§13.1 Analytic Function Elements

13.1 Analytic Function Elements

Definition 13.1.1.

An (analytic) function element is a pair

(𝑓, 𝑈)

, where

𝑈 \subseteq ℂ

is an open region and

𝑓

is a holomorphic function on

𝑈

Analytic elements serve as local representations of analytic functions. The process of extending these elements is formalized through analytic continuation. In the definition above for a function element, we let $𝑈$ be an arbitrary region. However, it is often more practical to consider the explicit case where $𝑈$ is a disk.

By the Identity Theorem (Theorem 3.3.3 $Theorem 3.3.3 Theorem 3.3.3$ ), such continuations are unique; if $(𝑓_{1}, 𝑈_{1})$ and $(𝑓_{2}, 𝑈_{2})$ are analytic elements with $𝑈_{1} \cap 𝑈_{2} \neq ⌀$ , and $𝑓_{1} \equiv 𝑓_{2}$ on $𝑈_{1} \cap 𝑈_{2}$ , then they are direct analytic continuations of each other. The combined function:

\tilde{𝑓} (𝑧) = {\begin{cases} 𝑓_{1} (𝑧) & if 𝑧 \in 𝑈_{1}, \\ 𝑓_{2} (𝑧) & if 𝑧 \in 𝑈_{2} ∖ 𝑈_{1}, \end{cases}

is holomorphic on $𝑈_{1} \cup 𝑈_{2}$ .

The most straightforward method of the derivation of analytic continuations uses power series. Let $𝑓 (𝑧) = \sum_{𝑛 = 0}^{\infty} 𝑐_{𝑛} {(𝑧 - 𝑧_{0})}^{𝑛}$ have radius of convergence $𝑅 > 0$ (by Theorem 2.3.4 $Theorem 2.3.4 Theorem 2.3.4$ ). For $𝑧_{1} \in 𝐷 (𝑧_{0}, 𝑅)$ , we can expand $𝑓$ at $𝑧_{1}$ :

𝑓 (𝑧) = \sum_{𝑘 = 0}^{\infty} \frac{𝑓^{(𝑘)} (𝑧_{1})}{𝑘!} {(𝑧 - 𝑧_{1})}^{𝑘} .

Let $𝜌$ be the radius of convergence of this series. Then:

𝜌 \geq 𝑅 - | 𝑧_{1} - 𝑧_{0} | .

If $𝜌 > 𝑅 - | 𝑧_{1} - 𝑧_{0} |$ , then $𝑓$ extends analytically to $𝐷 (𝑧_{0}, 𝑅) \cup 𝐷 (𝑧_{1}, 𝜌)$ . In the case that $𝜌 = 𝑅 - | 𝑧_{1} - 𝑧_{0} |$ , the disks $𝐷 (𝑧_{0}, 𝑅)$ and $𝐷 (𝑧_{1}, 𝜌)$ are tangent at a point $𝜁_{0}$ . Here, $𝜁_{0}$ is a singularity, and $𝑓$ cannot be continued beyond $𝜁_{0}$ .

Theorem 13.1.2.

Let

𝑓 (𝑧) = \sum_{𝑛 = 0}^{\infty} 𝑐_{𝑛} {(𝑧 - 𝑧_{0})}^{𝑛}

have radius of convergence

𝑅 > 0

. Then

𝜕 𝐷 (𝑧_{0}, 𝑅)

contains at least one singularity of

𝑓

Proof.

Assume $𝑓$ can be analytically continued from every $𝜁 \in 𝜕 𝐷 (𝑧_{0}, 𝑅)$ . Then for each $𝜁$ , there exists $𝑟_{𝜁} > 0$ and a holomorphic $𝑓_{𝜁}$ on $𝐷 (𝜁, 𝑟_{𝜁})$ agreeing with $𝑓$ on $𝐷 (𝜁, 𝑟_{𝜁}) \cap 𝐷 (𝑧_{0}, 𝑅)$ .

The disks ${𝐷 (𝜁, 𝑟_{𝜁})}_{𝜁 \in 𝜕 𝐷 (𝑧_{0}, 𝑅)}$ cover $𝜕 𝐷 (𝑧_{0}, 𝑅)$ . Then from compactness and the Heine–Borel Theorem (Theorem 1.1.9 $Theorem 1.1.9 Theorem 1.1.9$ ), the cover of disks admits a finite subcover ${𝐷 (𝜁_{𝑘}, 𝑟_{𝑘})}_{𝑘 = 1}^{𝑛}$ . Hence, $\exists 𝜌 > 0$ such that $𝐴 = {𝑧 : 𝑅 - 𝜌 \leq | 𝑧 - 𝑧_{0} | \leq 𝑅 + 𝜌} \subset 𝑉$ , where we let $𝑉 = ⋃_{𝑘 = 1}^{𝑛} 𝐷 (𝜁_{𝑘}, 𝑟_{𝑘})$ .

