# Jung-type Inequalities and Blaschke-Santaló Diagrams for Different Diameter
Variants

René Brandenberg  and  Mia Runge

###### Key words and phrases:

Diameter, Blaschke-Santaló diagram, Symmetrizations, Jung-type inequalities,
Completion, Asymmetric gauges

## Abstract

We study geometric inequalities for the circumradius and diameter with respect
to general gauges, partly also involving the inradius and the Minkowski
asymmetry. There are a number of options for defining the diameter of a convex
body that fall apart when we consider non-symmetric gauges. These definitions
correspond to different symmetrizations of the gauge, i.e. means of the gauge
$C$ and its origin reflection $-C$.

##  1\. Introduction and Notation

A compact convex set is called a _convex body_. The family of convex bodies is
denoted by $\mathcal{C}^{n}$ and those, which contain $0$ in their interior,
by $\mathcal{C}^{n}_{0}$. The _support function_ of $C\in\mathcal{C}^{n}$ is
denoted by $h_{C}(\cdot):\mathds{R}^{n}\to\mathds{R}$, $h_{C}(a):=\max_{x\in
C}a^{T}s$, while we write
$\|{\cdot}\|_{C}:\mathds{R}^{n}\to[0,\infty],x\mapsto\|{x}\|_{C}=\min\set{\lambda\geq
0:x\in\lambda C}$ for the corresponding _gauge function_. Hyperplanes are
denoted by $H_{(a,\beta)}:=\set{x\in\mathds{R}^{n}:a^{T}x=\beta}$ and
halfspaces by
$H^{\leq}_{(a,\beta)}:=\set{x\in\mathds{R}^{n}:a^{T}x\leq\beta}$. We say that
$H^{\leq}_{(a,\beta)}$ _supports_ $C\in\mathcal{C}^{n}$ in $p\in C$, if
$p^{T}a=\beta$ and $C\subset H^{\leq}_{(a,\beta)}$. In that case $a$ is called
an outer normal vector of $C$ in $p$. The _polar_ of $C\in\mathcal{C}^{n}$ is
defined as $C^{\circ}:=\\{a\in\mathds{R}^{n}:h_{C}(a)\leq 1\\}$. For any
$X\subset\mathds{R}^{n}$, the _positive, linear, affine_ and _convex hull_ are
denoted by $\operatorname{pos}(X)$,
$\operatorname{lin}(X),\operatorname{aff}(X)$ and $\operatorname{conv}(X)$,
respectively. We call the convex hull of two points $x$ and $y$ a _segment_
and abbreviate it by $[x,y]$. The boundary of $X$ is described by
$\operatorname{bd}(X)$ and the interior by $\operatorname{int}(X)$. For any
$X,Y\subset\mathds{R}^{n}$ and $\rho\in\mathds{R}$, let $X+Y:=\\{x+y:x\in
X,y\in Y\\}$ be the _Minkowski sum_ of $X$ and $Y$ and $\rho X:=\\{\rho x:x\in
X\\}$ be the $\rho$-_dilatation_ of $X$. We abbreviate $\set{x}+Y=:x+Y$ and
$(-1)X=:-X$. If $X=-X$, the set $X$ is called _0-symmetric_. If there exists
$t\in\mathds{R}^{n}$ such that $-(t+X)=t+X$, we say that $X$ is _symmetric_.

Inequalities between geometric functionals, such as the inradius,
circumradius, and diameter, form a central area of convex geometry. They play
an important role in many classical works such as [4, 13, 14, 19, 28] and are
still of interest today [3, 24, 29, 32]. These geometric inequalities have
proven to be useful for many results in convexity and also have many
applications such as providing bounds for approximation algorithms, for
example as bounds on the size of core-sets for containment under homothety
[11]. One of the first such inequalities has been given by Jung [34]. It
provides a bound for the diameter-circumradius ratio in euclidean spaces.
Variants and natural extensions, given e.g. in [2, 12, 17], are typically
subsumed under the term _Jung-type_.

Even when the gauge is not symmetric, the in- and circumradius have a unified
definition: The _circumradius_ of $K\in\mathcal{C}^{n}$ with respect to
$C\in\mathcal{C}^{n}$ is defined as

$R(K,C):=\min\\{\rho\geq 0:\exists t\in\mathds{R}^{n}\text{ such that
}K\subset t+\rho C\\}$

and the _inradius_ as

$r(K,C):=\max\\{\rho\geq 0:\exists t\in\mathds{R}^{n}\text{ such that }t+\rho
C\subset K\\}.$

One should recognize that the inradius can be expressed as a circumradius:
$r(K,C)=R(C,K)^{-1}$.

However, the diameter does not have such a unified definition. Maybe the first
diameter definition for non-symmetric gauges has been given by Leichtweiss
[35]. It differs from the one which later has been studied most [10, 12, 14,
27]. In the following we present four diameter definitions, including the two
mentioned above, all being identical if the gauge is symmetric. We show that
each belongs to a different symmetrization of the gauge.

This work expands upon Mia Runge’s master thesis [39].

As part of this investigation, we consider the following question: given a
gauge $C\in\mathcal{C}^{n}_{0}$ and values $(r,R,D)$, is there a convex body
$K\in\mathcal{C}^{n}$ such that its inradius w. r. t. $C$ is $r$, its
circumradius is $R$, and its diameter is $D$? This kind of question can be
answered by giving a system of inequalities such that for every triple
fulfilling these inequalities there exists such a convex body $K$. Such
systems have been considered first by Blaschke for the volume, surface area
and mean width in (euclidean) 3-space [1] and by Santaló for some triples of
functionals out of area, perimeter, circumradius, inradius, diameter and width
for the euclidean planar case [41]. Later, many of the missing triples from
Santaló’s list have been solved [5, 16, 30, 31] and also other functionals
such as the first Dirichlet eigenvalue [22] or the Cheeger constant [21] or
even four of these functionals [8, 42] have been taken into account. As for
the well known Blaschke-diagram [1, 40], no complete descriptions of such
diagrams of bodies in three or higher dimensional spaces are known so far, not
even for the triple of circumradius, inradius, and diameter. Partial results
can be derived from [15, 34, 43] (see [23] for an overview on the known
material).

In [9] a complete system of inequalities for the $(r,R,D)$-diagram with
triangular gauges and the most common diameter is given. We do the same for
the three previously mentioned diameter variants. Studying these diagrams for
triangular gauge bodies is interesting because they also provide us with some
generally valid inequalities (for all possible gauge bodies), as we expected
by the result in [9].

Moreover, we investigate on completeness and completion aspects for these
diameters. A convex body is _complete_ if we cannot add points without
increasing the diameter. A _completion_ of a convex body $K$ is a complete
body of the same diameter as $K$ containing $K$. Such sets often provide us
with extreme cases of geometric inequalities. While one might expect the gauge
to be complete, as it is the case with symmetric gauges or the common
diameter, this is not always true for non-symmetric gauges and the alternative
diameter definitions we consider. We show that the question, about the shape
of the completions of the gauge, is closely related to their symmetrizations.

##  2\. Preliminaries

When considering the containment between different convex bodies, such as the
mentioned symmetrizations, we use the following notation.

######  Definition 2.1 (Optimal Containment).

Let $K,C\in\mathcal{C}^{n}$. If $K\subset C$ and $R(K,C)=1$, we say that $K$
is _optimally contained_ in $C$, which is abbreviated by
$K\subset^{\operatorname{opt}}C$.

The next proposition from [11] characterizes optimal containment.

######  Proposition 2.2.

Let $K,C\in\mathcal{C}^{n}$. Then $K\subset^{\operatorname{opt}}C$ if and only
if

  1. i)

$K\subset C$ and

  2. ii)

for some $k\in\\{2,...,\dim(C)+1\\}$, there exist
$p^{1},...,p^{k}\in\operatorname{ext}(K)\cap\operatorname{bd}(C)$ and
halfspaces $H^{\leq}_{(a^{i},1)}$ supporting $C$ at $p^{i}$ with
$a^{i}\in\operatorname{aff}(C)\setminus\\{0\\}$, $i\in[k]$, affinely
independent, such that $0\in\operatorname{conv}(\\{a^{1},...,a^{k}\\})$.

The symmetrizations which will later correspond to the diameter definitions
are the minimum $C_{\operatorname{MIN}}:=C\cap-C$, the harmonic mean
$C_{\operatorname{HM}}:=\left(\frac{C^{\circ}-C^{\circ}}{2}\right)^{\circ}$,
the arithmetic mean $C_{\operatorname{AM}}:=\frac{C-C}{2}$, and the maximum
$C_{\operatorname{MAX}}:=\operatorname{conv}(C\cup-C)$. These symmetrizations
and their relations are studied for example in [20, 36] and later also in [6,
7]. The latter is motivated by the objective of achieving a better
understanding of these diameters, in particular by relating them to each other
through geometric inequalities.

Besides $C_{\operatorname{AM}}$, these symmetrizations depend decisively on
the position of $C$ and thus the diameters would do, too. However, as said
above, all symmetrization should coincide if $C$ is symmetric, which is
achieved if and only if in those cases $C$ is 0-symmetric, i.e. centered in 0.
Thus it seems natural to consider in general only “somehow” centered gauges
and the exact definition of the “somehow” should fit the theory.

Let $C\in\mathcal{C}^{n}$. The _Minkowski asymmetry_ of $C$ is defined as
$s(C):=R(C,-C)$ and we say that $C$ is _Minkowski-centered_ if
$C\subset^{\operatorname{opt}}-s(C)C$.

The range of the Minkowski asymmetry is $[1,n]$, where $s(C)=1$ if and only if
$C$ is symmetric and $s(C)=n$ if and only if C is an $n$-dimensional simplex
[26]. The symmetrizations can be ordered and the first and third containments
are always optimal [6, 20].

######  Proposition 2.3.

Let $C\in\mathcal{C}^{n}_{0}$. Then,

$C_{\operatorname{MIN}}\subset^{\operatorname{opt}}C_{\operatorname{HM}}\subset
C_{\operatorname{AM}}\subset^{\operatorname{opt}}C_{\operatorname{MAX}}.$

Figure 1. The equilateral triangle (black) and its symmetrizations: minimum
(blue), harmonic mean (purple), arithmetic mean (red), maximum (orange) ( cf.
[6]).

##  3\. Diameter Definitions

From now on, unless otherwise specified, we always assume that
$K\in\mathcal{C}^{n}$, that the gauge $C\in\mathcal{C}^{n}_{0}$ is a
Minkowski-centered full-dimensional convex body and
$s\in\mathds{R}^{n}\setminus\set{0}$.

There are several ways to interpret the diameter in the symmetric case and we
extend these ideas to the non-symmetric case. We will see that these
correspond to different symmetrizations of the gauge. First, one can try to
define the diameter of $K\in\mathcal{C}^{n}$ by finding a ”maximal” segment.
First, one can try to define the diameter of $K\in\mathcal{C}^{n}$ by finding
a ”maximal” segment. But how should this ”maximality” be defined? We could
measure the distance between two points $x,y\in K$ using the gauge function
$\|{x-y}\|_{C}$ and define the diameter as the maximal such distance
$\max_{x,y\in K}\|{x-y}\|_{C}$. Or, we define it as the maximal circumradius
of segments in $K$: $\max_{x,y\in K}2R([x,y],C)$. However, the diameter could
also be defined as the maximal distance between two parallel supporting
hyperplanes of $K$.

$\max_{s\in\operatorname{bd}(C^{\circ})}h_{K}(s)+h_{K}(-s)=\max_{s\in\mathds{R}^{n}\setminus\set{0}}\frac{h_{K}(s)+h_{K}(-s)}{h_{C}(s)}.$

As already mentioned, if the gauge $C$ is symmetric, all these definitions
lead to the same diameter.

$\max_{x,y\in K}\|{x-y}\|_{C}=\max_{x,y\in
K}2R([x,y],C)=\max_{s\in\mathds{R}^{n}\setminus\set{0}}\frac{h_{K}(s)+h_{K}(-s)}{h_{C}(s)}$

The most common diameter corresponds to the segement-radius definition and is
therefore equal to two times the first core-radius of the set $K$ [11]. We
call it the _arithmetic diameter_ (or standard diameter).

######  Definition 3.1.

  1. i)

The $s$-length $l_{s,\operatorname{AM}}$ is defined as

$l_{s,\operatorname{AM}}(K,C):=\max_{x-y\in(K-K)\cap\operatorname{lin}(s)}2R([x,y],C).$

  2. ii)

The $s$-breadth $b_{s,\operatorname{AM}}$ is defined as

$b_{s,\operatorname{AM}}(K,C):=2\cdot\frac{h_{K}(s)+h_{K}(-s)}{h_{C}(s)+h_{C}(-s)}.$

  3. iii)

The the arithmetic diameter is defined as the maximal $s$-length:

$D_{\operatorname{AM}}(K,C):=\max_{s\in\mathds{R}^{n}\setminus\set{0}}l_{s,\operatorname{AM}}(K,C).$

In [12] the following properties of $D_{\operatorname{AM}}$, which are well
known for symmetric gauges ( cf. [25]), are proven.

######  Proposition 3.2.

  1. i)

$D_{\operatorname{AM}}(K,C)=\max_{s\in\mathds{R}^{n}\setminus\set{0}}b_{s,\operatorname{AM}}(K,C)$

  2. ii)

$\begin{split}l_{s,\operatorname{AM}}(K,C)&=l_{s,\operatorname{AM}}\left(\frac{K-K}{2},C\right)=l_{s,\operatorname{AM}}\left(K,\frac{C-C}{2}\right)=l_{s,\operatorname{AM}}\left(\frac{K-K}{2},\frac{C-C}{2}\right)\end{split}$

  3. iii)

$\begin{split}b_{s,\operatorname{AM}}(K,C)&=b_{s,\operatorname{AM}}\left(\frac{K-K}{2},C\right)=b_{s,\operatorname{AM}}\left(K,\frac{C-C}{2}\right)=b_{s,\operatorname{AM}}\left(\frac{K-K}{2},\frac{C-C}{2}\right)\end{split}$

  4. iv)

$\begin{split}D_{\operatorname{AM}}(K,C)&=D_{\operatorname{AM}}\left(\frac{K-K}{2},C\right)=D_{\operatorname{AM}}\left(K,\frac{C-C}{2}\right)\\\
&=D_{\operatorname{AM}}\left(\frac{K-K}{2},\frac{C-C}{2}\right)=2R\left(\frac{K-K}{2},\frac{C-C}{2}\right)\end{split}$

  5. v)

$\min_{s\in\mathds{R}^{n}\setminus\set{0}}l_{s,\operatorname{AM}}(K,C)=\min_{s\in\mathds{R}^{n}\setminus\set{0}}b_{s,\operatorname{AM}}(K,C)=2r\left(\frac{K-K}{2},\frac{C-C}{2}\right)$

The fact that we can replace $C$ by its symmetrization $\frac{C-C}{2}$ is the
reason why this diameter is called $D_{\operatorname{AM}}$.

Arguably the most intuitive way to measure a diameter is using the gauge
function $\|{x-y}\|_{C}$. This diameter has been studied by Leichtweiss in
[35] for non-symmetric gauges.

######  Definition 3.3.

  1. i)

The asymmetric _$s$ -length_ $l^{\prime}_{s,\operatorname{MIN}}$ is defined as

$l^{\prime}_{s,\operatorname{MIN}}(K,C):=\max_{x-y\in(K-K)\cap\operatorname{pos}(s)}\|{x-y}\|_{C}.$

  2. ii)

The symmetric _$s$ -length_ $l_{s,\operatorname{MIN}}$ is defined as

$l_{s,\operatorname{MIN}}(K,C):=\max_{x-y\in(K-K)\cap\operatorname{lin}(s)}\|{x-y}\|_{C}.$

  3. iii)

The _minimum diameter_ is defined as the maximal $s$-length:

$D_{\operatorname{MIN}}(K,C):=\max_{s\in\mathds{R}^{n}\setminus\set{0}}l_{s,\operatorname{MIN}}(K,C).$

As we maximize over the $s$-lengths to obtain the diameter, both definitions
of the $s$-length lead to the same diameter:
$D_{\operatorname{MIN}}(K,C)=\max_{s\in\mathds{R}^{n}\setminus\set{0}}l_{s,\operatorname{MIN}}(K,C)=\max_{s\in\mathds{R}^{n}\setminus\set{0}}l^{\prime}_{s,\operatorname{MIN}}(K,C)$.

######  Lemma 3.4.

  1. i)

$\displaystyle l_{s,\operatorname{MIN}}(K,C)$
$\displaystyle=\max_{x-y\in(K-K)\cap\operatorname{lin}(s)}\max(\|{x-y}\|_{C},\|{x-y}\|_{-C})$
$\displaystyle=l_{s,\operatorname{MIN}}(K,C\cap(-C))=l_{s,\operatorname{AM}}(K,C\cap(-C))$
$\displaystyle=l_{s,\operatorname{MIN}}\left(\frac{K-K}{2},C\cap(-C)\right)=l_{s,\operatorname{MIN}}\left(\frac{K-K}{2},C\right)$

  2. ii)

$\begin{split}D_{\operatorname{MIN}}(K,C)&=D_{\operatorname{MIN}}(K,C\cap(-C))=D_{\operatorname{MIN}}\left(\frac{K-K}{2},C\cap(-C)\right)\\\
&=D_{\operatorname{MIN}}\left(\frac{K-K}{2},C\right)=D_{\operatorname{AM}}(K,C\cap(-C))\end{split}$

###### Proof.

Since $\|{v}\|_{C\cap-C}=\max(\|{v}\|_{C},\|{-v}\|_{C})$ for every
$v\in\mathds{R}^{n}$, we obtain

$\begin{split}\max_{x-y\in(K-K)\cap\operatorname{pos}(s)}\max(\|{x-y}\|_{C},\|{y-x}\|_{C})&=\max_{x-y\in(K-K)\cap\operatorname{lin}(s)}\|{x-y}\|_{C}\\\
=\max_{x-y\in(K-K)\cap\operatorname{lin}(s)}\|x-y\|_{C\cap-C}&=\max_{x-y\in(K-K)\cap\operatorname{lin}(s)}R([x,y],C\cap-C).\end{split}$

This proves i) and ii) follows obviously. ∎

The next two diameters have been first introduced in [7, Appendix]. For the
first, instead of taking the maximum of $\|{x-y}\|_{C}$ and $\|{x-y}\|_{-C}$
one takes the arithmetic mean of these two values in the definition of the
$s$-length. This diameter definition corresponds to the harmonic mean of the
gauge.

