Jung-type Inequalities and Blaschke-Santal\'o Diagrams for Different Diameter Variants

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 $-C$.

A compact convex set is called a convex body.The family of convex bodies in R n is denoted by C n , the convex bodies in R n excluding single points by Cn , and those, which contain 0 in their interior, by C n 0 .The support function of C ∈ C n is denoted by h C (•) : R n → R, h C (a) := max x∈C a T s, while we write ∥•∥ C : R n → [0, ∞], x → ∥x∥ C = min{λ ≥ 0 : x ∈ λC} for the corresponding gauge function.Hyperplanes are denoted by H (a,β) := x ∈ R n : a T x = β and halfspaces by H ≤ (a,β) := x ∈ R n : a T x ≤ β .We say that H (a,β) or H ≤ (a,β) support C ∈ C n in p ∈ C, if p T a = β and C ⊂ H ≤ (a,β) .In that case a is called an outer normal vector of C in p.The polar of C ∈ C n is defined as C • := {a ∈ R n : h C (a) ≤ 1}.For any X ⊂ R n , the positive, linear, affine and convex hull are denoted by pos(X), lin(X), aff(X) and 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 bd(X) and the interior by int(X).For any X, Y ⊂ R n and ρ ∈ R, let X + Y := {x + y : x ∈ X, y ∈ Y } be the Minkowski sum of X and Y and ρX := {ρx : x ∈ X} be the ρ-dilatation of X.We abbreviate {x} + Y =: x + Y and (−1)X =: −X.If X = −X, the set X is called 0-symmetric.If there exists t ∈ 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 ∈ C n with respect to C ∈ C n is defined as R(K, C) := inf{ρ ≥ 0 : ∃t ∈ R n such that K ⊂ t + ρC} and the inradius as r(K, C) := sup{ρ ≥ 0 : ∃t ∈ R n such that t + ρC ⊂ 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 0-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 ∈ C n 0 and values (r, R, D), is there a convex body K ∈ 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 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.
From now on, let us allways assume that n ≥ 2. When considering the containment between different convex bodies, such as the mentioned symmetrizations, we use the following notation.For K, C ∈ C n we say that K is optimally contained in C if K ⊂ C and R(K, C) = 1, which is abbreviated by K ⊂ opt C. The next proposition from [11] characterizes optimal containment.Proposition 1.1.Let K, C ∈ C n and C fulldimensional.Then K ⊂ opt C if and only if i) K ⊂ C and ii) for some k ∈ {2, ..., n + 1}, there exist p 1 , ..., p k ∈ ext(K) ∩ bd(C) and halfspaces H ≤ (a i ,h C (a i )) supporting C at p i with a i ∈ R n \ {0}, i ∈ [k], affinely independent, such that 0 ∈ conv({a 1 , ..., a k }).
The symmetrizations which will later correspond to the diameter definitions are the minimum 2 , and the maximum C MAX := conv(C ∪ −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.
Except C 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.
The Minkowski asymmetry of C ∈ C n is defined as s(C) := R(C, −C) and we say that C is Minkowski-centered if C ⊂ opt −s(C)C.For C ∈ Cn , 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].
Remark.From now on, unless otherwise specified, we always assume that K ∈ C n , that the gauge C ∈ C n 0 is a Minkowski-centered fulldimensional convex body and s ∈ R n \ {0}.

Diameter Definitions
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 ∈ C n by finding a "maximal" segment.But how should this "maximality" be defined?We could measure the distance between two points x, y ∈ K using the gauge function ∥x − y∥ C and define the diameter as the maximal such distance max x,y∈K ∥x − y∥ C .Or, we define it as the maximal circumradius of segments in K: max x,y∈K 2R([x, y], C).However, the diameter could also be defined as the maximal distance between two parallel supporting hyperplanes of K.
As already mentioned, if the gauge C is symmetric, all these definitions lead to the same diameter.max 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 2.1.i) The s-length l s,AM is defined as ii) The s-breadth b s,AM is defined as iii) The arithmetic diameter is defined as the maximal s-length: In [12] the following properties of D AM , which are well known for symmetric gauges (cf.[25]), are proven.
The fact that we can replace C by its symmetrization C−C 2 is the reason why this diameter is called D 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.
ii) The symmetric s-length l s,MIN is defined as iii) The minimum diameter is defined as the maximal symmetric s-length: As we maximize over the s-lengths to obtain the diameter, both definitions of the s-length lead to the same diameter: ii) 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 2.5.i) The s-length l s,HM is defined as ii) The harmonic diameter is defined as the maximal s-length: ii) Since C HM is symmetric, the first part follows.The second part follows again directly from the first.□

