%gmApndxA.tex

\chapter{Jordan Curves}

$\Gamma$ will always denote a Jordan curve that is at least $C^2$ and whose length is $L$. $\Gamma_i$ denotes the open, simply connected set bounded by $\Gamma$, and $\Gamma_e$ denotes the unbounded open set bounded by $\Gamma$. We will represent $\Gamma$ with a specific arc-length parameterization: Let $C_L$ be the circle at the origin with radius $r=L/2\pi$. Let the function $\fxc:C_L\to\Gamma$ denote the arc-length parameterization with the property that the point $\varepsilon\fxc'(s)^\perp\in\Gamma_i$ for all sufficiently small $\varepsilon>0$. It follows from Proposition 4.4 and the Jordan curve theorem that we can choose this parameterization. 

It is convenient to view $C_L$ as a compact Abelian group. One can think of $C_L$ as the interval $[0,L]$, where $0$ and $L$ are identified by ordinary addition modulo $L$. This allows us to write $s\pm t$ for $s,t\in C_L$ and have it make sense. This also means that a given point $\fxc(s)\in\Gamma$ is represented by a unique point $s\in C_L$.

The signed distance function $d:\RR^2\to\RR$ is defined as follows: If $\fxy\in\Gamma_i$, then by definition $d(\fxy)=d(\fxy,\Gamma)=\inf_{s\in\ C_L}|\fxy-\fxc(s)|$. If $\fxy\in\RR^2\setminus\Gamma_i$, then $d(\fxy)=-d(\fxy,\Gamma)$. The function $d$ is well defined, and, given $\fxy\in\RR^2$, there is always at least one $s\in C_L$ such that $d(\fxy)=|\fxy-\fxc(s)|$. Since 
the function $t\mapsto|\fxy-\fxc(t)|$ has a minimum at this $s$, $(\partial|\fxy-\fxc(t)|^2/\partial t)(s)=0$. This implies that $(\fxy-c(s))\cdot \fxc'(s)=0$, which means that the vector $\fxy-\fxc(s)$ is orthogonal to the unit tangent vector $\fxc'(s)$. Thus, there is a $\lambda\in\RR$ such that $\fxy=\fxc(s)+\lambda\fxc'(s)^\perp$. Furthermore, the definition of $d$ and the fact that $\fxc'(s)^\perp$ ``points inward'' imply that $\lambda=d(\fxy)$. One of the main results in this section is that the function $\fxy\mapsto d(\fxy)$ assumes the smoothness of $\Gamma$, less one derivative, if $\fxy$ is near $\Gamma$.

We have assumed that $\Gamma$ is $C^2$, so the scalar function $\kappa:C_L\to\RR$ defined by $\kappa(s)=\fxc''(s)\cdot\fxc'(s)^\perp$ is continuous and bounded. Define $M=\sup_{s\in C_L}\kappa(s)$ and define $\delta=1/M$. This notation is used throughout the section.

\begin{exercise}Show that $\int_0^L\kappa(s)\,\rmd s=2\pi$ and, hence, that $M>0$.
\end{exercise}

Consider function $\fxz:C_L\times\RR\to\RR^2$ defined by $\fxz(s,\lambda)=\fxc(s)+\lambda\fxc'(s)^\perp$. The Jacobian of this mapping is computed as follows:
\begin{align*}
\frac{\partial\fxz}{\partial s}&=\fxc'(s)+\lambda\fxc''(s)^\perp,\\
\frac{\partial\fxz}{\partial\lambda}&=\fxc'(s)^\perp.
\end{align*}
These two vectors are orthogonal, so the Jacobian is zero if and only $\fxc'(s)+\lambda\fxc''(s)^\perp=0$. Since $\fxc'(s)+\lambda\fxc''(s)^\perp$ is colinear with $\fxc'(s)$, taking the inner product with the unit tangent $\fxc'(s)$ shows that the Jacobian is zero if and only if $1+\lambda\fxc''(s)^\perp\cdot\fxc'(s)=0$. Note that
\[
1+\lambda\fxc''(s)^\perp\cdot\fxc'(s)=1-\lambda\fxc''(s)\cdot\fxc'(s)^\perp=1-\lambda\kappa(s).
\]
The next result states that $\fxz$ is locally one-to-one if $|\lambda|<\delta$.

