§13.2 Analytic Continuation Along a Curve

13.2 Analytic Continuation Along a Curve

Definition 13.2.1.

Let $𝛾 : [0, 1] \to ℂ$ be a (non-constant) curve. Let $𝑈$ be a disk centered at $𝛾 (0)$ and suppose $𝑓 : 𝑈 \to ℂ$ is holomorphic. An analytic continuation of $(𝑓, 𝑈)$ along $𝛾$ is defined to be a collection of analytic functions elements ${(𝑓_{𝑡}, 𝑈_{𝑡})}_{0 \leq 𝑡 \leq 1}$ where

We define $𝑈_{0} = 𝑈$ and $𝑓_{0} = 𝑓$ .
Each $𝑈_{𝑡}$ ( $0 \leq 𝑡 \leq 1$ ) is a disk centered at $𝛾 (𝑡)$ .
For each $𝑡_{0} \in [0, 1]$ , $\exists 𝛿 > 0$ such that $\forall 𝑡 \in [0, 1]$ satisfying $| 𝑡 - 𝑡_{0} | < 𝛿$ , $𝛾 (𝑡) \in 𝑈_{𝑡_{0}}$ and $𝑓_{𝑡} \equiv 𝑓_{𝑡_{0}}$ on $𝑈_{𝑡} \cap 𝑈_{𝑡_{0}} \neq ⌀$ .

For a fixed curve, such analytic continuations are unique in the following sense:

Lemma 13.2.2.

For a fixed curve

𝛾 : [0, 1] \to ℂ

, let

𝑓

be holomorphic on

𝑈

(a disk centered at

𝛾 (0)

). Then any two analytic continuations

{(𝑓_{𝑡}, 𝑈_{𝑡})}_{0 \leq 𝑡 \leq 1}

and

{({\tilde{𝑓}}_{𝑡}, {\tilde{𝑈}}_{𝑡})}_{0 \leq 𝑡 \leq 1}

along

𝛾

satisfy

𝑓_{1} \equiv {\tilde{𝑓}}_{1}

𝑈_{1} \cap {\tilde{𝑈}}_{1}

(where

(𝑓_{1}, 𝑈_{1})

and

({\tilde{𝑓}}_{𝑡}, {\tilde{𝑈}}_{𝑡})

are the respective terminal analytic function elements).

Proof.

Let $𝑆 \subseteq [0, 1]$ be the set of all $𝑡_{0}$ such that $\forall 0 \leq 𝑡 \leq 𝑡_{0}$ , $𝑓_{𝑡} \equiv {\tilde{𝑓}}_{𝑡}$ on $𝑈_{𝑡} \cap {\tilde{𝑈}}_{𝑡}$ (this intersection is nonempty since $𝛾 (𝑡) \in 𝑈_{𝑡}, {\tilde{𝑈}}_{𝑡}$ ). Since $0 \in 𝑆$ , it follows that $𝑆$ is nonempty.

Obviously, $𝑆$ is a connected set. Indeed, for any $𝑡_{0}$ in $𝑆$ , any $0 \leq 𝑡 < 𝑡_{0}$ also lies in $𝑆$ by definition.

Let $𝑡_{\infty} = sup (𝑆)$ , and choose an increasing sequence ${𝑡_{𝑛}}_{𝑛 \in ℕ} \subseteq 𝑆$ that converges to $𝑡_{\infty}$ . By Part 3 $Part 3 Part 3$ of Definition 13.2.1 $Definition 13.2.1 Definition 13.2.1$ , $\exists 𝛿 > 0$ such that $\forall 𝑛 \in ℕ$ satisfying $| 𝑡_{\infty} - 𝑡_{𝑛} | < 𝛿$ ,

𝑓_{𝑡_{𝑛}} \equiv 𝑓_{𝑡_{\infty}} on 𝑈_{𝑡_{𝑛}} \cap 𝑈_{𝑡_{\infty}} ∋ 𝛾 (𝑡_{𝑛}) .

Similarly, $\exists \tilde{𝛿} > 0$ such that $\forall 𝑛 \in ℕ$ satisfying $| 𝑡_{\infty} - 𝑡_{𝑛} | < \tilde{𝛿}$ ,

{\tilde{𝑓}}_{𝑡_{𝑛}} \equiv {\tilde{𝑓}}_{𝑡_{\infty}} on {\tilde{𝑈}}_{𝑡_{𝑛}} \cap {\tilde{𝑈}}_{𝑡_{\infty}} ∋ 𝛾 (𝑡_{𝑛}) .

