# Sharp inequalities involving the Cheeger constant of planar convex sets

Ilias Ftouhi, Alba Lia Masiello, Gloria Paoli  Friedrich-Alexander-Universität
Erlangen-Nürnberg, Department of Mathematics, Chair of Applied Analysis
(Alexander von Humboldt Professorship), Cauerstr. 11, 91058 Erlangen, Germany.
<EMAIL_ADDRESS>Dipartimento di Matematica e Applicazioni "R.
Caccioppoli", Universita degli Studi di Napoli "Federico II", Complesso
Universitario Monte S. Angelo, via Cintia - 80126 Napoli, Italy.
<EMAIL_ADDRESS>Dipartimento di Matematica e Applicazioni "R.
Caccioppoli", Universita degli Studi di Napoli "Federico II", Complesso
Universitario Monte S. Angelo, via Cintia - 80126 Napoli, Italy.
<EMAIL_ADDRESS>
###### Abstract.

We are interested in finding sharp bounds for the Cheeger constant $h$ via
different geometrical quantities, namely the area $|\cdot|$, the perimeter
$P$, the inradius $r$, the circumradius $R$, the minimal width $\omega$ and
the diameter $d$. We provide new sharp inequalities between these quantities
for planar convex bodies and enounce new conjectures based on numerical
simulations. In particular, we completely solve the Blaschke-Santaló diagrams
describing all the possible inequalities involving the triplets $(P,h,r)$,
$(d,h,r)$ and $(R,h,r)$ and describe some parts of the boundaries of the
diagrams of the triplets $(\omega,h,d)$, $(\omega,h,R)$, $(\omega,h,P)$,
$(\omega,h,|\cdot|)$, $(R,h,d)$ and $(\omega,h,r)$.

Keywords: Cheeger constant, convex sets, Blaschke–Santaló diagrams, sharp
inequalities.
MSC 2020: 52A10, 52A40, 65K15.

##  1\. Introduction

Let $\Omega$ be a bounded subset of $\mathbb{R}^{2}$. The Cheeger constant of
$\Omega$, introduced in and by Jeff Cheeger in [12] in connection with the
first eigenvalue of the Laplacian, is defined as

(1)
$h(\Omega):=\inf\left\\{\frac{P(E)}{{\left|E\right|}}\,:\,E\,\text{measurable,
}\,E\subseteq\Omega,\,{\left|E\right|}>0\right\\},$

where $P(E)$ is the perimeter of $E$ in the sense of De Giorgi and
${\left|E\right|}$ is the area of $E$. The minimum in (1) is achieved when
$\Omega$ has Lipschitz boundary, see as a reference [32], and the set $E$ that
realizes this minimum is called a _Cheeger set_ of $\Omega$. For the
properties of the Cheeger constant and for an introductory survey, see for
example [1, 27, 32]. In particular, in the case of convex sets, the authors in
[1] prove that the Cheeger set is unique and, in this case, we will denote it
by $C_{\Omega}$. At last, a complete characterization of the Cheeger sets of
planar convex bodies is provided in [27].

The problem of finding the Cheeger constant of a domain has been widely
considered and has several applications (see [32] for a general overview). One
of the possible interpretations of the Cheeger constant can be found for
instance in the context of maximal flow and minimal cut problems (see [40])
which has applications in medical images processing (see [4]). The Cheeger
problem appears also in the study of plate failure under stress (see [28]). It
is then useful to have estimates of the Cheeger constant in terms of geometric
quantities that can be easily computed.

In the present paper, we are interested in describing all the possible
inequalities involving the Cheeger constant of a given compact and convex set
$\Omega\subset\mathbb{R}^{2}$ with nonempty interior and the following
geometrical quantities: the area ${\left|\Omega\right|}$, the perimeter
$P(\Omega)$, the inradius $r(\Omega)$, the circumradius $R(\Omega)$, the
minimal width $\omega(\Omega)$ and the diameter $d(\Omega)$. We are then
aiming to study the Blaschke–Santaló diagrams involving those functionals and
collect them all in one single paper together with new conjectures.

A Blaschke–Santaló diagram is a tool that allows to visualize all the possible
inequalities between three geometric quantities. More precisely, we consider
three homogeneous shape functionals $(J_{1},J_{2},J_{3})$, that is to say that
for every $i\in\\{1,2,3\\}$ there exists $\alpha_{i}\in\mathbb{R}$ such that
$J_{i}(t\Omega)=t^{\alpha_{i}}J_{i}(\Omega)$ for every $t>0$, and we want to
find a system of inequalities describing the set

$\\{(J_{1}(\Omega),J_{2}(\Omega))|\
J_{3}(\Omega)=1,\,\Omega\in\mathcal{K}^{2}\\},$

where we denote by $\mathcal{K}^{2}$ the class of planar, compact and convex
sets with nonempty interior.

This kind of diagram was introduced by Blaschke in [5], in order to
investigate all the possible relations between the volume, the surface area
and the integral mean curvature in the class of compact convex sets in
$\mathbb{R}^{3}$. Following the idea of Blaschke, Santaló in [36] proposed the
study of these diagrams for all the triplets of the following geometrical
quantities: area, perimeter, inradius, circumradius, minimal width and
diameter. These diagrams were studied for the class of convex sets and six of
them are still not completely solved. We refer to the introduction in [13] for
an accurate and updated state of art. Moreover, for classical results about
Blaschke–Santaló diagrams, we refer for example to [9, 21, 22, 24, 25, 23, 36]
and for more recent results, we provide the following non-exhaustive list of
works [10, 11, 13, 15, 17, 18, 31].

In [14] and [16] the author studies Blaschke–Santaló diagram involving the
Cheeger constant. More precisely, in [14], it the Blaschke–Santaló diagram
involving the Cheeger constant, the area and the inradius is fully
characterized. It is proved that, if $\Omega$ in $\mathcal{K}^{2}$, then

(2) $\frac{1}{r(\Omega)}+\frac{\pi r(\Omega)}{|\Omega|}\leq
h(\Omega)\leq\frac{1}{r(\Omega)}+\sqrt{\frac{\pi}{|\Omega|}},$

where the upper bound in (2) is achieved by (and only by) sets that are
homothetic to their form bodies (see Definition 2.10), for instance, sets that
are homothetic to their form bodies, meanwhile the lower one is achieved by
(and only by) stadiums. On the other hand, in [16], the diagram involving the
Cheeger constant, the area and the perimeter is fully characterized and it is
proved that if $\Omega\in\mathcal{K}^{2}$, then

(3) $\frac{P(\Omega)+\sqrt{4\pi|\Omega|}}{2|\Omega|}\leq
h(\Omega)\leq\frac{P(\Omega)}{|\Omega|},$

where the upper bound is achieved by any set that is Cheeger of itself (in
particular stadiums), meanwhile the lower one is achieved, for example, by
circumscribed polygons.

Now, let us state the main results of the paper. In order to do so, we need to
define the following classes of admissible sets (we refer to [38, Table 2.1]
for the associated constraints):

  1. (1)

$\displaystyle{\mathcal{K}^{2}_{P,r}=\\{\Omega\in\mathcal{K}^{2}:\,P(\Omega)=P,\,r(\Omega)=r\\}}$,
where $P\geq 2\pi r$;

  2. (2)

$\mathcal{K}^{2}_{d,r}=\\{\Omega\in\mathcal{K}^{2}:\,d(\Omega)=d,\,r(\Omega)=r\\}$,
where $d\geq 2r$;

  3. (3)

$\mathcal{K}^{2}_{R,r}=\\{\Omega\in\mathcal{K}^{2}:\,R(\Omega)=R,\,r(\Omega)=r\\}$,
where $R\geq r$;

  4. (4)

$\mathcal{K}^{2}_{\omega,d}=\\{\Omega\in\mathcal{K}^{2}:\,\omega(\Omega)=\omega,\,d(\Omega)=d\\}$,
where $\omega\leq d$;

  5. (5)

$\mathcal{K}^{2}_{\omega,R}=\\{\Omega\in\mathcal{K}^{2}:\,\omega(\Omega)=\omega,\,R(\Omega)=R\\}$,
where $2R\geq\omega$;

  6. (6)

$\mathcal{K}^{2}_{\omega,P}=\\{\Omega\in\mathcal{K}^{2}:\,\omega(\Omega)=\omega,\,P(\Omega)=P\\}$,
where $P\geq\pi\omega$;

  7. (7)

$\mathcal{K}^{2}_{\omega,A}=\\{\Omega\in\mathcal{K}^{2}:\,\omega(\Omega)=\omega,\,{\left|\Omega\right|}=A\\}$,
where $\sqrt{3}A\geq\omega^{2}$;

  8. (8)

$\mathcal{K}^{2}_{R,d}=\\{\Omega\in\mathcal{K}^{2}:\,R(\Omega)=R,\,d(\Omega)=d\\}$,
where $\sqrt{3}R\leq d<2R$;

  9. (9)

$\mathcal{K}^{2}_{\omega,r}=\\{\Omega\in\mathcal{K}^{2}:\,\omega(\Omega)=\omega,\,r(\Omega)=r\\}$,
where $2r<\omega\leq 3r$;

  10. (10)

$\mathcal{K}^{2}_{R,A}=\\{\Omega\in\mathcal{K}^{2}:\,R(\Omega)=R,\,{\left|\Omega\right|}=A\\}$,
where $A\leq\pi R^{2}$;

  11. (11)

$\mathcal{K}^{2}_{P,R}=\\{\Omega\in\mathcal{K}^{2}:\,P(\Omega)=P,\,R(\Omega)=R\\}$,
where $4R<P\leq 2\pi R$;

  12. (12)

$\mathcal{K}^{2}_{P,d}=\\{\Omega\in\mathcal{K}^{2}:\,P(\Omega)=P,\,d(\Omega)=d\\}$,
where $2d<P\leq\pi d$;

  13. (13)

$\mathcal{K}^{2}_{d,A}=\\{\Omega\in\mathcal{K}^{2}:\,d(\Omega)=d,\,{\left|\Omega\right|}=A\\}$,
where $\pi d^{2}\geq 4A$.

Firstly, let us state the following existence result.

######  Theorem 1.1.

Let $\Omega\in\mathcal{K}^{2}$, then the minimization and the maximization
shape optimization problems of the Cheeger constant $h(\Omega)$ admit a
solution in the classes of sets defined in $(1)-(13)$.

In the following theorem, we consider the triplets of functionals for which we
are able to provide the complete description of the related Blaschke–Santaló
diagrams. For the precise definitions of the below-mentioned extremal sets,
see Section 2.2 and for the explicit bounds, see Propositions 4.1, 4.2 and
4.5. At last, for the description of the corresponding diagrams we refer to
Proposition 4.7.

######  Theorem 1.2.

The following results hold

  1. (i)

The maximum and the minimum of the Cheeger constant in $\mathcal{K}^{2}_{P,r}$
are respectively achieved by sets that are homothetic to their form bodies and
by stadiums.

  2. (ii)

The maximum of the Cheeger constant in $\mathcal{K}^{2}_{d,r}$ is achieved by
symmetrical two-cup bodies. On the other hand, there exists $D_{0}>0$ such
that if $d\geq rD_{0}$, then the minimum of the Cheeger constant in
$\mathcal{K}^{2}_{d,r}$ is achieved by symmetrical spherical slices, while, if
$d<rD_{0}$, the minimum is achieved by regular smoothed nonagons.

  3. (iii)

The maximum and the minimum of the Cheeger constant in $\mathcal{K}^{2}_{R,r}$
are respectively achieved by symmetrical two-cup bodies and symmetrical
spherical slices.

As far as the classes of sets $\mathcal{K}^{2}_{\omega,d}$
$\mathcal{K}^{2}_{R,\omega}$, $\mathcal{K}^{2}_{\omega,P}$ and
$\mathcal{K}^{2}_{A,\omega}$ are concerned, we are able to identify parts of
the boundary of the corresponding Blasche- Santaló diagrams, see Propositions
5.1, 5.3, 5.7 and 5.5 for the explicit bounds. For the class
$\mathcal{K}^{2}_{d,R}$, we are able to solve the maximization problem, see
Proposition 5.8, meanwhile, for the class $\mathcal{K}^{2}_{\omega,r}$ we are
able to solve the minimization one, see Proposition 5.9. Throughout the paper,
different strategies of proofs are used to obtain all the aforementioned
results.

As far as the classes
$\mathcal{K}^{2}_{A,R},\mathcal{K}^{2}_{R,P},\mathcal{K}^{2}_{P,d}$ and
$\mathcal{K}^{2}_{A,d}$ are concerned, we are not able to identify any parts
of the boundaries of the corresponding Blasche–Santaló diagrams. Nevertheless,
in the appendix we present the best bounds that we managed to obtain by
combining the known inequalities involving those functionals.

The paper is organized as follows: in Section 2, we state the preliminary
results, the definitions used throughout the paper and the known inequalities
relating the Cheeger constant to one of the geometric quantities taken into
consideration. Section 3 is dedicated to the description of the numerical
methods used to compute the functionals and to approximate the
Blaschke–Santaló diagrams. In Section 4, we prove the main results, that are
Theorem 1.1 and Theorem 1.2. Section 5 is dedicated to the results that we
have obtained for the classes of sets $\mathcal{K}^{2}_{\omega,d}$
$\mathcal{K}^{2}_{R,\omega}$, $\mathcal{K}^{2}_{\omega,P}$,
$\mathcal{K}^{2}_{A,\omega}$, $\mathcal{K}^{2}_{d,R}$ and
$\mathcal{K}^{2}_{\omega,r}$. Finally, in Section 6, we state some relevant
conjectures and collect all the inequalities proved in the paper in the
Appendix.

##  2\. Notations and Preliminaries

Throughout this article, ${\left\|\cdot\right\|}$ will denote the Euclidean
norm in $\mathbb{R}^{2}$, while $(\cdot)$ is the standard Euclidean scalar
product in $\mathbb{R}^{2}$. We denote by $P(\Omega)$ the perimeter of
$\Omega$ and by $|\Omega|$ the volume of $\Omega$. Moreover, $B_{r}$ is the
closed ball of radius $r>0$ centered at the origin, while $\mathbb{S}^{1}$ is
the unit sphere in $\mathbb{R}^{2}$. In the following, we work with the class
of sets $\mathcal{K}^{2}$, defined as

$\mathcal{K}^{2}:=\\{\Omega\>|\>\Omega\;\,\text{is a compact, bounded and
convex set with nonempty interior of
}\;\mathbb{R}^{2}\\}\setminus\\{\emptyset\\}.$

###  2.1. Classical results ad preliminary lemmas

We provide the classical definitions and results that we need in the
following.

