1}{𝑧}) |}_{𝑧 = 0} | \leq \frac{{sup}_{𝐷 (0, \frac{1}{𝑟})} | 𝜑 (\frac{1}{𝑧}) |}{dist {(0, 𝜕 𝐷 (0, \frac{1}{𝑟}))}^{2}} = 8 𝑟 .

Hence,

\begin{aligned} | 𝜑_{𝜁} (𝑧) | & \leq | 𝜑 (𝑧) | + | 𝜁 - 𝛽 | {| 𝜑 (𝑧) |}^{2} \leq | 𝜑 (𝑧) | + | 𝜁 - 𝛽 | {| 𝜑 (𝑧) |}^{2} \\ \leq \frac{8}{𝑟} + 9 𝑟 \frac{64}{𝑟^{2}} = \frac{584}{𝑟} . \end{aligned}

This is Part 1 $Part 1 Part 1$ . Suppose that $| 𝑧 - 𝜁 | > 2 𝑟$ . It follows from $| 𝜁 | < 𝑟$ that $| 𝑧 | > 𝑟$ (from the reverse triangle inequality) and hence disjoint from $𝐾$ and $𝜁$ . On this infinite annulus, we have the Laurent expansion (from Theorem 4.1.3 $Theorem 4.1.3 Theorem 4.1.3$ ) that

𝜑 (𝑧) = \sum_{𝑘 = 1}^{\infty} \frac{𝜇_{𝑘}}{{(𝑧 - 𝜁)}^{𝑘}} = \frac{1}{𝑧 - 𝜁} + \frac{𝜇}{{(𝑧 - 𝜁)}^{2}} + 𝒪︀ (\frac{1}{{(𝑧 - 𝜁)}^{3}})

where $𝜇_{1} = 1$ because $𝜑 \sim \frac{1}{𝑧}$ as $𝑧 \to \infty$ . Since $| 𝑧 | > 𝑟$ , we have the global Laurent expansion

𝜑 (𝑧) = \frac{1}{𝑧} + \frac{𝛽}{𝑧^{2}} + 𝒪︀ (\frac{1}{𝑧^{3}}) .

Hence,

\begin{aligned} \frac{1}{𝑧 - 𝜁} + \frac{𝜇}{{(𝑧 - 𝜁)}^{2}} & = \frac{1}{𝑧} + \frac{𝛽}{𝑧^{2}} + 𝒪︀ (\frac{1}{𝑧^{3}}) \\ 𝑧 + 𝜁 + 𝜇 & = 𝑧 + \frac{𝜁^{2}}{𝑧} + 𝛽 + \frac{𝛽 𝜁^{2}}{𝑧^{2}} - \frac{2 𝛽 𝜁}{𝑧} + 𝒪︀ (\frac{1}{𝑧}) = 𝑧 + 𝛽 + 𝒪︀ (\frac{1}{𝑧}) \\ 𝜇 & = 𝛽 - 𝜁 \end{aligned}

by letting $𝑧 \to \infty$ . Since

{𝜑 (𝑧)}^{2} = {(\frac{1}{𝑧 - 𝜁} + 𝒪︀ (\frac{1}{{(𝑧 - 𝜁)}^{2}}))}^{2} = \frac{1}{{(𝑧 - 𝜁)}^{2}} + 𝒪︀ (\frac{1}{{(𝑧 - 𝜁)}^{3}}),

from the definition of $𝜑_{𝜁}$ , we have

\begin{aligned} 𝜑_{𝜁} (𝑧) - \frac{1}{𝑧 - 𝜁} = 𝜑 - 𝜇 𝜑^{2} - \frac{1}{𝑧 - 𝜁} & = \frac{𝜇}{{(𝑧 - 𝜁)}^{2}} + 𝒪︀ (\frac{1}{{(𝑧 - 𝜁)}^{3}}) \\ \begin{matrix}  \end{matrix} - \frac{𝜇}{{(𝑧 - 𝜁)}^{2}} - 𝒪︀ (\frac{𝜇}{{(𝑧 - 𝜁)}^{3}}) \\ = 𝒪︀ (\frac{1}{{(𝑧 - 𝜁)}^{3}}) . \end{aligned}

Hence, there exists some $𝑀 > 0$ such that

| 𝜑_{𝜁} (𝑧) - \frac{1}{𝑧 - 𝜁} | < \frac{𝑀}{{| 𝑧 - 𝜁 |}^{3}} ⟺ | 𝜑_{𝜁} (𝑧) - \frac{1}{𝑧 - 𝜁} | {| 𝑧 - 𝜁 |}^{3} < 𝑀

for all $𝑧$ satisfying $| 𝑧 - 𝜁 | > 2 𝑟$ . By Theorem 3.2.9 $Theorem 3.2.9 Theorem 3.2.9$ , $(𝜑_{𝜁} (𝑧) - \frac{1}{𝑧 - 𝜁}) {(𝑧 - 𝜁)}^{3}$ has a removable singularity at $\infty$ . On the other hand, for $| 𝑧 - 𝜁 | \leq 2 𝑟$ such that $𝑧 \in \hat{ℂ} ∖ (𝐾 \cup {𝜁})$ , we have

\begin{aligned} | 𝜑_{𝜁} (𝑧) - \frac{1}{𝑧 - 𝜁} | {| 𝑧 - 𝜁 |}^{3} & \leq | 𝜑_{𝜁} (𝑧) | {| 𝑧 - 𝜁 |}^{3} + {| 𝑧 - 𝜁 |}^{2} \\ \leq \frac{584}{𝑟} {(2 𝑟)}^{3} + {(2 𝑟)}^{2} = 4676 𝑟^{2} \end{aligned}

from Part 1 $Part 1 Part 1$ . The Maximum Modulus Principle (Theorem 3.4.2 $Theorem 3.4.2 Theorem 3.4.2$ ) implies that

sup_{| 𝑧 - 𝜁 | > 2 𝑟} | 𝜑_{𝜁} (𝑧) - \frac{1}{𝑧 - 𝜁} | {| 𝑧 - 𝜁 |}^{3} \leq sup_{| 𝑧 - 𝜁 | = 2 𝑟} | 𝜑_{𝜁} (𝑧) - \frac{1}{𝑧 - 𝜁} | {| 𝑧 - 𝜁 |}^{3} \leq 4676 𝑟^{2}

and thus Part 2 $Part 2 Part 2$ follows. The joint continuity of $𝜑_{𝜁}$ is immediate from the definition.

