The continuity of curvature flow in Fréchet distance Shiyi Ma

###### Abstract

We study the evolution of a Jordan curve on the plane by curvature flow, also
known as curve shortening flow, and by level-set flow, which is a weak
formulation of curvature flow. We show that the evolution of the curve depends
continuously on the initial curve in Fréchet distance. This is an extension of
Michael Dobbins’ work on the evolution of Jordan curves bisecting the
$2$-sphere by curvature flow.

##  1 Introduction

Let $\mathfrak{J}$ denote the set of Jordan curves (which are also called
simple closed curves) with zero Lebesgue measure, on ${\mathbb{R}}^{2}$.
Michael Gage and Richard Hamilton proved that all smooth convex curves
eventually contract to a point without forming any other singularities [4],
and Matthew Grayson proved that every non-convex curve will eventually become
convex [3]. Let $\mathfrak{J}_{T}$ be a subset of $\mathfrak{J}$ for which the
curvature flow problem has a solution up to time $T$, for $0<T<\infty$. Define
a function
$\operatorname{cf}:\mathfrak{J}_{T}\times[0,T)_{{\mathbb{R}}}\to\mathfrak{J}$
by evolution by curvature flow. The main result of this article is that this
function is continuous in Fréchet distance.

######  Theorem 1.1.

Let $\gamma_{k}\in\mathfrak{J}_{T}$, and $t_{k}\in[0,T)_{{\mathbb{R}}}$ for
$k\in\\{1,...,\infty\\}$. If $\gamma_{k}\to\gamma_{\infty}$ in Fréchet
distance and $t_{k}\to t_{\infty}$, then
$\operatorname{cf}(\gamma_{k},t_{k})\to\operatorname{cf}(\gamma_{\infty},t_{\infty})$
in Fréchet distance.

The challenge of proving Theorem 1.1 is to prove Lemma 3.2 where given a curve
$\gamma$, we need to show the existence of a curve that always intersects
$\gamma$ at exactly two points until $\gamma$ shrinks to a point. To prove
Lemma 3.2, we construct a family of Jordan curves through homotopy, and prove
that curves in this family instantly evolve into three classes, among which
one class contains curves that intersect $\gamma$ at exactly two points
throughout the lifetime of $\gamma$.

###  1.1 Definitions and notations

We denote real intervals by $(a,b]_{{\mathbb{R}}}=\\{x\in R:a<x\leq b\\}$ for
bounds $a,b\in{\mathbb{R}}$ with any combination of round or square brackets
for (half) open or closed intervals. We also similarly denote segments in
${\mathbb{R}}^{2}$ by
$[a,b]_{{\mathbb{R}}^{2}}=\\{ta+(1-t)b:t\in[0,1]_{{\mathbb{R}}}\\}$ for
$a,b\in{\mathbb{R}}^{2}$. We denote the unit circle in ${\mathbb{R}}^{2}$ by
$\mathbf{S}^{1}$. Given a set $A\subset{\mathbb{R}}^{2}$, we denote the
interior of $A$ by $\operatorname{int}A$. For $X\subseteq{\mathbb{R}}^{2}$ and
$\delta>0$, let $X\oplus\delta$ be the set of points at most distance $\delta$
from $X$. Every Jordan curve on ${\mathbb{R}}^{2}$ is treated as a subset of
${\mathbb{R}}^{2}$. Note that for every Jordan curve $\gamma$,
${\mathbb{R}}^{2}\setminus\gamma$ consists of two components. We denote the
bounded component, which is the interior of the region enclosed by $\gamma$ by
$\mathcal{I}(\gamma)$, and the other unbounded component by
$\mathcal{O}(\gamma)$. Let $\gamma_{1},\gamma_{2}$ be two Jordan curves on
${\mathbb{R}}^{2}$. The Hausdorff distance is defined by

$\operatorname{dist}_{H}(\gamma_{1},\gamma_{2})=\inf\\{\delta\ |\
\gamma_{1}\subseteq\gamma_{2}\oplus\delta,\gamma_{2}\subseteq\gamma_{1}\oplus\delta\\}.$

The Fréchet distance on $\mathfrak{J}$ is defined by

$\operatorname{dist}_{F}(\gamma_{1},\gamma_{2})=\inf\limits_{\varphi_{1},\varphi_{2}}\sup\limits_{x}||\varphi_{1}(x)-\varphi_{2}(x)||,$

where $\varphi_{i}:{\mathbf{S}}^{1}\to\gamma_{i}$ are homeomorphisms. Let
$\mathfrak{J}_{c^{\infty}}$ be the set of smooth Jordan curves. Given a smooth
curve $\gamma\in\mathfrak{J}_{C^{\infty}}$ and a point $p\in\gamma$, let
$\mathrm{kn}(\gamma,p)\in{\mathbb{R}}^{2}$ denote the vector of curvature of
$\gamma$ at $p$. A solution to the curvature flow problem for a given initial
curve $\gamma_{0}\in\mathfrak{J}$ and stopping time
$T\in(0,\infty)_{{\mathbb{R}}}$, is a map
$\Gamma:{\mathbf{S}}^{1}\times(0,T)\to{\mathbb{R}}^{2}$ such that
$\partial_{t}\Gamma(x,t)=\mathrm{kn}(\Gamma({\mathbf{S}}^{1},t),\Gamma(x,t))$
and $\Gamma({\mathbf{S}}^{1},t)\to\gamma_{0}$ in Fréchet distance as $t\to 0$.
Let

$\operatorname{cf}(\gamma_{0},t)=\Gamma({\mathbf{S}}^{1},t),$

where $\Gamma$ is the solution to the curvature flow problem starting from
$\gamma_{0}$, provided that a unique solution exists.

###  1.2 Level-set flow

Given an oriented curve $\gamma\in\mathfrak{J}$, let
$\alpha_{k}\in\mathfrak{J}$ be a sequence of curves approaching $\gamma$ from
one side of $\gamma$, and $\beta_{k}\in\mathfrak{J}$ be a sequence of curves
approaching $\gamma$ from the other side of $\gamma$. Now, let $\\{A_{k}\\}$
be the sequence of nested annuli between $\alpha_{k}$ and $\beta_{k}$ and let
$A_{k}(t)$ be the annulus between $\alpha_{k}(t)$ and $\beta_{k}(t)$, the time
$t$ evolutions of $\alpha_{k}$ and $\beta_{k}$ by curvature flow. The
avoidance principle says that if two curves are initially disjoint then they
remain disjoint throughout their evolution, for as long as a solution exists.
Since $\gamma\subset A_{k}$ for all $k\in\\{1,...,\infty\\}$, the avoidance
principle implies that for any $t>0$,
$\operatorname{cf}(\gamma,t)\subset\bigcap\limits_{k=1}^{\infty}A_{k}(t)$. By
Lauer’s Proposition 8.3 [5], $\bigcap\limits_{k=1}^{\infty}A_{k}(t)$ is the
level-set flow of $\gamma$. Lauer also showed that if the Lebesgue measure of
the initial curve is zero, then the level-set flow immediately becomes a
smooth Jordan curve evolving by curvature flow [5, Theorem 1.2]. Therefore,
the level-set flow is the unique solution to the curvature flow problem in
this case, which implies
$\operatorname{cf}(\gamma,t)=\bigcap\limits_{k=1}^{\infty}A_{k}(t)$.

##  2 Previous results

In this section, we list out the lemmas about Jordan curves bisecting the
$2$-sphere, proved by Michael Dobbins [2]. These lemmas also apply to Jordan
curves in ${\mathbb{R}}^{2}$.

######  Lemma 2.1.

For each $\gamma\in\mathfrak{J}$ and $\epsilon>0$, there is
$\delta=\delta(\gamma,\epsilon)>0$ such that for every Jordan curve
$\eta\subset(\gamma\oplus\delta)$, there is a continuous map
$\varphi:\eta\to\gamma$ such that for all $x\in\eta,\ \
||\varphi(x)-x||<\epsilon$.