######  Definition 2.1.

Let $\Omega,K\subset\mathbb{R}^{2}$ two convex bounded sets. We define the
_Minkowski sum_ $(+)$ and _difference_ $(\sim)$ as

$\Omega+K:=\\{x+y\,:\,x\in\Omega,\,y\in K\\},$ $\Omega\sim
K:=\\{x\in\mathbb{R}^{2}\,:\,x+K\subseteq\Omega\\}.$

We now recall the definition of the Hausdorff distance.

######  Definition 2.2.

Let $\Omega,K\subset\mathbb{R}^{2}$ be two non-empty compact sets, we define
the Hausdorff distance between $\Omega$ and $K$ as follows:

$d_{\mathcal{H}}(\Omega,K):=\inf\left\\{\varepsilon>0\;:\;\Omega\subset
K+B_{\varepsilon},\;K\subset\Omega+B_{\varepsilon}\right\\},$

where $B_{\varepsilon}$ is the ball of radius $\varepsilon$ centered in the
origin.

Let $\\{\Omega_{k}\\}_{k\in\mathbb{N}}$ be a sequence of non-empty, compact,
bounded convex subsets of $\mathbb{R}^{2}$, we say that $\Omega_{k}$ converges
to $\Omega$ in the Hausdorff sense and we denote

$\Omega_{k}\stackrel{{\scriptstyle\mathcal{H}}}{{\longrightarrow}}\Omega,$

if and only if $d_{\mathcal{H}}(\Omega_{k},\Omega)\to 0$ as $k\to\infty$.

We recall that by Blaschke’s selection Theorem (see for example [37, Theorem
1.8.7]), every bounded sequence of nonempty compact convex sets has a
subsequence that converges in the Hausdorff sense to a convex set.

Let us now recall the following definitions:

######  Definition 2.3.

Let $\Omega\in\mathcal{K}^{2}$. The _distance function from the boundary of_
$\Omega$ is the function ${\rm
dist}(\cdot,\partial\Omega):\Omega\to[0,+\infty[$ defined as

${\rm dist}(x,\partial\Omega)=\inf_{y\in\partial\Omega}{\left\|x-y\right\|}.$

The _inradius_ of $\Omega$ is defined as

$r(\Omega):=\sup_{x\in\Omega}{\rm dist}(x,\partial\Omega),$

and the _circumradius_ of $\Omega$ is defined as

$R(\Omega):=\min_{x\in\Omega}\max_{y\in\partial\Omega}{\left\|x-y\right\|}.$

Let us now introduce the support function of a convex set:

######  Definition 2.4.

Let $\Omega\in\mathcal{K}^{2}$. The _support_ function of $\Omega$ is defined
as

$p_{\Omega}(y):=\max_{x\in\Omega}(x\cdot y),\qquad y\in\mathbb{R}^{2}.$

In this paper, we will also consider the minimal width (or thickness) of a
convex set, that is to say the minimal distance between two parallel
supporting hyperplanes. More precisely, we have

######  Definition 2.5.

Let $\Omega\in\mathcal{K}^{2}$. The width of $\Omega$ in the direction
$y\in\mathbb{S}^{1}$ is defined as

$\omega_{\Omega}(y):=p_{\Omega}(y)+p_{\Omega}(-y)$

and the _minimal width_ of $\Omega$ as

$\omega(\Omega):=\min\\{\omega_{\Omega}(y)\,|\;y\in\mathbb{S}^{1}\\}.$

We introduce the inner parallel set of a convex set $\Omega$.

######  Definition 2.6.

Let $\Omega$ be a bounded and convex set. The _inner parallel set_ of $\Omega$
at distance $t\in[0,r(\Omega)]$ is

$\Omega_{-t}:=\\{x\in\Omega\,:\,{\rm dist}(x,\partial\Omega)\geq t\\}.$

######  Remark 2.7.

We remark that, by Definition 2.6, we have

$\Omega_{-t}=\Omega\sim tB_{1}.$

Moreover, we observe that for any $y\in\mathbb{S}^{1}$ and for every
$\Omega,K\in\mathcal{K}^{2}$ , one has

$p_{\Omega\sim K}(y)\leq p_{\Omega}(y)-p_{K}(y),$

see e.g. [37, Section 3.1, page 148].

Therefore, in the case $K=tB_{1}$, this reads

(4) $p_{\Omega_{-t}}(y)\leq p_{\Omega}(y)-t.$

Moreover, as it is observed in [26, Proposition 3.2], one has

(5) $R(\Omega+K)\leq R(\Omega)+R(K)$

with equality if $K=tB_{1}$.

We are now in position to prove the following lemma:

######  Lemma 2.8.

Let $\Omega\in\mathcal{K}^{2}$. We have for every $t\in[0,r(\Omega)]$:

(6) $\displaystyle r(\Omega_{-t})=r(\Omega)-t,$ (7) $\displaystyle
d(\Omega_{-t})\leq d(\Omega)-2t,$ (8)
$\displaystyle\omega(\Omega_{-t})\leq\omega(\Omega)-2t,$ (9) $\displaystyle
R(\Omega_{-t})\leq R(\Omega)-t,$ (10) $\displaystyle P(\Omega_{-t})\leq
P(\Omega)-2\pi t.$

###### Proof.

  * •

The proof of (6) can be found in [30, Lemma 1.4].

  * •

Let us now prove (7). Let $x_{t},y_{t}\in\Omega_{-t}$ be two diametrical
points of $\Omega_{-t}$ (i.e., such that $\|x_{t}-y_{t}\|=d(\Omega_{-t})$). We
denote by $x,y\in\Omega$ the points corresponding to the intersection of the
line containing $x_{t}$ and $y_{t}$ with the boundary of $\Omega$. We have

$d(\Omega)\geq\|x-y\|=\|x-x_{t}\|+\|x_{t}-y_{t}\|+\|y_{t}-y\|=\|x-x_{t}\|+d(\Omega_{-t})+\|y_{t}-y\|\geq
d(\Omega_{-t})+2t,$

where the last inequality is a consequence of the fact that
$x_{t},y_{t}\in\Omega_{-t}=\\{x\in\Omega\ |\ {\rm dist}(x,\partial\Omega)\geq
t\\}$.

  * •

The proof of (8) follows directly from the definition of the minimal width
(Definition 2.5) and (4).

  * •

We prove now (9). As observed in Remark 2.7, and, in particular, by formula
(5), for every $\Omega\in\mathcal{K}^{2}$, we have that
$R(\Omega+tB_{1})=R(\Omega)+t$. Thus, we have

$R(\Omega_{-t})=R(\Omega_{-t}+tB_{1})-t\leq R(\Omega)-t.$

The last inequality follows from the inclusion
$\Omega_{-t}+tB_{1}\subset\Omega$ and the monotonicity of the circumradius
with respect to inclusions.

  * •

Formula (10) can be obtained as a consequence of the classical Steiner formula

$P(K+tB_{1})=P(K)+2\pi t,$

see for example [37, Section 4.1], and the fact that the perimeter is monotone
with respect to the inclusion for convex sets. Indeed, since
$\Omega_{-t}+tB_{1}\subset\Omega$, we have $P(\Omega_{-t}+tB_{1})\leq
P(\Omega)$, which is equivalent by the Steiner formula to $P(\Omega_{-t})+2\pi
t\leq P(\Omega)$.

∎

The following Lemma will play a key role in the proof of Theorem 1.2.

######  Lemma 2.9.

Let $\Omega\in\mathcal{K}^{2}$. We assume that there exists a continuous
function $g^{\Omega}:[0,r(\Omega)]\rightarrow\mathbb{R}$ such that

(11) $\forall t\in[0,r(\Omega)],\ \ \ \ |\Omega_{-t}|\leq g^{\Omega}(t),\ \ \
\text{(resp. $|\Omega_{-t}|\geq g^{\Omega}(t)$)},$

and that

(12) $G_{\Omega}:=\Set{t\in[0,r(\Omega)]:g^{\Omega}(t)=\pi
t^{2}}\neq\emptyset.$

We have

(13) $h(\Omega)\geq\frac{1}{t_{g^{\Omega}}}\ \ \ \ \ \text{(resp.
$h(\Omega)\leq\frac{1}{t_{g^{\Omega}}}$)},$

where $t_{g^{\Omega}}$ is the smallest (resp. largest) solution to the
equation $g^{\Omega}(t)=\pi t^{2}$ on $[0,r(\Omega)]$.

###### Proof.

From [27, Theorem 1], we know that there exists a unique $t=t_{\Omega}>0$ such
that ${\left|\Omega_{-t}\right|}=\pi t^{2}$ and
$\displaystyle{h(\Omega)=1/t_{\Omega}}$. It is then clear that, if there
exists a function $g(t)$ such that (11) and (12) hold, then, the smallest
(resp. largest) element $t_{g^{\Omega}}\in G_{\Omega}$ must satisfy (13) (see
Figure 1). ∎

Figure 1. The idea of the proof of Lemma 2.9.

###  2.2. Extremal sets and their properties

In this Section, we describe special planar shapes that appear in the
statement of the main results.

Firstly, let us recall the definition of the form body of a convex set
$\Omega$, following [37]: a point $x\in\partial\Omega$ is called _regular_ if
the supporting hyperplane at $x$ is uniquely defined. The set of all regular
points of $\partial\Omega$ is denoted by ${\rm reg}(\Omega)$. We also let
$U(\Omega)$ denote the set of all outward pointing unit normals to
$\partial\Omega$ at points of ${\rm reg}(\Omega)$.

######  Definition 2.10.

The form body $\Omega^{\star}$ of a set $\Omega\in\mathcal{K}^{2}$ is defined
as

$\Omega^{\star}=\bigcap_{u\in U(\Omega)}\\{x\in\mathbb{R}^{2}:\,(x,u)\leq
1\\}.$

Convex sets that are homothetic to their form bodies will appear in the
following as extremal sets. In particular, a polygon whose incircle touches
all its sides is homothetic to its form body.

######  Definition 2.11.

A _stadium_ $\mathcal{R}$ is defined as the convex hull of the union of two
balls in $\mathbb{R}^{2}$ with the same radius, see Figure 2.

Figure 2. A stadium.

######  Definition 2.12.

The _symmetrical spherical slice_ $\mathcal{S}$ of diameter $d$ and width
$\omega\leq d$ is the convex set obtained by the intersection of a ball of
radius $d/2$ and a strip of width $\omega$ centered at the center of the ball,
see Figure 3.

Figure 3. A symmetrical spherical slice.

######  Definition 2.13.

A _two-cup body_ $\mathcal{C}$ is the convex hull of a ball in
$\mathbb{R}^{2}$ with two points that are symmetric with respect to the center
of the ball, see Figure 4. In particular, a two-cup body is homothetic to its
form body.

Figure 4. A two-cup body.

######  Definition 2.14.

A _subequilateral triangle_ $T_{I}$ is an isosceles triangle with two equal
angles greater than $\pi/3$.

The following class of sets (introduced in [43], see also [36]) represents a
way to pass in a continuous manner with respect to the Hausdorff distance from
the equilateral triangle to the Reuleaux triangle.

######  Definition 2.15.

A _Yamanouti set_ $Y$ is the convex hull of an equilateral triangle and three
circular arcs of equal radius, whose centers are each of the vertices of this
triangle; the radius of these circular arcs is less than the length of the
edge of the triangle, see Figure 5.

Figure 5. A Yamanouti set.

In [13], the authors define the smoothed regular nonagon as follows:

######  Definition 2.16.

Let $r>0$ and $2r<d<2\sqrt{3}r$. The _smoothed regular nonagon_ of inradius
$r$ and diameter $d$, that we denote by $\mathcal{N}_{r,d}$, is the convex set
enclosed in an equilateral triangle $T_{E}$ with barycenter in the origin and
such that $r(T_{E})=r$, obtained following the construction below. Let
$\eta_{i}$ the normal angles to the sides of $T_{E}$ and let

$\tau:=(3+\sqrt{d^{2}-3r^{2}})/2,\,\,\,\,\text{and}\,\,\,\,h:=\sqrt{d^{2}-\tau^{2}}.$

We now define the points $A_{i},B_{i},M_{i}$, for $i=1,2,3$:

$A_{i}:=r\begin{pmatrix}\cos{\eta_{i}}+h\sin{\eta_{i}}\\\
\sin{\eta_{i}}-h\cos{\eta_{i}}\end{pmatrix},\,\,B_{i}:=r\begin{pmatrix}\cos{\eta_{i}}-h\sin{\eta_{i}}\\\
\sin{\eta_{i}}+h\cos{\eta_{i}}\end{pmatrix},\,\,\,M_{i}:=r(1-\tau)\begin{pmatrix}\cos{\eta_{i}}\\\
\sin{\eta_{i}}\end{pmatrix}.$

We obtain $\mathcal{N}_{r,d}$ as follows (see Figure 6):

  * •

the points $A_{i},B_{i}$ and $M_{i}$, for $i=1,2,3$, belong to
$\partial\mathcal{N}$;

  * •

¿ $\textstyle B_{1}M_{3}$ and  ¿ $\textstyle M_{1}A_{3}$ are diametrically
opposed arcs of the same circle of diameter $d$, the same for the pairs  ¿
$\textstyle B_{2}M_{1}$ and  ¿ $\textstyle M_{2}A_{1}$ ,  ¿ $\textstyle
M_{2}B_{3}$ and  ¿ $\textstyle M_{3}A_{2}$ ;

  * •

the boundary contains the segments $\overline{A_{i}B_{i}}$, for $i=1,2,3$, and
the contact point $I_{i}$ with the incircle is the middle of the corresponding
segment.

Figure 6. A smoothed regular nonagon.

Let us now recall some sharp inequalities that we will need in the sequel.
Those classic estimates are obtained via the study of Blaschke–Santaló
diagrams involving the following geometric quantities: the perimeter, the
area, the inradius, the circumradius, the diameter and the minimal width.

Firstly, let us consider the diagram $(A,d,r)$. We have the following two
theorems.

######  Theorem 2.17 ([24] and [13]).

Let $\Omega\in\mathcal{K}^{2}$. We have

(14) ${\left|\Omega\right|}\geq
r(\Omega)\sqrt{d^{2}(\Omega)-4r^{2}(\Omega)}+r^{2}(\Omega)\left(\pi-2\arccos{\left(\frac{2r(\Omega)}{d(\Omega)}\right)}\right),$

where the equality holds if and only if $\Omega$ is a two-cup body.