Lastly, if $𝑎 \neq 0$ , we may define $𝜑_{𝜁} (𝑧) = {\tilde{𝜑}}_{𝜁 - 𝑎} (𝑧 - 𝑎)$ where ${{\tilde{𝜑}}_{𝜁 - 𝑎}}$ is the family constructed above for the set ${𝑧 - 𝑎 : 𝑧 \in 𝐾} \subset 𝐷 (0, 𝑟)$ . ⁠ $□$

Proposition 6.2.4.

Suppose

𝜆 (𝑧) = {\begin{cases} {(1 - {| 𝑧 |}^{2})}^{2} & if | 𝑧 | < 1, \\ 0 & if | 𝑧 | \geq 1, \end{cases} 𝜆_{𝑟} (𝑧) = \frac{3}{π 𝑟^{2}} 𝜆 (\frac{𝑧}{𝑟}) \forall 𝑟 > 0 (6.2 . 4)

For fixed $𝑟$ , the function $𝜆_{𝑟}$ satisfies:

$\iint_{ℂ} 𝜆_{𝑟} (𝜁) d 𝜉 d 𝜂 = 1$ , where $𝜁 = 𝜉 + i 𝜂$ .
$𝜆_{𝑟} \in 𝐶^{1} (ℂ)$ and is compactly supported.
$\iint_{ℂ} \frac{𝜕 𝜆_{𝑟}}{𝜕 \overline{𝜁}} d 𝜉 d 𝜂 = 0$ .
$\iint_{ℂ} | \frac{𝜕 𝜆_{𝑟}}{𝜕 \overline{𝜁}} | d 𝜉 d 𝜂 \leq \frac{2 π}{𝑟}$ .
$‖ 𝛁 𝜆_{𝑟} (𝑧) ‖ \leq \frac{4}{𝑟^{3}}$ for all $𝑧$ , where $𝛁 = (\frac{𝜕}{𝜕 𝑥}, \frac{𝜕}{𝜕 𝑦})$ denotes the vector differential operator.
For any $𝑧 \in ℂ$ such that $𝑓$ is a holomorphic function on $𝐷 (𝑧, 𝑟)$ , we have the following integral formula:
$𝑓 (𝑧) = \iint_{𝐷 (0, 𝑟)} 𝑓 (𝑧 - 𝜁) 𝜆_{𝑟} (𝜁) d 𝜉 d 𝜂 . (6.2 . 5)$

Proof.

Let $𝜁 = 𝜌 e^{i 𝜃}$ . Then we have

\begin{aligned} \iint_{ℂ} 𝜆_{𝑟} (𝜁) d 𝜉 d 𝜂 & = \int_{0}^{2 π} \int_{0}^{𝑟} 𝜆_{𝑟} (𝜌 e^{i 𝜃}) 𝜌 d 𝑟 d 𝜃 \\ = \int_{0}^{2 π} \int_{0}^{𝑟} \frac{3 𝜌}{π 𝑟^{2}} {(1 - {(\frac{𝜌}{𝑟})}^{2})}^{2} d 𝑟 d 𝜃 \\ = \frac{6}{𝑟^{2}} \int_{0}^{𝑟} (𝜌 + \frac{𝜌^{5}}{𝑟^{4}} - 2 \frac{𝜌^{3}}{𝑟^{2}}) d 𝑟 \\ = \frac{6}{𝑟^{2}} (\frac{𝑟^{2}}{2} + \frac{𝑟^{6}}{6 𝑟^{4}} - \frac{𝑟^{4}}{2 𝑟^{2}}) = 1, \end{aligned}

which confirms Part 1 $Part 1 Part 1$ . Let $𝑧 \in ℂ$ be arbitrary. The integral in (6.2.5) $(6.2.5) (6.2.5)$ is equal to

\begin{aligned} \iint_{𝐷 (0, 𝑅)} 𝑓 (𝑧 - 𝜁) 𝜆_{𝑟} (𝜁) d 𝜉 d 𝜂 \\ \begin{matrix}  \end{matrix} = \int_{0}^{𝑟} \frac{3 𝜌}{π 𝑟^{2}} {(1 - {(\frac{𝜌}{𝑟})}^{2})}^{2} \int_{0}^{2 π} 𝑓 (𝑧 - 𝜌 e^{i 𝜃}) d 𝜃 d 𝑟 \\ \begin{matrix}  \end{matrix} = 2 π 𝑓 (𝑧) \int_{0}^{𝑟} \frac{3 𝜌}{π 𝑟^{2}} {(1 - {(\frac{𝜌}{𝑟})}^{2})}^{2} d 𝑟 = 𝑓 (𝑧) \end{aligned}

by the mean-value property (Lemma 3.4.1 $Lemma 3.4.1 Lemma 3.4.1$ ), proving Part 6 $Part 6 Part 6$ . For $𝑧 \in 𝔻$ , we have

\begin{aligned} ‖ 𝛁 𝜆 (𝑧) ‖ & = 2 (1 - {| 𝑧 |}^{2}) ‖ 𝛁 ({| 𝑧 |}^{2}) ‖ \\ = 2 (1 - {| 𝑧 |}^{2}) 2 | 𝑧 | ‖ 𝛁 \sqrt{𝑥^{2} + 𝑦^{2}} ‖ = 4 (1 - {| 𝑧 |}^{2}) | 𝑧 | . \end{aligned}

Hence,

\begin{aligned} ‖ 𝛁 𝜆_{𝑟} (𝑧) ‖ & = \frac{3}{π 𝑟^{2}} ‖ 𝛁 (𝜆 (\frac{𝑧}{𝑟})) ‖ = \frac{3}{π 𝑟^{2}} ‖ (𝛁 𝜆) (\frac{𝑧}{𝑟}) ‖ \frac{1}{𝑟} \\ = \frac{12}{π 𝑟^{3}} (1 - {| 𝑧 |}^{2}) | 𝑧 | < \frac{4}{𝑟^{3}}, \end{aligned}

which confirms Part 5 $Part 5 Part 5$ . Since $| \frac{𝜕 𝜆_{𝑟}}{𝜕 \overline{𝜁}} | = | \frac{1}{2} (\frac{𝜕 𝜆_{𝑟}}{𝜕 𝜉} + i \frac{𝜕 𝜆_{𝑟}}{𝜕 𝜂}) | = \frac{1}{2} ‖ 𝛁 𝜆_{𝑟} (𝜁) ‖ < \frac{2}{𝑟^{3}}$ , we have

\iint_{ℂ} | \frac{𝜕 𝜆_{𝑟}}{𝜕 \overline{𝜁}} | d 𝜉 d 𝜂 = \iint_{𝐷 (0, 𝑟)} | \frac{𝜕 𝜆_{𝑟}}{𝜕 \overline{𝜁}} | d 𝜉 d 𝜂 < π 𝑟^{2} \frac{2}{𝑟^{3}} = \frac{2 π}{𝑟}

since $supp (𝜆_{𝑟}) = \overline{𝐷 (0, 𝑟)}$ which verifies the inequality in Part 4 $Part 4 Part 4$ .