Note that $\varphi$ is not necessarily bijective, so this does not provide an
upper bound on Fréchet distance.

######  Lemma 2.2.

Let $\gamma_{k}\in\mathfrak{J}_{T}$ and $t_{k}\in[0,T)_{{\mathbb{R}}}$ for
$k\in\\{1,...,\infty\\}$. If $\gamma_{k}\to\gamma_{\infty}$ in Fréchet
distance and $t_{k}\to t_{\infty}$, then
$\operatorname{cf}(\gamma_{k},t_{k})\to\operatorname{cf}(\gamma_{\infty},t_{\infty})$
in Hausdorff distance.

######  Lemma 2.3.

Let $\gamma_{1},\gamma_{2}\in\mathfrak{J}_{S}$ be a pair of curves that
intersect at finitely many points, and let
$p_{T}\in(\operatorname{cf}(\gamma_{1},T)\cap\operatorname{cf}(\gamma_{2},T))$
be a point of intersection at time $T\in(0,S)_{{\mathbb{R}}}$. Then, there is
a continuous trajectory $p:[0,T]_{{\mathbb{R}}}\to{\mathbb{R}}^{2}$ such that
for all $t\in[0,T]_{{\mathbb{R}}}$, we have
$p(t)\in(\operatorname{cf}(\gamma_{1},T)\cap\operatorname{cf}(\gamma_{2},T))$
and $p(T)=p_{T}$.

######  Lemma 2.4.

Let $\gamma,\eta\in\mathfrak{J}$ be a pair of curves that intersect at only 2
points, and let $x$ be one of the points of intersection. By the time the two
points of intersection merge, there is a unique continuous trajectory
$p(\gamma,\eta,x):[0,\infty)\to{\mathbb{R}}^{2}$ such that for all $t$ we have
$p(\gamma,\eta,x;t)\in(\operatorname{cf}(\gamma,t)\cap\operatorname{cf}(\eta,t))$
and $p(\gamma,\eta,x;0)=x$. Furthermore, $p(\gamma,\eta,x,t)$ is continuous as
a function of $\gamma$ and $\eta$ in Fréchet distance and $x$ and $t$.

The proofs of these lemmas are analogous to Dobbins’s proofs. One difference
is that in Lemma 2.2 and 2.4, we need to define a closed circular region
$\mathbf{D}$ that contains $\gamma_{\infty}$, because we need to use the
compactness of $\mathbf{D}$ to show that some subsequence is convergent. The
arguments we need for the proofs are as follows:

Suppose the diameter of $\gamma_{\infty}$ is $d$. Let
$\mathbf{D}\in{\mathbb{R}}^{2}$ be a closed circular region with the diameter
$d+1$ such that $\gamma_{\infty}\subset\mathbf{D}$. Since $\gamma_{\infty}$
and $\partial\mathbf{D}$ are disjoint and $\partial\mathbf{D}$ encloses
$\gamma_{\infty}$, by the avoidance principle, the shrinking circle
$\operatorname{cf}(\partial\mathbf{D},t)$ encloses
$\operatorname{cf}(\gamma_{\infty},t)$ for all $t\in[0,T]_{{\mathbb{R}}}$.
Therefore, $\operatorname{cf}(\gamma_{\infty},t)\subset\mathbf{D}$ for all
$t\in[0,T]_{{\mathbb{R}}}$.

Dobbins proved these two lemmas for Jordan curves on the $2$-sphere by using
the compactness of the $2$-sphere. Another difference is that for Lemma 2.3,
we need to change geodesic curvature in Dobbins’s proof to curvature in the
argument that the transversal intersection point of two curves evolving by
curvature flow continues to be a point of intersection provided that the
intersection remains transversal.

##  3 Proof of continuity

######  Lemma 3.1.

Let $\gamma_{1},\gamma_{2}\in\mathfrak{J}$ be a pair of distinct curves that
have a common arc $\mathcal{A}$, i.e. $\mathcal{A}=\gamma_{1}\cap\gamma_{2}$.
Then for any $t>0$, $\operatorname{cf}(\gamma_{1},t)$ and
$\operatorname{cf}(\gamma_{2},t)$ are disjoint and if
$\gamma_{1}\setminus\mathcal{A}\subset\mathcal{I}(\gamma_{2})$ $(\text{or}\
\gamma_{2}\setminus\mathcal{A}\subset\mathcal{I}(\gamma_{1}))$, then
$\operatorname{cf}(\gamma_{1},t)\subset\mathcal{I}(\operatorname{cf}(\gamma_{2},t))$
$(\text{or}\
\operatorname{cf}(\gamma_{2},t)\subset\mathcal{I}(\operatorname{cf}(\gamma_{1},t)))$,
until one of them shrinks to a point.

###### Proof.

Figure 1: $\gamma_{1}$ and $\gamma_{2}$ immediately become disjoint, and
$\operatorname{cf}(\gamma_{1},t)$ remains in the region enclosed by
$\operatorname{cf}(\gamma_{2},t)$ up to some finite time $T$.

Without loss of generality, we assume that
$\gamma_{1}\subset\mathcal{I}(\gamma_{2})$; see Figure 1. Let $T$ be the first
time when one of the curves shrinks to a point as they evolve by curvature
flow. First, we want to show that for $t\in(0,T)_{{\mathbb{R}}}$, there is no
transversal intersection points of $\operatorname{cf}(\gamma_{1},t)$ and
$\operatorname{cf}(\gamma_{2},t)$. Suppose not. Let $P$ be a transversal
intersection point of $\operatorname{cf}(\gamma_{1},t_{\infty})$ and
$\operatorname{cf}(\gamma_{2},t_{\infty})$. Let $\\{\alpha_{k}\\}$ be a
sequence of Jordan curves such that $\alpha_{k}\subset\mathcal{I}(\gamma_{1})$
and $\alpha_{k}\to\gamma_{1}$ in Fréchet distance. Let $\\{\beta_{k}\\}$ be a
sequence of Jordan curves such that $\beta_{k}\subset\mathcal{O}(\gamma_{2})$
and $\beta_{k}\to\gamma_{2}$ in Fréchet distance. Then all $\alpha_{k}$’s are
disjoint from all $\beta_{k}$’s. Let $t_{k}\to t_{\infty}$ for $t_{\infty}<T$.
Since $\operatorname{cf}(\gamma_{i},t_{\infty})$ for $i=1,2$ are smooth, by
the implicit function theorem, there exist $\varepsilon_{1},\varepsilon_{2}$
such that the component of
$(\operatorname{cf}(\gamma_{1},t_{\infty})\oplus\varepsilon_{1})\cap(\operatorname{cf}(\gamma_{2},t_{\infty})\oplus\varepsilon_{2})$
with $P$ in it is a region that has a boundary consisting of 4 arcs; see
Figure 2. By Lemma 2.2,
$\operatorname{cf}(\alpha_{k},t_{k})\to\operatorname{cf}(\gamma_{1},t_{\infty})$
and
$\operatorname{cf}(\beta_{k},t_{k})\to\operatorname{cf}(\gamma_{2},t_{\infty})$
in Hausdorff distance, so
$\operatorname{cf}(\alpha_{k},t_{k})\subset\operatorname{cf}(\gamma_{1},t_{\infty})\oplus\varepsilon_{1}$
and
$\operatorname{cf}(\beta_{k},t_{k})\subset\operatorname{cf}(\gamma_{2},t_{\infty})\oplus\varepsilon_{2}$
for $k$ sufficiently large. Consequently, it is inevitable for
$\operatorname{cf}(\alpha_{k},t_{k})$ and $\operatorname{cf}(\beta_{k},t_{k})$
to intersect with each other, which contradicts the avoidance principle.
Therefore, for $t\in(0,T)_{{\mathbb{R}}}$, there is no transversal
intersection points of $\operatorname{cf}(\gamma_{1},t)$ and
$\operatorname{cf}(\gamma_{2},t)$.