######  Theorem 2.18 ([13], Theorem 2).

Let $\Omega\in\mathcal{K}^{2}$. We have

(15)
${\left|\Omega\right|}\leq\psi\left(d\left(\Omega\right),r(\Omega)\right),$

where

(16)
$\psi(d,r):=\begin{cases}\displaystyle{\frac{3\sqrt{3}r}{2}(\sqrt{d^{2}-3r^{2}}-r)+\frac{3d^{2}}{2}\left(\frac{\pi}{3}-\arccos{\left(\frac{\sqrt{3}r}{d}\right)}\right)},&\text{if}\,\,\,d\leq
rD^{*}\vspace{1mm}\\\
\displaystyle{r\sqrt{d^{2}-4r^{2}}+\frac{d^{2}}{2}\arcsin{\left(\frac{2r}{d}\right)}},&\text{if}\,\,\,d\geq
rD^{*}\end{cases}$

and $D^{*}$ is the unique number in $[2,2\sqrt{3}]$ for which the two
expressions of the function $\psi(d,1)$ are equal, i.e.,

$\displaystyle{\frac{3\sqrt{3}}{2}(\sqrt{(D^{\ast})^{2}-3}-1)+\frac{3(D^{\ast})^{2}}{2}\left(\frac{\pi}{3}-\arccos{\left(\frac{\sqrt{3}}{D^{\ast}}\right)}\right)}=\displaystyle{\sqrt{(D^{\ast})^{2}-4}+\frac{(D^{\ast})^{2}}{2}\arcsin{\left(\frac{2}{D^{\ast}}\right)}}.$

Moreover, if $d(\Omega)\leq r(\Omega)D^{*}$, we have the equality in (15) if
and only if $\Omega$ is a regular smoothed nonagon, while, if
$d(\Omega)>r(\Omega)D^{*}$, we have the equality if and only if $\Omega$ is a
symmetrical spherical slice.

As far as the diagram $(A,\omega,R)$ is concerned, we recall the following
theorems:

######  Theorem 2.19 ([24], Theorem 3).

Let $\Omega\in\mathcal{K}^{2}$. Then, it holds

(17) ${\left|\Omega\right|}\leq\chi(\omega(\Omega),R(\Omega)),$

where

(18)
$\chi(\omega(\Omega),R(\Omega)):=\frac{\omega(\Omega)}{2}\sqrt{4R(\Omega)^{2}-\omega(\Omega)^{2}}+2R(\Omega)^{2}\arcsin{\frac{\omega(\Omega)}{2R(\Omega)}},$

and the equality in (17) holds if and only if $\Omega$ is a symmetrical
spherical slice.

######  Theorem 2.20 ([24], Theorem 6).

Let $\Omega\in\mathcal{K}^{2}$. Then, if
$\omega(\Omega)\leq\frac{3}{2}R(\Omega)$, it holds

(19) $16{\left|\Omega\right|}^{6}\geq
R^{2}(\Omega)\omega^{2}(\Omega)\left(16{\left|\Omega\right|}^{4}-R^{2}(\Omega)\omega^{6}(\Omega)\right)$

and the equality holds if and only if $\Omega$ is a subequilateral triangle.

We recall the following inequality from the diagram $(R,r,\omega)$.

######  Theorem 2.21 ([23], Theorem 2).

Let $\Omega\in\mathcal{K}^{2}$. Then, it holds

(20)
$\left(4r(\Omega)-\omega(\Omega)\right)\left(\omega(\Omega)-2r(\Omega)\right)\leq\frac{2r^{3}(\Omega)}{R(\Omega)}$

and the equality holds if and only if $\Omega$ is an isosceles triangle.

######  Theorem 2.22 ([36], Section 10).

Let $\Omega\in\mathcal{K}^{2}$. Then, it holds

(21) $\omega(\Omega)\leq R(\Omega)+r(\Omega),$

where the equality is achieved by any set of constant width (i.e., the set
$\Omega$ is such that the function $\omega_{\Omega}$, defined in Definition
2.5, is constant).

The following theorem deals with the $(A,r,R)$ diagram.

######  Theorem 2.23 ([24], Theorems 1 and 2).

Let $\Omega\in\mathcal{K}^{2}$. Then, it holds

(22) ${\left|\Omega\right|}\geq
2r(\Omega)\left(\sqrt{R(\Omega)^{2}-r(\Omega)^{2}}+r\arcsin{\frac{r(\Omega)}{R(\Omega)}}\right),$

and the equality in (22) holds if and only if $\Omega$ is a two-cup body.
Moreover, we have

(23) ${\left|\Omega\right|}\leq\varphi(R(\Omega),r(\Omega)),$

where

(24)
$\varphi(R(\Omega),r(\Omega)):=2\left(r\sqrt{R(\Omega)^{2}-r(\Omega)^{2}}+R^{2}(\Omega)\arcsin{\frac{r(\Omega)}{R(\Omega)}}\right),$

and the equality in (24) holds if and only if $\Omega$ is a symmetrical
spherical slice.

Now we quote two inequalities concerning the $(A,\omega,r)$ and $(P,\omega,r)$
diagrams.

######  Theorem 2.24 ([24], Theorem 5).

Let $\Omega\in\mathcal{K}^{2}$. We have

(25)
$(\omega(\Omega)-2r(\Omega))^{2}(4r(\Omega)-\omega(\Omega)){\left|\Omega\right|}^{2}\leq
r^{4}(\Omega)\omega^{3}(\Omega)$

and

(26)
$(\omega(\Omega)-2r(\Omega))^{2}(4r(\Omega)-\omega(\Omega))P^{2}(\Omega)\leq
4r(\Omega)^{2}\omega^{3}(\Omega)$

In both inequalities, the equality holds if and only if $\Omega$ is a
subequilateral triangle.

Finally, we recall this result from the $(d,\omega,r)$ diagram.

######  Theorem 2.25 ([21], Theorem 1-2).

Let $\Omega\in\mathcal{K}^{2}$. We have

(27)
$d^{2}(\Omega)(\omega(\Omega)-2r(\Omega))^{2}(4r(\Omega)-\omega(\Omega))\leq
4r^{4}(\Omega)\omega(\Omega),$

where the equality holds if and only if $\Omega$ is a subequilateral triangle
$T_{I}$, and

(28) $\omega(\Omega)-r(\Omega)\leq\dfrac{\sqrt{3}}{3}d(\Omega),$

where the equality holds if $\Omega$ is a Yamanouti set.

###  2.3. Inequalities relating the Cheeger constant to one geometric quantity

In the following paragraph, we state the inequalities relating the Cheeger
constant to one of the geometric quantities taken into account, obtained by
combining classical results.

######  Proposition 2.26.

Let $\Omega\in\mathcal{K}^{2}$. We have

  1. (1)

$h(\Omega)\geq 2\sqrt{\frac{\pi}{|\Omega|}}$.

  2. (2)

$h(\Omega)\geq\frac{4\pi}{P(\Omega)}$.

  3. (3)

$h(\Omega)\geq\frac{4}{d(\Omega)}$.

  4. (4)

$\frac{1}{r(\Omega)}\leq h(\Omega)\leq\frac{2}{r(\Omega)}$.

  5. (5)

$h(\Omega)\geq\frac{2}{R(\Omega)}$.

In $(1)-(2)-(3)-(5)$, the equality is achieved by balls, while the one in the
upper bound in $(4)$ is achieved by balls and the lower bound is
asymptotically an equality for thinning vanishing stadiums.

###### Proof.

  * •

We have, by using the isoperimetric inequality $P(E)\geq 2\sqrt{\pi|E|}$,

(29) $h(\Omega)=\inf_{\underset{E\subset\Omega,\ |E|>0}{E\text{ is measurable
}}}\frac{P(E)}{|E|}\geq\inf_{\underset{E\subset\Omega,\ |E|>0}{E\text{ is
measurable
}}}\frac{2\sqrt{\pi}\sqrt{|E|}}{|E|}=\inf_{\underset{E\subset\Omega,\
|E|>0}{E\text{ is measurable
}}}\frac{2\sqrt{\pi}}{\sqrt{|E|}}=\frac{2\sqrt{\pi}}{\sqrt{|\Omega|}}.$

  * •

By using (29) and the isoperimetric inequality, we have

$h(\Omega)\geq\frac{2\sqrt{\pi}}{\sqrt{|\Omega|}}\geq\frac{4\pi}{P(\Omega)}.$

  * •

By using (29) and the isodiametric inequality
$|\Omega|\leq\frac{\pi}{4}d(\Omega)^{2}$, we have

$h(\Omega)\geq\frac{2\sqrt{\pi}}{\sqrt{|\Omega|}}\geq\frac{2\sqrt{\pi}}{\sqrt{\frac{\pi}{4}d(\Omega)^{2}}}=\frac{4}{d(\Omega)}.$

  * •

For the upper bound, we have

$h(\Omega)=\inf_{\underset{E\subset\Omega,\ |E|>0}{E\text{ is measurable
}}}\frac{P(E)}{|E|}\leq\frac{P(B_{r(\Omega)})}{|B_{r(\Omega)}|}=\frac{2}{r(\Omega)},$

where $B_{r(\Omega)}$ is a ball of radius $r(\Omega)$ inscribed in $\Omega$.
As, for the lower bound, we have

$h(\Omega)=\inf_{\underset{E\subset\Omega,\ |E|>0}{E\text{ is measurable
}}}\frac{P(E)}{|E|}\geq\inf_{\underset{E\subset\Omega,\ |E|>0}{E\text{ is
measurable }}}\frac{1}{r(E)}\geq\frac{1}{r(\Omega)},$

where we use the inequalities $|E|<r(E)P(E)$ (see [38]) and $r(E)\leq
r(\Omega)$.

  * •

By using (29) and the inequality $|\Omega|\leq\pi R(\Omega)^{2}$ (see [38]),
we have

$h(\Omega)\geq\frac{2\sqrt{\pi}}{\sqrt{|\Omega|}}\geq\frac{2}{R(\Omega)}.$

∎

##  3\. Numerical results and Blaschke–Santaló diagrams

In this Section, we introduce numerical tools, that we use to obtain more
information on the diagrams and state some conjectures.

###  3.1. Generation of random convex polygons

We want to provide a numerical approximation of the diagrams studied in
Section 5. To do so, a natural idea is to generate a large number of convex
bodies (more precisely convex polygons). Subsequently, for each of these sets,
we calculate the involved functionals. Nevertheless, the task of (properly)
generating random convex polygons is quite challenging and of intrinsic
interest. The main difficulty is that one wants to design an efficient and
fast algorithm that allows to obtain a uniform distribution of the generated
random convex polygons. For clarity, let us discuss two different (naive)
approaches:

  * •

one easy way to generate random convex polygons is by rejection sampling. We
generate a random set of points in a square; if they form a convex polygon, we
return it, otherwise we try again. Unfortunately, the probability of a set of
$n$ points uniformly generated inside a given square to be in convex position
is equal to $p_{n}=\left(\frac{\binom{2n-2}{n-1}}{n!}\right)^{2}$, see [41].
Thus, the random variable $X_{n}$ corresponding to the expected number of
iterations needed to obtain a convex distribution follows a geometric law of
parameter $p_{n}$, which means that its expectation is given by
$\mathbb{E}(X_{n})=\frac{1}{p_{n}}=\left(\frac{n!}{\binom{2n-2}{n-1}}\right)^{2}$.
For example, if $N=20$, the expected number of iterations is approximately
equal to $2.10^{9}$, and, since one iteration is performed in an average of
$0.7$ seconds, this means that the algorithm will need about $50$ years to
provide one convex polygon with $20$ sides;

  * •

another natural approach is to generate random points and take their convex
hull. This method is quite fast, as one can compute the convex hull of $N$
points in a $\mathcal{O}(N\log(N))$ time (see [2] for example), but it is not
quite relevant since it yields to a biased distribution.

In order to avoid the issues stated above, we use an algorithm presented in
[35], that is based on the work of P. Valtr [41], where the author computes
the probability of a set of $n$ points uniformly generated inside a given
square to be in convex position. The author remarks (in Section 4) that the
proof yields a fast and non-biased method to generate random convex sets
inside a given square. We also refer to [35] for a nice description of the
steps of the method and a beautiful animation where one can follow each step;
one can also find an implementation of Valtr’s algorithm in Java that we
decided to translate in Matlab.

To obtain an approximation of the Blaschke–Santaló diagram, we generate
$100.000$ random convex polygons of unit area and number of sides between $3$
and $30$, for which we compute the involved functionals. We then obtain clouds
of dots that provide approximations of the diagrams. This approach has been
used in several works, we refer for example to [3], [14] and [18]. For a new
and efficient method of the approximation of Blaschke–Santaló diagrams based
on centroidal Voronoi tessellations, we refer to the recent work [7].

###  3.2. About the computation of the functionals

Let us give few details on the numerical computation of the functionals
involved in the paper.

  * •

The Cheeger constant is computed by using a code implemented by Beniamin
Bogosel in [6] based on the characterization of the Cheeger sets of planar
convex sets given in [27] and the Clipper toolbox, a very good implementation
of polygon offset computation by Agnus Johnson.

  * •

The inradius is also computed by using the Clipper toolbox and the fact that
$r(\Omega)$ is the smallest solution to the equation $|\Omega_{-t}|=0$.

  * •

The diameter is computed via a fast method of computation, which consists in
finding all antipodal pairs of points and looking for the diametrical one
between them. This is classically known as Shamos algorithm [34].

  * •

The area is computed by using Matlab’s function "polyarea".

  * •

The minimum width of a polygon $\Omega$ of vertices $\\{A_{1},\cdots,A_{N}\\}$
is computed by using the following formula

$\omega(\Omega)=\min_{i\in\llbracket 1,N\rrbracket}\max_{j\in\llbracket
1,N\rrbracket}{\rm dist}(A_{j},(A_{i}A_{i+1})),$

where ${\rm dist}(A_{j},(A_{i}A_{i+1}))$ corresponds to the distance between
the point $A_{j}$ and the line $(A_{i}A_{i+1})$ (with the convention
$A_{N+1}:=A_{1}$).

  * •

The circumradius of a convex set $\Omega$ can be written as follows

$R(\Omega)=\min_{c\in\Omega}\max_{x\in\Omega}\|c-x\|.$

It is computed by using Matlab’s routine "fminmax".

##  4\. Proof of the main Results

###  4.1. Proof of Theorem 1.1

We start this Section by proving the existence results stated in Theorem 1.1.

###### Proof of the existence.

Let us consider the minimization problem of the Cheeger constant in the
classes of sets $(1)-(13)$; the maximization problem can be dealt with
similarly.

