# Reverse isoperimetric problems under curvature constraints

Kostiantyn Drach  and  Kateryna Tatarko  Institute of Science and Technology
Austria, Am Campus 1, 3400 Klosterneuburg, Austria
<EMAIL_ADDRESS><EMAIL_ADDRESS>Department of Pure
Mathematics, University of Waterloo, Waterloo, ON, N2L 3G1, Canada
<EMAIL_ADDRESS>
###### Abstract.

In this paper we solve several reverse isoperimetric problems in the class of
_$\lambda$ -convex_ bodies, i.e., convex bodies whose curvature at each point
of their boundary is bounded below by some $\lambda>0$.

We give an affirmative answer in $\mathbb{R}^{3}$ to a conjecture due to
Borisenko which states that the _$\lambda$ -convex lens_, i.e., the
intersection of two balls of radius $1/\lambda$, is the unique _minimizer_ of
volume among all $\lambda$-convex bodies of given surface area.

Also, we prove a _reverse inradius inequality_ : in model spaces of constant
curvature and arbitrary dimension, we show that the $\lambda$-convex lens
(properly defined in non-zero curvature spaces) has the smallest inscribed
ball among all $\lambda$-convex bodies of given surface area. This solves a
conjecture due to Bezdek on minimal inradius of isoperimetric ball-polyhedra
in $\mathbb{R}^{n}$.

Keywords:  $\lambda$-convexity; reverse isoperimetric inequality; inradius.

2020 Mathematics Subject Classification: 52A30, 52A38 (Primary); 52A27, 52A40,
52B60 (Secondary)

The first author is supported by ERC Advanced Grant “SPERIG” (#885707); the
second author is partially supported by NSERC Discovery Grant number
2022-02961

###### Contents

  1. 1 Introduction
    1. 1.1 Classical and reverse isoperimetric problems
    2. 1.2 Reverse isoperimetric problem under curvature constraints
    3. 1.3 Reverse inradius inequality for $\lambda$-convex bodies
    4. 1.4 Structure of the paper
  2. 2 Some background
    1. 2.1 Totally umbilical hypersurfaces in model spaces
    2. 2.2 $\lambda$-convex bodies
    3. 2.3 $\lambda$-convex polytopes and their properties
    4. 2.4 Inscribed balls and their properties
    5. 2.5 Inner parallel bodies in $\mathbb{R}^{n}$ and their properties
  3. 3 Proof of the reverse isoperimetric inequality in $\mathbb{R}^{3}$ (Theorem A)
    1. 3.1 Step I: $K$ is a $\lambda$-convex polytope
    2. 3.2 Step II: General case
  4. 4 Proof of the reverse inradius inequality (Theorem B)
    1. 4.1 Reduction to an equivalent statement (Theorem C)
    2. 4.2 Proof of Theorem C (assuming the Key Claim)
    3. 4.3 Proof of the Key Claim
  5. A Reverse isoperimetric inequality for $\lambda$-convex curves in $\mathbb{R}^{2}$, revisited
  6. B Lenses vs. spindles, an asymptotic competition

##  1\. Introduction

###  1.1. Classical and reverse isoperimetric problems

_What body has the largest volume for a given surface area?_

This classical _isoperimetric problem_ has been studied, generalized and
solved in a great variety of contexts, for example, in the spaces of constant
curvature, etc. (see, e.g., [Ro]). In the $n$-dimensional Euclidean space
$\mathbb{R}^{n}$, for instance, it leads to the following fundamental
inequality:

######  The Classical Isoperimetric Inequality.

Let $n\geqslant 2$. Let $K\subset\mathbb{R}^{n}$ be a domain with surface area
$|\partial K|$ and volume $|K|$, and let $B\subset\mathbb{R}^{n}$ be a ball.
If $|\partial K|=|\partial B|$, then $|K|\leqslant|B|$, and equality holds if
and only if $K$ is a ball.

At the same time, one may pose the following _reverse isoperimetric problem_ :

_What body has the smallest volume for a given surface area?_

The reverse isoperimetric problem has received a great deal of attention in
recent years. Observe that for a given surface area, the $n$-dimensional
volume can be arbitrary small (think of a flat pancake, i.e., an oblong and
thin domain). Hence, as stated, the reverse isoperimetric problem has a
trivial solution. In his celebrated work, Ball [Bal1, Bal2] overcame this
issue by restricting the problem to affine classes of convex bodies. In this
setting, the problem turns out to be non-trivial, and Ball showed the
following:

######  The Reverse Isoperimetric Inequality due to Ball.

Let $n\geqslant 2.$ Let $K\subset\mathbb{R}^{n}$ be a convex body and
$T\subset\mathbb{R}^{n}$ be a regular simplex. Then there is an affine image
$K^{\prime}$ of $K$ such that $|\partial K^{\prime}|=|\partial T|$ and
$|K^{\prime}|\geqslant|T|$.

The equality case in Ball’s reverse isoperimetric inequality was settled later
by Barthe [Ba]. We remark that for the classical isoperimetric problem
optimizers are balls, while this is no longer the case for the reverse
problem.

###  1.2. Reverse isoperimetric problem under curvature constraints

In this paper, we take a different approach and restrict the reverse
isoperimetric problem to the class of convex bodies with _uniform curvature
constraints_. Namely, we require that the boundary of a convex body is either
not too flat, or not too curved, in the following sense.

Let $K$ be a convex body in an $n$-dimensional model space $M^{n}(c)$ of
constant curvature $c\in\mathbb{R}$, $n\geqslant 2$. We say that $K$ is
_$\lambda$ -convex_ if there exists $\lambda>0$ so that the principal
curvatures $k_{i}$, $i\in\\{1,\ldots,n-1\\}$, with respect to inward pointing
normals of the boundary $\partial K$ satisfy

$k_{i}(p)\geqslant\lambda\quad\text{ for each point }p\in\partial K.$

If the boundary of $K$ is not smooth, then the bound above is understood in
the barrier sense, which we will make precise in Definition 2.1 (see also
Figure 2). In a similar way, $K$ is called _$\lambda$ -concave_ if the
inequality

$\lambda\geqslant k_{i}(p)\geqslant 0$

holds at each point $p\in\partial K$ and for every $i\in\\{1,\ldots,n-1\\}$.

It turns out that for $\lambda$-convex and $\lambda$-concave bodies the
reverse isoperimetric problem is well-posed and has a non-trivial solution in
spaces of constant curvature. In this paper, we will focus on its version in
the Euclidean space. Despite the fact that the notions of $\lambda$-concavity
and $\lambda$-convexity look dual, the progress of resolving the reverse
isoperimetric problem for these classes is not uniform.

For $\lambda$-concave bodies in $\mathbb{R}^{n}$, the reverse isoperimetric
problem was completely solved by Chernov and the authors in [CDT], where the
following sharp result was shown as part of a whole range of _reverse
quermassintegrals inequalities_ :

######  The Reverse Isoperimetric Inequality for $\lambda$-concave Bodies
([CDT]).

Let $n\geqslant 2$. Let $K~{}\subset~{}\mathbb{R}^{n}$ be a $\lambda$-concave
body, and let $S\subset\mathbb{R}^{n}$ be a _$\lambda$ -sausage_, i.e., the
convex hull of two balls of radius $1/\lambda$. If $|\partial K|=|\partial
S|$, then $|K|\geqslant|S|$, and equality holds if and only if $K$ is a
$\lambda$-sausage.

An alternative proof of this result was given later in [SY] using different
techniques (see also [Na] for an elegant proof of a weaker version of the
inequality).

Much less is known about the reverse isoperimetric problem for
$\lambda$-convex bodies in $\mathbb{R}^{n}$, $n\geqslant 2$. The following
conjecture is due to Borisenko (see [CDT]):

######  Conjecture (The Reverse Isoperimetric Inequality for $\lambda$-convex
bodies).

Let $n\geqslant 2$. Let $K\subset\mathbb{R}^{n}$ be a $\lambda$-convex body,
and let $L\subset\mathbb{R}^{n}$ be a $\lambda$-convex _lens_ , i.e., the
intersection of two balls of radius $1/\lambda$ (see Figure 1). If $|\partial
K|=|\partial L|$, then $|K|\geqslant|L|$, and equality holds if and only if
$K$ is a $\lambda$-convex lens.

For $n=2$, the conjecture was confirmed in [BD2] (see [FKV] for an alternative
proof). Further progress made in [BD3, Dr1, Bor] fully resolved an analogous
conjecture in all 2-dimensional spaces of constant curvature and their natural
generalizations (e.g., Alexandrov spaces of curvature bounded below; see also
[CFP] for a generalization of the inequality in $\mathbb{R}^{2}$ with
additional assumptions on convexity). For $n\geqslant 3$, partial progress was
made so far only in $\mathbb{R}^{3}$, where the conjecture was confirmed in a
restrictive class of surfaces of revolution [Dr3].

In this paper, we completely solve Borisenko’s conjecture in $\mathbb{R}^{3}$.
Our main result is

######  Theorem A (The Reverse Isoperimetric Inequality for $\lambda$-convex
bodies in $\mathbb{R}^{3}$).

Let $K\subset\mathbb{R}^{3}$ be a $\lambda$-convex body and
$L\subset\mathbb{R}^{3}$ be a $\lambda$-convex lens. If $|\partial
K|=|\partial L|$, then $|K|\geqslant|L|.$ Moreover, equality holds if and only
if $K$ is a $\lambda$-convex lens.

Figure 1. The $\lambda$-convex lens (or the UFO-body).

For the proof of Theorem A, we use some ideas of Nazarov [Naz]. Our approach
is different from [Dr3], where the optimal control theory and Pontryagin’s
Maximum Principle was used to solve the corresponding constrained optimization
problem. Our proof can be adapted to work also in $\mathbb{R}^{2}$, as we
illustrate in Appendix A. This gives an alternative proof of the result in
[BD2].

Recall that the _inradius_ of a convex body is the radius of the largest ball
contained in the body. As a by-product of the proof of Theorem A in Section 3,
we establish the inequality $r(K)\geqslant r(L)$ between the inradii $r(K)$
and $r(L)$ of a $\lambda$-convex body $K\subset\mathbb{R}^{3}$ and the lens
$L\subset\mathbb{R}^{3}$ satisfying $|\partial K|=|\partial L|$. This is an
instance of a reverse inradius inequality. As the second main result of the
paper (see Section 1.3), we show that the same inequality holds in all
constant curvature spaces of arbitrary dimension. In particular, this gives an
indication that Borisenko’s conjecture might be true for $n\geqslant 4$.

We also refer the reader to [HTr, Ga] where some partial progress on the
reverse isoperimetric problem under curvature constraints in the non-convex
setting was made. (In the non-convex setting, the problem does not always have
a well-defined solution. For example, in $\mathbb{R}^{2}$, there exists a
constant $a>0$ so that for any arbitrary large $\ell>0$ there is a domain $D$
with area $a$ such that $\partial D$ is a simple closed curve of length $\ell$
and curvature $|k|\leqslant 1$.)

###  1.3. Reverse inradius inequality for $\lambda$-convex bodies

In the spirit of our discussion in Section 1.1, one can ask the following
question:

_What convex body has the largest inradius for a given surface area?_

In $\mathbb{R}^{n}$, the answer is a ball, and it is a unique body with such
property. This is a classical result that immediately follows from the fact
that the surface area of the inscribed ball is no greater than the surface
area of the convex body in which it is inscribed. The reverse question

_What convex body has the smallest inradius for a given surface area?_

is well-defined and has a non-trivial solution for $\lambda$-convex bodies in
spaces of constant curvature, while this is not the case for general convex
bodies.

For $2$-dimensional $\lambda$-convex bodies in the Euclidean plane, i.e., the
space of constant zero curvature, Milka [Mi2] proved that the $\lambda$-convex
lens has the smallest inradius for a given length of the boundary. Much later,
this result was generalized in [Dr4] to arbitrary $\lambda$-convex domains in
$2$-dimensional Alexandrov spaces of curvature bounded below, thus closing
this question in dimension $2$. Recently, Bezdek [Be1] showed that among all
$\lambda$-convex bodies in $\mathbb{R}^{n}$, $n\geqslant 2$, the
$\lambda$-convex lens has the smallest inradius for a given _volume_. Bezdek
conjectured that the same result must be true if one considers
$\lambda$-convex bodies of a given surface area [Be1, Conjecture 5]. We
confirm this conjecture by establishing the following sharp and much more
general result:

######  Theorem B (The Reverse Inradius Inequality for $\lambda$-convex
bodies).

Let $n\geqslant 2$. Let $K\subset M^{n}(c)$ be a $\lambda$-convex body in a
model space $M^{n}(c)$ of constant curvature $c$, and let $L\subset M^{n}(c)$
be a $\lambda$-convex lens. If $|\partial K|=|\partial L|$, then the inradii
of the bodies satisfy $r(K)\geqslant r(L)$. Moreover, equality holds if and
only if $K$ is a $\lambda$-convex lens.

For the definition of a $\lambda$-convex lens in spaces of constant curvature,
see Definition 2.1.

The following result is an immediate corollary of Theorem B in the
$2$-dimensional space of constant positive curvature that can be obtained
using spherical duality (see [BD3, Section 4]):

######  Corollary 1.1 (The reverse outer radius inequality in
$\mathbb{S}^{2}$).

Let $K\subset\mathbb{S}^{2}$ be a $\lambda$-concave body and
$S\subset\mathbb{S}^{2}$ be a $\lambda$-sausage (i.e. the convex hull of two
circles of geodesic curvature equal to $\lambda$). If $|K|=|S|$, then the
circumradii $R(K),R(S)$ of the bodies (i.e. the radii of the smallest geodesic
disks containing the bodies) satisfy $R(K)\leqslant R(S)$. Moreover, equality
holds if and only if $K$ is a $\lambda$-sausage.∎

It is plausible that the same result holds for $\lambda$-concave bodies in
spheres of arbitrary dimension, but some additional tools might be required
for the proof.

Finally, we would like to point out that the reverse inradius problem in the
class of $\lambda$-concave bodies in $\mathbb{R}^{n}$ has a simple solution.
Indeed, $r(K)\geqslant 1/\lambda$ because every $\lambda$-concave body
$K\subset\mathbb{R}^{n}$ contains a ball of radius $1/\lambda$. This
inequality is the best possible and can be attained only by
$\lambda$-sausages.

###  1.4. Structure of the paper

In Section 2, we give the definition of $\lambda$-convexity and provide some
background on the notion. In Sections 3 and 4 we give the proof of Theorems A
and B, respectively. Using our method, we provide a self-contained proof of
the reverse isoperimetric inequality in $\mathbb{R}^{2}$ in Appendix A.
Finally, Appendix B contains an asymptotic computation in $\mathbb{R}^{n}$
showing that, as $n\to\infty$, the $\lambda$-convex lens has smaller volume
for given surface area compared to its most close ‘competitor’, a _$\lambda$
-convex spindle_ (will be defined in Appendix B).

Acknowledgement. The authors are grateful to Fedor Nazarov and Dmitry Ryabogin
for their invaluable help. The second author also wants to thank the Institute
for Computational and Experimental Research in Mathematics (ICERM) for the
hospitality. Part of the work on the paper was completed during her
participation at the program “Harmonic Analysis and Convexity”.

##  2\. Some background

We denote by $M^{n}(c)$ a complete simply connected Riemannian manifold of
dimension $n\geqslant 2$ and of constant sectional curvature equal to
$c\in\mathbb{R}$. We call these manifolds _model spaces_. It is well-known
that $M^{n}(c)$ is either the spherical, Euclidean, or hyperbolic space
depending on whether, respectively, $c>0$, $c=0$, or $c<0$.

