# On a Dowker-type problem for convex disks with almost constant curvature

Bushra Basit  and  Zsolt Lángi  Bushra Basit, Department of Algebra and
Geometry, Budapest University of Technology and Economics
Műegyetem rkp. 3., H-1111 Budapest, Hungary<EMAIL_ADDRESS>Zsolt
Lángi, Department of Algebra and Geometry, Budapest University of Technology
and Economics, and MTA-BME Morphodynamics Research Group
Műegyetem rkp. 3., H-1111 Budapest, Hungary<EMAIL_ADDRESS>
###### Abstract.

A classical result of Dowker (Bull. Amer. Math. Soc. 50: 120-122, 1944) states
that for any plane convex body $K$, the areas of the maximum (resp. minimum)
area convex $n$-gons inscribed (resp. circumscribed) in $K$ is a concave
(resp. convex) sequence. It is known that this theorem remains true if we
replace area by perimeter, or convex $n$-gons by disk-$n$-gons, obtained as
the intersection of $n$ closed Euclidean unit disks. It has been proved
recently that if $C$ is the unit disk of a normed plane, then the same
properties hold for the area of $C$-$n$-gons circumscribed about a $C$-convex
disk $K$ and for the perimeters of $C$-$n$-gons inscribed or circumscribed
about a $C$-convex disk $K$, but for a typical origin-symmetric convex disk
$C$ with respect to Hausdorff distance, there is a $C$-convex disk $K$ such
that the sequence of the areas of the maximum area $C$-$n$-gons inscribed in
$K$ is not concave. The aim of this paper is to investigate this question if
we replace the topology induced by Hausdorff distance with a topology induced
by the surface area measure of the boundary of $C$.

###### Key words and phrases:

Dowker’s theorems, inscribed polygon, area, Hausdorff distance

###### 2020 Mathematics Subject Classification:

52A40, 52A30, 52A27

Partially supported by the National Research, Development and Innovation
Office, NKFI, K-147544 grant

##  1\. Introduction

In the paper, by a convex disk we mean a compact, convex set with nonempty
interior in the Euclidean plane $\mathbb{R}^{2}$, and denote the closed,
Euclidean unit disk by $B^{2}$. Furthermore, we denote by $\mathcal{K}$ and
$\mathcal{K}_{o}$ the families of convex disks and origin-symmetric convex
disks, respectively. The families of the elements of $\mathcal{K}$ and
$\mathcal{K}_{o}$ with $C^{2}$-class boundary and strictly positive curvature
are denoted by $\mathcal{K}_{2+}$ and $\mathcal{K}_{o,2+}$, respectively. We
denote the origin by $o$.

For any integer $n\geq 3$ and plane convex body $K$, let $A_{n}(K)$ (resp.
$a_{n}(K)$) denote the the infimum (resp. supremum) of the areas of the convex
$n$-gons circumscribed about (resp. inscribed in) $K$. A famous result of
Dowker [5] states that for any convex disk $K$, the sequences $\\{A_{n}(K)\\}$
and $\\{a_{n}(K)\\}$ are convex and concave, respectively. It was proved
independently by L. Fejes Tóth [10], Molnár [15] and Eggleston [6] that the
same statements remain true if we replace area by perimeter, where the last
author also showed that these statements are false if we replace area by
Hausdorff distance. These results are known to be true also in any normed
plane [13]. Dowker’s theorems and their variants have became important in many
areas of discrete geometry, in particular in the theory of packing and
covering [9, 1, 11] and are often used even today (see e.g. [16, 3]).

We mention only one variant of these theorems. To introduce it, if $C$ is an
origin-symmetric convex disk in the Euclidean plane $\mathbb{R}^{2}$, we call
a set $X$ _$C$ -spindle convex_, or shortly _$C$ -convex_, if it is
$\mathbb{R}^{2}$, or there is a translate of $C$ with $X\subseteq C$ and for
any $p,q\in X$, $X$ contains the intersection of all translates of $C$
containing $\\{p,q\\}$. This concept was introduced by Mayer [14] in 1935, and
in modern times its systematic investigation, for the special case that $C$ is
the Euclidean closed unit disk $B^{2}$ centered at the origin $o$, began in
[4], and the general case in [12]. A result related to spindle convex sets is
due to G. Fejes Tóth and Fodor [8] who extended Dowker’s theorems, together
with their variants for perimeter, for spindle convex sets; in these theorems
the role of inscribed or circumscribed convex $n$-gons is played by the so-
called _disk- $n$-gons_, obtained as the intersections of $n$ closed Euclidean
unit disks. They also proved similar theorems for spindle convex sets in
hyperbolic or spherical plane.

These results were generalized in [2], where the authors prove that for any
$o$-symmetric convex disk $C$, the same statements hold for the areas of
inscribed $C$-$n$-gons, or the $C$-perimeters of inscribed or circumscribed
$C$-$n$-gons in any $C$-convex disk $K$, where a $C$-$n$-gon is defined to be
the intersection of $n$ translates of $C$. In addition, they show that for a
typical $C\in\mathcal{K}_{o}$, with respect to the topology induced by
Hausdorff distance, there is a $C$-convex disk $K$ such that the sequence of
the areas of the maximum area $C$-$n$-gons inscribed in $K$ is not concave. In
the following we say that $C\in\mathcal{K}_{o}$ satisfies the _Dowker
Property_ if the above sequence is concave for any $C$-convex disk $K$.

We remark that the proof of almost all Dowker-type theorems mentioned in the
introduction are based on different versions of the same inequality for convex
(resp. $C$-convex) quadrangles; we are going to define this inequality for the
areas of inscribed $C$-$n$-gons in Section 2, and say that an $o$-symmetric
convex disk satisfying this inequality for any $C$-convex quadrangle satisfies
the _Quadrangle Property_ , we note that any $C\in\mathcal{K}_{o}$ satisfying
the Quadrangle Property also satisfies the Dowker Property (for a more precise
and detailed discussion, see Section 2).

Our first result is a new proof of the theorem in [8] stating that the
Euclidean unit disk $B^{2}$ satisfies the Dowker Property. We prove Theorem 1
by showing that $B^{2}$ satisfies the Quadrangle Property. To state it, for
any $n\geq 3$, $C\in\mathcal{K}_{o}^{H}$, and $C$-convex disk $K$, we set
$\hat{a}_{n}^{C}(K)=\inf\\{\operatorname{area}(Q):Q=\hbox{ is a {C}-{n}-gon
inscribed in }K\\}$.

######  Theorem 1.

For any $n\geq 4$ and spindle convex disk $K$, we have
$\hat{a}_{n-1}^{B^{2}}(K)+\hat{a}_{n+1}^{B^{2}}(K)\leq
2\hat{a}_{n}^{B^{2}}(K)$.