\begin{lemma}\label{inverse}Assume that $\Gamma$ is $C^n$ and that$|\lambda|<\delta$. Then, given $(s,\lambda)\in C_L\times\RR$, there is a neighborhood $U$ of $(s,\lambda)$ and a neighborhood $V$ of $\fxz(s,\lambda)$ such that $\fxz:U\to V$ is one-to-one and onto. Furthermore, $\fxz^{-1}:V\to U$ is $C^{n-1}$.
\end{lemma}
\begin{proof}The Jacobian of the mapping $\fxz:C_L\times\RR\to\RR^2$ does not vanish at $(s,\lambda)$, so the inverse mapping theorem applies. The lemma is just a restatement of the inverse mapping theorem for the case at hand.  
\end{proof}

The next result is similar to Lemma \ref{inverse}, but it is more geometric and more specific about the size of the neighborhoods.

\begin{lemma}Assume that $\Gamma$ is $C^2$. Choose any $\alpha$ such that $0<\alpha<\delta$. Then there exists a positive number $\beta=\beta(\alpha)$ with the following property$:$ If $|s-t|<\beta$, $s\ne t$, and if $|\lambda|\leq\alpha$ and $|\mu|\leq\alpha$, then
\[
\fxc(s)+\lambda\fxc'(s)^\perp\ne\fxc(t)+\mu\fxc'(t)^\perp.
\]
\end{lemma}
\begin{proof}Let $\fxc(s_0)$ be any (fixed) point on $\Gamma$. For convenience of notation, and without loss of generality, we shift the parameterization and assume that $s_0=0$. Establish a local coordinate systems as follows: $\fxc(0)=(0,0)$, $\fxc'(0)=(1,0)$ is the unit vector in the $x$ direction and $\fxc'(0)^\perp$ is the unit vector in the $y$ direction. All of the following statements assume that the arc-length parameter $s$ is small and positive. Dual arguments for small negative values give the same results. 

Let $\varphi(s)$ denote the angle between the unit vector $\fxc'(s)$ and the $x$-axis, that is, the angle between $\fxc'(s)$ and $\fxc'(0)$. Then $\varphi'(s)=\kappa(s)$, and 
\[
\varphi(s)=\int_0^s\kappa(t)\,\rmd t, 
\]
which gives the inequality $|\varphi(s)|\leq sM$. 