We will use the notation $|A|$ for the $n$-dimensional volume of a measurable
set $A$ in $M^{n}(c)$. We will use the same notation $|\partial A|$ for the
$(n-1)$-dimensional volume of the boundary of $A$ (provided this volume is
well-defined).

As usual, a _convex body_ in a model space is a compact, geodesically convex
set with non-empty interior.

###  2.1. Totally umbilical hypersurfaces in model spaces

For a given $\lambda>0$, let $\mathcal{S}_{\lambda}\subset M^{n}(c)$ be a
complete totally umbilical hypersurface of constant normal curvature equal to
$\lambda$. We will use the notation $\mathcal{B}_{\lambda}$ for the convex
region bounded by $\mathcal{S}_{\lambda}$. Everywhere below we assume that
$\mathcal{B}_{\lambda}$ is closed.

The following classification of complete totally umbilical hypersurfaces is
well-known (see, e.g., [BZ, Section 35.2.4]):

  1. (1)

If $c=0$, i.e., the ambient model space is the Euclidean space
$\mathbb{R}^{n}$, then the sets $\mathcal{S}_{\lambda}$ and
$\mathcal{B}_{\lambda}$ are, respectively, spheres and balls of radius
$1/\lambda$.

  2. (2)

If $c>0$, i.e., $M^{n}(c)$ is the sphere $\mathbb{S}^{n}(c)$ of radius
$1/\sqrt{c}$, then $\mathcal{S}_{\lambda}$ is a geodesic sphere of radius
$1/\sqrt{c}\cdot\cot^{-1}(\lambda/\sqrt{c})$ and $\mathcal{B}_{\lambda}$ is
the corresponding ball.

  3. (3)

The situation is more interesting if $c<0$, i.e., when the model space is the
hyperbolic space $\mathbb{H}^{n}(c)$, and it depends on the relation between
$c$ and $\lambda$:

    1. (a)

If $\lambda>\sqrt{-c}$, then $\mathcal{S}_{\lambda}$ is a geodesic sphere of
radius $1/\sqrt{-c}\cdot\coth^{-1}(\lambda/\sqrt{-c})$, and
$\mathcal{B}_{\lambda}$ is the geodesic ball. In this case,
$\mathcal{B}_{\lambda}$ is compact.

    2. (b)

If $\lambda=\sqrt{-c}$, then $\mathcal{S}_{\lambda}$ is a _horosphere_ and
$\mathcal{B}_{\lambda}$ is a _horoball_. In this case, $\mathcal{B}_{\lambda}$
is not compact (however, it can be compactified by adding a single point at
infinity).

    3. (c)

Finally, if $\lambda<\sqrt{-c}$, then $\mathcal{S}_{\lambda}$ is an
_equidistant hypersurface_ (or _equidistant_ for short), i.e., the set of
points at a given distance from a totally geodesic hyperplane. By explicit
computation, one can see that for each $\mathcal{S}_{\lambda}$ there exists a
unique hyperplane $\mathcal{H}\subset\mathbb{H}^{n}(c)$ such that

$\mathcal{S}_{\lambda}=\left\\{x\in\mathbb{H}^{n}(c)\colon\operatorname{dist}_{\mathbb{H}^{n}(c)}(x,\mathcal{H})=\frac{1}{2\sqrt{-c}}\log\frac{\sqrt{-c}+\lambda}{\sqrt{-c}-\lambda}\right\\}.$

We will call
$\mathfrak{r}_{\lambda}:=\frac{1}{2\sqrt{-c}}\log\frac{\sqrt{-c}+\lambda}{\sqrt{-c}-\lambda}$
the _characteristic distance_ for the equidistant
$\mathcal{S}_{\lambda}\subset\mathbb{H}^{n}(c)$. In the equidistant case,
$\mathcal{B}_{\lambda}$ is not compact and there is a cone of directions to
infinity within $\mathcal{B}_{\lambda}$.

In the Poincaré unit ball model of $\mathbb{H}^{n}(c)$, the hypersurfaces
$\mathcal{S}_{\lambda}$ are Euclidean spheres that: lie inside the unit ball
if $\lambda>\sqrt{-c}$, touch the unit sphere if $\lambda=\sqrt{-c}$, and
intersect the unit sphere at an angle different from $\pi/2$ if
$\lambda<\sqrt{-c}$. Note that when the angle of intersection tends to
$\pi/2$, the corresponding equidistant hypersurface tends to a totally
geodesic hyperplane.

###  2.2. $\lambda$-convex bodies

Fix some $\lambda>0$. Recall that for every totally umbilical hypersurface
$\mathcal{S}_{\lambda}$ in a model space, we denote by $\mathcal{B}_{\lambda}$
the corresponding convex region bounded by $\mathcal{S}_{\lambda}$.

######  Definition 2.1 ($\lambda$-convex body).

A convex body $K\subset M^{n}(c)$ is _$\lambda$ -convex_ if for each
$p\in\partial K$ there exists a neighborhood $U_{p}\subset M^{n}(c)$ and a
totally umbilical hypersurface $\mathcal{S}_{\lambda}$ that passes through $p$
in such a way that

$U_{p}\cap\partial K\subset\mathcal{B}_{\lambda}$

(see Figure 2).

Figure 2. $K$ is a convex body, $\mathcal{S}_{\lambda}$ is a totally umbilical
hypersurface of constant normal curvature $\lambda>0$, e.g., a sphere. $K$ is
_$\lambda$ -convex_ if this local picture holds in some neighborhood $U_{p}$
around every point $p\in\partial K$. In this case, we say that the principal
curvatures of $\partial K$ are bounded below by $\lambda$ in the _barrier
sense_.

For a $\lambda$-convex body $K$, we call
$\mathcal{B}_{\lambda}=\mathcal{B}_{\lambda}(p)$ a _supporting convex set of
curvature $\lambda$_ (passing through $p\in\partial K$). If
$\mathcal{B}_{\lambda}$ is a geodesic ball, we will call it a _supporting
ball_ (and its boundary $\mathcal{S}_{\lambda}$ a _supporting sphere_) at the
point $p$.

The following classical theorem of Blaschke turns the local condition in
Definition 2.1 into a global one. It was first proven in [Bla] for
$\lambda$-convex bodies with smooth boundary in $\mathbb{R}^{2}$, and was
later extended in [Ra, BrS] for general $\lambda$-convex domains in
$\mathbb{R}^{n}$, and in [Ka, Mi1, Ho, BD4, Dr5] for $\lambda$-convex bodies
in other model spaces (and more generally, in some Riemannian manifolds of
bounded curvature; see [Dr5] and references therein).

######  Theorem 2.2 (Blaschke’s Rolling Theorem).

Let $K\subset M^{n}(c)$ be a $\lambda$-convex body. Then for every
$p\in\partial K$ there exists a supporting convex set
$\mathcal{B}_{\lambda}(p)$ of curvature $\lambda$ such that
$K\subseteq\mathcal{B}_{\lambda}(p)$. ∎

Finally, let us point out an equivalent definition of $\lambda$-convexity
which is, in some sense, dual to Definition 2.1. The equivalence can be seen
using Theorem 2.2. A convex body $K\subset M^{n}(c)$ is said to be
$\lambda$-convex if and only if any pair of points in $K$ can be connected by
the shortest arc of a curve of constant geodesic curvature $\lambda$ and this
arc lies entirely in $K$. From this point of view, $\lambda$-convexity is a
generalization of the standard notion of convexity where we say that a body is
convex if and only if the line segment that connects any two points lies
entirely inside the body. It follows that the whole spindle hypersurface, that
is the locus of all arcs connecting the two points, and the domain bounded by
this hypersurface must lie inside the body (this justifies the name “spindle
convexity” used by some authors; see Section 2.3). Spindle hypersurfaces will
appear in our discussion in Appendix B.

###  2.3. $\lambda$-convex polytopes and their properties

In this subsection, we define a special class of $\lambda$-convex bodies.

######  Definition 2.3 ($\lambda$-convex polytope).

A _$\lambda$ -convex polytope_ in $M^{n}(c)$ is a $\lambda$-convex body given
as the intersection of finitely many convex regions $\mathcal{B}_{\lambda}$.
Each _facet_ of the $\lambda$-convex polytope is the intersection of the
corresponding $\mathcal{B}_{\lambda}$ with the boundary of the polytope. A
facet is of dimension $n-1$. Two facets are either disjoint, or intersect. In
the latter case, their intersection is called a (lower-dimensional) _face_.
The faces of dimension $0$ and $1$ are called vertices and edges of the
polytope, respectively.

######  Definition 2.4 ($\lambda$-convex lens).

A $\lambda$-convex polytope with exactly two facets is called a _$\lambda$
-convex lens_ (or a _UFO body_) (see Figure 1).

Note that for $n\geqslant 3$, a $\lambda$-convex lens has exactly one
$(n-2)$-dimensional face and no faces of dimension less than $n-2$. For $n=2$,
there are exactly two $0$-dimensional faces (i.e., the vertices of the
$2$-dimensional lens). Each lens is reflection-symmetric with respect to the
totally geodesic hyperplane passing through its $(n-2)$-dimensional face(s).
In $\mathbb{R}^{n}$, a $\lambda$-convex lens is the intersection of two balls
of radius $1/\lambda$.

We would like to point out a direct correspondence between $\lambda$-convex
polytopes and _ball-polyhedra_ , i.e., bodies that are obtained as
intersections of finitely many geodesic balls of a fixed radius. These objects
from discrete geometry have been intensively studied, see e.g., [BLNP] and
references therein. The notion of $\lambda$-convexity in the context of ball-
polyhedra appears in a number of papers as “$r$-hyperconvexity”, “spindle
convexity”, “ball convexity”, and “$r$-duality”, e.g., [BLNP, Be2, FKV, LNT].
However, in our definition of $\lambda$-convexity in the hyperbolic space, we
allow for supporting hypersurfaces to be not only balls, but also horospheres
or equdistants. The former case is known as an instance of _$h$ -convexity_.
From this point of view, motivated by some questions in geometric probability,
$\lambda$-convexity was studied in Riemannian spaces, e.g., in [BGR, BM].

The following proposition is a direct corollary of Blaschke’s rolling theorem
(Theorem 2.2):

######  Proposition 2.5 (Approximation by $\lambda$-convex polytopes).

Let $K\subset M^{n}(c)$ be a $\lambda$-convex body. Then there exists a
sequence $(P_{i})_{i=1}^{\infty}$ of $\lambda$-convex polytopes such that
$K\subseteq...\subseteq P_{2}\subseteq P_{1}$ and
$\lim\limits_{i\rightarrow\infty}P_{i}=K$ in Hausdorff metric. ∎

Observe that an implicit assumption in Definition 2.3 is that the intersection
of balls in question is compact. In many cases, this is automatic:

######  Lemma 2.6 (Compact intersection of balls).

Let $\mathcal{B}_{1},\mathcal{B}_{2}\subset M^{n}(c)$ be a pair of convex
bodies bounded by totally umbilical hypersurfaces of curvature $\lambda$.
Assume $\mathcal{B}_{1}$ is not contained in $\mathcal{B}_{2}$, and
$\mathcal{B}_{2}$ is not contained in $\mathcal{B}_{1}$. If $c\geqslant 0$, or
$c<0$ and $\lambda\geqslant\sqrt{-c}$, then
$\mathcal{B}_{1}\cap\mathcal{B}_{2}$ is a compact set.

###### Proof.

It follows from the classification of totally umbilical hypersurfaces, see
Section 2.1. ∎

However, Lemma 2.6 is false in this generality in the hyperbolic equidistant
case, i.e., when $\lambda<\sqrt{-c}$. Since each $\mathcal{B}_{i}$ is
unbounded and contains a cone of directions to infinity, this causes extra
difficulty. Nonetheless, we can still get a compact intersection provided that
sufficiently many equidistant balls are taken and that they are sufficiently
‘close’ to each other. More precisely, we have:

######  Lemma 2.7 (Compact intersection of equidistant balls).

Let $B\subset\mathbb{H}^{n}(c)$, $c<0$, be a geodesic ball, and let
$\mathcal{B}_{1},\ldots,\mathcal{B}_{m}$ be a collection of $m$ convex regions
in $\mathbb{H}^{n}(c)$ bounded by equidistants of curvature
$\lambda\in(0,\sqrt{-c})$. Assume also that $B\subset\mathcal{B}_{i}$ and
$\partial B$ touches $\partial\mathcal{B}_{i}$ at some point $p_{i}$ for each
$i\in\\{1,\ldots,m\\}$. Suppose that the radius of $B$ is less than the
characteristic distance $\mathfrak{r}_{\lambda}$ and that the set of touching
points ${\bf p}=\\{p_{1},\ldots,p_{m}\\}$ does not lie in any open half-space
with respect to any hyperplane passing through the center of $B$. Then

$\mathcal{B}_{\bf p}:=\bigcap_{i=1}^{m}\mathcal{B}_{i}$

is a compact set.

######  Remark.

Lemma 2.7 is false if $r\geqslant\mathfrak{r}_{\lambda}$ (compare to Lemma 2.8
(3) below).

###### Proof of Lemma 2.7.

For the proof, we need an auxiliary construction as follows. Let
$\mathcal{H}\subset\mathbb{H}^{n}(c)$ be a hyperplane passing through the
center of $B$. Denote by $S=S(\mathcal{H}):=\partial B\cap\mathcal{H}$ the
corresponding great sphere. For $p\in S$, let $\mathcal{B}_{\lambda}(p)\supset
B$ be the convex region bounded by the equidistant of curvature $\lambda$ that
touches the ball $B$ at $p$. Let $r$ be the radius of $B$. The assumption
$r<\mathfrak{r}_{\lambda}$ is equivalent to the fact that

$\mathcal{Y}:=\bigcap_{p\in S}\mathcal{B}_{\lambda}(p)$

is a compact (convex) body. Equivalently, this means that for each such great
sphere $S$, one can find a point, say $q=q(S)\in\mathbb{H}^{n}(c)$, such that
each equidistant arc of curvature $\lambda$ joining $q$ with a point
$q^{\prime}\in S$ is perpendicular to $\mathcal{H}$ at $q^{\prime}$. By
symmetry, the point $q$ lies on the geodesic line $\gamma$ passing through the
center of $B$ perpendicular to $\mathcal{H}$. Moreover, there are exactly two
such points, one per each half-space with respect to $\mathcal{H}$. Call these
points $q^{+}=q^{+}(S)$ and $q^{-}=q^{-}(S)$. Now, for a point
$p\in\mathbb{H}^{n}(c)$, define $\mathcal{X}(B,p)$ to be the intersection of
all convex sets of the form $\mathcal{B}_{\lambda}$ containing both $B$ and
$p$. In this notation,

$\mathcal{X}(B,q^{+})=\mathcal{X}(B,q^{-})=\mathcal{Y}.$

In general, $\mathcal{X}(B,p)$ is not necessarily compact. However, since
$\mathcal{X}(B,q^{\pm})$ are compact, for every point $p$ sufficiently close
in the hyperbolic metric to either $q^{+}$ or $q^{-}$, the set
$\mathcal{X}(B,p)$ is also compact. Moreover, if, additionally, $p$ lies on
the geodesic $\gamma$ outside of the segment $[q^{-},q^{+}]$ and closer to,
say $q^{+}$, then both $\partial\mathcal{X}(B,p)\cap\partial B$ and $q^{-}$
are contained in the same open half-space with respect to $\mathcal{H}$. By
symmetry, the intersection $\partial\mathcal{X}(B,p)\cap\partial B$ lies in
some hyperplane $\mathcal{I}(p)$. Let $\mathcal{I}^{+}(p)$ be the open half-
space with respect to $\mathcal{I}(p)$ that contains $q^{+}$. Finally, set

$\mathcal{Y}^{+}(p):=B\cup\left(\mathcal{X}(B,p)\cap\mathcal{I}^{+}(p)\right).$

Let us return back to the proof of the lemma. Assume the contrary, and let
$\gamma\subset\mathcal{B}_{\bf p}$ be a geodesic ray that starts at the center
of $B$ and goes to infinity. Let $\mathcal{H}_{\gamma}$ be the hyperplane
passing through the center of $B$ perpendicular to $\gamma$, and let
$S_{\gamma}:=S(\mathcal{H}_{\gamma})$ be the corresponding great sphere. Since
$B\subset\mathcal{B}_{\bf p}$, Blaschke’s rolling theorem (Theorem 2.2)
guarantees that each equidistant arc of curvature $\lambda$ that joins a point
in $\gamma$ with a point in $\partial B$, lies in $\mathcal{B}_{\bf p}$. In
particular, $\mathcal{Y}^{+}(p)\subset\mathcal{B}_{\bf p}$ for some
$p\in\gamma$ sufficiently close to $q^{+}(S_{\gamma})\in\gamma$. But
$S_{\gamma}$ lies in the interior of $\mathcal{Y}^{+}(p)$ by construction.
This is a contradiction since the open half-space with respect to
$\mathcal{H}_{\gamma}$ that contains $p$ does not contain points in ${\bf p}$.
∎

###  2.4. Inscribed balls and their properties

Let $K\subset M^{n}(c)$ be a convex body. A geodesic ball of the largest
possible radius contained in $K$ is called an _inscribed ball for $K$_. The
_inradius of $K$_, denoted by $r(K)$, is the radius of an inscribed ball for
$K$. Note that, even though there might be several inscribed balls for $K$
(even when $K$ is strictly convex), they must have equal radii. Thus, $r(K)$
is well-defined. The question of uniqueness is addressed in the following
lemma:

######  Lemma 2.8 (On uniqueness of inscribed balls).

Let $K\subset M^{n}(c)$ be a strictly convex body.

