§6.1 Runge’s Theorem

6.1 Runge’s Theorem

In the earliest formulation by Carl Runge in 1885, he provided the sufficiency of the holomorphy of $𝑓$ on $𝐾$ (in effect, a neighborhood of $𝐾$ ).

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 $𝑓$ and the subsequent use of Riemann sums to approximate the complex line integral.

Proposition 6.1.1.

Let $𝐾 \subseteq ℂ$ be compact and suppose $𝑈 \supset 𝐾$ is a neighborhood of $𝐾$ that is relatively compact in $ℂ$ . Let $𝑓 : 𝑈 \to ℂ$ be an arbitrary holomorphic function. Then for fixed $𝜀 > 0$ , there exists a rational function $𝜓 (𝑧)$ with only simple poles (all of which lie in $ℂ ∖ 𝐾$ ) such that

lim_{𝑧 \to \infty} 𝜓 (𝑧) = 0, sup_{𝑧 \in 𝐾} | 𝑓 (𝑧) - 𝜓 (𝑧) | < 𝜀 .

Figure 6.1: The elements of

𝒢︀

, relative to

𝐾

and its neighborhood,

𝑈

Proof.

By assumption of relative compactness, $𝜎 = dist (𝜕 𝑈, 𝜕 𝐾)$ , or the distance (infimum) between $𝐾$ and $ℂ ∖ 𝑈$ , is positive and finite. More concretely, let

𝜎 = inf {| 𝑧_{1} - 𝑧_{2} | : 𝑧_{1} \in 𝐾, 𝑧_{2} \in ℂ ∖ 𝑈} > 0 .

The longest distance between two points in any square is the length of the diagonal. Hence, any square $𝑄$ that intersects $𝜕 𝐾$ with a side length less than $𝜎 \frac{\sqrt{2}}{2}$ will lie completely within $𝑈$ .

Choose $𝑚 \in ℕ$ to satisfy $2^{1 - 𝑚} < 𝜎$ and consider the grid generated by compact squares in the form of

{𝑥 + i 𝑦 : \frac{𝑗}{2^{𝑚}} \leq 𝑥 \leq \frac{𝑗 + 1}{2^{𝑚}}, \frac{𝑘}{2^{𝑚}} \leq 𝑦 \leq \frac{𝑘 + 1}{2^{𝑚}}}

(where $𝑗$ and $𝑘$ are integers) and let $𝒢︀$ be the collection of all such squares in this grid that intersect $𝐾$ , and it follows that $\tilde{𝐾} = ⋃_{𝑄 \in 𝒢︀} 𝑄 \subset 𝑈$ (refer to Figure 6.1 $Figure 6.1 Figure 6.1$ ).

As a consequence of Cauchy–Goursat (Theorem 3.1.10 $Theorem 3.1.10 Theorem 3.1.10$ ), we have

\frac{1}{2 π i} \oint_{𝜕 \tilde{𝐾}} \frac{𝑓 (𝜁) d 𝜁}{𝜁 - 𝑧} = 𝑓 (𝑧) (6.1 . 1)

in the case that $𝑧 \in \tilde{𝐾}$ . The boundary $𝜕 \tilde{𝐾}$ may be written as the union of $𝑛$ lines parameterized by $0 \leq 𝑡 \leq 1$ ; more concretely, we have $𝜕 \tilde{𝐾} = ⋃_{𝑗 \in ℕ_{\leq 𝑛}} 𝛾_{𝑗} ([0, 1])$ . Hence we have in equivalent formulation,

𝑓 (𝑧) = \frac{1}{2 π i} \sum_{𝑗 = 1}^{𝑛} \int_{𝛾_{𝑗} ([0, 1])} \frac{𝑓 (𝜁)}{𝜁 - 𝑧} d 𝜁 = \frac{1}{2 π i} \sum_{𝑗 = 1}^{𝑛} \int_{0}^{1} \frac{𝑓 (𝛾_{𝑗} (𝑡)) 𝛾_{𝑗}^{'} (𝑡)}{𝛾_{𝑗} (𝑡) - 𝑧} d 𝑡 .

