The Critical Set of a Convex Body

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1 Introduction

A convex set is, by definition, a set in which every segment connecting two points of the set lies in the set. In this paper I discuss convex sets which are both closed and bounded and with non-empty interior (in some finite- dimensional affine space over the real numbers) and I refer to such sets as convex bodies.

A special kind of convex bodies are the symmetric ones. They are char- acterized by the fact that they contain a point which divides every chord through the point in two equal parts. In the study of non-symmetric convex bodies, different measures of asymmetry are introduced. They are measures of how far the set is from being symmetric, and in this paper I use two such measures. With the help of these measures of asymmetry, an affine-invariant centre of a convex body is defined. This centre does not have to be unique, and the set of points in the centre is called the critical set. In this paper I mainly discuss convex bodies with non-trivial critical sets.

1.1 Notation and Terminology

The set of n-dimensional convex bodies will be denoted Cⁿ, and C will be the set of compact convex subsets of Rⁿ. When nothing else is said, C is used to denote a convex body of dimension n. Since the interior of a convex body is required to be non-empty, every closed and bounded convex subset of Rⁿ is a convex body in a unique affine space, the carrying space. The carrying space of C is denoted A(C).

The interior, boundary, dimension, diameter and convex hull of a set X will be denoted by X^o, F rX, dim(X), diam(X) and cvx(X) respectively.

Interior points of a set X with respect to the smallest affine space containing the set will be called equilibrium points. The set of equilibrium points will be denoted by E(X). The number of elements in a finite set X will be denoted

|X|.

1.2 Affine Spaces

An affine space A is a set A such that given a pair of two points, there is exactly one line passing through both of them, and such that every ordered pair of two points defines a unique vector in an associated vector space. Affine spaces are like vector spaces but with no point 0. This means that it is not possible to add points (or sets). The only linear combinations possible to do are combinations of the form (x1, x2, ..., xn) 7→ P

aixi where P

ai = 1. In this paper I consider finite-dimensional affine spaces over the real numbers.

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The set of affine functions from the affine space A to the affine space B will be denoted Af f (A, B). Given two points P and Q, the segment between them is denoted P Q, the distance between them is denoted |P Q| and the ray from P through Q is denoted −→

P Q. The closed n-dimensional ball of radius a centered in x is denoted B(x, a).

Remark An example of an n-dimensional affine space over the real numbers is a hyperplane in Rⁿ⁺¹ not containing 0. Conversely, it can be shown that every n-dimensional affine space over the real numbers can be represented as a subset of Rⁿ⁺¹ with codimension 1.

1.3 The Hausdorff Distance

In this paper we consider C and Cⁿ as metric spaces. To do that we regard them as subspaces of the space of compact subsets of Rⁿ equipped with the Hausdorff distance, d. The Hausdorff distance between two compact subsets C and D of a metric space (X, d⁰) is defined in the following way:

First let dC(x) = min_c∈Cd⁰(x, c) and dD(x) = min_d∈Dd⁰(x, d). Then dC,D(x) .

= |dC(x) − dD(x)| is a bounded function. Let the Hausdorff distance between C and D be d(C, D) = sup{|dC(x) − dD(x)|}.

a

C

D

Figure 1: d(C, D) = a

Note that with the Hausdorff metric, Cⁿ is not a complete metric space (but C is).

Sometimes it is convenient to identify a compact subset D of Rⁿ with its distance function dD(x) = min{|xy| : y ∈ D}. Since there is a 1-1- correspondence between compact sets and their distance functions, C and Cⁿ can be regarded as subspaces of the space of continuous functions with the sup-norm.

If nothing else is said, all concepts in C and Cⁿ involving the use of some topology (convergence, continuity, closedness etc.) are with respect to the Hausdorff distance.

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1.4 The Banach-Mazur Distance

Sometimes it is convenient to regard two convex bodies C and D, for which T(C) = D for some D and some affine map T , as the same. More specifically, let the equivalence relation ∼ be the following:

C₁ ∼ C₂ if and only if

C₂ = T (C₁) for some T ∈ Af f (Rⁿ, Rⁿ).

Then, for every C ∈ C, let EC be the equivalence class of C. Note that, for example, all triangles are in the same equivalence class. Finally, we let Eⁿ be the space consisting of all equivalence classes of n-dimensional convex bodies.

In this space, the Hausdorff metric does not work. Instead, the Banach- Mazur distance between convex sets can be used to obtain a metric. That distance is defined in the following way: If C, D ∈ Cⁿ and 0 ∈ C ⊂ D, let ρ(C, D) = min{λ|D ⊂ λC}. Then let the Banach-Mazur distance between EC and ED be given by, chosen a C ∈ EC containing 0,

dBM(EC, ED) .

= inf{ρ(C, D)|C ⊂ D ∈ ED}.

Observe that the distance defined is independent of the choice of C so we can also write dBM(EC, ED) = inf{ρ(C, D)|0 ∈ C ⊂ D ∈ ED with C ∈ EC}.