######  Definition 3.5.

  1. i)

The _$s$ -length_ $l_{s,\operatorname{HM}}$ is defined as

$l_{s,\operatorname{HM}}(K,C):=\max_{x-y\in(K-K)\cap\operatorname{lin}(s)}\frac{1}{2}(\|{x-y}\|_{C}+\|{x-y}\|_{-C})$

  2. ii)

The _harmonic diameter_ is defined as the maximal $s$-length

$D_{\operatorname{HM}}(K,C):=\max_{s\in\mathds{R}^{n}\setminus\set{0}}l_{s,\operatorname{HM}}(K,C).$

######  Lemma 3.6.

  1. i)

$\begin{split}l_{s,\operatorname{HM}}(K,C)&=\max_{x-y\in(K-K)\cap\operatorname{lin}(s)}\|{x-y}\|_{\left(\frac{C^{\circ}-C^{\circ}}{2}\right)^{\circ}}\\\
&=l_{s,\operatorname{HM}}\left(K,\left(\frac{C^{\circ}-C^{\circ}}{2}\right)^{\circ}\right)=l_{s,\operatorname{AM}}\left(K,\left(\frac{C^{\circ}-C^{\circ}}{2}\right)^{\circ}\right)\\\
&=l_{s,\operatorname{HM}}\left(\frac{K-K}{2},\left(\frac{C^{\circ}-C^{\circ}}{2}\right)^{\circ}\right)=l_{s,\operatorname{HM}}\left(\frac{K-K}{2},C\right)\end{split}$

  2. ii)

$\begin{split}D_{\operatorname{HM}}(K,C)&=D_{\operatorname{HM}}\left(K,\left(\frac{C^{\circ}-C^{\circ}}{2}\right)^{\circ}\right)=D_{\operatorname{HM}}\left(\frac{K-K}{2},\left(\frac{C^{\circ}-C^{\circ}}{2}\right)^{\circ}\right)\\\
&=D_{\operatorname{HM}}\left(\frac{K-K}{2},C\right)=D_{\operatorname{AM}}\left(K,\left(\frac{C^{\circ}-C^{\circ}}{2}\right)^{\circ}\right)\end{split}$

###### Proof.

We can use the fact that $h_{C^{\circ}}(a)=\|{a}\|_{C}$ for any
$a\in\mathds{R}^{n}$ and $C\in\mathcal{C}^{n}_{0}$ to obtain

$\begin{split}\|a\|_{\left(\frac{C^{\circ}-C^{\circ}}{2}\right)^{\circ}}&=h_{\frac{C^{\circ}-C^{\circ}}{2}}(a)=\frac{1}{2}(h_{C^{\circ}}(a)+h_{-C^{\circ}}(a))\\\
&=\frac{1}{2}(\|{a}\|_{C}+\|{a}\|_{-C})=\frac{1}{2}(\|{a}\|_{C}+\|{-a}\|_{C}).\end{split}$

Since $C_{\operatorname{HM}}$ is symmetric, the first part follows. The second
part follows again directly from the first. ∎

Instead of dividing by the mean $\frac{h_{C}(s)+h_{C}(-s)}{2}$ in the
definition of the $s$-breadth, one may also divide by the maximum of
$h_{C}(s)$ and $h_{C}(-s)$. With this idea, we obtain our last diameter, the
maximum diameter.

######  Definition 3.7.

  1. i)

The _$s$ -breadth_ $b_{s,\operatorname{MAX}}$ is defined as

$b_{s,\operatorname{MAX}}(K,C):=\frac{h_{K}(s)+h_{K}(-s)}{\max(h_{C}(s),h_{C}(-s))}.$

  2. ii)

The _maximum diameter_ is defined as the maximal $s$-breadth:

$D_{\operatorname{MAX}}(K,C):=\max_{s\in\mathds{R}^{n}\setminus\set{0}}b_{s,\operatorname{MAX}}(K,C).$

######  Lemma 3.8.

  1. i)

$\displaystyle b_{s,\operatorname{MAX}}(K,C)$
$\displaystyle=\frac{h_{K}(s)+h_{K}(-s)}{h_{\operatorname{conv}(C\cup(-C))}(s)}$
$\displaystyle=b_{s,\operatorname{MAX}}(K,\operatorname{conv}(C\cup(-C)))=b_{s,\operatorname{AM}}(K,\operatorname{conv}(C\cup(-C)))$
$\displaystyle=b_{s,\operatorname{MAX}}\left(\frac{K-K}{2},\operatorname{conv}(C\cup(-C))\right)=b_{s,\operatorname{MAX}}\left(\frac{K-K}{2},C\right)$

  2. ii)

$\begin{split}D_{\operatorname{MAX}}(K,C)&=D_{\operatorname{MAX}}\left(K,\operatorname{conv}(C\cup(-C))\right)=D_{\operatorname{MAX}}\left(\frac{K-K}{2},\operatorname{conv}(C\cup(-C))\right)\\\
&=D_{\operatorname{MAX}}\left(\frac{K-K}{2},C\right)=D_{\operatorname{AM}}(K,\operatorname{conv}(C\cup(-C))\end{split}$

###### Proof.

The first part follows directly from the fact that
$\max(h_{C}(s),h_{C}(-s))=h_{C_{\operatorname{MAX}}}(s)$ and the second part
again directly from the first. ∎

Since all definitions are equivalent for 0-symmetric gauges, we can always use
that
$D_{\operatorname{M}}(K,C)=D_{\operatorname{M}}(K,C_{\operatorname{M}})=D_{\operatorname{AM}}(K,C_{\operatorname{M}})$
for
$\operatorname{M}\in\\{\operatorname{MIN},\operatorname{HM},\operatorname{AM},\operatorname{MAX}\\}$and
results known about the arithmetic diameter. If we consider a symmetric gauge
we omit the index and denote the diameter by $D(K,C$. Moreover, for all
diameter definitions we say that $x,y\in K$ is a _diametral pair_ if
$D_{\operatorname{M}}(K,C)=D_{\operatorname{M}}([x,y],C)$.

######  Remark 3.9.

Using the different definitons of the $s$-lengths and -breadths width-
definitions can be done analogously to the diameters. For $g\in\set{l,b}$ such
that $g_{s,\operatorname{M}}$ is defined the _width_ is defined as

$w_{\operatorname{M}}(K,C):=\min_{s\in\mathds{R}^{n}\setminus\set{0}}g_{s,\operatorname{M}}(K,C).$

In the standard case $\operatorname{M}=\operatorname{AM}$ it does not make a
difference whether we minimize over the $s$-length or -breadth. By lemmas 3.4,
3.6, and 3.8 we can symmetrize the arguments of the width as well.

##  4\. Properties of the Diameters and Blaschke-Santaló Diagrams

In the following, we would like to study properties of the diameters and how
concepts such as completeness translate when using other definitions.
Furthermore, we compare the diameter to other functionals such as the
circumradius and inradius and introduce some theory on Blaschke-Santaló
diagrams.

Let
$\operatorname{M}\in\\{\operatorname{MIN},\operatorname{HM},\operatorname{AM},\operatorname{MAX}\\}$
be one of the symmetrizations.

######  Remark 4.1.

The inradius, circumradius and diameter are increasing and homogenious of
degree $+1$ in the first argument and decreasing and homogenious of degree
$-1$ in the second argument.

######  Lemma 4.2 (Linearity of $r,R,D_{\operatorname{M}}$).

Let $K_{1},K_{2}\in\mathcal{C}^{n}$.

  1. i)

If $r_{1}C\subset^{\operatorname{opt}}K_{1}$ and
$r_{2}C\subset^{\operatorname{opt}}K_{2}$, then

$r(\lambda K_{1}+(1-\lambda)K_{2},C)\geq\lambda r_{1}+(1-\lambda)r_{2}$

and equality holds if we can choose the same same outer normals of $K_{1}$ and
$K_{2}$ ( cf. Proposition 2.2).

  2. ii)

If $K_{1}\subset^{\operatorname{opt}}R_{1}C$ and
$K_{2}\subset^{\operatorname{opt}}R_{2}C$, then

$R(\lambda K_{1}+(1-\lambda)K_{2},C)\leq\lambda R_{1}+(1-\lambda)R_{2}.$

If we can choose the same touching points in the boundary of $C$, equality is
obtained.

  3. iii)

If $D_{1}=D_{\operatorname{M}}(K_{1},C)$ and
$D_{2}=D_{\operatorname{M}}(K_{2},C)$, then

$D_{\operatorname{M}}(\lambda K_{1}+(1-\lambda)K_{2},C)\leq\lambda
D_{1}+(1-\lambda)D_{2}.$

If the diameters are defined by the same $s$-breadth or $s$-length, we have
equality.

###### Proof.

  1. i)

Obviously, $(\lambda r_{1}+(1-\lambda)r_{2})C\subset\lambda
K_{1}+(1-\lambda)K_{2}$. To show optimal containment, we need to fulfill the
conditions of Proposition 2.2. Let $r_{1}p_{i}$, $r_{2}p_{i}$, with
$p_{i}\in\operatorname{bd}(C)$, be the touching points and $a_{i}$ the
corresponding outer normals. Then, $h_{\lambda
K_{1}+(1-\lambda)K_{2}}(a_{i})=\lambda
r_{1}p_{i}^{T}a_{i}+(1-\lambda)r_{2}p_{i}^{T}a_{i}=(\lambda
r_{1}+(1-\lambda)r_{2})p_{i}^{T}a_{i}=h_{(\lambda
r_{1}+(1-\lambda)r_{2})C}(a_{i})$. Thus, $p_{i}\in\operatorname{bd}((\lambda
r_{1}+(1-\lambda)r_{2})C)\cap\operatorname{bd}(\lambda
K_{1}+(1-\lambda)K_{2})$ and 0 is in the convex hull of the $a_{i}$.

  2. ii)

The statement for the circumradius follows analogously. Here, the outer body
is $C$ in both cases, so we automatically have the same supporting
hyperplanes.

  3. iii)

We know
$D_{\operatorname{M}}(K,C)=2R(K_{\operatorname{AM}},C_{\operatorname{M}})$.
Thus, the inequality follows from part $ii)$. If the diameters are attained by
the same $s$-length we have the same touching points in the containments
$\frac{K_{1}-K_{1}}{2}\subset^{\operatorname{opt}}\frac{D_{\operatorname{M}}(K_{1},C)}{2}C_{\operatorname{M}}$
and
$\frac{K_{2}-K_{2}}{2}\subset^{\operatorname{opt}}\frac{D_{\operatorname{M}}(K_{2},C)}{(2)}C_{\operatorname{M}}$
and the equality follows from part $ii)$. If the diameter is attained by the
same $s$-breadth, we have

$\begin{split}\lambda
D_{\operatorname{M}}(K_{1},C)+(1-\lambda)D_{\operatorname{M}}(K_{2},C)&=\lambda
b_{s,\operatorname{M}}(K_{1},C)+(1-\lambda)b_{s,\operatorname{M}}(K_{2},C)\\\
&=\lambda\frac{h_{K_{1}-K_{1}}(s)}{h_{C_{\operatorname{M}}}(s)}+(1-\lambda)\frac{h_{K_{2}-K_{2}}(s)}{h_{C_{\operatorname{M}}}(s)}\\\
&=\frac{h_{(\lambda K_{1}+(1-\lambda)K_{2})-(\lambda
K_{1}+(1-\lambda)K_{2})}(s)}{h_{C_{\operatorname{M}}}(s)}\\\
&=b_{s,\operatorname{M}}(\lambda K_{1}+(1-\lambda)K_{2},C)\\\ &\leq
D_{\operatorname{M}}(\lambda K_{1}+(1-\lambda)K_{2})\end{split}$

and equality follows.

∎

######  Lemma 4.3 (Invariance under transformations).

Let $A:\mathds{R}^{n}\to\mathds{R}^{n}$ be a non-singular affine
transformation and $L_{A}$ its corresponding linear transformation. Then

$\begin{split}&D_{\operatorname{M}}(A(K),L_{A}(C))=D_{\operatorname{M}}(K,C)\\\
&R(A(K),A(C))=R(K,C)\\\ &r(A(K),A(C))=r(K,C).\end{split}$

###### Proof.

It follows from Proposition 2.2 that the in- and circumradius are invariant
under affine transformations. All symmetrizations interchange with linear
transformations: $L_{A}(C)_{\operatorname{M}}=L_{A}(C_{\operatorname{M}})$
(see [6], Lemma 4). We must confine the transformation in the second argument
of the diameter to be linear, since we need the gauge to be Minkowski-
centered. Because we can interprete the diameter as a circumradius with
$D_{\operatorname{M}}(K,C)=2R(\frac{K-K}{2},C_{\operatorname{M}})$, the
invariance of the diameter follows. The position of $K$ does not change the
diameter and therefore we can apply a corresponding affine transformation to
$K$. ∎

To analyse properties as constant width and completeness we extend their
definitions to the different diameters.

######  Definition 4.4.

Let $K,K^{*}\in\mathcal{C}^{n}$, $K^{*}\supset K$, $C\in\mathcal{C}^{n}_{0}$.

  1. i)

$K$ is of _constant width_ if
$w_{\operatorname{M}}(K,C)=D_{\operatorname{M}}(K,C)$.

  2. ii)

$K$ is _complete_ if
$D_{\operatorname{M}}(K^{\prime},C)>D_{\operatorname{M}}(K,C)$ for all
$K^{\prime}\in\mathcal{C}^{n}$ such that $K^{\prime}\supsetneq K$.

  3. iii)

$K^{*}$ is a _completion_ of $K$ if it is complete and
$D_{\operatorname{M}}(K^{*},C)=D_{\operatorname{M}}(K,C)$.

######  Remark 4.5.

Since all diameters can be expressed by the arithmetic diameter $w.\,r.\,t.$
$C_{\operatorname{M}}$, we know the following [18]:

  1. i)

$K$ has constant width if and only if $K_{\operatorname{AM}}=\lambda
C_{\operatorname{M}}$ for some $\lambda\in\mathds{R}$.

  2. ii)

If $K$ is of constant width, it is complete.

  3. iii)

In the planar case, $K$ is complete if and only if it has constant width.

######  Lemma 4.6.

If $K^{*}$ is a completion of $K$, then $K\subset^{\operatorname{opt}}K^{*}$.

###### Proof.

By definition $K\subset K^{*}$ and
$D_{\operatorname{M}}(K^{*},C)=D_{\operatorname{M}}(K,C)$. Now, assuming there
exist $c\in\mathds{R}^{n}$ and $0\leq\rho<1$ such that
$K\subset^{\operatorname{opt}}c+\rho K^{*}$ implies
$D_{\operatorname{M}}(K,C)\leq D_{\operatorname{M}}(c+\rho
K^{*},C)<D_{\operatorname{M}}(K^{*},C)$, a contradiction. ∎

######  Remark 4.7.

Let us observe two things:

  1. i)

Whenever the gauge is symmetric, the only complete and symmetric set is the
gauge itself. Thus, when considering $D_{\operatorname{M}}$ with respect to a
possibly non-symmetric gauge $C$, the only complete and symmetric set is
always $C_{\operatorname{M}}$.

  2. ii)

In the case that $\operatorname{M}=\operatorname{AM}$, the gauge itself is
always complete. This is not always the case with other diameter definitions.
Therefore, in the following sections, we will characterize when the gauge is
complete and what the completion looks like.

In the following the containment factors between $C_{\operatorname{AM}}$ and
$C_{\operatorname{M}}$ will prove helpful to analyse the diameter
$D_{\operatorname{M}}$.

######  Notation.

By $\delta_{\operatorname{M}}:=\delta_{\operatorname{M}}(C)$ and
$\rho_{\operatorname{M}}:=\rho_{\operatorname{M}}(C)$ we denote the dilatation
factors needed, such that

$\rho_{\operatorname{M}}C_{\operatorname{M}}\subset^{\operatorname{opt}}C_{\operatorname{AM}}\subset^{\operatorname{opt}}\delta_{\operatorname{M}}C_{\operatorname{M}}.$

For better readability, we omit the argument $C$, the gauge body, whenever it
is clear from the context.

Segments $L$ optimally contained in $C$ with $D(L,C)=2\rho_{\operatorname{M}}$
are denoted by $L_{w}$ and in case of $D(L,C)=2\delta_{\operatorname{M}}$ by
$L_{D}$.

######  Lemma 4.8.

  1. i)

For any segment $L\subset^{\operatorname{opt}}C$:

$2\rho_{\operatorname{M}}\leq D_{\operatorname{M}}(L,C)\leq
2\delta_{\operatorname{M}}$

with equality on the right side iff $L$ is diametral and equality on the left
iff $L$ is a width chord of $C$.

All values in between are attained.

  2. ii)

The diameter of $C$ with respect to itself is
$D_{\operatorname{M}}(C,C)=2\delta_{\operatorname{M}}$ and the width of $C$
with respect to itself is
$w_{\operatorname{M}}(C,C)=2\rho_{\operatorname{M}}$.