Sigurd Angenent showed that the number of intersection points of two different
curves immediately becomes finite and then is non-increasing throughout their
evolution [1, Theorem 1.3]. The intersection is called a tangential
intersection point of two curves if the unit tangent vectors are dependent at
that point of intersection. If two curves initially have a common arc, then
they have infinitely many tangential intersection points. By Angenent’s
theorem, the number of intersection points of
$\operatorname{cf}(\gamma_{1},t)$ and $\operatorname{cf}(\gamma_{2},t)$ is
finite for $t\in(0,T)_{{\mathbb{R}}}$, and thus
$\operatorname{cf}(\gamma_{1},t)$ and $\operatorname{cf}(\gamma_{2},t)$ does
not share a common arc for $t\in(0,T)_{{\mathbb{R}}}$.

Now we want to show that for $t\in(0,T)_{{\mathbb{R}}}$, there is no
tangential intersection points of $\operatorname{cf}(\gamma_{1},t)$ and
$\operatorname{cf}(\gamma_{2},t)$. Suppose not. Then there is a tangential
intersection point of $\operatorname{cf}(\gamma_{1},t_{0})$ and
$\operatorname{cf}(\gamma_{2},t_{0})$ for some $t_{0}<T$. By Angenent’s
theorem, which says that the set of moments in time $t$ at which the evolution
of two curves have a tangential intersection points is discrete in
$(0,T)_{{\mathbb{R}}}$[1, Theorem 1.3], there exists $\delta>0$ such that
$\operatorname{cf}(\gamma_{1},t)$ and $\operatorname{cf}(\gamma_{2},t)$ are
not tangent when
$t\in(t_{0}-\delta,t_{0})_{{\mathbb{R}}}\cup(t_{0},t_{0}+\delta)_{{\mathbb{R}}}$.
Then $\operatorname{cf}(\gamma_{1},t)$ and $\operatorname{cf}(\gamma_{2},t)$
would either have transversal intersection points or be disjoint for
$t\in(t_{0}-\delta,t_{0})_{{\mathbb{R}}}\cup(t_{0},t_{0}+\delta)_{{\mathbb{R}}}$.
Since we proved that it is impossible for them to have transversal
intersection points for $t\in(0,T)_{{\mathbb{R}}}$,
$\operatorname{cf}(\gamma_{1},t)$ and $\operatorname{cf}(\gamma_{2},t)$ are
disjoint for
$t\in(t_{0}-\delta,t_{0})_{{\mathbb{R}}}\cup(t_{0},t_{0}+\delta)_{{\mathbb{R}}}$.
However, there is a contradiction because by the avoidance principle, if
$\operatorname{cf}(\gamma_{1},t)$ and $\operatorname{cf}(\gamma_{2},t)$ are
disjoint when $t<t_{0}$, then they remain disjoint when $t=t_{0}$. Therefore,
$\operatorname{cf}(\gamma_{1},t)$ and $\operatorname{cf}(\gamma_{2},t)$ are
disjoint for $t\in(0,T)_{{\mathbb{R}}}$.

Figure 2: Since $\operatorname{cf}(\gamma_{1},t_{\infty})$ and
$\operatorname{cf}(\gamma_{2},t_{\infty})$ intersect transversely, there exist
$\epsilon_{1}$ and $\epsilon_{2}$ such that
$(\operatorname{cf}(\gamma_{1},t_{\infty})\oplus\varepsilon_{1})\cap(\operatorname{cf}(\gamma_{2},t_{\infty})\oplus\varepsilon_{2})$
is a region that contains $P$ and has a boundary consisting of $4$ arcs.

Now we want to show that
$\operatorname{cf}(\gamma_{1},t)\subset\mathcal{I}(\operatorname{cf}(\gamma_{2},t))$
for $t\in(0,T)_{{\mathbb{R}}}$. Suppose not. Then,
$\operatorname{cf}(\gamma_{1},t)\subset\mathcal{O}(\operatorname{cf}(\gamma_{2},t))$
for $t\in(0,T)_{{\mathbb{R}}}$. Let $\\{\zeta_{k}\\}$ be a sequence of smooth
Jordan curves such that $\zeta_{k}\subset\mathcal{I}(\gamma_{1})$ and
$\zeta_{k}\to\gamma_{1}$ in Fréchet distance. Then
$\zeta_{k}\subset\mathcal{I}(\gamma_{2})$ for all $k\in\\{1,...,\infty\\}$.
Let $\varphi_{k}:\mathbf{S}^{1}\to\zeta_{k}$ and
$\psi:\mathbf{S}^{1}\to\gamma_{1}$ be two parameterizations such that
$\sup_{x\in\mathbf{S}^{1}}||\varphi_{k}(x)-\psi(x)||\to 0$. If we fix a point
$x_{0}\in\mathbf{S}^{1}$, then $||\varphi_{k}(x_{0})-\psi(x_{0})||\to 0$, i.e.
$\varphi_{k}(x_{0})$ converges to $\psi(x_{0})\in\gamma_{1}$. Suppose
$\zeta_{k}\in\mathfrak{J}_{T_{k}}$. Define
$\mathrm{g}_{k}:\varphi_{k}(x_{0})\times[0,T_{k})\to{\mathbb{R}}^{2}$ by the
trajectory of $\varphi_{k}(x_{0})$ as $\zeta_{k}$ evolves by curvature flow.
By Lemma 2.2,
$\operatorname{cf}(\zeta_{k},t_{k})\to\operatorname{cf}(\gamma_{1},t_{\infty})$
in Hausdorff distance, so for $k$ sufficiently large,
$\mathrm{g}_{k}(\varphi_{k}(x_{0}),t_{k})\in\operatorname{cf}(\gamma_{1},t_{\infty})\oplus\varepsilon$
for any $\varepsilon>0$. Since we assumed that
$\operatorname{cf}(\gamma_{1},t_{\infty})\subset\mathcal{O}(\operatorname{cf}(\gamma_{2},t_{\infty}))$,
we have
$\mathrm{g}_{k}(\varphi_{k}(x_{0}),t_{k})\in\mathcal{O}(\operatorname{cf}(\gamma_{2},t_{\infty}))$,
which implies that $\operatorname{cf}(\zeta_{k},t_{k})$ crosses
$\operatorname{cf}(\gamma_{2},t_{k})$ at some point as $t_{k}\to t_{\infty}$.
However, since each $\zeta_{k}$’s was initially disjoint from $\gamma_{2}$,
their intersection contradicts the avoidance principle. ∎

######  Lemma 3.2.

Let $\gamma\in\mathfrak{J}_{T}$, and let $x\in\gamma$ be a point. Then, there
is a curve $\eta\in\mathfrak{J}$ such that it does not shrink to a point
before $\gamma$ does, only intersects $\gamma$ at two points, which are $x$
and some other point, and the number of intersection points of
$\operatorname{cf}(\gamma,t)$ and $\operatorname{cf}(\eta,t)$ remains 2 for
any $t\in(0,T)_{{\mathbb{R}}}$. Furthermore, $\eta$ is continuous in Fréchet
distance as a function of $x$.