  1. (1)

If $c\geqslant 0$, then $K$ has a unique inscribed ball;

  2. (2)

If $c<0$ and $K$ is $\lambda$-convex with $\lambda\geqslant\sqrt{-c}$, then
$K$ has a unique inscribed ball;

  3. (3)

For each $c<0$ and $\lambda^{\prime}\in(0,\sqrt{-c})$, there is a
$\lambda^{\prime}$-convex body $K$ such that $K$ has exactly two inscribed
balls; in particular, in the hyperbolic space, a $\lambda$-convex body can
have several inscribed balls.

###### Proof.

The argument in the Euclidean space is well-known and goes, for example, as
follows. If there are two distinct balls $B_{1},B_{2}\subset
K\subset\mathbb{R}^{n}$, necessarily of the same radius $r:=r(K)$, then the
convex hull $H$ of $B_{1}$ and $B_{2}$ lies entirely in $K$. Moreover, since
$K$ is strictly convex, any ball $B\subset H$ of radius $r$ different from
$B_{1}$ and $B_{2}$ must lie in the interior of $K$. Hence, one can enlarge
$B$ (keeping its center the same) so that the larger ball still lies in $K$.
This contradicts the maximality of $r$.

We can reinterpret this argument so that it also works in $\mathbb{S}^{n}$.
Again, by contradiction, we assume that there are two different inscribed
balls $B_{1}$ and $B_{2}$ for $K\subset\mathbb{S}^{n}$.

Let $H^{\prime}$ be the locus of balls of radius $r$ centered on the geodesic
segment $I$ connecting the centers of $B_{1}$ and $B_{2}$. Equivalently,
$H^{\prime}$ is the set of points at distance at most $r$ from $I$. In the
Euclidean space, $H^{\prime}=H$. In the spherical case, we have that
$H^{\prime}\subsetneq H\subset K$, and hence the argument above applied to
$H^{\prime}$ yields that a strictly convex body $K\subset\mathbb{S}^{n}$ has
the unique inscribed ball. This concludes (1).

Now we consider the hyperbolic case. For simplicity, assume $c=-1$ and write
$\mathbb{H}^{n}:=\mathbb{H}^{n}(-1)$. The inclusion $H^{\prime}\subsetneq H$
does not hold in $\mathbb{H}^{n}$. Observe, however, that $\partial
H^{\prime}\setminus(\partial B_{1}\cup\partial B_{2})$ is $\mu$-convex, with
$\mu=\tanh(2r)\in(0,1)$ (i.e., $r=\mathfrak{r}_{\mu}$, see (3c) in Section
2.1). If we are in the assumption of case (2), then $\mu<\lambda$. By
Blaschke’s rolling theorem it follows that $H^{\prime}\subsetneq K$, and hence
we can reach a contradiction in the same way as in $\mathbb{R}^{n}$ and
$\mathbb{S}^{n}$. This concludes (2).

Figure 3. In the Poincaré disk model of $\mathbb{H}^{2}$: the body
$H^{\prime\prime}$ is the concatenation of two circular arcs on the boundary
of disks $B_{1}$ and $B_{2}$ (in blue), and two green arcs of curves
$\mathcal{S}_{\lambda^{\prime}}$ of constant curvature
$\lambda^{\prime}\in(0,1)$ (in green). The outer parallel set $H^{\prime}$ of
the segment $I$ is the concatenation of two orange arcs
$\mathcal{S}_{\lambda}$ and two blue circular arcs on the boundary of the
disks. For comparison, the disk centered at the origin has the same radius as
$B_{1}$ and $B_{2}$. In this example, $B_{1}$ and $B_{2}$ are the only
inscribed disks for $H^{\prime\prime}$.

In general, there is no guarantee that $H^{\prime}\subset K$. Also,
$H^{\prime}$ provides an example of a $\mu$-convex set, with $\mu\in(0,1)$,
for which all balls of radius $r$ inside $H^{\prime}$ are inscribed, and there
is a continuous family of such balls. To conclude (3), one can modify this
example as follows (see Figure 3 for a construction in $\mathbb{H}^{2}$).
Given $\lambda^{\prime}\in(0,1)$, pick a geodesic segment
$I\subset\mathbb{H}^{n}$ and let $H^{\prime}$ be defined as above for
$r>\mathfrak{r}_{\lambda^{\prime}}$. Consider all possible arcs of curves of
curvature $\lambda^{\prime}$ joining points in $B_{1}$ and $B_{2}$. These arcs
will sweep out a $\lambda^{\prime}$-convex body $H^{\prime\prime}$ for which
$B_{1}$ and $B_{2}$ are the only inscribed balls. ∎

The following proposition gives a necessary and some sufficient conditions
under which a ball inside a convex body is an inscribed ball. Parts of this
proposition are well-known (and are not the most general) and we outline the
proof for completeness of exposition.

######  Proposition 2.9 (Necessary and sufficient conditions for inscribed
balls).

Let $K\subset M^{n}(c)$ be a convex body, and let $B\subset K$ be a geodesic
ball centered at some point $o\in K$.

  1. (1)

If $B$ is an inscribed ball for $K$, then $\partial B\cap\partial K$ is not
contained in any open half-space with respect to a hyperplane passing through
$o$.

  2. (2)

If $K$ is a $\lambda$-convex polytope, $\partial B\cap\partial K$ is not
contained in any open half-space with respect to a hyperplane passing through
$o$, and $B$ touches all facets of $K$, then $B$ is an inscribed ball for $K$.
Moreover, it is unique.

###### Proof.

We start with part (1). Towards a contradiction, suppose that $S:=\partial
B\cap\partial K$ lies in some open half space $\mathcal{H}^{+}$. Write
$\mathcal{H}:=\partial\mathcal{H}^{+}$, $o\in\mathcal{H}$. Let
$\mathcal{H}^{-}:=M^{n}(c)\setminus\mathcal{H}^{+}$ be the complement of
$\mathcal{H}^{+}$. Pick a small $\varepsilon>0$, and consider a geodesic
segment $I_{\varepsilon}\subset\mathcal{H}^{-}$ of length $\varepsilon$,
starting at $o$ and perpendicular to $\mathcal{H}$. Let $r=r(K)$ be the radius
of $B$. Define $H^{\prime}_{\varepsilon}$ to be the locus of balls of radius
$r=r(K)$ centered at $I_{\varepsilon}$. Since $B$ and $K$ are compact and $S$
is contained in $\mathcal{H}^{+}$, the distance

$\operatorname{dist}_{M^{n}(c)}\left(\partial K,\mathcal{H}^{-}\cap B\right)$

is strictly positive. Therefore, there exists $\varepsilon$ such that
$H^{\prime}_{\varepsilon}\subset K$. To reach a contradiction, we can apply
the argument that is similar to the one that was used in the proof of Lemma
2.8 (1).

Now we deal with part (2). We will give an argument in the hyperbolic case; in
other cases, the argument is similar and we omit it.

Assume first that $\partial B\cap\partial K$ contains at least $n+1$ distinct
points $p_{1},\ldots,p_{n+1}$ such that the smallest totally geodesic subspace
containing this set of points is equal to $M^{n}(c)$. By assumption, each
$p_{i}$ is the tangent point of $B$ and some of the facets. In the Poincaré
disk model of $\mathbb{H}^{n}(c)$, each $\mathcal{S}_{\lambda}(p_{i})$ is a
Euclidean sphere. Furthermore, $\partial B$ is a Euclidean sphere too. But in
$\mathbb{R}^{n}$, there exists only one sphere that lies in the interior and
touches any given collection of $n+1$ mutually distinct spheres. If
$B^{\prime}$ is the inscribed ball for $K$, then $\partial
B^{\prime}\cap\partial K$ is not contained in any open half space with respect
to the origin (part (1)). By the uniqueness described above, $B=B^{\prime}$.

In general, let $1\leqslant m<n$ be the smallest dimension of a totally
geodesic subspace $Y\subset M^{n}(c)$ such that $o\in Y$ and $\partial
B\cap\partial K\subset Y$. Since $\partial B\cap\partial K$ is not contained
in any open half-space with respect to a hyperplane passing through $o$, the
body $K$ must be symmetric with respect to $Y$. Hence, any inscribed ball for
$K$ is also symmetric with respect to $Y$. Note that within $Y$, the points in
$\partial B\cap\partial K$ cannot lie in any open half-space of $Y$ with
respect to a plane of dimension $m-1$ passing through $o$. By minimality of
$m$, we can choose at least $m+1$ distinct points $p_{1},\ldots,p_{m+1}$ in
$\partial B\cap\partial K$ so that the geodesic rays $op_{1},\ldots,op_{m+1}$
span $Y$. We can now apply a similar uniqueness argument as above. For every
$m<n$, there exists a unique Euclidean sphere in $\mathbb{R}^{n}$ that lies
inside and touches any $m$ given Euclidean spheres, and is symmetric with
respect to an $m$-dimensional Euclidean plane (we can assume that $o$ is the
center of the unit ball model). The proposition now follows in the same way as
in the case of $n+1$ distinct points. ∎

The following proposition relates the extreme properties of the inradius to
the structure of the boundary of a $\lambda$-convex body.

######  Proposition 2.10 (Extreme properties of the inradius).

Let $K\subset M^{n}(c)$ be a $\lambda$-convex body. (In the case $c<0$ and
$\lambda<\sqrt{-c}$, we additionally assume that
$r(K)<\mathfrak{r}_{\lambda}$.)

If $K\subset M^{n}(c)$ is not a $\lambda$-convex polytope, then one can find a
$\lambda$-convex polytope $K^{\prime}$ such that

(2.1) $r(K^{\prime})=r(K)\quad\text{ and }\quad|\partial K^{\prime}|>|\partial
K|.$

Equivalently, there exists a $\lambda$-convex polytope $K^{\prime\prime}$ such
that

(2.2) $r(K^{\prime\prime})<r(K)\quad\text{ and }\quad|\partial
K^{\prime\prime}|=|\partial K|.$

###### Proof.

We start with (2.1).

Let $B$ be an inscribed ball for $K$, and let $o$ be its center. By
Proposition 2.9 (1), $\partial B\cap\partial K$ is not contained in any open
half space with $o$ on its boundary. Pick a subset $T\subset\partial
B\cap\partial K$ consisting of at most $n+1$ points that have the same
property, i.e., they do not lie in any open half space with respect to
hyperplanes through $o$. Since $K$ is $\lambda$-convex, for each $p\in T$
there exists a supporting convex set $\mathcal{B}_{\lambda}(p)$ of curvature
$\lambda$ passing through $p$. Define the following intersection:

$K^{\prime}=\bigcap_{p\in T}\mathcal{B}_{\lambda}(p).$

By Lemmas 2.6 and 2.7, $K^{\prime}$ is a $\lambda$-convex polytope. By
Blaschke’s rolling theorem (Theorem 2.2), $K^{\prime}\supset K$. Hence,
$|\partial K^{\prime}|>|\partial K|$. This inequality is strict because $K$ is
not a $\lambda$-convex polytope by assumption. Furthermore, $B\subset
K^{\prime}$ and $T\subset\partial B$ by construction. By part (2) of
Proposition 2.9, $B$ is the inscribed ball for $K^{\prime}$. Therefore,
$r(K^{\prime})=r(K)$. This concludes (2.1).

To prove (2.2), we use (2.1) and a monotonicity argument. Indeed, let
$K^{\prime}$ be the polytope that satisfies (2.1), and let $B$ be its
inscribed ball of radius $r=r(K)$. For $s\leqslant r$, let $B^{\prime}$ be a
ball of radius $s$ concentric with $B$. For each $p\in T$, let $p^{\prime}$ be
the intersection of the geodesic ray $op$ with $\partial B^{\prime}$. Let
$T^{\prime}$ be the set of all such $p^{\prime}$’s. For each $p^{\prime}$, let
$\mathcal{S}_{\lambda}(p^{\prime})$ be the $\lambda$-convex hypersurface that
touches $B^{\prime}$ at $p^{\prime}$ in such a way that
$\mathcal{B}_{\lambda}(p^{\prime})\supset B^{\prime}$. Finally, define

$K^{\prime\prime}(s):=\bigcap_{p^{\prime}\in
T^{\prime}}\mathcal{B}_{\lambda}(p^{\prime}).$

Since $s\leqslant r$, $K^{\prime\prime}(s)$ is a $\lambda$-convex polytope. By
Proposition 2.9 (2), $r\left(K^{\prime\prime}(s)\right)=s$. By construction,
$K^{\prime\prime}(r)=K^{\prime}$, and hence by assumption, $|\partial
K^{\prime\prime}(r)|>|\partial K|$. By continuity, there exists $s_{0}<r$ such
that $|\partial K^{\prime\prime}(s_{0})|=|\partial K|$, and the proposition
follows with $K^{\prime\prime}=K^{\prime\prime}(s_{0})$. ∎

######  Remark.

We will use a similar monotonicity argument in the proof of Theorem B.

###  2.5. Inner parallel bodies in $\mathbb{R}^{n}$ and their properties

Let $A$ be a convex body in $\mathbb{R}^{n}$. The _inner parallel body_ at
distance $t\geqslant 0$ is defined as

$A_{t}=A-t{\bm{B}}:=\\{x\in\mathbb{R}^{n}:x+t{\bm{B}}\subset A\\},$

where ${\bm{B}}$ is the unit Euclidean ball. The greatest number $t$ for which
$A_{t}$ is not empty is the inradius $r(A)$ of $A$. Observe that if $A\subset
A^{\prime}\subset\mathbb{R}^{n}$ are two convex bodies, then $A_{t}\subset
A^{\prime}_{t}$.

