Single Point Characterization of Simple Closed Contour/Lemma 2

Lemma
Let $C$ be a simple closed contour in the complex plane $\C$ with parameterization $\gamma: \closedint a b \to \C$.

Let $\set { c_0, \ldots , c_N }$ be a subdivision of $\closedint a b$ with $N \in \N$ such that $\gamma$ is complex-differentiable at all $t \in \openint {c_{n-1} }{c_n}$ for $n \in \set {1 , \ldots, N}$.

Let $S \in \set {-1,1}$.

For all $\epsilon \in \R_{>0}$ and $t \in \openint a b$ for which $\gamma$ is complex-differentiable at $t$, define:


 * $\map v { t, \epsilon } := \map \gamma {t} + \epsilon i S \map {\gamma '}{ t }$

Let $k \in \set {1, \ldots, N}$ such that for all $t \in \openint {c_{k-1} }{c_k}$, there exists $r \in \R_{>0}$ such that for all $\epsilon \in \openint 0 {r}$:


 * $\map v { t, \epsilon } \in \Int \gamma$

Set $l := \begin {cases} k+1 & : k < N \\ 1 & : k = N \end {cases}$.

Then for all $t \in \openint {c_{l-1} }{c_l}$, there exists $r \in \R_{>0}$ such that for all $\epsilon \in \openint 0 {r}$:


 * $\map v { t, \epsilon } \in \Int \gamma$

Construct two open disks with center $\map {\gamma_k}{c_k}$
Let $\gamma_n := \gamma {\restriction_{ \closedint {c_{n-1} }{c_n} } }$ denote the restriction of $\gamma$ to $\closedint {c_{n-1} }{c_n}$.

From Continuous Image of Compact Space is Compact and Finite Union of Compact Sets is Compact, it follows that


 * $\ds \bigcup_{ n \mathop \in \set {1, \ldots, N } \setminus \set {k, l} } \Img {\gamma_n} $

is compact.

Set $r_0 := \ds \map d { \bigcup_{ n \mathop \in \set {1, \ldots, N } \setminus \set {k, l} }, \map {\gamma_k}{c_k} } / 2$, where $\map d {X,y}$ denotes the distance between the set $X \subseteq \R^2$ and $y \in \R^2$.

From Distance between Disjoint Compact Set and Closed Set in Metric Space is Positive, it follows that $r_0 > 0$.

Set:


 * $\ds \tilde t_1 := \max \set {t \mathop \in \closedint {c_{k-1} }{ c_k } : \map {\gamma_k} t \in \map { B_{r_0}^- }{ \map {\gamma_k}{c_k} } }$
 * $\ds \tilde t_2 := \min \set {t \mathop \in \closedint {c_{l-1} }{ c_l } : \map {\gamma_l} t \in \map { B_{r_0}^- }{ \map {\gamma_k}{c_k} } }$

where $\map { B_{r_0}^- }{ \map {\gamma_k}{c_k} }$ denotes a closed disk.

Set $r_1 := \map d { \gamma_k \sqbrk{ \closedint {c_{k-1} }{\tilde t_1} } \cup \gamma_l \sqbrk{ \closedint {\tilde t_2}{c_l} }, \map {\gamma_k}{c_k} } / 2$.

From Distance between Disjoint Compact Set and Closed Set in Metric Space is Positive, it follows that $r_1 > 0$.

Set :


 * $\tilde t_3 := \min \set{t \in \closedint {\tilde t_1}{c_k} : \exists \delta > 0 : \forall \epsilon \in \openint 0 \delta : \map {\gamma_k}{t + \epsilon} \in \map { B_{r_1} }{ \map {\gamma_k}{c_k} } }$
 * $\tilde t_4 := \max \set{t \in \closedint {c_{l-1} }{\tilde t_2} : \exists \delta > 0 : \forall \epsilon \in \openint 0 \delta : \map {\gamma_l}{t - \epsilon} \in \map { B_{r_1} }{ \map {\gamma_k}{c_k} } }$

where $\map { B_{r_1} }{ \map {\gamma_k}{c_k} }$ denotes an open disk.

As $r_1 < r_0$, this implies that:

where $n \notin \set {k, l}$.

From Complex Plane is Homeomorphic to Real Plane, it follows that we can identify the complex plane $\C$ with the Euclidean plane $\R^2$ by the homeomorphism $\phi : \R^2 \to \C$, defined by:


 * $\map \phi {x, y} = x + i y$

In the following, we will use definitions and theorems which refer to $\R^2$, which we translate to $\C$.

Construct a Jordan arc $\tilde \gamma_2$ that will smoothen the corner of the contour at $\map {\gamma_k}{c_k}$
Define two Jordan arcs $\tilde \gamma_0, \tilde \gamma_1 : \closedint 0 1 \to \C$ by:


 * $\tilde \gamma_0 := \begin {cases} \gamma_l {\restriction_{\closedint {\tilde t_4}{c_l} } } * \gamma_{l+1} * \gamma_{l+2} * \ldots * \gamma_n * \gamma_1 * \gamma_2 * \ldots * \gamma_{k-1} * \gamma_k {\restriction{ \closedint {c_{k-1} }{\tilde t_3} } }

& : k < N \\ \gamma_l {\restriction_{\closedint {\tilde t_4}{c_l} } } * \gamma_2 * \ldots * \gamma_{k-1} * \gamma_k {\restriction{ \closedint {c_{k-1} }{\tilde t_3} } } & : k = N \end {cases}$


 * $\tilde \gamma_1 := \gamma_k {\restriction{ \closedint {\tilde t_3}{c_k} } }* \gamma_l {\restriction{ \closedint {c_{l-1} }{\tilde t_4} } }$

where $\gamma_1 * \gamma_2$ denotes the concatenation of $\gamma_1$ and $\gamma_2$.