For all of these classes of sets, in order to prove the existence of the
solution, we consider a minimizing sequence $(\Omega_{k})_{k\in\mathbb{N}}$
and we prove that it satisfies the hypothesis of the Blaschke Selection
Theorem (see [37, Theorem 1.8.7]), that is to say, its boundedness up to
translations. Concerning the class of sets involving a diameter or a
circumradius constraint, it is clear that, up to a translation, the minimizing
sequence is contained in a sufficiently big ball. So it remains to study the
problem in $\mathcal{K}^{2}_{P,r}$, $\mathcal{K}^{2}_{\omega,A}$,
$\mathcal{K}^{2}_{r,\omega}$ and $\mathcal{K}^{2}_{\omega,P}$. Concerning
$\mathcal{K}^{2}_{P,r}$ and $\mathcal{K}^{2}_{\omega,P}$, we know from [38]
that $P=P(\Omega_{k})>2d(\Omega_{k})$, for every $k$, so the sequence of the
diameters $d(\Omega_{k})$ is equibounded and, consequently, there exists a
sufficiently big ball containing the sequence $(\Omega_{k})_{k}$. As far as
$\mathcal{K}^{2}_{r,\omega}$ is concerned, it is possible to prove the
boundness of the minimizing sequence whenever
$\omega(\Omega_{k})>2r(\Omega_{k})$, indeed it holds (see [38])

$d(\Omega_{k})\leq\frac{\omega^{2}(\Omega_{k})}{2(\omega(\Omega_{k})-2r(\Omega_{k}))}.$

For the last class $\mathcal{K}^{2}_{\omega,A}$, from [38], we know that, if
$2\omega(\Omega_{k})\leq\sqrt{3}d(\Omega_{k})$, then

$2{\left|\Omega_{k}\right|}\geq\omega(\Omega_{k})d(\Omega_{k}),$

and, also in this case, the boundedness follows.

So, for every class of sets considered, the Blaschke selection theorem ensures
us that, up to a subsequence, $(\Omega_{k})$ converges with respect to the
Hausdorff distance to a convex set $\Omega^{*}$; it remains only to prove that
this set belongs to the relative class of admissible sets. We observe that all
the considered constraints are stable for the Hausdorff convergence. In
particular, the stability of the inradius is proved in in [13] and the one of
the diameter is proved in [20], meanwhile the stability of the area, the
perimeter and the width may be found in [37].

It only remains to show that the circumradius is continuous with respect to
the Hausdorff distance in the class of admissible sets having a circumradius
constraint. Since $R(\Omega_{k})=R$, for all $k\in\mathbb{N}$, we have
$\Omega_{k}\subseteq B_{R}$. Using the stability of the Hausdorff convergence
for the inclusion (see [20, Proposition 2.2.17]), we have that
$\Omega^{*}\subseteq B_{R}$, and consequently $R(\Omega^{*})\leq R$. By
contradiction, let us suppose that $R(\Omega^{*})<R$, so there exists
$\overline{R}>0$ such that $R(\Omega^{*})<\overline{R}<R$ and so
$\Omega^{*}\subseteq B_{\overline{R}}$. Therefore, by the Hausdorff
convergence, for sufficiently large $k$, $\Omega_{k}\subseteq
B_{\overline{R}}$, but this would imply $R\leq\overline{R}$, which is absurd.

In order to conclude, we observe that in all the above cases, the set
$\Omega^{*}$ cannot be a segment, that is to say, that the minimizing sequence
$(\Omega_{k})_{k\in\mathbb{N}}$ cannot degenerate loosing one dimension. If we
are working in a class of sets involving an inradius or width constraint,
then, it is clear that, thanks to the continuity of the inradius and width
under the Hausdorff convergence, there exists, up to a translation, a
sufficiently small ball contained in the minimizing sequence.

Moreover, in the case $\mathcal{K}^{2}_{d,A}$ and $\mathcal{K}^{2}_{R,A}$, the
non-degeneration is ensured by the continuity of the area under Hausdorff
distance and the equiboundedness of the diameter. On the other hand, if we
consider the minimization problem in $\mathcal{K}^{2}_{R,d}$,
$\mathcal{K}^{2}_{P,d}$ and $\mathcal{K}^{2}_{P,R}$, the inradius can be
bounded from below by a positive quantity. In [36, Section 9], it is proved
that

$r(\Omega_{k})\geq\dfrac{d^{2}(\Omega_{k})\sqrt{4R^{2}(\Omega_{k})-d^{2}(\Omega_{k})}}{2R(\Omega_{k})\left(2R(\Omega_{k})+\sqrt{4R^{2}(\Omega_{k})-d^{2}(\Omega_{k})}\right)}.$

In [19, Section 3], it is proved that

$r(\Omega_{k})\geq\frac{P(\Omega_{k})}{4}-\frac{d(\Omega_{k})}{2},$

which also yields to

$r(\Omega_{k})\geq\dfrac{P(\Omega_{k})}{4}-R(\Omega_{k}),$

as $d(\Omega_{k})\leq 2R(\Omega_{k})$.

Recalling now that the Cheeger constant is continuous with respect to the
Hausdorff convergence when the sets do not degenerate to a segment (see [33,
Proposition 3.1]), the existence part of the theorem is proved. ∎

###  4.2. Proof of Theorem 1.2

The following paragraphs of this Section are dedicated to the proof of the
explicit bounds and their sharpness.

####  4.2.1. The triplet $(P,h,r)$

######  Proposition 4.1.

Let $\Omega\in\mathcal{K}^{2}$. Then, it holds

(30) $\frac{1}{r(\Omega)}+\frac{\pi}{P(\Omega)-\pi r(\Omega)}\leq
h(\Omega)\leq\frac{1}{r(\Omega)}+\sqrt{\frac{2\pi}{P(\Omega)r(\Omega)}},$

where the equality is achieved by sets that are homothetic to their form
bodies in the upper bound and by the stadiums in the lower bound.

###### Proof.

We combine the following classical convex geometric inequalities that can be
found in [38]

(31) $\frac{P(\Omega)r(\Omega)}{2}\leq|\Omega|\leq r(\Omega)(P(\Omega)-\pi
r(\Omega)),$

with the estimates (2) to obtain the optimal inequalities (30).

The upper bound in (2) is an equality for sets that are homothetic to their
form bodies, since both the upper bound in (2) and the lower one in (31) are
sharp for such sets.

The lower bound an equality for stadiums, since both the lower bound in (2)
and the upper one in (31) are sharp for those sets. ∎

####  4.2.2. The triplet $(d,h,r)$

In the following, we will denote by $\mathcal{S}_{r,d}$ the symmetrical
spherical slice of inradius $r$ and diameter $d$ and by $\mathcal{N}_{r,d}$
the regular smoothed nonagon with the same inradius and diameter, see
Definitions 2.16 and 2.12.

######  Proposition 4.2.

Let $\Omega\in\mathcal{K}^{2}$. Then, it holds

(32)
$h(\Omega)\leq\frac{1}{r(\Omega)}+\sqrt{\frac{\pi}{r(\Omega)\sqrt{d^{2}(\Omega)-4r^{2}(\Omega)}+r^{2}(\Omega)\left(\pi-2\arccos{\left(\frac{2r(\Omega)}{d(\Omega)}\right)}\right)}},$

where the equality is achieved if and only if $\Omega$ is a symmetric two-cup
body. Moreover, we have

(33) $h(\Omega)\geq\frac{1}{t_{g_{1}^{\Omega}}},$

where $t_{g_{1}^{\Omega}}$ is the smallest solution to the equation

(34) $g_{1}^{\Omega}(t):=\psi(d(\Omega)-2t,r(\Omega)-t)=\pi t^{2}$

on the interval $[0,r(\Omega)]$ and the function $\psi$ is defined in (16).

Moreover, there exists $D_{0}$ such that the problem

$\min\\{h(\Omega)\ |\ \Omega\in\mathcal{K}^{2}_{d,r}\\}$

is solved by the smoothed nonagon $\mathcal{N}_{r,d}$ if $d<rD_{0}$ and by the
slice $\mathcal{S}_{r,d}$ if $d\geq rD_{0}$.

###### Proof.

To prove (32), one has just to combine the upper bound in (2), which is an
equality for sets that are homothetic to their form bodies, and (14), which is
an equality for and only for symmetric two-cup bodies, that are also
homothetic to their form bodies.

Let us now prove (33). As in Theorem 2.18 (see [13]), for any constant
parameter $r>0$, we consider the function

$\Psi_{r}(x)=\begin{cases}f_{r}(x):=\displaystyle{\frac{3\sqrt{3}r}{2}(\sqrt{x^{2}-3r^{2}}-r)+\frac{3x^{2}}{2}\left(\frac{\pi}{3}-\arccos{\left(\frac{\sqrt{3}r}{x}\right)}\right)},&\text{if}\,\,\,x\leq
rD^{*}\vspace{1mm}\\\
g_{r}(x):=\displaystyle{r\sqrt{x^{2}-4r^{2}}+\frac{x^{2}}{2}\arcsin{\left(\frac{2r}{x}\right)}},&\text{if}\,\,\,x\geq
rD^{*}\end{cases}$

defined for $x\geq 2r$. The function $\Psi_{r}$ is strictly increasing.
Indeed, we have for every $x<rD^{*}$

$f_{r}^{\prime}(x)=\frac{3\sqrt{3}rx}{2\sqrt{x^{2}-3r^{2}}}+3x\left(\frac{\pi}{3}-\arccos\left(\frac{\sqrt{3}r}{x}\right)\right)-\frac{3\sqrt{3}r}{2\sqrt{1-\frac{3r^{2}}{x^{2}}}}=3x\left(\frac{\pi}{3}-\arccos\left(\frac{\sqrt{3}r}{x}\right)\right),$

and for every $x>rD^{*}$,

$g_{r}^{\prime}(x)=-\frac{r}{\sqrt{1-\frac{4r^{2}}{x^{2}}}}+\frac{rx}{\sqrt{x^{2}-4r^{2}}}+x\arcsin\left(\frac{2r}{x}\right)=x\arcsin\left(\frac{2r}{x}\right)>0.$

Thus, the function $f_{r}$ is increasing on $[2r,2\sqrt{3}r]$ and is
decreasing on $[2\sqrt{3}r,+\infty)$ and the function $g_{r}$ is increasing on
$[2r,+\infty)$. Moreover, we have by Theorem 2.18 that $D^{*}\leq 2\sqrt{3}$.
So, the function $f_{r}$ is increasing on the sub-interval $[2r,rD^{*}]$ and,
since $f_{r}(rD^{*})=g_{r}(rD^{*})=\Psi_{r}(D^{*})$, the continuous function
$\Psi_{r}$ is increasing on $[2r,+\infty)$.

Let $t\in[0,r(\Omega)]$, by applying the result of Theorem 2.18 on the convex
set $\Omega_{-t}$, we have

$|\Omega_{-t}|\leq\Psi_{r(\Omega_{-t})}(d(\Omega_{-t}))=\Psi_{r(\Omega)-t}(d(\Omega_{-t}))\leq\Psi_{r(\Omega)-t}(d(\Omega)-2t)=:g_{1}^{\Omega}(t),$

where, we use the monotonicity of the function $\Psi_{r(\Omega)-t}$ and the
estimates (6) and (7).

Now, using Lemma 2.9, we have the following bound for the Cheeger constant

(35) $h(\Omega)\geq\frac{1}{t_{g_{1}^{\Omega}}},$

where $t_{g_{1}^{\Omega}}$ is the smallest solution to the equation
$g_{1}^{\Omega}(t)=\pi t^{2}$ on the interval $[0,r(\Omega)]$.

It remains to prove that for every $r>0$ and $d\geq 2r$, there exists a convex
set of inradius $r$ and diameter $d$ such that (35) is an equality. If $d=2r$
then $\Omega$ is a ball and thus the equality is trivial. Let us now consider
the following two cases:

  * •

If $d\geq rD^{*}$, we have, for every $t\in[0,r)$,

(36) $|(\mathcal{S}_{r,d})_{-t}|=|\mathcal{S}_{r-t,d-2t}|=\Psi_{r-t}(d-2t),$

where the first equality is a consequence of the equality
$(\mathcal{S}_{r,d})_{-t}=\mathcal{S}_{r-t,d-2t}$ and the second one is a
consequence of [13, Theorem 2] and of the following estimate

$d((\mathcal{S}_{r,d})_{-t})=d-2t>rD^{*}-2t>rD^{*}-tD^{*}=(r-t)D^{*}=r((\mathcal{S}_{r,d})_{-t})D^{*},$

where we used $D^{*}\approx 2,3888>2$ (see [13, Theorem 2]). Thus, we have by
(36)

$h(\mathcal{S}_{r,d})=\frac{1}{t_{g_{1}^{\Omega}}},$

with

$r(\mathcal{S}_{r,d})=r\ \ \text{and}\ \ d(\mathcal{S}_{r,d})=d.$

  * •

If $d\in(2r,rD^{*}]$, we consider $t^{*}:=\frac{D^{*}r-d}{D^{*}-2}$, that is
the value for which the graphs of the functions
$t\longmapsto|(\mathcal{N}_{r,d})_{-t}|$ and
$t\longmapsto|(\mathcal{S}_{r,d})_{-t}|$ intersect each other, see Figure 7.
We note that $(\mathcal{N}_{r,d})_{-t}=\mathcal{N}_{r-t,d-2t}$ for every
$t\in[0,t^{*}]$.

We introduce the quantity $D_{0}\in(2,D^{*})$ as the (unique) value in the
interval $(2,D^{*})$ for which the graph of the (decreasing) function
$x\longmapsto\frac{D^{*}-x}{D^{*}-2}$ intersects the graph of the
(increasing111The function $x\longmapsto\frac{1}{h(\mathcal{N}_{1,x})}$ is
increasing because $x\longmapsto\mathcal{N}_{1,x}$ is increasing for the
inclusion, meanwhile the Cheeger constant is decreasing with respect to the
inclusion.) one $x\longmapsto\frac{1}{h(\mathcal{N}_{1,x})}$. As shown in
Figure 7, we have the following cases:

    * –

If $\frac{d}{r}<D_{0}$, i.e., $t^{*}>\frac{1}{h(\mathcal{N}_{r,d})}$, we have
$h(\mathcal{N}_{r,d})=\frac{1}{t_{g_{1}^{\Omega}}}$, which means that in this
case the smoothed nonagon $\mathcal{N}_{r,d}$ provides the equality in (33).

    * –

If $\frac{d}{r}>D_{0}$, i.e., $t^{*}<\frac{1}{h(\mathcal{N}_{r,d})}$, we have
$h(\mathcal{S}_{r,d})=\frac{1}{t_{g_{1}^{\Omega}}}$, which means that in this
case the slice $\mathcal{S}_{r,d}$ provides the equality in (33).

    * –

If $\frac{d}{r}=D_{0}$, i.e., $t^{*}=\frac{1}{h(\mathcal{N}_{r,d})}$, we have
$h(\mathcal{N}_{r,d})=h(\mathcal{S}_{r,d})=\frac{1}{t_{g_{1}^{\Omega}}}$,
which means that in this case both the smoothed nonagon $\mathcal{N}_{r,d}$
and the slice $\mathcal{S}_{r,d}$ provide the equality in (33).

So, the proof is concluded.

