All Normal Vectors of Simple Closed Contour Cannot Point into Interior

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

Let $t \in \openint a b$ such that $\gamma$ is complex-differentiable at $t$.

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

Let $\Int C$ denote the interior of $C$.

Then there exists $r_0 \in \R_{>0}$ such that:


 * for all $\epsilon \in \openint 0 {r_0}$, we have $\map \gamma {t} + \epsilon i S \map {\gamma '}{ t } \in \Int C$

there exists $r_1 \in \R_{>0}$ such that:


 * for all $\epsilon \in \openint 0 {r_1}$, we have $\map \gamma {t} - \epsilon i S \map {\gamma '}{ t } \notin \Int C$.

Proof
From Normal Vectors Form Space around Simple Complex Contour, it follows that there exists $\tilde r, \tilde R \in \R_{>0}$ such that:


 * for all $s \in \openint { t - \tilde R }{ t + \tilde R }$ and for all $\epsilon \in \openint 0 {\tilde r}$: $\map \gamma s + \epsilon i \map {\gamma'} s \notin \Img C$, and $\map \gamma s - \epsilon i \map {\gamma'} s \notin \Img C$

where $\Img C$ denotes the image of $C$.

From the Lemma, we find an open set $V_{r,R} \subseteq \C$ with $r \le \tilde r, R \le \tilde R$.

Let $z_1, z_2 \in V_{r,R}$.

Then there exist $s_1, s_2 \in \openint { t-R }{ t+R }$ and $\epsilon_1 , \epsilon_2 \in \openint 0 r$ with $S_0 \in \set {-1,1}$ such that:


 * $z_1 = \map \gamma {s_1} + \epsilon_1 i S_0 \map {\gamma'}{ s_1 } \in V_{r,R}, z_2 = \map \gamma {s_2} + \epsilon_2 i S_0 \map {\gamma'}{ s_2 } \in V_{r,R}$.

We now define a path $f: \closedint 0 1 \to V_{r,R}$ with endpoints $z_1$ and $z_2$ by:


 * $\map f {\hat t} = \map \gamma { s_1 + \paren{ s_2 - s_1} \hat t } + \paren{ \epsilon_1 + \paren{ \epsilon_2 - \epsilon_1 } \hat t } i S_0 \map {\gamma'}{ s_1 + \paren{ s_2 - s_1} \hat t  }$

As $\paren{ \epsilon_1 + \paren{ \epsilon_2 - \epsilon_1 } \hat t } \in \openint 0 r$, it follows that $\map f {\hat t} \notin \Img C$ for all $\hat t \in \closedint 0 1$.

Complex Plane is Homeomorphic to Real Plane shows that we can identify the complex plane $\C$ with the real plane $\R^2$ by the homeomorphism $\map \phi {x, y} = x + i y$.

Interior of Simple Closed Contour is Well-Defined shows that $\Img C$ can be identified with the image of a Jordan curve $g: \R^2 \to \R^2$.

From the same theorem, it follows that $\Int C$ can be identified with the interior of $g$.

From the Jordan Curve Theorem, it follows that $\C \setminus \Img C$ is a union of two open connected components, which are $\Int C$ and the exterior of $g$, which we denote $\Ext C$.

From Connected Open Subset of Euclidean Space is Path-Connected, it follows that $\Int C$ and $\Ext C$ are path components of $\C \setminus \Img C$.

Sufficient condition
By assumption, it follows that:


 * $z_0 := \map \gamma {t} + \dfrac {r_0} 2 i S \map {\gamma '}{ s } \in \Int C$.

For $z \in V_{r,R}$ such that $ z = \map \gamma s + \epsilon i S \map {\gamma'} s$, we have shown that there is a path $f$ in $V_{r,R} \setminus \Img C$ between $z_0$ and $z$.

By definition of path component, it follows that $z \in \Int C$.

Set $z_1 := \map \gamma {t} - \dfrac r 2 i S \map {\gamma '}{ s }$.

Suppose $z_1 \in \Int C$.

For $z \in V_{r,R}$ such that $ z = \map \gamma s - \epsilon i S \map {\gamma'} s$, it follows as above that $z \in \Int C$.

For $z \in V_{r,R}$ such that $ z = \map \gamma s$, we have $z \in \Img C$.

It follows that $V_{r,R} \subseteq \Int C \cup \Img C$.

From the Jordan Curve Theorem, it follows that $\Img C$ is the common boundary of $\Int C$ and $\Ext C$.

This leads to a contradiction, as $\map \gamma t \in \Img C$, but $V_{r,R}$ contains no points of $\Ext C$.

It follows that $z_1 \in \Ext C$.

For $z \in V_{r,R}$ such that $ z = \map \gamma s - \epsilon i S \map {\gamma'} s$, it now follows that $z \in \Ext C$.

Specifically, $\map \gamma {t} - \epsilon i S \map {\gamma '}{ t } \notin \Int C$.

It follows that we can set $r_1 := r$.

Necessary condition
By assumption, it follows that:


 * $z_0 := \map \gamma {t} + \dfrac {r_1} 2 i S \map {\gamma '}{ s } \notin \Int C$.

As $\C \setminus \Img C$ consists of two path components, it follows that $z_0 \in \Ext C$.

For $z \in V_{r,R}$ such that $ z = \map \gamma s + \epsilon i S \map {\gamma'} s$, we have shown that there is a path $f$ in $V_{r,R} \setminus \Img C$ between $z_0$ and $z$.

By definition of path component, it follows that $z \in \Ext C$.

Set $z_1 := \map \gamma {t} - \dfrac r 2 i S \map {\gamma '}{ s }$.

Suppose $z_1 \notin \Int C$.

For $z \in V_{r,R}$ such that $ z = \map \gamma s - \epsilon i S \map {\gamma'} s$, it follows as above that $z \notin \Int C$.

For $z \in V_{r,R}$ such that $ z = \map \gamma s$, we have $z \in \Img C$.

It follows that $V_{r,R} \subseteq \Ext C \cup \Img C$.

From the Jordan Curve Theorem, it follows that $\Img C$ is the common boundary of $\Int C$ and $\Ext C$.

This leads to a contradiction, as $\map \gamma t \in \Img C$, but $V_{r,R}$ contains no points of $\Int C$.

It follows that $z_1 \in \Int C$.

For $z \in V_{r,R}$ such that $ z = \map \gamma s - \epsilon i S \map {\gamma'} s$, it now follows that $z \in \Int C$.

It follows that we can set $r_0 := r$.