The Hausdorff distance of convex bodies $A,A^{\prime}$ in $\mathbb{R}^{n}$ is
defined as

$d_{H}(A,A^{\prime})=\min\\{\mu\geqslant 0:\ A\subset A^{\prime}+\mu{\bm{B}},\
A^{\prime}\subset A+\mu{\bm{B}}\\}.$

For a convex body $A\subset\mathbb{R}^{n}$ and $0\leqslant\varepsilon\leqslant
r(A)$, define

(2.3) $T_{A}(\varepsilon):=\min\left\\{\mu\geqslant
0:A-\varepsilon{\bm{B}}+\mu{\bm{B}}\supset A\right\\}.$

Clearly, $T_{A}(\varepsilon)\geqslant\varepsilon$, but this inequality can be
strict. Observe that $T_{A}(\varepsilon)=d_{H}(A,A_{\varepsilon})$.

######  Lemma 2.11 (Dependence on $\varepsilon$).

$T_{A}(\varepsilon)\to 0\quad\text{as}\quad\varepsilon\searrow 0.$

###### Proof.

For monotonically decreasing $\varepsilon$, the bodies in the sequence
$A_{\varepsilon}=A-\varepsilon{\bm{B}}$ are nested, i.e.,
$A_{\varepsilon}\subsetneq A_{\varepsilon^{\prime}}$ if
$\varepsilon>\varepsilon^{\prime}$. Hence, by Blaschke’s Selection Theorem
[Sch, p. 63], there exists the Hausdorff limit of this sequence and it is
equal to the convex body

$A^{\prime}:=\operatorname{Cl}\left(\bigcup_{\varepsilon>0}A_{\varepsilon}\right),$

where $\operatorname{Cl}$ denotes the closure of a set. Note that the interior
of $A^{\prime}$, denoted by $\operatorname{int}A^{\prime}$, is equal to
$\bigcup_{\varepsilon>0}A_{\varepsilon}$. Finally, observe that for each
$x\in\operatorname{int}A$ there exists $\delta=\delta(x)$ (which can be
arbitrary small) so that $x+\delta{\bm{B}}\subset A$. For example, $\delta$
can be chosen to be the half of the distance of $x$ to $\partial A$. By
definition of inner parallel bodies, $x\in A_{\delta}$. Thus,
$x\in\operatorname{int}A^{\prime}$ for every $x\in\operatorname{int}A$, which
implies that $\operatorname{int}A^{\prime}=\operatorname{int}A$, and hence
$A=A^{\prime}$. By construction, $d_{H}(A^{\prime},A_{\varepsilon})\to 0$ as
$\varepsilon\to 0$. Thus, $T_{A}(\varepsilon)=d_{H}(A,A_{\varepsilon})\to 0$
as $\varepsilon\to 0$, which is the claim of the lemma. ∎

######  Lemma 2.12 (Dependence on $A$).

Let $A$ and $A^{\prime}$ be convex bodies in $\mathbb{R}^{n}$ such that
$A^{\prime}\supset A$. Then

$T_{A^{\prime}}(\varepsilon)\leqslant T_{A}(\varepsilon)+d_{H}(A,A^{\prime}).$

###### Proof.

We have the following implications:

$\displaystyle A^{\prime}\supset A\ \Rightarrow\
A^{\prime}-\varepsilon{\bm{B}}\supset A-\varepsilon{\bm{B}}\ \Rightarrow\
A^{\prime}-\varepsilon{\bm{B}}+T_{A}(\varepsilon){\bm{B}}\supset
A-\varepsilon{\bm{B}}+T_{A}(\varepsilon){\bm{B}}\supset A,$

where the last inclusion follows by (2.3). Using the definition of the
Hausdorff distance, we obtain

$\displaystyle
A^{\prime}-\varepsilon{\bm{B}}+T_{A}(\varepsilon){\bm{B}}\supset A\
\Rightarrow\
A^{\prime}-\varepsilon{\bm{B}}+T_{A}(\varepsilon){\bm{B}}+d_{H}(A,A^{\prime}){\bm{B}}\supset
A+d_{H}(A,A^{\prime}){\bm{B}}\supset A^{\prime}.$

Therefore,
$A^{\prime}-\varepsilon{\bm{B}}+\left(T_{A}(\varepsilon)+d_{H}(A,A^{\prime})\right){\bm{B}}\supset
A^{\prime}$, and hence $T_{A^{\prime}}(\varepsilon)\leqslant
T_{A}(\varepsilon)+d_{H}(A,A^{\prime})$, as desired. ∎

######  Lemma 2.13 (Hausdorff limits of nested sequences).

If $A^{1}\supset A^{2}\supset...\supset C$ is a nested sequence of convex
bodies, then

$\mathcal{A}:=\bigcap_{j=1}^{\infty}A^{j}$

is a convex body and

$d_{H}\left(A^{j},\mathcal{A}\right)\to 0\quad\text{ as }\quad j\to\infty.$

###### Proof.

It is immediate that $\mathcal{A}$ is a closed convex set. Since each $A^{j}$
contains the convex body $C$, the interior of $\mathcal{A}$ is not empty.
Hence, $\mathcal{A}$ is a convex body. The convergence follows by [Sch, Lemma
1.8.2]. ∎

Using the three lemmas above, we establish the following auxiliary result.

######  Lemma 2.14 (Approximation descends to inner parallel bodies).

If $A\subset\mathbb{R}^{n}$ is a convex body, and $(A^{j})_{j=1}^{\infty}$ is
a sequence of convex bodies such that

$A^{j}\supset A^{j+1}\supset A\quad\text{ and
}\quad\lim_{j\to\infty}d_{H}\left(A,A^{j}\right)=0,$

then for the inner parallel bodies at distance $0\leqslant t<r(A)$, we have

$\lim_{j\to\infty}d_{H}\left(A_{t},A_{t}^{j}\right)=0.$

###### Proof.

Since $(A^{j})_{j=1}^{\infty}$ is a nested sequence that approximates $A$,
there is a monotone sequence $\varepsilon_{j}\searrow 0$ such that

$d_{H}(A,A^{j})=\varepsilon_{j}.$

Hence,

$A+\varepsilon_{j}{\bm{B}}\supset A^{j}\ \ \Rightarrow\ \ A\supset
A^{j}-\varepsilon_{j}{\bm{B}}\ \ \Rightarrow\ \ A-t{\bm{B}}\supset
A^{j}-\varepsilon_{j}{\bm{B}}-t{\bm{B}}.$

In our notation, the last inclusion also reads as $A_{t}\supset
A_{t}^{j}-\varepsilon_{j}{\bm{B}}$. Since by (2.3)

$A_{t}^{j}-\varepsilon_{j}{\bm{B}}+T_{A_{t}^{j}}(\varepsilon_{j}){\bm{B}}\supset
A_{t}^{j},$

we conclude that

(2.4) $A_{t}+T_{A_{t}^{j}}(\varepsilon_{j}){\bm{B}}\supset A_{t}^{j}.$

By assumption of the lemma, $A^{j}\supset A$, so $A^{j}_{t}\supset A_{t}$.
Hence, (2.4) implies that

$d_{H}\left(A_{t},A_{t}^{j}\right)\leqslant T_{A_{t}^{j}}(\varepsilon_{j}).$

It remains to show that $T_{A_{t}^{j}}(\varepsilon_{j})\to 0$ as $j\to\infty$.

Since the sequence $\left(A^{j}\right)_{j=1}^{\infty}$ is decreasing, so is
the sequence $\left(A^{j}_{t}\right)_{j=1}^{\infty}$. Moreover, each
$A^{j}_{t}$ contains $A_{t}$. The latter set is a (non-empty) convex body by
our choice of $0\leqslant t<r(A)$. Hence, by Lemma 2.13,

$\mathcal{A}:=\bigcap_{j=1}^{\infty}A_{t}^{j}$

is a convex body and is the Hausdorff limit of $A_{t}^{j}$ as
$j\rightarrow\infty$. Since $A_{t}^{j}\supset\mathcal{A}$, Lemma 2.12
guarantees that

$T_{A_{t}^{j}}(\varepsilon_{j})\leqslant
T_{\mathcal{A}}(\varepsilon_{j})+d_{H}(\mathcal{A},A_{t}^{j})$

for each $j\geqslant 1$. By Lemma 2.13,

$d_{H}(\mathcal{A},A_{t}^{j})\to 0\quad\text{ as }\quad j\to\infty.$

Since $\varepsilon_{j}\searrow 0$ as $j\to\infty$, Lemma 2.11 guarantees that

$T_{\mathcal{A}}(\varepsilon_{j})\to 0\quad\text{ as }\quad j\to\infty.$

Combining everything together, we get that
$\lim\limits_{j\to\infty}T_{A_{t}^{j}}(\varepsilon_{j})=0$. This concludes the
proof. ∎

##  3\. Proof of the reverse isoperimetric inequality in $\mathbb{R}^{3}$
(Theorem A)

The proof breaks into two steps. In the first step, we establish the theorem
when $K$ is a $\lambda$-convex polytope. In the second step, we use
approximation (Proposition 2.5) and the result of the first step to prove the
theorem in general: the reverse isoperimetric inequality will follow by
continuity, and thus the only thing that we would have to deal with is the
equality case.

###  3.1. Step I: $K$ is a $\lambda$-convex polytope

Let $K$ be a $\lambda$-convex polytope (see Definition 2.3). Without loss of
generality, we set $\lambda=1$. We can also assume that $K$ is not a lens (as
for lenses the claim of the theorem is true). It is known (see, e.g., [Ma])
that $\frac{d}{dt}|K_{t}|=|\partial K_{t}|$ at any $t\geqslant 0$, where
$K_{t}$ is the inner parallel body of $K$ at distance $t$. Thus,

$|K|=\int\limits_{0}^{r(K)}|\partial K_{t}|dt.$

Observe that $f_{K}\colon t\mapsto|\partial K_{t}|$ is a strictly decreasing,
absolutely continuous function, and hence the integral makes sense (see [Sch,
p. 439] for the question about differentiability of such functions for general
convex bodies).

For the lens $L$, let $f_{L}\colon t\mapsto|\partial L_{t}|$ be an analogous
function that measures the area of the inner parallel bodies of $L$. By
assumption of the theorem, $f_{K}(0)=f_{L}(0)$. The following key proposition
relates the values of $f_{K}$ and $f_{L}$ in a small right neighborhood of
$0$.

######  Proposition 3.1 (Areas of inner parallel bodies are smaller for
lenses).

$f_{K}(t)>f_{L}(t)$ for every sufficiently small $t>0$.

For the proof of Proposition 3.1, we need the following auxiliary result which
says that locally the area of a small spherical triangle behaves as the area
of a Euclidean triangle.

######  Lemma 3.2 (Area of small triangles).

Let $\Delta(t)=aob\subset\mathbb{S}^{2}$ be a spherical geodesic triangle such
that $|oa|=|ob|=O(t)$ as $t\to 0$. Then

$|\Delta(t)|=O\left(t^{2}\right).$

###### Proof.

By the triangle inequality, $\Delta(t)$ lies in a disk, say $U$, centered at
$o$ of radius $O(t)$.

For the normal coordinates centered at $o$ in the neighborhood $U$, we have
the following expansion of the metric tensor of the sphere (see, for example,
[Pet, Lemma 5.5.7]):

$g_{ij}(p)=\delta_{ij}+O(|op|^{2}),$

where $\delta_{ij}$ is the Kronecker delta.

Therefore, computing the area of $\Delta$ in normal coordinates, we obtain:

$|\Delta(t)|=|\Delta(t)|_{Euc}+O\left(|\text{diam}(U)|^{2}\right)$

Here, $|\Delta(t)|_{Euc}$ is the area of $\Delta$ computed in the Euclidean
metric (using the same coordinates). In our case,
$|\Delta(t)|_{Euc}=O(t^{2})$. The lemma follows. ∎

###### Proof of Propostion 3.1.

Our goal is to estimate $f_{K}(t)=|\partial K_{t}|$ for small positive $t$.

We start with introducing some notation (see Figure 4). Let $F_{i}$ be the
$i$-th facet of $K$ and let $o_{i}$ be the center of the unit sphere
containing $F_{i}$. Denote by $\beta_{i}$ the area of $F_{i}$, by $l_{ij}$ the
length of the common edge $E_{ij}$ for a pair of adjacent facets $F_{i}$ and
$F_{j}$, and by $\gamma_{ij}$ the angle between the unit spheres containing
$F_{i}$ and $F_{j}$.

$F_{i}$$F_{j}$$o_{j}$$o_{i}$$\gamma_{ij}$$E_{ij}$$a$ Figure 4. The notation in
the proof of Proposition 3.1: the facets $F_{i}$ and $F_{j}$ are adjacent with
$E_{ij}=F_{i}\cap F_{j}$ being the common edge; $|F_{i}|=\beta_{i}$,
$|F_{j}|=\beta_{j}$, $|E_{ij}|=l_{\ij}$. The unit spheres containing $F_{i}$,
respectively $F_{j}$, are centered at $o_{i}$, respectively $o_{j}$, and
$\gamma_{ij}=\angle o_{i}ao_{j}$ is the angle between $F_{i}$ and $F_{j}$,
where $a\in E_{ij}$ is an arbitrary point.

In this notation,

$f_{K}(0)=|\partial K|=\sum\limits_{i}\,\beta_{i}.$

We want to show that

(3.1) $\displaystyle f_{K}(t)$
$\displaystyle=\left(1-t\right)^{2}\sum\limits_{i}\,\beta_{i}-2t\sum\limits_{i,j}l_{ij}\tan{\frac{\gamma_{ij}}{2}}+O\left(t^{2}\right).$

Since $K$ is a $1$-convex polytope, it follows that $K_{t}$ is a
$(1-t)^{-1}$-convex polytope. Observe that for sufficiently small $t$, the
polytope $K_{t}$ has the same combinatorial structure as $K$; in particular,
it has the same number of facets as $K$. Denote by $\widetilde{F_{i}}(t)$ the
$i$-th facet of $K_{t}$ (corresponding to $F_{i}$), and by
$\widetilde{E}_{ij}=\widetilde{E}_{ij}(t)$ the edge of $K_{t}$ corresponding
to the edge $E_{ij}$ of $K$. Then $|\partial
K_{t}|=\sum_{i}|\widetilde{F}_{i}(t)|$. Our goal is to compute
$|\widetilde{F}_{i}(t)|$.

Define $\Pi_{t}$ to be the radial projection from the unit sphere centered at
$o_{i}$ onto the concentric sphere of radius $1-t$. Denote
$F_{i}(t):=\Pi_{t}(F_{i})$ to be the projection of the $i$-th facet $F_{i}$.
Note that $F_{i}=F_{i}(0)=\widetilde{F}_{i}(0)$. At the same time, for small
positive $t$ we have $F_{i}(t)\supset\widetilde{F}_{i}(t)$. Therefore,

$|\widetilde{F}_{i}(t)|=|F_{i}(t)|-|F_{i}(t)\backslash\widetilde{F}_{i}(t)|.$

The difference $F_{i}(t)\backslash\widetilde{F}_{i}(t)$ is the disjoint union
of spherical quadrilaterals $Q_{ij}(t)$, where $j$ ranges over indices of all
adjacent to $F_{i}$ facets of $K$. For each $j$, the quadrilateral $Q_{ij}(t)$
is bounded by $\widetilde{E}_{ij}$, $\Pi_{t}(E_{ij})$ and a pair of arcs of
big circles that join the respective endpoints of $\widetilde{E}_{ij}$ and
$\Pi_{t}(E_{ij})$ (see Figure 5). We would like to emphasize that the
endpoints of $\widetilde{E}_{ij}$ are vertices of $K_{t}$, while the endpoints
of $\Pi_{t}(E_{ij})$ are the radial projections of the vertices of $K$. We
claim that by construction, both sides $\widetilde{E}_{ij}$ and
$\Pi_{t}(E_{ij})$ of $Q_{ij}(t)$ lie on parallel planes in $\mathbb{R}^{3}$.
Indeed, if $o_{j}$ is the center of the unit sphere containing $F_{j}$, then
the plane $\mathcal{H}_{ij}\subset\mathbb{R}^{3}$ that contains $E_{ij}$ is
perpendicular to the Euclidean geodesic $o_{i}o_{j}$. The edge
$\widetilde{E}_{ij}$ also lies in $\mathcal{H}_{ij}$ (as well as in the
intersection of the two spheres of radius $1-t$ centered at $o_{i}$ and
$o_{j}$). Finally, since the projection $\Pi_{t}$ is centered at $o_{i}$ and
$\Pi_{t}(E_{ij})$ lies in a plane parallel to $\mathcal{H}_{ij}$, this implies
the claim above about the sides of the quadrilateral $Q_{ij}(t)$.