###### Proof.

We begin by showing part $ii)$: By the definition of $\delta_{M}$ as well as
the diameter properties collected in Proposition 3.2 and Lemmas 3.4, 3.6, and
3.8, we have

$\displaystyle D_{\operatorname{M}}(C,C)$
$\displaystyle=D_{\operatorname{AM}}(C_{\operatorname{AM}},C_{\operatorname{M}})=2R(C_{\operatorname{AM}},C_{\operatorname{M}})=2\delta_{\operatorname{M}}$
as well as $\displaystyle w_{\operatorname{M}}(C,C)$
$\displaystyle=w_{\operatorname{AM}}(C_{\operatorname{AM}},C_{\operatorname{M}})=2r(C_{\operatorname{AM}},C_{\operatorname{M}})=2\rho_{\operatorname{M}}.$

Now, for part $i)$, let $L\subset C$ be a segment. It follows that
$D_{\operatorname{M}}(L,C)\leq 2\delta_{\operatorname{M}}$. If $L$ is the
convex hull of a diametral pair,
$D_{\operatorname{M}}(L,C)=D_{\operatorname{M}}(C,C)=2\delta_{\operatorname{M}}$.
Segments with circumradius $1$ have diameter $2$ when considering the
arithmetic mean. Hence,
$2=D_{\operatorname{AM}}(L,C)\leq\frac{1}{\rho_{\operatorname{M}}}D_{\operatorname{M}}(L,C)$.
If $L$ provides us with the minimal $s$-length or -breadth, we obtain by Lemma
4.8:
$D_{\operatorname{M}}(L,C)=\min_{s\in\mathds{R}^{n}\setminus\\{0\\}}l_{s,\operatorname{M}}(C,C_{\operatorname{M}})=2\rho_{\operatorname{M}}$
or the analogue for the $s$-breadth. All values in between are attained since
the $s$-length and $s$-breadth are continuous as a function of $s$ on
$\mathds{R}^{n}\setminus\\{0\\}$. ∎

######  Definition 4.9.

The set

$K^{\sup}=\bigcap_{x\in K}x+D_{\operatorname{M}}(K,C)C_{\operatorname{M}}.$

is called the _supercompletion_ of $K$.

Let us remark that Moreno and Schneider [38] call $K^{\sup}$ the _wide
spherical hull_. It is shown in [18] for arbitrary Minkowski spaces (i.e. for
0-symmetric $C$) that $K$ is complete if and only if $K^{\sup}=K$ and in [37]
that $K^{\sup}$ is the union of all completions of $K$. All the above were
previously only defined for symmetric $C$, but it is obvious that these
properties stay true in the general case.

######  Definition 4.10.

  1. i)

A _supporting slab_ of $K$ is the intersection of two antipodal parallel
supporting halfspaces of $K$.

  2. ii)

A supporting slab is _regular_ if at least one of the bounding hyperplanes
contains a smooth boundary point of $K$.

  3. iii)

We say that $s\in\mathds{R}^{n}\setminus\set{0}$ defines a regular slab if
there exists a supporting slab such that the defining halfspaces have outer
normals $\pm s$.

Every convex body is the intersection of its regular slabs and completeness
can be characterized by using these slabs [38, Theorem 1].

######  Proposition 4.11.

The following are equivalent:

  1. i)

$K$ is complete.

  2. ii)

For every outer normal $s$ defining a regular supporting slab we have
$g_{s,\operatorname{M}}(K,C)=D_{\operatorname{M}}(K,C)$, where $g\in\set{l,b}$
as in Remark 3.9.

######  Remark 4.12.

As mentioned after the definition of the supercompletion, $K^{*}=K^{\sup}$
implies uniqueness for the completion $K^{*}$ of $K$. Thus, defining
$K_{X}:=\bigcap_{x\in X}x+D_{\operatorname{M}}(K,C)C_{\operatorname{M}}$ for
some $X\subset K$ we obviously have $K^{*}\subset K^{\sup}\subset K_{X}$. This
means that describing properties for such subsets $X$ in the following, that
imply $K^{*}=K_{X}$ implicitely guarantee uniqueness of the completion
$K^{*}$.

######  Lemma 4.13.

Let $X\subset K$. Then, the following are equivalent:

  1. i)

$K_{X}:=\bigcap_{x\in X}x+D_{\operatorname{M}}(K,C)C_{\operatorname{M}}$ is a
completion of $K$.

  2. ii)

For every $s\in\mathds{R}^{n}\setminus\set{0}$, which defines a regular slab
of $K_{X}$ there exists $\tilde{s}\in\set{s,-s}$ and $p\in X$ such that
$p^{T}(-\tilde{s})=h_{K_{X}}(-\tilde{s})$ and
$h_{K_{X}}(\tilde{s})=h_{p+DC_{\operatorname{M}}}(\tilde{s})$.

###### Proof.

Let us abbreviate $D:=D_{\operatorname{M}}(K,C)$ for the proof.

$ii)\Rightarrow i)$: Let $s\in\mathds{R}^{n}\setminus\set{0}$ define a regular
slab of $K_{X}$ and $p\in X$ be defined as in $ii)$. Then

$\begin{split}b_{s,\operatorname{M}}(K_{X},C)&=\frac{h_{K_{X}}(s)+h_{K_{X}}(-s)}{h_{C_{\operatorname{M}}}(s)}=\frac{p^{T}(-\tilde{s})+h_{p+DC_{\operatorname{M}}}(\tilde{s})}{h_{C_{\operatorname{M}}}(\tilde{s})}\\\
&=\frac{p^{T}(-\tilde{s})+p^{T}\tilde{s}+Dh_{C_{\operatorname{M}}}(\tilde{s})}{h_{C_{\operatorname{M}}}(\tilde{s})}=D=D_{\operatorname{M}}(K,C).\end{split}$

By Proposition 4.11 this proves that $K_{X}$ is a completion of $K$.

$i)\Rightarrow ii)$: Assume $K_{X}$ is a completion of $K$ and there exists
some $s\in\mathds{R}^{n}\setminus\set{0}$ that defines a regular slab such
that there is no $p$ as defined in $ii)$. By the construction of $K_{X}$ there
exist $p^{1},p^{2}\in X$ such that
$h_{K_{X}}(s)=(p^{1})^{T}s+h_{DC_{\operatorname{M}}}(s)$ and
$h_{K_{X}}(-s)=(p^{2})^{T}(-s)+h_{DC_{\operatorname{M}}}(s)$. By our
assumption $(p^{2})^{T}s<(p^{1})^{T}s+h_{DC_{\operatorname{M}}}(s)$, otherwise
we could choose $p=p^{2}$. Then,

$\begin{split}b_{s}(K_{X},C_{\operatorname{M}})&=\frac{h_{K_{X}}(s)+h_{K_{X}}(-s)}{h_{C_{\operatorname{M}}}(s)}=\frac{(p^{1})^{T}s+h_{DC_{\operatorname{M}}}(s)+(p^{2})^{T}(-s)+h_{DC_{\operatorname{M}}}(-s)}{h_{C_{\operatorname{M}}}(s)}\\\
&>\frac{(p^{2})^{T}s+(p^{2})^{T}(-s)+h_{DC_{\operatorname{M}}}(-s)}{h_{C_{\operatorname{M}}}(s)}=\frac{h_{DC_{\operatorname{M}}}(-s)}{h_{C_{\operatorname{M}}}(s)}=D=D_{\operatorname{M}}(K,C).\end{split}$

which implies that $K_{X}$ is not a completion of $K$. ∎

Now, we consider the special case where $X$ is a simplex.

######  Definition 4.14.

We say that a subset $X\subset K\in\mathcal{C}^{n}$ is a _diametric simplex_
of $K$ if

  1. i)

$X$ is a simplex and

  2. ii)

$D_{\operatorname{M}}([x,y],K)=D_{\operatorname{M}}(K,C)$ for all pairs of
vertices $x,y$ of $X$.

######  Lemma 4.15.

Let $X$ be a diametric triangle of $K\in\mathcal{C}^{2}$. Then, $K_{X}$ is the
unique completion of $K$. As a consequence, any triangle $T$ for which $X=T$
is diametric has a unique completion.

###### Proof.

We show that property $ii)$ of Lemma 4.13 is fulfilled. Assume w. l. o. g.
that $D_{\operatorname{M}}(X,C)=D_{\operatorname{M}}(K,C)=1$ and let
$X=\operatorname{conv}(\set{p^{1},p^{2},p^{3}})$. Then, the translations
$-p^{i}+X$ with $i\in\set{1,2,3}$ are subsets of $C_{\operatorname{M}}$, all
with one vertex in the origin and the other two on the boundary of
$C_{\operatorname{M}}$ ( cf. Figure 2). Since $p_{i}-p_{j}$ and $p_{j}-p_{i}$
are each other’s negative, we have three pairs of touching points. For each,
we choose an outer normal $a_{k}$, $k\in\set{1,2,3}$ ordered as given in
Figure 2. Then, if $s\in\operatorname{pos}(\set{a_{i},-a_{j}})$, $i\neq j$,
property $ii)$ of Lemma 4.13 holds with $p=p_{k}$, $k\neq i,j$. Hence, for all
$s\in\mathds{R}^{n}\setminus\set{0}$, property $ii)$ is fulfilled and it
follows that $K_{X}$ is a completion. Using Remark 4.12 we obtain the
uniqueness. ∎

$a^{1}$$-a^{1}$$-a^{2}$$a^{2}$$-a^{3}$$a^{3}$$p^{1}-p^{2}$$p^{3}-p^{2}$$p^{2}-p^{1}$$p^{2}-p^{3}$

$p^{1}$$p^{2}$$p^{3}$$p^{1}+DC_{\operatorname{M}}$$p^{2}+DC_{\operatorname{M}}$$p^{3}+DC_{\operatorname{M}}$

Figure 2. If $K$ contains a diametric triangle, its completion is constructed
similar to the Reuleaux triangle in the euclidean case since it suffices to
consider the extreme points, i. e. the vertices of a diametric triangle.

From the containment chain in Proposition 2.3 we know

$D_{\operatorname{MAX}}(K,C)\leq D_{\operatorname{AM}}(K,C)\leq
D_{\operatorname{HM}}(K,C)\leq D_{\operatorname{MIN}}(K,C)$

The containment factors between the symmetrizations of the gauge can be used
to improve this chain and to formulate new inequalities.

######  Lemma 4.16.

  1. i)

$\displaystyle\rho_{\operatorname{M}}D_{\operatorname{AM}}(K,C)\leq
D_{\operatorname{M}}(K,C)\leq\delta_{\operatorname{M}}D_{\operatorname{AM}}(K,C)$,

  2. ii)

$\displaystyle\delta_{\operatorname{M}}r(K,C)\leq\frac{D_{\operatorname{M}}(K,C)}{2}$,

  3. iii)

$\displaystyle\frac{D_{\operatorname{M}}(K,C)}{2}\leq\delta_{\operatorname{M}}R(K,C)$,
and

  4. iv)

$\displaystyle\rho_{\operatorname{M}}(s(C)r(K,C)+R(K,C))\leq(s(C)+1)\frac{D_{\operatorname{M}}(K,C)}{2}$.

  5. v)

$\displaystyle r(K,C)+R(K,C)\leq
R(C_{\operatorname{M}},C)D_{\operatorname{M}}(K,C)$.

###### Proof.

  1. i)

Follows directly from Remark 4.1 and Proposition 2.3.

  2. ii)

Follows directly from (Notation) and Remark 4.1.

  3. iii)

Since $r(K,C)C$ is contained in a translate of $K$, we obtain

$\delta_{\operatorname{M}}r(K,C)=\frac{1}{2}D_{\operatorname{M}}(C,C)r(K,C)=\frac{1}{2}D_{\operatorname{M}}(r(K,C)C,C)\leq\frac{D_{\operatorname{M}}(K,C)}{2}.$

  4. iv)

By [10, Theorem 1.1] we have

$s(C)r(K,C)+R(K,C)\leq\frac{s(C)+1}{2}D_{\operatorname{AM}}(K,C)$

and

$(s(C)+1)\frac{D_{\operatorname{AM}}(K,C)}{2}\leq(s(C)+1)\frac{D_{\operatorname{M}}(K,C)}{2\rho_{\operatorname{M}}}$

follows directly from the definition of $\rho_{\operatorname{M}}$.

  5. v)

For symmetric gauges $r(K,C)+R(K,C)\leq D(K,C)$. Thus,

$D_{\operatorname{M}}(K,C)\geq
r(K,C_{\operatorname{M}})+R(K,C_{\operatorname{M}})\geq\frac{1}{R(C_{\operatorname{M}},C)}\left(r(K,C)+R(K,C)\right).$

∎

We would like to describe the values the inradius, circumradius, and diameter
of sets $K\in\mathcal{C}^{n}$ may have, when we consider a fixed, Minkowski-
centered $C\in\mathcal{C}^{n}_{0}$. To do so, we study the following Blaschke-
Santaló diagrams.

######  Definition 4.17.

Let $f_{\operatorname{M}}$ be the following mapping.

(1)
$f_{\operatorname{M}}:\mathcal{C}^{n}\times\mathcal{C}^{n}_{0}\to\mathds{R}^{2},\>f_{\operatorname{M}}(K,C)=\left(\frac{r(K,C)}{R(K,C)},\frac{D_{\operatorname{M}}(K,C)}{2R(K,C)}\right)$

The set $f_{\operatorname{M}}(\mathcal{C}^{n},C)$ is called the _Blaschke-
Santaló diagram_ for the inradius, circumradius, and diameter (depending on
the respective definitions) with regard to the gauge $C$ – the
($r,R,D_{\operatorname{M}}$)-diagram.

As for the diameter we only write $f$ if the gauge is symmetric. In [9]
$f_{\operatorname{AM}}(\mathcal{C}^{2},S)$ is described and it is shown that
this diagram is equal to the union of the diagrams over all possible gauges.

######  Proposition 4.18.

For every triangle $S\in\mathcal{C}^{2}$, the diagram
$f_{\operatorname{AM}}(\mathcal{C}^{2},S)$ is fully desribed by the
inequalities

$\displaystyle D_{\operatorname{AM}}(K,S)$ $\displaystyle\leq 2R(K,C)$
$\displaystyle 4r(K,S)+2R(K,S)$ $\displaystyle\leq
3D_{\operatorname{AM}}(K,S)$
$\displaystyle\frac{D_{\operatorname{AM}}(K,C)}{2R(K,C)}\left(1-\frac{D_{\operatorname{AM}}(K,C)}{2R(K,C)}\right)$
$\displaystyle\leq\frac{r(K,C)}{R(K,C)}.$

Moreover,
$f_{\operatorname{AM}}(\mathcal{C}^{2},S)=f_{\operatorname{AM}}(\mathcal{C}^{2},\mathcal{C}^{2}_{0})$.

The diagrams $f(\mathcal{C}^{2},\mathds{B}^{2}_{2})$ [41] (name giving) and
$f_{\operatorname{AM}}(\mathcal{C}^{2},S)$ [9] can be seen in Figure 3.

$\frac{r}{R}$$\frac{D}{2R}$.$\frac{\sqrt{3}}{2}$.$1$.$\sqrt{3}-1$.$\frac{1}{2}$.$1$.$\mathds{B}^{2}_{2}$.$L$.$T$.$\operatorname{RT}$

$\frac{r}{R}$$\frac{D_{\operatorname{AM}}}{2R}$.$\frac{1}{2}$.$1$.$\frac{1}{4}$.$1$.$S$.$L$.$-S$

Figure 3. The $(r,R,D)$-diagram w. r. t. $\mathds{B}^{2}_{2}$ (left) and the
$(r,R,D_{\operatorname{AM}})$-diagram w. r. t. a triangle $S$ (right).

It is shown in [9] that $f_{\operatorname{AM}}(\mathcal{C}^{n},C)$ is star-
shaped with respect to the vertex $f_{\operatorname{AM}}(C,C)=(1,1)$. This
means that these diagrams can be fully described by characterizing the
boundaries of the set. In the following, we prove similar (slightly weaker,
but sufficient for our purposes) results for the other diameters. One may note
that all diagrams with respect to triangles that are described in the
following chapters are still star-shaped w.r.t. $(1,1)$.

However, we will show that the ($r,R,D_{\operatorname{M}}$)-Blaschke-Santaló
diagram is simply connected, i.e. it does not have any holes.

######  Theorem 4.19.

The diagram $f_{\operatorname{M}}(\mathcal{C}^{n},C)$ is closed and simply
connected.

###### Proof.

Assume there is a sequence $(K_{n})_{n\in\mathds{N}}\subset\mathcal{C}^{n}$
such that $R(K_{n},C)=1$ for all $n\in\mathds{N}$ and $r(K_{n},C)\to r^{*}$
and $D_{\operatorname{M}}(K_{n},C)\to D^{*}$ for $n\to\infty$. The sequence
$(K_{n})_{n\in\mathds{N}}$ is bounded as all sets are contained in $C$. Thus,
by the Blaschke-Selection-Theorem there exists a converging subsequence
$K_{n_{k}}\to K^{*}$ for $k\to\infty$. The inradius and diameter are
continuous and therefore $r(K^{*},C)=r^{*}$ and
$D_{\operatorname{M}}(K^{*},C)=D^{*}$. Hence,
$f_{\operatorname{M}}(\mathcal{C}^{n},C)$ is closed.
As a consequence, we know that $f_{\operatorname{M}}(\mathcal{C}^{n},C)$ can
only have open holes and therefore only full-dimensional holes. For
$K\in\mathcal{C}^{n}$ such that $K\subset^{\operatorname{opt}}C$, define
$K_{t}:=(1-t)K+tC$ for $t\in[0,1]$.
Then by Lemma 4.2,

$r(K_{t},C)=(1-t)r(K,C)+t,$ $R(K_{t},C)=(1-t)R(K,C)+t=1$

and

$D_{\operatorname{M}}(K_{t},C)\leq(1-t)D_{\operatorname{M}}(K,C)+tD_{\operatorname{M}}(C,C).$

In the case $\operatorname{M}=\operatorname{AM}$, we also have equality for
the diameter, but this does not hold for the other diameters. Since
$R(\cdot,C),r(\cdot,C)$ and $D_{\operatorname{M}}(\cdot,C)$ are continuous
with respect to the Hausdorff distance and $t\in[0,1]\mapsto(1-t)K+tC$ is
continuous in $t$, the composition
$\Gamma_{K}:[0,1]\to\mathds{R}^{2},\>t\mapsto\left(r(K_{t},C),\frac{D_{\operatorname{M}}(K_{t},C)}{2}\right)$
is continuous as well. Thus, for every such $K$ there is a continuous curve
$\Gamma_{K}$ in the diagram from $f_{\operatorname{M}}(K,C)$ to
$f_{\operatorname{M}}(C,C)$.
Let $(K^{n})_{n\in\mathds{N}}$ be a sequence of bodies on the boundary
converging to $K$ on the boundary. We show that the functions $\Gamma_{K^{n}}$
converge uniformly to $\Gamma_{K}$. We can consider the components separately.
For the inradius, we know

$\begin{split}\norm{r(K_{t},C)-r(K^{n}_{t},C)}&=\norm{(1-t)r(K,C)+t-(1-t)r(K^{n},C)-t}\\\
&=(1-t)\norm{r(K,C)-r(K^{n},C)}\\\ &\leq\norm{r(K,C)-r(K^{n},C)}.\end{split}$

Let $\epsilon>0$. Since $\norm{r(K,C)-r(K^{n},C)}\to 0$ for $n\to\infty$,
there exists an $N$ such that for all $n\geq N$,
$\norm{r(K_{t},C)-r(K^{n}_{t},C)}<\epsilon$ for all $t\in[0,1]$. It is known
that when convex, continuous functions $f_{n}:[a,b]\to\mathds{R}$ converge
pointwise to a convex and continuous function $f$, the convergence is uniform
[33, Lemma 21]. The functions $g_{n}:[0,1]\to\mathds{R}$, $t\mapsto
D_{\operatorname{M}}(K^{n}_{t},C)$ are convex and continuous in $t$ and they
converge pointwise to the convex and continuous function
$g:[0,1]\to\mathds{R}$, $t\mapsto D_{\operatorname{M}}(K_{t},C)$. Thus, this
convergence is also uniform and the curves converge uniformly.