###### Proof.

Suppose the diameter of $\gamma$ is $2r$, for some $r>0$. Let
$\mathbf{D}\in{\mathbb{R}}^{2}$ be a closed circular region with diameter $4r$
such that $\gamma\in\mathbf{D}$. Then we can find another closed circular
region $\mathbf{E}$ with diameter $d>4\sqrt{2T}+4r$ such that $\mathbf{E}$ and
$\mathbf{D}$ have the same center. Let $\mathbf{\Gamma}$ be the closed region
enclosed by $\gamma$. Let $A$ be the closed annulus between two concentric
circles with radii $1$ and $2$ in ${\mathbb{R}}^{2}$. Let
$\varphi:A\to\overline{\mathbf{D}\setminus\mathbf{\Gamma}}$ be a
homeomorphism. For each $x\in\gamma$, we have points
$\varphi^{-1}(x),-\varphi^{-1}(x),2\varphi^{-1}(x),-2\varphi^{-1}(x)\in\partial
A$; see Figure 3. Let $\mathbf{U}$ and $\mathbf{V}$ denote two arcs of
$\gamma$ divided by $x$ and $\varphi(-\varphi^{-1}(x))$. Then there is a
homotopy $\sigma(\cdot,s)=sf+(1-s)g$ between the path $f:[0,1]\to\mathbf{U}$
and the path $g:[0,1]\to\mathbf{V}$ such that $\sigma(\cdot,0)=g$ and
$\sigma(\cdot,1)=f$. Given a point $x\in\gamma$, we denote each path generated
by the homotopy $\sigma$ by $\eta_{1}(x,s)$ for some $s\in[0,1]$. Let
$\eta_{2}(x)=\varphi([\varphi^{-1}(x),2\varphi^{-1}(x)]_{{\mathbb{R}}^{2}})$,
and
$\eta_{3}(x)=\varphi([-\varphi^{-1}(x),-2\varphi^{-1}(x)]_{{\mathbb{R}}^{2}})$.

Given two distinct points, $v_{1},v_{2}\in\partial\mathbf{D}$, let
$\pi_{1}:\partial\mathbf{D}\times\partial\mathbf{D}\to\partial\mathbf{E}$ be a
map defined by $\pi_{1}(v_{1},v_{2})=w$ such that $v_{1},v_{2},w$ are
collinear, and $||w-v_{1}||<||w-v_{2}||$. Likewise, let
$\pi_{2}:\partial\mathbf{D}\times\partial\mathbf{D}\to\partial\mathbf{E}$ be a
map defined by $\\{v_{1},v_{2}\\}\mapsto w$ such that $v_{1},v_{2},w$ are
collinear, and $||w-v_{1}||>||w-v_{2}||$. Let
$\eta_{4}(x)=[\varphi(2\varphi^{-1}(x)),\pi_{1}(\varphi(2\varphi^{-1}(x)),\varphi(-2\varphi^{-1}(x)))]_{{\mathbb{R}}^{2}}$,
$\eta_{5}(x)=[\varphi(-2\varphi^{-1}(x)),\pi_{2}(\varphi(2\varphi^{-1}(x)),$
$\varphi(-2\varphi^{-1}(x)))]_{{\mathbb{R}}^{2}}$, and
$\eta_{6}(x)\subset\partial\mathbf{E}$ be an arc of $\partial\mathbf{E}$ going
from $\pi_{1}(\varphi(2\varphi^{-1}(x)),\varphi(-2\varphi^{-1}(x)))$ to
$\pi_{2}(\varphi(2\varphi^{-1}(x)),\varphi(-2\varphi^{-1}(x)))$ clockwise.

Now we can construct a Jordan curve, $\eta(x,s)$, consisting of
$\eta_{1}(x,s),\eta_{2}(x),\eta_{3}(x),\eta_{4}(x),\eta_{5}(x)$ and
$\eta_{6}(x)$. Note that $\eta(x,s)$ only intersects $\gamma$ at two points,
$x$ and $\varphi(-\varphi^{-1}(x))$. Let the closed region enclosed by
$\eta(x)$ be $\mathbf{H}$. Note that a circle with initial radius $r_{0}$ can
be seen to evolve by curvature flow to a circle with radius
$r(t)=\sqrt{r_{0}^{2}-2t}$, and thus the circle disappears at $t=r_{0}^{2}/2$.
Since $d>4\sqrt{2T}+4r$, the difference between the radii of $\mathbf{E}$ and
$\mathbf{D}$ is greater than $2\sqrt{2T}$. Therefore, a circle with radius
$\sqrt{2T}$ can be fitted in the interior of $\mathbf{H}\setminus\mathbf{D}$,
and its lifetime is $T$. By avoidance principle, the lifetime of $\eta(x)$ is
greater than $T$.

Figure 3: We use a homeomorphism $\varphi$ and a homotopy $\sigma$ to define a
Jordan curve, $\eta(x,s)$ consisting of $6$ arcs.

Now we want to show that $\eta$ is continuous. Since $\varphi$ is a
homeomorphism, as $x_{k}\to x_{\infty}$, we have
$\varphi^{-1}(x_{k})\to\varphi^{-1}(x_{\infty})$,
$-\varphi^{-1}(x_{k})\to-\varphi^{-1}(x_{\infty})$, $2\varphi^{-1}(x_{k})\to
2\varphi^{-1}(x_{\infty})$, and
$-2\varphi^{-1}(x_{k})\to-2\varphi^{-1}(x_{\infty})$, so
$[\varphi^{-1}(x_{k}),2\varphi^{-1}(x_{k})]_{{\mathbb{R}}^{2}}\to[\varphi^{-1}(x_{\infty}),2\varphi^{-1}(x_{\infty})]_{{\mathbb{R}}^{2}}$
and
$[-\varphi^{-1}(x_{k}),-2\varphi^{-1}(x_{k})]_{{\mathbb{R}}^{2}}\to[-\varphi^{-1}(x_{\infty}),-2\varphi^{-1}(x_{\infty})]_{{\mathbb{R}}^{2}}$
in Fréchet distance. By the Heine-Cantor theorem, $\varphi$ is uniformly
continuous, so $\eta_{i}(x_{k})\to\eta_{i}(x_{\infty})$ in Fréchet distance
for $i\in\\{2,3\\}$. Since
$\varphi(2\varphi^{-1}(x_{k}))\to\varphi(2\varphi^{-1}(x_{\infty}))$,
$\varphi(-2\varphi^{-1}(x_{k}))\to\varphi(-2\varphi^{-1}(x_{\infty}))$ and how
$\pi_{1}$ and $\pi_{2}$ are defined, we have
$\eta_{i}(x_{k})\to\eta_{i}(x_{\infty})$ in Fréchet distance, for
$i\in\\{4,5,6\\}$.