The Banach-Mazur distance is a multiplicative metric, which means that:

1. dBM(EC, ED) ≥ 1 with equality iff EC = ED

2. dBM(EC, ED) = dBM(ED, EC)

3. dBM(EC, ED) ≤ dBM(EC, EF)^.dBM(EF, ED)

1 and 2 are obvious. To prove 3, we may choose representatives C ∈ EC, D ∈ ED and F ∈ EF such that, if dBM(EC, EF) = λ₂ and dBM(EF, ED) = λ₃, then C ⊂ F ⊂ λ2C and F ⊂ D ⊂ λ3F. But this means that C ⊂ D ⊂ λ₂λ₃C. Thus dBM(C, D) ≤ λ2λ₃.

Letting dB(EC, ED) .

= ln dBM(EC, ED), we get an ordinary additive metric on Eⁿ.

In dealing with the space Eⁿ, it is often practical to choose a representative of an equivalence class EC. When doing that, the following theorem by John is useful:

Theorem 1.1 If C is the ellipsoid of maximal volume contained in D (such an ellipsoid exists and is unique), and if c is the centre of C, then D ⊂ (1 − n)c + nC.

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Since all ellipsoids are in the same equivalence class as the unit ball, a consequence of Theorem 1.1 is that every equivalence class has a representative which contains the ball B(0, 1) and is contained in B(0, n). Therefore it is of interest to study the subspace Dⁿ of Cⁿ consisting of convex bodies C such that B(0, 1) ⊂ C ⊂ B(0, n).

1.5 The First Measure of Asymmetry

One way of defining a measure of the asymmetry of a convex body C is the following:

Let c be a point in C^o and let pq be a chord through c with p and q in F rC. Let ρ(c, θ) = ^|pc|_|cq|, where θ is the direction of pq. With c fixed, ρ is

c q p

Figure 2:

a continuous function of the direction θ, and therefore assumes a maximum ρ(c). This maximum ρ(c) is a continuous function of c, and it approaches infinity as c approaches F rC. Hence ρ(c) takes a smallest value somewhere in C^o. Let as₁(C) = mincρ(c) = mincmaxθρ(c, θ).

Remark With this definition, a convex body C is symmetric if and only if as1(C) = 1. If C is non-symmetric, then as1(C) > 1.

There is an alternative way of defining as₁. An affine space does not contain an origin, but we can force a linear structure on it by choosing an origin. If we choose an origin c ∈ C^o, we can define the Minkowski functional

kxkC,c .

= inf{t > 0|x ∈ tC}.

Now let

k − Idk^C_c .

= sup

x∈C

k − xkC,c.

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Observe that kxk^C_c 6= k − xk^C_c in general, and hence ψC(c) .

= k − Idk^C_c ≥ 1.

Since ψC(c) is a continuous function of c which tends to infinity near F rC, it assumes a minimal value for some c ∈ C^o.

Since the functional kxkC,c is a continuous function of x and since C is compact, the supremum k − Idk^C_c = sup_x∈Ck − xkC,c is actually attained at some point. This point gives rise to a chord divided by c in the ratio k − Idk^C_c. Conversely, every chord which is divided by c in the ratio ρ(c) gives a point x with k − xkC,c = ρ(c). Therefore ψC(c) = ρ(c), and we get that as1(C) = minc∈C^oψC(c).

1.6 The Second Measure of Asymmetry

An equivalent measure is defined by taking a minimum of maximums of affine functions. Namely, let

A = {f ∈ Af f (A(C), R)|f∗(C) = [−1, 1]},

i.e. A is the set of affine (real-valued) functions mapping C onto the interval [-1,1]. Then define the function ϕ on A(C) by, for every x ∈ A(C)

ϕ(x) = sup

f∈A

f(x).

Since ϕ is a supremum of convex and continuous functions, it is itself convex and lower semi-continuous. Since the set A is equi-continuous, ϕ is also continuous. Since ϕ ≡ 1 on the boundary of C, ϕ attains a minimum somewhere inside C. Define as₂(C) to be the minimal value of ϕ.

Remark With this definition, a body C is symmetric if and only if as₂(C) = 0. If C is non-symmetric, then as2(C) > 0.

1.7 Various Results Concerning the Measures of Asym- metry

Both measures of asymmetry are affine-invariant, i.e.

as₁(C) = as1(T (C)) and

as2(C) = as2(T (C))

for every T ∈ Af f (A(C), A(C)). This means, for example, that with these measures of asymmetry, all triangles are equally symmetric (since to every

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pair of triangles C1, C₂ there always exists an affine function T such that C₁ = T (C₂)). In general, different representatives of an equivalence class EC

are equally symmetric.

Moreover, our two measures are equivalent. That follows from the fact that ψ and ϕ are two equivalent functions:

Theorem 1.2 For every x ∈ C^o we have that ϕC(x) = ψC(x) − 1

ψC(x) + 1 and

ψC(x) = 1 + ϕC(x) 1 − ϕC(x).

Proof Since the two equalities only are equalities with real numbers, it is easy to check that they are equivalent. Let x ∈ C^o.

