All Normal Vectors of Simple Closed Contour Cannot Point into Interior/Lemma 1

Lemma
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 $r_0, R_0 \in \R_{>0}$ such that:


 * for all $s \in \openint { t - R_0 }{ t + R_0 }$ and for all $\epsilon \in \openint 0 {r_0}$: $\map \gamma s + \epsilon i \map {\gamma'} s \notin \Img C$, and $\map \gamma s - \epsilon i \map {\gamma'} s \notin \Img C$.

There exist $r, R \in \R_{>0}$ with $r \le r_0, R \le R_0$ such that:


 * $V_{r,R} = \set { \map \gamma {s} + \epsilon i \map {\gamma '}{ s } : s \in \openint { t-R }{ t+R }, \epsilon \in \openint {-r}{r} }$

is an open set in $\C$.

Proof
This proof assumes that $\gamma '$ is continuously differentiable.

Let $x, y : \openint a b \to \R$ be the real functions defined by:


 * $\map \gamma t = \map x t + i \map y t$

Let $f: \openint {t - R_0}{t + R_0} \times \openint {-r_0}{ r_0 } \to \R^2$ be defined by $\map f {s, \epsilon} = \tuple{ \map x s - \epsilon \map {y'} s, \map y {s} + \epsilon \map {x'} s }$.

The Jacobian matrix of $f$ at $\tuple{s, \epsilon}$ is :


 * $\mathbf J_f = \begin{pmatrix}

\map {x'}{ s } - \epsilon \map {y' '}{ s } & - \map {y'}{ s } \\ \map {y'}{ s } + \epsilon \map {x' '}{ s } & \map {x'}{ s } \end{pmatrix}$

and the Jacobian determinant of $f$ evaluated at $\tuple{t, 0}$ is:


 * $\map { \map \det {\mathbf J_f } }{ t, 0 }= \map {x'}{ t }^2 + \map {y'}{ t }^2$

As $\map {\gamma'} t \neq 0$ by definition of smooth path, it follows that $\map { \map \det {\mathbf J_f } }{ t, 0 } \ne 0$.

From Matrix is Invertible iff Determinant has Multiplicative Inverse, it follows that the Jacobian matrix of $f$ evaluated at $\tuple{t, 0}$ is invertible.

From Inverse Function Theorem for Real Functions, it follows that there exist open sets:


 * $U \subseteq \openint {t - R_0}{t + R_0} \times \openint {-r_0}{ r_0 }, V \subseteq \R^2$

such that the restriction of $f$ to $U \times V$ is bijective, and the inverse $f^{-1}$ is continuous.

Equip $\R^2$ with the direct product norm.

Norms on Finite-Dimensional Real Vector Space are Equivalent shows that the direct product norm is equivalent to the Euclidean norm on $\R^2$.

From Open Sets in Vector Spaces with Equivalent Norms Coincide, it follows that $U$ is open in $\R^2$ equipped with the direct product norm.

Then there exist an open ball $\map {B_r} t \subseteq U$ with center $t$ of the form:


 * $\map {B_r} t = \openint {t - r}{t + r} \times \openint {-r}{r}$

From Bijection is Open iff Inverse is Continuous, it follows that $f \sqbrk { \map {B_r} t }$ is open in $\R^2$.

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

We identify $\map f {s,\epsilon}$ with $\map \phi { \map f {s,\epsilon} } = \map \gamma {s} + \epsilon i \map {\gamma '}{ s }$.

Then $V_{r,r} := \phi \sqbrk{ f \sqbrk { \map {B_r} t } }$ fulfils the claim of the theorem.