Note that every pair of $\eta(x,0)$, $\eta(x,1)$ and $\gamma$ share a common
arc. By Lemma 3.1, $\operatorname{cf}(\eta(x,0),t)$
$\operatorname{cf}(\eta(x,1),t)$ and $\operatorname{cf}(\gamma,t)$ are
disjoint from each other for $t\in(0,T)_{{\mathbb{R}}}$. Without the loss of
generality, we assume that
$\eta(x,0)\setminus(\eta(x,0)\cap\eta(x,1))\subset\mathcal{I}(\eta(x,1))$.
Then $\gamma\setminus(\gamma\cap\eta(x,0))\subset\mathcal{O}(\eta(x,0))$, and
$\gamma\setminus(\gamma\cap\eta(x,1))\subset\mathcal{I}(\eta(x,1))$. By Lemma
3.1, both $\operatorname{cf}(\gamma,t)$ and $\operatorname{cf}(\eta(x,0),t)$
are lying in $\mathcal{I}(\operatorname{cf}(\eta(x,1),t))$, with
$\operatorname{cf}(\gamma,t)\subset\mathcal{O}(\operatorname{cf}(\eta(x,0),t))$.
For each $t_{\infty}\in(0,T)_{{\mathbb{R}}}$, the family of Jordan curves
$\mathcal{F}=\\{\eta(x,s)\ |\ 0\leq s\leq 1\\}$ can be divided into three
classes, $\mathcal{C}_{1}$, $\mathcal{C}_{2}$, and $\mathcal{C}_{3}$. Class
$\mathcal{C}_{1}$ contains curves $\eta(x,s)\in\mathcal{F}$ such that
$\operatorname{cf}(\gamma,t_{\infty})\subset\mathcal{O}(\operatorname{cf}(\eta(x,s),t_{\infty}))$.
Class $\mathcal{C}_{2}$ contains curves $\eta(x,s)\in\mathcal{F}$ such that
$\operatorname{cf}(\gamma,t_{\infty})\cap\operatorname{cf}(\eta(x,s),t_{\infty})\neq\emptyset$.
Class $\mathcal{C}_{3}$ contains curves $\eta(x,s)\in\mathcal{F}$ such that
$\operatorname{cf}(\gamma,t_{\infty})\subset\mathcal{I}(\operatorname{cf}(\eta(x,s),t_{\infty}))$.
Since $\eta(x,0)\in\mathcal{C}_{1}$ and $\eta(x,1)\in\mathcal{C}_{3}$,
$\mathcal{C}_{1}$ and $\mathcal{C}_{3}$ are both non-empty. Let $R$ be the
radius of the largest circle that can be inscribed within
$\operatorname{cf}(\gamma,t_{\infty})$. By the definition of $\mathcal{C}_{1}$
and $\mathcal{C}_{3}$, given two curves $\mu\in\mathcal{C}_{1}$ and
$\mu^{\prime}\in\mathcal{C}_{3}$, $\operatorname{cf}(\mu,t_{\infty})$ and
$\operatorname{cf}(\mu^{\prime},t_{\infty})$ are at least $R$ apart; see
Figure 4. We want to show that class $\mathcal{C}_{2}$ is also non-empty.
Suppose not. Let $s_{0}=\inf\\{s\ |\ \eta(x,s)\in\mathcal{C}_{3},\ 0\leq s\leq
1\\}$.

Case I. Suppose $\eta(x,s_{0})\in\mathcal{C}_{3}$. There is a sequence of
curves $\eta(x,s_{k})\to\eta(x,s_{0})$ in Fréchet distance as $s_{k}\to s_{0}$
from below such that $\eta(x,s_{k})\in\mathcal{C}_{1}$. Then,
$\operatorname{cf}(\eta(x,s_{k}),t_{\infty})$ and
$\operatorname{cf}(\eta(x,s_{0}),t_{\infty})$ are at least $R$ apart. Since,
by Lemma 2.2,
$\operatorname{cf}(\eta(x,s_{k}),t_{\infty})\to\operatorname{cf}(\eta(x,s_{0}),t_{\infty})$
in Hausdorff distance, $\operatorname{cf}(\eta(x,s_{k}),t_{\infty})$ and
$\operatorname{cf}(\eta(x,s_{0}),t_{\infty})$ are within a Hausdorff distance
of $\varepsilon$, for any $\varepsilon>0$ and all $k$ sufficiently large. For
$\varepsilon<R$, we have $\operatorname{cf}(\eta(x,s_{k}),t_{\infty})$ and
$\operatorname{cf}(\eta(x,s_{0}),t_{\infty})$ less than $R$ apart, and thus
there is a contradiction.

Case II. Suppose $\eta(x,s_{0})\in\mathcal{C}_{1}$. There is a sequence of
curves $\zeta(s_{k})\to\zeta(s_{0})$ in Fréchet distance as $s_{k}\to s_{0}$
from above such that $\eta(x,s_{k})\in\mathcal{C}_{3}$. Then,
$\operatorname{cf}(\eta(x,s_{k}),t_{\infty})$ and
$\operatorname{cf}(\eta(x,s_{0}),t_{\infty})$ are at least $R$ apart. Since,
by Lemma 2.2,
$\operatorname{cf}(\eta(x,s_{k}),t_{\infty})\to\operatorname{cf}(\eta(x,s_{0}),t_{\infty})$
in Hausdorff distance, $\operatorname{cf}(\eta(x,s_{k}),t_{\infty})$ and
$\operatorname{cf}(\eta(x,s_{0}),t_{\infty})$ are within a Hausdorff distance
of $\varepsilon$, for any $\varepsilon>0$ and all $k$ sufficiently large. For
$\varepsilon<R$, we have $\operatorname{cf}(\eta(x,s_{k}),t_{\infty})$ and
$\operatorname{cf}(\eta(x,s_{0}),t_{\infty})$ less than $R$ apart, and thus
there is a contradiction.

Therefore, class $\mathcal{C}_{2}$ is also non-empty. Now let us show that,
based on the value of $s$, these three classes correspond to three intervals
on $[0,1]_{{\mathbb{R}}}$; see Figure 4. We want to show that for any
$a\in[0,1]_{{\mathbb{R}}}$, if $\eta(x,a)\in\mathcal{C}_{1}$, then
$\eta(x,b)\in\mathcal{C}_{1}$ for any $0\leq b<a$. Let $\mathcal{A}$ be the
common arc of $\eta(x,a)$ and $\eta(x,b)$. Since
$\eta(x,b)\setminus\mathcal{A}\subset\mathcal{I}(\eta(x,a))$, by Lemma 3.1,
$\operatorname{cf}(\eta(x,b),t_{\infty})\subset\mathcal{I}(\eta(x,a),t_{\infty}))$.
Since $\eta(x,a)\in\mathcal{C}_{1}$,
$\operatorname{cf}(\gamma,t_{\infty})\subset\mathcal{O}(\operatorname{cf}(\eta(x,a),t_{\infty}))$.
Therefore,
$\operatorname{cf}(\gamma,t_{\infty})\subset\mathcal{O}(\operatorname{cf}(\eta(x,b),t_{\infty}))$,
and thus $\eta(x,b)\in\mathcal{C}_{1}$. Similarly, we can show that for any
$a\in[0,1]_{{\mathbb{R}}}$, if $\eta(x,a)\in\mathcal{C}_{3}$, then
$\eta(x,b)\in\mathcal{C}_{3}$ for any $a<b\leq 1$.

Let $\nu_{1}(x)=\inf\\{s\ |\ \eta(x,s)\in\mathcal{C}_{2},0\leq s\leq 1\\}$,
and $\nu_{2}(x)=\sup\\{s\ |\ \eta(x,s)\in\mathcal{C}_{2},0\leq s\leq 1\\}$. We
want to show that $\eta(x,\nu_{1}(x))\in\mathcal{C}_{2}$ for any $x\in\gamma$.
Suppose not. Then $\eta(x,\nu_{1}(x))\in\mathcal{C}_{1}$, and there exists
$\varepsilon_{1}>0$ such that
$(\operatorname{cf}(\eta(x,\nu_{1}(x)),t_{\infty})\oplus\varepsilon_{1})\cap\operatorname{cf}(\gamma,t_{\infty})=\emptyset$,
for $t_{\infty}\in(0,T)_{{\mathbb{R}}}$. There is a sequence of curves
$\eta(x,s_{k})\to\eta(x,\nu_{1}(x))$ in Fréchet distance as
$s_{k}\to\nu_{1}(x)$ from above such that $\eta(x,s_{k})\in\mathcal{C}_{2}$.
Let $t_{k}\to t_{\infty}$. By Lemma 2.2,
$\operatorname{cf}(\eta(x,s_{k}),t_{k})\to\operatorname{cf}(\eta(x,\nu_{1}(x)),t_{\infty})$
in Hausdorff distance, implying that
$\operatorname{cf}(\eta(x,s_{k}),t_{k})\subset\operatorname{cf}(\eta(x,\nu_{1}(x)),t_{\infty})\oplus\varepsilon_{1}$
for $k$ sufficiently large, and consequently
$\operatorname{cf}(\eta(x,s_{k}),t_{\infty})\cap\operatorname{cf}(\gamma,t_{\infty})=\emptyset$
for $k$ sufficiently large. This contradicts that
$\eta(x,s_{k})\in\mathcal{C}_{2}$.