First, let a, b ∈ F rC such that ^|ax|_|xb| = ψ(x). Let Hb be a supporting plane to C at b. Then Ha .

= Hb + a − b is a supporting hyperplane to C at a.

Define an affine functional f to be 1 at Hb and -1 at Ha. Clearly f ∈ AC, and hence ϕ(x) ≥ f (x) = ^ψ(x)−1_ψ(x)+1.

Now, let g ∈ AC such that g(x) = ϕC(x), and choose y ∈ C such that g(y) = −1. Let the planes where g is −1, ϕC(x) and 1 be H−1, Hk and H1

respectively. Assume that the distance between H−1 and Hk is c, and that the distance between H−1and H1 is d. Then ψC(x) ≥ k−ykC,x ≥ −1+_d−c^c =

1+ϕC(x)

1−ϕC(x). But this inequality is equivalent with ϕC(x) ≤ ^ψ(x)−1_ψ(x)+1. Hence the theorem is proved.

Corollary 1.3 Our two measures of asymmetry are equivalent, i.e.

as₂(C) = as₁(C) − 1 as1(C) + 1. and

as₁(C) = 1 + as2(C) 1 − as2(C)

Even though the two measures of aymmetry are equivalent, it is very useful to have both. In some cases it is easier to work with chords, and in other cases affine functionals are more convenient.

Neumann showed in the case n = 2, and S¨uss and Hammer in the general case the following:

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Theorem 1.4 If C is a convex body, then 1 ≤ as1(C) ≤ n.

Moreover, as1(C) = n if and only if C is an n-simplex.

Proof See for example S¨uss.

Remark The corresponding inequalities for the second measure of asymmetry is

0 ≤ as2(C) ≤ n− 1 n+ 1.

The second part of Theorem 1.4 means that the triangle is the most asymmetric convex body in the plane, the tetraeder the most asymmetric body in three-space, and so on.

Finally, both measures of asymmetry are continuous in Dⁿ and Eⁿ (but not in C). This is shown later in Theorem 1.13.

1.8 The Critical Set

For a convex body C, consider the set FC(t) .

= {p ∈ C^o|ϕ(p) ≤ t}. For every t < as₂(C), FC(t) = ∅, and for t ≥ as₂(C), FC(t) is convex and compact. In particular, the set FC(as2(C)) = {p ∈ C|ϕ(p) = as2(C)} is convex and compact. This set is called the critical set of C, and we denote it C^∗. Observe that C^∗ is an affine invariant. In Section 1.11 we show that the map ^∗ : Dⁿ→ C is upper semi-continuous.

Since the critical set of a body is defined as the set of all points fulfilling a special condition involving the critical value, the critical set and the critical value are closely tied together. If we know the critical value, the critical set is (theoretically) easy to find according to the formula

C^∗ = {x|x − C ⊆ as1(C)(C − x)}.

We can also go from the critical set to find the critical value: If x ∈ C^∗, then as₁(C) = inf{t|x − C ⊆ t(C − x)}.

Of course a harder question is how to determine if a given point is a critical point, without knowing the critical value. Is there a simple algorithm of find- ing them? The easiest case is when we have some sort of symmetry involved.

If a body C is symmetric (with respect to a point), then C^∗ consists of only

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one point, namely the point of symmetry. But we can say more. If C is symmetric with respect to a hyperplane, then C^∗ intersected with the plane has to be non-empty (due to the symmetry and the convexity of C^∗). And since C^∗ is an affine invariant, it is enough if T (C) is symmetric for some T ∈ Af f (A(C), A(C)).

If a set contains only one point, we say that the set is trivial. Neumann proved that the critical set of a planar convex body is always trivial, but Hammer/Sobczyk showed that this is not the case in general. Indeed, the triangular prism (Figure 3) is an easy example of a set with non-trivial critical set.

C

^*

Figure 3: The triangular prism The triangular prism is an example of a more general rule:

Theorem 1.5 Let C1 and C2 be two convex bodies. Assume that as2(C1) ≥ as2(C2). Then

as₂(C1× C₂) = as2(C1)

and the critical set of the cartesian product of C₁ and C₂ is (C1× C2)^∗ = C₁^∗× FC2(as2(C1)).

Proof It is enough to establish the equality

ϕC₁×C2((x1, x2)) = max{ϕC₁(x1), ϕC₂(x2)}, since this yields the wanted equalities.

So, fix an x = (x1, x2) and choose an f ∈ AC1 such that f (x1) = ϕC₁(x1). Now form f⁰ by f⁰(x1, x2) = f (x1). Clearly, f⁰ ∈ AC₁×C2. Hence ϕC₁×C2(x1, x₂) ≥ f⁰(x1, x₂) = f (x1) = ϕC₁(x1). A similar argument shows

that ϕC1×C2(x1, x2) ≥ ϕC2(x2), and hence ϕC1×C2((x1, x2)) ≥ max{ϕC1(x1), ϕC2(x2)}.