The Part 3 $Part 3 Part 3$ is also true since

\begin{aligned} \iint_{ℂ} \frac{𝜕 𝜆_{𝑟}}{𝜕 \overline{𝜁}} d 𝜉 d 𝜂 & = \frac{1}{2} \int_{- 𝑟}^{𝑟} \int_{- 𝑟}^{𝑟} \frac{𝜕 𝜆_{𝑟}}{𝜕 𝜉} d 𝜉 d 𝜂 + \frac{i}{2} \int_{- 𝑟}^{𝑟} \int_{- 𝑟}^{𝑟} \frac{𝜕 𝜆_{𝑟}}{𝜕 𝜂} d 𝜂 d 𝜉 \\ = \frac{1}{2} \int_{- 𝑟}^{𝑟} [𝜆_{𝑟} (𝑟 + i 𝜂) - 𝜆 (- 𝑟 + i 𝜂)] d 𝜂 \\ \begin{matrix}  \end{matrix} + \frac{i}{2} \int_{- 𝑟}^{𝑟} [𝜆_{𝑟} (𝜉 + i 𝑟) - 𝜆_{𝑟} (𝜉 - i 𝑟)] d 𝜉 = 0 . \end{aligned}

Trivially, $𝜆_{𝑟}$ is continuous on $𝐷 (0, 𝑟)$ and $ℂ ∖ \overline{𝐷 (0, 𝑟)}$ . Thus, we only need to prove the joint continuity of $𝜆$ (the continuity of $𝜆_{𝑟}$ implies that of $𝜆$ ) on an open neighborhood of $𝜕 𝐷 (0, 𝑟)$ .

Let $𝜆 (𝑥, 𝑦) = {(1 - 𝑥^{2} - 𝑦^{2})}^{2}$ . By simple calculation, we have

\frac{𝜕 𝜆}{𝜕 𝑥} = - 4 (1 - 𝑥^{2} - 𝑦^{2}) 𝑥, \frac{𝜕 𝜆}{𝜕 𝑦} = - 4 (1 - 𝑥^{2} - 𝑦^{2}) 𝑦 .

At $𝑥^{2} + 𝑦^{2} = 1$ , both partial derivatives vanish, and hence, they match the vanishing derivative on the complement of $supp (𝜆)$ , completing the proof of Part 2 $Part 2 Part 2$ . ⁠ $□$

Theorem 6.2.5 (Tietze–Urysohn–Brouwer).

Let

𝐾 \subseteq ℂ

be compact and

𝑓 : 𝐾 \to ℝ

be continuous. Then

\exists 𝑔 \in 𝐶 (ℂ)

such that

𝑔 \equiv 𝑓

𝐾

Proof.

For any two disjoint closed $𝐴, 𝐵 \subseteq ℂ$ , consider the continuous separation function

𝜂_{𝐴, 𝐵} (𝑧) = \frac{dist (𝑧, 𝐴) - dist (𝑧, 𝐵)}{dist (𝑧, 𝐴) + dist (𝑧, 𝐵)}

so that $𝜂_{𝐴, 𝐵} (𝐴) = {- 1}$ and $𝜂_{𝐴, 𝐵} (𝐵) = {1}$ .

For simplicity, by the boundedness of $𝑓$ , we may assume that $𝑓 (𝐾) = [- 1, 1]$ (by a scaling and shift). We now aim to construct a sequence ${𝑔_{𝑛}}_{𝑛 \in ℕ_{\geq 0}}$ inductively such that

| 𝑔_{𝑛} | \leq \frac{2^{𝑛}}{3^{𝑛 + 1}} on ℂ, | 𝑓 - \sum_{𝑘 = 0}^{𝑛} 𝑔_{𝑘} | \leq {(\frac{2}{3})}^{𝑛 + 1} on 𝐾 \forall 𝑛 \in ℕ .

In the case that $𝑛 = 0$ , define the disjoint closed sets

𝐴_{0} = {𝑧 \in 𝐾 : 𝑓 (𝑧) \leq - \frac{1}{3}} and 𝐵_{0} = {𝑧 \in 𝐾 : 𝑓 (𝑧) \geq \frac{1}{3}} .

Let $𝑔_{0} (𝑧) = \frac{1}{3} 𝜂_{𝐴_{0}, 𝐵_{0}} (𝑧)$ . It is clear that $| 𝑔_{0} | \leq \frac{1}{3}$ on $ℂ$ . If $𝑧 \in 𝐴_{0}$ , then $- 1 \leq 𝑓 (𝑧) \leq - \frac{1}{3}$ , $𝑔_{0} (𝑧) = - \frac{1}{3}$ , and hence $| 𝑓 - 𝑔_{0} | \leq \frac{2}{3}$ . If $𝑧 \in 𝐵_{0}$ , then $\frac{1}{3} \leq 𝑓 (𝑧) \leq 1$ , $𝑔_{0} (𝑧) = \frac{1}{3}$ , and thus $| 𝑓 - 𝑔_{0} | \leq \frac{2}{3}$ . If $𝑧 \notin 𝐴_{0} \cup 𝐵_{0}$ , then $- \frac{1}{3} < 𝑓 (𝑧) < \frac{1}{3}$ and $| 𝑓 - 𝑔_{0} | \leq | 𝑓 | + | 𝑔_{0} | < \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$ . Thus, $\forall 𝑧 \in 𝐾$ ,

| 𝑓 (𝑧) - 𝑔_{0} (𝑧) | \leq \frac{2}{3} .

This proves the base case. For the inductive step, assume the claim holds for each $𝑔_{0}, 𝑔_{1}, \dots, 𝑔_{𝑛 - 1}$ . Define

ℎ_{𝑛} (𝑧) = 𝑓 (𝑧) - \sum_{𝑘 = 0}^{𝑛 - 1} 𝑔_{𝑘} (𝑧)

for $𝑧 \in 𝐾$ . By the inductive hypothesis, we have $| ℎ_{𝑛} | \leq {(\frac{2}{3})}^{𝑛}$ on $𝐾$ . Define the disjoint closed sets

𝐴_{𝑛} = {𝑧 \in 𝐾 : - \frac{2^{𝑛}}{3^{𝑛}} \leq ℎ_{𝑛} (𝑧) \leq - \frac{2^{𝑛}}{3^{𝑛 + 1}}}

and

𝐵_{𝑛} = {𝑧 \in 𝐾 : \frac{2^{𝑛}}{3^{𝑛}} \geq ℎ_{𝑛} (𝑧) \geq \frac{2^{𝑛}}{3^{𝑛 + 1}}} .