Consider the two straight lines $\fxc(0)+\mu\fxc'(0)^\perp$, $\mu\in\RR$, (which is the $y$-axis in our coordinate system), and $\fxc(s)+\lambda\fxc'(s)^\perp$, $\lambda\in\RR$. If $\varphi(s)\ne0$, then these lines intersect in a point $(0,Y)$. We are going to estimate the value of $\lambda=\lambda(s)$ for which 
\[
\fxc(0)+Y(s)\fxc'(0)=\fxc(s)+\lambda(s)\fxc'(s)^\perp.
\]
Write $x(s)=\fxc(s)\cdot\fxc'(0)$ and $y(s)=\fxc(s)\cdot\fxc'(0)^\perp$. Then
\[
\lambda(s)=\frac{x(s)}{\sin\varphi(s)}.
\]
For small $s$, $\varphi(s)$ is small, and $|\sin\varphi(s)|\leq|\varphi(s)|$. We have already mehtioned that $\varphi(s)\leq sM$. Thus, we see that
\begin{equation}\label{lambda}
|\lambda(s)|=\frac{x(s)}{|\sin\varphi(s)|}\geq \frac{x(s)}{|\varphi(s)|}\geq \frac{x(s)}{sM}.
\end{equation}
The next step is to get bounds for $x(s)/s$. For this, note that $x'(s)=\fxc'(s)\cdot\fxc'(0)=\cos\varphi(s)$, so we have
\[
x(s)=\int_0^s\cos\varphi(t)\,\rmd t.
\]
We get the inequality 
\[
\int_0^s\Big(1-\frac{\varphi^2(t)}{2}\Big)\,\rmd t\leq x(s)\leq s 
\]
by using the fact that $1-t^2/2\leq\cos t\leq 1$ for small $t$. Since we are assuming here that $s>0$, this shows that
\[
1-\frac1{2s}\int_0^s \varphi^2(t)\,\rmd t\leq\frac{x(s)}{s}\leq 1. 
\]
Now use the bound $|\varphi(s)|\leq sM$ to deduce that 
\[
\frac1{2s}\int_0^s \varphi^2(t)\,\rmd t\leq \frac1{6}s^2M^2,
\]
and hence that 
\[
1-\frac1{6}s^2M^2\leq \frac{x(s)}{s}\leq 1.
\]
For $0<s<\sqrt{6}\delta$, the term on the left in positive, and we can combine the last inequality with \eqref{lambda} to see that
\[
|\lambda|\geq \delta-\frac1{6}s^2M.
\]
It is important to note that this estimate does not depend on properties of the points $\fxc(0)$ and $\fxc(s)$; it depends only on the arc-length distance between these points, that is, on $s$ and on $M$. Of course, $s$ must be small enough so the trigonometric estimates hold, and this depends on $\varphi$. But these estimates are all based on the bound $|\varphi(s)|\leq sM$.  

This is what we have shown: Given two points $\fxc(s)$ and $\fxc(t)$, $s\ne t$, if
$|s-t|<\sqrt{6}\delta$ and if the lines $\fxc(s)+\lambda\fxc'(s)^\perp$ and $\fxc(t)+\mu\fxc'(t)^\perp$ intersect at a point $\fxx$ then
\begin{align*}
|\fxx-\fxc(s)|&=|\lambda|>\delta-|s-t|^2M/6,\\ 
|\fxx-\fxc(t)|&=|\mu|>\delta-|s-t|^2M/6.
\end{align*}
This proves the result: Given $0<\alpha<\delta$, let $\beta=\sqrt{6}(\min\{\delta, \sqrt{(\delta-\alpha)\delta}\})$. Then $|s-t|<\beta$, $|\lambda|<\alpha$, and $|\mu|<\alpha$, imply that $\fxc(s)+\lambda\fxc'(s)^\perp\ne\fxc(t)+\mu\fxc'(t)^\perp $.
\end{proof}

These last two results are strictly local result; in fact, it is possible that the point $\fxz(s,\lambda)$ has two or more representations of the form $\fxz(t,\mu)=\fxc(t)+\mu\fxc'(t)^\perp$. To eliminate this ambiguity, we analyze the ambiguous sets. Define $\Delta_i$ to be the set of those points in $\Gamma_i$ with the property that there are (at least) two points $s,t\in C_L$, $s\not=t$ such that $d(\fxy)=|\fxy-\fxc(s)|$ and $d(\fxy)=|\fxy-\fxc(t)|$. Similarly, let $\Delta_e$ be the set of those points in $\Gamma_e$ such that there are (at least) two points $s,t\in C_L$, $s\not=t$ such that $d(\fxy)=-|\fxy-\fxc(s)|$ and $d(\fxy)=-|\fxy-\fxc(t)|$. If $\Gamma$ is a circle, then $\Delta_i$ consists of the center point and $\Delta_e$ is the empty set. Our first result is that $\Delta_i$ is never empty.