Define $Φ : 𝑉 \to ℂ$ by $Φ (𝑧) = 𝑓_{𝜁_{𝑘}} (𝑧)$ if $𝑧 \in 𝐷 (𝜁_{𝑘}, 𝑟_{𝑘})$ . This is well-defined: If $𝑧 \in 𝐷 (𝜁_{𝑖}, 𝑟_{𝑖}) \cap 𝐷 (𝜁_{𝑗}, 𝑟_{𝑗})$ , then $𝐷 (𝜁_{𝑖}, 𝑟_{𝑖}) \cap 𝐷 (𝜁_{𝑗}, 𝑟_{𝑗}) \cap 𝐷 (𝑧_{0}, 𝑅) \neq ⌀$ , and $𝑓_{𝜁_{𝑖}} \equiv 𝑓_{𝜁_{𝑗}} \equiv 𝑓$ there. By the Identity Theorem (Theorem 3.3.3 $Theorem 3.3.3 Theorem 3.3.3$ ), $𝑓_{𝜁_{𝑖}} \equiv 𝑓_{𝜁_{𝑗}}$ on $𝐷 (𝜁_{𝑖}, 𝑟_{𝑖}) \cap 𝐷 (𝜁_{𝑗}, 𝑟_{𝑗})$ .

Since $Φ$ is holomorphic on $𝑉$ and agrees with $𝑓$ on $𝐷 (𝑧_{0}, 𝑅) \cap 𝑉$ , the function:

\tilde{𝑓} (𝑧) = {\begin{cases} 𝑓 (𝑧) & if 𝑧 \in 𝐷 (𝑧_{0}, 𝑅), \\ Φ (𝑧) & if 𝑧 \in 𝑉, \end{cases}

is holomorphic on $𝐷 (𝑧_{0}, 𝑅) \cup 𝑉 \supseteq 𝐷 (𝑧_{0}, 𝑅 + 𝜌)$ , contradicting the maximality of $𝑅$ . ⁠ $□$

Definition 13.1.3.

A maximal analytic continuation

{(𝑓, 𝑈)}

of an analytic function element

(\tilde{𝑓}, \tilde{𝑈})

is obtained by all possible succesive analytic continuations of

(\tilde{𝑓}, \tilde{𝑈})

. The union of every

𝑈

is known as the domain of holomorphy of

\tilde{𝑓}

. The boundary set

𝜕 𝑈

is known as a natural boundary. The continuation defines a function

𝑓

, known a global analytic function, which can be multi-valued.

Example 13.1.4.

The series

𝑓 (𝑧) = \sum_{𝑛 = 0}^{\infty} 𝑧^{𝑛!}

has

𝔻

as its disk of convergence, and every point on

𝜕 𝔻

is a singularity.

Proof.

By Cauchy–Hadamard (Theorem 2.3.5 $Theorem 2.3.5 Theorem 2.3.5$ ), ${lim sup}_{𝑛 \to \infty} \sqrt[𝑛]{| 𝑐_{𝑛} |} = 1$ since $𝑐_{𝑛} = 1$ if $𝑛 = 𝑘!$ and $0$ otherwise. Thus, $𝑅 = 1$ .

Fix $𝜁 \in 𝜕 𝔻$ . Suppose $𝑓$ extends analytically to a disk $𝐷 (𝜁, 𝛿)$ . Since $exp (2 π i ℚ)$ is dense in $𝜕 𝔻$ , there exists $𝜁^{'} = exp (2 π i \frac{𝑝}{𝑞}) \in 𝐷 (𝜁, 𝛿) \cap 𝜕 𝔻$ for coprime integers $𝑝, 𝑞$ . The extension $𝑔$ of $𝑓$ to $𝐷 (𝜁, 𝛿)$ would satisfy:

lim_{𝑟 \to 1^{-}} 𝑓 (𝑟 𝜁^{'}) = 𝑔 (𝜁^{'}) .

However, for $0 < 𝑟 < 1$ :

𝑓 (𝑟 𝜁^{'}) = \sum_{𝑘 = 0}^{𝑞 - 1} {(𝑟 𝜁^{'})}^{𝑘!} + \sum_{𝑘 = 𝑞}^{\infty} 𝑟^{𝑘!} {(𝜁^{'})}^{𝑘!} .