Next, suppose that a = (a1, a2), b = (b1, b2) such that ab is a chord in C₁× C₂ with the property that _|ax|^|bx| = ψC1×C2(x), where x = (x₁, x₂). Since F r(C1× C₂) = (F rC1× C₂)S

(C1× F rC₂), either a1 ∈ F rC₁ or a2 ∈ F rC₂.

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Assume that a1 ∈ F rC₁. Then ψC₁(x1) ≥ ^|b_|a¹₁^x_x¹₁^|_| = _|ax|^|bx| = ψC₁×C2(x). But this is equivalent to ϕC1×C2((x1, x2)) ≤ ϕC1(x1).

Hence the wanted equality is established.

Remark Since as₁ and as₂are equivalent, a similar result holds for as₁. Since FC2(as2(C1)) in general is non-trivial (as soon as as2(C1) > as2(C2)), also (C1× C₂)^∗ is in general non-trivial. From now on, C⁰, Cⁿ⁰, Eⁿ⁰ and Dⁿ⁰ will denote the subsets of C, Cⁿ, Eⁿ and Dⁿ, respectively, consisting of all convex bodies with non-trivial critival sets.

By using Theorem 1.5, it is easy to construct a convex body C with dim(C^∗) = n − 2. This is also the highest-dimensional critical set which is possible to construct. That follows from a theorem by Klee. He sharpened the result in Theorem 1.4 by showing that a convex body with a critical set of high dimension can not be very asymmetric:

Theorem 1.6 The inequality as1(C) + dim(C^∗) ≤ n holds.

Proof See Klee.

Corollary 1.7 If C is a convex body, then dim(C^∗) ≤ n − 2.

Proof If as1(C) > 1, the inequality holds in view of the theorem, and if as₁(C) = 1, C is symmetric and hence C^∗ is trivial.

In one sense, the critical set can be very large. Namely, for every convex body A, let diam(A) = sup{|xy| : x, y ∈ A}, that is the diameter of A. Set µ(A) = ^diam(A_diam(A)^∗⁾, and µn = sup{µ(A) : A ∈ Cⁿ}. Trivially, µn ≤ 1 (since A^∗ ⊆ A), but when the dimension n increases, µn will grow closer to 1.

Theorem 1.8 The constant µn defined above equals 1-_n² (for n ≥ 2).

Proof Let C1 be an (n-1)-simplex, and let Ia be an interval of length a. If C = C1 × Ia, then as1(C) = as1(C1) = n − 1 and C^∗ = C₁^∗ × FIa(1 − ²_n) (Theorem 1.4 and Theorem 1.5). Hence diam(C^∗) = a−^2a_n. Since diam(C) ≤ diam(C1) + a, we have that µ(C) ≥ _diam(C^a−^2aⁿ

1)+a. By letting a tend to infinity, we get that µn≥ 1 − _n².

To produce the other inequality, we let ` be a longest intervall inside A^∗. We assume that the length of ` is x > 0 (otherwise A^∗ is a single point, and then we are ready) and that the shortest distance from a point in ` to F rA along the direction of ` is b (Figure 4). Theorem 1.6 tells us that as₁(A) ≤ n − 1 (since dim(A^∗) ≥ 1), so we have ^x+b_b ≤ n − 1, which yields x≤ b(n − 2). Now µ(A) ≤ _x+2b^x ≤ _b(n−2)+2b^b(n−2) = _n+2ⁿ = 1 − ²_n.

Hence µn= 1 − _n².

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b C^*

Figure 4:

1.9 Critical Chords

Chords which are divided by a critical point in the critical ratio will be called critical chords. For every x ∈ C^o, let SC(x) = {p ∈ F rC|pq is a chord through x such that ^|px|_|xq| = ρ(x) for some q ∈ F rC}.

Theorem 1.9 Suppose that x ∈ E(C^∗) and y ∈ SC(x). Then y+as₁(C) + 1

as₁(C) (C^∗− y) ⊆ F rC and

y∈ SC(p) for every p ∈ C^∗.

Proof The first statement follows from the second one. So, let y ∈ SC(x) for some x ∈ E(C^∗), and let p ∈ C^∗. Let b be the intersection of the ray −yx→ and F rC, let c be the intersection of the ray −→yp and F rC, and let Hb be a supporting hyperplane to C at b. Since x is an equilibrium point of C^∗, xp has to be parallel to Hb. Since Hb is a supporting plane, c can not be on the side of it away from C. But still c can not be on the same side of Hb as C, because then ^|yp|_|pc| > ^|yx|_|xb| = as₁(C), contradicting the assumptions. Hence c ∈ Hb. But since xp is parallel to Hb, ^yp_pc = ^yx_xb = as1(C), i.e. y ∈ SC(p).

The theorem shows that SC(x) does not vary as x ranges over E(C^∗). There- fore we will denote this set C†. Another important consequence of Theo- rem 1.9 is that if C^∗ is k-dimensional, then F rC contains a k-dimensional ball. In particular, if C^∗ is non-trivial, then F rC contains line-segments.