\begin{lemma}If $\Gamma$ is $C^2$, then $\Delta_i\not=\emptyset$. 
\end{lemma}
\begin{proof}Let $\fxy_m\in\Gamma_i$ be a point where $d$ assumes its maximum value. We will show that $\fxy_m\in\Delta_i$. Suppose that this is not the case. Let $s_m\in C_L$ be the unique point such that $d(\fxy_m)=|\fxy_m-\fxc(s_m)|$. Then $\fxy_m=\fxc(s_m)+d(\fxy_m)\fxc'(s_m)^\perp$, and the closed disk $D(\fxy_m,d(\fxy_m)$ meets $\Gamma$ in only one point. This implies that, for a sufficiently small $\varepsilon>0$, the disk $\fxy=\fxc(s_m)+(d(\fxy_m)+\varepsilon)\fxc'(s_m)^\perp$ does not intersect $\Gamma$, and hence that $d(\fxy)>d(\fxy_m)+\varepsilon$, contradicting the fact that $d(\fxy_m)$ was a maximum. The conclusion is that $\fxy_m\in\Delta_i$.
\end{proof}

\begin{lemma}\label{delta}$\Gamma\cap(\overline\Delta_i\cup\overline\Delta_e)=\emptyset$.
\end{lemma}
\begin{proof}The proof is indirect, so assume that there is a point $s'\in C_L$ such that  $\fxc(s')\in\overline\Delta_i\cap\Gamma$. Then there is a sequence of points $\fxy_n\in\Delta_i$ that $\fxy_n\to\fxc(s')$ as $n\to\infty$. Since $\fxy_n\in\Delta_i$, there are at least two different points $s_n,t_n\in C_L$ such that 
$d(y_n)=|\fxy_n-\fxc(s_n)|$ and $d(\fxy_n)=|\fxy_n-\fxc(t_n)|$. Furthermore, the points $\fxy_n$ have the two representations
\begin{align*}
\fxy_n=\fxz(s_n,d(\fxy_n))&=\fxc(s_n)+d(\fxy_n)\fxc'(s_n)^\perp,\\
\fxy_n=\fxz(t_n,d(\fxy_n))&=\fxc(t_n)+d(\fxy_n)\fxc'(t_n)^\perp.
\end{align*}
The sequences $\{s_n\}$ and $\{t_n\}$ each have at least one accumulation point. If they have two different accumulation points $s$ and $t$, then it follows from the representations above that $\fxc(s)=\fxc(t)$, which contradicts that fact that the mapping $\fxc:C_L\to\Gamma$ is one-to-one. The alternative is that the sequences $\{s_n\}$ and $\{t_n\}$ have only one accumulation point $s$. Then  $s_n\to s$ and  $t_n\to s$ as $n\to\infty$. This implies that $s=s'$: Since $d(\fxy_n)\to0$, and we are left with $\fxy=\fxc(s)=\fxc(s')$, and since $\fxc$ is one-to-one, $s=s'$. Eventually, $d(\fxy_n)<\delta$ and we have two representations of $\fxy_n$ with $s_n$ and $t_n$ arbitrarily close to $s'$. This contradicts Lemma \ref{inverse}. The conclusion is that $\overline\Delta_i\cap\Gamma=\emptyset$. The same proof works for $\Delta_e$, if $\Delta_e\not=\emptyset$.
\end{proof}

Now define the open set $\Omega$ as $\Omega=\RR^2\setminus(\overline\Delta_i\cup\overline\Delta_e)$. This is a global uniqueness set, in the sense that if $\fxy\in\Omega$, then there is a unique point $s\in C_L$ such that $d(\fxy,\Gamma)=|\fxy-\fxc(s)|$. By Lemma \ref{delta}, $\Omega$ is a neighborhood of $\Gamma$, and since $\Delta_i$ is not empty, the constant $\gamma=d(\Gamma,\overline\Delta_i\cup\overline\Delta_e)$ is finite and positive. 