The distance between $𝐾$ and $𝜕 \tilde{𝐾}$ is strictly positive. Suppose instead that the distance were zero. Then some point of $𝐾$ would lie on the boundary of a square $𝑄 \in 𝒢︀$ and also lie on $𝜕 \tilde{𝐾}$ . If this point lies on an edge of $𝑄$ (but not at a vertex), then the square adjacent along that edge must also intersect $𝐾$ , and hence belong to $𝒢︀$ , contradicting the assumption that the point lies on $𝜕 \tilde{𝐾}$ . If the point lies at a vertex of $𝑄$ , then all three adjacent squares also intersect $𝐾$ , so they too belong to $𝒢︀$ , leading to the same contradiction. Thus, the distance must be positive.

Hence, each integrand as defined in (6.1.1) $(6.1.1) (6.1.1)$ is jointly continuous for $𝑡 \in [0, 1]$ and $𝑧 \in 𝐾$ . By compactness of the product, it is in fact uniformly continuous by Heine–Cantor (Theorem 1.2.26 $Theorem 1.2.26 Theorem 1.2.26$ ).

Hence, $\forall 𝜀 > 0$ , $\exists 𝛿 > 0$ such that $\forall 𝑧 \in 𝐾$ , $\forall 1 \leq 𝑗 \leq 𝑛$ (uniform in $𝑗$ as we can take the minimum of each $𝛿_{𝑗}$ ), and $\forall 𝑡_{1}, 𝑡_{2} \in [0, 1]$ satisfying $| 𝑡_{1} - 𝑡_{2} | < 𝛿$ ,

| \frac{𝑓 (𝛾_{𝑗} (𝑡_{1})) 𝛾_{𝑗}^{'} (𝑡_{1})}{𝛾_{𝑗} (𝑡_{1}) - 𝑧} - \frac{𝑓 (𝛾_{𝑗} (𝑡_{2})) 𝛾_{𝑗}^{'} (𝑡_{2})}{𝛾_{𝑗} (𝑡_{2}) - 𝑧} | < \frac{𝜀}{𝑛} .

Partition $[0, 1]$ by $0 = 𝑡_{0} < 𝑡_{1} < \dots < 𝑡_{𝑚} = 1$ such that $\forall 0 \leq 𝑘 < 𝑚$ , $Δ 𝑡_{𝑘} = 𝑡_{𝑘 + 1} - 𝑡_{𝑘} < 𝛿$ . It follows that

\begin{aligned} | 𝑓 (𝑧) - \frac{1}{2 π i} \sum_{𝑗 = 1}^{𝑛} \sum_{𝑘 = 0}^{𝑚 - 1} \frac{𝑓 (𝛾_{𝑗} (𝑡_{𝑘})) 𝛾_{𝑗}^{'} (𝑡_{𝑘})}{𝛾_{𝑗} (𝑡_{𝑘}) - 𝑧} Δ 𝑡_{𝑘} | \\ \begin{matrix}  \end{matrix} = \frac{1}{2 π} | \sum_{𝑗 = 1}^{𝑛} \sum_{𝑘 = 0}^{𝑚 - 1} \int_{𝑡_{𝑘}}^{𝑡_{𝑘 + 1}} [\frac{𝑓 (𝛾_{𝑗} (𝑡)) 𝛾_{𝑗}^{'} (𝑡)}{𝛾_{𝑗} (𝑡) - 𝑧} - \frac{𝑓 (𝛾_{𝑗} (𝑡_{𝑘})) 𝛾_{𝑗}^{'} (𝑡_{𝑘})}{𝛾_{𝑗} (𝑡_{𝑘}) - 𝑧}] d 𝑡 | \\ \begin{matrix}  \end{matrix} \leq \frac{1}{2 π} \sum_{𝑗 = 1}^{𝑛} \sum_{𝑘 = 0}^{𝑚 - 1} \int_{𝑡_{𝑘}}^{𝑡_{𝑘 + 1}} | \frac{𝑓 (𝛾_{𝑗} (𝑡)) 𝛾_{𝑗}^{'} (𝑡)}{𝛾_{𝑗} (𝑡) - 𝑧} - \frac{𝑓 (𝛾_{𝑗} (𝑡_{𝑘})) 𝛾_{𝑗}^{'} (𝑡_{𝑘})}{𝛾_{𝑗} (𝑡_{𝑘}) - 𝑧} | d 𝑡 \\ \begin{matrix}  \end{matrix} \leq \frac{𝜀}{2 𝑛 π} \sum_{𝑗 = 1}^{𝑛} \sum_{𝑘 = 0}^{𝑚 - 1} Δ 𝑡_{𝑘} = \frac{𝜀}{2 π} < 𝜀 \end{aligned}