Instead of dividing by the mean
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 2.7.i) The s-breadth b s,MAX is defined as .
ii) The maximum diameter is defined as the maximal s-breadth: ii) Proof.The first part follows directly from the fact that max(h C (s), h C (−s)) = h CMAX (s) and the second part again directly from the first.□ Since all definitions are equivalent for 0-symmetric gauges, we can always use that for M ∈ {MIN, HM, AM, 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 Remark 2.9.Using the different definitons of the s-lengths and -breadths, width-definitions can be done analogously to the diameters.For g ∈ {l, b} such that g s,M is defined, the width is defined as w M (K, C) := min g s,M (K, C).
In the standard case M = AM it does not make a difference whether we minimize over the s-length or -breadth.By lemmas 2.4, 2.6, and 2.8 we can symmetrize the arguments of the width as well.

Properties of the Diameters and Blaschke-Santaló Diagrams
In the following, we 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 M ∈ {MIN, HM, AM, MAX} be one of the symmetrizations.
Remark 3.1.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.
and equality holds if we can choose the same outer normals of If we can choose the same (up to dilatation) touching points p i in the boundary of C as in in Proposition 1.1, equality is obtained. iii If the diameters are defined by the same s-breadth or s-length, we have equality.

Proof. i) Obviously, (λr
be the touching points and a i the corresponding outer normals as in Proposition 1.1.Then, and 0 is in the convex hull of the a i , which shows that we have optimal containment.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.iii) We know D M (K, C) = 2R(K AM , C 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 C M and the equality follows from part ii).If the diameter is attained by the same s-breadth, we have and equality follows.□ Lemma 3.3 (Invariance under transformations).Let A : R n → R n be a non-singular affine transformation and L A its corresponding linear transformation.Then Proof.It follows from Proposition 1.1 that the in-and circumradius are invariant under affine transformations.All symmetrizations interchange with linear transformations: (L A (C)) M = L A (C M ) (see [6], Lemma 4).We must confine the transformation in the second argument of the diameter to be linear, since the symmetrizations (besides the arithmetic) are not invariant under translations.Because we can interprete the diameter as a circumradius with D M (K, C) = 2R( K−K 2 , C 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 such as constant width and completeness we extend their definitions to the different diameters.
Remark 3.5.Since all diameters can be expressed by the arithmetic diameter w. r. t.C M , we know the following [18]: i) K has constant width if and only if K AM = λC M for some λ ∈ R.
ii) If K is of constant width, it is complete.iii) In the planar case, K is complete if and only if it has constant width.
Let us observe two things: i) Whenever the gauge is symmetric, the only (up to translation and dilatation) complete and symmetric set is the gauge itself.Thus, when considering D M with respect to a possibly nonsymmetric gauge C, the only complete and symmetric set is always C M .ii) In the case that M = 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 AM and C M will prove helpful to analyse the diameter D M .
Notation.By δ M := δ M (C) and ρ M := ρ M (C) we denote the dilatation factors needed, such that These factors always exist since all symmetrizations are 0-symmetric and fulldimensional.For better readability, we omit the argument C, the gauge body, whenever it is clear from the context.Proof.We begin by showing part ii): By the definition of δ M as well as the diameter properties collected in Proposition 2.2 and Lemmas 2.4, 2.6, and 2.8, we have as well as Segments with circumradius 1 have diameter 2 when considering the arithmetic mean.Hence, If L provides us with the minimal s-length or -breadth, we obtain by part ii): D M (L, C) = min s∈R n \{0} l s,M (C, C M ) = 2ρ 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 R n \ {0}.□ Definition 3.9.The set 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 a set K is complete w. r. t. a symmetric gauge C 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 3.10.i) A supporting slab of K is the intersection of two antipodal parallel supporting halfspaces of K. ii) A boundary point of K is called smooth if the supporting hyperplane of C in this point is unique.iii) A supporting slab is regular if at least one of the bounding hyperplanes contains a smooth boundary point of K. iv) We say that s ∈ R n \ {0} defines a regular slab if there exists a supporting slab such that the defining halfspaces have outer normals ±s.
It is easy to argue that a subdimensional convex body is never complete.On the the other hand, every fulldimensional convex body is the intersection of its regular slabs and completeness can be characterized by using these slabs [38, Theorem 1].
Proposition 3.11.Let K be fulldimensional.Then the following are equivalent: ii) For every outer normal s defining a regular supporting slab of K we have h K (s)+h −K (s) Remark 3.12.As mentioned after the definition of the supercompletion, K * = K sup implies uniqueness for the completion K * of K. Thus, defining 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 3.13.Let X be a closed subset of K. Then the following are equivalent: ii) For every s ∈ R n \ {0} that defines a regular slab of C M there exist s ∈ {s, −s} and p ∈ X such that p T (−s) = h K X (−s) and h K X (s) = h p+DM(K,C)CM (s).
Proof.Let us abbreviate D := D M (K, C) for the proof.ii) ⇒ i): In [38] it is shown that the diameter is the supremum of the breadthes b s (K X , C M ) where s defines a regular slab of C M .For any such s and p ∈ X as defined in ii) we have To show completeness of K X using Proposition 3.11, we need that all the regular slabs of K X are of diametral breadth.However, by the construction of K X , every s which defines a regular slab of K X also defines a regular slab of C M .
i) ⇒ ii): Assume K X is a completion of K and there exists some s ∈ R n \ {0} that defines a regular slab such that there is no p as defined in ii).By the construction of K X there exist which implies that K X is not a completion of K. □ Now, we consider the special case where X is a simplex.
Definition 3.14.We say that a subset ) for all pairs of vertices x, y of X.
Lemma 3.15.Let X be a diametric triangle of K ∈ 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 3.13 is fulfilled.Assume w. l. o. g. that D M (X, C) = D M (K, C) = 1 and let X = conv p 1 , p 2 , p 3 .Then, the translations −p i + X with i ∈ {1, 2, 3} are subsets of C M , all with one vertex in the origin and the other two on the boundary of C M (cf. Figure 2).Since p i − p j and p j − p i are each other's negative, we have three pairs of points in the boundary of C M .For each, we choose an outer normal a k , k ∈ {1, 2, 3} ordered as given in Figure 2. Now, the boundary of K X = K ext(X) consists of three parts which are built by parts of the boundary of C M (colored in blue in the left part of Figure 2).Then, if s ∈ pos({a i , −a j }), i ̸ = j, property ii) of Lemma 3.13 holds for K X with p = p k , k ̸ = i, j.Hence, for all s ∈ R n \ {0}, property ii) is fulfilled and it follows that K X is a completion.Using Remark 3.12 we obtain the uniqueness.
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 1.2 we know The containment factors between the symmetrizations of the gauge can be used to improve this chain and to formulate new inequalities.
Proof.i) Follows directly from Remark 3.1 and Proposition 1.2.
ii) Since r(K, C)C is contained in a translate of K, we obtain iii) Follows directly from Lemma 3.8 and Remark 3.1.iv) By [10, Theorem 1.1] we have