Now, assume, the diagram has a hole. For $K$ on the boundary of the diagram we
say that the curve $\Gamma_{K}$ lies above the hole, if the set enclosed by
$[f_{\operatorname{M}}(L_{D},C),f_{\operatorname{M}}(C,C)]$, $\Gamma_{K}$ and
the boundary between $f_{\operatorname{M}}(K,C)$ and
$f_{\operatorname{M}}(L_{D},C)$ which does not contain the segment
$[f_{\operatorname{M}}(L_{D},C),f_{\operatorname{M}}(C,C)]$ does not contain
the hole. Analogously, we say that $\Gamma_{K}$ lies below the hole if the set
contains the hole. Thus, $\Gamma_{L_{D}}$ lies above the hole and $\Gamma_{C}$
below. Then, there exists a converging sequence $(K_{n})_{n\in\mathds{N}}$
with $K_{n}\to K$ of bodies on the boundary such that all $\Gamma_{K^{n}}$ lie
above the hole and $\Gamma_{K}$ below or vice versa. This contradicts the fact
that the curves converge uniformly.

$\frac{r}{R}$$\frac{D_{\operatorname{M}}}{2R}$.$K$.$K_{n}$.$C$.$L_{D}$.$L_{w}$
Figure 4. Proof of Theorem 4.19: $K_{n}\to K$ but $\Gamma_{K}$ lies below the
hole and $\Gamma_{K_{n}}$ above.

∎

In the standard diameter case it was sufficient to describe the Blaschke-
Santaló diagram w. r. t. a triangle to obtain
$f_{\operatorname{AM}}(\mathcal{C}^{2},\mathcal{C}^{2}_{0})$. Thus, it seems
reasonable to look at the diagrams for the three other diameters
$D_{\operatorname{MIN}}$, $D_{\operatorname{MAX}}$ and $D_{\operatorname{HM}}$
in terms of triangular gauges first, which we do in the remaining sections.

##  5\. The diameter $D_{\operatorname{MAX}}$

When we use the notions “equilateral”, “regular”, and “isosceles” in the
following, then it is meant in the euclidean sense. Unless otherwise
specified, we fix the _equilateral triangle_ to be
$T:=\operatorname{conv}(\\{p^{1},p^{2},p^{3}\\})\subset\mathds{R}^{2}$ with
$p^{1}=(0,1)^{T}$, $p^{2}=(-\sqrt{3}/2,-1/2)^{T}$ and
$p^{3}=(\sqrt{3}/2,-1/2)^{T}$. It is Minkowski-centered with $s(T)=2$.
Moreover, in this case $T_{\operatorname{MAX}}$ is the regular hexagon
$\operatorname{conv}(\\{p^{1},p^{2},p^{3},-p^{1},-p^{2},-p^{3}\\})$.

$p^{1}$$p^{2}$$p^{3}$$-p^{1}$$-p^{2}$$-p^{3}$ Figure 5. The equilateral
triangle $T$ and its maximum $T_{\operatorname{MAX}}$

In the case of the maximum both factors $\rho_{\operatorname{MAX}}$ and
$\delta_{\operatorname{MAX}}$ are known [7] and depend at most on $s(C)$:

######  Proposition 5.1.

$C_{\operatorname{AM}}{\subset}^{\operatorname{opt}}C_{\operatorname{MAX}}\subset^{\operatorname{opt}}\frac{2s(C)}{s(C)+1}C_{\operatorname{AM}},$

i. e. $\rho_{\operatorname{MAX}}=\frac{s(C)+1}{2s(C)}$ and
$\delta_{\operatorname{MAX}}=1$.

The inequalities from Lemma 4.16 have the form:

(2) $\frac{D_{\operatorname{MAX}}(K,C)}{2}\leq R(K,C),$

(3) $r(K,C)\leq\frac{D_{\operatorname{MAX}}(K,C)}{2},$

(4) $s(C)r(K,C)+R(K,C)\leq s(C)D_{\operatorname{MAX}}(K,C),$

and

(5) $0\leq r(K,C).$

In case of $K=C$ (2) and (3) become tight while (3) and (4) become tight for
$K=C_{\operatorname{MAX}}$, since $R(C_{\operatorname{MAX}},C)=s(C)$,
$r(C_{\operatorname{MAX}},C)=1$ and
$D_{\operatorname{MAX}}(C_{\operatorname{MAX}},C)=2$. Moreover, the last one
is tight for segments as described in Lemma 4.8.

Asymmetric gauges are not complete, but a completion is easy to find.

######  Definition 5.2.

Let
$A_{C}^{\operatorname{oss}}:=\operatorname{bd}(C^{\circ})\cap\operatorname{bd}(-C^{\circ})$.

We define the _outer symmetric support_ :

$C^{\operatorname{oss}}:=\bigcap_{a\in
A_{C}^{\operatorname{oss}}}H_{(a,1)}^{\leq}$

######  Lemma 5.3.

  1. i)

$C_{\operatorname{MAX}}$ is always a completion of $C$ with maximal
circumradius $R(C_{\operatorname{MAX}},C)=s(C)$,

  2. ii)

$C_{\operatorname{MAX}}\subset C^{\sup}\subset C^{\operatorname{oss}}$,

  3. iii)

$C_{\operatorname{MAX}}=C^{\sup}$ if and only if
$C_{\operatorname{MAX}}=C^{\operatorname{oss}}$, and

  4. iv)

$C_{\operatorname{MAX}}$ is always the unique 0-symmetric completion of $C$.

One should recognize that we need a scaling factor of up to $n$ to cover the
completion $C_{\operatorname{MAX}}$ by $C$ here, while with the arithmetic
diameter $C$ is always already complete itself.

###### Proof.

$C_{\operatorname{MAX}}$ is a completion of $C$ since $C\subset
C_{\operatorname{MAX}}$ and
$D_{\operatorname{MAX}}(C,C)=2=D_{\operatorname{MAX}}(C_{\operatorname{MAX}},C)$,
while for all $K\supset C_{\operatorname{MAX}}$ we have
$D_{\operatorname{MAX}}(K,C)=D_{\operatorname{MAX}}(K,C_{\operatorname{MAX}})=2R(K,C_{\operatorname{MAX}})>2$.

The maximality of the circumradius can be seen as follows: Let $K$ be any
completion of $C$. Then, from (4) and Lemma 4.6 we obtain

$\displaystyle R(K,C)$ $\displaystyle\leq
s(C)(D_{\operatorname{MAX}}(K,C)-r(K,C))=s(C)(D_{\operatorname{MAX}}(C_{\operatorname{MAX}},C)-r(C_{\operatorname{MAX}},C))$
$\displaystyle=s(C)=R(C_{\operatorname{MAX}},C).$

Next, we show that $C_{\operatorname{MAX}}\subset C^{\sup}\subset
C^{\operatorname{oss}}$. The first containment follows from the fact that
$C^{\sup}$ is the union of all completions of $C$. For the second containment
it suffices to show that $h_{C^{\sup}}(a)\leq h_{C^{\operatorname{oss}}}(a)$
for all $a\in A^{\operatorname{oss}}(C)$. Let $a\in
A^{\operatorname{oss}}(C)$. Then, there exists $p\in-C\cap H_{(a,1)}$. Since
$-p\in C$ we obtain $C^{\sup}\subset-p+2C_{\operatorname{MAX}}$ and therefore

$h_{C^{\sup}}(a)\leq
h_{-p+2C_{\operatorname{MAX}}}(a)=-p^{T}a+2h_{C_{\operatorname{MAX}}}(a)=1=h_{C^{\operatorname{oss}}}(a).$

The containment chain directly shows the backward direction of part of $iii)$.

To show the forward direction, assume $C_{\operatorname{MAX}}\neq
C^{\operatorname{oss}}$. Since a fulldimensional convex body is the
intersection of its regular slabs, there must exist some
$a\in\operatorname{bd}(C^{\circ})$ which defines a regular slab of
$C_{\operatorname{MAX}}$ but $h_{C}(a)>h_{C}(-a)$. Thus, there exists a smooth
boundary point $x\in\operatorname{bd}(C)\cap H_{(a,1)}$ of
$C_{\operatorname{MAX}}$. Now, assume $x\in\operatorname{bd}(C^{\sup})$. Then,
there exist an outer normal
$a_{x}\in\operatorname{bd}(C_{\operatorname{MAX}})^{\circ}$ and a point
$p_{x}\in C$ such that $x^{T}a_{x}=h_{C^{\sup}}(a_{x})$ and
$(a_{x})^{T}(x-p_{x})=2$. Since $x$ is a smooth boundary point of
$C_{\operatorname{MAX}}$ it has to be a smooth boundary point of $C^{\sup}$,
which implies $a_{x}=a$. But since $h_{C}(-a)<h_{C}(a)$ we cannot choose
$p_{x}\in C$. Thus, $x$ is not contained in the boundary of $C^{sup}$ and
therefore, $C_{\operatorname{MAX}}\neq C^{\sup}$.

Finally, any 0-symmetric completion of $C$ must contain $C$ and $-C$ and
therefore $C_{\max}$. ∎

######  Example 5.4.

Trapezoids within the following family have completions besides their maximum:

$Z_{\lambda}:=\operatorname{conv}\left((\sqrt{3}/2,-1/2)^{T},(\sqrt{3}/2,-1/2)^{T},(\lambda\sqrt{3}/2,1-\lambda/2)^{T},(-\lambda\sqrt{3}/2,1-\lambda/2)^{T}\right)$

with $\lambda\in(0,1)$. $Z_{\lambda}$ is Minkowski-centered with Minkowski
asymmetry $2-\lambda$ and
$(Z_{\lambda})^{\operatorname{oss}}\neq{(Z_{\lambda})}_{\operatorname{MAX}}$,
which because of Lemma 5.3 means that $(Z_{\lambda})_{\operatorname{MAX}}$ is
not the unique completion of $Z_{\lambda}$. In the extreme cases
$\lambda\in\set{0,1}$, $Z_{\lambda}$ is a triangle or a rectangle and
$(Z_{\lambda})_{\operatorname{MAX}}=(Z_{\lambda})^{\sup}=(Z_{\lambda})^{\operatorname{oss}}$.

$(Z_{\lambda})^{\operatorname{oss}}$$(Z_{\lambda})^{\sup}$$Z_{\lambda}$ Figure
6. For trapezoids $Z_{\lambda}$ the maximum
$(Z_{\lambda})_{\operatorname{MAX}}$ is not their unique completion since
$(Z_{\lambda})_{\operatorname{MAX}}\neq(Z_{\lambda})^{\operatorname{oss}}$.

The first new inequality we provide is a lower bound for the diameter-
circumradius ratio, a so called Jung-type inequality, which stays true
independently of the gauge $C$.

######  Theorem 5.5.

Let $K,C\in\mathcal{C}^{2}$, s. t. $C$ is Minkowski-centered. Then

$D_{\operatorname{MAX}}(K,C)\geq R(K,C)$

###### Proof.

Assume $K\subset^{\operatorname{opt}}C$, which implies $R(K,C)=1$. Then, there
exist touching points
$q^{1},\dots,q^{k}\in\operatorname{bd}(K)\cap\operatorname{bd}(C)$ with
$k\in\set{2,3}$ and corresponding outer normals $a^{i}$ as described in
Proposition 2.2. If $k=2$ is possible, there exists a segment $L\subset K$
with the same circumradius as $K$ and by Lemma 4.8
$D_{\operatorname{MAX}}(K,C)\geq D_{\operatorname{MAX}}(L,C)\geq
2\rho_{\operatorname{MAX}}(C)=\frac{s(C)+1}{s(C)}>1$. Thus, we can assume that
$K$ is a proper triangle $K=\operatorname{conv}(\set{q^{1},q^{2},q^{3}})$.

Let $S:=\bigcap_{i=1}^{3}H_{(a^{i},1)}^{\leq}$ be the intersection of the
three supporting halfspaces of $C$ s. t. $q^{i}\in H_{(a^{i},1)}$. Denote the
vertex opposing the edge defined by $a^{i}$ by $\tilde{p}^{i}$. Then,
$R(K,S)=R(K,C)=1$ and
$D_{\operatorname{MAX}}(K,C)=D_{\operatorname{MAX}}(K,S)$. By invariance under
linear transformations we can assume that $S=c+T$ where $T$ is the Minkowski-
centered equilateral triangle as described above: $\tilde{p}^{i}=c+p^{i}$,
$i=1,2,3$. Let $\alpha_{i}\in[0,1]$ be, s. t.
$q^{i}=\alpha_{i}\tilde{p}^{i+1}+(1-\alpha)\tilde{p}^{i+2}$. Since $C$ is
Minkowski-centered, $0$ lies in the interior of $C$.

We split the proof into two parts. First, we consider the case where the
origin is close to the center $c$, i. e.
$0\in\operatorname{int}(\operatorname{conv}(\set{c-\frac{1}{2}p^{i},\quad
i=1,2,3}))$. Afterwards, we care about the case, where $c$ is further apart
from the origin.

Let us start with the case where
$0\in\operatorname{int}(\operatorname{conv}(\set{c-\frac{1}{2}p^{i},\quad
i=1,2,3}))$, which is equivalent to
$c\in\operatorname{int}(\operatorname{conv}(\set{\frac{1}{2}p^{i},\quad
i=1,2,3}))$. Define $\lambda_{1},\lambda_{2},\lambda_{3}>0$ with
$\sum_{i=1}^{3}=\frac{1}{2}$ such that $c=\sum_{i=1}^{3}\lambda_{i}p^{i}$. In
the following indices are to be understood modulo 3. Let
$z^{i}\in\mathds{R}^{2}$ be the direction such that

$(z^{i})^{T}\tilde{p}^{i+1}=-1\quad\text{and}\quad(z^{i})^{T}\tilde{p}^{i+2}=1.$

Since $c=\frac{1}{3}\sum_{i=1}^{3}\tilde{p}^{i}$ we have

$(z^{i})^{T}\tilde{p}^{i}=3(z^{i})^{T}c.$

Inserting $c=\sum_{i=1}^{3}\lambda_{i}p^{i}$ yields

$\displaystyle(z^{i})^{T}c$
$\displaystyle=(z^{i})^{T}\sum_{i=1}^{3}\lambda_{i}(\tilde{p}^{i}-c)=-\frac{1}{2}(z^{i})^{T}c+3\lambda_{i}(z^{i})^{T}c-\lambda_{i+1}+\lambda_{i+2}$

which implies

$\displaystyle
3(z^{i})^{T}c=\frac{\lambda_{i+2}-\lambda_{i+1}}{\frac{1}{2}-\lambda_{i}}=\frac{\lambda_{i+2}-\lambda_{i+1}}{\lambda_{i+2}+\lambda_{i+1}}.$

Thus,

$1+3(z^{i})^{T}c=\frac{2\lambda_{i+2}}{\lambda_{i+2}+\lambda_{i+1}}\geq
0\quad\text{and}\quad
1-3(z^{i})^{T}c=\frac{2\lambda_{i+1}}{\lambda_{i+2}+\lambda_{i+1}}\geq 0,$

which shows that $z^{i}$ is an outer normal of
$\operatorname{conv}(S\cup(-S))$ with
$h_{\operatorname{conv}(S\cup(-S))}(z^{i})=1$.

.$0$.$c$$\tilde{p}^{1}$$\tilde{p}^{2}$$\tilde{p}^{3}$$-\tilde{p}^{1}$$-\tilde{p}^{2}$$-\tilde{p}^{3}$$q^{1}$$q^{2}$$q^{3}$$z^{1}$$z^{2}$$z^{3}$
Figure 7. Proof of Theorem 5.5. If
$0\in\operatorname{int}(\operatorname{conv}(\set{c-\frac{1}{2}p^{i},\quad
i=1,2,3}))$, then $\max_{i=1,2,3}b_{z^{i}}(K,S\cup(-S))\geq 1$.