uniformly in $𝑧 \in 𝐾$ . The summation

𝜓 (𝑧) = \frac{1}{2 π i} \sum_{𝑗 = 1}^{𝑛} \sum_{𝑘 = 0}^{𝑚 - 1} \frac{𝑓 (𝛾_{𝑗} (𝑡_{𝑘})) 𝛾_{𝑗}^{'} (𝑡_{𝑘})}{𝛾_{𝑗} (𝑡_{𝑘}) - 𝑧} Δ 𝑡_{𝑘}

defines a rational function with simple poles at each $𝛾_{𝑗} (𝑡_{𝑘}) \in 𝜕 \tilde{𝐾}$ , which is disjoint from $𝐾$ . ⁠ $□$

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 6.1.2 (Pole-Pushing Lemma).

Let $𝛼, 𝛽 \in ℂ$ and let $𝑓 (𝑧)$ be a rational function with a single singularity, a pole at $𝑧 = 𝛼$ , whose Laurent expansion consists solely of its principal part. Then $\forall 𝑟 > | 𝛼 - 𝛽 |$ , $\forall 𝜀 > 0$ , there exists a rational function $𝜓 (𝑧)$ whose only singularity is a pole at $𝑧 = 𝛽$ such that

sup_{𝑧 \in \hat{ℂ} ∖ 𝐷 (𝛽, 𝑟)} | 𝑓 (𝑧) - 𝜓 (𝑧) | < 𝜀 .

Proof.

By assumption, $𝑓$ can be expressed as a polynomial of

\begin{aligned} {(𝑧 - 𝛼)}^{- 1} & = {(𝑧 - 𝛽)}^{- 1} \frac{1}{(𝑧 - 𝛼) {(𝑧 - 𝛽)}^{- 1}} \\ = {(𝑧 - 𝛽)}^{- 1} \frac{1}{1 - (𝛼 - 𝛽) {(𝑧 - 𝛽)}^{- 1}} \\ = {(𝑧 - 𝛽)}^{- 1} \sum_{𝑘 = 0}^{\infty} {(\frac{𝛼 - 𝛽}{𝑧 - 𝛽})}^{𝑘} . \end{aligned}

This series locally uniformly converges on

{𝑧 \in \hat{ℂ} : | 𝛼 - 𝛽 | {| 𝑧 - 𝛽 |}^{- 1} < 1} = \hat{ℂ} ∖ \overline{𝐷 (𝛽, | 𝛼 - 𝛽 |)}

and uniformly converges on $\hat{ℂ} ∖ 𝐷 (𝛽, 𝑟)$ . Hence, for $𝑚 \in ℕ$ , we have

𝑓 (𝑧) = \sum_{𝑗 = 1}^{𝑚} 𝑎_{- 𝑗} {({(𝑧 - 𝛽)}^{- 1} \sum_{𝑘 = 0}^{\infty} {(\frac{𝛼 - 𝛽}{𝑧 - 𝛽})}^{𝑘})}^{𝑗} .