□
We would like to describe the values the inradius, circumradius, and diameter of sets K ∈ C n may have, when we consider a fixed, Minkowski-centered C ∈ C n 0 .To do so, we study the following Blaschke-Santaló diagrams.Definition 3.17.Let f M be the following mapping. (1) The set f M ( Cn , 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 M )-diagram.
As for the diameter we only write f if the gauge is symmetric.In [9] f AM ( C2 , S) is described and it is shown that this diagram is equal to the union of the diagrams over all possible gauges.
Proposition 3.18.For every triangle S ∈ C 2 , the diagram f AM ( C2 , S) is fully described by the inequalities 2 ) [41] (name giving) and f AM ( C2 , S) [9] can be seen in Figure 3.It is shown in [9] that f AM ( Cn , C) is star-shaped with respect to the vertex f 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.f M (C, C).
As a consequence, we know that f M ( Cn , C) can only have open holes and therefore only fulldimensional holes.For In the case M = AM, we also have equality for the diameter, but this does not necessarily hold for the other diameters.Since R(•, C), r(•, C) and D M (•, C) are continuous with respect to the Hausdorff distance and t ∈ is continuous as well.Thus, for every such K there is a continuous curve Γ K in the diagram from f M (K, C) to f M (C, C).Let (K n ) n∈N be a sequence of bodies on the boundary converging to K on the boundary.We show that the functions Γ K n converge uniformly to Γ K .We can consider the components separately.For the inradius, we know ) are convex and continuous in t and they converge pointwise to the convex and continuous function g 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 Γ K lies above the hole, if the set enclosed by [f M (L D , C), f M (C, C)], Γ K and the boundary between f M (K, C) and f M (L D , C) which does not contain the segment [f M (L D , C), f M (C, C)] does not contain the hole.Analogously, we say that Γ K lies below the hole if the set contains the hole.Thus, Γ L D lies above the hole and Γ C below.Then, there exists a converging sequence (K n ) n∈N with K n → K of bodies on the boundary such that all Γ K n lie above the hole and Γ K below or vice versa.This contradicts the fact that the curves converge uniformly.).Thus, it seems reasonable to look at the diagrams for the three other diameters D MIN , D MAX and D HM in terms of triangular gauges first, which we do in the remaining sections.
The inequalities from Lemma 3.16 have the form: (2)
In case of K = C (2) and ( 3) become tight while (3) and ( 4) become tight for K = C MAX .
Asymmetric gauges are not complete, but a completion is easy to find.
We define the outer symmetric support: One should recognize that we need a scaling factor of up to n to cover the completion C MAX by C here, while with the arithmetic diameter C is always already complete itself.
The maximality of the circumradius can be seen as follows: Let K be any completion of C.Then, from (4) and Lemma 3.6 we obtain Next, we show that C MAX ⊂ C sup ⊂ C 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) ≤ h C oss (a) for all a ∈ A oss C .Let a ∈ A oss C .Then, there exists p ∈ −C ∩ H (a,1) .Since −p ∈ C we obtain C sup ⊂ −p + 2C MAX and therefore The containment chain directly shows the backward direction of part iii).
To show the forward direction, assume C MAX ̸ = C oss .Since C MAX is the intersection of its regular slabs, there must exist some a ∈ R n \ {0} which defines a regular slab of with λ ∈ (0, 1).Z λ is Minkowski-centered with Minkowski asymmetry 2 − λ and (Z λ ) oss ̸ = (Z λ ) MAX , which because of Lemma 4.3 means that (Z λ ) MAX is not the unique completion of Z λ .In the extreme cases λ ∈ {0, 1}, Z λ is a triangle or a rectangle and 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.
. Thus, we can assume K ∈ C2 and K ⊂ opt C, which implies R(K, C) = 1.Then, there exist touching points q 1 , . . ., q k ∈ bd(K)∩bd(C) with k ∈ {2, 3} and corresponding outer normals a i as described in Proposition 1.1.If k = 2 is possible, there exists a segment L ⊂ K with the same circumradius as K and by Lemma 3.8 For k = 3, the triangle conv( q 1 , q 2 , q 3 ) has the same circumradius as K and D MAX (K, C) ≥ D MAX (conv( q 1 , q 2 , q 3 ), C).Thus, it suffices to prove the claim for the case that K is a proper triangle K = conv( q 1 , q 2 , q 3 ).
Let S := 3 i=1 H ≤ (a i ,1) be the intersection of the three supporting halfspaces of C s. t. q i ∈ H (a i ,1) .Denote the vertex opposing the edge defined by a i by pi .Then, R(K, S) = R(K, C) = 1 and D MAX (K, C) ≥ D 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 before: pi = c + p i , i = 1, 2, 3.In the following indices are to be understood modulo 3. Let α i ∈ [0, 1] be, s. t.
We split the proof into two parts.First, we consider the case where the origin is close to the center c, i. e. 0 ∈ int(conv( c − 1 2 p i , 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 ∈ int(conv( c − 1 2 p i , i = 1, 2, 3 )), which is equivalent to c ∈ int(conv( 1 2 p i , i = 1, 2, 3 )).Define λ 1 , λ 2 , λ 3 > 0 with Thus, which shows that z i is an outer normal of conv(S ∪ (−S)) with h conv(S∪(−S)) Next, we show that the last term is larger than or equal to 1.If b z i (K, T MAX ) ≥ 1 for i ∈ {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 By adding these inequalities we obtain Now, consider the case that 0 / ∈ int(conv( c − 1 2 p i , i = 1, 2, 3 )).Because of the symmetries of T we can assume that 0 = 3 i=1 β i pi with 8).