We know

(6)
$\begin{split}D_{\operatorname{MAX}}(K,S)&\geq\max_{i=1,2,3}b_{z^{i}}(K,\operatorname{conv}(S\cup(-S)))\\\
&\geq\max_{i=1,2,3}\frac{(z^{i})^{T}q^{i+1}-(z^{i})^{T}q^{i+2}}{h_{\operatorname{conv}(S\cup(-S))}(z^{i})}\\\
&=\max_{i=1,2,3}(z^{i})^{T}(\alpha_{i+1}\tilde{p}^{i+2}+(1-\alpha_{i+1})\tilde{p}^{i})-(z^{i})^{T}(\alpha_{i+2}\tilde{p}^{i}+(1-\alpha_{i+2})\tilde{p}^{i+1})\\\
&=\max_{i=1,2,3}\alpha_{i+1}(1-3(z^{i})^{T}c)+(1-\alpha_{i+2})(1+3(z^{i})^{T}c)\\\
&=\max_{i=1,2,3}\frac{2\alpha_{i+1}\lambda_{i+1}+2(1-\alpha_{i+2})\lambda_{i+2}}{\lambda_{i+2}}\end{split}$

Next, we show that the last term is larger than or equal to 1. If
$b_{z^{i}}(K,T_{\operatorname{MAX}})\geq 1$ for $i\in\set{1,2}$ there is
nothing to show. Thus let us assume that those breadths are smaller than 1. We
show that the latter implies $b_{z^{3}}(K,T_{\operatorname{MAX}})\geq 1$.

$\displaystyle b_{z^{1}}(K,S_{\operatorname{MAX}})<1\text{\quad implies
\quad}2\alpha_{2}\lambda_{2}+2(1-\alpha_{3})\lambda_{3}$
$\displaystyle<\lambda_{2}+\lambda_{3}\text{\quad and}$ $\displaystyle
b_{z^{2}}(K,S_{\operatorname{MAX}})<1\text{\quad implies
\quad}2\alpha_{3}\lambda_{3}+2(1-\alpha_{1})\lambda_{1}$
$\displaystyle<\lambda_{3}+\lambda_{1}.$

By adding these inequalities we obtain

$\displaystyle 2\lambda_{3}+2\alpha_{2}\lambda_{2}+2(1-\alpha_{1})\lambda_{1}$
$\displaystyle<\lambda_{2}+2\lambda_{3}+\lambda_{1}$ or equivalently
$\displaystyle 2\alpha_{1}\lambda_{1}+2(1-\alpha_{2})\lambda_{2}$
$\displaystyle>\lambda_{1}+\lambda_{2},$

which proves $b_{z^{3}}(K,T_{\operatorname{MAX}})\geq 1$. Thus,
$D_{\operatorname{MAX}}(K,C)\geq D_{\operatorname{MAX}}(K,S)\geq 1$.

Now, consider the case that
$0\notin\operatorname{int}(\operatorname{conv}(\set{c-\frac{1}{2}p^{i},\quad
i=1,2,3}))$. Because of the symmetries of $T$ we can assume that
$0=\sum_{i=1}^{3}\beta_{i}\tilde{p^{i}}$ with $\sum_{i=1}^{3}\beta_{i}=1$,
$\beta_{3}\geq\frac{1}{2}$ and $\beta_{2}\geq\beta_{1}\geq 0$ ( cf. Figure 8).

$a^{3}$$a^{1}$$a^{2}$$0$.$c$$\tilde{p}^{1}$$\tilde{p}^{2}$$\tilde{p}^{3}$$q^{1}$$q^{2}$$q^{3}$
Figure 8. Proof of Theorem 5.5. If
$0\notin\operatorname{int}(\operatorname{conv}(\set{c-\frac{1}{2}p^{i},\quad
i=1,2,3}))$ we may assume, w. l. o. g., that the origin is within the colored
area. Then, $\max_{i=1,3}b_{a^{i}}(K,\operatorname{conv}(C\cup(-C)))\geq 1$.

By the above conditions we have

$h_{C}(-a^{3})\leq
h_{S}(-a^{3})=(-a^{3})^{T}\tilde{p}^{3}=\frac{\beta_{1}+\beta_{2}}{\beta_{3}}\leq
1=h_{C}(a^{3}).$

Thus, using $\alpha_{1},\alpha_{2},\alpha_{3}$ as above, we know, if
$\alpha_{1}<1-\beta_{3}$ or $\alpha_{2}>\beta_{3}$, then

$\displaystyle h_{K}(-a^{3})$
$\displaystyle=\max\set{(-a^{3})^{T}q^{1},(-a^{3})^{T}q^{1}}$
$\displaystyle=\max\set{(-a^{3})^{T}(\alpha_{1}\tilde{p}^{2}+(1-\alpha_{1})\tilde{p}^{3}),(-a^{3})^{T}(\alpha_{2}\tilde{p}^{3}+(1-\alpha_{2})\tilde{p}^{1})}$
$\displaystyle=\max\set{-\alpha_{1}+(1-\alpha_{1})\frac{1-\beta_{3}}{\beta_{3}},\alpha_{2}\frac{1-\beta_{3}}{\beta_{3}}-(1-\alpha_{2})}>0$

and therefore,

$D_{\operatorname{MAX}}(K,C)\geq
b_{a^{3}}(K,\operatorname{conv}(C\cup(-C)))\geq\frac{h_{K}(a^{3})+h_{K}(-a^{3})}{h_{C}(a^{3})}>1.$

Now, let $\alpha_{1}\geq 1-\beta_{3}$ and $\alpha_{2}\leq\beta_{3}$. Since
$\beta_{1}\leq\beta_{2}$ we have $\beta_{1}\leq\frac{1-\beta_{3}}{2}$. Thus,

$\displaystyle(-{a^{1}})^{T}q^{2}$
$\displaystyle=\alpha_{2}((-{a^{1}})^{T}\tilde{p^{3}})+(1-\alpha)((-a^{1})^{T}\tilde{p}^{1})=-\alpha_{2}+(1-\alpha_{2})\left(\frac{\beta_{2}}{\beta_{1}}+\frac{\beta_{3}}{\beta_{1}}\right)$
$\displaystyle\geq-\beta_{3}+(1-\beta_{3})(1+\frac{2\beta_{3}}{1-\beta_{3}})=1.$

Together with $h_{C}(-a^{1})\leq s(C)h_{C}(a^{1})=s(C)$ we obtain

$\displaystyle D_{\operatorname{MAX}}(K,C)$ $\displaystyle\geq
b_{a^{1}}(K,\operatorname{conv}(C\cup(-C)))\geq\frac{(a^{1})^{T}q^{1}-(a^{1})^{T}q^{2}}{\max(h_{C}(a^{1}),h_{C}(-a^{1}))}$
$\displaystyle\geq\frac{1+1}{s(C)}=\frac{2}{s(C)}\geq 1.$

∎

Since $D_{\operatorname{MAX}}$ is the smallest diameter, this provides a lower
bound for all four diameter definitions.

######  Corollary 5.6.

Let $K,C\in\mathcal{C}^{2}$, $C$ Minkowski-centered. Then

$D_{\operatorname{M}}(K,C)\geq R(K,C).$

If $\operatorname{M}\neq\operatorname{MAX}$ it follows directly from
Proposition 2.3 and the translation invariance in the arithmetic case that
Corollary 5.6 stays true for non-Minkowski-centered $C\in\mathcal{C}^{2}$.
However, since $D_{\operatorname{AM}}(K,C)=R(K,C)$ is obtained if and only if
$C$ is a triangle and $K=-C$ (up to translation), we will see in the next
section that this bound cannot be reached if
$\operatorname{M}\in\set{\operatorname{MIN},\operatorname{HM}}$.

If we omit the restriction of $C$ being Minkowski-centered Theorem 5.5 is not
true. However, for gauges that still contain the origin, we obtain the
following Jung-type inequality.

######  Theorem 5.7.

Let $K,C\in\mathcal{C}^{2}$, s. t. $0\in C$. Then,

$D_{\operatorname{MAX}}(K,C)\geq\frac{2}{3}R(K,C).$

Moreover, equality can be attained only if $C$ is a triangle with one vertex
at the origin.

###### Proof.

We use the same notation as in the previous proof. As in Theorem 5.5, we can
assume that $K$ is a triangle and $K\subset^{\operatorname{opt}}C$. It
suffices to show $D(K,\operatorname{conv}(S\cup(-S)))\geq\frac{2}{3}$ for $S$
as given in Theorem 5.5 since
$D_{\operatorname{MAX}}(K,C)=D(K,\operatorname{conv}(C\cup(-C)))\geq
D(K,\operatorname{conv}(S\cup(-S)))$.

If there exists $\alpha_{i}\notin\left[\frac{1}{3},\frac{2}{3}\right]$, there
is an $a^{j}$ with
$\frac{{a^{j}}^{T}q^{j}-{a^{j}}^{T}q^{i}}{h_{C}(a^{j})+h_{C}(-a^{j})}>\frac{2}{3}$
( cf. Figure 9) and therefore,

$\displaystyle D(K,\operatorname{conv}(S\cup(-S)))$ $\displaystyle\geq
b_{a^{j}}(K,\operatorname{conv}(S\cup(-S)))\geq\frac{{a^{j}}^{T}q^{j}-{a^{j}}^{T}q^{i}}{\max(h_{C}(a^{j}),h_{C}(-a^{j}))}$
$\displaystyle\geq\frac{{a^{j}}^{T}q^{j}-{a^{j}}^{T}q^{i}}{h_{C}(a^{j})+h_{C}(-a^{j})}>\frac{2}{3}.$

.$0$.$c$$\tilde{p}^{1}$$\tilde{p}^{2}$$\tilde{p}^{3}$$-\tilde{p}^{1}$$-\tilde{p}^{2}$$-\tilde{p}^{3}$$q^{1}$$q^{2}$$q^{3}$$z^{3}$$a^{3}$$-a^{3}$

Figure 9. Proof of Theorem 5.7. If $\alpha_{1}<\frac{1}{3}$ or
$\alpha_{2}>\frac{2}{3}$, we have
$b_{a^{3}}(K,\operatorname{conv}(S\cup(-S)))>\frac{2}{3}$.

Now, consider the case where
$\alpha_{i}\in\left[\frac{1}{3},\frac{2}{3}\right]$ for all $i\in\set{1,2,3}$.
There exists $i\in\set{1,2,3}$ such that
$h_{\operatorname{conv}(S\cup(-S))}(z^{i})=1$, w. l. o. g. $i=3$. Then, using
(6) we obtain

(7) $\begin{split}D(K,\operatorname{conv}(S\cup(-S)))&\geq
b_{z^{3}}(K,\operatorname{conv}(S\cup(-S)))\geq\alpha_{1}(1-3(z^{3})^{T}c)+(1-\alpha_{2})(1+3(z^{3})^{T}c)\\\
&\geq\frac{1}{3}(1-3(z^{3})^{T}c)+\frac{1}{3}(1+3(z^{3})^{T}c)=\frac{2}{3}\end{split}$

Reaching equality in the last inequality chain we need
$\alpha_{1}=(1-\alpha_{2})=\frac{1}{3}$. In this case we need
$0\in\set{\tilde{p}^{3}}\cup[\tilde{p}^{1},\tilde{p}^{2}]$ for $b_{a^{3}}$ not
to be larger than $\frac{2}{3}$. However, if
$0\in[\tilde{p}^{1},\tilde{p}^{2}]$ we have
$h_{\operatorname{conv}(S\cup(-S))}(z^{i})=1$ for $i=1$ or $i=2$. For this
$z^{i}$ we do not have the correct $\alpha$-values to obtain equality in (7).
Thus, $0$ remains to be a vertex of $S$ to reach equality. For
$D_{\operatorname{MAX}}(K,C)=\frac{2}{3}$ to be true, we need

$b_{z^{3}}(K,\operatorname{conv}(C\cup(-C)))=b_{z^{3}}(K,\operatorname{conv}(S\cup(-S)))=\frac{2}{3},$

which is only possible if $\tilde{p}^{1},\tilde{p}^{2}\in C$, which means
$C=S$.

All in all, we see that equality can be obtained for
$C=\operatorname{conv}(\set{0,\tilde{p}^{1},\tilde{p}^{2}})$ and
$\alpha_{1}=(1-\alpha_{2})=\frac{1}{3}$ and
$\alpha_{3}\in\left[\frac{1}{3},\frac{2}{3}\right]$ only. In that case

$D_{\operatorname{MAX}}(K,C)=\max\set{b_{a^{3}}(K,\operatorname{conv}(C\cup(-C))),b_{z^{3}}(K,\operatorname{conv}(C\cup(-C)))}=\frac{2}{3}$

(see Figure 10), which is achieved, e.g. if we choose
$K=\operatorname{conv}\left(\set{\frac{1}{3}\tilde{p}^{1},\frac{1}{3}\tilde{p}^{2},\frac{1}{2}(\tilde{p}^{1}+\tilde{p}^{2})}\right)$.

.$\tilde{p}^{1}$$\tilde{p}^{2}$$0$$q^{1}$$q^{2}$$q^{3}$$z^{3}$$a^{3}$ Figure
10. Equality case in Theorem 5.7:
$C=\operatorname{conv}(\set{0,\tilde{p}^{1},\tilde{p}^{2}})$ and
$K=\operatorname{conv}\left(\set{\frac{1}{3}\tilde{p}^{1},\frac{1}{3}\tilde{p}^{2},\frac{1}{2}(\tilde{p}^{1}+\tilde{p}^{2})}\right)$.

∎

For fixed $C$, we refer to any $\tilde{K}$ that fulfills
$\frac{D_{\operatorname{M}}(\tilde{K},C)}{R(\tilde{K},C)}=\min_{K\in\mathcal{C}^{n}}\frac{D_{\operatorname{M}}(K,C)}{R(K,C)}$
as _Jung-extremal_. If $C=T$, we have equality in Theorem 5.5 for the
following family of triangles.

######  Example 5.8.

Let $T$ be the equilateral triangle as defined above and

$T_{\alpha}:=\operatorname{conv}(\set{\alpha p^{1}+(1-\alpha)p^{2},\alpha
p^{2}+(1-\alpha)p^{3},\alpha p^{3}+(1-\alpha)p^{1}})$

for $\alpha\in\left[\frac{1}{3},\frac{2}{3}\right]$ ( cf. Figure 11). Then,

$\displaystyle
D_{\operatorname{MAX}}(T_{\alpha},T)=R(T_{\alpha},T)\quad\text{and}\quad
r(T_{\alpha},T)$ $\displaystyle=(1-3\alpha+3\alpha^{2})R(T_{\alpha},T)$

The triangles $T_{\alpha}$ are all equilaterals. Special cases are
$\alpha=\frac{1}{2}$, where $T_{\frac{1}{2}}=-\frac{1}{2}T$ and
$\alpha\in\set{\frac{1}{3},\frac{2}{3}}$, rotations of $T$ by
$\pm\frac{\pi}{6}$.

$p^{1}$$p^{2}$$p^{3}$$q^{1}$$q^{2}$$q^{3}$$w^{1}$$w^{2}$$w^{3}$ Figure 11. All
the equilateral triangles $T_{\alpha}$ are Jung-extremal ( w. r. t. $T$).

###### Proof.

By construction $T_{\alpha}\subset^{\operatorname{opt}}T$. The diameter of
$T_{\alpha}$ is attained between two of its vertices and since
$T_{\operatorname{MAX}}$ is a regular hexagon which has rotational symmetry of
order six, is does not matter which of the edges of $T_{\alpha}$ we consider.
Let $q^{i}=\alpha_{i}p^{i+1}+(1-\alpha)p^{i+2}$. By using $p^{3}=-p^{2}-p^{1}$
we obtain

$\displaystyle q^{2}-q^{1}$ $\displaystyle=\alpha p^{3}+(1-\alpha)p^{1}-\alpha
p^{2}-(1-\alpha)p^{3}$ $\displaystyle=(2-3\alpha)p^{1}+(3\alpha-1)(-p^{2})$
$\displaystyle\subset[p^{1},-p^{2}]\subset\operatorname{bd}\left(T_{\operatorname{MAX}}\right)$

Hence,
$D_{\operatorname{MAX}}(T_{\alpha},T)=\|{q^{2}-q^{1}}\|_{T_{\operatorname{MAX}}}=1$.

The triangle $\operatorname{conv}(\set{w^{1},w^{2},w^{3}})$ with
$w^{i}:=\alpha q^{i+2}+(1-\alpha)q^{i+1}$ is optimally contained in
$T_{\alpha}$ by Proposition 2.2. Furthermore,

$\displaystyle w^{i}$ $\displaystyle=\alpha q^{i+2}+(1-\alpha)q^{i+1}$
$\displaystyle=\alpha\left(\alpha
p^{i}+(1-\alpha)p^{i+1}\right)+(1-\alpha)\left(\alpha
p^{i+2}+(1-\alpha)p^{i}\right)$
$\displaystyle=(1-2\alpha+2\alpha^{2})p^{i}+\alpha(1-\alpha)p^{i+1}+\alpha(1-\alpha)p^{i+2}$

Using this we compute

$\displaystyle w^{i+1}-w^{i}$
$\displaystyle=(1-2\alpha+2\alpha^{2})p^{i+1}+\alpha(1-\alpha)p^{i}-(1-2\alpha+2\alpha^{2})p^{i}-\alpha(1-\alpha)p^{i+1}$
$\displaystyle=(1-3\alpha-3\alpha^{2})(p^{i+1}-p^{i}).$

Thus, $\operatorname{conv}(\set{w^{1},w^{2},w^{3}})$ is a translate of
$(1-3\alpha-3\alpha^{2})T$ and the inradius is
$r(T_{\alpha},T)=1-3\alpha-3\alpha^{2}$. ∎

$T_{\frac{2}{3}}$ is especially interesting: it is complete since its
arithmetic mean is a dilatation of $T_{\operatorname{MAX}}$. Since the
symmetrization $T_{\operatorname{MAX}}$ of $T$ is also Jung-extremal, there
are at least two complete Jung-extremal bodies, $T_{\operatorname{MAX}}$ and
$T_{\frac{2}{3}}$ with different inradius-circumradius ratio.

Now, we compute the values of the functionals for a second family of
triangles. We will see that these lie on the boundary of the diagram w. r. t.
triangles as well.