For fixed $𝑗$ (where $𝑎_{- 𝑗} \neq 0$ ), we aim to prove the existence of an $𝑁 \in ℕ$ such that $\forall 𝑛 > 𝑁$ , we have at least

\begin{aligned} | & {[{(𝑧 - 𝛽)}^{- 1} \sum_{𝑘 = 0}^{\infty} {(\frac{𝛼 - 𝛽}{𝑧 - 𝛽})}^{𝑘}]}^{𝑗} \\ \begin{matrix}  \end{matrix} - {[{(𝑧 - 𝛽)}^{- 1} \sum_{𝑘 = 0}^{𝑛} {(\frac{𝛼 - 𝛽}{𝑧 - 𝛽})}^{𝑘}]}^{𝑗} | < \frac{𝜀}{𝑚 | 𝑎_{- 𝑗} |}, (6.1 . 2) \end{aligned}

where $𝑧 \in \hat{ℂ} ∖ 𝐷 (𝛽, 𝑟)$ . Since $| \frac{1}{𝑧 - 𝛽} | < \frac{1}{𝑟}$ , we can restrict (6.1.2) $(6.1.2) (6.1.2)$ further with

| {(\sum_{𝑘 = 0}^{\infty} {(\frac{𝛼 - 𝛽}{𝑧 - 𝛽})}^{𝑘})}^{𝑗} - {(\sum_{𝑘 = 0}^{𝑛} {(\frac{𝛼 - 𝛽}{𝑧 - 𝛽})}^{𝑘})}^{𝑗} | < 𝑟^{𝑗} \frac{𝜀}{𝑚 | 𝑎_{- 𝑗} |} .

Additionally, the difference on the left-hand side is also equal to

\begin{aligned} | \sum_{𝑘 = 𝑛 + 1}^{\infty} {(\frac{𝛼 - 𝛽}{𝑧 - 𝛽})}^{𝑘} | \\ \begin{matrix}  \end{matrix} \cdot | \sum_{𝑙 = 0}^{𝑗 - 1} {(\sum_{𝑘 = 0}^{𝑛} {(\frac{𝛼 - 𝛽}{𝑧 - 𝛽})}^{𝑘})}^{𝑙} {(\sum_{𝑘 = 0}^{\infty} {(\frac{𝛼 - 𝛽}{𝑧 - 𝛽})}^{𝑘})}^{𝑗 - 𝑙 - 1} | . (6.1 . 3) \end{aligned}

For any $𝑛 \in ℕ$ , we have

| \sum_{𝑘 = 0}^{𝑛} {(\frac{𝛼 - 𝛽}{𝑧 - 𝛽})}^{𝑘} | \leq \sum_{𝑘 = 0}^{𝑛} {| \frac{𝛼 - 𝛽}{𝑧 - 𝛽} |}^{𝑘} \leq \sum_{𝑘 = 0}^{\infty} {| 𝛼 - \frac{𝛽}{𝑟} |}^{𝑘} \leq \frac{1}{1 - | 𝛼 - \frac{𝛽}{𝑟} |} .

Since the dominating sequence of partial sums are monotonically increasing, it follows that the sequence of partial sums is uniformly bounded by

𝑀 = \frac{𝑟}{𝑟 - | 𝛼 - 𝛽 |}

on $\hat{ℂ} ∖ 𝐷 (𝛽, 𝑟)$ . Thus, (6.1.3) $(6.1.3) (6.1.3)$ is bounded by $𝑀^{𝑗 - 1} 𝑗 | \sum_{𝑘 = 𝑛 + 1}^{\infty} {(\frac{𝛼 - 𝛽}{𝑧 - 𝛽})}^{𝑘} |$ , and we apply further restriction by setting this to be bounded by $𝑟^{𝑗} \frac{𝜀}{𝑚 | 𝑎_{- 𝑗} |}$ . By uniform convergence, for any $𝜀 > 0$ , $\exists 𝑁_{𝑗} \in ℕ$ such that $\forall 𝑛 > 𝑁_{𝑗}$ ,

| \sum_{𝑘 = 𝑛 + 1}^{\infty} {(\frac{𝛼 - 𝛽}{𝑧 - 𝛽})}^{𝑘} | < 𝑟^{𝑗} \frac{𝜀}{𝑀^{𝑗 - 1} 𝑗 𝑚 | 𝑎_{- 𝑗} |} .