By the above conditions we have
Thus, using α 1 , α 2 , α 3 as above, we know, if and therefore, Using that C is Minkowski-centered, we know h C (−a 1 ) ≤ s(C)h C (a 1 ) = s(C) and we obtain

□
Since D MAX is the smallest diameter, this provides a lower bound for all four diameter definitions.If we omit the restriction of C being Minkowski-centered, Theorem 4.5 is not necessarily true.However, for gauges that still contain the origin, we obtain the following Jung-type inequality.
Theorem 4.7.Let K, C ∈ C 2 , s. t. 0 ∈ C and dim(C) = 2.Then, Moreover, equality can be attained for some K if and 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 4.5, we can assume that K is a triangle and K ⊂ opt C. It suffices to show D(K, conv(S ∪ (−S))) ≥ 2 3 for S as given in Theorem 4.5 since If there exists α i / ∈ 1 3 , 2 3 , there is an a j with a j T q j −a j T q i h C (a j )+h C (−a j ) > 2 3 (cf.Figure 9) and therefore, Now, consider the case where α i ∈ 1 3 , 2 3 for all i ∈ {1, 2, 3}.There exists i ∈ {1, 2, 3} such that z i defined as in the previous proof defines a supporting hyperplane of conv (S ∪ (−S)) with h conv(S∪(−S)) (z i ) = 1, w. l. o. g. i = 3.Then, we know 1 = h conv(S∪(−S)) (z 3 ) ≥ (z 3 ) T (±p 3 ) = ±3(z 3 ) T c and using ( 6) we obtain Reaching equality in the last inequality chain we need α 1 = (1 − α 2 ) = 1 3 .In this case we need 0 ∈ p3 ∪ [p 1 , p2 ] for b a 3 not to be larger than 2  3 .However, if 0 ∈ [p 1 , p2 ] we have that z 3 cannot define a supporting hyperplane as described above.Thus, 0 remains to be a vertex of S to reach equality.For D MAX (K, C) = 2  3 to be true, we need 3 .Thus, p1 , p2 ∈ C, which means C = S.All in all, we see that equality can be obtained for C = conv( 0, p1 , p2 ) and conv( q 1 , q 2 , q 3 ) with In that case (see Figure 10), which is achieved, e.g. if we choose K = conv For fixed C, we refer to any K that fulfills DM( K,C) R( K,C) = min K∈C n DM(K,C) R(K,C) as Jung-extremal .If C = T , we have equality in Theorem 4.5 for the following family of triangles.
. p1 p2 0 Example 4.8.Let T be the equilateral triangle as defined above and for α ∈ 1 3 , 2 3 (cf.Figure 11).Then, The triangles T α are all equilaterals.Special cases are α = 1 2 , where T 1 2 = − 1 2 T , and α ∈ 1 3 , 2 3 , rotations of dilatated T by ± π 6 .Proof.By construction T α ⊂ opt T .The diameter of T α is attained between two of its vertices and since T MAX is a regular hexagon which has rotational symmetry of order six, is does not matter which of the edges of T α we consider.Let q i = αp i+1 + (1 − α)p i+2 .By using p 3 = −p 2 − p 1 we obtain The triangle conv( w 1 , w 2 , w 3 ) with w i := αq i+2 + (1 − α)q i+1 is optimally contained in T α by Proposition 1.1.Furthermore, Using this we compute Thus, conv( w 1 , w 2 , w 3 ) is a translate of (1 − 3α − 3α 2 )T and the inradius is 3 is especially interesting: it is complete since its arithmetic mean is a dilatation of T MAX .Since the symmetrization T MAX of T is also Jung-extremal, there are at least two complete Jung-extremal bodies, T MAX and T 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.Lemma 4.9.Let T be the equilateral triangle as defined above and S λ = conv({q 1 , q 2 , q 3 }) with Proof.By construction, R(S λ , T ) = 1.Obviously, q 2 − q 3 TMAX ≤ 1 and q 2 − q 1 TMAX = q 3 − q 1 TMAX .Using p 3 = −(p 1 + p 2 ) we can compute: 2 .Now we show the formula for the inradius (cf. Figure 12).Let conv( w 1 , w 2 , w 3 ) be the inner triangle.By axial symmetry of T and S λ we know that w 1 = 1 2 (q 2 + q 3 ).Denote the euclidean edge length of the inner triangle by a. Since T has edge length √ 3, we obtain r(S λ , T ) = a √ 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 conv( w 1 , w 3 , q 2 ) and conv( p 3 , q 2 , q 1 ) are similar, implying and therefore r(S λ , T ) = a Theorem 4.10.Let K ∈ Cn and T an equilateral Minkowski-centered triangle.Then with equality for the triangles S λ , λ ∈ [ 1 2 , 1], as described in Lemma 4.9.