The second summation is unbounded since

\sum_{𝑘 = 𝑞}^{\infty} 𝑟^{𝑘!} {(𝜁^{'})}^{𝑘!} = \sum_{𝑘 = 𝑞}^{\infty} 𝑟^{𝑘!} > \sum_{𝑘 = 𝑞}^{𝑁} 𝑟^{𝑘!} > (𝑁 - 𝑞 + 1) 𝑟^{𝑁!}

for any integer with $𝑁 > 𝑞$ . Hence, as $𝑟 \to 1^{-}$ , $\sum_{𝑘 = 𝑞}^{\infty} 𝑟^{𝑘!} \to \infty$ . Hence, $𝜁$ is a singularity. ⁠ $□$

Example 13.1.5.

Show that

𝑓 (𝑧) = \sum_{𝑛 = 0}^{\infty} 𝑧^{2^{𝑛}}

cannot be analytically continued to the outside of

𝔻

Proof.

Trivially, at $𝑧 = 1$ , the series diverges. Therefore, $𝔻$ is its convergence disk. Observe that $𝑓 (𝑧) = \sum_{𝑛 = 0}^{\infty} {(𝑧^{2})}^{2^{𝑛 - 1}} = \sum_{𝑛 = 0}^{\infty} {(𝑧^{2})}^{2^{𝑛}} + 𝑧 = 𝑓 (𝑧^{2}) + 𝑧$ . Hence, we have

𝑓 (𝑧) = 𝑓 (𝑧^{2}) + 𝑧 = 𝑓 (𝑧^{4}) + 𝑧^{2} + 𝑧 = 𝑓 (𝑧^{8}) + 𝑧^{4} + 𝑧^{2} + 𝑧 \dots,

which diverges at each

𝑧^{2}, 𝑧^{4}, 𝑧^{8}, \dots = 1

. The solutions form a dense set in

𝜕 𝔻

. By the same reasoning as Example 13.1.4

Example 13.1.4 Example 13.1.4

𝑓

cannot be analytically continued to the outside of

𝔻

. ⁠

□

Example 13.1.6.

Let $Log (𝑧)$ denote the principal branch of $log (𝑧)$ , with $- π < Arg (𝑧) \leq π$ . The analytic function elements

(Log, 𝐷 (1, 1)) and (Log + 2 π i, 𝐷 (1, 1))

are analytic continuations of each other.

Example 13.1.7.

Show that the analytic functions defined by the series

𝑓 (𝑧) = \sum_{𝑛 = 0}^{\infty} 𝛼^{𝑛} 𝑧^{𝑛}

and

\tilde{𝑓} (𝑧) = \sum_{𝑛 = 0}^{\infty} \frac{{(𝛼 - 1)}^{𝑛} 𝑧^{𝑛}}{{(1 - 𝑧)}^{𝑛 + 1}}

are analytic continuations of each other.

Proof.

The analytic function element $(𝑓, 𝐷 (0, | \frac{1}{𝛼} |))$ can be directly continued to the analytic function element $(𝑧 \mapsto \frac{1}{1 - 𝛼 𝑧}, ℂ ∖ {\frac{1}{𝛼}})$ . The function element $(\tilde{𝑓}, 𝐷 (0, \frac{1 - 𝑧}{𝛼 - 1}))$ can be analytically continued to

(𝑧 \mapsto \frac{1}{(1 - 𝑧) (1 - \frac{𝑧 𝛼 - 𝑧}{1 - 𝑧})}, ℂ ∖ {1, \frac{1}{𝛼}}) = (𝑧 \mapsto \frac{1}{1 - 𝛼 𝑧}, ℂ ∖ {1, \frac{1}{𝛼}}),

which is a direction analytic continuation of

(𝑧 \mapsto \frac{1}{1 - 𝛼 𝑧}, ℂ ∖ {\frac{1}{𝛼}})

. Therefore,

𝑓

and

\tilde{𝑓}

are analytic continuations of each other. ⁠

□

In Example 13.1.6 $Example 13.1.6 Example 13.1.6$ , we showed that two analytic function elements can on the same domain can be analytic continuations even if they do not agree on the entire domain. In this case, the two elements are on different branches of the function. Hence, depending on the chain of function elements chosen, we may obtain two different analytic function elements that have the same domain.

This is a common issue when it comes to the problem of analytic continuation. This question of non-ambiguity can be explained by planar topology; specifically the concept of homotopy. We will now introduce the concept of analytic continuation along a given curve.