For $𝑛 > 𝑁_{𝑗}$ , (6.1.2) $(6.1.2) (6.1.2)$ is satisfied, and $\forall 𝑛 > {max}_{\begin{matrix} 𝑗 \in ℕ_{\leq 𝑚} \\ 𝑎_{- 𝑗} \neq 0 \end{matrix}} (𝑁_{𝑗})$ , $𝑧 \in \hat{ℂ} ∖ 𝐷 (𝛽, 𝑟)$ , we have

\begin{aligned} | 𝑓 (𝑧) - \sum_{𝑗 = 1}^{𝑚} 𝑎_{- 𝑗} ({(\frac{1}{𝑧 - 𝛽} \sum_{𝑘 = 0}^{𝑛} {(\frac{𝛼 - 𝛽}{𝑧 - 𝛽})}^{𝑘})}^{𝑗}) | \\ \begin{matrix}  \end{matrix} \leq \sum_{\begin{matrix} 𝑗 = 1 \\ 𝑎_{- 𝑗} \neq 0 \end{matrix}}^{𝑚} | 𝑎_{- 𝑗} | | {(\sum_{𝑘 = 0}^{\infty} {(\frac{𝛼 - 𝛽}{𝑧 - 𝛽})}^{𝑘})}^{𝑗} - {(\sum_{𝑘 = 0}^{𝑛} {(\frac{𝛼 - 𝛽}{𝑧 - 𝛽})}^{𝑘})}^{𝑗} | \\ \begin{matrix}  \end{matrix} \leq \sum_{\begin{matrix} 𝑗 = 1 \\ 𝑎_{- 𝑗} \neq 0 \end{matrix}}^{𝑚} | 𝑎_{- 𝑗} | \frac{𝜀}{𝑚 | 𝑎_{- 𝑗} |} \leq 𝜀, \end{aligned}

which completes the proof as

𝜓 (𝑧) = \sum_{𝑗 = 1}^{𝑚} 𝑎_{- 𝑗} {({(𝑧 - 𝛽)}^{- 1} \sum_{𝑘 = 0}^{𝑛} {(\frac{𝛼 - 𝛽}{𝑧 - 𝛽})}^{𝑘})}^{𝑗}

is rational with a pole at

𝑧 = 𝛽

. ⁠

□

Lemma 6.1.3 (Generalized Pole-Pushing Lemma).

Let $𝐾 \subseteq ℂ$ be compact and choose $𝑎 \in ℂ ∖ 𝐾$ . Let $𝑈$ be the connected component of $\hat{ℂ} ∖ 𝐾$ containing $𝑎$ . Then $\forall 𝜀 > 0$ , $\forall 𝜁 \in 𝑈$ , there exists a rational function $𝜓$ with a pole only at $𝜁$ such that

sup_{𝑧 \in 𝐾} | \frac{1}{𝑧 - 𝑎} - 𝜓 (𝑧) | < 𝜀 .

Proof.

Define the set

𝑆 = {𝜁 \in 𝑈 ∖ {\infty} : (\forall 𝜀 > 0) (\exists 𝜓) [\begin{matrix} 𝜓 is rational, \\ 𝜓 (ℂ ∖ {𝜁}) \subseteq ℂ \land 𝜓 (𝜁) = \infty \\ {sup}_{𝑧 \in 𝐾} | \frac{1}{𝑧 - 𝑎} - 𝜓 (𝑧) | < 𝜀 \end{matrix}]} .

Since $𝑎 \in 𝑈$ satisfies the condition with $𝜓 (𝑧) = \frac{1}{𝑧 - 𝑎}$ , it follows that $𝑎 \in 𝑆$ , ensuring that $𝑆$ is nonempty.

Consider $𝜁 \in 𝑆$ , where $𝜁$ lies in the complement of $𝐾$ . The distance from $𝜁$ to $𝐾$ , denoted $dist (𝜁, 𝐾)$ , is positive, and the open disk $𝐷 (𝜁, dist (𝜁, 𝐾))$ is disjoint from $𝐾$ . Let $𝜁^{'}$ be an arbitrary point in this disk. By the definition of $𝑆$ , for every $𝜀 > 0$ , there exists a rational function $𝜓$ with a pole only at $𝜁$ such that

sup_{𝑧 \in 𝐾} | \frac{1}{𝑧 - 𝑎} - 𝜓 (𝑧) | < \frac{𝜀}{2} .