Note that by the results in [2], there is no ‘stability’ version of the result
of G. Fejes Tóth and Fodor [8] about the areas of disk-$n$-gons inscribed in a
spindle convex disk $K$. In other words, if $\mathcal{K}_{o}^{H}$ denotes the
topological space induced by Hausdorff distance on $\mathcal{K}_{o}$, $B^{2}$
has no neighborhood $U$ in $\mathcal{K}_{o}^{H}$ such that any $C\in U$
satisfies the Dowker Property. On the other hand, one may observe that if the
boundary of some $C\in\mathcal{K}_{o}^{H}$ has ‘almost unit curvature’ at
every point, then $C$ is ‘close’ to $B^{2}$ with respect to Hausdorff
distance, but the converse does not hold even if $C$ is assumed to have
$C^{2}$-class boundary with strictly positive curvature. Thus, it is a natural
question to ask if any $C\in\mathcal{K}_{o}^{H}$ with almost constant
curvature satisfies the Dowker Property or not.

To investigate this question, we introduce a new topology on the family
$\mathcal{K}_{o}$. This topology is induced by a generalized distance function
$d_{PM}:\mathcal{K}_{o}\times\mathcal{K}_{o}\to\mathbb{R}\cup\\{\infty\\}$,
where $d_{PM}(C_{1},C_{2})$ is defined by the surface area measures of $C_{1}$
and $C_{2}$. We call this generalized distance _perimeter measure distance_ or
_PM-distance_ , and denote the topological space induced by it on
$\mathcal{K}_{o}$ by $\mathcal{K}_{o}^{PM}$. We show that
$\mathcal{K}_{o}^{PM}$ is a refinement of $\mathcal{K}_{o}^{H}$. We observe
also that if $\mathcal{K}_{o,2+}^{PM}$ denotes the subfamily of
$\mathcal{K}_{o}^{PM}$ consisting of the elements having $C^{2}$-class
boundary and strictly positive curvature, then the property that
$C_{1}\in\mathcal{K}_{o,2+}^{PM}$ is ‘close’ to
$C_{2}\in\mathcal{K}_{o,2+}^{PM}$ is equivalent to the property that the
curvatures of $\operatorname{bd}(C_{1})$ and $\operatorname{bd}(C_{2})$ at
points with the same outer normal vector are ‘almost equal’.

Our second result is the following.

######  Theorem 2.

A typical element $C\in\mathcal{K}_{o,2+}^{PM}$ does not satisfy the
Quadrangle Property.

We ask the following.

######  Problem 1.

Prove or disprove that a typical element $C\in\mathcal{K}_{o}^{PM}$ does not
satisfy the Dowker Property. What happens if we replace $\mathcal{K}_{o}^{PM}$
by $\mathcal{K}_{o,2+}^{PM}$, or the Dowker Property by the Quadrangle
Property?

The structure of the paper is as follows. In Section 2, we give precise
definitions of the concepts used in the introduction, investigate their
properties, and prove some lemmas necessary to prove our theorems. Finally, in
Sections 3 and 4, we prove Theorems 1 and 2, respectively.

In the paper, for any $p,q\in\mathbb{R}^{2}$, we denote by $[p,q]$ the closed
segment with endpoints $p,q$, by $||p-q||$ the Euclidean distance of $p$ and
$q$, and by $|p,q|$ the value of the $2\times 2$ determinant whose columns are
$p$ and $q$.

##  2\. Preliminaries

We start with a formal definition of a $C$-spindle.

######  Definition 1.

Let $C\in\mathcal{K}_{o}$, and consider two (not necessarily distinct) points
$p,q\in\mathbb{R}^{2}$ such that a translate of $C$ contains both $p$ and $q$.
Then the _$C$ -spindle_ of $p$ and $q$, denoted as $[p,q]_{C}$, is the
intersection of all translates of $C$ that contain $p$ and $q$. If no
translate of $C$ contains $p$ and $q$, we set $[p,q]_{C}=\mathbb{R}^{2}$. We
call a set $K\in\mathcal{K}$ _$C$ -spindle convex_, or shortly _$C$ -convex_,
if for any $p,q\in K$, we have $[p,q]_{C}\subset K$.

We recall from [12, Corollary 3.13] that a closed set in $\mathbb{R}^{2}$,
different from $\mathbb{R}^{2}$, is $C$-convex if and only if it is the
intersection of some translates of $C$. Furthermore, we recall that for any
set $X\subseteq\mathbb{R}^{2}$, the _$C$ -convex hull_, or shortly _$C$ -hull_
is the intersection of all $C$-convex sets that contain $X$. We denote it by
$\operatorname{conv}_{C}(X)$, and note that it is $C$-convex, and if $X$ is
compact, then it coincides with the intersection of all translates of $C$
containing $X$ [12].

######  Definition 2.

The intersection of $n$ translates of $C$ is called a $C$-$n$-gon.

Note that any $C$-$n$-gon is the $C$-hull of at most $n$ points, and vice
versa [2].

Let $K$ denote a $C$-convex disk, and for any distinct points
$p,q\in\operatorname{bd}(K)$, let $\widehat{pq}$ denote the connected arc of
$\operatorname{bd}(K)$ such that running along it from $p$ to $q$ we move in
counterclockwise direction in $\operatorname{bd}(K)$. Furthermore, let
$r_{K}^{C}(p,q)$ denote the region bounded by $\widehat{pq}$ and the $C$-arc
in the boundary of $[p,q]_{C}$ that runs from $p$ to $q$ in counterclockwise
direction; in this notation we may omit the symbol $C$ if it clear which
$o$-symmetric convex disk it is referred to.

For any $p\in\operatorname{bd}(K)$, a point $q\in\operatorname{bd}(K)$ is
_antipodal_ to $p$ if $K$ has parallel supporting lines at $p$ and $q$. If the
arc $\widehat{pq}\subset\operatorname{bd}(K)$ contains no point antipodal to
$p$, we say that _$\widehat{pq}$ has turning angle strictly less than $\pi$_.

######  Definition 3.

Let $C\in\mathcal{K}_{o}$. Assume that for any $C$-convex disk $K$, and for
any boundary points $x_{1},x_{2},x_{3},x_{4}$ of $K$ in this counterclockwise
order, if $\widehat{x_{1}x_{4}}$ has turning angle strictly less than $\pi$,
then

$\operatorname{area}(r_{K}(x_{1},x_{4}))+\operatorname{area}(r_{K}(x_{2},x_{3}))\geq\operatorname{area}(r_{K}(x_{1},x_{3}))+\operatorname{area}(r_{K}(x_{2},x_{4})).$

Then we say that $C$ satisfies the _Quadrangle Property_ (see Figure 1).

Figure 1. An illustration for the Quadrangle Property. According to it, the
area of the dashed region is not less than the area of the dotted region.

######  Definition 4.

Let $C\in\mathcal{K}_{o}$. Assume that for any $C$-convex disk $K$ and $n\geq
4$, we have

$\hat{a}_{n-1}^{C}(K)+\hat{a}_{n+1}^{C}(K)\leq 2\hat{a}_{n}^{C}(K).$

Then we say that $C$ satisfies the _Dowker Property_.

Our next remark follows by a straightforward modification of [2, Lemma 4]. The
same statement for inscribed convex polygons or disk-polygons is the main idea
of the proofs in [5] and [8], respectively.