Calculating the inradius of S λ .
To prepare the proof of Theorem 4.10 we need the following lemma.
Lemma 4.11.Let K = 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 q 3 − p 1 2 ≤ q 1 − p 3 2 .Let K := conv({q 1 , q 3 , q2 }) ⊂ opt T be a triangle such that q2 is on the same edge of T as q 2 and q 2 − p 1 2 ≤ q2 − p 1 2 .If (q 1 , q 3 ) is steep, the inradius of 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 we can generalize this result to all choices i, j ∈ {1, 2, 3}: At least one of the ordered pairs (q i , q j ) or (q j , q i ) is always steep.
Proof of Lemma 4.11.Let H 1 and H 2 be the two lines parallel to [p 1 , p 3 ] supporting the inner triangle conv( 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 w 2 intersects K.By the intercept theorem with the two parallel lines H 1 and H 2 and the points q 2 or q2 the segment of H 1 contained in K is greater or equal than the one contained in K. Thus, we can move the inner triangle of K within the slab between H 1 and H 2 until it touches [q 1 , q2 ].The resulting translations of w 1 , w 2 , w 3 are all contained in K, the translation of w 2 due to the steepness of (q 1 , q 3 ).Thus, r(K, T ) ≤ r( K, T ).□ Proof of Theorem 4.10.The equality case follows directly from Lemma 4.9.So we only need to prove the correctness of the inequality.
Let K ⊂ opt T .As shown in the proof of Theorem 4.5, we either have D MAX (K, T ) ≥ 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 4.10 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 := conv({q 1 , q 2 , q 3 }), where q i ∈ 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 q 1 − p 3  2 ≤ q 1 − p 2 2 .Our goal is to show that there exists a triangle S λ ⊂ opt T , λ ∈ [ 1  2 , 1], as defined in Lemma 4.9 with at most the same diameter and inradius as S. Using the fact that . Proof of Lemma 4.11.Since (q 1 , q 3 ) is steep, the inner triangle can be translated along the orange hyperplanes.
then conclude 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.
Case 1 This case is depicted in Figure 14.If D MAX (S, T ) = D MAX ([q 1 , q 2 ], T ) our assumption q 1 − p 3 2 ≤ q 1 − p 2 2 implies that q 2 − p 1 2 ≤ q 2 − p 3 2 .Let us assume otherwise.Then we would have [q 1 , q 2 ] ⊂ conv( p 3 , 1 2 (p 2 + p 3 ), 1 2 (p 1 + p 3 ) = 1 2 (p 3 + T ) with [q 1 , q 2 ] not being an edge of 1  2 (T + p 3 ), which implies , which implies the steepness of (q 1 , q 2 ).Since [q 1 , q 2 ] is diametral, q 3 lies inside q 1 + D MAX (S, T )T MAX .However, q 1 − p 3  2 ≤ 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 MAX (S, T )T 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 4.11).Doing so, [q 1 , q 3 ] becomes diametral, too.Case 2 If D MAX (S, T ) = D MAX ([q 1 , q 3 ], T ), we need to consider three subcases: a) If q 3 − p 2 2 ≤ 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.b) If q 3 − p 2 2 ≥ 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.c) If q 3 − p 2 2 ≥ 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 q 2 − q 1 TMAX = q 3 − q 1 TMAX is possible and doing so the points q 2 and q 3 do not only lie on the boundary of q 1 + D MAX (S, T )T MAX , they essentialy lie on the (translated and dilatated) edges [p 1 , −p 3 ] or [p 1 , −p 2 ] of T MAX .For q 2 this follows from the fact that it has to lie closer to p 1 than to p 3 .As described in the proof of Case 1, the boundary of q 1 + D MAX (S, T )T MAX intersects [p 1 , p 2 ] once or twice, but it is not possible that it only intersects with the segment q 1 + D MAX (S, T )[p 2 , −p 3 ] as this would contradict our assumption Proof of Case 1 of Theorem 4.10.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.