We have

$|F_{i}(t)\backslash\widetilde{F}_{i}(t)|=\sum_{j}|Q_{ij}(t)|,$

and hence our goal is to estimate each $|Q_{ij}(t)|$. For this, we decompose
$Q_{ij}(t)$ into several sets, $Q_{ij}^{e}(t)$, $Q_{ij}^{v}(t)$ and
$Q_{ij}^{v^{\prime}}(t)$, defined as follows. The set $Q_{ij}^{e}(t)$ is the
spherical quadrilateral that is bounded by $\Pi_{t}(E_{ij})$, by the extension
of the circular arc $\widetilde{E}_{ij}$, and by a pair of arcs of the big
circles passing through the endpoints of $\Pi_{t}(E_{ij})$ orthogonal to it
(see Figure 6). Note that $Q_{ij}^{e}(t)$ is not a geodesic quadrilateral,
since it has only two sides that are geodesic segments. At the same time, each
angle of $Q_{ij}^{e}(t)$ is equal to $\pi/2$. It follows that
$Q_{ij}^{e}(t)\setminus Q_{ij}(t)$ is the disjoint union of two triangular
regions, which we call $Q_{ij}^{v}(t)$ and $Q_{ij}^{v^{\prime}}(t)$. Each of
them is attached to an endpoint of $\Pi_{t}(E_{ij})$ denoted by $v$ and
$v^{\prime}$, respectively. Each such region has two sides that are spherical
geodesic segments and one angle which is equal to $\pi/2$. Therefore,
$Q_{ij}^{v}(t)$ and $Q_{ij}^{v^{\prime}}(t)$ are not geodesic triangles, but
each of them is contained in a geodesic triangle; we denote these triangles
$\widehat{Q}_{ij}^{v}(t)$ and $\widehat{Q}_{ij}^{v^{\prime}}(t)$,
respectively.

In terms of our decomposition,
$|Q_{ij}(t)|=|Q_{ij}^{e}(t)|-|Q_{ij}^{v}(t)|-|Q_{ij}^{v^{\prime}}(t)|$. Thus,

$\displaystyle f_{K}(t)$
$\displaystyle=\sum_{i}|F_{i}(t)|-\sum_{i}|F_{i}(t)\backslash\widetilde{F}_{i}(t)|$
(3.2)
$\displaystyle=\left(1-t\right)^{2}\sum\limits_{i}\,\beta_{i}-\sum_{i}\sum_{j}\left(|Q_{ij}^{e}(t)|-|Q_{ij}^{v}(t)|-|Q_{ij}^{v^{\prime}}(t)|\right).$

We show that the area of the triangular regions $Q_{ij}^{v}(t)$ and
$Q_{ij}^{v^{\prime}}(t)$ is of order at least $t^{2}$ for sufficiently small
$t$, and only $|Q_{ij}^{e}(t)|$ contributes to terms of order $t$ in (3.1).

$\Pi_{t}(E_{ij})$$E_{ij}$$\widetilde{E}_{ij}$$o_{i}$ Figure 5. The
quadrilateral $Q_{ij}(t)$ is bounded by the edge $\widetilde{E}_{ij}$ of
$K_{t}$ (in violet), the radial projection $\Pi_{t}(E_{ij})$ (in yellow) of
the edge $E_{ij}$ of $K$ (in orange), and by the pair of bright green arcs of
big circles that connect the endpoints of $\widetilde{E}_{ij}$ and
$\Pi_{t}(E_{ij})$ (one of these arcs is dashed). The green sphere is the
sphere of radius $1-t$ centered at $o_{i}$ and defined by the facet $F_{i}$.

Estimation of $|Q_{ij}^{v}(t)|$ and $|Q_{ij}^{v^{\prime}}(t)|$. We estimate
$|\widehat{Q}^{v}_{ij}(t)|$ using Lemma 3.2 and show that

(3.3) $|\widehat{Q}^{v}_{ij}(t)|=O(t^{2}).$

Since $\widehat{Q}^{v}_{ij}(t)\supset Q^{v}_{ij}(t)$, we will get the same
estimate for the area of this triangular region; an analogous estimate will
hold true for the triangular region attached to $v^{\prime}$.

Suppose $p,p^{\prime}$ and $v$ are the vertices of $\widehat{Q}^{v}_{ij}(t)$
labeled so that $pv$ is orthogonal to $\Pi_{t}(E_{ij})$ (see Figure 6). Let
$x(t)$ be the length of the spherical segment $pv$, $x^{\prime}(t)$ be the
length of the spherical segment $p^{\prime}v$. From the construction it
follows that

$x(0)=x^{\prime}(0)=0,$

hence $x(t)=x^{\prime}(t)=O(t)$. Hence, (3.3) follows by Lemma 3.2 applied to
the triangle $pvp^{\prime}$.

Estimation of $|Q_{ij}^{e}(t)|$. Now we compute a non-zero contribution of
$|Q_{ij}^{e}(t)|$ to the term of order $t$.

Since $Q_{ij}^{e}(t)$ is a spherical quadrilateral bounded by two parallels of
the sphere of radius $1-t$,

$|Q_{ij}^{e}(t)|=\left(1-t\right)\cdot\vartheta\cdot|o^{\prime}_{i}h|,$

where $o_{i}^{\prime}$ is the center of the Euclidean circle spanning
$\widetilde{E}_{ij}$, $h$ is the center of the Euclidean circle spanning
$\Pi_{t}(E_{ij})$, and $\vartheta=\angle vhv^{\prime}$ is the angle that
subtends $\Pi_{t}(E_{ij})$ (see Figure 6). We observe that
$|o^{\prime}_{i}h|=|o^{\prime}_{i}o_{i}|-|o_{i}h|$. Let $q$ be the endpoint of
$E_{ij}$ such that $\Pi_{t}(q)=v$. Since $|o_{i}q|=1$ and $\angle
o_{i}qo^{\prime}_{i}={\gamma_{ij}}/{2}$, we get
$|o_{i}o^{\prime}_{i}|=\sin({\gamma_{ij}}/{2})$. Using similarity of the
Euclidean triangles $\Delta o_{i}qo^{\prime}_{i}$ and $\Delta o_{i}vh$, we
obtain $|o_{i}h|=(1-t)\sin({\gamma_{ij}}/{2})$. Thus,

$|o^{\prime}_{i}h|=t\sin\frac{\gamma_{ij}}{2}\qquad\textup{and}\qquad|vh|=(1-t)\cos\frac{\gamma_{ij}}{2}.$

Recall that the length of $E_{ij}$ is equal to $l_{ij}$; by homothety, the
length of $\Pi_{t}(E_{ij})$ is equal to $(1-t)l_{ij}$. Therefore,

$\displaystyle\vartheta=\frac{(1-t)l_{ij}}{|vh|}=\frac{l_{ij}}{\cos\frac{\gamma_{ij}}{2}}.$

Hence,

$\displaystyle|Q^{e}_{ij}(t)|=(1-t)\frac{l_{ij}}{\cos\frac{\gamma_{ij}}{2}}\cdot
t\sin\frac{\gamma_{ij}}{2}=t\cdot l_{ij}\tan{\frac{\gamma_{ij}}{2}}-t^{2}\cdot
l_{ij}\tan{\frac{\gamma_{ij}}{2}},$

which confirms expansion (3.1). From this expansion it follows that the right
derivative $\dot{f}_{K}(0)$ of $f_{K}(t)$ at $t=0$ has the form

(3.4)
$\dot{f}_{K}(0)=-2\left(\sum\limits_{i}\beta_{i}+\sum\limits_{i,j}l_{ij}\tan{\frac{\gamma_{ij}}{2}}\right)=-2f_{K}(0)-2\sum\limits_{i,j}l_{ij}\tan{\frac{\gamma_{ij}}{2}}.$

Observe that (3.4) similarly holds for the lens $L$.

We want to show that the quantity in (3.4) is minimized for the lens $L$.
Since $f_{K}(0)=f_{L}(0)$, it is enough to show that the term
$\sum_{i,j}\tan\frac{\gamma_{ij}}{2}l_{ij}$ is maximal for the lens $L$ under
the constraint coming from the Gauss–Bonnet theorem in $\mathbb{R}^{3}$:

(3.5) $\int\limits_{\partial K}\kappa(z)\,\text{d}\sigma_{K}(z)=4\pi.$

Here $\kappa(z)$ denotes the Gaussian curvature of $\partial K$ at the point
$z$ and the integration is done with respect to the area measure on the
boundary of $K$.

$\Pi_{t}(E_{ij})$$\widetilde{E}_{ij}$$o_{i}$$o_{i}^{\prime}$$h$$v$$v^{\prime}$$p^{\prime}$$p$$q$
Figure 6. The quadrilateral $Q_{ij}^{e}(t)$ is bounded by the extension of
$\widetilde{E}_{ij}$ (in violet and violet, dashed), $\Pi_{t}(E_{ij})$ (in
yellow) and a pair of arcs of the big circles passing through the endpoints of
$\Pi_{t}(E_{ij})$ orthogonal to it (in blue). The triangular region
$Q_{ij}^{v}(t)$ is bounded by the geodesic segments $pv$ (blue), $p^{\prime}v$
(bright green), and the arc $pp^{\prime}$ (violet, dashed) of a small circle
centered at $o_{i}^{\prime}$ that contains $\widetilde{E}_{ij}$.

Since $\partial K$ is not smooth, this constraint in the case of
$\lambda$-convex polytopes should be understood as follows. Recall that, for a
point $z\in\partial K$, the _normal cone_ of $K$ at $z$ is a subset of
$\mathbb{R}^{3}$ consisting of $0$ and all normal vectors to all possible
supporting planes at $z$ for $K$. We denote this cone by $N(K,z)$; this is a
closed convex set (see [Sch, p. 81]). For a subset $\omega\subset\partial K$,
we denote by $u(K,\omega)$ the _spherical image_ of $\omega$, that is,

$u(K,\omega)=\underset{z\in\omega}{\bigcup}N(K,x)\cap S^{2},$

where $S^{2}\subset\mathbb{R}^{3}$ is the unit sphere centered at the origin
(see [Sch, p. 88]). Let us denote by $\sigma$ the area measure on $S^{2}$.
Finally, if $\mathcal{V}$, $\mathcal{E}$ and $\mathfrak{F}$ denote the sets of
vertices, edges and facets of the polytope $K$, then the Gauss–Bonnet theorem
for $K$ reads as follows:

(3.6) $\displaystyle\int\limits_{\partial
K}\kappa(z)\text{d}\sigma_{K}(z)=\sum\limits_{F\in\mathfrak{F}}\int\limits_{F}\kappa(z)\text{d}\sigma_{K}(z)+\sum\limits_{e\in\mathcal{E}}\sigma\left(u(K,e)\right)+\sum\limits_{v\in\mathcal{V}}\sigma\left(u(K,v)\right)=4\pi.$

To finish the proof, we note that the Gaussian curvature $\kappa$ on each
$F\in\mathfrak{F}$ is constant and is equal to $1$. Introduce the standard
coordinates $(\vartheta,\varphi)$ on $S^{2}$ such that
$\text{d}\sigma=\cos\vartheta\,\text{d}\vartheta\text{d}\varphi$. These
coordinates can be further rotated so that the spherical image of the edge
$E_{ij}\in\mathcal{E}$ is the piece of a strip in $S^{2}$ given by
$-{\gamma_{ij}}/2\leqslant\vartheta\leqslant{\gamma_{ij}}/2,\
0\leqslant\varphi\leqslant{l_{ij}}/\cos({\gamma_{ij}}/{2})$ in coordinates
$(\vartheta,\varphi)$. Therefore,

$\displaystyle\sigma\left(u(K,E_{ij})\right)=\int\limits_{0}^{\frac{l_{ij}}{\cos{\frac{\gamma_{ij}}{2}}}}\int\limits_{-\frac{\gamma_{ij}}{2}}^{\frac{\gamma_{ij}}{2}}\cos\vartheta\,\text{d}\vartheta\text{d}\varphi=\frac{l_{ij}}{\cos{\frac{\gamma_{ij}}{2}}}\cdot
2\sin{\frac{\gamma_{ij}}{2}}=2l_{ij}\tan{\frac{\gamma_{ij}}{2}}.$

Thus, using (3.6),

$\displaystyle\sum_{i}\beta_{i}+2\sum\limits_{i,j}l_{ij}\tan{\frac{\gamma_{ij}}{2}}+\sum\limits_{v\in\mathcal{V}}\sigma\left(u(K,v)\right)=4\pi;$
$\displaystyle
2\sum\limits_{i,j}l_{ij}\tan{\frac{\gamma_{ij}}{2}}=4\pi-\sum_{i}\beta_{i}-\sum\limits_{v\in\mathcal{V}}\sigma\left(u(K,v)\right).$

As before, $\sum_{i}\beta_{i}$ is the same for $K$ and for the lens $L$. On
the other hand, $\sum_{v\in\mathcal{V}}\sigma\left(u(K,v)\right)$ is always
positive and equals to zero if and only if $K$ has no vertices. The only
finite intersection of balls without vertices is the lens $L$. Therefore,
$\sum_{i,j}l_{ij}\tan{\frac{\gamma_{ij}}{2}}$ is maximal for the lens $L$.
This finishes the proof of Proposition 3.1. ∎

Now we finish the proof of Theorem A in Step I.

By Proposition 3.1, $f_{K}(t)>f_{L}(t)$ in an open right neighborhood of
$t=0$. We claim that the latter implies that $f_{K}(t)\geqslant f_{L}(t)$ for
any $t\in(0,r(K))$, and there are open intervals where this inequality is
strict. If $f_{K}(t)>f_{L}(t)$ for every $t\in(0,r(K))$, then we are done.
Otherwise, there exists $t_{0}\in(0,r(K))$ such that
$f_{K}(t_{0})=f_{L}(t_{0})$, i.e., $|\partial K_{t_{0}}|=|\partial
L_{t_{0}}|$. Pick the smallest $t_{0}$ with such property. With this choice,
$f_{K}(t)>f_{L}(t)$ for each $t\in(0,t_{0})$. Observe that $K_{t_{0}}$ cannot
be a lens: otherwise, since a lens is uniquely defined by its surface area, we
would get that the lenses $K_{t_{0}}$ and $L_{t_{0}}$ are equal, and thus at
the starting moment we must have had $|\partial K|<|\partial L|$, which is
impossible. But since $K_{t_{0}}$ is not a lens, we can repeat the argument of
Proposition 3.1 and show that $f_{K}(t)>f_{L}(t)$ in an open right
neighborhood of $t_{0}$. Then we can continue inductively and the claim
follows.

We have shown that $|\partial K_{t}|\geqslant|\partial L_{t}|$ for any
$t\geqslant 0$, and this inequality is strict on some open intervals of values
of $t$ provided $K$ is not a lens. Therefore,

$|K|=\int\limits_{0}^{r(K)}|\partial K_{t}|dt>\int\limits_{0}^{r(L)}|\partial
L_{t}|dt=|L|.$

This finishes the proof in Step I.