######  Remark 1.

If $C\in\mathcal{K}_{o}$ satisfies the Quadrangle Property, then it satisfies
the Dowker Property.

Our next lemma can be proved by a straightforward modification of the proof of
[2, Lemma 5].

######  Lemma 1.

Let $C\in\mathcal{K}_{o}$, and let $K$ be a $C$-convex disk. Let $\Gamma(s)$
be the arclength-parametrization of the boundary of $K$. Then, for any
$x_{i}=\Gamma(s_{i})$ with $i=1,2,3,4$ in this counterclockwise order where
$\widehat{x_{1}x_{4}}$ has turning angle strictly less than $\pi$, we have

$\operatorname{area}(r_{K}(x_{1},x_{4}))+\operatorname{area}(r_{K}(x_{2},x_{3}))\geq\operatorname{area}(r_{K}(x_{1},x_{3}))+\operatorname{area}(r_{K}(x_{2},x_{4}))$

if and only if
$\left.\partial_{s}\partial_{t}\operatorname{area}(r_{K}(\Gamma(s),\Gamma(t)))\right|_{(s,t)=(\bar{s},\bar{t})}\geq
0$ for all $\Gamma(\bar{s})=p\neq\Gamma(\bar{t})=q$ where $\widehat{pq}$ has
turning angle strictly less than $\pi$.

In the remaining part of this section we introduce a new metric on
$\mathcal{K}_{o}$ and examine its properties. For any convex disk $K$, and for
any $X\in\mathbb{S}^{1}$, let $N_{K}(X)$ denote the set of the boundary points
$p$ of $X$ with the property that an element of $X$ is an outer unit normal
vector of $K$ at $p$. Then, for any Borel set $\sigma\subseteq\mathbb{S}^{1}$,
the _surface area measure_ of $\sigma$, denoted by $A_{K}(\sigma)$ is defined
as the Hausdorff measure of $N_{K}(\sigma)$. We note that if $K$ has a
$C^{2}$-class boundary with positive curvature, then for any Borel set
$\sigma\subseteq\mathbb{S}^{1}$, $A_{K}(\sigma)$ is the integral of the radius
of curvature of $K$ at $p(u)$ over $u\in\sigma$, where $p(u)$ denotes the
unique point of $\operatorname{bd}(K)$ with outer unit normal vector $u$ [17].

For any Borel set $\sigma\subset\mathbb{S}^{1}$, let $\mathcal{H}(\sigma)$
denote the Hausdorff measure of $\sigma$.

######  Definition 5.

Let $C,D$ be $o$-symmetric convex bodies. We define the _perimeter measure
distance_ or _PM-distance_ of $C$ and $D$ as

(1) $d_{PM}(C,D)=\sup\left\\{\tau\geq
0:|A_{C}(\sigma)-A_{D}(\sigma)|\leq\tau\cdot\mathcal{H}(\sigma)\hbox{ for any
Borel-set }\sigma\subseteq\mathbb{S}^{1}\right\\}$

if the supremum exists, and otherwise we set $d_{PM}(C,D)=\infty$.

As we have remarked in the introduction, the function $d_{PM}(\cdot,\cdot)$
induces a topology on $\mathcal{K}_{o}$, and the induced topological space is
denoted by $\mathcal{K}_{o}^{PM}$. It is worth noting that this topology can
be naturally extended to the family $\mathcal{K}$, or to families of convex
bodies in $d$-dimensional Euclidean space for any value of $d$.

######  Proposition 1.

Let $\\{C_{n}\\}$ be a sequence of convex disks in $\mathcal{K}_{o}$ and let
$C\in\mathcal{K}_{o}$. If $C_{n}\to C$ with respect to the topology induced by
$d_{PM}(\cdot,\cdot)$, then $C_{n}\to C$ with respect to the topology induced
by $d_{H}(\cdot,\cdot)$. Furthermore, there is a sequence $\\{D_{n}\\}$ of
convex disks in $\mathcal{K}_{o}$ and a convex disk $D\in\mathcal{K}_{o}$ such
that $D_{n}\to D$ with respect to $d_{H}(\cdot,\cdot)$, but $D_{n}\not\to D$
with respect to $d_{PM}(\cdot,\cdot)$.

###### Proof.

Assume that $C_{n}\to C$ with respect to $d_{PM}(\cdot,\cdot)$, and let
$\varepsilon>0$ be a sufficiently small fixed value. Then, if $n$ is
sufficiently large, for any Borel set $\sigma\in\mathbb{S}^{1}$, we have

$|A_{C_{n}}(\sigma)-A_{C}(\sigma)|\leq\varepsilon\cdot\mathcal{H}(\sigma)\leq
2\pi\varepsilon.$

In particular, we obtain that if $n$ is sufficiently large, then
$|\operatorname{perim}(C_{n})-\operatorname{perim}(C)|\leq 2\pi\varepsilon$,
implying that for some $0<r<R$, each $C_{n}$ and $C$ contains $rB^{2}$ and is
contained in $RB^{2}$. Thus, by [17, Theorem 8.5.1], we have that if
$\varepsilon$ is sufficiently small, then, for some translate $C_{n}^{\prime}$
of $C_{n}$, we have $d_{H}(C_{n}^{\prime},C)\leq\gamma\varepsilon^{1/2}$ for
all sufficiently large values of $n$, where $\gamma$ depends only on $r$ and
$R$. On the other hand, since $C_{n}$ and $C$ are $o$-symmetric, this implies
$d_{H}(C_{n},C)\leq\gamma\varepsilon^{1/2}$, and hence, $C_{n}\to C$ with
respect to Hausdorff distance.

Now, let $D=B^{2}$, and $D_{n}=\operatorname{conv}(K\cup\\{p_{n},-p_{n}\\})$,
where $p_{n}=\left(1+\frac{1}{n},0\right)$. Then, clearly, $D_{n}\to D$ with
respect to Hausdorff distance. On the other hand, setting
$\sigma=\\{(0,1)\\}$, we have that $A_{D_{n}}(\sigma)>0$, and
$A_{D}(\sigma)=\mathcal{H}(\sigma)=0$, showing that $d_{PM}(D_{n},D)=\infty$
for all values of $n$. ∎

######  Corollary 1.

The topology on $\mathcal{K}_{o}$ induced by $d_{PM}(\cdot,\cdot)$ is a
refinement of the topology induced by $d_{H}(\cdot,\cdot)$.

Based on the proof of Proposition 1, we remark that if $D_{n}\to D$ with
respect to $d_{PM}(\cdot,\cdot)$, then we have that for sufficiently large
values of $D$, the family of the outer unit normal vectors of the smooth
points of $D_{n}$ coincides with the same set for $D$. Furthermore, we note
that Proposition 1 indicates that a set in $\mathcal{K}_{o}$ which is residual
with respect to $d_{H}(\cdot,\cdot)$ is not necessarily residual in the
topology induced by $d_{PM}(\cdot,\cdot)$.