\begin{exercise}Sketch an example illustrating that the condition $|\lambda|<\delta$ is not sufficient to ensure that $z(s,\lambda)\in\Omega$. 
\end{exercise}

\begin{lemma}\label{onetoone}Assume that $\Gamma$ is $C^n$, $n\geq2$, and that  $|\lambda|<\min\{\gamma,\delta\}$ is fixed. Then $s\mapsto \fxz(s,\lambda)$ defines a $C^{n-1}$ Jordan curve $\Gamma_\lambda$, and $\Gamma_\lambda=\{\fxy\mid d(\fxy)=\lambda\}$. 
\end{lemma}
\begin{proof}For a fixed $\lambda$, $s\mapsto \fxz(s,\lambda)$ is a $C^{n-1}$ mapping of $C_L$ onto a plane curve $\Gamma_\lambda$. Since $|\lambda|<\gamma$, $\fxz(s,\lambda)\in\Omega$, and it has the unique representation $\fxz(s,\lambda)=\fxc(s)+\lambda\fxc'(s)^\perp$, where $d(\fxz(s,\lambda))=\lambda$. This means that
the mapping $s\mapsto \fxz(s,\lambda)$ is one-to-one and that $\Gamma_\lambda=\{\fxy\mid d(\fxy)=\lambda\}$. It remains to show that the curve $\Gamma_\lambda$ has an arc-length parameterization that is $C^{n-1}$. This follows from the assumption that $|\lambda|<\delta$:
\[
\fxz'(s,\lambda)=\fxc'(s)+\lambda\fxc'(s)^\perp=(1-\lambda\kappa(s))\fxc'(s),
\]
where $\kappa(s)=\fxc''(s)\cdot\fxc'(s)^\perp$. Since $1-\lambda\kappa(s)\ne0$ for $s\in C_L$, $\Gamma_\lambda$ has an arc-length parameterization that is $C^{n-1}$ (see Exercise 4.3).
\end{proof}

\begin{theorem}Assume that $\Gamma$ is $C^n$, $n\geq2$. Then the signed distance function $d$ is $C^{n-1}$ on the open set $\{\fxy\mid|d(\fxy)|<\min\{\gamma,\delta\}\}$.
\end{theorem}
\begin{proof}Assume that $|d(\fxy)|<\min\{\gamma,\delta\}$. This implies there is a unique $s\in C_L$ such that $d(\fxy)=\pm|\fxy-\fxc(s)|$ and  $\fxy$ has the unique representation 
\[
\fxz(s,d(\fxy))=\fxc(s)+d(\fxy)\fxc'(s)^\perp. 
\]
This is a global statement base on the fact that $|d(\fxy)|<\gamma$. Lemma \ref{inverse} states that $\fxz^{-1}$ is locally $C^{n-1}$. Since $\fxz^{-1}(\fxy)=(s,d(\fxy))$, we conclude that $d$ is $C^{n-1}$.
\end{proof}

There are two more results that are used when applying the shape local inclusion principle. The first one involves ``trapping'' a $C^2$ Jordan arc locally between the boundaries of disks, and the second one shows the existence of the disks $B(\fxx,r)$ that appear in the definition of local inclusion. It was easy to establish \eqref{eq221aa}, including the definition of the disk $B$,  because we knew everything about the shape of the ellipse $E(1)$. If, instead of having $E(1)$, we have a set $X\in\mathcal D$ whose boundary is $C^2$ in some neighborhood of $\fxx\in\partial X$, then we can still ``trap'' the boundary of $X$ between the boundaries of two disks. 