$t$$\frac{1}{h(\mathcal{S}_{r,d})}$$\frac{1}{h(\mathcal{N}_{r,d})}$$t^{*}$$|\mathcal{N}_{r,d}|$$|\mathcal{S}_{r,d}|$$r$$\bullet$Graph
of the function $t\longmapsto\Psi_{r-t}(d-2t)$.Graph of the function
$t\longmapsto\pi t^{2}$.Graph of the function
$t\longmapsto|(\mathcal{N}_{r,d})_{-t}|=|\mathcal{N}_{r-t,d-2t}|$.Graph of the
function $t\longmapsto|(\mathcal{S}_{r,d})_{-t}|=|\mathcal{S}_{r-t,d-2t}|$.

$t$$\frac{1}{h(\mathcal{S}_{r,d})}$$\frac{1}{h(\mathcal{N}_{r,d})}$$t^{*}$$|\mathcal{N}_{r,d}|$$|\mathcal{S}_{r,d}|$$r$$\bullet$

Figure 7. Different cases of equality in inequality (33).

∎

######  Remark 4.3.

We note that the symmetrical slices and the smoothed nonagons are not the only
sets solving the shape optimization problem $\min\\{h(\Omega)\ |\
\Omega\in\mathcal{K}^{2}_{d,r}\\}$. Indeed, if for example we consider a
spherical slice $\mathcal{S}$ and denote by $C_{\mathcal{S}}$ its Cheeger set,
we have $h(\mathcal{S})=h(C_{\mathcal{S}})$ and, by the explicit
characterization of the Cheeger sets given in [27, Theorem 1], we have

$r(C_{\mathcal{S}})=r\left(\mathcal{S}_{-\frac{1}{h(\mathcal{S})}}+\frac{1}{h(\mathcal{S})}B_{1}\right)=r\left(\mathcal{S}_{-\frac{1}{h(\mathcal{S})}}\right)+\frac{1}{h(\mathcal{S})}=r(\mathcal{S})-\frac{1}{h(\mathcal{S})}+\frac{1}{h(\mathcal{S})}=r(S)$

and

$d(C_{\mathcal{S}})=d(\mathcal{S}),$

meanwhile $\mathcal{S}\neq C_{\mathcal{S}}$, which proves the non-uniqueness
of the solution of the minimization problem

$\min\\{h(\Omega)\ |\ \Omega\in\mathcal{K}^{2}_{d,r}\\}.$

######  Remark 4.4.

We give the following explicit lower bound. In [38], it is proved that

$|\Omega|<2d(\Omega)r(\Omega).$

By applying the strategy of Lemma 2.9, we obtain that

(37)
$h(\Omega)\geq\dfrac{4-\pi}{d(\Omega)+2r(\Omega)-\sqrt{(d(\Omega)+2r(\Omega))^{2}-2(4-\pi)d(\Omega)r(\Omega)}}.$

This estimate is asymptotically achieved by spherical slices with increasing
diameter, as shown in Figure 8.

Figure 8. The explicit inequality (37) and the lower part of the boundary of
the diagram $(d,h,r)$.

####  4.2.3. The triplet $(R,h,r)$

######  Proposition 4.5.

Let $\Omega\in\mathcal{K}^{2}$. We have

(38) $h(\Omega)\geq\frac{1}{t_{g_{2}^{\Omega}}},$

where $t_{g_{2}^{\Omega}}$ is the smallest solution of the equation

(39)
${g_{2}^{\Omega}}(t):=2\left(\left(r-t\right)\sqrt{(R(\Omega)-t)^{2}-(r(\Omega)-t)^{2}}+(R(\Omega)-t)^{2}\arcsin{\left(\frac{r(\Omega)-t}{R(\Omega)-t}\right)}\right)=\pi
t^{2}.$

The equality in (38) is achieved if and only if $\Omega$ is a symmetrical
spherical slice. Moreover,

(40)
$h(\Omega)\leq\frac{1}{r(\Omega)}+\sqrt{\frac{\pi}{2r(\Omega)\left(\sqrt{R(\Omega)^{2}-r(\Omega)^{2}}+r(\Omega)\arcsin\left(\frac{r(\Omega)}{R(\Omega)}\right)\right)}},$

where the equality in (40) is achieved by two-cup bodies.

###### Proof.

In order to prove (38), we apply the result of Lemma 2.9. Let us introduce the
function

$\varphi:(R,r)\longmapsto
2\left(r\sqrt{R^{2}-r^{2}}+R^{2}\arcsin{\frac{r}{R}}\right),$

which is increasing with respect to the first variable, indeed

$\frac{\partial\varphi}{\partial R}(R,r)=2R\arcsin\left(\frac{r}{R}\right)>0.$

By applying (23) (where the equality holds only for symmetrical spherical
slices), we have, for every $t\in[0,r(\Omega)]$,

$|\Omega_{-t}|\leq\varphi(R(\Omega_{-t}),r(\Omega_{-t}))=\varphi(R(\Omega_{-t}),r(\Omega)-t)\leq\varphi(R(\Omega)-t,r(\Omega)-t)=:{g_{2}^{\Omega}}(t),$

where the last inequality is a consequence of the monotonicity of the function
$R\longmapsto\varphi(R,r)$ and of the fact that $R(\Omega_{-t})\leq
R(\Omega)-t$ (see Lemma 2.8). Finally, we conclude by applying the result of
Lemma 2.9.

In order to prove (40), we combine the upper bound in (2) and inequality (22).
As far as the sharpness of (40) is concerned, we observe that (2) is sharp on
sets that are homothetic to their form bodies and (22) is attained by
symmetric two-cup bodies, that are also sets that are homothetic to their form
bodies. ∎

######  Remark 4.6.

We have the following explicit lower bound

(41)
$h(\Omega)\geq\frac{4-\pi}{2(R(\Omega)+r(\Omega))-\sqrt{4(R(\Omega)+r(\Omega))^{2}-4(4-\pi)R(\Omega)r(\Omega)}}.$

The inequality can be obtained by combining

${\left|\Omega\right|}\leq 4R(\Omega)r(\Omega),$

see [19], and the strategy from Lemma 2.9. Moreover, the inequality is
asymptotically achieved by spherical slices with increasing circumradius, see
Figure 9.

Figure 9. The explicit inequality (41) and the lower part of the boundary of
the diagram $(R,h,r)$.

###  4.3. Explicit description of the Blaschke-Santaló diagrams

We denote by $\mathcal{D}_{1}$, $\mathcal{D}_{2}$ and $\mathcal{D}_{3}$ the
Blaschke Santaló diagram respectively corresponding to the triplets $(P,h,r)$,
$(d,h,r)$ and $(R,h,r)$. We have defined the quantities $t_{g_{1}^{\Omega}}$
and $t_{g_{2}^{\Omega}}$ respectively as the smallest solution on
$[0,r(\Omega)]$ of equations (34) and (39). We observe that
$t_{g_{1}^{\Omega}}$ depends on $r(\Omega)$ and $d(\Omega)$, and
$t_{g_{2}^{\Omega}}$ depends on $r(\Omega)$ and $R(\Omega)$.

In the following proposition, we are keeping the inradius $r(\Omega)=1$ and
consider different values of the remaining variables: diameter for the diagram
$\mathcal{D}_{2}$ and circumradius for $\mathcal{D}_{3}$. We then use the
notation:

$t_{g_{1}^{\Omega}}=t_{g_{1}}(x)\ \ \text{when $d(\Omega)=x$\ \ and}\quad
t_{g_{2}^{\Omega}}=t_{g_{2}}(x)\ \ \text{when $R(\Omega)=x$}.$

######  Proposition 4.7.

We obtain the following description of the Blaschke-Santaló diagrams

(42) $\mathcal{D}_{1}:=\left\\{\left(P(\Omega),h(\Omega)\right)\ |\
\Omega\in\mathcal{K}^{2},\,r(\Omega)=1\right\\}=\left\\{(x,y)\ |\ x\geq 2\pi\
\ \text{and}\ \ 1+\frac{\pi}{x-\pi}\leq y\leq
1+\sqrt{\frac{2\pi}{x}}\right\\}.$

(43) $\mathcal{D}_{2}:=\left\\{\left({R(\Omega)},h(\Omega)\right)\ |\
\Omega\in\mathcal{K}^{2},\,r(\Omega)=1\right\\}=\left\\{(x,y)\ |\ x\geq 1\ \
\text{and}\ \ \frac{1}{t_{g_{2}}(x)}\leq y\leq
1+\sqrt{\frac{\pi}{2\left(\sqrt{x^{2}-1}+\arcsin{\frac{1}{x}}\right)}}\right\\}.$

(44) $\mathcal{D}_{3}:=\left\\{\left({d(\Omega)},h(\Omega)\right)\ |\
\Omega\in\mathcal{K}^{2},\,r(\Omega)=1\right\\}=\left\\{(x,y)\ |\ x\geq 2\
\text{and}\ \frac{1}{t_{g_{1}}(x)}\leq y\leq
1+\sqrt{\frac{\pi}{\sqrt{x^{2}-4}+\left(\pi-2\arccos{\frac{2}{x}}\right)}}\right\\}.$

###### Proof.

  * •

The diagram $\mathcal{D}_{1}$ of the triplet $(P,h,r)$:
Let us prove that (42) holds. The bounds proved in Proposition 4.1 ensure that

$D_{1}\subseteq\left\\{(x,y)\ |\ x\geq 2\pi\ \ \text{and}\ \
1+\frac{\pi}{x-\pi}\leq y\leq 1+\sqrt{\frac{2\pi}{x}}\right\\}=:\mathcal{P}.$

It remains to prove the converse inclusion. First, we observe that the
boundary of $\mathcal{P}$ is included in $\mathcal{D}_{1}$. Indeed, we can
explicitly construct a family of convex sets which fill the lower part of the
boundary of $\mathcal{P}$: let us consider the family of stadiums
$\Set{R_{l}}_{l\geq 0}$ obtained as the convex hull of two balls of radius $1$
and centered in $(0,-l/2)$ and $(0,l/2)$. We have

$P(R_{l})=2\pi+2l,\quad r(R_{l})=1\ \ \text{and}\ \quad
h(R_{l})=\frac{P(R_{l})}{{\left|R_{l}\right|}}=\frac{2\pi+2l}{\pi+2l}=1+\frac{\pi}{P(R_{l})-\pi}.$

Let us now construct a family of convex sets filling the upper part of the
boundary of $\mathcal{P}$. We consider the family of two-cup bodies
$\Set{C_{k}}_{k\geq 1}$ obtained as the convex hull of the ball $B_{1}$
centered at the origin and the points $(-k,0)$ and $(k,0)$. By [24], we have

$P(C_{k})=2\left(\sqrt{4k^{2}-4}+2\arcsin{\frac{1}{k}}\right),\quad
r(C_{k})=1\ \ \text{and}\ \
{\left|C_{k}\right|}=\left(\sqrt{4k^{2}-4}+2\arcsin{\frac{1}{k}}\right)$

and, as it is shown in [14],

$h(C_{k})=\frac{1}{r(\mathcal{C}_{k})}+\sqrt{\frac{\pi}{{\left|C_{k}\right|}}}=1+\sqrt{\frac{2\pi}{P(C_{k})}}.$

In order to conclude, we show that the diagram $\mathcal{D}_{1}$ is vertically
convex, i.e., that we can always construct continuous and vertical paths
included in the diagram and connecting the upper and the lower parts of the
boundary of $\mathcal{D}_{1}$, covering, in this way, all the area between the
two curves.

Let $x_{0}\geq 2\pi$. There exist $R_{l_{0}}\in\Set{R_{l}}$ and
$C_{k_{0}}\in\Set{C_{k}}$ such that $P(R_{l_{0}})=P(C_{k_{0}})=x_{0}$. Let us
define, via the Minkowski sum, the set

$K_{t}=tR_{l_{0}}+(1-t)C_{k_{0}}.$

By the linearity of the perimeter with respect to the Minkowski sum, we have

$\forall t\in[0,1],\ \ \
P(K_{t})=P(tR_{l_{0}}+(1-t)C_{k_{0}})=tP(R_{l_{0}})+(1-t)P(C_{k_{0}})=tx_{0}+(1-t)x_{0}=x_{0}.$

As for the inradius, let us consider the unit ball $B_{1}$ centred at the
origin and the rectangle $Q$ of vertices $(-M,-1)$, $(-M,1)$, $(M,1)$ and
$(M,-1)$, where $M>0$ is sufficiently large such that both $R_{l_{0}}$ and
$C_{k_{0}}$ are contained in $Q$. We then have for any $t\in[0,1]$,

$B_{1}\subset tR_{l_{0}}+(1-t)C_{k_{0}}\subset Q,$

which yields by the monotonicity of the inradius with respect to inclusions to

$\forall t\in[0,1],\ \ \ 1=r(B_{1})\leq r(tR_{l_{0}}+(1-t)C_{k_{0}})\leq
r(Q)=1.$

Thus,

$\forall t\in[0,1],\ \ \ r(K_{t})=1.$

On the other hand, by the continuity of the Cheeger constant with respect to
the Hausdorff distance, we have by the intermediate value theorem

$[h(R_{t_{0}}),h(C_{k_{0}})]\subset\\{h(K_{t})\ |\ t\in[0,1]\\}.$

Thus, since $\\{h(K_{t})\ |\ t\in[0,1]\\}\subset[h(R_{t_{0}}),h(C_{k_{0}})]$
(because $R_{t_{0}}$ and $C_{k_{0}}$ respectively correspond to points laying
on the lower and the upper parts of the boundary of the diagram
$\mathcal{D}_{1}$), we have the equality

$[h(R_{t_{0}}),h(C_{k_{0}})]=\\{h(K_{t})\ |\ t\in[0,1]\\}.$

This proves that the diagram $\mathcal{D}_{1}$ is vertically convex and that
$\mathcal{D}_{1}=\mathcal{P}$.

  * •

The diagram $\mathcal{D}_{2}$ of the triplet $(R,h,r)$:

The proof of (43) follows the same scheme. Indeed, one can prove that the
diagram is vertically convex by considering vertical paths constructed via the
Minkowski sums of the extremal sets (those corresponding to points laying on
the upper and lower parts of the boundary of the diagram). In the present
case, by Proposition 4.5, the upper boundary is filled by points corresponding
to two-cup bodies $(C_{x})_{x\geq 1}$ such that $R(C_{x})=x$ and $r(C_{x})=1$,
for all $x\geq 1$. Meanwhile, the lower boundary is filled by points
corresponding to spherical slices $(S_{x})_{x\geq 1}$ such that $R(S_{x})=x$
and $r(S_{x})=1$, for all $x\geq 1$. If one assumes (without loss of
generality) that for all $x\geq 1$, both $C_{x}$ and $S_{x}$ are centered at
the origin and that their diameters are colinear, then, it is clear that
$C_{x}\subset S_{x}$. Thus, it is straightforward that
$C_{x}\subset(1-t)C_{x}+tS_{x}\subset S_{x}$, for all $t\in[0,1]$. Therefore,
we have by the monotonicity of the circumradius and the inradius with respect
to inclusions that

$\forall t\in[0,1],\ \ \ 1=r(C_{x})\leq r((1-t)C_{x}+tS_{x})\leq r(S_{x})=1$

and

$\forall t\in[0,1],\ \ \ x=R(C_{x})\leq R((1-t)C_{x}+tS_{x})\leq R(S_{x})=x.$

Thus,

$\forall t\in[0,1],\ \ \ r((1-t)C_{x}+tS_{x})=1\ \ \text{and}\ \
R((1-t)C_{x}+tS_{x})=x,$

which allows to conclude the vertical convexity of the diagram as in the
previous case.