Let $𝑔_{𝑛} (𝑧) = \frac{2^{𝑛}}{3^{𝑛 + 1}} 𝜂_{𝐴_{𝑛}, 𝐵_{𝑛}} (𝑧)$ , so that $| 𝑔_{𝑛} | \leq \frac{2^{𝑛}}{3^{𝑛 + 1}}$ on $ℂ$ , and

| ℎ_{𝑛} (𝑧) - 𝑔_{𝑛} (𝑧) | \leq \frac{2^{𝑛 + 1}}{3^{𝑛 + 1}}

for all $𝑧 \in 𝐾$ by the same argument as in the base case. Hence,

| 𝑓 (𝑧) - \sum_{𝑘 = 0}^{𝑛} 𝑔_{𝑘} (𝑧) | = | ℎ_{𝑛} (𝑧) - 𝑔_{𝑛} (𝑧) | \leq {(\frac{2}{3})}^{𝑛 + 1}

for all $𝑧 \in 𝐾$ , completing the induction. Because

| 𝑔 (𝑧) | \leq \sum_{𝑛 = 0}^{\infty} | 𝑔_{𝑛} (𝑧) | \leq \frac{1}{3} \sum_{𝑛 = 0}^{\infty} \frac{2^{𝑛}}{3^{𝑛}} = 1 \forall 𝑧 \in ℂ,

the Weierstrass $𝑀$ –Test (Theorem 2.3.3 $Theorem 2.3.3 Theorem 2.3.3$ ) implies that the series $\sum_{𝑛 = 0}^{\infty} 𝑔_{𝑛} (𝑧)$ converges uniformly on $ℂ$ to $𝑔$ . Since each $𝑔_{𝑛}$ is continuous, Theorem 2.3.6 $Theorem 2.3.6 Theorem 2.3.6$ gives the continuity of $𝑔$ on $ℂ$ . Finally, for any $𝑧 \in 𝐾$ , we have

| 𝑓 (𝑧) - 𝑔 (𝑧) | \leq lim_{𝑛 \to \infty} \frac{2^{𝑛 + 1}}{3^{𝑛 + 1}} = 0 . □

Corollary 6.2.5.1.

𝐾 \subseteq ℂ

is compact and

𝑓 : 𝐾 \to ℂ

is continuous, then

\exists 𝑔 \in 𝐶 (ℂ)

such that

𝑔 \equiv 𝑓

𝐾

and has compact support.

Proof.

Let $𝑓 = 𝑢 + i 𝑣$ where $𝑢, 𝑣 : 𝐾 \to ℝ$ are continuous. By Tietze–Urysohn–Brouwer (Theorem 6.2.5 $Theorem 6.2.5 Theorem 6.2.5$ ), $\exists \tilde{𝑢}, \tilde{𝑣} \in 𝐶 (ℂ)$ such that $\tilde{𝑢} \equiv 𝑢$ and $\tilde{𝑣} \equiv 𝑣$ on $𝐾$ . Let $𝑅 > 0$ be such that $𝐾 \subset 𝐷 (0, 𝑅)$ , provided by compactness. Define the piecewise-linear function

𝜓 (𝑧) = {\begin{cases} 1 & if | 𝑧 | \leq 𝑅, \\ 2 - \frac{| 𝑧 |}{𝑅} & if 𝑅 < | 𝑧 | < 2 𝑅, \\ 0 & if | 𝑧 | \geq 2 𝑅 \end{cases}

such that

𝜓 \in 𝐶 (ℂ)

and is compactly supported. Let

𝑔 (𝑧) = (\tilde{𝑢} (𝑧) + i \tilde{𝑣} (𝑧)) 𝜓 (𝑧)

, and the assertion follows. ⁠

□

Let $𝑓 \in 𝐶 (𝐾)$ be holomorphic on $\overset{⚬}{𝐾}$ . Then $𝑓$ has a continuous extension to all of $ℂ$ by virtue of Corollary 6.2.5.1 $Corollary 6.2.5.1 Corollary 6.2.5.1$ . Define the modulus of continuity of $𝑓$ to be the function $𝜔_{𝑓} : ℝ_{\geq 0} \to ℝ_{\geq 0}$ with

𝜔_{𝑓} (𝛿) = sup_{\begin{matrix} 𝑧, 𝜁 \in ℂ \\ | 𝑧 - 𝜁 | \leq 𝛿 \end{matrix}} | 𝑓 (𝑧) - 𝑓 (𝜁) | .

Because $𝑓$ has compact support, it must be uniformly continuous; hence we have ${lim}_{𝛿 \to 0^{+}} 𝜔_{𝑓} (𝛿) = 0$ .

For $𝑟 > 0$ , define

Φ (𝑧) = \iint_{ℂ} 𝜆_{𝑟} (𝑧 - 𝜁) 𝑓 (𝜁) d 𝜉 d 𝜂 where 𝜁 = 𝜉 + i 𝜂, (6.2 . 6)

where $𝜆_{𝑟}$ employs the same definition as in (6.2.4) $(6.2.4) (6.2.4)$ .

Proposition 6.2.6.

The function $Φ$ as in (6.2.6) $(6.2.6) (6.2.6)$ satisfies:

$Φ \in 𝐶^{1} (ℂ)$ and has compact support.
$Φ \equiv 𝑓$ on $𝑈 = {𝑧 \in 𝐾 : dist (𝑧, ℂ ∖ 𝐾) > 𝑟}$ .
$| 𝑓 (𝑧) - Φ (𝑧) | \leq 𝜔_{𝑓} (𝑟)$ for all $𝑧 \in ℂ$ .
For all $𝑧 \in ℂ$ , $| \frac{𝜕 Φ}{𝜕 \overline{𝑧}} (𝑧) | \leq \frac{4 π 𝜔_{𝑓} (𝑟)}{𝑟}$ .
$Φ (𝑧) = - \frac{1}{π} \iint_{𝐻} \frac{𝜕 Φ}{𝜕 \overline{𝜁}} (𝜁) \frac{d 𝜉 d 𝜂}{𝜁 - 𝑧}$ for $𝑧 \in ℂ$ , where $𝐻 = supp (Φ) ∖ 𝑈$ .

Proof.