\begin{lemma}\label{trap}Suppose that a Jordan arc $\Gamma$ is $C^2$ in a neighborhood of a point $\fxc(s)$. Then for every sufficiently small $\varepsilon>0$, there is a neighborhood $N=N(\varepsilon)$ of $s$ such that the following conditions hold$:$
\begin{enumerate}[$(i)$]\setlength\itemsep{2pt}  
\item $\fxc''(s)\not=0$. For $s\in N$, the arc $\fxc(s)$ lies between the boundaries of the two disks $D_1=D_1(\fxc_1, r_1)$ and $D_2=D_2(\fxc_2, r_2)$, where 
\begin{alignat}{2}
\fxc_1&=\fxc(s)+(\kappa+\varepsilon)^{-1}\frac{\fxc''(s)}{\kappa},  
&&r_1=(\kappa+\varepsilon)^{-1},\notag\\
\fxc_2&=\fxc(s)+(\kappa-2\varepsilon)^{-1}\frac{\fxc''(s)}{\kappa}, \;\;
&&r_2=(\kappa-2\varepsilon)^{-1},\notag
\end{alignat}
where $\kappa=|\fxc''(s)|$.
\item $\fxc''(s)=0$. For $s\in N$, the arc $\fxc(s)$ lies between the boundaries of the two disks $D_1=D_1(\fxc_1, r_1)$ and $D_2=D_2(\fxc_2, r_2)$, where 
\begin{alignat}{2}
\fxc_1&=\fxc(s)+\varepsilon^{-1}\fxc'(s)^\perp,  
&&r_1=\varepsilon^{-1},\notag\\
\fxc_2&=\fxc(s)-\varepsilon^{-1}\fxc'(s)^\perp, \;\;
&&r_2=\varepsilon^{-1}.\notag
\end{alignat}
\end{enumerate}
\end{lemma}
\begin{proof}The proof is completely straightforward and is based on the continuity of the curvature. We may assume that the boundary of $X$ at $x$ is represented by the graph of a $C^2$ function $f$, where $f(0)=0$ and $f'(0)=0$. The curvature of $\partial X$ at $\fxx$ is $f''(0)$. There are two cases: $f''(0)\not=0$ and $f''(0)=0$. We begin by assuming that $f''(0)>0$. Choose $\varepsilon$ such that $0<2\varepsilon<f''(0)$. By the continuity of $f''$, there is a neighborhood $N=N(\varepsilon)$ of $x=0$ such that  
\[
f''(0)-\varepsilon\leq f''(x)\leq f''(0)+\varepsilon
\]
for all $x\in N$. This implies by integration that 
\[
(f''(0)-\varepsilon)x\leq f'(x)\leq (f''(0)+\varepsilon)x
\]
and that
\[
\frac{x^2}{2}(f''(0)-\varepsilon)\leq f(x)\leq \frac{x^2}{2}(f''(0)+\varepsilon)
\]
for all $x\in N$. Thus, the graph of $f$ is trapped between the graphs of two parabolas as long as $x\in N$. The next step is to trap the graph of $f$ between two circles.

The osculating circle of the graph of the parabola $y=(x^2/2)(f''(0)+\varepsilon)$ has radius $r=1/(f''(0)+\varepsilon)$ and center at $(0,r)$. Furthermore, the parabola lies below the circle for all $x$ in the interval $-r<x<r$. Stated analytically, this is the inequality
\[
\frac{x^2}{2r}\leq r-\sqrt{r^2-x^2},
\]
which holds for all $|x|<r$.

We now want a circle that lies below the parabola $y=(x^2/2)(f''(0)-\varepsilon)$. For this, we take the circle whose radius is $R=1/(f''(0)-2\varepsilon)$, which is slightly larger than the radius of the osculating circle of the graph of $y=(x^2/2)(f''(0)-\varepsilon)$. A straightforward computation shows that there is a constant $C>0$ such that  
\[
R-\sqrt{R^2-x^2}\leq \frac{x^2}{2}(f''(0)-\varepsilon)
\]
for all $x$ such that $x^2\leq C\varepsilon$. Thus, for all sufficiently small $|x|$, the graph of $f$ is trapped between two circles:
\[
R-\sqrt{R^2-x^2}\leq f(x)\leq r-\sqrt{r^2-x^2},
\]
where $R=1/(f''(0)-2\varepsilon)$ and $r=1/(f''(0)+\varepsilon)$.