By Lemma 6.1.2 $Lemma 6.1.2 Lemma 6.1.2$ , there exists a rational function $𝜙$ with a pole only at $𝜁^{'}$ such that

sup_{𝑧 \in \hat{ℂ} ∖ 𝐷 (𝜁, dist (𝜁, 𝐾))} | 𝜙 (𝑧) - 𝜓 (𝑧) | < \frac{𝜀}{2},

which implies

sup_{𝑧 \in 𝐾} | 𝜙 (𝑧) - 𝜓 (𝑧) | < \frac{𝜀}{2} .

Thus,

sup_{𝑧 \in 𝐾} | \frac{1}{𝑧 - 𝑎} - 𝜙 (𝑧) | \leq sup_{𝑧 \in 𝐾} | \frac{1}{𝑧 - 𝑎} - 𝜓 (𝑧) | + sup_{𝑧 \in 𝐾} | 𝜓 (𝑧) - 𝜙 (𝑧) | < 𝜀,

and by definition, $𝜁^{'} \in 𝑆 ⟹ 𝐷 (𝜁, dist (𝜁, 𝐾)) \subseteq 𝑆$ . Hence, $𝑆$ is relatively open in $𝑈 ∖ {\infty}$ .

Now, consider $𝜁 \in 𝑈 ∖ (𝑆 \cup {\infty})$ . Suppose there exists $𝜁^{'} \in 𝐷 (𝜁, dist (𝜁, 𝐾)) \cap 𝑆$ . By repeated application of the preceding argument, this would imply $𝜁 \in 𝑆$ , contradicting the assumption that $𝜁 \in 𝑈 ∖ (𝑆 \cup {\infty})$ . Therefore, no such $𝜁^{'}$ exists, and $𝑆$ is relatively closed in $𝑈 ∖ {\infty}$ .

Since $𝑈 ∖ {\infty}$ is connected and $𝑆$ is both relatively open and closed in $𝑈 ∖ {\infty}$ , it follows from Theorem 3.2.1.19 $Theorem 3.2.1.19 Theorem 3.2.1.19$ that $𝑆 = 𝑈 ∖ {\infty}$ , completing the proof under the assumption that $\infty \notin 𝑈$ .

Now suppose $\infty \in 𝑈$ . In essence, we pole push to a point outside a disk on which we can make approximations by Taylor polynomials. Let $𝑅 > 0$ satisfy $𝐾 \subset 𝐷 (0, 𝑅)$ and let $𝑏 \in 𝑈 ∖ ({\infty} \cup \overline{𝐷 (0, 𝑅)})$ be an arbitrary point. By Lemma 6.1.2 $Lemma 6.1.2 Lemma 6.1.2$ , there exists some rational function $\tilde{𝜓} (𝑧)$ with a pole at $𝑏$ such that

sup_{𝑧 \in 𝐾} | \tilde{𝜓} (𝑧) - \frac{1}{𝑧 - 𝑎} | < \frac{𝜀}{2} .

Since $\tilde{𝜓}$ is holomorphic on some neighborhood of $\overline{𝐷 (0, 𝑅)}$ , we have

\tilde{𝜓} (𝑧) = \sum_{𝑘 = 0}^{\infty} 𝑎_{𝑘} 𝑧^{𝑘} on \overline{𝐷 (0, 𝑅)},

and it uniformly converges on $\overline{𝐷 (0, 𝑅)}$ . Hence, $\exists 𝑁 \in ℕ$ such that

sup_{𝑧 \in \overline{𝐷 (0, 𝑅)}} | \tilde{𝜓} (𝑧) - \sum_{𝑘 = 0}^{𝑁} 𝑎_{𝑘} 𝑧^{𝑘} | < \frac{𝜀}{2} .