  * •

The diagram $\mathcal{D}_{3}$ of the triplet $(d,h,r)$:

As for the proof of (44), let $x_{0}\geq 2$ and $C_{x_{0}}$ be a two-cup body
such that $r(C_{x_{0}})=1$ and $d(C_{x_{0}})=x_{0}$ (corresponding to a point
on the upper boundary of $\mathcal{D}_{3}$, see Proposition 4.2) and
$L_{x_{0}}$ be an extremal shape corresponding to a point on the lower
boundary such that $r(C_{x_{0}})=1$ and $d(C_{x_{0}})=x_{0}$. In this case, we
have to be more careful as we should distinguish two cases:

    * –

If $x_{0}\geq D_{0}$, then by Proposition 4.2, $L_{x_{0}}$ is a symmetrical
spherical slice whose diameter can be assumed to be colinear to the diameter
of $C_{x_{0}}$. In this case, we can conclude exactly as in the previous case
by considering the convex Minkowski combination of $C_{x_{0}}$ and
$L_{x_{0}}$.

    * –

If $x_{0}\in[2,D_{0})$, then $L_{x_{0}}$ can be taken as a smoothed nonagon of
inradius $1$ and diameter $x_{0}$ (see Proposition 4.2). In this case, the
Minkowski sum does no longer allow to construct a vertical line included in
the diagram and joining the upper and the lower parts of its boundary.
Therefore, to prove the vertical convexity, we use the following construction
introduced in [13, Section 1.2]: Starting from the smoothed nonagon
$L_{x_{0}}$, we fix one of its diameter, say $[A,B]$ and we shrink
continuously $L_{x_{0}}$ to the set $L_{AB}$ defined as the convex hull of the
points $A$, $B$ and the disk of radius $1$ contained in $L_{x_{0}}$. Then, we
continuously move the points $A$ and $B$ to the points $A_{0}$ and $B_{0}$ at
distance $x_{0}$, oppositely located with respect to the center of the disk
(in the sense that the center is the middle of $[A_{0},B_{0}]$) by keeping the
convex hull with the disk through the perturbation process, obtaining the two-
cup body $C_{x_{0}}$ as a final shape. By doing so, we constructed a
continuous family of convex shapes (of inradius $1$ and diameter $x_{0}$)
connecting $L_{x_{0}}$ and $C_{x_{0}}$, yielding a vertical and continuous
line connecting the upper and the lower parts of the boundary of the
Blaschke–Santaló diagram $\mathcal{D}_{3}$.

∎

######  Remark 4.8.

As observed in this section, proving sharp bounds corresponding to the
boundary of the Blaschke–Santaló diagram is not equivalent to completely
characterizing it. Indeed, once we managed to identify the boundary, it
remains to show that the diagram is simply connected, which can be a difficult
task, see for example [3, Conjecture 5], [31, Open problem 2] and [42, Problem
3]. We refer to [18] for a generic method of proof of the simple connectedness
of a Blaschke–Santaló diagram.

##  5\. The remaining diagrams

For the remaining triplets of shape functionals, we have proved the existence
of a maximum and minimum to the relative shape optimization problems (see
Theorem 1.1). Moreover, we are able to identify parts of the boundaries of the
corresponding Blaschke–Santaló diagrams. In the following, we state and prove
the results that we have obtained.

###  5.1. The triplet $(\omega,h,d)$

######  Proposition 5.1.

Let $\Omega\in\mathcal{K}^{2}$. We have

(45) $h(\Omega)\geq\frac{1}{t_{g_{3}^{\Omega}}},$

where $t_{g_{3}^{\Omega}}$ is the smallest solution to

${g_{3}^{\Omega}}(t):=\frac{\omega(\Omega)-2t}{2}\sqrt{(d(\Omega)-2t)^{2}-(\omega(\Omega)-2t)^{2}}+\frac{(d(\Omega)-2t)^{2}}{2}\arcsin\left(\frac{\omega(\Omega)-2t}{d(\Omega)-2t}\right)=\pi
t^{2}.$

The equality in (45) is achieved by symmetrical spherical slices. Moreover,

  * •

if $\omega(\Omega)\leq\frac{\sqrt{3}}{2}d(\Omega)$, we have

(46) $h(\Omega)\leq h(T_{I}),$

where $T_{I}$ is the subequilateral triangle such that
$\omega(T_{I})=\omega(\Omega)$ and $d(T_{I})=d(\Omega)$. The equality in (46)
is achieved by the isosceles triangle $T_{I}$;

  * •

and if $\frac{\sqrt{3}}{2}d(\Omega)\leq\omega(\Omega)\leq d(\Omega)$, we have

(47)
$h(\Omega)\leq\frac{\sqrt{3}}{\sqrt{3}\omega(\Omega)-d(\Omega)}+\sqrt{\frac{2\pi}{\pi\omega(\Omega)^{2}-\sqrt{3}d(\Omega)^{2}+6\omega(\Omega)(\tan\left(\arccos(\frac{\omega(\Omega)}{d(\Omega)})\right)-\arccos(\frac{\omega(\Omega)}{d(\Omega)}))}}.$

The equality in (47) is achieved by equilateral triangles.

###### Proof.

  * •

Let us start by proving the lower bound (45), by using the strategy given in
Lemma 2.9. For every $\Omega\in\mathcal{K}^{2}$, it holds

${\left|\Omega\right|}\leq\frac{\omega(\Omega)}{2}\sqrt{d^{2}(\Omega)-\omega^{2}(\Omega)}+\frac{d^{2}(\Omega)}{2}\arcsin\left(\frac{\omega(\Omega)}{d(\Omega)}\right),$

see [29, 38], with equality if and only if $\Omega$ is a symmetrical spherical
slice. If we denote

$f(d,w):=\frac{w}{2}\sqrt{d^{2}-w^{2}}+\frac{d^{2}}{2}\arcsin\left(\frac{w}{d}\right),$

we have

$\frac{\partial f}{\partial d}(d,w)=d\arcsin\left(\frac{w}{d}\right)>0\ \ \
\text{and}\ \ \ \frac{\partial f}{\partial w}(d,w)=\sqrt{d^{2}-w^{2}}>0.$

Thus,

${\left|\Omega_{-t}\right|}\leq f(d(\Omega_{-t}),\omega(\Omega_{-t}))\leq
f(d(\Omega)-2t,\omega(\Omega)-2t).$

Therefore, by using Lemma 2.9, we have

$h(\Omega)\geq\frac{1}{t_{g_{3}^{\Omega}}},$

where $t_{g_{3}^{\Omega}}$ is the smallest solution of

${g_{3}^{\Omega}}(t):=f(d(\Omega)-2t,\omega(\Omega)-2t)=\pi t^{2}.$

  * •

In order to prove the upper bound (46), we consider the following minimization
problem for the area in the class of convex sets with given diameter and
width, studied in [38] and [39]:

    1. (i)

if $2\omega(\Omega)\leq\sqrt{3}d(\Omega)$, then

(48) $2{\left|\Omega\right|}\geq\omega(\Omega)d(\Omega),$

where the equality is achieved by triangles;

    2. (ii)

if $\sqrt{3}d(\Omega)\leq 2\omega(\Omega)\leq 2d(\Omega)$, then

(49)
$\displaystyle{2{\left|\Omega\right|}\geq\pi\omega^{2}(\Omega)-\sqrt{3}d^{2}(\Omega)+6\omega^{2}(\Omega)\left(\tan\left(\arccos\left(\frac{\omega(\Omega)}{d(\Omega)}\right)\right)-\arccos\left(\frac{\omega(\Omega)}{d(\Omega)}\right)\right)}=|T_{Y}|,$

where the equality is achieved by the Yamanouti set $T_{Y}$ such that
$\omega(T_{Y})=\omega(\Omega)$ and $d(T_{Y})=d(\Omega)$.

Moreover, if we consider the minimization problem of the inradius in the class
of convex sets with given diameter and width, we have from Theorem 2.25

(50) $r(\Omega)\geq\begin{cases}r(T_{I}),&\text{if}\
\,2\omega(\Omega)\leq\sqrt{3}d(\Omega),\\\ r(T_{Y}),&\text{if}\
\,\sqrt{3}d(\Omega)\leq 2\omega(\Omega)\leq 2d(\Omega).\end{cases}$

So combining (2) with the estimates (48), (49) and (50), we obtain

$h(\Omega)\leq\begin{cases}\frac{1}{r(T_{I})}+\sqrt{\frac{\pi}{|T_{I}|}}=h(T_{I}),&\text{if}\
\,2\omega(\Omega)\leq\sqrt{3}d(\Omega),\\\
\frac{1}{r(T_{Y})}+\sqrt{\frac{\pi}{|T_{Y}|}},&\text{if}\
\,\sqrt{3}d(\Omega)\leq 2\omega(\Omega)\leq 2d(\Omega).\end{cases}$

The explicit formula given in inequality (47) is then obtained by using (49)
and $r(T_{Y})=\omega(T_{Y})-\frac{d(T_{Y})}{\sqrt{3}}$, see [21, Theorem 2].

∎

Figure 10. Blaschke–Santaló diagram of the triplet $(\omega,h,d)$.

We are also able to give an explicit sharp lower bound for the Cheeger
constant in terms of the width and the diameter. We note that although being
sharp, this inequality does not correspond to a part of the boundary of the
Blaschke–Santaló diagram as shown in Figure 11.

######  Remark 5.2.

Let $\Omega\in\mathcal{K}^{2}$. We have

(51)
$h(\Omega)>\frac{1}{\omega(\Omega)}+\frac{1}{d(\Omega)}+\sqrt{\left(\frac{1}{\omega(\Omega)}+\frac{1}{d(\Omega)}\right)^{2}-\frac{4-\pi}{\omega(\Omega)d(\Omega)}},$

where the equality is asymptotically achieved by a sequence of thin collapsing
rectangles or spherical slices as shown in Figure 11.

In order to prove (51), it is enough to consider the inequality

${\left|\Omega\right|}\leq\omega(\Omega)d(\Omega)$

and to use the strategy of Lemma 2.9.

Figure 11. The explicit inequality (51) and the lower part of the boundary of
the diagram $(\omega,h,d)$.

###  5.2.  The triplet $(\omega,h,R)$

######  Proposition 5.3.

Let $\Omega\in\mathcal{K}^{2}$. We have

(52) $h(\Omega)\geq\frac{1}{t_{g_{4}^{\Omega}}},$

where $t_{g_{4}^{\Omega}}$ is the smallest solution of the equation

$g_{4}^{\Omega}(t):=\frac{(\omega(\Omega)-2t)}{2}\sqrt{4\left(R(\Omega)-t\right)^{2}-\left(\omega(\Omega)-2t\right)^{2}}+2(R(\Omega)-t)^{2}\arcsin{\left(\frac{\omega(\Omega)-2t}{2(R(\Omega)-t)}\right)}=\pi
t^{2}$

on $[0,r(\Omega)]$. The equality in (52) is achieved by symmetrical spherical
slices.

Moreover, if $\omega(\Omega)\leq\frac{3}{2}R(\Omega)$, then

(53) $h(\Omega)\leq h(T_{I}),$

where $T_{I}$ is the subequilateral triangle such that $R(T_{I})=R(\Omega)$
and $\omega(T_{I})=\omega(\Omega)$. The equality in (53) is achieved by the
subequilateral triangle $T_{I}$.

###### Proof.

  * •

Let us start by proving the lower bound (52), by using the strategy given in
Lemma 2.9. Let us recall the function defined in (18)

$\chi:(\omega,R)\longmapsto\frac{\omega}{2}\sqrt{4R^{2}-\omega^{2}}+2R^{2}\arcsin{\frac{\omega}{2R}}.$

We have, for every $R,\omega>0$,

$\frac{\partial\chi}{\partial R}(\omega,R)=4R\arcsin{\frac{\omega}{2R}}\geq 0\
\ \ \text{and}\ \ \ \
\frac{\partial\chi}{\partial\omega}(\omega,R)=\sqrt{4R^{2}-\omega^{2}}\geq 0.$

Thus, using Theorem 2.19, we have, for every $t\in[0,r(\Omega))$,

$|\Omega_{-t}|\leq\chi(\omega(\Omega_{-t}),R(\Omega_{-t}))\leq\chi(\omega(\Omega)-2t,R(\Omega)-t)=:g_{4}^{\Omega}(t),$

where we used (8) and (9). By Lemma 2.9, we have

$h(\Omega)\geq\frac{1}{t_{g_{4}^{\Omega}}},$

where $t_{g_{4}^{\Omega}}$ is the smallest solution to the equation
$g_{4}^{\Omega}(t)=\pi t^{2}$ on the interval $[0,r(\Omega)]$.

  * •

Let us now prove the upper bound (53). We recall that for all
$\Omega\in\mathcal{K}^{2}$ such that $\omega(\Omega)\leq\frac{3}{2}R(\Omega)$,
we have

${\left|\Omega\right|}\geq{\left|T_{I}\right|}\ \ \text{and}\ \ \
r(\Omega)\geq r(T_{I}),$

where $T_{I}$ is a subequilateral triangle such that
$\omega(T_{I})=\omega(\Omega)$ and $R(T_{I})=R(\Omega)$, see respectively (19)
and (20).

To conclude, one just has to combine those estimates with the inequality

$h(\Omega)\leq\frac{1}{r(\Omega)}+\sqrt{\frac{\pi}{{\left|\Omega\right|}}},$

see [14], which is an equality for sets that are homothetic to their form
bodies, in particular, subequilateral triangles.

∎

Figure 12. Blaschke–Santaló diagram of the triplet $(\omega,h,R)$.

In the following remark, we give some explicit sharp bounds, that do not
correspond to parts of the boundary of the Blaschke–Santaló diagram.