######  Remark 2.

Let $K\in\mathcal{K}_{o}$ contain a segment $[p,q]$ on its boundary. Assume
that $L\in\mathcal{K}_{o}$ satisfies $d_{PM}(K,L)<\infty$. Then, from the
definition of $d_{PM}(\cdot,\cdot)$, we obtain that $L$ contains a translate
of $[p,q]$ on its boundary of the same length. In particular, it follows that
if $P$ and $Q$ are two polygons with $d_{PM}(P,Q)<\infty$, then $Q$ is a
translate of $P$. Thus, the topology induced by $d_{PM}(\cdot,\cdot)$ on the
family of $o$-symmetric convex polygons is the discrete topology.

##  3\. Proof of Theorem 1

Let $C\in\mathcal{K}_{o,2+}$, and let $K$ be a $C$-convex disk. Without loss
of generality, we assume that the curvature of a point of
$\operatorname{bd}(K)$ is strictly greater than the curvature of
$\operatorname{bd}(C)$ at the point with the same outer unit normal.

######  Remark 3.

By our conditions about the curvatures at points of $\operatorname{bd}(K)$ and
$\operatorname{bd}(C)$, it is easy to see that for any points
$p,q\in\operatorname{bd}(K)$, the set $[p,q]_{C}\setminus\\{p,q\\}$ is
contained in $\operatorname{int}(K)$.

Let us parametrize $\operatorname{bd}(K)$ and $\operatorname{bd}(C)$ as the
curves $\Gamma:[0,\operatorname{perim}(K)]\to\mathbb{R}^{2}$ and
$\Theta:[0,\operatorname{perim}(C)]\to\mathbb{R}^{2}$, respectively, using
arclength-parametrization and running along the boundaries of $C$ and $K$ in
counterclockwise direction. We choose the starting points $\Gamma(0)$ and
$\Theta(0)$ in such a way that $\operatorname{bd}(K)$ and
$\operatorname{bd}(C)$ have the same outer unit normal. For any two points
$p=\Gamma(s_{1}),q=\Gamma(s_{2})$, let us denote the arc $\widehat{pq}$ of
$\Gamma$ by $\Gamma|_{[s_{1},s_{2}]}$, and set
$A(s_{1},s_{2})=\operatorname{area}(r_{K}(\Gamma(s_{1}),\Gamma(s_{2})))$.

Our main tool will be the following. First, we compute the mixed partial
$(\partial_{s}\partial_{t}A)(\bar{s},\bar{t})$ for any points
$p=\Gamma(\bar{s}),q=\Gamma(\bar{t})$, where $\widehat{pq}$ has turning angle
strictly less than $\pi$. We note that our formula for
$(\partial_{s}\partial_{t}A)(\bar{s},\bar{t})$ will be valid for any
$C\in\mathcal{K}_{o,2+}$, not only for $B^{2}$. In the second part, we show
that if $C=B^{2}$, then this derivative is nonnegative, and apply Lemma 1.

###  3.1. Deriving a formula for the mixed partial
$(\partial_{s}\partial_{t}A)(\bar{s},\bar{t})$

Let $p=\Theta(\bar{S}),q=\Theta(\bar{T})$, where we assume that
$0<\bar{s}<\bar{t}<\operatorname{perim}(K)$ and
$0<\bar{S}<\bar{T}<\operatorname{perim}(C)$.

Note that by Remark 3, for any $0<s<t<\operatorname{perim}(K)$, there are
unique values $0<S^{\prime}<T^{\prime}<\operatorname{perim}(C)$ with the
property that $\Theta(T^{\prime})-\Theta(S^{\prime})=\Gamma(t)-\Gamma(s)$.
Thus, if we set $H=\\{(s,t):0<s<t<\operatorname{perim}(K)\\}$, we may define
the functions $S,T:H\to\mathbb{R}$ to identically satisfy
$\Theta(T(s,t))-\Theta(S(s,t))=\Gamma(t)-\Gamma(s)$. Note that
$\bar{S}=S(\bar{s},\bar{t})$ and $\bar{T}=T(\bar{s},\bar{t})$. More
specifically, consider the function
$f(s,t,S^{\prime},T^{\prime})=\Theta(S^{\prime})-\Theta(T^{\prime})+\Gamma(s)-\Gamma(t)$.
This is a twice continuously differentiable function in a neighborhood of
$(\bar{s},\bar{t},\bar{S},\bar{T})$, and satisfies
$f(\bar{s},\bar{t},\bar{S},\bar{T})=0$. Its Jacobian is the $2\times 4$ matrix
$[\Gamma^{\prime}(s)\quad-\Gamma^{\prime}(t)|-\Theta^{\prime}(S^{\prime})\quad\Theta^{\prime}(T^{\prime})]$.
Since $\Theta^{\prime}(\bar{S})$ and $\Theta^{\prime}(\bar{T})$ are not
parallel, the matrix $[-\Theta(S^{\prime})\quad\Theta^{\prime}(T^{\prime})]$
is invertible. Thus, by the Implicit Function Theorem, the functions $S(s,t)$
and $T(s,t)$ exist and are twice continuously differentiable in a neighborhood
of $(\bar{s},\bar{t})$.

Our next remark will play an important role in the proof.

######  Remark 4.

Since $K$ is $C$-convex, and $[p,q]_{C}$ is contained in
$\operatorname{int}(K)$ apart from $p,q$, we immediately have that the
following vectors are in this counterclockwise order around $o$:
$p-q,-\Theta^{\prime}(\bar{T}),-\Gamma^{\prime}(\bar{t}),\Gamma^{\prime}(\bar{s}),\Theta^{\prime}(\bar{S}),q-p,\Theta^{\prime}(\bar{T}),\Gamma^{\prime}(\bar{t}),-\Gamma^{\prime}(\bar{s}),-\Theta^{\prime}(\bar{S}),p-q$.

Figure 2. The vectors described in Remark 4. The $C$-spindle $[p,q]_{C}$ is
illustrated by a dashed region.

We denote the oriented angles of the vectors listed in Remark 4, measured from
the positive half of the $x$-axis, by
$\alpha,\theta_{q}-\pi,\gamma_{p},\theta_{p},\alpha+\pi,\theta_{q},\gamma_{q},\theta_{p}+\pi,\alpha+2\pi$,
respectively.