Because $supp (𝜆_{𝑟} (𝑧 - 𝜁)) = \overline{𝐷 (𝑧, 𝑟)}$ and $supp 𝑓$ is compact, for sufficiently large $𝑧$ , the two supports will be disjoint and hence the integrand vanishes for all $𝜁$ . We can explicitly find that

\begin{aligned} \frac{𝜕 Φ}{𝜕 𝑥} & = lim_{\begin{matrix} Δ 𝑥 \to 0 \\ Δ 𝑥 \in ℝ \end{matrix}} \frac{Φ (𝑧 + Δ 𝑥) - Φ (Δ 𝑥)}{Δ 𝑥} \\ = lim_{\begin{matrix} Δ 𝑥 \to 0 \\ Δ 𝑥 \in ℝ \end{matrix}} \int_{ℂ} \frac{𝜆_{𝑟} (𝑧 + Δ 𝑥 - 𝜁) - 𝜆_{𝑟} (𝑧 - 𝜁)}{Δ 𝑥} 𝑓 (𝜁) d 𝜉 \land d 𝜂 . \end{aligned}

Because $𝑓$ is continuous and vanishes on a compact set, it is bounded. Similarly, Part 2 $Part 2 Part 2$ of Proposition 6.2.4 $Proposition 6.2.4 Proposition 6.2.4$ implies that $\frac{𝜕 𝜆_{𝑟}}{𝜕 𝑥}$ is bounded. Hence, by Lebesgue’s Dominated Convergence Theorem, we have

\frac{𝜕 Φ}{𝜕 𝑥} = \int_{ℂ} \frac{𝜕 𝜆_{𝑟}}{𝜕 𝑥} (𝑧 - 𝜁) 𝑓 (𝜁) d 𝜉 \land d 𝜂,

and similarly,

\frac{𝜕 Φ}{𝜕 𝑦} = \int_{ℂ} \frac{𝜕 𝜆_{𝑟}}{𝜕 𝑦} (𝑧 - 𝜁) 𝑓 (𝜁) d 𝜉 \land d 𝜂 .

Hence, $Φ \in 𝐶^{1} (ℂ)$ and this is Part 1 $Part 1 Part 1$ . Because

Φ (𝑧) = \iint_{ℂ} 𝜆_{𝑟} (𝑧 - 𝜁) 𝑓 (𝜁) d 𝜉 d 𝜂 = \iint_{ℂ} 𝜆_{𝑟} (𝜁) 𝑓 (𝑧 - 𝜁) d 𝜉 d 𝜂,

by Part 1 $Part 1 Part 1$ of Proposition 6.2.4 $Proposition 6.2.4 Proposition 6.2.4$ , we have

\begin{aligned} | 𝑓 (𝑧) - Φ (𝑧) | & \leq | \int_{ℂ} 𝑓 (𝑧) 𝜆_{𝑟} (𝜁) d 𝜉 \land d 𝜂 - \int_{ℂ} 𝑓 (𝑧 - 𝜁) 𝜆_{𝑟} (𝜁) d 𝜉 \land d 𝜂 | \\ = | \int_{ℂ} 𝜆_{𝑟} (𝜁) (𝑓 (𝑧) - 𝑓 (𝑧 - 𝜁)) d 𝜉 \land d 𝜂 | \\ \leq \int_{𝐷 (0, 𝑟)} 𝜆_{𝑟} (𝜁) | 𝑓 (𝑧) - 𝑓 (𝑧 - 𝜁) | d 𝜉 \land d 𝜂 \leq 𝜔_{𝑓} (𝑟), (6.2 . 7) \end{aligned}

which implies Part 3 $Part 3 Part 3$ . For $𝑧 \in 𝑈$ , $𝜁 \in 𝐷 (0, 𝑟)$ now implies that $𝑧 - 𝜁 \in \overset{⚬}{𝐾}$ and hence $𝑓 (𝑧) - 𝑓 (𝑧 - 𝜁)$ is holomorphic in $𝜁$ on $𝐷 (𝑧, 𝑟)$ . By Part 6 $Part 6 Part 6$ of Proposition 6.2.4 $Proposition 6.2.4 Proposition 6.2.4$ , (6.2.7) $(6.2.7) (6.2.7)$ becomes

| \int_{ℂ} 𝜆_{𝑟} (𝜁) (𝑓 (𝑧) - 𝑓 (𝑧 - 𝜁)) d 𝜉 \land d 𝜂 | = | 𝑓 (𝑧) - 𝑓 (𝑧 - 0) | = 0,

which proves Part 2 $Part 2 Part 2$ . Because $\forall 𝑧 \in ℂ$ ,

\begin{aligned} \frac{𝜕 Φ}{𝜕 \overline{𝑧}} (𝑧) & = \frac{1}{2} (\frac{𝜕 Φ}{𝜕 𝑥} + i \frac{𝜕 Φ}{𝜕 𝑦}) = \int_{ℂ} \frac{𝜕 𝜆_{𝑟}}{𝜕 \overline{𝑧}} (𝑧 - 𝜁) 𝑓 (𝜁) d 𝜉 \land d 𝜂 \\ = \int_{ℂ} \frac{𝜕 𝜆_{𝑟}}{𝜕 \overline{𝜁}} (𝜁) 𝑓 (𝑧 - 𝜁) d 𝜉 \land d 𝜂 \\ = \int_{ℂ} \frac{𝜕 𝜆_{𝑟}}{𝜕 \overline{𝜁}} (𝜁) 𝑓 (𝑧 - 𝜁) d 𝜉 \land d 𝜂 - 𝑓 (𝑧) \int_{ℂ} \frac{𝜕 𝜆_{𝑟}}{𝜕 \overline{𝜁}} d 𝜉 \land d 𝜂 \\ = \int_{ℂ} \frac{𝜕 𝜆_{𝑟}}{𝜕 \overline{𝜁}} (𝜁) (𝑓 (𝑧 - 𝜁) - 𝑓 (𝑧)) d 𝜉 \land d 𝜂 \end{aligned}

by Part 3 $Part 3 Part 3$ of Proposition 6.2.4 $Proposition 6.2.4 Proposition 6.2.4$ . Hence,