######  Remark 5.4.

We can prove that

(54)
$h(\Omega)\leq\frac{3}{\omega(\Omega)}+\sqrt{\frac{\pi}{\sqrt{3}R(\Omega)\omega(\Omega)}}.$

We recall the following inequalities, proved in [38],

(55) ${\left|\Omega\right|}\geq\sqrt{3}R(\Omega)\omega(\Omega)\ \ \
\text{and}\ \ \ \omega(\Omega)\leq 3r(\Omega).$

By combining these estimates and the upper bound in (2), we have

$h(\Omega)\leq\frac{3}{\omega(\Omega)}+\sqrt{\frac{\pi}{\sqrt{3}R(\Omega)\omega(\Omega)}}.$

Since the equality in (2) is achieved by sets that are homothetic to their
form bodies, while the equalities in (55) are achieved by equilateral
triangles, that are particular sets that are homothetic to their form bodies,
we have the equality in (54) for equilateral triangles.

Moreover, another sharp lower bound can be obtained by using the strategy from
Lemma 2.9, starting from

${\left|\Omega\right|}<2R(\Omega)\omega(\Omega),$

which is asymptotically achieved by a sequence of rectangles or spherical
slices with circumradius that goes to infinity (see [19]), as shown in Figure
13. We get ${\left|\Omega_{-t}\right|}\leq
2R(\Omega_{-t})\omega(\Omega_{-t})\leq 2(R(\Omega)-t)(\omega(\Omega)-2t)$ and,
consequently,

(56)
$h(\Omega)\geq\frac{4-\pi}{(2R(\Omega)+\omega(\Omega))-\sqrt{(2R(\Omega)+\omega(\Omega))^{2}-(4-\pi)(2R(\Omega)\omega(\Omega))}}.$

Figure 13. The explicit inequality (56) and the lower part of the boundary of
the diagram $(\omega,h,d)$.

###  5.3. The triplet $(\omega,h,|\cdot|)$

######  Proposition 5.5.

Let $\Omega\in\mathcal{K}^{2}$. We have

(57)
$h(\Omega)\geq\frac{2}{\omega(\Omega)}+\frac{\pi\omega(\Omega)}{2{\left|\Omega\right|}},$

where the equality is achieved by stadiums. Moreover, we have

(58) $h(\Omega)\leq h(T_{I}),$

where $T_{I}$ is a subequilateral triangle such that $|\Omega|=|T_{I}|$ and
$\omega(\Omega)=\omega(T_{I})$. The equality in (58) is achieved by the
subequilateral triangle $T_{I}$.

###### Proof.

  * •

Let us prove inequality (57). We recall the lower bound in (2)

$h(\Omega)\geq\frac{1}{r(\Omega)}+\frac{\pi r(\Omega)}{2|\Omega|},$

which is an equality if and only if $\Omega$ is a stadium. Inequality (57) is
then a consequence of the fact that the function
$r\longmapsto\frac{1}{r}+\frac{\pi r}{2|\Omega|}$ is strictly decreasing and
$r(\Omega)\leq\frac{\omega(\Omega)}{2}$ (where the equality holds for
stadiums).

  * •

Let us now prove inequality (58). We start by recalling inequality (25):

$|\Omega|\leq\sqrt{\frac{r^{4}(\Omega)\omega^{3}(\Omega)}{(\omega(\Omega)-2r(\Omega))^{2}(4r(\Omega)-\omega(\Omega))}}=:f_{\omega(\Omega)}(r(\Omega)).$

By direct computations, we prove that the continuous function

$f_{\omega(\Omega)}:r\longmapsto\sqrt{\frac{r^{4}\omega(\Omega)^{3}}{(\omega(\Omega)-2r)^{2}(4r-\omega(\Omega))}}$

is strictly increasing on
$\left[\frac{\omega(\Omega)}{3},\frac{\omega(\Omega)}{2}\right)$. Let us
denote by $g_{\omega(\Omega)}$ the inverse function of $f_{\omega(\Omega)}$,
which is also continuous and strictly increasing. We have

$r(\Omega)\geq g_{\omega(\Omega)}(|\Omega|)=r(T_{I}),$

where $T_{I}$ is any subequilateral triangle such that
$\omega(T_{I})=\omega(\Omega)$ and $|T_{I}|=|\Omega|$. Thus, we have

$h(\Omega)\leq\frac{1}{r(\Omega)}+\sqrt{\frac{\pi}{|\Omega|}}\leq\frac{1}{r(T_{I})}+\sqrt{\frac{\pi}{|T_{I}|}}=h(T_{I}),$

with equality if and only if $\Omega=T_{I}$, where we used the upper bound in
(2).

∎

Figure 14. Blaschke–Santaló diagram of the triplet $(\omega,h,|\cdot|)$.

######  Remark 5.6.

One may use classical convex geometry inequalities to obtain simpler bounds
than the implicit one given in (58). Indeed, if we combine the inequalities in
(2) and the following classical ones

$\frac{2}{\omega(\Omega)}\leq\frac{1}{r(\Omega)}\leq\frac{2}{\omega(\Omega)}+\frac{\omega(\Omega)}{\sqrt{3}|\Omega|},$

where the lower bound is realized in particular by stadiums, and the upper one
by equilateral triangles (see for example [38] and the references therein), we
obtain the following inequalities

(59)
$h(\Omega)\leq\frac{2}{\omega(\Omega)}+\frac{\omega(\Omega)}{\sqrt{3}|\Omega|}+\sqrt{\frac{\pi}{|\Omega|}}$

and

(60)
$h(\Omega)\leq\frac{2}{\omega(\Omega)-\frac{\omega(\Omega)^{3}}{4|\Omega|}}+\sqrt{\frac{\pi}{|\Omega|}}.$

The bound (59) is attained for equilateral triangles and (60) is
asymptotically attained for a sequence of thin subequilateral triangles.

###  5.4. The triplet $(\omega,h,P)$

######  Proposition 5.7.

Let $\Omega\in\mathcal{K}^{2}$. We have

(61)
$h(\Omega)\geq\frac{2}{\omega(\Omega)}+\frac{2\pi}{2P(\Omega)-\pi\omega(\Omega)},$

where the equality is achieved by stadiums.

Moreover, if $P(\Omega)\geq 2\sqrt{3}\omega(\Omega)$, then,

(62) $h(\Omega)\leq h(T_{I}),$

where $T_{I}$ is a subequilateral triangle such that $P(T_{I})=P(\Omega)$ and
$\omega(T_{I})=\omega(\Omega)$. The equality in (62) is achieved by the
subequilateral triangle $T_{I}$.

###### Proof.

  * •

Inequality (61) is a consequence of (57) and the inequality

$|\Omega|\leq\frac{\omega(\Omega)}{2}\left(P(\Omega)-\frac{\pi\omega(\Omega)}{2}\right),$

which is an equality for stadiums, see for example [38].

  * •

Let us now assume that $P(\Omega)\geq 2\sqrt{3}\omega(\Omega)$. In order to
prove (62), we recall inequality (26):

$P(\Omega)\leq\sqrt{\frac{4r^{2}(\Omega)\omega^{3}(\Omega)}{(\omega(\Omega)-2r(\Omega))^{2}(4r(\Omega)-\omega(\Omega))}}=:f_{\omega(\Omega)}(r(\Omega)).$

By direct computations, we prove that the continuous function

$f_{\omega(\Omega)}:r\longmapsto\sqrt{\frac{4r^{2}\omega(\Omega)^{3}}{(\omega(\Omega)-2r)^{2}(4r-\omega(\Omega))}}$

is strictly increasing on
$\left[\frac{\omega(\Omega)}{3},\frac{\omega(\Omega)}{2}\right)$. Let us
denote by $g_{\omega(\Omega)}$ the inverse function of $f_{\omega(\Omega)}$,
which is also continuous and strictly increasing. We have

$r(\Omega)\geq g_{\omega(\Omega)}(P(\Omega))=r(T_{I}),$

where $T_{I}$ is any subequilateral triangle such that
$\omega(T_{I})=\omega(\Omega)$ and $P(T_{I})=P(\Omega)$. Moreover, since
$P(\Omega)\geq 2\sqrt{3}\omega(\Omega)$, we have by the results contained in
[43],

$|\Omega|\geq|T_{I}|,$

(see also [38] as a reference). Finally, we obtain

$h(\Omega)\leq\frac{1}{r(\Omega)}+\sqrt{\frac{\pi}{|\Omega|}}\leq\frac{1}{r(T_{I})}+\sqrt{\frac{\pi}{|T_{I}|}}=h(T_{I}).$

∎

Figure 15. Blaschke–Santaló diagram of the triplet $(\omega,h,P)$.

###  5.5. The triplet $(R,h,d)$

######  Proposition 5.8.

Let $\Omega\in\mathcal{K}^{2}$. We have

(63)
$h(\Omega)\leq\frac{2R(\Omega)\left(2R(\Omega)+\sqrt{4R(\Omega)^{2}-d(\Omega)^{2}}\right)}{d(\Omega)^{2}\sqrt{4R(\Omega)^{2}-d(\Omega)^{2}}}+\sqrt{\frac{4\pi
R(\Omega)^{2}}{d(\Omega)^{3}\sqrt{4R(\Omega)^{2}-d(\Omega)^{2}}}},$

where the equality is achieved by subequilateral triangles.

###### Proof.

Inequality (63) is obtained by combining

$h(\Omega)\leq\frac{1}{r(\Omega)}+\sqrt{\frac{\pi}{|\Omega|}},$

see [14], which is an equality for sets that are homothetic to their form
bodies (in particular subequilateral triangles), and

$|\Omega|\geq\frac{d(\Omega)^{3}\sqrt{4R(\Omega)^{2}-d(\Omega)^{2}}}{4R(\Omega)^{2}}\
\ \text{and}\ \ \
r(\Omega)\geq\frac{d(\Omega)^{2}\sqrt{4R(\Omega)^{2}-d(\Omega)^{2}}}{2R(\Omega)\left(2R(\Omega)+\sqrt{4R(\Omega)^{2}-d(\Omega)^{2}}\right)},$

respectively proved in [22] and [36], where the equalities hold only for
subequilateral triangles. ∎

Figure 16. Blaschke–Santaló diagram of the triplet $(R,h,d)$, See Conjecture 3
and Figure 18 for the definition of constant width nonagons.

###  5.6. The triplet $(\omega,h,r)$

######  Proposition 5.9.

Let $\Omega\in\mathcal{K}^{2}$. We have

(64)
$h(\Omega)\geq\frac{1}{r(\Omega)}+\frac{1}{r(\Omega)}\sqrt{\pi\left(1-\frac{2r(\Omega)}{\omega(\Omega)}\right)\sqrt{\frac{4r(\Omega)}{\omega(\Omega)}-1}},$

where the equality is achieved by subequilateral triangles.

###### Proof.

The present demonstration is inspired by the proof of [24, Theorem 5].

It is known that the incircle of a set $\Omega$ meets the boundary of $\Omega$
either in two points contained in two parallel lines, or in (at least) three
points that form the vertices of a triangle, see [8]. In the first case, we
have $\omega(\Omega)=2r(\Omega)$, thus inequality (64) is equivalent to
$h(\Omega)\geq\frac{1}{r(\Omega)}$, which is proved in Proposition 2.26. In
the second case, let us denote by $T$ a triangle formed by three lines of
support common to $\Omega$ and the incircle. We have $r(\Omega)=r(T)$ and, by
$\Omega\subset T$ and the monotonicity with respect to the inclusion, we get

(65) $h(\Omega)\geq h(T)$

and

(66) $\omega(\Omega)\leq\omega(T).$

So, inequality (64) is equivalent to the following one

$\frac{1}{r(\Omega)h(\Omega)}f\left(\frac{r(\Omega)}{\omega(\Omega)}\right)\leq
1,$

where $f:x\in\left[\frac{1}{3},\frac{1}{2}\right]\longmapsto
1+\sqrt{\pi\left(1-2x\right)\sqrt{4x-1}}$. We observe that the function
$g:x\in\left[\frac{1}{3},\frac{1}{2}\right]\longmapsto\left(1-2x\right)\sqrt{4x-1}$
is decreasing. Indeed,

$g^{\prime}(x)=\frac{4(1-3x)}{\sqrt{4x-1}}\leq 0.$

Thus, $f$ is also decreasing on $\left[\frac{1}{3},\frac{1}{2}\right]$. Then,
since $\frac{r(\Omega)}{\omega(\Omega)}\geq\frac{r(T)}{\omega(T)}$ by (66), we
have

$f\left(\frac{r(\Omega)}{\omega(\Omega)}\right)\leq
f\left(\frac{r(T)}{\omega(T)}\right).$

Moreover, we get by (65),

$\frac{1}{r(\Omega)h(\Omega)}\leq\frac{1}{r(T)h(T)}.$

Thus, we obtain

$\frac{1}{r(\Omega)h(\Omega)}f\left(\frac{r(\Omega)}{\omega(\Omega)}\right)\leq\frac{1}{r(T)h(T)}f\left(\frac{r(T)}{\omega(T)}\right)=\frac{1}{1+r(T)\sqrt{\frac{\pi}{|T|}}}f\left(\frac{r(T)}{\omega(T)}\right),$

where we used the equality $h(T)=\frac{1}{r(T)}+\sqrt{\frac{\pi}{|T|}}$, which
holds because $T$ is a triangle, see [14, Theorem 1.3]. Now, we use the
inequality

$|T|\leq\frac{r(T)^{2}}{\left(1-\frac{2r(T)}{\omega(T)}\right)\sqrt{\frac{4r(T)}{\omega(T)}-1}},$

which is an equality if and only if $T$ is a subequilateral triangle, see [24,
Theorem 5]. We then have

$\frac{1}{r(\Omega)h(\Omega)}f\left(\frac{r(\Omega)}{\omega(\Omega)}\right)\leq\frac{1}{1+r(T)\sqrt{\frac{\pi}{|T|}}}f\left(\frac{r(T)}{\omega(T)}\right)\leq
1,$

which ends the proof. ∎

Figure 17. Blaschke–Santaló diagram of the triplet $(\omega,h,r)$.

######  Remark 5.10.

It is possible to prove the following upper bound

$h(\Omega)\leq\frac{1}{r(\Omega)}+\frac{\sqrt{\pi\sqrt{3}}}{\omega(\Omega)},$

where the equality holds for equilateral triangles, see Figure 17.

In order to prove it, one may combine the upper bound in (2) and the
inequality $\omega(\Omega)^{2}\leq\sqrt{3}|\Omega|$, see [38].

##  6\. Conclusions and conjectures

In this Section, we collect the conjectures that we deduced from the numerical
approximation of the Blaschke–Santaló diagrams that we have plotted in the
previous sections.

######  Conjecture 1.

We consider the diagram $(\omega,h,d)$ plotted in Figure 10. Let
$\Omega\in\mathcal{K}^{2}$, we conjecture that if
$\frac{\sqrt{3}}{2}d(\Omega)\leq\omega(\Omega)\leq d(\Omega)$, then

$h(\Omega)\leq h(Y)$

where $Y$ is a Yamanouti set (see Definition 2.15) such that
$\omega(Y)=\omega(\Omega)$ and $d(Y)=d(\Omega)$.