Furthermore, $\operatorname{cf}(\eta(x,\nu_{1}(x)),t_{\infty})$ only
intersects $\operatorname{cf}(\gamma,t_{\infty})$ at one point. Suppose not.
Then $\operatorname{cf}(\eta(x,\nu_{1}(x)),t_{\infty})$ intersects
$\operatorname{cf}(\gamma,t_{\infty})$ at two points. There exists
$\varepsilon_{2}>0$ such that the boundary of
$(\operatorname{cf}(\eta(x,\nu_{1}(x)),t_{\infty})\oplus\varepsilon_{2})$
consists of two Jordan curves that both intersect
$\operatorname{cf}(\gamma,t_{\infty})$ at two points. There is a sequence of
curves $\eta(x,s_{k})\to\eta(x,\nu_{1}(x))$ in Fréchet distance as
$s_{k}\to\nu_{1}(x)$ from below such that $\eta(x,s_{k})\in\mathcal{C}_{1}$.
Let $t_{k}\to t_{\infty}$. By Lemma 2.2,
$\operatorname{cf}(\eta(x,s_{k}),t_{k})\to\operatorname{cf}(\eta(x,\nu_{1}(x)),t_{\infty})$
in Hausdorff distance, implying that
$\operatorname{cf}(\eta(x,s_{k}),t_{k})\subset\operatorname{cf}(\eta(x,\nu_{1}(x)),t_{\infty})\oplus\varepsilon_{2}$
for $k$ sufficiently large, and consequently
$\operatorname{cf}(\eta(x,s_{k}),t_{\infty})\cap\operatorname{cf}(\gamma,t_{\infty})\neq\emptyset$
for $k$ sufficiently large. This contradicts that
$\eta(x,s_{k})\in\mathcal{C}_{1}$. We can use analogous arguments to show that
$\eta(x,\nu_{2}(x))\in\mathcal{C}_{2}$ and
$\operatorname{cf}(\eta(x,\nu_{2}(x)),t_{\infty})$ only intersects
$\operatorname{cf}(\gamma,t_{\infty})$ at one point.

Let $\nu_{0}(x)=(\nu_{1}(x)+\nu_{2}(x))/2$. Note that
$\operatorname{cf}(\eta(x,\nu_{0}(x)),t_{\infty})\subset\mathcal{O}(\operatorname{cf}(\eta(x,\nu_{1}(x)),t_{\infty}))$
and
$\operatorname{cf}(\eta(x,\nu_{0}(x)),t_{\infty})\subset\mathcal{I}(\operatorname{cf}(\eta(x,\nu_{2}(x)),t_{\infty}))$.
Since $\operatorname{cf}(\eta(x,\nu_{1}(x)),t_{\infty})$ and
$\operatorname{cf}(\eta(x,\nu_{2}(x)),t_{\infty})$ are both tangent to
$\operatorname{cf}(\gamma,t_{\infty})$ with
$\operatorname{cf}(\eta(x,\nu_{1}(x)),t_{\infty})\subset\mathcal{I}(\operatorname{cf}(\eta(x,\nu_{2}(x)),t_{\infty}))$,
$\operatorname{cf}(\eta(x,\nu_{0}(x)),t_{\infty})$ intersects
$\operatorname{cf}(\gamma,t_{\infty})$ at at least $2$ points. Since Angenent
proved that the point of intersection does not increase throughout the
evolution by curvature flow [1], and $\eta(x,\nu_{0}(x))$ initially intersects
$\gamma$ at two points, $\operatorname{cf}(\eta(x,\nu_{0}(x)),t_{\infty})$
must intersect $\operatorname{cf}(\gamma,t_{\infty})$ at two points.

Figure 4: On the left, the black curve represents
$\operatorname{cf}(\gamma,t)$, and the red, blue, purple curves are curves in
$\mathcal{C}_{1},\mathcal{C}_{2},\mathcal{C}_{3}$, evolving by curvature flow
at time $t\in(0,T)_{{\mathbb{R}}}$. On the right are three intervals
corresponding to $\mathcal{C}_{1},\mathcal{C}_{2},\mathcal{C}_{3}$. The
interval corresponding to $\mathcal{C}_{2}$ is closed.

Now we want to show that $\nu_{1}$ is continuous. Suppose not. Let
$x_{\infty}\in\gamma$. Then there exists a sequence $x_{k}\to x_{\infty}$ such
that $\nu_{1}(x_{k})$ is bounded away $\nu_{1}(x_{\infty})$ by some
$\varepsilon>0$. By the compactness of the unit interval, we can restrict
$\\{\nu_{1}(x_{k})\\}$ to a subsequence $\\{\nu_{1}(x_{i})\\}$ that converges
to $p$.

Case I. Assume that $p<\nu_{1}(x_{\infty})$. Then there is
$s_{0}\in[0,1]_{{\mathbb{R}}}$ such that $p<s_{0}<\nu_{1}(x_{\infty})$ and
$\eta(x_{i},s_{0})\in\mathcal{C}_{2}$ for $i$ sufficiently large. Let
$t_{\infty}\in(0,T)_{{\mathbb{R}}}$. Since $s_{0}<\nu_{1}(x_{\infty})$,
$\eta(x_{\infty},s_{0})\in\mathcal{C}_{1}$, and then
$\operatorname{cf}(\eta(x_{\infty},s_{0}),t_{\infty})\cap\operatorname{cf}(\gamma,t_{\infty})=\emptyset$.
Therefore, there exists $\delta_{1}>0$ such that
$(\operatorname{cf}(\eta(x_{\infty},s_{0}),t_{\infty})\oplus\delta_{1})\cap\operatorname{cf}(\gamma,t_{\infty})=\emptyset$.
Let $t_{i}\to t_{\infty}$. Since $x_{i}\to x_{\infty}$,
$\eta(x_{i},s_{0})\to\eta(x_{\infty},s_{0})$ in Fréchet distance. By Lemma
2.2,
$\operatorname{cf}(\eta(x_{i},s_{0}),t_{i})\to\operatorname{cf}(\eta(x_{\infty},s_{0}),t_{\infty})$
in Hausdorff distance, implying that
$\operatorname{cf}(\eta(x_{i},s_{0}),t_{i})\subset\operatorname{cf}(\eta(x_{\infty},s_{0}),t_{\infty})\oplus\delta_{1}$
for $i$ sufficiently large, and consequently
$\operatorname{cf}(\eta(x_{i},s_{0}),t_{\infty})\cap\operatorname{cf}(\gamma,t_{\infty})=\emptyset$.
This contradicts that $\eta(x_{i},s_{0})\in\mathcal{C}_{2}$.

Case II Assume that $p>\nu_{1}(x_{\infty})$. Then there is
$s_{1}\in[0,1]_{{\mathbb{R}}}$ such that
$\nu_{1}(x_{\infty})<s_{1}<\nu_{1}(x_{i})$ and
$\operatorname{cf}(\eta(x_{\infty},s_{1}),t_{\infty})$ intersects
$\operatorname{cf}(\gamma,t_{\infty})$ at two points. Therefore, there exists
$\delta_{2}>0$ such that the boundary of
$\operatorname{cf}(\eta(x_{\infty},s_{0}),t_{\infty})\oplus\delta_{2}$
consists of two Jordan curves that both intersect
$\operatorname{cf}(\gamma,t_{\infty})$ at two points. Since $\nu_{1}(x_{i})\to
p$ and $s_{1}<p$, $\eta(x_{i},s_{1})\in\mathcal{C}_{1}$ for $i$ sufficiently
large. Let $t_{i}\to t_{\infty}$. Since $x_{i}\to x_{\infty}$,
$\eta(x_{i},s_{1})\to\eta(x_{\infty},s_{1})$ in Fréchet distance. By Lemma
2.2,
$\operatorname{cf}(\eta(x_{i},s_{1}),t_{i})\to\operatorname{cf}(\eta(x_{\infty},s_{1}),t_{\infty})$
in Hausdorff distance, implying that
$\operatorname{cf}(\eta(x_{i},s_{1}),t_{i})\subset\operatorname{cf}(\eta(x_{\infty},s_{1}),t_{\infty})\oplus\delta_{2}$
for $i$ sufficiently large, and consequently
$\operatorname{cf}(\eta(x_{i},s_{1}),t_{\infty})$ intersects
$\operatorname{cf}(\gamma,t_{\infty})$ at two points. This contradicts that
$\eta(x_{i},s_{1})\in\mathcal{C}_{1}$.