######  Lemma 2.

Let $\kappa_{p},\kappa_{q}$ denote the curvatures of $\operatorname{bd}(C)$ at
$p$ and $q$, respectively. Then

$(\partial_{s}S)(\bar{s},\bar{t})=\frac{\sin(\theta_{q}-\gamma_{p})}{\sin(\theta_{q}-\theta_{p})},\,(\partial_{t}S)(\bar{s},\bar{t})=\frac{\sin(\gamma_{q}-\theta_{q})}{\sin(\theta_{q}-\theta_{p})},$
$(\partial_{s}T)(\bar{s},\bar{t})=\frac{\sin(\theta_{p}-\gamma_{p})}{\sin(\theta_{q}-\theta_{p})},\,(\partial_{t}T)(\bar{s},\bar{t})=\frac{\sin(\gamma_{q}-\theta_{p})}{\sin(\theta_{q}-\theta_{p})},\hbox{
and}$
$(\partial_{s}\partial_{t}S)(\bar{s},\bar{t})=\frac{\kappa_{p}\sin(\gamma_{q}-\theta_{q})\cos(\theta_{q}-\theta_{p})\sin(\theta_{q}-\gamma_{p})-\kappa_{q}\sin(\theta_{p}-\gamma_{p})\sin(\gamma_{q}-\theta_{p})}{\sin^{3}(\theta_{q}-\theta_{p})}.$

###### Proof.

By the Implicit Function Theorem, in a neighborhood of $(\bar{s},\bar{t})$ we
have

$\left[\begin{array}[]{cc}\partial_{s}S&\partial_{t}S\\\
\partial_{s}T&\partial_{t}T\end{array}\right]=-[-\Theta^{\prime}(S)\quad\Theta^{\prime}(T)]^{-1}[\Gamma^{\prime}(s)\quad-\Gamma^{\prime}(t)].$

For $*\in\\{S,T\\}$, let us denote the $x$ and $y$ coordinates of
$\Theta^{\prime}(*)$ with $\Theta^{\prime}_{x}(*)$ and
$\Theta^{\prime}_{y}(*)$, respectively. Then

$[-\Theta^{\prime}(S)\quad\Theta^{\prime}(T)]^{-1}=\frac{1}{|\Theta^{\prime}(S),\Theta^{\prime}(T)|}\left[\begin{array}[]{cc}\Theta^{\prime}_{y}(T)&-\Theta^{\prime}_{x}(T)\\\
\Theta^{\prime}_{y}(S)&-\Theta^{\prime}_{x}(S)\end{array}\right].$

Computing the product of the two matrices, we obtain that
$\partial_{s}S=\frac{|\Gamma^{\prime}(s),\Theta^{\prime}(T)|}{|\Theta^{\prime}(S),\Theta^{\prime}(T)|}$.
Since $\Gamma$ and $\Theta$ are parametrized by arclength, the vectors
appearing in this expression are unit vectors. Since for any vectors
$u,v\in\mathbb{R}^{2}$, the determinant $|u,v|$ is the signed area of the
parallelogram spanned by them, we obtain that
$|\Gamma^{\prime}(\bar{s}),\Theta^{\prime}(\bar{T})|=\sin(\theta_{q}-\gamma_{p})$
and
$|\Theta^{\prime}(\bar{S}),\Theta^{\prime}(\bar{T})|=\sin(\theta_{q}-\theta_{p})$.
We obtain the other three first partials in a similar way; in particular we
have
$\partial_{t}S=\frac{|\Theta^{\prime}(T),\Gamma^{\prime}(t)|}{|\Theta^{\prime}(S),\Theta^{\prime}(T)|}$.
To compute $(\partial_{s}\partial tS)(\bar{s},\bar{t})$, we differentiate this
expression, and obtain that

(2)
$\partial_{s}\partial_{t}S=\frac{1}{{|\Theta^{\prime}(S),\Theta^{\prime}(T)|^{2}}}\bigl{(}|\Theta^{\prime\prime}(T),\Gamma^{\prime}(t)|(\partial_{s}T)|\Theta^{\prime}(S),\Theta^{\prime}(T)|-\\\
-|\Theta^{\prime}(T),\Gamma^{\prime}(t)|\left(|\Theta^{\prime\prime}(S),\Theta^{\prime}(T)|(\partial_{s}S)+|\Theta^{\prime}(S),\Theta^{\prime\prime}(T)|(\partial_{s}T)\right)\bigr{)}.$

Note that by arclength-parametrization, we have

(3)
$\Theta^{\prime\prime}(\bar{S})=\kappa_{p}\left[\begin{array}[]{c}\cos\left(\theta_{p}+\frac{\pi}{2}\right)\\\
\sin\left(\theta_{p}+\frac{\pi}{2}\right)\end{array}\right],\quad\Theta^{\prime\prime}(\bar{T})=\kappa_{q}\left[\begin{array}[]{c}\cos\left(\theta_{q}+\frac{\pi}{2}\right)\\\
\sin\left(\theta_{q}+\frac{\pi}{2}\right)\end{array}\right].$

Thus, to derive the expression for
$(\partial_{s}\partial_{t}S)(\bar{s},\bar{t})$ in Lemma 2, we substitute these
expressions and our formulas for the first partials of $S$ and $T$ into (2),
and use trigonometric identities. ∎

Now, let $\bar{A}(s,t)=\operatorname{area}(K)-A(s,t)$, and we prove that
$(\partial_{s}\partial_{t}\bar{A})(\bar{s},\bar{t})\geq 0$. Let $(s,t)$ be
sufficiently close to $(\bar{s},\bar{t})$. We compute
$\bar{A}(s,t)-\bar{A}(\bar{s},\bar{t})$.

Note that for any $s,t$, there is a unique vector $x(s,t)$ such that
$x(s,t)+C$ contains the oriented $C$-arc from $\Gamma(s)$ to $\Gamma(t)$. To
compute this vector, we observe that $\Gamma(s)=x(s,t)+\Theta(S(s,t))$ and
$\Gamma(t)=x(s,t)+\Theta(T(s,t))$, i.e.

(4) $x(s,t)=\Gamma(s)-\Theta(S(s,t))=\Gamma(t)-\Theta(T(s,t)).$

By the properties of $S(s,t)$ and $T(s,t)$, the function $x(s,t)$ has
continuous second partials.

Let $R$ denote the region swept by the $C$-arc from $p$ to $q$ as it is
translated in the direction of $x$ to its translate by $x$ (see Figure 3). We
denote the signed area of $R$ by $A(R)$, where the sign is defined such that
the part of $R$ inside $K$ has negative area, and the part outside it has
positive area.

Let $X_{p}$ denote the closed oriented curve obtained by joining the oriented
segment from $p$ to $x+p$, the arc $x+\Theta([\bar{S},S(s,t)])$ oriented from
$x+p$ to $\Gamma(s)$, and the arc $\Gamma([\bar{s},s])$ oriented from
$\Gamma(s)$ to $p$. Similarly, let $X_{q}$ denote the closed oriented curve
obtained by joining the oriented segment from $q$ to $x+q$, the arc
$x+\Theta([\bar{T},T(s,t)])$ oriented from $x+q$ to $\Gamma(t)$, and the arc
$\Gamma([\bar{t},t])$ oriented from $\Gamma(t)$ to $q$. Let $R_{p}$ and
$R_{q}$ denote the region enclosed by $X_{p}$ and $X_{q}$, respectively. For
$*\in\\{p,q\\}$, we say that the signed area $A(R_{*})$ of the region $R_{*}$
is positive or negative if the curve bounding it runs in counterclockwise or
clockwise direction on the boundary. Then

(5) $\bar{A}(s,t)-\bar{A}(\bar{s},\bar{t})=A(R)-A(R_{p})+A(R_{q}).$

Figure 3. An illustration for the notation in the proof of Theorem 1. The
region $R$ is denoted by skew lines.

In the following part we compute the second mixed derivatives of $A(R)$,
$A(R_{p})$ and $A(R_{q})$.