Neumann showed that there are always at least three different critical chords to a planar convex body. Klee generalized to higher dimensional convex bodies:

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Theorem 1.10 |C † | ≥ as₁(C) + 1.

Proof See Klee.

Corollary 1.11 |C † | ≥ 3.

Proof If as1(C) = 1, then C is symmetric, and hence the statement holds.

If as1(C) > 1, then the theorem tells us that |C † | ≥ 3.

Corollary 1.12 If C has non-trivial critical set, then F rC contains three different segments parallel with a segment contained in C^∗.

Proof Combine Theorem 1.9 and Corollary 1.11.

Klee also showed the following: if M = {f ∈ A|f (x) = −1 for some x ∈ C†}

and for every f ∈ M the set Mf .

= {x|f (x) ≥ 1}, then T

f∈MMf = ∅. A consequence of this and Corollary 1.12 is that if C is a polyhedra with the property that no three different faces have n − 2 directions in common, then C^∗ is trivial. This fact we use in Section 2.1.

1.10 A Description of the Critical Set Due to Hammer

Given a boundary point b of a convex body C, define Cb(r) to be the set C shrinked with a factor r in all directions, b kept fixed, for 0 ≤ r ≤ 1. Observe that Cb(1) = C and if r 6= 1, then Cb(r) ⊆ C. Then define the associated sets C(r) by C(r) =T

b∈F rC Cb(r). Then C(1) = C and for r 6= 1, C(r) ⊆ C.

If r⁰ is the smallest r such that C(r) is non-empty, then as1(C) = _1−r^r⁰0, and C^∗ = C(r⁰).

b

C C_b^(r)

Figure 5: Cb(r)

Remark If we in the definition of as₂ take the interval [0,1] (instead of [-1,1]), then we get r⁰.

Associated sets are best thought of as being the intersection of all possible rC which lie totally inside C. That a point p belongs to C† means that the set Cp(r⁰) “touches” equilibrium points of E(C^∗).

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1.11 Stability in the Critical Set

The critical set appears to be a bit unstable, especially the dimension of it.

It is rather easy to find an example showing that the map^∗ : C → C, which to every convex body associates its critical set, is not continuous with respect to the Hausdorff metric. However, as long as one only considers sequences of bodies which tend to a body of the same dimension (for example sequences in Dⁿ), the map is actually upper semi-continuous, meaning that if Ck tends to C and xk ∈ C_k^∗ tends to x, then x is a critical point of C. (Theorem 1.14.) We start the discussion with an example showing that the condition on the convex bodies to be of the same dimension is necessary.

Example Let Ckbe the right-angled triangle with catheti of length 1 and ¹_k. Then limk→∞Ck is the unit interval, but C_k^∗ does not converge to the mid-

Figure 6:

point but to a point at distance ¹₃ from one of the endpoints of the interval (Figure 6).

The reason why the critical sets of a non-collapsing convergent sequence (meaning that the dimensions of the bodies are the same) do not converge to the critical set of the limit is that the dimension of the critical set seems to be very unstable. Loosely speaking, it is easy to, by small means, destroy a non-trivial critical set. To show that ^∗ : Dⁿ → C is not continuous it sufficies to construct a sequence of polyhedras with no three different faces having a direction in common, converging to a triangular prism. Such a sequence is a sequence of convex bodies with trivial critical sets, but with a limit with non-trivial critical set. Hence ^∗ : Dⁿ→ C can not be continuous.

To prove that the map^∗ : Dⁿ→ C is upper semi-continuous we first prove that our asymmetry measures are continuous.

Theorem 1.13 Suppose that C, D ∈ Cⁿboth contain the same n-dimensional ball B(0, r). Then |as₁(C) − as₁(D)| ≤ ³⁽ⁿ⁺¹⁾_r ²d(C, D).

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Proof It is sufficient to show that |as2(C) − as2(D)| ≤ ⁶_rd(C, D). Let a = d(C, D) and f ∈ EC. Suppose x ∈ A(C) is such that f (x) = b and that the distance between the hyperplanes on which f is -1 and 1, respectively, is d.

Let f⁰ be the affine map in ED which has the same direction as f . Then f⁰(x) ≥ ²⁽

(b+1)d 2 −a)

d+2a − 1 = b −^2a(b+2)_d+2a ≥ b − ^2a(b+2)_r ≥ b − ^6a_r = f (x) −^6a_r. Thus, if y ∈ D^∗ and f is such that f (y) = ϕC(y), then as2(C) ≤ ϕC(y) = f (y) ≤ f⁰(y) + ^6a_r ≤ ϕ(y) + ^6a_r = as2(D) + ^6a_r. Replacing C with D, we get that

|as₂(C) − as2(D)| ≤ ⁶_rd(C, D).

Observe that Theorem 1.13 really tells us that our measures of asymmetry are continuous in Dⁿ. Now we are ready to prove that ^∗ : Dⁿ → C is upper semi-continuous.