\begin{aligned} | \frac{𝜕 Φ}{𝜕 \overline{𝑧}} | & \leq \iint_{𝐷 (0, 𝑟)} | \frac{𝜕 𝜆_{𝑟}}{𝜕 \overline{𝜁}} | | 𝑓 (𝑧 - 𝜁) - 𝑓 (𝑧) | d 𝜉 d 𝜂 \\ \leq 𝜔_{𝑓} (𝑟) \iint_{𝐷 (0, 𝑟)} ‖ 𝛁 𝜆_{𝑟} ‖ d 𝜉 d 𝜂 \\ \leq \frac{4 𝜔_{𝑓} (𝑟)}{𝑟^{3}} \iint_{𝐷 (0, 𝑟)} d 𝜉 d 𝜂 \leq \frac{4 𝜔_{𝑓} (𝑟)}{𝑟^{3}} \cdot π 𝑟^{2} = \frac{4 π 𝜔_{𝑓} (𝑟)}{𝑟}, \end{aligned}

by Part 5

Part 5 Part 5

of Proposition 6.2.4

Proposition 6.2.4 Proposition 6.2.4

, confirming Part 4

Part 4 Part 4

. Finally, Part 5

Part 5 Part 5

follows from Corollary 3.1.3.1

Corollary 3.1.3.1 Corollary 3.1.3.1

(since outside the support the integral trivially vanishes and within

𝑈

\frac{𝜕 Φ}{𝜕 \overline{𝜁}}

vanishes as a consequence of holomorphy). ⁠

□

Theorem 6.2.7 (Mergelyan).

Let $𝐾 \subseteq ℂ$ be compact such that $\hat{ℂ} ∖ 𝐾$ has finitely many connected components. Let $𝐸 \subseteq \hat{ℂ} ∖ 𝐾$ contain exactly one point from each of the connected components of $\hat{ℂ} ∖ 𝐾$ . Suppose $𝑓 \in 𝐶 (𝐾)$ is holomorphic on $\overset{⚬}{𝐾}$ . Then $\forall 𝜀 > 0$ , there exists a rational function $𝜓 (𝑧)$ with poles in $𝐸$ such that

sup_{𝑧 \in 𝐾} | 𝜓 (𝑧) - 𝑓 (𝑧) | < 𝜀 .

Figure 6.2: The striped region bounds

𝐾

, while the thin lines bound the dotted region

𝑈

Proof.

Let $𝐹 = {𝑝_{𝑘}}_{1 \leq 𝑘 \leq 𝑛}$ contain precisely one point from each connected component of $\hat{ℂ} ∖ 𝐾$ (such that each $𝑝_{𝑘} \neq \infty$ is finite). Suppose that $𝑟$ is chosen such that $0 < \frac{3}{4} 𝑟 < dist (𝐾, 𝐹)$ so that for each $𝑝_{𝑘} \in 𝐹$ not equal to $\infty$ ,

\overline{𝐷 (𝑝_{𝑘}, \frac{3}{4} 𝑟)} \subset \hat{ℂ} ∖ 𝐾 .

Figure 6.3: The striped region represents

𝐻

while the unshaded regions represent the set which

𝜁_{𝑘}

can be in, each of which are denoted by small stars. The dotted disks represent the finite subcover of

𝐻

: notice that every striped region is also dotted.

Define the extension of $𝑓$ , $Φ$ , $𝑈$ , and $𝐻$ as in the previous results (see Figure 6.2 $Figure 6.2 Figure 6.2$ ). Hence, (see Figure 6.3 $Figure 6.3 Figure 6.3$ )

{𝐷 (𝜁_{𝑘}, \frac{5}{4} 𝑟) : (\forall) 𝜁_{𝑘} \in \hat{ℂ} ∖ (𝐾 \cup \overline{𝐷 (𝑝_{𝑘}, \frac{3}{4} 𝑟)}), 1 \leq 𝑘 \leq 𝑛}

covers a (compact) $𝑟$ -neighborhood of $\hat{ℂ} ∖ 𝐾$ (so that each $𝜁_{𝑘} \notin 𝐾$ , and is labeled so that each $𝜁_{𝑘}$ is in the same connected component as $𝑝_{𝑘}$ ) (in the case that $𝑝_{𝑘} = \infty$ , let the disk inside be the empty set). Thus, the collection also covers $𝐻$ . A finite subcover ${𝐷 (𝜁_{𝑘}^{(𝑗)}, \frac{5}{4} 𝑟)}_{\begin{matrix} 1 \leq 𝑗 \leq 𝑚_{𝑘} \\ 1 \leq 𝑘 \leq 𝑛 \end{matrix}}$ covering $𝐻$ exists by the Heine–Borel Theorem (Theorem 1.1.9 $Theorem 1.1.9 Theorem 1.1.9$ ).

By the connectivity of each component of $\hat{ℂ} ∖ 𝐾$ , there exists a piecewise-linear simple curve $𝛾_{𝑘}^{(𝑗)}$ for all $1 \leq 𝑘 \leq 𝑛$ , $1 \leq 𝑗 \leq 𝑚_{𝑘}$ , joining $𝜁_{𝑘}^{(𝑗)}$ and $𝑝_{𝑘}$ , which lies entirely within $\hat{ℂ} ∖ 𝐾$ . Since each $𝜁_{𝑘}^{(𝑗)} \notin \overline{𝐷 (𝑝_{𝑘}, \frac{3}{4} 𝑟)}$ by definition, it follows that $| 𝑝_{𝑘} - 𝜁_{𝑘}^{(𝑗)} | > \frac{3}{4} 𝑟$ .

Figure 6.4: The construction of

𝐸_{𝑘}^{𝑗}

. The entire polyline from

𝑝_{𝑘}

𝜁_{𝑘}^{(𝑗)}

𝛾_{𝑘}^{(𝑗)}

Figure 6.5: A conceptual construction of

{𝐻_{𝑘}^{(𝑗)}}

, which unions to

𝐻

Hence, the intersection $\overline{𝐷 (𝜁_{𝑘}^{(𝑗)}, \frac{3}{4} 𝑟)} \cap 𝛾_{𝑘}^{(𝑗)}$ consists of at least one connected component joining $𝜁_{𝑘}^{(𝑗)}$ to a point on $𝜕 𝐷 (𝜁_{𝑘}^{(𝑗)}, \frac{3}{4} 𝑟)$ . Denote the connected component of this intersection by $𝐸_{𝑘}^{(𝑗)}$ , satisfying $diam 𝐸_{𝑘}^{(𝑗)} \geq \frac{3}{4} 𝑟 > \frac{𝑟}{2}$ and $𝐸_{𝑘}^{(𝑗)} \cap 𝐾 = ⌀$ (see Figure 6.4 $Figure 6.4 Figure 6.4$ ).