######  Remark.

Note that the above implies that $r(K)\geqslant r(L)$. Thus, approximating
$\lambda$-convex bodies with $\lambda$-convex polytopes, we obtain the
following result: among all $\lambda$-convex bodies of given surface area in
$\mathbb{R}^{3}$, the lens has the smallest inradius. This is a partial case
of the Theorem B, which we prove in Section 4 by using different methods.

###  3.2. Step II: General case

Assume that $K$ is not a $\lambda$-convex polytope. By Proposition 2.5, we can
find a nested sequence $(K^{j})_{j=1}^{\infty}$ of $\lambda$-convex polytopes
such that

$K\subset K^{j}\quad\text{ and }\quad\lim_{j\to\infty}d_{H}(K,K^{j})=0.$

Since $K$ is not a polytope, we can assume that none of the approximating
polytopes $K^{j}$ is a lens. By continuity of the volume functionals under
Hausdorff limits for convex bodies,

$\lim_{j\to\infty}|K^{j}|=|K|\quad\text{ and }\quad\lim_{j\to\infty}|\partial
K^{j}|=|\partial K|.$

By Step I, if $L^{j}$ is a $\lambda$-convex lens such that $|\partial
K^{j}|=|\partial L^{j}|$, then

(3.7) $|K^{j}|>|L^{j}|.$

The Hausdorff limit of $(L^{j})_{j=1}^{\infty}$ is the lens $L$ with the
surface area equal to $|\partial K|$. Thus, the required inequality between
the $n$-dimensional volumes of $K$ and $L$ follows from (3.7) by passing to
the limit. To study the equality case, it is enough to show that we must have
strict inequality in the limit.

For each $j\geqslant 1$, consider the graph of the function
$f_{K^{j}}(t)=|\partial K^{j}_{t}|$, $t<r(K)$. By Lemma 2.14,

$\lim_{j\to\infty}d_{H}(K^{j}_{t},K_{t})=0.$

Since

$\lim_{j\to\infty}|\partial K_{t}^{j}|=|\partial K_{t}|$

by continuity of volumes, the graph of the function $t\mapsto f_{K^{j}}(t)$
converges pointwise to the graph of the function $t\mapsto f_{K}(t)=|\partial
K_{t}|$. The same is true for the functions $f_{L^{j}}$ and $f_{L}$ associated
to lenses. From Step I, $f_{K^{j}}(t)>f_{L^{j}}(t)$ unless $t=0$ (in which
case we have equality by assumption). Hence,

(3.8) $f_{K}(t)\geqslant f_{L}(t)$

for all $t$ in the common domain of definition.

The graph of $f_{K}$ intersects the $t$-axis only at $r(K)$. Since $K$ is not
a $\lambda$-convex polytope, Proposition 2.10 ensures that there exists a
$\lambda$-convex polytope $K^{\prime\prime}$ so that $|\partial K|=|\partial
K^{\prime\prime}|$ and $r(K)>r(K^{\prime\prime})$. Since $|\partial
K|=|\partial L|$, we have $|\partial L|=|\partial K^{\prime\prime}|$, and thus
$r(K^{\prime\prime})\geqslant r(L)$ by the remark at the end of Step I.
Combining this all together, we conclude that $r(K)$ must be strictly larger
than $r(L)$. Therefore, the graphs $f_{K}$ and $f_{L}$ cannot coincide.
Because of (3.8), this means that the area under $f_{K}$, i.e., $|K|$, must be
strictly larger than the area under $f_{L}$, i.e., $|L|$. We showed that
$|K|>|L|$, and this concludes the proof of the equality case and the proof of
Theorem A.

##  4\. Proof of the reverse inradius inequality (Theorem B)

In this section we work in model spaces $M^{n}(c)$. In each model space, the
proof of Theorem B follows the same lines, except for the equidistant case in
the hyperbolic space, where an additional argument is required. We describe
this argument in the next subsection, and after that reduce Theorem B to an
equivalent statement.

###  4.1. Reduction to an equivalent statement (Theorem C)

Assume $c<0$ and $0<\lambda<\sqrt{-c}$, i.e., we are in the equidistant case
(see (3c), Section 2.1). Since a $\lambda$-convex lens $L$ is compact by
definition, we must have $r(L)<\mathfrak{r}_{\lambda}$, where
$\mathfrak{r}_{\lambda}$ is the characteristic distance in the equidistant
case. If $r(K)\geqslant\mathfrak{r}_{\lambda}$, then $r(K)\geqslant r(L)$ and
the claim of the theorem follows.

Therefore, in the rest of this section we consider $\lambda$-convex bodies
$K\subset M^{n}(c),\ c\in\mathbb{R}$ such that

(4.1) either $c\geqslant 0$; or $c<0$ and $\lambda\geqslant\sqrt{-c}$; or
$c<0$ and $\lambda\in(0,\sqrt{-c})$ with $r(K)<\mathfrak{r}_{\lambda}.$

Observe that for every $n\geqslant 2$, every $c\in\mathbb{R}$ and
$\lambda$-convex body $K\subset M^{n}(c)$ as in (4.1), there exists a
$\lambda$-convex lens $L\subset M^{n}(c)$ such that $r(K)=r(L)$ (and it is
unique up to rigid motions in the space).

######  Theorem C.

Let $c\in\mathbb{R}$ and $K\subset M^{n}(c)$ be a $\lambda$-convex body as in
(4.1). If $L\subset M^{n}(c)$ is the $\lambda$-convex lens such that
$r(L)=r(K)$, then $|\partial L|\geqslant|\partial K|$. Moreover, equality
holds if and only if $K$ is a $\lambda$-convex lens.

In fact, we have the following:

###### Claim 4.1.

For a $\lambda$-convex body $K\subset M^{n}(c)$ and $c\in\mathbb{R}$ as in
(4.1), Theorems B and C are equivalent.

###### Proof of Claim 4.1.

We use the monotonicity argument. Suppose Theorem B is true. Let $K$ and $L$
be as in the assumption of Theorem C, i.e., $r(L)=r(K)$. Let $L^{\prime}$ be
the $\lambda$-convex lens such that $|\partial L^{\prime}|=|\partial K|$. By
Theorem B, $r(L^{\prime})\leqslant r(K)$. Hence, $r(L^{\prime})\leqslant
r(L)$. The surface area of a lens grows monotonically as its inradius
increases. Thus, $|\partial L^{\prime}|\leqslant|\partial L|$. We conclude
that $|\partial K|\leqslant|\partial L|$, which yields Theorem C.

In the other direction, suppose Theorem C is true. Let $K$ and $L$ be as in
the assumption of Theorem B, i.e., $|\partial L|=|\partial K|$. Let
$L^{\prime\prime}$ be a $\lambda$-convex lens such that
$r(L^{\prime\prime})=r(K)$. Since (4.1) holds, such a lens $L^{\prime\prime}$
always exists. By Theorem C, $|\partial L^{\prime\prime}|\geqslant|\partial
K|$. Hence, $|\partial L^{\prime\prime}|\geqslant|\partial L|$, and
$r(L^{\prime\prime})\geqslant r(L)$ by monotonicity. Therefore, $r(K)\geqslant
r(L)$, as asserted by Theorem B. ∎

In view of Claim 4.1 and our discussion at the beginning of the section, to
prove Theorem B it is enough to establish Theorem C. The rest of the section
will be dedicated to the proof of Theorem C. The proof will be based on three
claims: Claims 4.2, 4.3, and 4.4 (Key Claim).

###  4.2. Proof of Theorem C (assuming the Key Claim)

By the first part of Proposition 2.10, we can assume that $K$ is a
$\lambda$-convex polytope.

Fix the inradius $r=r(K)=r(L)>0$. Let $B\subset K$ be an inscribed ball for
$K$ (of radius $r$), and denote by $o$ the center of $B$. Consider the radial
projection

(4.2) $\Pi\colon M^{n}(c)\setminus\\{o\\}\to\partial B$

from the center of $B$ onto its boundary (see Figure 9). By symmetry, $\Pi$
maps each edge of $K$ to piece of a great sphere in $\partial B$.

We enumerate the facets of $K$ as $F_{1},\ldots,F_{\ell}$, and let
$\mathcal{S}_{1},\ldots,\mathcal{S}_{\ell}$ be the totally umbilical
hypersurfaces containing the corresponding facets,
$F_{i}\subset\mathcal{S}_{i}$. As usual, we denote by $\mathcal{B}_{i}$ the
convex region bounded by $\mathcal{S}_{i}$. We have

$K=\bigcap_{i=1}^{\ell}\mathcal{B}_{i}.$

###### Claim 4.2.

Assume there exist facets of $K$ that do not touch $B$, and let
$\mathcal{N}\subset\\{1,\dots,\ell\\}$ be the set of their indices. If

$K^{\prime}=\bigcap_{i=1,\,i\not\in\mathcal{N}}^{\ell}\mathcal{B}_{i},$

then

$r(K^{\prime})=r(K)\quad\text{ and }\quad|\partial K|<|\partial K^{\prime}|.$

###### Proof.

Firstly, observe that $K^{\prime}$ is compact. This follows by Lemma 2.7 in
the equidistant case, and by Lemma 2.6 in all other cases. Hence, $K^{\prime}$
is a $\lambda$-convex polytope. By construction, $K\subset K^{\prime}$, so we
obtain $|\partial K|<|\partial K^{\prime}|$.

Finally, since $B$ is the inscribed ball for $K$, the set $\partial
B\cap\partial K$ is not contained in any open half space with respect to a
hyperplane passing through $o$ (by part (1) of Proposition 2.9). By
construction, $B\subset K^{\prime}$ and $\partial B\cap\partial K=\partial
B\cap\partial K^{\prime}$. Hence, $\partial B\cap\partial K^{\prime}$ does not
lie in any open half space through $o$. Moreover, each facet of $K^{\prime}$
touches $B$. By part (2) of Proposition 2.9, $B$ is the inscribed ball of
$K^{\prime}$. Thus, $r(K^{\prime})=r(K)$. ∎

To formulate the next two claims, we introduce some additional notation. By
Claim 4.2, we can assume that each $F_{i}$ touches $B$ at some point $p_{i}$.
Fix $i\in\\{1,\ldots,\ell\\}$. Let $\delta_{i}$ be a totally geodesic
hyperplane passing through $o$ perpendicularly to the geodesic $op_{i}$, and
let $\mathcal{H}_{i}$ be the closed half-space bounded by $\delta_{i}$ and
containing the point $p_{i}$. Consider a _natural radial extension_ $F$ of the
facet $F_{i}$ defined as $F:=\mathcal{S}_{i}\cap\mathcal{H}_{i}$.

Observe that the natural radial extension $F$ is compact. Indeed, for
$c\geqslant 0$, or $c<0$ and $\lambda>\sqrt{-c}$, this is true since in these
cases $\mathcal{S}_{i}$ is a geodesic sphere, and hence compact. If $c<0$ and
$\lambda=\sqrt{-c}$, then $\mathcal{S}_{i}$ is a horosphere. In this case,
choose the coordinates in the Poincaré unit ball model so that
$\mathcal{H}_{i}$ is a closed upper half-ball centered around the north pole
and $\mathcal{S}_{i}$ is a Euclidean sphere inside the unit ball touching the
south pole. It becomes clear that $\mathcal{S}_{i}\cap\mathcal{H}_{i}$ is a
compact set.

$Q$$Q$$T$$\mathcal{S}_{i}$$\mathcal{H}_{i}$$r$$\mathfrak{r}_{\lambda}-r$$M^{n}(c),c<0$
Figure 7.

Finally, if $c<0$ and $\lambda<\sqrt{-c}$, then $\mathcal{S}_{i}$ is an
equidistant. Again, choose the coordinates in the ball model so that
$\mathcal{H}_{i}$ is a closed upper half-ball centered around the north pole
and $\mathcal{S}_{i}$ the Euclidean sphere that intersects the boundary of the
unit ball along the small sphere $Q$ parallel to the equatorial sphere (see
Figure 7). Let $T$ be the hyperbolic hyperplane spanning $Q$. Then
$\mathcal{S}_{i}$ is a connected component of the equidistant set for $T$ at
distance $\mathfrak{r}_{\lambda}$. Since $r<\mathfrak{r}_{\lambda}$, $Q$ must
lie in the open southern hemisphere of the unit ball. Therefore,
$\mathcal{S}_{i}\cap\mathcal{H}_{i}$ is compact.

Since $F$ is compact, the $(n-1)$-dimensional volume of $F$ is well-defined
and we observe that

(4.3) $|F|=\frac{1}{2}\cdot|\partial L|.$

(Here and everywhere below, we emphasize that $|\cdot|$ denotes the
$(n-1)$-dimensional volume.) Define

$\widetilde{F}_{i}=\Pi(F_{i}),\quad\widetilde{F}=\Pi(F).$

Note that $\widetilde{F}$ is the half-sphere.

###### Claim 4.3.

(4.4) $\sum_{i=1}^{\ell}|\widetilde{F}_{i}|=\left|\partial
B\right|=2\cdot|\widetilde{F}|.$

###### Proof.

It follows from the fact that for each $i$ and $j\neq i$, the lens
$\mathcal{B}_{i}\cap\mathcal{B}_{j}$ is reflection-symmetric with respect to
the totally geodesic hyperplane that passes through $o$ and is perpendicular
to the plane spanned by $o$, $p_{i}$ and $p_{j}$. ∎

###### Claim 4.4 (Key Claim).

For each $i\in\\{1,\ldots,\ell\\}$,

(4.5)
$\frac{\left|F_{i}\right|}{|\widetilde{F}_{i}|}\leqslant\frac{\left|F\right|}{|\widetilde{F}|},$

and the equality is possible if and only if $F_{i}=F$.

We postpone the proof of the Key Claim to the next subsection. Instead, we
show how Claim 4.4 together with Claim 4.3 imply Theorem C. We have,

$\displaystyle|\partial K|$
$\displaystyle=\sum_{i=1}^{\ell}|F_{i}|\overset{\eqref{Eq:KeyClaim}}{\leqslant}\sum_{i=1}^{\ell}|\widetilde{F}_{i}|\cdot\frac{|F|}{|\widetilde{F}|}$
$\displaystyle\leqslant\frac{|F|}{|\widetilde{F}|}\cdot\sum_{i=1}^{\ell}|\widetilde{F}_{i}|\overset{{\eqref{Eq:Claim3}}}{=}\frac{|F|}{|\widetilde{F}|}\cdot
2|\widetilde{F}|=2|F|$
$\displaystyle\overset{\eqref{Eq:Half}}{=}\left|\partial L\right|.$

To finish the proof, it remains to analyze the equality case. If $|\partial
K|=|\partial L|$, then there must be equality in the chain of the inequalities
above. By Claim 4.4, it is only possible if all the facets of $K$ are equal to
one of the facets of the lens. This implies that $K$ must be a lens itself,
and thus Theorem C follows.

###  4.3. Proof of the Key Claim

For the proof of Claim 4.4, we need the following auxiliary result.

######  Lemma 4.5.

If $f\colon[0,1]\to\mathbb{R}$ is a continuous, strictly increasing function
with zero mean, i.e.,

$\int\limits_{0}^{1}f(t)dt=0,$

then

(4.6) $\int\limits_{0}^{x}f(t)dt\leqslant 0\quad\text{for all }x\in[0,1],$

and this inequality is strict unless $x=1$.