From the definitions of Jordan curve and simple closed contour, it follows that $\tilde \gamma_0 * \tilde \gamma_1$ is a Jordan curve.

From their definition, it follows that $\Img {\tilde \gamma_0 * \tilde \gamma_1} = \Img {\gamma}$.

As the Jordan Curve Theorem says that the interior of a Jordan curve only depends on its image, it follows that $\Int {\tilde \gamma_0 * \tilde \gamma_1} = \Int {\gamma}$.

Let $\tilde t_5 \in \R$ be defined such that $\map {\tilde \gamma_1}{\tilde t_5} = \map {\gamma_k}{c_k}$.

Let $\tilde \gamma_2: \closedint 0 1$ be the injective complex-differentiable function defined in Single Point Characterization of Simple Closed Contour/Lemma 3 with $\map {\tilde \gamma_2} t \in \map {B_{r_1} }{ \map {\gamma_k}{c_k} }$ for all $t \in \openint 0 1$.

By the definitions of $\tilde t_3$ and $\tilde t_4$, it follows that $\map {\tilde \gamma_2} t \notin \Img {\tilde \gamma_0}$ for all $t \in \openint 0 1$.

By definition of Jordan curve, it follows that $\tilde \gamma_0 * \tilde \gamma_2$ is a Jordan curve.

By definition of contour, it follows that $\tilde \gamma_0 * \tilde \gamma_2$ is a parameterization of a simple closed contour.

Show that the normal vectors along the new curve point into the interior
By the definition of $\tilde t_1$, there exists some $t_0 \in \openint {c_{k-1} }{\tilde t_1}$ such that $\map {\gamma_k}{t_0} \notin \map {B_{r_0} }{ \map {\gamma_k}{c_k} }$.

By the definition of $\tilde t_2$, there exists some $t_1 \in \openint {\tilde t_2}{c_l}$ such that $\map {\gamma_l}{t_1} \notin \map {B_{r_0} }{ \map {\gamma_k}{c_k} }$.

By assumption, there exists $r \in \R_{>0}$ such that $\map v {\epsilon, t_0} \in \Int {\gamma}$ for all $\epsilon \in \openint 0 r$.

From Interior and Exterior of Partially Disjoint Jordan Curves, it follows that exists $\overline r \in \R_{>0}$ such that:


 * for all $z \in \map {B_{\overline r} }{\map {\gamma_k}{t_0} }$: $z \in \Int { \gamma }$, $z \in \Int { \tilde \gamma_0 * \tilde \gamma_2 }$
 * for all $z \in \map {B_{\overline r} }{\map {\gamma_l}{t_1} }$: $z \in \Int { \gamma }$, $z \in \Int { \tilde \gamma_0 * \tilde \gamma_2 }$

Set $\rho := \min \set {r, \overline r}$.

It follows that $\map v {\epsilon, t_0} \in \Int {\tilde \gamma_0 * \tilde \gamma_2}$ for all $\epsilon \in \openint 0 \rho$.

From Derivative of Concatenation of Complex Paths, it follows that there exists $K_0 \in \R_{>0}$ such that:


 * $\map {\gamma_0 * \gamma_2} {t_0} + \dfrac {\epsilon}{K_0} i S \map {\paren{ \tilde \gamma_0 * \tilde \gamma_2}'}{ t_0 } \in \Int { \tilde \gamma_0 * \tilde \gamma_2 }$

Let $\tilde c_{k-1}, \tilde c_l \in \closedint 0 1$ such that $\map {\gamma_0 * \gamma_2}{\tilde c_{k-1} } = \map {\gamma_k}{c_{k-1} } , \map {\gamma_0 * \gamma_2}{\tilde c_l} = \map {\gamma_l}{c_l}$.

By definition of $\gamma_0 * \gamma_2$, it follows that $\gamma_0 * \gamma_2$ is complex-differentiable at all $t \in \closedint {\tilde c_{k-1} }{\tilde c_l}$.

From Lemma 1, it follows that there exists $K_1 \in \R_{>0}$ such that:


 * $\map {\gamma_0 * \gamma_2} {t_1} + \epsilon i S \map {\paren{ \tilde \gamma_0 * \tilde \gamma_2}'}{ t_1 } \in \Int { \tilde \gamma_0 * \tilde \gamma_2 }$ for all $\epsilon \in \openint 0 {\dfrac{\rho}{k_1} }$

which implies that $\map v {\epsilon, t_1} \in \Int {\gamma}$ for all $\epsilon \in \openint 0 \rho$.

As $t_1 \in \openint {c_{l-1} }{c_l}$, it follows from Lemma 1 that for all $t \in \openint {c_{l-1} }{c_l}$, there exists $r \in \R_{>0}$ such that:


 * $\map v {\epsilon, t} \in \Int {\gamma}$ for all $\epsilon \in \openint 0 r$

which is what we wanted to prove.