Therefore, $\nu_{1}$ is continuous. Similarly, we can show that $\nu_{2}$ is
also continuous. For each $x\in\gamma$, since $\nu_{1},\nu_{2}$ are
continuous, and $\gamma$ is compact, by Heine-Cantor theorem, $\nu_{1}$ and
$\nu_{2}$ are uniformly continuous. Therefore, $\eta(x,\nu_{0}(x))$ is
continuous in Fréchet distance. Therefore,
$\operatorname{cf}(\eta(x,\nu_{0}(x)),t_{\infty})$ is the curve that has the
desired properties. ∎

######  Lemma 3.3.

Let $\gamma\in\mathfrak{J}_{T}$, and
$z\in\operatorname{cf}(\gamma,t_{\infty})$ for $t_{\infty}\in(0,T]$. Then
there is a Jordan curve that intersects $\gamma$ at exactly $2$ points, and
evolves to pass through $z$ at time $t_{\infty}$, i.e.
$z\in\operatorname{cf}(\eta,t_{\infty})$.

###### Proof.

For any $x\in\gamma$, let $\eta(x)=\eta(x,\nu_{0}(x))$ be the Jordan curved
defined in Lemma 3.2. Let
$p(\gamma,\eta,x,t)\in\operatorname{cf}(\gamma,t)\cap\operatorname{cf}(\eta(x),t)$
be the trajectory starting from $p(\gamma,\eta,x,0)=x$ as in Lemma 2.4. Since
$\eta$ is continuous in Fréchet distance, by Lemma 2.4, $p$ is continuous. The
map $p$ defines a homotopy from the identity map on $\gamma$ to the map
$p(\gamma,\eta,\cdot,t_{\infty}):\gamma\to\operatorname{cf}(\gamma,t_{\infty})$,
so $p(\gamma,\eta,\cdot,t_{\infty})$ winds once around
$\operatorname{cf}(\gamma,t_{\infty})$, which implies there is some
$x\in\gamma$ such that $p(\gamma,\eta,x,t_{\infty})=z$. Therefore, $\eta(x)$
has the desired property. ∎

###### Proof of Theorem 1.1.

This proof is analogous to Michael Dobbins’s proof of Theorem 1.1 in [2]. For
the reader’s convenience, the proof is shown here.

Let $\gamma_{k}$ satisfy the hypothesis of the theorem, and assume for the
sake of contradiction that
$\operatorname{cf}(\gamma_{k},t_{k})\not\to\operatorname{cf}(\gamma_{\infty},t_{\infty})$
in Fréchet distance. By Theorem 1.2 (1) and Corollary 1.6 of Lauer [5],
$\operatorname{cf}(\gamma_{k},t_{k})$ is smooth and has length bounded by some
constant for all $k$ sufficiently large, provided that $t_{\infty}>0$.
Therefore, the sequence of constant speed parameterizations
$\psi_{k}:{\mathbf{S}}^{1}\to\operatorname{cf}(\gamma_{k},t_{k})$ is uniformly
equicontinuous, so by the Arzelà-Ascoli theorem, we can restrict to a sequence
that converges uniformly to a map $\psi_{\infty}$. Moreover, by Lemma 2.2, the
range of the limit $\psi_{\infty}$ is
$\operatorname{cf}(\gamma_{\infty},t_{\infty})$. We can also let
$\omega:\mathbf{S}^{1}\to\operatorname{cf}(\gamma_{\infty},t_{\infty})$ be a
constant speed parameterization. Let
$\varphi:[0,2\pi]_{{\mathbb{R}}}\to{\mathbb{R}}$ by
$\varphi(\theta)=-ilog(\omega^{-1}\circ\varphi_{\infty}(e^{i\theta}))$, which
is just $\omega^{-1}\circ\varphi_{\infty}$ lifted by the standard
parameterization of the circle by angle. Since the map
$(\theta\mapsto\omega^{-1}\circ\varphi_{\infty}(e^{i\theta}))$ is periodic,
$\varphi(2\pi)-\varphi(0)$ is a multiple of $2\pi$. We may choose the
direction of $\omega$ so that $\varphi(2\pi)-\varphi(0)\geq 0$. If we had
$\varphi(2\pi)-\varphi(0)=0$, then the region on one side of
$\operatorname{cf}(\gamma_{k},t_{k})$ would converge to a subset of
$\operatorname{cf}(\gamma_{\infty},t_{\infty})$, but $\gamma_{k}$ continues to
be a Jordan curve as it evolves, so that cannot happen. Hence,
$\varphi(2\pi)-\varphi(0)\neq 0$. If we had $\varphi(2\pi)-\varphi(0)>2\pi$,
then $\operatorname{cf}(\gamma_{k},t_{k})$ would wind more than once around a
tubular neighborhood of $\operatorname{cf}(\gamma_{\infty},t_{\infty})$, which
is impossible for a simple closed curve, so $\varphi(2\pi)-\varphi(0)=2\pi$.

If $\varphi$ were weakly increasing, then $\varphi$ would be the limit of some
sequence of strictly increasing functions $\varphi_{k}\to\varphi$, and we
would have homeomorphisms
$\tau_{k}:\mathbf{S}^{1}\to\operatorname{cf}(\gamma_{\infty},t_{\infty})$
given by $\tau_{k}(x)=\omega(e^{i\varphi_{k}(-ilog(x))}))$ that converge to
$\psi_{\infty}$, but then the Fréchet distance between
$\operatorname{cf}(\gamma_{k},t_{k})$ and
$\operatorname{cf}(\gamma_{\infty},t_{\infty})$ would be bounded by
$\sup_{x\in\mathbf{S}^{1}}||\psi_{k}(x)-\tau_{k}(x)||\to 0$, which contradicts
our assumption that
$\operatorname{cf}(\gamma_{k},t_{k})\not\to\operatorname{cf}(\gamma_{\infty},t_{\infty})$
in Fréchet distance. Hence, $\varphi$ must decrease somewhere, i.e. there is
$w<y$ such that $\varphi(y)<\varphi(w)$, and we may choose the period of
$\varphi$ so that $\varphi(w)$ is between $\varphi(0)$ and
$\varphi(2\pi)=\varphi(0)+2\pi$, and may choose $w,y$ arbitrarily close
together. Then, there are $v<w<x<y<z$ where
$\varphi(v)=\varphi(y)<\varphi(x)<\varphi(w)=\varphi(z)$.

Let
$v_{k}=\psi_{k}(e^{iv}),w_{k}=\psi_{k}(e^{iw}),x_{k}=\psi_{k}(e^{ix}),y_{k}=\psi_{k}(e^{iy}),z_{k}=\psi_{k}(e^{iz}).$
The situation so far is that as we traverse $cf(\gamma_{k},t_{k})$, we pass
through $v_{k},w_{k},x_{k},y_{k},z_{k}$ in that order, and these points
respectively converge to
$y_{\infty},w_{\infty},x_{\infty},y_{\infty},w_{\infty}$; see Figure 5. By
Lemma 3.3, there is a Jordan curve $\eta$ that intersects $\gamma$ at exactly
$2$ points and such that $\operatorname{cf}(\eta,t_{\infty})$ intersects
$\operatorname{cf}(\gamma_{\infty},t_{\infty})$ at $x_{\infty}$. Also, we can
choose $w,y$ sufficiently close together so that the $2$ points
$\operatorname{cf}(\eta,t_{\infty})\cap\operatorname{cf}(\gamma_{\infty},t_{\infty})$
are not on the same arc from $w_{\infty}$ to $y_{\infty}$. That is,
$w_{\infty}$ and $y_{\infty}$ are on opposite sides of
$\operatorname{cf}(\eta,t_{\infty})$.