Now for each $𝑗$ and $𝑘$ , Proposition 6.2.3 $Proposition 6.2.3 Proposition 6.2.3$ now provides the existence of a family of holomorphic functions $𝜑_{𝜁, 𝑘}^{(𝑗)} : \hat{ℂ} ∖ 𝐸_{𝑘}^{(𝑗)} \to ℂ$ given with $𝜁 \in 𝐷 (𝜁_{𝑘}^{(𝑗)}, \frac{5}{4} 𝑟)$ such that

| 𝜑_{𝜁, 𝑘}^{(𝑗)} (𝑧) | \leq \frac{584}{𝑟}, | 𝜑_{𝜁, 𝑘}^{(𝑗)} (𝑧) - \frac{1}{𝑧 - 𝜁} | \leq \frac{4676}{{| 𝑧 - 𝜁 |}^{3}}, \forall 𝑧 \in \hat{ℂ} ∖ 𝐸_{𝑘}^{(𝑗)} . (6.2 . 8)

Let ${\tilde{𝐻}}_{𝑘}^{(𝑗)} = 𝐻 \cap 𝐷 (𝜁_{𝑘}^{(𝑗)}, \frac{5}{4} 𝑟)$ , for each $𝑗, 𝑘$ and construct the disjoint sets

𝐻_{𝑘}^{(𝑗)} = {\tilde{𝐻}}_{𝑘}^{(𝑗)} ∖ (⋃_{𝑗^{'} < 𝑗} {\tilde{𝐻}}_{𝑘}^{(𝑗^{'})} \cup ⋃_{𝑘^{'} < 𝑘} ⋃_{𝑗^{'} \leq 𝑚_{𝑘^{'}}} {\tilde{𝐻}}_{𝑘^{'}}^{(𝑗^{'})}) if (𝑗, 𝑘) \neq (1, 1), 𝐻_{1}^{(1)} = {\tilde{𝐻}}_{1}^{(1)} .

as done in Figure 6.5 $Figure 6.5 Figure 6.5$ . Thus the union

⋃_{𝑘 = 1}^{𝑛} ⋃_{𝑗 = 1}^{𝑚_{𝑘}} 𝐻_{𝑘}^{(𝑗)} = 𝐻 \cap (⋃ 𝐷 (𝜁_{𝑘}^{(𝑗)}, \frac{5}{4} 𝑟)) = 𝐻

since the set of all $𝐷 (𝜁_{𝑘}^{(𝑗)}, \frac{5}{4} 𝑟)$ covers $𝐻$ . Let

Ψ (𝑧) = \frac{1}{π} \sum_{𝑘 = 1}^{𝑛} \sum_{𝑗 = 1}^{𝑚_{𝑘}} \int_{𝐻_{𝑘}^{(𝑗)}} \frac{𝜕 Φ}{𝜕 \overline{𝜁}} 𝜑_{𝜁, 𝑘}^{(𝑗)} (𝑧) d 𝜉 \land d 𝜂,

where $𝜁 = 𝜉 + i 𝜂, \forall 𝑧 \notin ⋃ 𝐸_{𝑘}^{(𝑗)}$ . Because

\frac{Ψ (𝑧 + Δ 𝑥) - Ψ (𝑧)}{Δ 𝑥} = \frac{1}{π} \sum_{𝑘 = 1}^{𝑛} \sum_{𝑗 = 1}^{𝑚_{𝑘}} \int_{𝐻_{𝑘}^{(𝑗)}} \frac{𝜕 Φ}{𝜕 \overline{𝜁}} \frac{𝜑_{𝜁, 𝑘}^{(𝑗)} (𝑧 + Δ 𝑥) - 𝜑_{𝜁, 𝑘}^{(𝑗)} (𝑧)}{Δ 𝑥} d 𝜉 \land d 𝜂,

and both $𝜕 Φ / 𝜕 \overline{𝜁}$ and the integrand is continuous on a set (we only need to consider the factors involving $𝜑_{𝜁, 𝑘}^{(𝑗)}$ since $𝜕 Φ / 𝜕 \overline{𝜁}$ is independent from $𝑧$ ) by Cauchy’s Estimates and the first bound of (6.2.8) $(6.2.8) (6.2.8)$ , Lebesgue’s Dominated Convergence gives that

\frac{𝜕 Ψ}{𝜕 𝑥} = \frac{1}{π} \sum_{𝑘 = 1}^{𝑛} \sum_{𝑗 = 1}^{𝑚_{𝑘}} \int_{𝐻_{𝑘}^{(𝑗)}} \frac{𝜕 Φ}{𝜕 \overline{𝜁}} \frac{𝜕 𝜑_{𝜁, 𝑘}^{(𝑗)}}{𝜕 𝑥} (𝑧) d 𝜉 \land d 𝜂,

and in analogous fashion,

\frac{𝜕 Ψ}{𝜕 𝑦} = \frac{1}{π} \sum_{𝑘 = 1}^{𝑛} \sum_{𝑗 = 1}^{𝑚_{𝑘}} \int_{𝐻_{𝑘}^{(𝑗)}} \frac{𝜕 Φ}{𝜕 \overline{𝜁}} \frac{𝜕 𝜑_{𝜁, 𝑘}^{(𝑗)}}{𝜕 𝑦} (𝑧) d 𝜉 \land d 𝜂 .

Hence, $Ψ$ is holomorphic on $\hat{ℂ} ∖ ⋃_{𝑘 = 1}^{𝑛} ⋃_{𝑗 = 1}^{𝑚_{𝑘}} 𝐸_{𝑘}^{(𝑗)}$ , a neighborhood of $𝐾$ . Since $\forall 𝑧 \in \hat{ℂ} ∖ ⋃ 𝐸_{𝑘}^{(𝑗)}$ , by Part 2 $Part 2 Part 2$ of Proposition 6.2.3 $Proposition 6.2.3 Proposition 6.2.3$ ,