First, we compute the value of $\partial_{s}\partial_{t}A(R)$ at
$(\bar{s},\bar{t})$. By Fubini’s theorem, $A(R)$ is equal to the signed area
of the parallelogram spanned by $x$ and $q-p$, or more precisely,
$A(R)=|x,q-p|=|p-q,x|$. Since $p-q$ is a constant, we have
$\partial_{s}\partial_{t}A(R)=|p-q,\partial_{s}\partial_{t}x|$. Now, by (4),
we have

$\partial_{s}\partial_{t}x=\partial_{s}\left(-\Theta^{\prime}(S)(\partial_{t}S)\right)=-\Theta^{\prime\prime}(S)(\partial_{s}S)(\partial_{t}S)-\Theta^{\prime}(S)(\partial_{s}\partial_{t}S).$

The formulas in (3) yield that

$\frac{|p-q,(\partial_{s}\partial_{t}x)(\bar{s},\bar{t})|}{||p-q||}=-\kappa_{p}\sin\left(\theta_{p}+\frac{\pi}{2}-\alpha\right)(\partial_{s}S)(\partial_{t}S)-\sin(\theta_{q}-\alpha)(\partial_{s}\partial_{t}S).$

Thus, by Lemma 2 and using trigonometric identities, we obtain that

(6) $\frac{|p-q,(\partial_{s}\partial_{t}x)(\bar{s},\bar{t})|}{||p-q||}=\\\
=\frac{\kappa_{p}\sin(\alpha-\theta_{q})\sin(\theta_{q}-\gamma_{p})\sin(\gamma_{q}-\theta_{q})+\kappa_{q}\sin(\theta_{p}-\alpha)\sin(\theta_{p}-\gamma_{p})\sin(\gamma_{q}-\theta_{p})}{\sin^{3}(\theta_{q}-\theta_{p})}.$

Note that by Remark 4, all angles in (6) are convex angles. This and the
inequalities $\kappa_{p},\kappa_{q},||p-q||>0$ imply that
$|p-q,(\partial_{s}\partial_{t}x)(\bar{s},\bar{t})|>0$.

Next, we compute the value of $\partial_{s}\partial_{t}A(R_{p})$ at
$(\bar{s},\bar{t})$. Recall the well-known formula that if
$G:[s_{1},s_{2}]\to\mathbb{R}^{2}$ is an arclength-parametrized curve
enclosing a region and running on its boundary in counterclockwise direction,
then the area of the enclosed region is
$\frac{1}{2}\int_{s_{1}}^{s_{2}}|G(s),G^{\prime}(s)|\,ds$. Using this, we
obtain that

(7)
$2A(R_{p})=\int_{\bar{S}}^{S(s,t)}|x(s,t)+\Theta(u)-\Gamma(\bar{s}),\Theta^{\prime}(u)|\,du+\int_{s}^{\bar{s}}|\Gamma(u)-\Gamma(\bar{s}),\Gamma^{\prime}(u)|\,du.$

Note that
$\partial_{t}\int_{s}^{\bar{s}}|\Gamma(u)-\Gamma(\bar{s}),\Gamma^{\prime}(u)|\,du=0$.
Furthermore, $S(s,t)$ has continuous second partials, and the function
$\partial_{t}|x+\Theta(u)-\Gamma(\bar{s}),\Theta^{\prime}(u)|$ is continuous
in an open neighborhood of the set $\\{(\bar{s},u):u\in[\bar{S},S(s,t)]\\}$.
Thus, we may apply Leibniz’s integral rule and obtain that

$\partial_{t}2A(R_{p})=|x(s,t)+\Theta(S(s,t))-\Gamma(\bar{s}),\Theta^{\prime}(S(s,t))|(\partial_{t}S)+\int_{\bar{S}}^{S(s,t)}|(\partial_{t}x),\Theta^{\prime}(u)|\,du$

Again applying Leibniz’s integral rule and the identities
$x(\bar{s},\bar{t})=0$, $\Theta(\bar{S})=\Gamma(\bar{s})$, we have

$\left.\partial_{s}\partial_{t}2A(R_{p})\right|_{(s,t)=(\bar{s},\bar{t})}=\\\
=|(\partial_{s}x)(\bar{s},\bar{t})+\Theta^{\prime}(\bar{S})(\partial_{s}S)(\bar{s},\bar{t}),\Theta^{\prime}(\bar{S})|(\partial_{t}S)(\bar{s},\bar{t})+|0,\Theta^{\prime\prime}(\bar{S})|(\partial_{s}S)(\bar{s},\bar{t})(\partial_{t}S)(\bar{s},\bar{t})+\\\
+|0,\Theta^{\prime}(\bar{S})|(\partial_{s}\partial_{t}S)(\bar{s},\bar{t})+|(\partial_{t}x)(\bar{s},\bar{t}),\Theta^{\prime}(\bar{S})|(\partial_{s}S)(\bar{s},\bar{t})+\int_{\bar{S}}^{\bar{S}}|(\partial_{s}\partial_{t}x)(\bar{s},\bar{t}),\Theta^{\prime}(u)|\,du.$

This yields that

$\left.\partial_{s}\partial_{t}2A(R_{p})\right|_{(s,t)=(\bar{s},\bar{t})}=|(\partial_{s}x)(\bar{s},\bar{t}),\Theta^{\prime}(\bar{S})|(\partial_{t}S)(\bar{s},\bar{t})+|(\partial_{t}x)(\bar{s},\bar{t}),\Theta^{\prime}(\bar{S})|(\partial_{s}S)(\bar{s},\bar{t}).$

From (4), we obtain
$(\partial_{s}x)(\bar{s},\bar{t})=-\Theta^{\prime}(\bar{T})(\partial_{s}T)(\bar{s},\bar{t})$
and
$(\partial_{t}x)(\bar{s},\bar{t})=-\Theta^{\prime}(\bar{S})(\partial_{t}S)(\bar{s},\bar{t})$.
Substituting these into the above expression yields that

$\left.\partial_{s}\partial_{t}A(R_{p})\right|_{(s,t)=(\bar{s},\bar{t})}=-\frac{1}{2}|\Theta^{\prime}(\bar{T}),\Theta^{\prime}(\bar{S})|(\partial_{s}T)(\bar{s},\bar{t})(\partial_{t}S)(\bar{s},\bar{t}).$

Thus, applying Lemma 2, we have

(8)
$\left.\partial_{s}\partial_{t}A(R_{p})\right|_{(s,t)=(\bar{s},\bar{t})}=\frac{\sin(\gamma_{q}-\theta_{q})\sin(\theta_{p}-\gamma_{p})}{2\sin(\theta_{q}-\theta_{p})}$

Finally, using a similar argument, we obtain that

(9)
$2A(R_{q})=\int_{\bar{T}}^{T(s,t)}|x(s,t)+\Theta(u)-\Gamma(\bar{t}),\Theta^{\prime}(u)|\,du+\int_{t}^{\bar{t}}|\Gamma(u)-\Gamma(\bar{t}),\Gamma^{\prime}(u)|\,du,\hbox{
and}$ (10)
$\left.\partial_{s}\partial_{t}A(R_{q})\right|_{(s,t)=(\bar{s},\bar{t})}=-\frac{\sin(\gamma_{q}-\theta_{q})\sin(\theta_{p}-\gamma_{p})}{2\sin(\theta_{q}-\theta_{p})}$