######  Lemma 5.9.

Let $T$ be the equilateral triangle as defined above and
$S_{\lambda}=\operatorname{conv}(\\{q^{1},q^{2},q^{3}\\})$ with
$q^{1}=\frac{1}{2}(p^{2}+p^{3})$, $q^{2}=\lambda p^{1}+(1-\lambda)p^{3}$ and
$q^{3}=\lambda p^{1}+(1-\lambda)p^{2}$ for some $\lambda\in[\frac{1}{2},1]$.
Then,
$R(S_{\lambda},T)=1,D_{\operatorname{MAX}}(S_{\lambda},T)=\lambda+\frac{1}{2}$
and $r(S_{\lambda},T)=\lambda(1-\lambda)$.

###### Proof.

Obviously, $\norm{q^{2}-q^{3}}_{T_{\operatorname{MAX}}}\leq 1$ and
$\norm{q^{2}-q^{1}}_{T_{\operatorname{MAX}}}=\norm{q^{3}-q^{1}}_{T_{\operatorname{MAX}}}$.
Using $p^{3}=-(p^{1}+p^{2})$ we can compute:

$\begin{split}q^{2}-q^{1}&=\lambda
p^{1}+(1-\lambda)p^{3}-\frac{1}{2}(p^{2}+p^{3})=\lambda
p^{1}+\frac{1}{2}(-p^{2})+(\frac{1}{2}-\lambda)(-p^{1}-p^{2})\\\
&=(2\lambda-\frac{1}{2})p^{1}+(1-\lambda)(-p^{2})\in(\lambda+\frac{1}{2})\operatorname{conv}(\\{p^{1},-p^{2}\\})\\\
&\subset(\lambda+\frac{1}{2})\operatorname{bd}(T_{\operatorname{MAX}}).\end{split}$

Since $\lambda+\frac{1}{2}\geq 1$ for $\lambda\in[\frac{1}{2},1]$, it follows
$D_{\operatorname{MAX}}(S_{\lambda},T)=\lambda+\frac{1}{2}$.

Now we show the formula for the inradius ( cf. Figure 12). Let
$\operatorname{conv}(\set{w^{1},w^{2},w^{3}})$ be the inner triangle. By axial
symmetry of $T$ and $S_{\lambda}$ we know that
$w^{1}=\frac{1}{2}(q^{2}+q^{3})$. Denote the euclidean edge length of the
inball by $a$. Since $T$ has edge length $\sqrt{3}$, we obtain
$r(S_{\lambda},T)=\frac{a}{\sqrt{3}}$. The segments $[q^{3},q^{2}]$ and
$[p^{2},p^{3}]$ are parallel and the corresponding edges of the inner and
outer triangle are parallel as well. Thus, the triangles
$\operatorname{conv}(\set{w^{1},w^{3},q^{2}})$ and
$\operatorname{conv}(\set{p^{3},q^{2},q^{1}})$ are similar, implying
$\frac{a}{\norm{q^{2}-p^{3}}_{2}}=\frac{q^{2}_{1}}{p^{3}_{1}}$. We obtain

$a=\frac{q^{2}_{1}}{p^{3}_{1}}\cdot\norm{q^{2}-p^{3}}_{2}=(1-\lambda)\norm{q^{2}-p^{3}}_{2}=(1-\lambda)\lambda\norm{p^{1}-p^{3}}_{2}=\sqrt{3}\lambda(1-\lambda),$

and therefore $r(S_{\lambda},T)=\frac{a}{\sqrt{3}}=\lambda(1-\lambda)$. ∎

$p^{1}$$p^{2}$$p^{3}$$q^{1}$$q^{2}$$q^{3}$$w_{1}$$w^{2}$$w^{3}$$a$ Figure 12.
Calculating the inradius of $S_{\lambda}$.

$S_{\frac{1}{2}}=-\frac{1}{2}T$ is Jung-extremal and $S_{1}$ is the segment
$L_{w}$.

######  Theorem 5.10.

Let $K\in\mathcal{C}^{n}$ and $T$ an equilateral Minkowski-centered triangle.
Then

$\left(\frac{D_{\operatorname{MAX}}(K,T)}{R(K,T)}-\frac{1}{2}\right)\left(\frac{3}{2}-\frac{D_{\operatorname{MAX}}(K,T)}{R(K,T)}\right)\leq\frac{r(K,T)}{R(K,T)}$

with equality for the triangles $S_{\lambda}$, $\lambda\in[\frac{1}{2},1]$, as
described in Lemma 5.9.

To prepare the proof of Theorem 5.10 we need the following lemma.

######  Lemma 5.11.

Let $K=\operatorname{conv}\\{q^{1},q^{2},q^{3}\\}$ be a triangle that is
optimally contained in $T$, s. t. $q^{i}$ belongs to the edge of $T$ opposing
$p^{i}$. We say that $(q^{1},q^{3})$ is _steep_ if
$\norm{q^{3}-p^{1}}_{2}\leq\norm{q^{1}-p^{3}}_{2}$. Let
$\tilde{K}:=\operatorname{conv}(\\{q^{1},q^{3},\tilde{q}^{2}\\})\subset^{\operatorname{opt}}T$
be a triangle such that $\tilde{q}^{2}$ is on the same edge of $T$ as $q^{2}$
and $\norm{q^{2}-p^{1}}_{2}\leq\norm{\tilde{q}^{2}-p^{1}}_{2}$. If
$(q^{1},q^{3})$ is steep, the inradius of $\tilde{K}$ w. r. t. $T$ is greater
or equal than the one of $K$.

We call this property ”steep” since the segment $[q^{1},q^{3}]$ is steeper
than $[p^{3},p^{1}]$. By symmetry of $T$ can generalize this result to all
choices $i,j\in\set{1,2,3}$: $(q^{i},q^{j})$, $i,j\in\\{1,2,3\\}$, $i\neq j$
is steep if $\norm{q^{j}-p^{i}}_{2}\leq\norm{q^{i}-p^{j}}_{2}$. At least one
of the ordered pairs $(q^{i},q^{j})$ or $(q^{j},q^{i})$ is always steep.

###### Proof of Lemma 5.11.

Let $H_{1}$ and $H_{2}$ be the two lines parallel to $[p^{1},p^{3}]$
supporting the inball $\operatorname{conv}(\set{w^{1},w^{2},w^{3}})$ of $K$
(that necessarily touches all three edges of $K$), s.t. $H_{1}$ contains the
edge $[w^{1},w^{3}]$ and $H_{2}$ the opposing vertex $w^{2}$ ( cf. Figure 13).
Since $(q^{1},q^{3})$ is steep the part of $H_{2}$ below the touching point
intersects $\tilde{K}$. By the intercept theorem with the two parallel lines
$H_{1}$ and $H_{2}$ and the points $q^{2}$ or $\tilde{q}^{2}$ the segment of
$H_{1}$ contained in $\tilde{K}$ is greater or equal than the one contained in
$K$. Thus, we can move the inball of $K$ within the slab between $H_{1}$ and
$H_{2}$ until it touches $[q^{1},\tilde{q}^{2}]$. The resulting translations
of $w^{1},w^{2},w^{3}$ are all contained in $\tilde{K}$, the translation of
$w^{2}$ due to the steepness of $(q^{1},q^{3})$. Thus, $r(K,T)\leq
r(\tilde{K},T)$. ∎

$p^{1}$$p^{2}$$p^{3}$$q^{1}$$q^{2}$$q^{3}$$\tilde{q}^{2}$$w^{1}$$w^{2}$$w^{3}$$H_{1}$$H_{2}$
Figure 13. Proof of Lemma 5.11. Since $(q^{1},q^{3})$ is steep, the inner
triangle can be translated along the orange hyperplanes.

###### Proof of Theorem 5.10.

The equality case follows directly from Lemma 5.9. So we only need to prove
the correctness of the inequality.

Let $K\subset^{\operatorname{opt}}T$. As shown in the proof of Theorem 5.5, we
either have $D_{\operatorname{MAX}}(K,T)\geq\frac{3}{2}$ or three touching
points of $K$ to the boundary of $T$, each situated on a different edge of
$T$.

In the first case, the left side of the inequality in Theorem 5.10 is non-
positive while the right side is always non-negative. Hence, in this case the
inequality is fulfilled.

In the other case we consider the triangle
$S:=\operatorname{conv}(\\{q^{1},q^{2},q^{3}\\})$, where $q^{i}\in K$ belongs
to the edge of $T$ opposing $p^{i}$, $i=1,2,3$. Assume w. l. o. g. that the
diameter of $S$ is attained between $q^{1}$ and one of the other points and
that $\norm{q^{1}-p^{3}}_{2}\leq\norm{q^{1}-p^{2}}_{2}$. Our goal is to show
that there exists a triangle $S_{\lambda}\subset^{\operatorname{opt}}T$,
$\lambda\in[\frac{1}{2},1]$, as defined in Lemma 5.9 with at most the same
diameter and inradius as $S$. Using the fact that
$\left(D_{\operatorname{MAX}}-\frac{1}{2}\right)\left(\frac{3}{2}-D_{\operatorname{MAX}}\right)$
is decreasing in $D_{\operatorname{MAX}}$, we may then conclude

$\begin{split}\left(\frac{D_{\operatorname{MAX}}(K,T)}{R(K,T)}-\frac{1}{2}\right)\left(\frac{3}{2}-\frac{D_{\operatorname{MAX}}(K,T)}{R(K,T)}\right)&\leq\left(\frac{D_{\operatorname{MAX}}(S,T)}{R(S,T)}-\frac{1}{2}\right)\left(\frac{3}{2}-\frac{D_{\operatorname{MAX}}(S,T)}{R(S,T)}\right)\\\
&\leq\left(\frac{D_{\operatorname{MAX}}(S_{\lambda},T)}{R(S_{\lambda},T)}-\frac{1}{2}\right)\left(\frac{3}{2}-\frac{D_{\operatorname{MAX}}(S_{\lambda},T)}{R(S_{\lambda},T)}\right)\\\
&=\frac{r(S_{\lambda},T)}{R(S_{\lambda},T)}\leq\frac{r(S,T)}{R(S,T)}\leq\frac{r(K,T)}{R(K,T)}.\end{split}$

We distinguish between the two cases if the diameter is attained by
$[q^{1},q^{2}]$ or $[q^{1},q^{3}]$ and show that we may always assume that
both segments are diametral.

  1. Case 1

This case is depicted in Figure 14. If
$D_{\operatorname{MAX}}(S,T)=D_{\operatorname{MAX}}([q^{1},q^{2}],T)$ our
assumption $\norm{q^{1}-p^{3}}_{2}\leq\norm{q^{1}-p^{2}}_{2}$ implies that
$\norm{q^{2}-p^{1}}_{2}\leq\norm{q^{2}-p^{3}}_{2}$. Let us assume otherwise.
Then we would have
$[q^{1},q^{2}]\subset\operatorname{conv}(\set{p^{3},\frac{1}{2}(p^{2}+p^{3}),\frac{1}{2}(p^{1}+p^{3})}=\frac{1}{2}(p^{3}+T)$
with $[q^{1},q^{2}]$ not being an edge of $\frac{1}{2}(T+p^{3})$, which
implies
$D_{\operatorname{MAX}}([q^{1},q^{2}],T)=2R([q^{1},q^{2}],T_{\operatorname{MAX}})<2R(\frac{1}{2}T,T_{\operatorname{MAX}})=1=R(K,T)$,
contradicting Theorem 5.5.

Hence, $\norm{q^{2}-p^{1}}_{2}\leq\sqrt{3}\leq\norm{q^{1}-p^{2}}_{2}$, which
implies the steepness of $(q^{1},q^{2})$. Since $[q^{1},q^{2}]$ is diametral,
$q^{3}$ lies inside $q^{1}+D_{\operatorname{MAX}}(S,T)T_{\operatorname{MAX}}$.
However, $\norm{q^{1}-p^{3}}_{2}\leq\norm{q^{1}-p^{2}}_{2}$ now implies the
existence of an intersection point between $[p^{1},p^{2}]$ and the boundary of
$q^{1}+D_{\operatorname{MAX}}(S,T)T_{\operatorname{MAX}}$ which is not further
from $p^{1}$ than $q^{3}$. Choosing this point as our new $q^{3}$ neither
increases the diameter nor the inradius (the latter because of Lemma 5.11).
Doing so, $[q^{1},q^{3}]$ becomes diametral, too.

$p^{1}$$p^{2}$$p^{3}$$q^{2}$$q^{1}$$q^{3}_{new}$${q}^{3}$$q^{1}+D_{\operatorname{MAX}}(S,T)T_{\operatorname{MAX}}$
Figure 14. Proof of Case 1 of Theorem 5.10. The diameter is attained between
$q^{1}$ and $q^{2}$. We can replace $q^{3}$ such that it is attained between
$q^{1}$ and $q^{3}$ as well since $(q^{1},q^{2})$ is steep.

  2. Case 2

If $D_{\operatorname{MAX}}(S,T)=D_{\operatorname{MAX}}([q^{1},q^{3}],T)$, we
need to consider three subcases:

    1. a)

If $\norm{q^{3}-p^{2}}_{2}\leq\norm{q^{3}-p^{1}}_{2}$ this corresponds to Case
1 with $q^{3}$ in the role of $q^{1}$ and $q^{2}$ being the vertex that is
moved.

    2. b)

If $\norm{q^{3}-p^{2}}_{2}\geq\norm{q^{3}-p^{1}}_{2}$ and $(q^{1},q^{3})$ is
steep one can move $q^{2}$ in the direction of $p^{1}$ such that
$[q^{1},q^{2}]$ becomes diametral, too.

    3. c)

If $\norm{q^{3}-p^{2}}_{2}\geq\norm{q^{3}-p^{1}}_{2}$ and $(q^{3},q^{1})$ is
steep this corresponds to Case 2b) with roles of $q^{1}and$$q^{3}$
interchanged, which means that we may move $q^{2}$ towards $p^{3}$.

Altogether we see that assuming
$\norm{q^{2}-q^{1}}_{T_{\operatorname{MAX}}}=\norm{q^{3}-q^{1}}_{T_{\operatorname{MAX}}}$
is possible and doing so the points $q^{2}$ and $q^{3}$ do not only lie on the
boundary of $q^{1}+D_{\operatorname{MAX}}(S,T)T_{\operatorname{MAX}}$, they
essentialy lie on the (translated and dilatated) edges $[p^{1},-p^{3}]$ or
$[p^{1},-p^{2}]$ of $T_{\operatorname{MAX}}$. For $q^{2}$ this follows from
the fact that it has to lie closer to $p^{1}$ than $p^{3}$. As described in
the proof of Case 1, the boundary of
$q^{1}+D_{\operatorname{MAX}}(S,T)T_{\operatorname{MAX}}$ intersects
$[p^{1},p^{2}]$ once or twice, but it is not possible that it only intersects
with the segment $q^{1}+D_{\operatorname{MAX}}(S,T)[p^{2},-p^{3}]$ as this
would contradict our assumption
$D_{\operatorname{MAX}}(S,T)=D_{\operatorname{MAX}}([q^{1},q^{2}],T)$. An
analogous argument works for $q^{3}$ being situated on the corresponding edge
of the homothet of $T_{\operatorname{MAX}}$.

For the next part of the proof we now assume that $q^{3}\in
q^{1}+D_{\operatorname{MAX}}(S,T)[p^{1},-p^{3}]$ and $q^{2}\in
q^{1}+D_{\operatorname{MAX}}(S,T)[p^{1},-p^{2}]$ is true. This is not the case
if $q^{1}_{1}>q^{2}_{1}$ (see Figure 15). But then, we know
$q^{3}_{2}<q^{2}_{2}$ and $(q^{2},q^{3})$ is steep. Thus, replacing $q^{1}$ by
$\tilde{q}^{1}$ such that $\tilde{q}^{1}_{1}=q^{2}_{1}$ does not increase the
diameter or the inradius and we still have
$\|{q^{3}-q^{1}}\|_{T_{\operatorname{MAX}}}=\|{q^{2}-q^{1}}\|_{T_{\operatorname{MAX}}}$.
This shows that we can assume $q^{1}_{1}\leq q^{2}_{1}$ and that $q^{2}\in
q^{1}+D_{\operatorname{MAX}}(S,T)[-p^{2},p^{1}]$.

$p^{1}$$p^{2}$$p^{3}$$\tilde{q}^{1}$$q^{2}$$q^{1}$$q^{3}$$q^{1}+D_{\operatorname{MAX}}(S,T)T_{\operatorname{MAX}}$$\tilde{q}^{1}+D_{\operatorname{MAX}}(\tilde{S},T)T_{\operatorname{MAX}}$
Figure 15. Proof of Theorem 5.10. We can assume that $q^{3}$ lies on
$q^{1}+D_{\operatorname{MAX}}(S,T)[p^{1},-p^{3}]$ and $q^{2}$ lies on
$q^{1}+D_{\operatorname{MAX}}(S,T)[p^{1},-p^{2}]$. Otherwise we can consider
the triangle $\tilde{S}=\operatorname{conv}(\set{\tilde{q}^{1},q^{2},q^{3}})$
which has smaller or equal inradius and diameter.

Now, we consider the triangle
$S_{\lambda}=\operatorname{conv}(\\{\tilde{q}^{1},\tilde{q}^{2},\tilde{q}^{3}\\})$
with the same diameter and $\lambda$ as in Lemma 5.9. Due to symmetry reasons
and our assumptions about the positions of $q^{2}$ and $q^{3}$,
$\norm{q^{1}-\tilde{q}^{1}}_{2}=\norm{q^{2}-\tilde{q}^{2}}_{2}=\norm{q^{3}-\tilde{q}^{3}}_{2}=:\kappa$
(see Figure 16).

$p^{1}$$p^{2}$$p^{3}$$q^{2}$$\tilde{q}^{1}$$q^{1}$$q^{3}$$\tilde{q}^{2}$$\tilde{q}^{3}$$\kappa$$\kappa$$\kappa$
Figure 16. Proof of Theorem 5.10. Transformation of $S$ into $S_{\lambda}$
with the same diameter. The distances $\|{q^{i}-\tilde{q}^{i}}\|_{2}$ are
equal.