This covers the case where $f''(0)>0$. The case where $f''(0)<0$ is similar, the function $f$ being trapped between circles with centers on the vegetive $y$-axis:
\[
-r+\sqrt{r^2-x^2}\leq f(x)\leq -R+\sqrt{R^2-x^2}.
\]

If $f''(0)=0$, the situation is slightly different. By the continuity of $f''$, given $\varepsilon>0$, there is a neighborhood $N$ such that $|f''(x)|\leq\varepsilon$ for all $x\in N$. Thus,
\[
-x\varepsilon\leq f'(x)\leq x\varepsilon
\]
and
\[
-\frac{x^2}{2}\varepsilon\leq f(x)\leq\frac{x^2}{2}\varepsilon
\]
for all $x\in N$. The graph of $f$ is then trapped between the two osculating circles of the two parabolas:
$
-r+\sqrt{r^2-x^2}\leq f(x)\leq r-\sqrt{r^2-x^2},
$
where $r=1/\varepsilon$. Again, this holds for all $x\in N$. The statement of the results in the lemma is just a restatement of the results for $f$ in terms of the vector representation of $\partial X$. 
\end{proof}

Lemma \ref{trap} tells us exactly how to trap a $C^2$ arc between two disks. The next, and last result of the section, tells us how to select the disks $B$ needed to apply the local inclusion principle.

\begin{lemma}\label{diskB}Suppose that a Jordan curve $\Gamma$ is $C^2$ in a neighborhood of a point $\fxc(s)$. Then there is a closed disk $D(\fxc(s),r)$ centered at $\fxc(s)$ with radius $r$ that has the following property$:$ For every closed disk $D'(\fxc(s),r')$, $0<r'\leq r$, $\Gamma\cap D'$ is a Jordan arc that intersects the boundary of $D'$ in exactly two points, and $D'\setminus\Gamma$ consists of two disjoint sets $D'_i\subset\Gamma_i$ and $D'_e\subset\Gamma_e$.       
\end{lemma}
\begin{proof}As in Lemma \ref{trap}, we represent $\Gamma$ near $\fxc(s)$ as the graph of a function $f$, where $f(0)=0$ and $f'(0)=0$. We wish to formalize the idea that, for all sufficiently small $r$, the graph of a function $f$ intersects the circles $C(0,r)$ in only two points, at one point for $x<0$ and one for $x>0$. Assume that $x>0$. Then what we mean analytically is that the tangent vector $(1,f'(x)$ has a positive component in the direction of the radial vector $(x,f(x))$. Thus we need to show that $(x,f(x))\cdot(1,f'(x)=x+f(x)f'(x)>0$ for all sufficiently small $x>0$. A direct consequence of the computation in the proof of Lemma \ref{trap} is that $|f(x)f'(x)|=O(x^3)$ in some neighborhood of $x=0$. Thus, for all sufficiently small $x>0$, $x+f(x)f'(x)>0$, and the piece of $\Gamma$ represented locally by the graph of $f$ intersects the circles $C(0,r)$ once in the right-half plane for all sufficiently small $r$. A similar analysis shows that the piece of $\Gamma$ represented locally by the graph of $f$ intersects the circles $C(0,r)$ once in the left-half plane for all sufficiently small $r$. 

Let $\Gamma_r$ denote the graph of $f$ that lies in the disk $D(0,r)$. We must show that 
\[
D(0,r)\cap\Gamma=D(0,r)\cap\Gamma_r
\]
for all sufficiently small $r$. If this were not the case, there would be a sequence of points $\fxc(t_n)$ such that $\fxc(t_n)\to\fxc(s)$. It should be clear from what we have just done that the points $t_n\in C_L$ are bounded away from $s\in C_L$. Then if $t$ is any accumulation point of $\{t_n\}$, which must exist, $t\not=s$, but $\fxc(t)=\fxc(s)$. This contradicts the fact that $\fxc$ is one-to-one.

We have shown that there is an $r>$ such that for all disks $D'=D'(0,r')$, $0<r'\leq r$, $D'\setminus\Gamma_r$ consists of two arc-wise connected sets that do not intersect $\Gamma$ but whose boundaries intersect $\Gamma$. By the Jordan curve theorem, one of the sets must lie in $\Gamma_i$ and the other must be in $\Gamma_e$.
\end{proof}