Theorem 1.14 Let C, Ck ∈ Cⁿ, k ∈ N, be such that lim_k→∞Ck = C. If C⁰ = {x|∃{xkl}^∞_l=1, xkl ∈ C_k^∗_l such that xkl → x as l → ∞}, then C⁰ ⊆ C^∗. Proof Let > 0. If xk ∈ C_k^∗, xk → x and Ck → C, then x ∈ C_k^o and x ∈ C^o if k > N1 for some large N1. Thus

|ψCk(xk) − ψCk(x)| <

2 if k > N2

for some large N2, and

|ψCk(x) − ψC(x)| <

2 if k > N3

for some large N3. This means that

ψCk(xk) → ψC(x) = as1(C).

But we also have that ψCk(xk) = as1(Ck) which tends to as1(C) according to Theorem 1.13. Hence x ∈ C^∗ and C⁰ ⊆ C^∗.

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2 About Non-Trivial Critical Sets

Interesting cases of convex bodies are the ones with non-trivial critical sets.

In this section we concentrate on the question of how large the set of convex bodies with non-trivial critical sets is. Two questions arise: Is the set of convex bodies with trivial critical sets dense in some space of convex bodies?

Are the convex bodies with non-trivial critical sets dense somewhere?

In the last section, we deduce that Dⁿ⁰ is of the first (Baire) category in Dⁿ, and that Eⁿ⁰ is of the first category in Eⁿ.

2.1 The First Question

In this section we show that C \ C⁰, i.e. the subset of C consisting of convex bodies with trivial critical sets, is dense in C. To do that, we first prove that the polyhedras are dense in C, and then we deduce that close to every polyhedra with non-trivial critical set there is a polyhedra with trivial critical set.

Lemma 2.1 Given > 0 and C ∈ C, there exist finitely many points p₁...pN

in F rC such that d(C, cvx({p1...pN})) < .

Proof For every b ∈ F rC, let Ub, be the open ball with centre in b and radius . Then {Ub,}_{b∈F rC} is an open cover of F rC. Since F rC is compact, there exists a finite subcover, say {Upi,}^N_i=1. Now the points p₁, p₂...pN have the property that no point in F rC has distance more than to every pi. Hence d(C, cvx({p1...pN})) < .

Theorem 2.2 Given > 0 and C ∈ Cⁿ, there exists a convex body A ∈ Cⁿ with d(A, C) < , such that A^∗ is trivial.

Proof Our strategy is to approximate C with a polygon with the property that every triple of faces fail to have n − 2 directions in common.

So, choose points p1...pN in F rC such that d(C, cvx({p1...pN})) < (This is possible according to Lemma 2.1). Let A⁰ = cvx({p1...pN}). Now make, if necessary, a small deformation of A⁰ so that no three different faces have n− 2 directions in common. This can be done by enumerating the (finitely many) faces, say F₁⁰...F_M⁰ . Then, follow the algorithm:

1. Let F1 and F2 be F₁⁰ and F₂⁰ respectively.

2. Let i = 3.

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3. If F_i⁰ does not have n − 2 directions in common with Fj and Fk for any j, k < i, let Fi = F_i⁰.

4. If F_i⁰ have n − 2 directions in common with Fj and Fk for some j, k < i, then let Fi be the face F_i⁰ rotated in a way so that still no point on it is more than from F rC and so that it has not n − 2 directions in common with two faces Fl, Fm for any l, m < i. This can be done since there are only finitely many faces, and hence there are only finitely many “forbidden” directions.

5. Go to 3. with i + 1 instead of i.

Now we have by small deformations of A⁰ created a polyhedra, A, such that d(A, C) < and with the property that no three different faces have n − 2 directions in common. Now Theorem 1.12 tells us that A^∗ is trivial. Hence we are done.

Another way to prove Theorem 2.2 is to show that arbitrarily close to every convex body C there is a convex body D with the property that F rD does not contain any line segments. That can be done by approximating C with an intersection of balls of large radius. The theorem then follows from Theorem 1.9.

2.2 The Second Question

Now to the question whether the set of convex bodies with non-trivial critical sets is dense in some space of convex bodies. It is easily seen that the set C⁰ is not dense in C. To see this, we use Theorem 1.7:

Let C be a convex body with as1(C) > (n − 1). If C⁰ would be dense in C, we could find sets Ck ∈ C⁰ converging to C. Then, according to Theorem 1.13, as1(Ck) → as1(C) > n − 1. But Theorem 1.7 tells us that as1(Ck) ≤ (n − 1), so we have a contradiction.

However, if we allow embeddings of a convex body in a higher-dimensional space, we can find convex bodies with non-trivial critical sets arbitrarily close.

Indeed, given a C, let A = [0, ]. Then D = C × A has non-trivial critical set (Theorem 1.5) (if C is not symmetric; in the symmetric case we can use some asymmetric body C⁰ close to C), and d(C, D) can be made arbitrarily small by choosing small enough.

What we can say about bodies in Cⁿ is that in every -neighborhood of a symmetric body in Cⁿ, n > 2, there is another body in Cⁿ with non-trivial critical set. This is Theorem 2.7. I do not know if there is a convex body with non-trivial critical set close to every C with 1 < as₁(C) ≤ n − 1.