###### Proof.

Observe that $f(0)\leqslant 0$ and $f(1)\geqslant 0$. Indeed, if $f(0)>0$,
then $f(t)>0$ for all $t\in[0,1]$ because $f$ is increasing. This is a
contradiction to the zero mean condition. Similarly with $f(1)\geqslant 0$.
Again, since $f$ is strictly increasing, there exists $\alpha\in(0,1)$ so that
$f(\alpha)=0$, $f(t)<0$ for all $t<\alpha$ and $f(t)>0$ for all $t>\alpha$.
Now (4.6) is clear if $x\leqslant\alpha$. If $x>\alpha$, then

$\int\limits_{0}^{x}f(t)dt=-\int\limits_{x}^{1}f(t)dt\leqslant 0$

by the choice of $\alpha$ and the zero mean condition. ∎

Now we turn to the proof of Claim 4.4. It proceeds in three steps:

  * •

Step I: we establish an inequality similar to (4.5) for certain conical
sectors (defined below);

  * •

Step II: we approximate each $F_{i}$ by such sectors (as indicated in Figure
8); we get the required inequality using this approximation;

  * •

Step III: we analyze the approximation procedure in order to conclude the
equality case.

Step I. Consider a triple of objects: two hyperplanes $\xi$ and $\xi^{\prime}$
passing through $op_{i}$ and, for $x\in[0,1]$, a solid cone $\mathcal{C}_{x}$
with angle $\pi x/2$ and centered around $op_{i}$ with vertex at $o$. Define
$C_{x}$ to be a connected component of $\mathcal{S}_{i}\cap\mathcal{C}_{x}$
lying between $\xi$ and $\xi^{\prime}$. We call $C_{x}$ a _conical sector_.
Let $\widetilde{C}_{x}=\Pi(C_{x})$ be the radial projection of $C_{x}$. Then,
in the polar coordinates centered at $o$, we can write

$|C_{x}|=\int\limits_{\widetilde{C}_{x}}g(t,\vartheta)\,{\text{d}}\sigma,$

where $\sigma=(t,\vartheta)$ is a point in $\widetilde{C}_{x}\subset\partial
B$, $\text{d}\sigma=\text{d}t\text{d}\vartheta$ is the area measure on
$\partial B$, $t$ is the angle between $op_{i}$ and the direction of $\sigma$,
and $g$ is the density.

Observe that, for $x=1$, we have

(4.7)
$\frac{|C_{1}|}{|\widetilde{C}_{1}|}=\frac{|F|}{|\widetilde{F}|}=:\mathcal{F}.$

Indeed, both $|C_{1}|$ and $|\widetilde{C}_{1}|$ are linearly proportional to
the angle between the hyperplanes $\xi$ and $\xi^{\prime}$, as it immediately
follows by integrating in polar coordinates centered at $p_{i}$. Therefore,
the quotient of these volumes is independent of the angle, and equal to the
right hand side in (4.7) (this corresponds to the case when the angle between
$\xi$ and $\xi^{\prime}$ is zero).

Define the function

$\mathcal{R}(x):=\left|C_{x}\right|-|\widetilde{C}_{x}|\cdot\mathcal{F}=\int\limits_{\widetilde{C}_{x}}\left(g(t,\vartheta)-\mathcal{F}\right)\,\text{d}\sigma.$

By (4.7), $\mathcal{R}(1)=0$. Furthermore, we can rewrite

$\mathcal{R}(x)=\int\limits_{0}^{x}\left(\int_{\partial(\mathcal{C}_{t}\cap\widetilde{C}_{x})}\left(g(t,\vartheta)-\mathcal{F}\right)\text{d}\vartheta\right)\text{d}t.$

Since $t\mapsto g(t,\vartheta)$ is a continuous and strictly increasing
function, the integrant

$\widehat{\mathcal{R}}\colon
t\mapsto\int_{\partial(\mathcal{C}_{t}\cap\widetilde{C}_{x})}\left(g(t,\vartheta)-\mathcal{F}\right)\text{d}\vartheta$

is also continuous and strictly increasing on $[0,1]$.

Applying Lemma 4.5 to the function $\widehat{\mathcal{R}}$, we conclude that
$\mathcal{R}(x)=\int_{0}^{x}\widehat{\mathcal{R}}(t)\text{d}t\leqslant 0$ for
$x\in[0,1]$. This translates into the inequality

(4.8) $\frac{\left|C_{x}\right|}{|\widetilde{C}_{x}|}\leqslant\mathcal{F},$

which is true for every angle $x$ and every choice of hyperplanes $\xi$ and
$\xi^{\prime}$. Moreover,

(4.9) inequality (4.8) is strict unless $x=1$.

This concludes Step I.

Step II. Now we approximate the facet $F_{i}$ by conical sectors $C_{x}$ as
follows (see Figure 8).

Figure 8. Approximation of the facet $F_{i}$ with conical sectors $C_{x_{j}}$.

For an index $s\in\\{1,2,\ldots,\\}$, consider a sequence of $J_{s}$ totally
geodesic hyperplanes $\left(\xi_{s,j}\right)_{j=1}^{J_{s}}$ passing through
$op_{i}$ and labeled cyclically around $op_{i}$. Denote by $\alpha_{s,j}$ the
angle between $\xi_{s,j}$ and $\xi_{s,j+1}$, and define

$\overline{\alpha}_{s}:=\max_{j\in\\{1,\ldots,J_{s}\\}}\alpha_{s,j}$

to be the maximal angle.

Let $G_{i,(s,j)}$ be the connected component of $F_{i}$ that lies between
$\xi_{s,j}$ and $\xi_{s,j+1}$. We can order $G_{i,(s,j)}$ cyclically as well.
For each ‘slice’ $G_{i,(s,j)}$, consider the largest angle $x_{j}$ such that
$C_{x_{j}}\subset G_{i,(s,j)}$, where $C_{x_{j}}$ is defined using the
hyperplanes $\xi_{s,j}$ and $\xi_{s,j+1}$ (see Figure 9). It follows that

$|F_{i}|=\lim\limits_{\overline{\alpha}_{s}\to
0}\sum_{j=1}^{J_{s}}\left|C_{x_{j}}\right|\text{ and
}|\widetilde{F}_{i}|=\lim\limits_{\overline{\alpha}_{s}\to
0}\sum_{j=1}^{J_{s}}|\widetilde{C}_{x_{j}}|.$

Applying (4.8) to each pair $C_{x_{j}},\widetilde{C}_{x_{j}}$, summing up and
passing to the limit as $\overline{\alpha}_{s}\to 0$, we obtain required
inequality (4.5). This finishes Step II, i.e., the proof of the inequality in
Claim 4.4.

Figure 9. Projection onto $\partial B$ and the inscribed conical sector
$C_{x_{j}}$.

Step III. Now we establish the equality case in Claim 4.4. For this, let
$\overline{x}$, respectively $\underline{x}$, be the smallest, respectively
the largest angle $x\in[0,1]$ such that
$\mathcal{C}_{x}\cap\mathcal{S}_{i}\supset F_{i}$, respectively
$\mathcal{C}_{x}\cap\mathcal{S}_{i}\subset F_{i}$. By definition,
$\mathcal{C}_{\overline{x}}$ is the ‘circumscribed’ cone around $F_{i}$, while
$\mathcal{C}_{\underline{x}}$ is the ‘inscribed’ cone for $F_{i}$. In this
notation,

$\underline{x}\leqslant x_{j}\leqslant\overline{x}\qquad\text{ for each
}j\in\\{1,\ldots,J_{s}\\}.$

If $\underline{x}=1$, then $\overline{x}=\underline{x}=1$, and thus $F_{i}=F$.
This yields equality in (4.5). Now we show that if $\underline{x}<1$, then one
must have strict inequality in (4.5). This will conclude Step III.

If $\underline{x}<1$, then by continuity of our approximation, there exist
$\nu,\mu\in(0,1)$ (depending only on $\underline{x}$) such that for each
sequence of cones $\left(C_{x_{j}}\right)_{j=1}^{J_{s}}$, we can find a sub-
range of indices $\mathcal{J}^{\prime}=\\{j^{\prime},\ldots,j^{\prime}+k\\}$
satisfying the following conditions:

  * •

the angle between $\xi_{s,j^{\prime}}$ and $\xi_{s,j^{\prime}+k}$ is equal to
$\nu\cdot 2\pi$, i.e.,

$\sum_{j\in\mathcal{J}^{\prime}}\alpha_{s,j}=\nu\cdot 2\pi;$

  * •

$x_{j}<\mu<0$ for each $j\in\mathcal{J}^{\prime}$.

Since $\nu$ does not depend on $s$, the cones in the sequence $C_{x_{j}}$ for
$j\in\mathcal{J}^{\prime}$ approximate a definite proportion of the area of
$F_{i}$, i.e., there exists a constant $\eta=\eta(\nu,\mu)\in(0,1)$ such that

$\lim_{s\to\infty}\sum_{j\in\mathcal{J}^{\prime}}|C_{x_{j}}|=\eta\cdot|F_{i}|.$

The same is true for the corresponding projections (with the same constant
$\eta$). By construction, for $j\in\mathcal{J}^{\prime}$, we have $x_{j}<1$,
and by (4.9), $|C_{x_{j}}|<\mathcal{F}\cdot|\widetilde{C}_{x_{j}}|$. Summing
up these inequalities over $j\in\mathcal{J}^{\prime}$, we obtain that

$\sum_{j\in\mathcal{J}^{\prime}}|C_{x_{j}}|<\mathcal{F}\cdot\sum_{j\in\mathcal{J}^{\prime}}|\widetilde{C}_{x_{j}}|$

Since $x_{j}<\mu<1$ is true for every $s$, and $\mu$ does not depend on $s$,
we can pass to the limit in the inequality above while keeping the strict
inequality sign. Thus, in the case $\underline{x}<1$, we conclude that
$|F_{i}|<\mathcal{F}\cdot|\widetilde{F}_{i}|$. As mentioned above, this
implies that equality in (4.5) is possible if and only if $F_{i}=F$.

##  Appendix A Reverse isoperimetric inequality for $\lambda$-convex curves in
$\mathbb{R}^{2}$, revisited

In this appendix, we give an alternative proof of the result in [BD2] and
[FKV] on the reverse isoperimetric problem in $\mathbb{R}^{2}$ by using the
method exploited in this paper (see Section 3).

######  Theorem A.1 ($2$-dimensional reverse isoperimetric inequality).

Let $K\subset\mathbb{R}^{2}$ be a $\lambda$-convex body and
$L\subset\mathbb{R}^{2}$ be a $\lambda$-convex lens. If $|\partial
K|=|\partial L|$, then $|K|\geqslant|L|$, and the equality is only possible if
$K$ is a lens.

###### Proof of Theorem A.1.

As in the proof of Theorem A, using Blaschke’s rolling theorem we can
approximate the $\lambda$-convex body $K$ by $\lambda$-convex polygons. We
will establish the required inequality for polygons, and the inequality in the
general case follows by the limiting argument. The equality case will be
analyzed separately at the end of the proof.

Assume now that $K$ is a $\lambda$-convex polygon. As in the proof of Theorem
A, we have

$|K|=\int_{0}^{r(K)}|\partial K_{t}|dt.$

In order to establish the result for $\lambda$-convex polygons it is enough to
show that one-sided derivatives at $0$ satisfy

(A.1) $\left.\frac{d}{dt}\right|_{t=0^{+}}|\partial
K_{t}|\geqslant\left.\frac{d}{dt}\right|_{t=0^{+}}|\partial L_{t}|,\text{ with
equality if and only if }K\text{ is a $\lambda$-convex lens}.$

Here $L_{t}$ is the corresponding inner parallel body at distance $t$ for the
lens $L$ isoperimetric to $K$:

(A.2) $|\partial K|=|\partial K_{0}|=|\partial L_{0}|=|\partial L|.$

Note that $L_{t}$ is a $(1/\lambda-t)^{-1}$-convex lens.

$\gamma_{i}$$\beta_{i+1}$$\beta_{i}$$\partial K_{t}$$\partial K$$t$ Figure 10.
The notation in the proof of Theorem A.1. The corresponding two pairs of
adjacent arcs on the boundary of $\partial K$ and $\partial K_{t}$ are shown
in green and violet, respectively. The pink pieces are of length of order
$t^{2}$ for sufficiently small $t$.

Without loss of generality, assume $\lambda=1$. Denote by $\beta_{i}$,
$i\in\\{1,\ldots,m\\}$, the angles that support the arcs of unit circles
forming $\partial K$, and denote by $\gamma_{i}$ the angles between the
adjacent arcs (see Figure 10). Then

$|\partial K|=\sum\limits_{i=1}^{m}\beta_{i}$

and by a computation similar to the one in the proof of Theorem A, we get

$|\partial
K_{t}|=\sum\limits_{i=1}^{m}(1-t)\,\beta_{i}-2\cdot\sum\limits_{i=1}^{m}t\cdot\tan{\frac{\gamma_{i}}{2}}+O\left(t^{2}\right)=|\partial
K|-t\cdot\sum\limits_{i=1}^{m}\left(\beta_{i}+2\tan{\frac{\gamma_{i}}{2}}\right)+O\left(t^{2}\right).$

Therefore

$\left.\frac{d}{dt}\right|_{t=0^{+}}|\partial
K_{t}|=-\sum\limits_{i=1}^{m}\left(\beta_{i}+2\tan{\frac{\gamma_{i}}{2}}\right)=-|\partial
K|-2\cdot\sum\limits_{i=1}^{m}\tan\frac{\gamma_{i}}{2}.$

In particular, in the case of the lens, this expression reads as

$\left.\frac{d}{dt}\right|_{t=0^{+}}|\partial L_{t}|=-|\partial
L|-4\tan\frac{\gamma_{*}}{2},$

where $\gamma_{*}$ is the angle at the vertex of the lens.

Under assumption (A.2), in order to establish (A.1), we need to show that

(A.3) $\sum\limits_{i=1}^{m}\tan\frac{\gamma_{i}}{2}\leqslant
2\tan\frac{\gamma_{*}}{2},\quad\text{ with equality only if }m=2\text{ and
}\gamma_{1}=\gamma_{2}=\gamma_{*}.$

Observe that we have the following constrains:

(A.4)
$\max\gamma_{i}\leqslant\gamma_{*}\qquad\textup{and}\qquad\sum\limits_{i=1}^{m}\gamma_{i}=2\gamma_{*}.$

The first constraint follows by Blaschke’s rolling theorem. The second
constraint follows by the fact that the spherical image of a convex curve is
equal to $2\pi$ and is a sum of the spherical images of the smooth boundary
components (summing up to the length of the boundary in our case) plus the
angles at the vertices.

Now, using (A.4) and the fact that $(\tan x)/x$ is an increasing function on
$[0,\pi/2)$, we estimate:

(A.5)
$\sum\limits_{i=1}^{m}\tan{\frac{\gamma_{i}}{2}}=\sum\limits_{i=1}^{m}\frac{\gamma_{i}}{2}\cdot\frac{\tan{\frac{\gamma_{i}}{2}}}{\frac{\gamma_{i}}{2}}\leqslant\frac{\tan{\frac{\gamma_{*}}{2}}}{\frac{\gamma_{*}}{2}}\sum\limits_{i=1}^{m}\frac{\gamma_{i}}{2}=2\tan{\frac{\gamma_{*}}{2}}.$

Thus, (A.3) follows. Observe that we have established Theorem A.1 for
$1$-polygons and proved the inequality for arbitrary $1$-convex bodies in
$\mathbb{R}^{2}$. Let us now conclude the equality case for general bodies.

Assume that $\partial K$ has no vertices. It will be clear how to adopt the
proof to the case when $\partial K$ has vertices (in some sense, they only
help).

Let $K^{j}$ be a sequence of $1$-convex polygons approximating $K$, i.e.,
$K^{j}\to K$ as $j\rightarrow\infty$ in Hausdorff metric. Let $L^{j}$ be the
corresponding sequence of $1$-convex lenses so that

$|\partial K^{j}|=|\partial L^{j}|\quad\text{for every }j.$

By continuity, it follows that $L^{j}\to L$, where $L$ is the lens such that
$|\partial K|=|\partial L|$.