Since polynomials have poles at $\infty$ , we have

\begin{aligned} sup_{𝑧 \in \overline{𝐷 (0, 𝑅)}} | \frac{1}{𝑧 - 𝑎} - \sum_{𝑘 = 0}^{𝑁} 𝑎_{𝑘} 𝑧^{𝑘} | & \leq sup_{𝑧 \in \overline{𝐷 (0, 𝑅)}} | \frac{1}{𝑧 - 𝑎} - \tilde{𝜓} (𝑧) | \\ \begin{matrix}  \end{matrix} + sup_{𝑧 \in \overline{𝐷 (0, 𝑅)}} | \tilde{𝜓} (𝑧) - \sum_{𝑘 = 0}^{𝑁} 𝑎_{𝑘} 𝑧^{𝑘} | < 𝜀 . □ \end{aligned}

Theorem 6.1.4 (Runge).

Let $𝐾 \subseteq ℂ$ be compact such that $\hat{ℂ} ∖ 𝐾$ has finitely many connected components and suppose $𝑓$ is holomorphic on a neighborhood of $𝐾$ . Let $𝐸$ be a subset of $\hat{ℂ} ∖ 𝐾$ containing one point from each of its connected components. Then $\forall 𝜀 > 0$ , there is a rational function $𝜓$ whose poles lie in $𝐸$ such that

sup_{𝑧 \in 𝐾} | 𝑓 (𝑧) - 𝜓 (𝑧) | < 𝜀 .

Proof.

By Proposition 6.1.1 $Proposition 6.1.1 Proposition 6.1.1$ , there is a rational function $𝜙$ with simple poles in $ℂ ∖ 𝐾$ satisfying $𝜙 (\infty) = 0$ such that

sup_{𝑧 \in 𝐾} | 𝑓 (𝑧) - 𝜙 (𝑧) | < \frac{𝜀}{2} . (6.1 . 4)

Let the poles of $𝜙$ be ${𝛽_{𝑘}}_{𝑘 \in ℕ_{\leq 𝑛}} \subseteq ℂ ∖ 𝐾$ , and as a consequence, we have $𝜙 (𝑧) = \sum_{𝑘 = 1}^{𝑛} \frac{𝑎_{𝑘}}{𝑧 - 𝛽_{𝑘}} + 𝜑 (𝑧)$ where $𝜑 (𝑧)$ is entire. Since $𝜙 (\infty) = 0$ , we have $𝜑 \equiv 0$ by Liouville’s Theorem (Theorem 3.2.5 $Theorem 3.2.5 Theorem 3.2.5$ ). By Lemma 6.1.3 $Lemma 6.1.3 Lemma 6.1.3$ , there exist rational functions ${𝜓_{𝑘}}_{𝑘 \in ℕ_{\leq 𝑛}}$ whose only poles lie in $𝐸$ such that $\forall 𝑘$ ,

sup_{𝑧 \in 𝐾} | \frac{1}{𝑧 - 𝛽_{𝑘}} - 𝜓_{𝑘} (𝑧) | < \frac{𝜀}{2 𝑛 | 𝑎_{𝑘} |}

and it follows that

sup_{𝑧 \in 𝐾} | 𝜙 (𝑧) - \sum_{𝑘 = 1}^{𝑛} 𝑎_{𝑘} 𝜓_{𝑘} (𝑧) | \leq sup_{𝑧 \in 𝐾} \sum_{𝑘 = 1}^{𝑛} | \frac{𝑎_{𝑘}}{𝑧 - 𝛽_{𝑘}} - 𝑎_{𝑘} 𝜓_{𝑘} (𝑧) | < \frac{𝜀}{2} .

Let $𝜓 (𝑧) = \sum_{𝑘 = 1}^{𝑛} 𝑎_{𝑘} 𝜓_{𝑘} (𝑧)$ . From (6.1.4) $(6.1.4) (6.1.4)$ , we have

sup_{𝑧 \in 𝐾} | 𝑓 (𝑧) - 𝜓 (𝑧) | \leq sup_{𝑧 \in 𝐾} | 𝑓 (𝑧) - 𝜙 (𝑧) | + sup_{𝑧 \in 𝐾} | 𝜙 (𝑧) - 𝜓 (𝑧) | < 𝜀 . □