, T ).If we have two intersection points, we can replace q 3 if necessary by the upper one without increasing the inradius since (q 1 , q 2 ) is steep.Thus, we can assume For the next part of the proof we now assume that q 3 ∈ q 1 + D MAX (S, T )[p 1 , −p 3 ] and q 2 ∈ q 1 + D 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 q1 such that q1 1 = q 2 1 does not increase the diameter or the inradius and we still have ∥q 3 − q 1 ∥ TMAX = ∥q 2 − q 1 ∥ TMAX .This shows that we can assume q 1 1 ≤ q 2 1 and that Proof of Theorem 4.10.We can assume that q 3 lies on q 1 + D MAX (S, T )[p 1 , −p 3 ] and q 2 lies on q 1 + D MAX (S, T )[p 1 , −p 2 ].Otherwise we can consider the triangle S = conv( q1 , q 2 , q 3 ) which has smaller or equal inradius and diameter.Now, we consider the triangle S λ = conv({q 1 , q2 , q3 }) with λ as in Lemma 4.9 such that it has the same diameter.Due to symmetry reasons and our assumptions about the positions of q 2 and q 3 , q 1 − q1 2 = q 2 − q2 2 = q 3 − q3 2 =: κ (see Figure 16).By using the intercept theorem and the law of sines we will show that, possibly after a suitable translation, all vertices of the inner triangle of S λ are contained in S.This way we see that the inradius of S λ is at most the one of S. We denote the vertices of the inner triangle of S λ by w i (see Figure 18) and the euclidean distance in the horizontal direction of w i to the segment [q j , q k ], {i, j, k} = {1, 2, 3}, by l i .If κ ̸ = 0, every side of S intersects the corresponding side of S λ exactly once.Let us denote these intersection points by v i (cf. Figure 17).We will show that if we shift the inner triangle of S λ by l 3 to the left it is completely contained in S. To do so we compute all the values l i , i = 1, 2, 3 and prove that we have l 3 ≤ l j , j = 1, 2. Computation of l 1 (cf.Figure 17): We use the intercept theorem for [q 2 , q3 ] and the two lines parallel to this segment through q 2 and q 3 , respectively.Since T is an equilateral triangle, conv( p 1 , q3 , q2 ) is also equilateral with an edge length of (1 − λ) √ 3. Furthermore, It follows that w 1 is always contained in S and Computation of l 2 (cf. Figure 18): Let α 2 = ∠v 2 q 3 q3 , β 2 = ∠v 2 q 1 p 2 and γ 2 = ∠q 1 v 2 q 1 .We compute using the law of sines, first for the triangles conv( v 2 , q 1 , q1 ) and conv( v 2 , q 3 , q3 ), and then for conv( p 2 , q 1 , q 3 ).
Furthermore, we know from Lemma 4.9 that r(S λ , T ) = λ(1 − λ).Together with the intercept theorem we obtain , v 2 is closer to q1 than w 2 and therefore also w 2 ∈ S.
The distance of w 2 to [q 1 , q 3 ] in the direction of (−1, 0) T is by the intercept theorem Computation of l 3 (cf.Figure 18): If w 3 is also contained in S, we have that the complete inner triangle of S λ is contained in S and we are done.
Theorem 4.12.For every Minkowski-centered triangle S the diagram f MAX ( C2 , S) is fully described by the inequalities 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 4.
which enables us to give a bound for the union of the diagrams f MAX ( C2 , C) with C Minkowskicentered (depicted in red within Figure 19).