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In the following we often use a construction using a triangular prism circumscribed round a convex body, and a short term for this is convenient.

Definition Let a touching prism to a convex body C mean an intersection of three closed halfspaces H1, H2 and H3 with boundary planes P1, P2 and P3 such that

i) The Pi:s have n − 2 directions in common.

ii) Every Pi is a supporting plane to C.

iii) C ⊆T3 i=1Hi

iv) PjT

Pk⊆T3

i=1Hi for j 6= k.

In the case of n = 2, I use the expression touching triangle.

Condition iv) assures us that a touching prism is infinite in only n-2 directions. In the case n = 3, a touching prism is an infinitely long triangular prism circumscribed round the body (Figure 7).

Figure 7: A touching prism to a 3-dimensional convex body.

We begin the discussion with an illustrative example in three dimensions.

Example Let C ⊂ R³ be the closed three-dimensional ball centred in the origin and with radius 1. Let T3

i=1Hi be a touching prism with boundary planes P1, P2 and P3.

Let the set C be defined by C = ((1 − )(T3

i=1Hi))T

C. So, C is the set C “cut” along the planes (1 − )Pi (Figure 8).

If is small enough, 0 will be a point in the critical set of C, and the critical value will then be as1(C) = ₁₋¹ . This means that the critical set will consist of a short line-segment in the direction which is parallel to all the three planes of the touching prism.

Example 2.2 shows that arbitrarily close to a three-dimensional ball, we can find a set with non-trivial critical set. In proving Theorem 2.7 we use the same construction as in the example, i.e. we shrink a suitable touching prism to a symmetric convex body. A problem is however that the touching prism might, when being shrinked, “cut off” too much from the original convex

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Figure 8: C.

body. We do not want to cut away anything from the critical chords that we create by cutting the body. This means that we need a touching prism with the property that the supports of the six planes ±Pi are seperated in some sense. Lemma 2.4 assures us of the existence of a triangular prism “decent enough”. Lemma 2.4 is easily proved with the help of Lemma 2.3.

Lemma 2.3 For every planar symmetric body C ⊂ R² with C^∗ = {0}, we can find a touching triangle T3

i=1Hi (with F rHi = `i) with the following properties:

i) (−`iT

C) \ (`jT

C) 6= ∅ for i < j ii) (`j

TC) \ (`i

TC) 6= ∅ for i < j.

Proof First, suppose that there are (at least) two non-parallel supporting lines to C which have non-trivial intersection with C. Choose two such lines and name them `1 and `2 respectively. Then let `3 be any supporting line such that the `i:s form the boundary of a touching triangle. (There will be such lines). Since the three lines are pairwise non-parallel, they have the wanted properties.

1 2

3

Figure 9: The case with two non-parallel supporting lines.

Next, suppose that there are at most two supporting lines which have non-trivial intersection with C (such lines come in parallel pairs). Let `1 be

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such that no other supporting line has strictly longer support on C, i.e. if F rC contains a line-segment, `₁ is a supporting line to C with the segment as support. Otherwise `1 is a line with a one-point support on C. Let p1

and p2 be the two points in the intersection which are at the longest distance from each other (clearly, they may coincide). Due to the assumption, neither p1 nor −p2 lies on the supporting line (say `2) parallel to the line through p1

and −p2. Let p3 be the point of touch of `2. Then choose a point p4 on the boundary of C which lies strictly between −p1 and −p3, and let `3 be the supporting line through p4. Then `1, `2 and `3 form a touching triangle with the wanted properties. (See Figure 10.)

p2

p₁

-

p3 -p

2 p4

-p

3

0

2

1 3 1

Figure 10:

Lemma 2.4 For every symmetric body C ⊂ Rⁿ, n ≥ 2, with C^∗ = {0}, we can find a touching prism T3

i=1Hi with the following properties:

i) −(Pi ∩ C) \ (Pj ∩ C) 6= ∅ for i < j.

ii) (Pj

TC) \ (Pi

TC) 6= ∅ for i < j.

Proof If n = 2 this is Lemma 2.3. So, suppose n > 2. Let P r be the orthogonal projection of Rⁿ to the x₁x₂-plane. Identify this plane with R². Since C is symmetric, P r(C) is also symmetric, and 0 ∈ R² is the critical point of P r(C). Find a touching triangleT3

i=1H_i⁰ to P r(C) with the property in Lemma 2.3. Then let Hi = {x∈ Rⁿ|P r(x) ∈ H_i⁰}.

T3

i=1Hi is then a touching prism to C with the wanted properties.

We need two more lemmas.

Lemma 2.5 If in a convex body C, ϕ is larger than ϕ(p) on every point which is not in the affine subspace L (with dim(L) < n), then as₂(C) = ϕ(p), and C^∗ ⊆ L.

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Proof This follows from the continuity of ϕ.

Lemma 2.6 If for some C ∈ Cⁿ, n ≥ 3, there are only three different affine functionals in AC, say f1, f₂ and f3, such that fi(x) = as2(C) for some x∈ C^∗, then C^∗ is non-trivial.