Thus,

$\left.\partial_{s}\partial_{t}A(R_{q})-A(R_{p})\right|_{(s,t)=(\bar{s},\bar{t})}=-\frac{\sin(\gamma_{q}-\theta_{q})\sin(\theta_{p}-\gamma_{p})}{\sin(\theta_{q}-\theta_{p})},$

implying that

(11)
$(\partial_{s}\partial_{t}A)(\bar{s},\bar{t})=-\frac{\sin(\gamma_{q}-\theta_{q})\sin(\theta_{p}-\gamma_{p})}{\sin(\theta_{q}-\theta_{p})}+\\\
||p-q||\frac{\kappa_{p}\sin(\alpha-\theta_{q})\sin(\theta_{q}-\gamma_{p})\sin(\gamma_{q}-\theta_{q})+\kappa_{q}\sin(\theta_{p}-\alpha)\sin(\theta_{p}-\gamma_{p})\sin(\gamma_{q}-\theta_{p})}{\sin^{3}(\theta_{q}-\theta_{p})}.$

###  3.2. Determining the sign of the derivative if $C$ is a Euclidean disk

In the following, we assume that $C$ is a Euclidean disk of radius $r$, and
compute the value of the expression in (11). Using the notation in this
expression, we can assume that $\theta_{p}=\pi-\theta$ and
$\theta_{q}=\pi+\theta$ for some $0<\theta<\frac{\pi}{2}$, which, by Remark 4,
implies that $\theta<\gamma_{p}<\pi-\theta$ and
$\pi+\theta<\gamma_{q}<\gamma_{p}+\pi$. Moreover, we have
$||p-q||\kappa_{p}=||p-q||\kappa_{q}=2\sin\theta$. Substituting these into
(11), we obtain

(12) $(\partial_{s}\partial_{t}A)(\bar{s},\bar{t})=\\\
=\frac{2\sin^{2}{\theta}[\sin(\gamma_{p}-\theta)\sin(\theta-\gamma_{q})-\sin(\theta+\gamma_{p})\sin(\gamma_{q}+\theta)]}{\sin^{3}(2\theta)}-\frac{\sin(\theta-\gamma_{q})\sin(\theta+\gamma_{p})}{\sin(2\theta)}.$

Using the trigonometric identity
$\sin^{2}2\theta=4\sin^{2}\theta\cos^{2}\theta$, we obtain that

(13)
$2\cos^{2}\theta\sin(2\theta)(\partial_{s}\partial_{t}A)(\bar{s},\bar{t})=\\\
=\sin(\gamma_{p}-\theta)\sin(\theta-\gamma_{q})-\sin(\theta+\gamma_{p})\sin(\gamma_{q}+\theta)-2\cos^{2}\theta\sin(\theta-\gamma_{q})\sin(\theta+\gamma_{p}).$

By additional trigonometric identities, this expression simplifies to

(14)
$2\cos^{2}\theta\sin(2\theta)(\partial_{s}\partial_{t}A)(\bar{s},\bar{t})=-\sin^{2}\theta\cos(\gamma_{p}+\gamma_{q})-\cos(\gamma_{p}-\gamma_{q})+\cos^{2}\theta\cos(2\theta+\gamma_{p}-\gamma_{q}).$

Now, a simple computation shows that under our conditions and considering
$\gamma_{p}-\gamma_{q}$ fixed, the quantity in (14) is minimal if and only if
$\gamma_{p}+\gamma_{q}=2\pi$, implying that

(15)
$2\cos^{2}\theta\sin(2\theta)(\partial_{s}\partial_{t}A)(\bar{s},\bar{t})\geq-\sin^{2}\theta-\cos(2\gamma_{p})+\cos^{2}\theta\cos(2\theta+2\gamma_{p}).$

Note that $\gamma_{p}+\gamma_{q}=2\pi$ also yields that
$\frac{\pi}{2}<\gamma_{p}$. If we set $\gamma=\gamma_{p}-\frac{\pi}{2}$, then
the right-hand side expression in (15) can be written as

$-\sin^{2}\theta+\cos(2\gamma)-\cos^{2}\theta\cos(2\theta+2\gamma)=2\cos(\gamma+\theta)\left(\cos(\gamma-\theta)-\cos^{2}\theta\cos(\gamma+\theta)\right),$

where our conditions for the angles are transformed into the inequalities
$0<\theta,\gamma$ and $\gamma+\theta<\frac{\pi}{2}$. Thus, we have that
$\cos(\gamma+\theta)>0$, and also that
$\cos(\gamma-\theta)>\cos(\gamma+\theta)>\cos^{2}\theta\cos(\gamma+\theta)$.
This yields Theorem 1.

##  4\. Proof of Theorem 2

Let $\mathcal{K}_{-}$ denote the family of $o$-symmetric convex disks in
$\mathcal{K}_{o,2+}$ for which there is a $C$-convex disk $K$ that does not
satisfy the Quadrangle Property. We show that $\mathcal{K}_{-}$ is residual in
$\mathcal{K}_{o,2+}^{PM}$.

First, we show that $\mathcal{K}_{-}$ is open in $\mathcal{K}_{0,2+}^{H}$,
which, by Proposition 1, yields that it is open in $\mathcal{K}_{0,2+}^{PM}$.
Let $C\in\mathcal{K}_{-}$, and suppose for contradiction that there is a
sequence $C_{m}\notin\mathcal{K}_{-}$ with $C_{m}\to C$. Since
$C\in\mathcal{K}_{-}$, there is a $C$-convex disk $K$ and points
$x_{1},x_{2},x_{3},x_{4}$ in this counterclockwise order in
$\operatorname{bd}(K)$, with turning angle strictly less than $\pi$, such that
$\operatorname{area}(r_{K}^{C}(x_{1},x_{4}))+\operatorname{area}(r_{K}^{C}(x_{2},x_{3}))<\operatorname{area}(r_{K}^{C}(x_{1},x_{3}))+\operatorname{area}(r_{K}^{C}(x_{2},x_{4}))$.
As $K$ is $C$-convex, it is the intersection of some translates of $C$, i.e.
there is some $X\subset\mathbb{R}^{2}$ such that $K=\bigcap_{x\in X}(x+C)$.
Let $K_{m}=\bigcap_{x\in X}(x+C_{m})$. Then, clearly, $K_{m}$ is
$C_{m}$-convex, and $K_{m}\to K$. For $i=1,2,3,4$, let
$x_{i}^{m}\in\operatorname{bd}(K_{m})$ be a point with $x_{i}^{m}\to x_{i}$.
Note that if $m$ is sufficiently large,
$x_{1}^{m},x_{2}^{m},x_{3}^{m},x_{4}^{m}$ are in this counterclockwise order
in $\operatorname{bd}(K^{m})$, and the turning angle of
$\widehat{x_{1}^{m}x_{4}^{m}}$ is strictly less than $\pi$. Thus,
$\operatorname{area}(r_{K_{m}}^{C_{m}}(x_{1}^{m},x_{4}^{m}))+\operatorname{area}(r_{K_{m}}^{C_{m}}(x_{2}^{m},x_{3}^{m}))\geq\operatorname{area}(r_{K_{m}}^{C_{m}}(x_{1}^{m},x_{3}^{m}))+\operatorname{area}(r_{K_{m}}^{C_{m}}(x_{2}^{m},x_{4}^{m}))$.
But then [2, Remark 2] implies that
$\operatorname{area}(r_{K}^{C}(x_{1},x_{4}))+\operatorname{area}(r_{K}^{C}(x_{2},x_{3}))\geq\operatorname{area}(r_{K}^{C}(x_{1},x_{3}))+\operatorname{area}(r_{K}^{C}(x_{2},x_{4}))$,
a contradiction.