The diameter D HM
In the case of the harmonic mean, the factors ρ HM and δ HM are not bound to the Minkowski asymmetry s(C).However, in [7] the following bounds are proven: Proposition 5.1.
. Furthermore, the inequalities from Lemma 3.16 have the form: then C HM fulfills the first inequality in (11) with equality, which in this case can be rewritten as ( 14) Unlike with D MAX , (a dilatate of) the symmetrization C HM is not always a completion of the gauge.
Lemma 5.2.The following are equivalent: Proof.We know from [7] where the p i are the vertices of T .For the Reuleaux triangle one obtains (omitting the detailed calculations) s(RT) ≈ 1.025.Thus, by Lemma 5.2 this is a case where the (dilatated) harmonic mean RT HM is not a completion of the gauge RT.
Since T ⊂ RT is diametric we obtain from Lemma 3.15 that the unique completion of RT is RT * := 3 i=1 p i + 2δ HM RT HM (cf. Figure 20).
The following system of inequalities provides an upper bound for the union of the diagrams f HM ( C2 , C) over all Minkowski-centered gauges C ∈ C 2 0 (cf.Remark 5.6.Since δ 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 HM (C, C) = (1, δ HM ) and this is always the only combination where inradius and circumradius coincide.Thus, we cannot find a single gauge C which defines the union of the diagrams.