\begin{aligned} | Ψ (𝑧) - Φ (𝑧) | & = | \frac{1}{π} \sum_{𝑘 = 1}^{𝑛} \sum_{𝑗 = 1}^{𝑚_{𝑘}} \iint_{𝐻_{𝑘}^{(𝑗)}} \frac{𝜕 Φ}{𝜕 \overline{𝜁}} 𝜑_{𝜁, 𝑘}^{(𝑗)} (𝑧) d 𝜉 d 𝜂 - \frac{1}{π} \iint_{𝐻} \frac{𝜕 Φ}{𝜕 \overline{𝜁}} \frac{d 𝜉 d 𝜂}{𝑧 - 𝜁} | \\ = \frac{1}{π} | \sum_{𝑘 = 1}^{𝑛} \sum_{𝑗 = 1}^{𝑚_{𝑘}} \iint_{𝐻_{𝑘}^{(𝑗)}} \frac{𝜕 Φ}{𝜕 \overline{𝜁}} (𝜑_{𝜁, 𝑘}^{(𝑗)} (𝑧) - \frac{1}{𝑧 - 𝜁}) d 𝜉 d 𝜂 | \\ \leq \frac{1}{π} \sum_{𝑘 = 1}^{𝑛} \sum_{𝑗 = 1}^{𝑚_{𝑘}} \iint_{𝐻_{𝑘}^{(𝑗)}} | \frac{𝜕 Φ}{𝜕 \overline{𝜁}} | | 𝜑_{𝜁, 𝑘}^{(𝑗)} (𝑧) - \frac{1}{𝑧 - 𝜁} | d 𝜉 d 𝜂 \\ \leq \frac{1}{π} \sum_{𝑘 = 1}^{𝑛} \sum_{𝑗 = 1}^{𝑚_{𝑘}} (\iint_{𝐻_{𝑘}^{(𝑗)} \cap 𝐷 (𝑧, 2 𝑟)} + \iint_{𝐻_{𝑘}^{(𝑗)} ∖ 𝐷 (𝑧, 2 𝑟)}) \\ \begin{matrix}  \end{matrix} \cdot | \frac{𝜕 Φ}{𝜕 \overline{𝜁}} | | 𝜑_{𝜁, 𝑘}^{(𝑗)} (𝑧) - \frac{1}{𝑧 - 𝜁} | d 𝜉 d 𝜂 \end{aligned}

The estimates in Part 4 $Part 4 Part 4$ of Proposition 6.2.6 $Proposition 6.2.6 Proposition 6.2.6$ , in tandem with those from (6.2.8) $(6.2.8) (6.2.8)$ now give that

\begin{aligned} | Ψ (𝑧) - Φ (𝑧) | & \leq 18704 𝑟 𝜔_{𝑓} (𝑟) \sum_{𝑘 = 1}^{𝑛} \sum_{𝑗 = 1}^{𝑚_{𝑘}} \iint_{𝐻_{𝑘}^{(𝑗)} ∖ 𝐷 (𝑧, 2 𝑟)} \frac{1}{{| 𝑧 - 𝜁 |}^{3}} d 𝜉 d 𝜂 \\ \begin{matrix}  \end{matrix} + \frac{4 𝜔_{𝑓} (𝑟)}{𝑟} \sum_{𝑘 = 1}^{𝑛} \sum_{𝑗 = 1}^{𝑚_{𝑘}} \iint_{𝐻_{𝑘}^{(𝑗)} \cap 𝐷 (𝑧, 2 𝑟)} (\frac{584}{𝑟} + \frac{1}{| 𝑧 - 𝜁 |}) d 𝜉 d 𝜂 \\ \leq 18704 𝑟 𝜔_{𝑓} (𝑟) \iint_{| 𝜁 | > 2 𝑟} \frac{d 𝜉 d 𝜂}{{| 𝜁 |}^{3}} \\ \begin{matrix}  \end{matrix} + \frac{4 𝜔_{𝑓} (𝑟)}{𝑟} \iint_{| 𝜁 | < 2 𝑟} (\frac{584}{𝑟} + \frac{1}{| 𝜁 |}) d 𝜉 d 𝜂 \end{aligned}

through a linear change of variables. Now evaluation via polar coordinates (with $𝜌 e^{i 𝜃} = 𝜉 + i 𝜂$ , $d 𝜉 \land d 𝜂 = 𝜌 d 𝑟 \land d 𝜃$ ) yields a revised upper bound of

\begin{aligned} 18704 𝑟 𝜔_{𝑓} (𝑟) \int_{0}^{2 π} \int_{2 𝑟}^{\infty} \frac{d 𝑟 d 𝜃}{𝜌^{2}} \\ \begin{matrix}  \end{matrix} + \frac{4 𝜔_{𝑓} (𝑟)}{𝑟} (\frac{584}{𝑟} area 𝐷 (0, 2 𝑟) + \int_{0}^{2 π} \int_{0}^{2 𝑟} d 𝑟 d 𝜃) \\ \begin{matrix}  \end{matrix} = {18704 𝑟 𝜔_{𝑓} (𝑟) 2 π \frac{1}{𝜌} |}_{\infty}^{2 𝑟} + \frac{9344 π 𝜔_{𝑓} (𝑟) 𝑟^{2}}{𝑟^{2}} + \frac{4 𝜔_{𝑓} (𝑟)}{𝑟} 4 𝑟 π \\ \begin{matrix}  \end{matrix} = 18704 π 𝜔_{𝑓} (𝑟) + 9344 π 𝜔_{𝑓} (𝑟) + 16 π 𝜔_{𝑓} (𝑟) = 28064 π 𝜔_{𝑓} (𝑟) . \end{aligned}

Runge’s Theorem (Theorem 6.1.4 $Theorem 6.1.4 Theorem 6.1.4$ ) provides the existence of some rational function $𝜓$ with poles in $𝐸$ such that

sup_{𝑧 \in 𝐾} | 𝜓 (𝑧) - Ψ (𝑧) | \leq π 𝜔_{𝑓} (𝑟)

since $Ψ$ is holomorphic on a neighborhood of $𝐾$ (to assure this bound is positive, we assume $𝑓$ is not identically zero, otherwise the assertion is trivial). Therefore, for all $𝑧 \in 𝐾$ , we have (the third supremum term coming from Part 3 $Part 3 Part 3$ of Proposition 6.2.6 $Proposition 6.2.6 Proposition 6.2.6$ )

\begin{aligned} sup_{𝑧 \in 𝐾} | 𝜓 (𝑧) - 𝑓 (𝑧) | & \leq sup_{𝑧 \in 𝐾} | 𝜓 (𝑧) - Ψ (𝑧) | + | Ψ (𝑧) - Φ (𝑧) | + | Φ (𝑧) - 𝑓 (𝑧) | \\ \leq 28065 π 𝜔_{𝑓} (𝑟) + 𝜔_{𝑓} (𝑟) \leq 28066 π 𝜔_{𝑓} (𝑟) . \end{aligned}

Because ${lim}_{𝑟 \to 0^{+}} 𝜔_{𝑓} (𝑟) = 0$ , for any $𝜀 > 0$ , there exists a $𝑟 > 0$ such that

𝜔_{𝑓} (𝑟) < \frac{𝜀}{28066 π} .

Hence for any such $𝜀$ , we now construct $𝜓$ in accordance with an $𝑟$ satisfying $28066 π 𝜔_{𝑓} (𝑟) < 𝜀$ . ⁠ $□$