By using the intercept theorem and the law of sines we show that, possibly
after a suitable translation, all vertices of the inball of $S_{\lambda}$ are
contained in $S$. This way we see that the inradius of $S_{\lambda}$ is at
most the one of $S$. We denote the vertices of the inball of $S_{\lambda}$ by
$w^{i}$ (see Figure 18) and the euclidean distance in the horizontal direction
of $w^{i}$ to the segment $[q^{j},q^{k}]$, $\set{i,j,k}=\set{1,2,3}$, by
$l_{i}$. If $q\neq 0$, every side of $S$ intersects the corresponding side of
$S_{\lambda}$ exactly once. Let us denote these intersection points by $v^{i}$
( cf. Figure 17). We will show that if we shift the inball of $S_{\lambda}$ by
$l_{3}$ it is completely contained in $S$. To do so we compute all the values
$l_{i}$, $i=1,2,3$, in dependence of $\lambda$ and prove that we have
$l_{3}\leq l_{j}$, $j=1,2$.

$p^{1}$$\tilde{q}^{2}$$\tilde{q}^{3}$$q^{2}$$q^{3}$$w^{1}$$v^{1}$$l_{1}$$r$
Figure 17. Proof of Theorem 5.10. Computation of $l_{1}$. The triangles
defined by $p^{1}$ and the intersection points of the parallel lines are all
three equilateral.

Computation of $l_{1}$ ( cf. Figure 17): We use the intercept theorem for
$[\tilde{q}^{2},\tilde{q}^{3}]$ and the two lines parallel to this segment
through $q^{2}$ and $q^{3}$, respectively. Since $T$ is an equilateral
triangle, $\operatorname{conv}(\set{p^{1},\tilde{q}^{3},\tilde{q}^{2}})$ is
also equilateral with an edge length of $(1-\lambda)\sqrt{3}$. Furthermore,

$\|{\tilde{q}^{3}-v^{1}}\|_{2}=\frac{1}{2}\|{r-q^{2}}\|_{2}=\frac{1}{2}\|{p^{1}-q^{2}}\|_{2}=\frac{1}{2}(\|{p^{1}-\tilde{q}^{2}}\|_{2}-\kappa).$

It follows

$l_{1}=\frac{1}{2}\|{p^{1}-\tilde{q}^{2}}\|_{2}-\|{\tilde{q}^{3}-v^{1}}\|_{2}=\frac{1}{2}\kappa$

and that $w^{1}$ is always contained in $S$.

Computation of $l_{2}$ ( cf. Figure 18): Let $\alpha_{2}=\angle
v^{2}q^{3}\tilde{q}^{3}$, $\beta_{2}=\angle v^{2}q^{1}p^{2}$ and
$\gamma_{2}=\angle\tilde{q}^{1}v^{2}q^{1}$. We compute
$\frac{\|{\tilde{q}^{3}-v^{2}}\|_{2}}{\|{\tilde{q}^{1}-v^{2}}\|_{2}}$ using
the law of sines, first for the triangles
$\operatorname{conv}(\set{v^{2},q^{1},\tilde{q}^{1}})$ and
$\operatorname{conv}(\set{v^{2},q^{3},\tilde{q}^{3}})$ and then for
$\operatorname{conv}(\set{p^{2},q^{1},q^{3}})$.

$\begin{split}\frac{\|{\tilde{q}^{3}-v^{2}}\|_{2}}{\|{\tilde{q}^{1}-v^{2}}\|_{2}}&=\frac{\sin{(\gamma_{2})}}{\kappa\sin{(\beta_{2})}}\cdot\frac{\kappa\sin{(\alpha_{2})}}{\sin{(\gamma_{2})}}=\frac{\sin{(\alpha_{2})}}{\sin{(\beta_{2})}}\\\
&=\frac{\sin{(\pi-\alpha_{2})}}{\sin{(\beta_{2})}}=\frac{\frac{\sqrt{3}}{2}+\kappa}{\sqrt{3}\lambda-\kappa}\end{split}$

Furthermore, we know from Lemma 5.9 that
$r(S_{\lambda},T)=\lambda(1-\lambda)$. Together with the intercept theorem we
obtain

$\displaystyle\frac{\|{\tilde{q}^{3}-w^{2}}\|_{2}}{\|{\tilde{q}^{1}-w^{2}}\|_{2}}=\frac{w^{1}_{2}-w^{2}_{2}}{(\tilde{q}^{3}_{2}-\tilde{q}^{1}_{2})-(w^{1}_{2}-w^{2}_{2})}=\frac{1-\lambda}{\lambda}.$

Since
$\frac{\frac{\sqrt{3}}{2}+\kappa}{\sqrt{3}\lambda-\kappa}\geq\frac{1-\lambda}{\lambda}$
for $\lambda\in[\frac{1}{2},1]$, $v^{2}$ is closer to $\tilde{q}^{1}$ than
$w^{2}$ and therefore also $w^{2}\in S$.

We compute the distance of $w^{2}$ to $[q^{1},q^{3}]$ in the direction of
$(-1,0)^{T}$. By the intercept theorem

$\begin{split}l_{2}&=\kappa\cdot\frac{\|{v^{2}-w^{2}}\|_{2}}{\|{\tilde{q}^{1}-v^{2}}\|_{2}}=q\cdot\frac{\|{\tilde{q}^{3}-v^{2}}\|_{2}-\|{\tilde{q}^{3}-w^{2}}\|_{2}}{\|{\tilde{q}^{1}-v^{2}}\|_{2}}\\\
&=\kappa\cdot\left(\frac{\frac{\sqrt{3}}{2}+\kappa}{\sqrt{3}\lambda-\kappa}-\frac{1-\lambda}{\lambda}\right).\end{split}$

Computation of $l_{3}$ ( cf. Figure 18): If $w^{3}$ is also contained in $S$,
we have that the complete inball of $S_{\lambda}$ is contained in $S$ and we
are done.

Otherwise we need to show that we can translate the inball to be in $S$. To do
so, we compute the distance $l_{3}$ of $w^{3}$ to $[q^{1},q^{2}]$ in the
direction of $(-1,0)^{T}$, which can be done completely analogously to
$l_{2}$: Let $\alpha_{3}=\angle q^{1}q^{2}p^{3}$, $\beta_{3}=\angle
v^{3}q^{1}p^{2}$ and $\gamma_{3}=\angle\tilde{q}^{1}v^{3}q^{1}$. Then

$\frac{\|{\tilde{q}^{2}-v^{3}}\|_{2}}{\|{\tilde{q}^{1}-v^{3}}\|_{2}}=\frac{\sin{(\gamma_{3})}}{\kappa\sin{(\beta_{3})}}\cdot\frac{\kappa\sin{(\alpha_{3})}}{\sin{(\gamma_{3})}}=\frac{\sin{(\alpha_{3})}}{\sin{(\pi-\beta_{3})}}=\frac{\frac{\sqrt{3}}{2}-\kappa}{\sqrt{3}\lambda+\kappa}.$

and
$\frac{\|{\tilde{q}^{2}-w^{3}}\|_{2}}{\|{\tilde{q}^{1}-w^{3}}\|_{2}}=\frac{1-\lambda}{\lambda}$.
Thus,

$l_{3}=\kappa\cdot\frac{\|{v^{3}-w^{3}}\|_{2}}{\|{\tilde{q}^{1}-v^{3}}\|_{2}}=\kappa\cdot\left(\frac{\frac{\sqrt{3}}{2}-\kappa}{\sqrt{3}\lambda+\kappa}-\frac{1-\lambda}{\lambda}\right)$

and we easily see that $l_{3}\leq l_{2}$. Moreover, since $q\geq 0$ and
$\lambda\in[\frac{1}{2},1]$ it follows

$\frac{\frac{\sqrt{3}}{2}-\kappa}{\sqrt{3}\lambda+\kappa}-\frac{1-\lambda}{\lambda}\leq\frac{1}{2\lambda}-\frac{1-\lambda}{\lambda}=1-\frac{1}{2\lambda}\leq\frac{1}{2}.$

which proves $l_{3}\leq l_{1}$. Hence, if we translate the inball of
$S_{\lambda}$ by $(l_{3},0)^{T}$, it is contained in $S$, which proves
$r(S_{\lambda},T)\leq r(S,T)$.

$p^{2}$$p^{3}$$\tilde{q}^{1}$$\tilde{q}^{2}$$\tilde{q}^{3}$$q^{1}$$q^{2}$$q^{3}$$w^{1}$$w^{2}$$w^{3}$$l_{3}$$l_{2}$$v^{3}$$v^{2}$$\gamma_{2}$$\beta_{2}$$\alpha_{2}$$\pi-\alpha_{2}$$\gamma_{3}$$\beta_{3}$$\pi-\beta_{3}$$\alpha_{3}$$p^{2}$$p^{3}$$\tilde{q}^{1}$$\tilde{q}^{2}$$\tilde{q}^{3}$$q^{1}$$q^{2}$$q^{3}$$w^{1}$$w^{2}$$w^{3}$$l_{3}$$l_{2}$$v^{3}$$v^{2}$$\gamma_{2}$$\beta_{2}$$\alpha_{2}$$\pi-\alpha_{2}$$\gamma_{3}$$\beta_{3}$$\pi-\beta_{3}$$\alpha_{3}$

Figure 18. Proof of Theorem 5.10. Computation of $l_{2}$ and $l_{3}$.

∎

The collected inequalities are now sufficient to provide a full description of
the Blaschke-Santaló-diagram $f_{\operatorname{MAX}}(C^{2},T)$ ( cf. Figure
19).

######  Theorem 5.12.

For every Minkowski-centered triangle $S$ the diagram
$f_{\operatorname{MAX}}(\mathcal{C}^{2},S)$ is fully described by the
inequalities

$\displaystyle\frac{D_{\operatorname{MAX}}(K,S)}{2}$ $\displaystyle\leq
R(K,S)$ $\displaystyle r(K,S)$
$\displaystyle\leq\frac{D_{\operatorname{MAX}}(K,S)}{2}$ $\displaystyle 0$
$\displaystyle\leq r(K,S)$ $\displaystyle R(K,S)$ $\displaystyle\leq
D_{\operatorname{MAX}}(K,S)$
$\displaystyle\left(\frac{D_{\operatorname{MAX}}(K,S)}{R(K,S)}-\frac{1}{2}\right)$
$\displaystyle\left(\frac{3}{2}-\frac{D_{\operatorname{MAX}}(K,S)}{R(K,S)}\right)\leq\frac{r(K,S)}{R(K,S)}.$

###### Proof.

Since all Minkowski-centered triangles can be linearly transformed into the
equilateral triangle $T$ it suffices to show the claim for $T$. We give a
continuous description of the boundaries described by the inequalities (2),
(3), and (5), as well as those given by Theorem 5.5 and Theorem 5.10. First,
(2) is attained with equality for $(1-\lambda)L_{D}+\lambda T$,
$\lambda\in[0,1]$ where $L_{D}$ and $T$ are the extreme cases. Second, (3) is
attained with equality for $(1-\lambda)T+\lambda T_{\operatorname{MAX}}$,
$\lambda\in[0,1]$ by Lemma 4.2 where $T$ and $T_{\operatorname{MAX}}$ are the
extreme cases. The boundary induced by (5) is filled by segments from $L_{w}$
to $L_{D}$. Next, since $T_{\operatorname{MAX}}$ ist the completion of $-T$,
the boundary from Theorem 5.5 is filled by the sets $-T_{+}$ with $-T\subset-
T_{+}\subset T_{\operatorname{MAX}}$. Finally, the inequality in Theorem 5.10
is fulfilled with equality by the triangles $S_{\lambda}$ as introduced in
Lemma 5.9. Since we have presented a continuous description of the boundary we
can apply Theorem 4.19 and follow that the diagram is simply connected. ∎

While the inequalities (2), (3), (5) and the one given by Theorem 5.5 are
valid for all choices of planar, Minkowski-centered gauges, the inequality
from Theorem 5.10 is only proven for triangles. Using the result from
Proposition 4.18 and
$D_{\operatorname{AM}}(K,C)\leq\frac{4}{3}D_{\operatorname{MAX}}(K,C)$ which
follows from
$\frac{2s(C)}{s(C)+1}C_{\operatorname{MAX}}\subset^{\operatorname{opt}}C_{\operatorname{AM}}$
[7], we are able to give another general inequality

(8)
$\frac{2}{3}\frac{D_{\operatorname{MAX}}(K,C)}{R(K,C)}\left(1-\frac{2}{3}\frac{D_{\operatorname{MAX}}(K,C)}{R(K,C)}\right)\leq\frac{r(K,C)}{R(K,C)},$

which enables us to give a bound for the union of the diagrams
$f(\mathcal{C}^{2},C)$ with $C$ Minkowski-centered (depicted in red within
Figure 19).

######  Conjecture 5.13.

The diagram of the triangle is dominating, i.e. for every Minkowski-centered
$C$ we have $f_{\operatorname{MAX}}(\mathcal{C}^{2},C)\subset
f_{\operatorname{MAX}}(\mathcal{C}^{2},S)$.

$\frac{r}{R}$$\frac{D_{\operatorname{MAX}}}{2R}$.$\frac{1}{2}$.$1$.$\frac{1}{4}$.$\frac{1}{2}$.$1$.$\frac{3}{4}$.$S$.$L_{D}$.$L_{w}$.$S_{\operatorname{MAX}}$.$-S$
Figure 19. The diagram $f_{\operatorname{MAX}}(\mathcal{C}^{2},S)$ w. r. t. a
Minkowski-centered triangle $S$ (black) and an upper bound for the union over
all Minkowski-centered gauges bounded by (8) (additional purple region).

##  6\. The diameter $D_{\operatorname{HM}}$

In the case of the harmonic mean, the factors $\rho_{\operatorname{HM}}$ and
$\delta_{\operatorname{HM}}$ are not bound to the Minkowski asymmetry $s(C)$.
However, in [7] the following bounds are proven:

######  Proposition 6.1.

$1\leq\rho_{\operatorname{HM}}\leq\frac{(s(C)+1)^{2}}{4s(C)}\leq\delta_{\operatorname{HM}}\leq\frac{s(C)+1}{2}.$

Furthermore, the inequalities from Lemma 4.16 have the form:

(9) $\displaystyle\delta_{\operatorname{HM}}r(K,C)$
$\displaystyle\leq\frac{D_{\operatorname{HM}}(K,C)}{2}$ and (10)
$\displaystyle s(C)r(K,C)+R(K,C)$
$\displaystyle\leq\frac{s(C)+1}{2\rho_{\operatorname{HM}}}D_{\operatorname{HM}}(K,C)\leq\frac{s(C)+1}{2}D_{\operatorname{HM}}(K,C)$
and (11) $\displaystyle r(K,C)+R(K,C)$
$\displaystyle\leq\frac{2s(C)}{s(C)+1}D_{\operatorname{HM}}(K,C)$

Taking $K=C_{\operatorname{HM}}$, we know
$r(C_{\operatorname{HM}},C)=\frac{2}{s(C)+1}$,
$R(C_{\operatorname{HM}},C)=\frac{2s(C)}{s(C)+1}$, and
$D_{\operatorname{HM}}(C_{\operatorname{HM}},C)=2$. Thus, if
$\delta_{\operatorname{HM}}=\frac{s(C)+1}{2}$ then $C_{\operatorname{HM}}$
fulfills (9) with equality and if
$\rho_{\operatorname{HM}}=\frac{(s(C)+1)^{2}}{4s(C)}$ then
$C_{\operatorname{HM}}$ fulfills the first inequality in (10) with equality,
which in this case can be rewritten as

(12) $s(C)r(K,C)+R(K,C)\leq\frac{2s(C)}{s(C)+1}D_{\operatorname{HM}}(K,C).$

Unlike with $D_{\operatorname{MAX}}$ (a dilatate of) the symmetrization
$C_{\operatorname{HM}}$ is not always a completion of the gauge.

######  Lemma 6.2.

Let $C\in\mathcal{C}^{n}_{0}$ be Minkowski-centered. The following are
equivalent:

  1. i)

$\delta_{\operatorname{HM}}=\frac{s(C)+1}{2}$,

  2. ii)

$\frac{s(C)+1}{2}C_{\operatorname{HM}}$ is a completion of $C$ w. r. t. $C$,
and

  3. iii)

$D(C,C_{\operatorname{HM}})=2R(C,C_{\operatorname{HM}})$.

###### Proof.

We know from [7] that
$C\subset^{\operatorname{opt}}\frac{s(C)+1}{2}C_{\operatorname{HM}}$ and
therefore $\frac{s(C)+1}{2}C_{\operatorname{HM}}$ is a complete set containing
$C$ with $R(C,C_{\operatorname{HM}})=\frac{s(C)+1}{2}$.

  1. $i)\Rightarrow ii)$:

If $\delta_{\operatorname{HM}}=\frac{s(C)+1}{2}$ then
$D_{\operatorname{HM}}\left(\frac{s(C)+1}{2}C_{\operatorname{HM}},C\right)=s(C)+1=2\delta_{\operatorname{HM}}=D_{\operatorname{HM}}(C,C)$,
implying that $\frac{s(C)+1}{2}C_{\operatorname{HM}}$ is a completion of $C$.

  2. $ii)\Rightarrow iii)$:

If $\frac{s(C)+1}{2}C_{\operatorname{HM}}$ is a completion of $C$,
$D(C,C_{\operatorname{HM}})=D_{\operatorname{HM}}(C,C)=s(C)+1=2R(C,C_{\operatorname{HM}})$.

  3. $iii)\Rightarrow i)$:

Assuming (iii) it follows
$2\delta_{\operatorname{HM}}=2R(C_{\operatorname{AM}},C_{\operatorname{HM}})=D(C,C_{\operatorname{HM}})=2R(C,C_{\operatorname{HM}})=s(C)+1$.

∎

If $\frac{s(C)+1}{2}C_{\operatorname{HM}}$ is a completion of the gauge $C$,
then it is also a completion of $-C$ and since
$R(-C,C)=s(C)=R(\frac{s(C)+1}{2}C_{\operatorname{HM}},C)$ for $-C$ it is even
a Scott-completion.