Choose $𝑛$ arbitrarily to satisfy $| 𝑡_{\infty} - 𝑡_{𝑛} | < min (𝛿, \tilde{𝛿})$ . By the conditions on $𝛿, \tilde{𝛿}$ set by Part 3 $Part 3 Part 3$ of Definition 13.2.1 $Definition 13.2.1 Definition 13.2.1$ ,

𝛾 (𝑡_{𝑛}) \in 𝑈_{𝑡_{\infty}} and 𝛾 (𝑡_{𝑛}) \in {\tilde{𝑈}}_{𝑡_{\infty}} .

Hence, ${\tilde{𝑈}}_{𝑡_{\infty}} \cap 𝑈_{𝑡_{\infty}} \cap {\tilde{𝑈}}_{𝑡_{𝑛}} \cap 𝑈_{𝑡_{𝑛}} \neq ⌀$ . Since $𝑡_{𝑛} \in 𝑆$ , it follows that ${\tilde{𝑓}}_{𝑡_{𝑛}} \equiv 𝑓_{𝑡_{𝑛}}$ on ${\tilde{𝑈}}_{𝑡_{𝑛}} \cap 𝑈_{𝑡_{𝑛}}$ . Thus,

𝑓_{𝑡_{\infty}} \equiv {\tilde{𝑓}}_{𝑡_{\infty}} on {\tilde{𝑈}}_{𝑡_{\infty}} \cap 𝑈_{𝑡_{\infty}} \cap {\tilde{𝑈}}_{𝑡_{𝑛}} \cap 𝑈_{𝑡_{𝑛}} .

By the Identity Theorem (Theorem 3.3.3 $Theorem 3.3.3 Theorem 3.3.3$ ), this equality holds on the entire intersection ${\tilde{𝑈}}_{𝑡_{\infty}} \cap 𝑈_{𝑡_{\infty}}$ . It follows that $𝑡_{\infty} \in 𝑆$ and thus $𝑆$ is closed.

Let $\tilde{𝑆} = [0, 1] ∖ 𝑆$ , and assume that it is nonempty (if not, our result is proven). Suppose ${𝑡_{𝑛}}_{𝑛 \in ℕ} \subset \tilde{𝑆}$ is an arbitrary sequence that converges to $𝑡_{\infty}$ . For each $𝑛 \in ℕ$ , since $𝑡_{𝑛} \notin 𝑆$ , by definition, there exists $0 \leq 𝑠_{𝑛} \leq 𝑡_{𝑛}$ such that $𝑓_{𝑠_{𝑛}} ≢ {\tilde{𝑓}}_{𝑠_{𝑛}}$ on $𝑈_{𝑠_{𝑛}} \cap {\tilde{𝑈}}_{𝑠_{𝑛}}$ (otherwise $𝑡_{𝑛} \in 𝑆$ would be satisfied).

By the Bolzano–Weierstrass Theorem (Theorem 1.1.10 $Theorem 1.1.10 Theorem 1.1.10$ ), the sequence ${𝑠_{𝑛}}_{𝑛 \in ℕ}$ has a convergent subsequence ${𝑠_{𝑛_{𝑘}}}_{𝑘 \in ℕ}$ that converges to $𝑠_{\infty}$ . Since $𝑡_{𝑛_{𝑘}} \to 𝑡_{\infty}$ and $𝑠_{𝑛_{𝑘}} \leq 𝑡_{𝑛_{𝑘}}$ for all $𝑘$ , it follows that $𝑠_{\infty} \leq 𝑡_{\infty}$ . By definition, there exists $𝛿 > 0$ (choose it to be the minimum of the two $𝛿$ values, similar to in the previous section) such that $\forall 𝑡 \in [0, 1]$ satisfying $| 𝑡 - 𝑠_{\infty} | < 𝛿$ , $𝑓_{𝑡} \equiv 𝑓_{𝑠_{\infty}}$ and ${\tilde{𝑓}}_{𝑡} \equiv {\tilde{𝑓}}_{𝑠_{\infty}}$ on $𝑈_{𝑡} \cap 𝑈_{𝑠_{\infty}}$ and on ${\tilde{𝑈}}_{𝑡} \cap {\tilde{𝑈}}_{𝑠_{\infty}}$ respectively, and $𝛾 (𝑡) \in 𝑈_{𝑠_{\infty}} \cap {\tilde{𝑈}}_{𝑠_{\infty}}$ .