The diameter D MIN
In the case of the minimum δ MIN depends solely on the asymmetry of C but ρ MIN can only be bounded in terms of the asymmetry.[7].
Proposition 6.1.For the minimum diameter we obtain the same bound.As in the case of the harmonic diameter, since δ 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 MIN ( C2 , C) over all Minkowski-centered gauges C ∈ C 2 0 (cf.

Segments
L optimally contained in C with D(L, C) = 2ρ M are denoted by L w and in case of D(L, C) = 2δ M by L D .Lemma 3.8.i) For any segment L ⊂ opt C: 2ρ M ≤ D M (L, C) ≤ 2δ 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.ii) The diameter of C with respect to itself is D M (C, C) = 2δ M , and the width of C with respect to itself is w M (C, C) = 2ρ M .

Figure 4 .
Figure 4. Proof of Lemma 3.19: K n → K but Γ K lies below the hole and Γ Kn above.

Figure 5 .
Figure 5.The equilateral triangle T and its maximum T MAX In the case of the maximum both factors ρ MAX and δ MAX are known [7] and depend at most on s(C): Proposition 4.1.
if and only if C MAX = C oss , and iv) C MAX is always the unique 0-symmetric completion of C.

Example 4 . 4 .
Assuming a ∈ bd(C • ), there exists a smooth boundary point x of C MAX supported by the hyperplane H (a,1) .Because of h C (a) > h C (−a), the point x must belong to bd(C) ∩ H (a,1) .Now, assume x ∈ bd(C sup ) as well.Then, there exist an outer normal a x ∈ bd(C • MAX ) and a point p x ∈ C such that x T a x = h C sup (a x ) and (a x ) T (x − p x ) = 2.Then, H (ax,h C sup (ax)) also supports C MAX , which implies a X = a since x is a smooth boundary point of C MAX .But since h C (−a) < h C (a) we cannot choose p x ∈ C. Thus, x is not contained in the boundary of C sup and therefore, C MAX ̸ = C sup .Finally, any 0-symmetric completion of C must contain C and −C and therefore C max .□Trapezoids within the following family have completions besides their maximum:

Corollary 4 . 6 .
Let K, C ∈ C 2 , C Minkowski-centered.Then D M (K, C) ≥ R(K, C).If M ̸ = MAX it follows directly from Proposition 1.2 and the translation invariance in the arithmetic case that Corollary 4.6 stays true for non-Minkowski-centered C ∈ C 2 .However, since D AM (K, C) = R(K, C) is obtained if and only if C is a triangle and K is a homothet of −C[12,10], we will see in the next section that this bound cannot be reached if M ∈ {MIN, HM}.

Figure 16 .
Figure 16.Proof of Theorem 4.10.Transformation of S into S λ with the same diameter.The distances ∥q i − qi ∥ 2 are equal.

1 v 1 l 1 rFigure 17 .
Figure 17.Proof of Theorem 4.10.Computation of l 1 .The triangles defined by p 1 and the intersection points of the parallel lines are all three equilateral.
5 and Theorem 4.10.First, (2) is attained with equality for (1 − λ)L D + λT , λ ∈ [0, 1] where L D and T are the extreme cases.Second, (3) is attained with equality for (1 − λ)T + λT MAX , λ ∈ [0, 1] by Lemma 3.2 where T and T MAX are the extreme cases.The boundary induced by (5) is filled by segments from L w to L D .Next, since T MAX is the completion of −T , the boundary from Theorem 4.5 is filled by the sets −T + with −T ⊂ −T + ⊂ T MAX .Finally, the inequality in Theorem 4.10 is fulfilled with equality by the triangles S λ as introduced in Lemma 4.9.Since we have presented a continuous description of the boundary we can apply Lemma 3.19 and follow that the diagram is simply connected.□ While the inequalities (2), (3), (5) and the one given by Theorem 4.5 are valid for all choices of planar, Minkowski-centered gauges, the inequality from Theorem 4.10 is only proven for triangles.Using the result from Proposition 3.18 and D AM (K, C) ≤ 4 3 D MAX (K, C) which follows from s(C)+1 2s(C) C MAX ⊂ opt C AM [7], we are able to give another general inequality (8) 2 3

Figure 19 .
Figure 19.The diagram f MAX ( C2 , 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).

2 C
HM is a completion of C w. r. t.C, and iii) D(C, C HM ) = 2R(C, C HM ).

2 C 2 C
HM is a completion of the gauge C, then it is also a completion of −C and since R(−C, C) = s(C) = R( s(C)+1 HM , C) for −C it is even a Scott-completion.Example 5.3.The Reuleaux triangle RT is the completion of the equilateral triangle T in the euclidean case:
Figure 22).0 ≤ r(K, C)r(K, C) ≤ R(K, C) D HM (K, C) ≤ 3R(K, C) 2r(K, C) ≤ D HM (K, C) 9R(K, C) ≤ 8D HM (K, C) D HM (K, C) 2R(K, C) 1 − D HM (K, C) 2R(K, C) ≤ r(K, C) R(K, C) .Moreover, the following parts of the boundary described by the above inequalities are reached:i) r(K, C) = 0 for segments K = L D for gauges C with s(C) ∈ [1, 2] ii) r(K, C) = R(K, C) for K = C for gauges C with s(C) ∈ [1, 2] iii) D HM (K, C) = 3R(K, C) with C being a triangle as in f HM ( C2 , S)The first two inequalities are trivial.The third and forth follow from Lemma 3.16 and the fifth from Remark 5.5.The last inequality follows from Proposition 3.18 and Remark 3.1.The equality cases follow from Lemma 3.2, Lemma 3.8 and Corollary 5.4.

Figure 22 .
Figure 22.The diagram f 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 ∈ C 2 0 (right).