######  Example 6.3.

The Reuleaux triangle $\operatorname{RT}$ is the completion of the equilateral
triangle $T$ in the euclidean case:

$\operatorname{RT}:=\bigcap_{i=1}^{3}p^{i}+\sqrt{3}\mathds{B}^{2}_{2},$

where the $p_{i}$ are the vertices of $T$. For the Reuleaux triangle one
obtains $s(\operatorname{RT})=\frac{1}{\sqrt{3}-1}\approx 1.366$,
$\delta_{\operatorname{HM}}=\frac{\sqrt{3}}{\sqrt{11}-\sqrt{3}}\approx
1.093<\frac{s(\operatorname{RT})+1}{2}$ and
$\rho_{\operatorname{HM}}=\frac{(s(\operatorname{RT})+1)^{2}}{4s(\operatorname{RT})}=\frac{3(\sqrt{3}+1)}{8}\approx
1.025$. Thus, by Lemma 6.2 this is a case where the (dilatated) harmonic mean
$\operatorname{RT}_{\operatorname{HM}}$ is not a completion of the gauge
$\operatorname{RT}$.

Since $T$ is diametric we obtain from Lemma 4.15 that the unique completion of
$\operatorname{RT}$ is

$\operatorname{RT}^{*}:=\bigcap_{i=1}^{3}p_{i}+2\delta_{\operatorname{HM}}\operatorname{RT}_{\operatorname{HM}}\quad\text{(
cf.\leavevmode\nobreak\ \lx@cref{creftypecap~refnum}{reultrianglehm})}.$

Figure 20. Reuleaux Triangle $\operatorname{RT}$ (dashed), its completion
$\operatorname{RT}^{*}$ (left) and its harmonic symmetrization
$\operatorname{RT}_{\operatorname{HM}}$ (right).

Using the symmetries of the Reuleaux triangle we obtain
$R(\operatorname{RT}^{*},\operatorname{RT})=(s(\operatorname{RT})+1)\cdot\frac{\delta_{\operatorname{HM}}}{\rho_{\operatorname{HM}}}-s(\operatorname{RT})$
and thus $\operatorname{RT}^{*}$ fulfills (12) with equality, which is also
true for $\operatorname{RT}_{\operatorname{HM}}$.

The diagram $f_{\operatorname{HM}}(\mathcal{C}^{2},\operatorname{RT})$ is
given in Figure 21. In addition to known inequalities and points, two
conjectured inequalities are shown.

$\frac{r}{R}$$\frac{D_{\operatorname{HM}}}{2R}$.$\operatorname{RT}$.$L_{D}$.$L_{w}$.$\operatorname{RT}^{*}$$\operatorname{RT}_{\operatorname{HM}}$.$-T$.$-\operatorname{RT}$.$-\operatorname{RT}^{*}$.$\frac{\delta_{\operatorname{HM}}}{s(\operatorname{RT})}$.$1$.$\frac{1}{2s(\operatorname{RT})}\quad$.$\frac{1}{s(\operatorname{RT})}$.$\frac{1}{s(\operatorname{RT})^{2}}$.$\delta_{\operatorname{HM}}$.$\rho_{\operatorname{HM}}$
Figure 21. The Blaschke-Santaló diagram
$f_{\operatorname{HM}}(\mathcal{C}^{2},\operatorname{RT})$ with conjectured
inequalities (purple). Keep in mind that the (dilatated) harmonic
symmetrization of the Reuleaux triangle is not its completion.

For the equilateral triangle we have
$\delta_{\operatorname{HM}}=\frac{s(T)+1}{2}=\frac{3}{2}$,
$\rho_{\operatorname{HM}}=\frac{(s(T)+1)^{2}}{4s(T)}=\frac{9}{8}$ and
$T_{\operatorname{HM}}=\frac{2}{3}T_{\operatorname{MAX}}$. Thus,
$D_{\operatorname{HM}}(K,T)=\frac{3}{2}D_{\operatorname{MAX}}(K,T)$ for all
$K\in\mathcal{C}^{2}$ and we can transfer all inequalities concerning
triangles from the previous section.

######  Corollary 6.4.

For every Minkowski-centered triangle $S$ the diagram
$f_{\operatorname{HM}}(\mathcal{C}^{2},S)$ is fully described by the
inequalities

$\displaystyle D_{\operatorname{HM}}(K,S)$ $\displaystyle\leq 3R(K,S)$
$\displaystyle 3r(K,S)$ $\displaystyle\leq D_{\operatorname{HM}}(K,S)$
$\displaystyle 0$ $\displaystyle\leq r(K,S)$ $\displaystyle R(K,S)$
$\displaystyle\leq\frac{2}{3}D_{\operatorname{HM}}(K,S)$
$\displaystyle\frac{4}{9}\left(\frac{D_{\operatorname{HM}}(K,S)}{R(K,S)}-\frac{3}{4}\right)$
$\displaystyle\left(\frac{9}{4}-\frac{D_{\operatorname{HM}}(K,S)}{R(K,S)}\right)\leq\frac{r(K,S)}{R(K,S)}$

( cf. Figure 22).

######  Remark 6.5.

The bound $D_{\operatorname{HM}}(K,C)\geq R(K,C)$ derived from Corollary 5.6
for the planar case cannot be reached in case of the harmonic diameter. If
$D_{\operatorname{HM}}(K,C)=2$, we obtain from the containment chain

$\frac{1}{2}\left(1+\frac{1}{s(K)}\right)K\subset\frac{K-K}{2}\subset
C_{\operatorname{HM}}\subset\frac{2s(C)}{s(C)+1}C,$

that

(13)
$R(K,C)\leq\frac{4s(K)s(C)}{(s(K)+1)(s(C)+1)}\leq\frac{4n^{2}}{(n+1)^{2}}.$

Thus,

$D_{\operatorname{HM}}(K,C)\geq\frac{(n+1)^{2}}{2n^{2}}R(K,C)$

, which for $n=2$ gives
$D_{\operatorname{HM}}(K,C)\geq\frac{9}{8}R(K,C)>R(K,C)$. Furthermore,
equailty in (13) can only be reached if $K$ and $C$ are simplices, but we have
shown that for Minkowski-centered triangles
$D_{\operatorname{HM}}(K,S)\geq\frac{3}{2}R(K,S)$. Finally, one may recognize
that if we consider symmetric planar gauges, (13) also yields
$D_{\operatorname{HM}}(K,C)\geq\frac{3}{2}R(K,C)$.

The following system of inequalities provides an upper bound for the union of
the diagrams $f_{\operatorname{HM}}(\mathcal{C}^{2},C)$ over all Minkowski-
centered gauges $C\in\mathcal{C}^{2}_{0}$ ( cf. Figure 22).

$\displaystyle 0$ $\displaystyle\leq r(K,C)$ $\displaystyle r(K,C)$
$\displaystyle\leq R(K,C)$ $\displaystyle D_{\operatorname{HM}}(K,C)$
$\displaystyle\leq 3R(K,C)$ $\displaystyle 2r(K,C)$ $\displaystyle\leq
D_{\operatorname{HM}}(K,C)$ $\displaystyle 8R(K,C)$ $\displaystyle\leq
9D_{\operatorname{MIN}}(K,C)$
$\displaystyle\frac{D_{\operatorname{HM}}(K,C)}{2R(K,C)}\left(1-\frac{D_{\operatorname{HM}}(K,C)}{2R(K,C)}\right)$
$\displaystyle\leq\frac{r(K,C)}{R(K,C)}.$

Moreover, the following parts of the boundary described by the above
inequalities are reached:

  1. i)

$r(K,C)=0$ for segments $K=L_{D}$ for gauges $C$ with $s(C)\in[1,2]$

  2. ii)

$r(K,C)=R(K,C)$ for $K=C$ for gauges $C$ with $s(C)\in[1,2]$

  3. iii)

$D_{\operatorname{MIN}}(K,C)=3R(K,C)$ with $C$ being a triangle as in
$f_{\operatorname{HM}}(\mathcal{C}^{2},S)$

The first two inequalities are trivial. The third follows from Lemma 4.16 and
the fourth from Remark 6.5. The last inequality follows from Proposition 4.18
and Remark 4.1. The equality cases follow from Lemma 4.2, Lemma 4.8 and
Corollary 6.4.

$\frac{r}{R}$$\frac{D_{\operatorname{HM}}}{2R}$.$\frac{3}{4}$.$\frac{1}{4}$.$\frac{1}{2}$.$\frac{3}{2}$.$\frac{9}{8}$.$S$.$L_{D}$.$L_{w}$.$S_{\operatorname{HM}}$.$-S$

$\frac{r}{R}$$\frac{D_{\operatorname{HM}}}{2R}$.$\frac{9}{16}$.$1$.$\frac{1}{4}$.$\frac{9}{16}$.$\frac{3}{2}$.$(S,S$.$(S_{\operatorname{HM}},S_{\operatorname{HM}}$.$L_{D}$

Figure 22. The diagram $f_{\operatorname{HM}}(C^{2},S)$ w. r. t. a Minkowski-
centered triangle $S$ (left) and an upper bound for the union of the diagrams
over all Minkowski-centered gauges $C\in\mathcal{C}^{2}_{0}$ (right).

######  Remark 6.6.

Since $\delta_{\operatorname{HM}}$ is not the same for every gauge as it is in
the arithmetic case, the diagram for the equilateral triangle cannot be
dominating. For every $C$,
$f_{\operatorname{HM}}(C,C)=(1,\delta_{\operatorname{HM}})$ and this is always
the only combination where inradius and circumradius coincide. so we cannot
find a single gauge $C$ which defines the union of the diagrams.

##  7\. The diameter $D_{\operatorname{MIN}}$

In the case of the minimum $\delta_{\operatorname{MIN}}$ depends solely on the
asymmetry of $C$ but $\rho_{\operatorname{MIN}}$ can only be bounded in terms
of the asymmetry. [7].

######  Proposition 7.1.

$\displaystyle 1\leq$
$\displaystyle\rho_{\operatorname{MIN}}\leq\frac{s(C)+1}{2}=\delta_{\operatorname{MIN}}$

The inequalities from Lemma 4.16 have the form:

$D_{\operatorname{MIN}}(K,C)\leq(s(C)+1)R(K,C),$

$(s(C)+1)r(K,C)\leq D_{\operatorname{MIN}}(K,C),$

$s(C)r(K,C)+R(K,C)\leq\rho_{\operatorname{M}}(s(C)r(K,C)+R(K,C))\leq(s(C)+1)\frac{D_{\operatorname{M}}(K,C)}{2},$

and

$r(K,C)+R(K,C)\leq D_{\operatorname{MIN}}(K,C).$

Contrary to the other diameters the (dilatated) symmetrization only yields a
completion of the gauge $C$ in the trivial case.

######  Lemma 7.2.

$\rho C_{\operatorname{MIN}}$ is a completion of $C$ if and only if
$\rho=s(C)=1$, which means $C$ is $0$-symmetric.

###### Proof.

We have $C\subset^{\operatorname{opt}}s(C)C_{\operatorname{MIN}}$, which shows
that $\rho$ must equal $s(C)$. However,

$D_{\operatorname{MIN}}(s(C)C_{\operatorname{MIN}},C)=2s(C)\geq
s(C)+1=2\delta_{\operatorname{MIN}}=D_{\operatorname{MIN}}(C,C),$

with equality if and only if $s(C)=1$. ∎

Next, we prove a Jung-bound. Bohnenblust [2] shows for symmetric gauges

######  Proposition 7.3.

(14) $\frac{n+1}{2n}\leq\frac{D(K,C)}{2R(K,C)}.$

For the minimum diameter we obtain the same bound.

######  Corollary 7.4.

(15) $\frac{n+1}{2n}\leq\frac{D_{\operatorname{MIN}}(K,C)}{2R(K,C)}$

Moreover, equality can be obtained only if K is a simplex. For every
$s_{C}\in[1,n]$, there exists $C\in\mathcal{C}^{n}_{0}$, Minkowski-centered,
with $s(C)=s_{C}$ and a simplex $K$ such that the inequality is tight.

###### Proof.

Using Proposition 7.3 for the gauge $C_{\operatorname{MIN}}$ and Remark 4.1,
it follows

(16)
$\frac{n+1}{2n}\leq\frac{D(K,C_{\operatorname{MIN}})}{2R(K,C_{\operatorname{MIN}})}\leq\frac{D(K,C_{\operatorname{MIN}})}{2R(K,C)}=\frac{D_{\operatorname{MIN}}(K,C)}{2R(K,C)}.$

By [12, Theorem 4.1] equailty in Bohnenblust’s inequality can only be attained
if $K$ is a simplex. Let $S$ be a Minkowski-centered simplex. Then, $C=S\cap
s_{C}(_{S})$ with $s_{C}\in[1,n]$ has Minkowski asymmetry $s(C)=s_{C}$, and
for $K=-S$, we have $R(K,C)=n$ and
$D_{\operatorname{MIN}}(K,C)=D(K,C_{\operatorname{MIN}})=D(S,S\cap(-S))=n+1$.
∎

Since for Minkowski-centered triangles
$S_{\operatorname{MIN}}=\frac{2}{3}S_{\operatorname{AM}}$, we know
$D_{\operatorname{MIN}}(K,S)=\frac{3}{2}D_{\operatorname{AM}}(K,S)$ and the
inequalities from Proposition 4.18 can be transferred to describe
$f_{\operatorname{MIN}}(\mathcal{C}^{2},S)$.

######  Corollary 7.5.

For every Minkowski-centered triangle $S\in\mathcal{C}^{2}$, the diagram
$f_{\operatorname{MIN}}(\mathcal{C}^{2},S)$ is fully desribed by the
inequalities

$\displaystyle D_{\operatorname{MIN}}(K,S)$ $\displaystyle\leq 3R(K,C)$
$\displaystyle 2r(K,S)+R(K,S)$ $\displaystyle\leq D_{\operatorname{MIN}}(K,S)$
$\displaystyle\frac{D_{\operatorname{MIN}}(K,C)}{3R(K,C)}\left(1-\frac{D_{\operatorname{MIN}}(K,C)}{3R(K,C)}\right)$
$\displaystyle\leq\frac{r(K,C)}{R(K,C)}.$

As in the case of the harmonic diameter, since $\delta_{\operatorname{MIN}}$
is not the same for every gauge, the diagram for the equilateral triangle
cannot be dominating. There does not exist a single gauge which defines the
union of the diagrams. The following system of inequalities provides an upper
bound for the union of the diagrams
$f_{\operatorname{MIN}}(\mathcal{C}^{2},C)$ over all Minkowski-centered gauges
$C\in\mathcal{C}^{2}_{0}$ ( cf. Figure 23).

$\displaystyle 0$ $\displaystyle\leq r(K,C)$ $\displaystyle r(K,C)$
$\displaystyle\leq R(K,C)$ $\displaystyle D_{\operatorname{MIN}}(K,C)$
$\displaystyle\leq 3R(K,C)$ $\displaystyle r(K,C)+R(K,C)$ $\displaystyle\leq
D_{\operatorname{MIN}}(K,C)$ $\displaystyle 3R(K,C)$ $\displaystyle\leq
2D_{\operatorname{MIN}}(K,C)$
$\displaystyle\frac{D_{\operatorname{MIN}}(K,C)}{2R(K,C)}\left(1-\frac{D_{\operatorname{MIN}}(K,C)}{2R(K,C)}\right)$
$\displaystyle\leq\frac{r(K,C)}{R(K,C)}.$

Moreover, the following parts of the boundary described by the above
inequalities are reached:

  1. i)

$r(K,C)=0$ for segments $K=L_{D}$ for gauges $C$ with $s(C)\in[1,2]$

  2. ii)

$r(K,C)=R(K,C)$ for $K=C$ for gauges $C$ with $s(C)\in[1,2]$

  3. iii)

$D_{\operatorname{MIN}}(K,C)=3R(K,C)$ with $C$ being a triangle as in
$f_{\operatorname{MIN}}(\mathcal{C}^{2},S)$

  4. iv)

$3R(K,C)=2D_{\operatorname{MIN}}(R,C)$ for $K=-S$ and $C=S\cap s(-S)$ with
$s\in[1,2]$

  5. v)

$r(K,C)+R(K,C)=D_{\operatorname{MIN}}(K,C)$ for $K=\lambda
S+(1-\lambda)S_{\operatorname{MIN}}$ with $\lambda\in[0,1]$ and
$C=S_{\operatorname{MIN}}$

The first two inequalities are trivial. The third and the fourth follow from
Lemma 4.16 and the fifth from Corollary 7.4. The last inequality follows from
Proposition 4.18 and Remark 4.1. The equality cases follow from Lemma 4.2,
Lemma 4.8, Corollary 7.5, and the fact that $R(-S,S_{\operatorname{MIN}})=2$.

$\frac{r}{R}$$\frac{D_{\operatorname{MIN}}}{2R}$.$\frac{3}{4}$.$1$.$\frac{1}{4}$.$\frac{3}{2}$.$S$.$L$.$-S$

$\frac{r}{R}$$\frac{D_{\operatorname{MIN}}}{2R}$.$\frac{3}{4}$.$1$.$1$.$\frac{1}{2}$.$\frac{3}{2}$.$(S,S)$.$L_{D}$.$L_{D}$.$(-S,S$.$(-S,S_{\operatorname{MIN}}$.$(S_{\operatorname{MIN}},S_{\operatorname{MIN}})$

Figure 23. The diagram $f_{\operatorname{MIN}}(C^{2},S)$ w. r. t. a Minkowski-
centered triangle $S$ (left) and an upper bound for the union of the diagrams
over all Minkowski-centered gauges $C\in\mathcal{C}^{2}_{0}$ (right).

$\mathrm{true}$moreauthor $\mathrm{true}$morelabelname

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René Brandenberg – Technical University of Munich, Department of Mathematics,
Germany.
<EMAIL_ADDRESS>
Mia Runge – Technical University of Munich, Department of Mathematics,
Germany.
<EMAIL_ADDRESS>