Since $𝑠_{𝑛_{𝑘}} \to 𝑠_{\infty}$ , we have $| 𝑠_{𝑛_{𝑘}} - 𝑠_{\infty} | < 𝛿$ for sufficiently large $𝑘$ . Thus, $𝑓_{𝑠_{𝑛_{𝑘}}} \equiv 𝑓_{𝑠_{\infty}}$ and ${\tilde{𝑓}}_{𝑠_{𝑛_{𝑘}}} \equiv {\tilde{𝑓}}_{𝑠_{\infty}}$ on $𝑈_{𝑠_{𝑛_{𝑘}}} \cap 𝑈_{𝑠_{\infty}}$ and ${\tilde{𝑈}}_{𝑠_{𝑛_{𝑘}}} \cap {\tilde{𝑈}}_{𝑠_{\infty}}$ respectively, and $𝛾 (𝑠_{𝑛_{𝑘}}) \in 𝑈_{𝑠_{\infty}} \cap {\tilde{𝑈}}_{𝑠_{\infty}}$ .

For the sake of contradiction, assume that $𝑠_{\infty} \in 𝑆$ , and it follows that $𝑓_{𝑠_{\infty}} \equiv {\tilde{𝑓}}_{𝑠_{\infty}}$ on $𝑈_{𝑠_{\infty}} \cap {\tilde{𝑈}}_{𝑠_{\infty}}$ . Because

𝑓_{𝑠_{𝑛_{𝑘}}} \equiv 𝑓_{𝑠_{\infty}} \equiv {\tilde{𝑓}}_{𝑠_{\infty}} \equiv {\tilde{𝑓}}_{𝑠_{𝑛_{𝑘}}} on 𝑈_{𝑠_{\infty}} \cap {\tilde{𝑈}}_{𝑠_{\infty}} \cap 𝑈_{𝑠_{𝑛_{𝑘}}} \cap {\tilde{𝑈}}_{𝑠_{𝑛_{𝑘}}} ∋ 𝛾 (𝑠_{𝑛_{𝑘}}),

by the Identity Theorem (Theorem 3.3.3 $Theorem 3.3.3 Theorem 3.3.3$ ), this implies that $𝑓_{𝑠_{𝑛_{𝑘}}} \equiv {\tilde{𝑓}}_{𝑠_{𝑛_{𝑘}}}$ on $𝑈_{𝑠_{𝑛_{𝑘}}} \cap {\tilde{𝑈}}_{𝑠_{𝑛_{𝑘}}}$ . This contradicts the assumption that $𝑓_{𝑠_{𝑛}} ≢ {\tilde{𝑓}}_{𝑠_{𝑛}}$ on $𝑈_{𝑠_{𝑛}} \cap {\tilde{𝑈}}_{𝑠_{𝑛}}$ for all $𝑛$ , and hence, $𝑠_{\infty} \in \tilde{𝑆}$ , implying that $𝑡_{\infty} \in \tilde{𝑆}$ (if not, then $𝑠_{\infty} \notin \tilde{𝑆}$ , which we derived was a contradiction). It follows that $\tilde{𝑆}$ is closed as it contains all of its accumulation points. By the connectivity argument (Theorem 3.2.1.19 $Theorem 3.2.1.19 Theorem 3.2.1.19$ ), we have $\tilde{𝑆}$ is either $[0, 1]$ or $⌀$ . Obviously, the former is an impossibility and thus $𝑆 = [0, 1]$ . Therefore, $𝑓_{1} \equiv {\tilde{𝑓}}_{1}$ on $𝑈_{1} \cap {\tilde{𝑈}}_{1}$ . ⁠ $□$

Provided by the trivial fact, under the assumption that an analytic continuation along a fixed curve exists, it is unique in the defined sense.

To avoid being pedantic, we will refer to two analytic continuations on a fixed curve as equivalent if $𝑓_{𝑡} \equiv {\tilde{𝑓}}_{𝑡}$ on $𝑈_{𝑡} \cap {\tilde{𝑈}}_{𝑡}$ for all $0 \leq 𝑡 \leq 1$ .