######  Conjecture 2.

We consider the diagram $(\omega,h,P)$ plotted in Figure 15. Let
$\Omega\in\mathcal{K}^{2}$, we conjecture that, if $\pi\omega(\Omega)\leq
P(\Omega)\leq 2\sqrt{3}\omega(\Omega)$, then

$h(\Omega)\leq h(Y),$

where $Y$ is a Yamanouti set (see Definition 2.15) such that $P(Y)=P(\Omega)$
and $\omega(Y)=\omega(\Omega)$.

######  Conjecture 3.

We consider the diagram $(R,h,d)$ plotted in Figure 16. Let
$\Omega\in\mathcal{K}^{2}$, we conjecture that

$h(\Omega)\geq h(N),$

where $N$ refers to a nonagon of constant width such that $d(N)=d(\Omega)$ and
$R(N)=R(\Omega)$ described in [22] as follows: let $\Gamma$ and $\gamma$ be
the circumcircle and the incircle of a constant width set $K$ that are known
to be concentric and such that $d(K)=\omega(K)=R(K)+r(K)$. The extremal set
can be constructed in the following way: an equilateral triangle $ABC$ is
inscribed in the circle $\Gamma$, and now we take the circular arcs of radius
$R(K)+r(K)$ drawn about the three vertex points. These arcs touch $\gamma$ at
the opposite points $\bar{A},\bar{B},\bar{C}$ of $A,B,C$, respectively.
Furthermore, we construct three circles of radius $(R(K)+r(K))/2$ that have
the sides of the triangle as chords and whose centers lie inside the triangle.
The required constant width set has 3-fold symmetry and is formed by nine arcs
of the six constructed circles, see Figure 18.

Figure 18. A nonagon of constant width.

######  Conjecture 4.

We consider the diagram $(\omega,h,R)$ plotted in Figure 12. Let
$\Omega\in\mathcal{K}^{2}$, we conjecture that:

  * •

If $\omega(\Omega)\in[\frac{3}{2}R(\Omega),\sqrt{3}R(\Omega)]$, then

$h(\Omega)\leq h(Y),$

where $Y$ is a Yamanouti set (see Definition 2.15) such that
$\omega(Y)=\omega(\Omega)$ and $R(Y)=R(\Omega)$.

  * •

If $\omega(\Omega)\geq\sqrt{3}R(\Omega)$, then

$h(\Omega)\leq h(N),$

where $N$ refers to a nonagon of constant width such that
$\omega(N)=\omega(\Omega)$ and $R(N)=R(\Omega)$.

##  7\. Appendix: Summary tables with the results

In this first table, we summarize the results relative to the diagrams that
are completely solved.

Param. | Condition | Inequality | Extremal sets | Ref.
---|---|---|---|---
$P,h,A$ |  | $\displaystyle h\leq\frac{P}{A}$ | Cheeger of itself | [16]
|  | $\displaystyle h\geq\frac{P+\sqrt{4\pi A}}{2A}$ | sets that are homothetic to their form bodies | [16]
$r,h,A$ |  | $\displaystyle h\leq\frac{1}{r}+\sqrt{\frac{\pi}{A}}$ | sets that are homothetic to their form bodies | [14]
|  | $\displaystyle h\geq\frac{1}{r}+\frac{\pi r}{A}$ | stadiums | [14]
$P,h,r$ |  | $\displaystyle{h\leq\frac{1}{r}+\sqrt{\frac{2\pi}{Pr}}}$ | sets that are homothetic to their form bodies |  Prop. 4.1
|  | $\displaystyle{h\geq\frac{1}{r}+\frac{\pi}{P-\pi r}}$ | stadiums |  Prop. 4.1
$d,h,r$ |  | $\displaystyle{h\leq\frac{1}{r}+\sqrt{\frac{\pi}{r\sqrt{d^{2}-4r^{2}}+r^{2}(\pi-2\arccos{\left(\frac{2r}{d}\right)})}}}$ | two-cup bodies |  Prop. 4.2
|  |  ${h\geq\frac{1}{t_{g_{1}^{\Omega}}}}$ $(i)$ | spherical slices/smoothed nonagons |  Prop. 4.2
|  | $h>\dfrac{4-\pi}{d+2r-\sqrt{(d+2r)^{2}-2(4-\pi)dr}}$ | thinning rectangles |  Remark 4.4
$R,h,r$ |  | $\displaystyle h\leq\frac{1}{r}+\sqrt{\frac{\pi}{2r\left(\sqrt{R^{2}-r^{2}}+r\arcsin\left(\frac{r}{R}\right)\right)}}$ | two-cup bodies |  Prop. 4.5
|  |  $\displaystyle h\geq\frac{1}{t_{g_{2}^{\Omega}}}$ $(ii)$ | spherical slices |  Prop. 4.5
|  | $h>\frac{4-\pi}{2(R+r)-\sqrt{4(R+r)^{2}-4(4-\pi)Rr}}$ | thinning rectangles |  Remark 4.6

$(i)$ $t_{g_{1}^{\Omega}}$ is the smallest solution on $[0,r(\Omega)]$ to

$g_{1}^{\Omega}(t):=\psi(d(\Omega)-2t,r(\Omega)-t)=\pi t^{2},$

where

$\psi(d,r):=\begin{cases}\displaystyle{\frac{3\sqrt{3}r}{2}(\sqrt{d^{2}-3r^{2}}-r)+\frac{3d^{2}}{2}\left(\frac{\pi}{3}-\arccos{\left(\frac{\sqrt{3}r}{d}\right)}\right)},&\text{if}\,\,\,d\leq
rD^{*}\vspace{1mm}\\\
\displaystyle{r\sqrt{d^{2}-4r^{2}}+\frac{d^{2}}{2}\arcsin{\left(\frac{2r}{d}\right)}},&\text{if}\,\,\,d\geq
rD^{*}.\end{cases}$

and $D^{*}$ is the unique number in $[2,2\sqrt{3}]$ for which the two
expression of the function $\psi(d,r)$ are equal.
$(ii)$ $t_{g_{2}^{\Omega}}$ is the smallest solution on $[0,r(\Omega)]$ to

${g_{2}^{\Omega}}(t):=2\left((r-t)\sqrt{(R(\Omega)-t)^{2}-(r(\Omega)-t)^{2}}+(R(\Omega)-t)^{2}\arcsin{\left(\frac{r(\Omega)-t}{R(\Omega)-t}\right)}\right)=\pi
t^{2}.$

In this second table, we summarize the results of the partially solved
Blaschke–Santaló diagrams.

Param. | Condition | Inequality | Extremal sets | Ref.
---|---|---|---|---
$\omega,h,d$ | $\omega\leq\sqrt{3}/2d$ |  $h\leq\frac{1}{r(\omega,d)}+\sqrt{\frac{2\pi}{\omega d}}$ $(iii)$ | subequilateral triangles | Prop. 5.1
| $\sqrt{3}/2d\leq\omega\leq d$ | $h\leq\frac{\sqrt{3}}{\sqrt{3}\omega-d}+\sqrt{\frac{2\pi}{\pi\omega^{2}-\sqrt{3}d^{2}+6\omega(\tan\left(\arccos(\frac{\omega}{d})\right)-\arccos(\frac{\omega}{d}))}}$ | equilateral triangles |
|  |  $h\geq\frac{1}{t_{g_{3}^{\Omega}}}$ $(iv)$ | spherical slices |
|  | $h>\frac{1}{\omega}+\frac{1}{d}+\sqrt{\left(\frac{1}{\omega}+\frac{1}{d}\right)^{2}-\frac{4-\pi}{\omega d}}$ | thinning rectangles | Remark 5.2
$\omega,h,R$ | $\omega\leq 3/2R$ |  $h\leq\frac{1}{r(\omega,R)}+\sqrt{\frac{\pi}{A(\omega,R)}}$ $(v)$ | subequilateral triangles | Prop. 5.3
|  |  $h\geq\frac{1}{t_{g_{4}^{\Omega}}}$ $(vi)$ | spherical slices |
|  | $h\geq\frac{4-\pi}{(2R+\omega)-\sqrt{(2R+\omega)^{2}-2(4-\pi)R\omega}}$ | thinning rectangles | Remark 5.4
$\omega,h,P$ | $P\geq 2\sqrt{3}\omega$ |  $h\leq\frac{1}{r(\omega,P)}+\sqrt{\frac{\pi}{A(\omega,P)}}$ $(vii)$ | subequilateral triangles | Prop. 5.7
|  | $h\geq\frac{2}{\omega}+\frac{2\pi}{2P-\pi\omega}$ | stadiums |
$\omega,h,A$ |  |  $h\leq\frac{1}{r(\omega,A)}+\sqrt{\frac{\pi}{A}}$ $(viii)$ | subequilateral triangles | Prop. 5.5
|  | $h\geq\frac{2}{\omega}+\frac{\pi\omega}{2A}$ | stadiums |
$R,h,d$ |  | $h\leq\frac{2R(2R+\sqrt{4R^{2}-d^{2}})}{d^{2}\sqrt{4R^{2}-d^{2}}}+\sqrt{\frac{4\pi R^{2}}{d^{3}\sqrt{4R^{2}-d^{2}}}}$ | subequilateral triangles | Prop. 5.8
$\omega,h,r$ |  | $h\geq\frac{1}{r}+\frac{1}{r}\sqrt{\pi\left(1-\frac{2r}{\omega}\right)\sqrt{\frac{4r}{\omega}-1}}$ | subequilateral triangles | Prop. 5.9

$(iii)$ $r(\omega,d)$ is given by

$d^{2}\left(\omega-2r(\omega,d)\right)^{2}(4r(\omega,d)-\omega)=4r^{4}(\omega,d)\;\omega.$

$(iv)$ $t_{g_{3}^{\Omega}}$ is the smallest solution to

${g_{3}^{\Omega}}(t):=f(d(\Omega)-2t,\omega(\Omega)-2t)=\pi t^{2},$

where

$f(d,w)=\frac{w}{2}\sqrt{d^{2}-w^{2}}+\frac{d^{2}}{2}\arcsin\left(\frac{w}{d}\right)$

$(v)$ $r(\omega,R)$ is given by

$\left(4r(\omega,R)-\omega\right)\left(\omega-2r(\omega,R)\right)=\frac{2r^{3}(\omega,R)}{R}$

and $A(\omega,R)$ is given by

$16A(\omega,R)^{6}=R^{2}\omega^{2}\left(16A(\omega,R)^{4}-R^{2}\omega^{6}\right).$

$(vi)$ $t_{g_{4}^{\Omega}}$ is the smallest solution on $[0,r(\Omega)]$ to

$g_{4}^{\Omega}(t):=\chi(\omega(\Omega)-2t,R(\Omega)-t)=\pi t^{2},$

where

$\chi(\omega,R):=\frac{\omega}{2}\sqrt{4R^{2}-\omega^{2}}+2R^{2}\arcsin{\frac{\omega}{2R}}.$

$(vii)$ $r(\omega,P)$ is given by

$(\omega-2r(\omega,P))^{2}(4r(\omega,P)-\omega)P^{2}=4r(\omega,P)^{2}\omega^{3}$

and $A(\omega,P)$ is the middle root of the equation

$128PA(\omega,P)^{3}-16\omega(5P^{2}+\omega^{2})A(\omega,P)^{2}+16\omega^{2}P^{3}A(\omega,P)-\omega^{3}P^{4}=0$

$(viii)$ $r(\omega,A)$ is given by

$(\omega-2r(\omega,A))^{2}(4r(\omega,A)-\omega)A^{2}=r^{4}(\omega,A)\;\omega^{3}.$

Finally, in this last table, we have summarized the inequalities that we have
found and that do not correspond to parts of the boundaries of the
corresponding Blaschke–Santaló diagrams.

Param. | Condition | Inequality | Extremal sets | Ref.
---|---|---|---|---
$R,h,A$ |  | $h<\frac{1}{R}+\frac{4R}{A}$ | thinning rectangles | [16, 38]
|  | $h\geq\frac{1}{2R}+\frac{\pi R}{2A}+\sqrt{\frac{\pi}{A}}$ | balls |
$P,h,R$ |  | $h<\frac{P}{R(P-4R)}$ | thinning rectangles | [16, 36]
|  |  $h\geq\frac{4\operatorname{arcsinc}{\left(\frac{4R}{P}\right)}}{P-4R\cos{\left(\operatorname{arcsinc}{\left(\frac{4R}{P}\right)}\right)}}+\sqrt{\frac{8\pi\operatorname{arcsinc}{\left(\frac{4R}{P}\right)}}{P\left(P-4R\cos{\left(\operatorname{arcsinc}{\left(\frac{4R}{P}\right)}\right)}\right)}}$ $(ix)$ | balls |
$P,h,d$ | $2d<P<3d$ | $h<\frac{4}{P-2d}+\sqrt{\frac{4\pi}{(P-2d)\sqrt{P(4d-P)}}}$ |  | [14, 16, 38]
| $3d\leq P\leq\pi d$ | $h<\frac{4}{P-2d}+\sqrt{\frac{4\pi}{\sqrt{3}d(P-2d)}}$ |  |
|  |  $h\geq\frac{4\operatorname{arcsinc}{\left(\frac{2d}{P}\right)}}{P-2d\cos{\left(\operatorname{arcsinc}{\left(\frac{2d}{P}\right)}\right)}}+\sqrt{\frac{8\pi\operatorname{arcsinc}{\left(\frac{2d}{P}\right)}}{(P-2d\cos{\left(\operatorname{arcsinc}{\left(\frac{2d}{P}\right)}\right)}}}$ $(ix)$ | balls |
$d,h,A$ |  | $h\leq\frac{4}{d}+\frac{2d}{A}$ |  | [14, 16, 38]
|  | $h<\frac{2d}{A}+\sqrt{\frac{\pi}{A}}$ |  |
|  | $h>\frac{d}{A}+\sqrt{\frac{\pi}{A}}$ | thinning two-cup |

$(ix)$ where the cardinal-arcsine is defined as

${\rm sinc}:x\in\mathbb{R}\longmapsto{\rm sinc}(x)=\frac{\sin(x)}{x},\quad{\rm
arcsinc}(x)={\rm sinc}^{-1}(x)$

Acknowledgements: The author would also like to thank the anonymous referees
for their careful reading and useful comments that helped to improve the
manuscript.

The authors would also like to thank Jimmy Lamboley for useful discussions.

I. Ftouhi and G. Paoli are supported by the Alexander von Humboldt Foundation
through Alexander von Humboldt grants for Postdocs.

A. L. Masiello and G. Paoli are partially supported by GNAMPA of INdAM.

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