Since level-set flow gives a solution to the curvature flow problem, we can
find a nested sequence of smooth annuli $\\{A_{i}(0)\\}_{i=1}^{\infty}$ such
that $A_{1}(0)\supset A_{2}(0)\supset\cdot\cdot\cdot$ such that
$\operatorname{cf}(\eta,t)=\bigcap_{i=1}^{\infty}Ai(t)$ where the boundary of
$A_{i}(t)$ evolves by curvature flow starting $A_{i}(0)$. Choose one of the
annuli $A=A_{i}$ such that $A(t)$ remains an annulus up to time
$s\in(t_{\infty},T)$ and close enough to $\eta$ that $A(t_{\infty})$ does not
contain $w_{\infty}$ or $y_{\infty}$; see Figure 5. Let $\alpha$ and $\beta$
be the curves on the boundary of $A(0)$. Note that
$\operatorname{cf}(\alpha,t)$ and $\operatorname{cf}(\beta,t)$ are on either
side of $\operatorname{cf}(\eta,t)$ for all $t\in[0,s]_{{\mathbb{R}}}$. Hence,
each arc of $\gamma_{\infty}$ subdivided by $\alpha\cup\beta$ that goes from a
point on $\alpha$ to a point on $\beta$ must cross $\eta$, and since
$\gamma_{\infty}$ meets $\eta$ at only $2$ points, there can be at most $2$
such arcs.

Since $\alpha,\eta,\beta$ are compact and pairwise disjoint, we may choose
$\varepsilon_{1}>0$ small enough that $\alpha\oplus\varepsilon_{1}$,
$\zeta\oplus\varepsilon_{1}$, $\beta\oplus\varepsilon_{1}$ are pairwise
disjoint. Let us choose $k$ sufficiently large that $\gamma_{k}$ is at most
Fréchet distance $\varepsilon_{1}$ from $\gamma_{\infty}$, and $v_{k}$,
$w_{k}$, $y_{k}$, and $z_{k}$ are each outside of $A(t_{k})$. To see that the
later condition can be satisfied, recall that these points each approach
either $w_{\infty}$ or $y_{\infty}$, which are bounded away from
$A(t_{\infty})$, and $A(t_{k})\to A(t_{\infty})$ in Hausdorff distance.

Consider an arc $\xi$ of $\gamma_{k}\cap A(0)$ from $\alpha$ to $\beta$. Since
$\gamma_{k}$ is within Fréchet distance $\varepsilon_{1}$ from
$\gamma_{\infty}$ there is some map $\psi:\xi\to\gamma_{\infty}$ such that
$||\psi(x)-x_{k}||\leq\varepsilon_{1}$ for all $x\in\xi$, so $\psi(\xi)$ must
be an arc of $\gamma_{\infty}$ from $\alpha\oplus\varepsilon_{1}$ to
$\beta\oplus\varepsilon_{1}$. Hence, $\psi(\xi)$ must intersect $\zeta$, and
since $\gamma_{\infty}$ only intersects $\zeta$ at $2$ points, there are only
$2$ arcs of $\gamma_{k}\cap A(0)$ from one boundary of $A(0)$ to the other.
Therefore, we can find a curve $\eta$ of area $0$ that winds once around the
interior of $A(0)$ and intersects $\gamma_{k}$ at only $2$ points; see Figure
5.

Figure 5:  Assuming
$\operatorname{cf}(\gamma_{k},t_{k})\to\operatorname{cf}(\gamma_{\infty},t_{\infty})$
in Hausdorff distance but not in Fréchet distance, $\gamma_{k}$ evolves to
pass close by a point $x_{\infty}$three times, and the curve $\zeta$ only
intersects $\gamma_{\infty}$ twice and evolves to pass through $x_{\infty}$.
The curve $\eta$ crosses $\gamma_{k}$ only twice, but evolves to cross
$\operatorname{cf}(\gamma_{k},t_{k})$ more than twice, contradicting
Angenent’s theorem.

By Angenent’s theorem, $\operatorname{cf}(\eta,t)$ remains disjoint from
$\operatorname{cf}(\alpha,t)$ and $\operatorname{cf}(\beta,t)$, and crosses
$\operatorname{cf}(\gamma_{k},t)$ at most twice, for as long as a solution to
curvature flow exists [1], and by Lauer’s theorem, the evolving curves
$\operatorname{cf}(\alpha,t),\operatorname{cf}(\beta,t)$ on either side of
$\eta$ ensures that a unique solution exists for $\operatorname{cf}(\eta,t)$
up to time $s$ [5].

Since $\operatorname{cf}(\gamma_{k},t_{k})$ passes though
$v_{k},w_{k},y_{k},z_{k}$ in that order, and $v_{k}$ and $y_{k}$ are on one
side of $A(t_{k})$, and $w_{k}$ and $z_{k}$ are on the other side of
$A(t_{k})$, $\operatorname{cf}(\gamma_{k},t_{k})$ must cross from one side of
$A(t_{k})$ to the other at least $4$ times. Since $\operatorname{cf}(\eta,t)$
never intersects the boundary of $A(t)$, $\operatorname{cf}(\eta,t_{k})$ winds
once around $A(t_{k})$. Therefore, $\operatorname{cf}(\eta,t_{k})$ intersects
$\operatorname{cf}(\gamma_{k},tk)$ at least at $4$ points, but $\eta$
intersects $\gamma_{k}$ at only $2$ points, which contradicts Angenent’s
theorem that the number of intersection points does not increase; see Figure
5. Thus, our assumption that $\gamma_{k}\not\to\gamma_{\infty}$ in Fréchet
distance cannot hold. ∎

##  4 Conclusions

We showed that the evolution of a Jordan curve on the plane by curvature flow
depends continuously on the initial curve in Fréchet distance. The theoretical
results we obtain in this work helps to connect this geometric evolution to
the broader framework of functional analysis, providing a deeper understanding
of the underlying mathematical structures. The proof of Theorem 1.1 relies on
Joseph Lauer’s theorem that a nonsmooth curve in ${\mathbb{R}}^{2}$ instantly
becomes smooth and has length bounded by some constant [5]. Therefore, it
remains an open question whether our results can be extended to Jordan curves
on other surfaces. Another promising research direction can be the study of
the continuity of surfaces in ${\mathbb{R}}^{3}$, where the challenge in
establishing continuity arises from the potential formation of singularities,
such as the neckpinch singularity.

## References

  * [1] Sigurd Angenent Parabolic Equations for Curves on Surfaces: Part II. Intersections, Blow-up and Generalized Solutions. Annals of Mathematics 133, no. 1 (1991): 171–215.
  * [2] Michael Gene Dobbins. Continuous Dependence of Curvature Flow on Initial Conditions. arXiv:2106.08907, 2021.
  * [3] Matthew A. Grayson. The heat equation shrinks embedded plane curves to round points. J. Differential Geom. 26(2): 285-314 (1987).
  * [4] M. Gage & R.S. Hamilton The Heat Equation Shrinking Convex Plane Curves. J. Differential Geom. 23(1): 69-96 (1986).
  * [5] Joseph Lauer. A New Length Estimate for Curve Shortening Flow and Low Regularity Initial Data. Geom. Funct. Anal. 23, 1934–1961 (2013).