In the remaining part of the proof, we show that $\mathcal{K}_{-}$ is
everywhere dense in $\mathcal{K}_{o,2+}^{PM}$. Note that for any
$C\in\mathcal{K}_{o,2+}^{PM}$ and Borel set $\sigma\subset\mathbb{S}^{1}$,
$A_{C}(\sigma)=\int R_{C}(u)\,d\sigma$, where $R_{C}(u)$ denotes the radius of
curvature at the point of $\operatorname{bd}(C)$ with outer unit normal $u$.
In particular, if for $C,C^{\prime}\in\mathcal{K}_{o,2+}$ we have that for any
$u\in\mathbb{S}^{1}$, $|R_{C}(u)-R_{C^{\prime}}(u)|\leq\varepsilon$, then
$d_{PM}(C,C^{\prime})\leq\varepsilon$.

Based on this, we observe that for any $\lambda_{0}>0$, the subfamily of
$\mathcal{K}_{o,2+}$ of the elements $C$ satisfying the property that there is
a point $p\in\operatorname{bd}(C)$ where

  * •

the curvature of $\operatorname{bd}(C)$ is locally maximal,

  * •

if $\Theta(s)$ is the arclength-parametrization of $\operatorname{bd}(C)$ with
$p=\Theta(0)$, then $\Theta$ is a $C^{\infty}$-class function in a
neighborhood of $0$,

  * •

If $\kappa(s)$ denotes the curvature of $\Theta(s)$, then
$\kappa(s)=\kappa_{0}-\lambda s^{2}+O(s^{3})$ for some $\kappa_{0}>0$ and any
given $\lambda>\lambda_{0}$,

is everywhere dense in $\mathcal{K}_{o,2+}^{PM}$. Indeed, let $\Theta_{0}(s)$
be the arclength-parametrization of the boundary of an arbitrary element
$C_{0}$ of $\mathcal{K}_{o,2+}$. Without loss of generality, assume that the
curvature $\kappa_{0}(s)$ of $\Theta_{0}(s)$ has a local maximum at $s=0$. Let
$\varepsilon>0$ and $\lambda>0$ be fixed, and choose some $\delta>0$ such that
$|\kappa_{0}(s)-\kappa_{0}(0)|\leq\frac{\varepsilon}{2}$ for any
$|s|\leq\delta$. Replace the function $\kappa_{0}(s)$ by the function
$\kappa(s)=\kappa_{0}(0)-\lambda s^{2}$ in a small neighborhood
$[-s_{0},s_{0}]$ of $0$ in $(-\delta,\delta)$. We extend this function to the
interval $[-\delta,\delta]$ in such a way that the obtained function
$\kappa(s)$ is continuous and strictly positive, satisfies
$|\kappa_{0}(s)-\kappa(s)|\leq\frac{\varepsilon}{2}$ for any
$s\in[-s_{0},s_{0}]$, and for $\bar{s}\in\\{-s_{0},s_{0}\\}$, it satisfies

  * (i)

$\int_{0}^{\bar{s}}\kappa_{0}(t)\,dt=\int_{0}^{\bar{s}}\kappa(t)\,dt$;

  * (ii)

$\int_{0}^{\bar{s}}\cos\int_{0}^{t}\kappa_{0}(u)\,du\,dt=\int_{0}^{\bar{s}}\cos\int_{0}^{t}\kappa(u)\,du\,dt$;
and

  * (iii)

$\int_{0}^{\bar{s}}\sin\int_{0}^{t}\kappa_{0}(u)\,du\,dt=\int_{0}^{\bar{s}}\sin\int_{0}^{t}\kappa(u)\,du\,dt$.

We observe that then $\kappa(s)$ is the curvature function of a convex disk in
$\mathcal{K}_{2+}$. Modifying also a neighborhood of the point
$-\Gamma_{0}(0)$ in the same way yields an $o$-symmetric convex disk $C$ at
PM-distance at most $\varepsilon$ from $C_{0}$ with a curvature function
$\kappa(s)=\kappa_{0}-\lambda s^{2}$ in a neighborhood of $s=0$.

Thus, in the following we assume that the above conditions hold for $C$ with
$\lambda>\frac{9\kappa_{0}^{3}}{2}$. Let $\Theta(0)=(0,0)$ such that
$\Theta^{\prime}(0)$ points in the direction of the negative half of the
$x$-axis. As before, let $\kappa(s)$ denote the curvature of the curve
$\Theta$ at the point $\Theta(s)$. We set $p=\Theta(-\bar{s})$ and
$q=\Theta(\bar{s})$ for some sufficiently small value of $\bar{s}$, and set
$\gamma_{p}=\frac{\pi}{2}+\kappa_{0}\bar{s}$ and
$\gamma_{q}=\frac{3\pi}{2}-\kappa_{0}\bar{s}$. Note that then
$\theta_{p}=\pi-\kappa_{0}\bar{s}+\frac{\lambda}{3}\bar{s}^{3}$ and
$\theta_{q}=\pi+\kappa_{0}\bar{s}-\frac{\lambda}{3}\bar{s}^{3}$.

Then, computing the coordinates of $p$ and $q$ as in (ii) and (iii), we obtain

$||p-q||=2\bar{s}-\frac{\kappa_{0}^{2}}{3}\bar{s}^{3}+\frac{\kappa_{0}(8\lambda+\kappa_{0}^{3})}{60}\bar{s}^{5}+O(\bar{s}^{7}),$

and, from $p_{y}=q_{y}$ we obtain that $p-q$ points in the direction of the
positive half of the $x$-axis. Substituting these into (11), we obtain that

$(\partial_{s}\partial_{t}A)(-\bar{s},\bar{s})=\frac{4\kappa_{0}(9\kappa_{0}^{3}-2\lambda)}{3}\bar{s}^{4}+O(\bar{s}^{6}).$

This shows that the mixed partial derivative
$(\partial_{s}\partial_{t}A)(-\bar{s},\bar{s})$ is negative if $\bar{s}>0$ is
sufficiently small, implying that $C$ does not satisfy the Quadrangle
Property.

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