Let $\gamma_{i}^{j}$ be the angles of $K^{j}$, $\gamma_{*}^{j}$ be the angle
of $L^{j}$, and denote $\alpha^{j}:=\max\limits_{i}\gamma_{i}^{j}$. Since
$\partial K$ is smooth, we can choose an approximating sequence so that

$\alpha^{j}\searrow 0.$

We also have

$\gamma_{*}^{j}\nearrow\gamma_{*}.$

(Note that $\alpha^{j}\leqslant\gamma_{*}^{j}$.) Because of our choice of
sequences, we can find $N>0$ so that

(A.6)
$\alpha^{j}<\frac{\gamma_{*}}{3}<\frac{2\gamma_{*}}{3}<\gamma_{*}^{j}\quad\text{
for all }\quad j\geqslant N.$

We claim that for each $j\geqslant N$,

(A.7) $\left.\frac{d}{dt}\right|_{t=0^{+}}|\partial
K^{j}_{t}|\geqslant\left.\frac{d}{dt}\right|_{t=0^{+}}|\partial L^{j}_{t}|+C,$

where $C=C(\gamma_{*})>0$ depends only on $\gamma_{*}$. First observe that a
more detailed analysis of (A.5) gives the following bound:

$\left.\frac{d}{dt}\right|_{t=0^{+}}|\partial
K^{j}_{t}|\geqslant\left.\frac{d}{dt}\right|_{t=0^{+}}|\partial
L^{j}_{t}|+\mathcal{E}_{j},$

where

$\mathcal{E}_{j}=\gamma_{*}^{j}\cdot\left(\frac{\tan\frac{\gamma_{*}^{j}}{2}}{\frac{\gamma_{*}^{j}}{2}}-\frac{\tan\frac{\alpha^{j}}{2}}{\frac{\alpha^{j}}{2}}\right)$

By the Mean Value Theorem applied to $f(x):=\tan(x)/x$ and using (A.6), for
some $\xi\in[\alpha^{j}/2,\gamma_{*}^{j}/2]$, we have the following uniform
estimate for $j\geqslant N$:

$\mathcal{E}_{j}=\gamma^{j}_{*}\cdot
f^{\prime}(\xi)\cdot\frac{\gamma_{*}^{j}-\alpha^{j}}{2}\geqslant\frac{1}{9}(\gamma_{*})^{2}\cdot
f^{\prime}(\xi).$

The value $f^{\prime}(\xi)$ coming from the Mean Value Theorem has its minimum
on $[0,\gamma_{*}]$ for our choice of $N$. This minimum is strictly positive
and depends only on $\gamma_{*}$. Hence, again for our choice (A.6), there
exists $C>0$, depending only on $\gamma_{*}$, such that

$\mathcal{E}_{j}\geqslant C\quad\text{for all}\quad j\geqslant N.$

This translates into the assertion of the claim.

Passing to the limit in (A.7), as $C$ does not depend on $j$, we obtain that

$\left.\frac{d}{dt}\right|_{t=0^{+}}|\partial
K_{t}|\geqslant\left.\frac{d}{dt}\right|_{t=0^{+}}|\partial L_{t}|+C.$

Therefore, $|K|>|L|$, which proves _strict_ inequality for $1$-convex bodies
with smooth boundaries. If $\partial K$ has vertices, then we can choose an
approximating sequence of $1$-convex polygons that does not change the
vertices. Repeating the argument of the claim above, we will get a strict
inequality unless $K$ is a $1$-convex lens. This concludes the proof of
Theorem A.1 in the general case. ∎

######  Remark.

The argument as in Step II of the proof of Theorem A can be also applied in
$\mathbb{R}^{2}$ to study the equality case. Above, we presented a different
proof in order to illustrate another approach of how one can control
inequalities under limits.

######  Remark.

In $\mathbb{R}^{2}$ and $\mathbb{R}^{3}$, it is possible to explicitly compute
the volume of a $\lambda$-convex lens as a function of its surface area. Using
these explicit formulas, we obtain the following reverse isoperimetric
inequalities (RIP) as corollaries to Theorem A and Theorem A.1:

######  Corollary A.2 (Explicit RIP in dimension 2).

If $K\subset\mathbb{R}^{2}$ is a $\lambda$-convex body, then

$\quad|K|\geqslant\frac{|\partial
K|}{2\lambda}-\frac{1}{\lambda^{2}}\sin\left(\frac{|\partial
K|\lambda}{2}\right),$

and equality holds if and only if $K$ is a $\lambda$-convex lens. ∎

######  Corollary A.3 (Explicit RIP in dimension 3).

If $K\subset\mathbb{R}^{3}$ is a $\lambda$-convex body, then

$\quad|K|\geqslant\frac{|\partial
K|^{2}\lambda}{96\pi^{2}}\cdot\left(12\pi-|\partial K|\lambda^{2}\right),$

and equality holds if and only if $K$ is a $\lambda$-convex lens. ∎

##  Appendix B Lenses vs. spindles, an asymptotic competition

Here we discuss an extra evidence in support of the Borisenko conjecture. By
Theorem A.1, a $\lambda$-convex lens is the solution for the $2$-dimensional
reverse isoperimetric problem in $\mathbb{R}^{2}$. If one would try to
conjecture a potential solution for dimension $n\geqslant 3$, a natural
candidate would be a body obtained by rotating this lens. However, there are
two ways one can rotate it in order to preserve convexity. The result of
rotating the $2$-dimensional lens with respect to the axis of symmetry which
realizes the smallest width is the $n$-dimensional lens. By Theorem A, this
body is indeed the solution in $\mathbb{R}^{3}$. On the other hand, one can
rotate the lens with respect to the axis of symmetry which realizes its
diameter. The resulting body, which we call a _spindle_ and define more
formally below, is a natural ‘competitor’ to the lens for the optimal shape.
As we will see below, in $\mathbb{R}^{n}$, the lens ‘wins’ this competition
asymptotically as $n\to\infty$. Spindles also appear as solutions to other
optimization problems with curvature constraints (see, e.g., [Be1, BD1]).

We say that a convex body $\textup{Sp}\subset\mathbb{R}^{n}$ is a _$\lambda$
-convex spindle_ if Sp is the intersection of all balls of radius $1/\lambda$
containing a pair of given points. The boundary of Sp is obtained by taking a
smaller arc of a circle of radius $1/\lambda$ in the $x_{1}x_{n}$-plane, with
both endpoints on the $x_{1}$-axis, and rotating it in $\mathbb{R}^{n}$ about
the $x_{1}$-axis (here and below, $x_{1},\ldots,x_{n}$ are the Cartesian
coordinates in $\mathbb{R}^{n}$).

The next proposition shows that asymptotically a lens has smaller volume than
a spindle assuming that their surface areas are the same.

######  Proposition B.1 (Lens wins at infinity).

Let $L$ be a $\lambda$-convex lens and Sp be a $\lambda$-convex spindle in
$\mathbb{R}^{n}$. If $|\partial L|=|\partial\textup{Sp}|$, then
$|L|<|\textup{Sp}|$ as $n\rightarrow\infty.$

###### Proof.

Without loss of generality, assume $\lambda=1$. We start with computing
surface areas of a lens and a spindle. Consider the circle of radius $1$ in
the $x_{1}x_{n}$-plane centered at the point $(0,\dots,0,-h_{1})$ for
$0<h_{1}<1$. Let $C_{1}$ be the arc of this circle above the $x_{1}$-axis
given by the following equation

$x_{n}=\sqrt{1-x_{1}^{2}}-h_{1},$

where $x_{1}$ changes between $0$ and $\sqrt{1-h_{1}^{2}}$. By rotating
$C_{1}$ about the $x_{1}$-axis, we obtain the part of Sp in the hyperplane
$x_{n}\geqslant 0$. Now we compute the surface area of Sp using the
corresponding formula for the surface area of the body of revolution:

$\displaystyle|\partial\textup{Sp}|=2\int_{0}^{\sqrt{1-h_{1}^{2}}}\left(\sqrt{1-x_{1}^{2}}-h_{1}\right)^{n-2}\frac{1}{\sqrt{1-x_{1}^{2}}}\,dx_{1}.$

We will use Laplace’s method to obtain the asymptotic behavior of
$|\partial\textup{Sp}|$ as $n\rightarrow\infty$ (see, e.g., [BH]). We claim
that

(B.1) $\displaystyle|\partial\textup{Sp}|$
$\displaystyle\approx\frac{\sqrt{\pi}}{\sqrt{2}}(1-h_{1})^{n-\frac{5}{2}}\frac{1}{\sqrt{n}}\left(1+O\left(\frac{1}{n}\right)\right).$

The dominant contribution in the asymptotic expansion of the integral above
comes from the neighborhood of $x_{1}=0$, so we can consider the integral over
$(0,\,\infty)$. Also, using the fact that
$\sqrt{1-x_{1}^{2}}-h_{1}=(1-h_{1})\left(1-\frac{1}{2(1-h_{1})}x_{1}^{2}+O(x_{1}^{4})\right)$
and that the integral over $(\sqrt{1-h_{1}^{2}},\,\infty)$ is exponentially
small, we have

$\displaystyle|\partial\textup{Sp}|$ $\displaystyle\approx
2(1-h_{1})^{n-2}\int_{0}^{\infty}\left(1-\frac{1}{2(1-h_{1})}x_{1}^{2}+O(x_{1}^{4})\right)^{n-2}e^{\frac{1}{2}x_{1}^{2}+O(x_{1}^{4})}\,dx_{1}$
$\displaystyle=2(1-h_{1})^{n-2}\int_{0}^{\infty}e^{-(n-2)\left(\frac{1}{2(1-h_{1})}x_{1}^{2}+O(x_{1}^{4})\right)}e^{\frac{1}{2}x_{1}^{2}+O(x_{1}^{4})}\,dx_{1}$
$\displaystyle=2(1-h_{1})^{n-2}\int_{0}^{\infty}e^{\frac{-(n-3)-h_{1}}{2(1-h_{1})}x_{1}^{2}-(n-3)O(x_{1}^{4})}\,dx_{1}$
$\displaystyle\approx
2(1-h_{1})^{n-2}\int_{0}^{\infty}e^{-(n-3)\left(\frac{1}{2(1-h_{1})}x_{1}^{2}+O(x_{1}^{4})\right)}\,dx_{1}.$

Making the change of variables
$y^{2}=\frac{1}{2(1-h_{1})}x_{1}^{2}+O(x_{1}^{4})$ in the last integral, and
using that $y=\frac{x_{1}}{\sqrt{2(1-h_{1})}}\left(1+O(x_{1}^{2})\right)$ and
$dx_{1}=\left(\frac{1}{\sqrt{2(1-h_{1})}}+O(y^{2})\right)dy,$ we obtain

$\displaystyle|\partial\textup{Sp}|$ $\displaystyle\approx
2(1-h_{1})^{n-2}\int_{0}^{\infty}e^{-(n-3)\left(\frac{1}{2(1-h_{1})}x_{1}^{2}+O(x_{1}^{4})\right)}\,dx_{1}$
$\displaystyle=2(1-h_{1})^{n-2}\left[\frac{1}{\sqrt{2(1-h_{1})}}\int\limits_{0}^{\infty}e^{-(n-3)y^{2}}dy+\int\limits_{0}^{\infty}e^{-(n-3)y^{2}}O(y^{2})dy\right].$

The first integral is just the Gaussian integral

$\int\limits_{0}^{\infty}e^{-(n-3)y^{2}}dy=\frac{\sqrt{\pi}}{2\sqrt{n-3}}.$

The second integral we estimate using the change of variables
$y=\sqrt{n-3}\,t$ as follows

$\int\limits_{0}^{\infty}e^{-(n-3)y^{2}}O(y^{2})dy=\frac{1}{(n-3)^{\frac{3}{2}}}\int\limits_{0}^{\infty}e^{-t^{2}}O(t^{2})dt.$

Therefore, we get (B.1).

Now we can find the surface area of a lens $L$. For this, consider the circle
of radius $1$ in the $x_{1}x_{n}$-plane centered at the point
$(-\sqrt{1-h_{2}^{2}},\dots,0,0)$ for $0<h_{2}<1$. Let $C_{2}$ be the arc of
this circle given by the following equation

$x_{n}=\sqrt{1-\left(x_{1}+\sqrt{1-h_{2}^{2}}\right)^{2}}$

where $x_{1}$ changes between $0$ and $1-\sqrt{1-h_{2}^{2}}$. We claim that

(B.2) $\displaystyle|\partial L|$ $\displaystyle\approx
2\frac{h_{2}^{n-1}}{\sqrt{1-h_{2}^{2}}}\frac{1}{n}\left(1+O\left(\frac{1}{n}\right)\right).$

Neglecting the exponentially small tails, the surface area of $L$ is

$\displaystyle|\partial L|$
$\displaystyle=2\int\limits_{0}^{1-\sqrt{1-h_{2}^{2}}}\left(1-\left(x_{1}+\sqrt{1-h_{2}^{2}}\right)^{2}\right)^{\frac{n-3}{2}}dx_{1}$
$\displaystyle\approx
2h_{2}^{n-3}\int\limits_{0}^{\infty}\left(1-\frac{x_{1}^{2}}{h_{2}^{2}}-2\frac{\sqrt{1-h_{2}^{2}}}{h_{2}^{2}}x_{1}\right)^{\frac{n-3}{2}}dx_{1}$
$\displaystyle=2h_{2}^{n-3}\int\limits_{0}^{\infty}e^{-\frac{n-3}{2}\left(2\frac{\sqrt{1-h_{2}^{2}}}{h_{2}^{2}}x_{1}+O(x_{1}^{2})\right)}dx_{1}.$

Making the change of variables
$y=2\frac{\sqrt{1-h_{2}^{2}}}{h_{2}^{2}}x_{1}+O(x_{1}^{2})$ and using that

$\displaystyle
dx_{1}=\left(\frac{h_{2}^{2}}{2\sqrt{1-h_{2}^{2}}}+O(y)\right)dy,$

we get

$\displaystyle|\partial L|$ $\displaystyle\approx
2\,h_{2}^{n-3}\int\limits_{0}^{\infty}e^{-\frac{n-3}{2}y}\left(\frac{h_{2}^{2}}{2\sqrt{1-h_{2}^{2}}}+O(y)\right)dy.$

By the argument similar to the one above, we obtain (B.2).

In the same manner, we obtain the expression for volumes:

$\displaystyle|\textup{Sp}|$
$\displaystyle\approx\frac{\sqrt{\pi}}{\sqrt{2}}(1-h_{1})^{n-\frac{3}{2}}\frac{1}{\sqrt{n}}\left(1+O\left(\frac{1}{n}\right)\right);$
$\displaystyle|L|$ $\displaystyle\approx
2\frac{h_{2}^{n+1}}{\sqrt{1-h_{2}^{2}}}\frac{1}{n}\left(1+O\left(\frac{1}{n}\right)\right),$

where $h_{1},h_{2}\in(0,1)$.

Since $|\partial\textup{Sp}|=|\partial L|$, we have

$\displaystyle(1-h_{1})^{n-\frac{5}{2}}\left(1+O\left(\frac{1}{n}\right)\right)=\frac{2\sqrt{2}}{\pi}\frac{h_{2}^{n-1}}{\sqrt{1-h_{2}^{2}}}\frac{1}{\sqrt{n}}\left(1+O\left(\frac{1}{n}\right)\right).$

This implies that $1-h_{1}=h_{2}$ when $n\rightarrow\infty$. Thus,

$\frac{|\textup{Sp}|}{|\partial\textup{Sp}|}=1-h_{1}>h_{2}^{2}=\frac{|L|}{|\partial
L|}$

as $n\rightarrow\infty$ which confirms that a lens has smaller volume than a
spindle.

∎

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