Proof This follows from the fact that C^∗ =T

f∈AC critical{x|f (x) ≤ as²(C)}.

If there are only three different critical affine functionals in AC, then C^∗ is infinite (since the intersection of three halfspaces cannot be bounded if n >2). Hence there is at least one more critical affine functional, and hence C^∗ 6= {x}.

Now we are ready to prove the theorem:

Theorem 2.7 Let > 0 and let C ∈ Cⁿ with n ≥ 3 be a symmetric convex body. Then there exists a convex body D ∈ Cⁿ⁰ such that d(C, D) < .

Proof Our strategy is to create, with the help of a suitable touching prism, a body D (close to C) which fulfills the conditions in Lemma 2.6.

If C^∗ is non-trivial, there is nothing to prove. So, assume that C^∗ is consists of a single point. Without loss of generality we can assume that C ⊂ Rⁿ and C^∗ = {0}. Let T3

i=1Hi be a touching prism to C with the properties in Lemma 2.4. Choose a small ₁ > 0 and let C₁ be the set ((1 − 1)H1)T

C. Then the affine functional f1 which is -1 at −P1 and 1 at (1 − 1)P1, is ¹⁺₁₋¹₁ in the origin. Let the supporting plane to C1 which is

C C₁

C₂ C₃=D

Figure 11:

parallel to P2 be called P₂⁰. Now, let C2 be the set which is obtained by, with the help from P2, cutting off sufficiently much from C1. Namely, choose 2

so that the affine functional f2 which is -1 at P₂⁰ and 1 at (1 − 2)P2 is ¹⁺₁₋¹₁ in the origin. Let C2 be the set ((1 − 2)H2)T

C1.

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Let the supporting plane to C2 which is parallel to P3 be called P₃⁰. Now, construct C₃ in the same way as C₂ was constructed. Namely, let ₃ be such that the affine functional f3 which is -1 at P₃⁰ and 1 at (1 − 3)P3 is ¹⁺₁₋¹

1 in the origin. Let C3 = C2T

(1 − 3)H3.

Now suppose that ₁ is chosen small enough, that is so small so that we do not cut away anything that we do not want to cut away. This means that f₁, f₂, f₃ ∈ AC₃. The two properties of the touching prism guarantess that this is possible. Now there are three different affine functionals on C3 which in the origin all take the value ¹⁺₁₋¹

1. These functionals are constructed so that at least one of them are strictly greater than ¹⁺₁₋¹

1 on every point which is not on the linear subspace L (which is the intersection of the three hyperplanes Li on which fi = ¹⁺₁₋¹₁). Lemma 2.5 tells us that 0 ∈ C^∗. Moreover, these three affine functionals are the only ones in the set AC₃ which in the origin take the value ¹⁺₁₋¹

1. Now Lemma 2.6 gives that dim(C3) ≥ 1. Letting 1 be small enough and letting D = C3, D is the set we are looking for.

2.3 More About Non-Trivial Critical Sets

Let, as in Section 1.4, Dⁿ be the space consisting of n-dimensional convex bodies that contain the ball B(0, 1) and are contained in B(0, n), and let Dⁿ⁰ be the subset of Dⁿ consisting of convex bodies with non-trivial critical sets.

(Dⁿ with the Hausdorff metric is then a complete metric space.)

Also recall that Eⁿ is the space of equivalence classes of n-dimensional convex bodies equipped with the metric dE = ln dBM, and Eⁿ⁰ is the subset consisting of bodies with non-trivial critical sets. Our final goal is to show that Eⁿ⁰ is of the first (Baire) category in Eⁿ, i.e. it is a countable union of nowhere dense sets. To do that we consider convex bodies with non-trivial critical sets in the less complicated space Dⁿ. On our way, we also will be able to conclude that Dⁿ⁰ is of the first category in Dⁿ.

First, let D_tⁿ = {C ∈ Dⁿ⁰|diam(C^∗) ≥ t}.

Theorem 2.8 D_tⁿ is closed in Dⁿ for every t > 0.

Proof Let t > 0, and let {Ck}^∞_k=1 be such that Ck ∈ Dⁿ_t and Ck → C as k → ∞, with C ∈ Dⁿ. We now want to show that C ∈ D_tⁿ. To do that we need to show that diam(C^∗) ≥ t.

For every Ck, choose two critical points ak and bk such that |akbk| ≥ t.

Since all ak:s lie inside a compact set, {ak}^∞_k=1 has a convergent subsequence, say {akl}^∞_l=1, such that akl → a as l tends to infinity. But then {bkl}^∞_l=1 also has a convergent subsequence, say bk_lm → b as m tends to infinity.

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Now, consider the sequence {Ck_lm}^∞_m=1. Since ak_lm → a, bk_lm → b and

|aklmbklm| ≥ t, we have that |ab| ≥ t. Theorem 1.14 yields that a, b ∈ C^∗, and hence C ∈ Dⁿ_t.

Corollary 2.9 Dⁿ⁰ is of the first category in Dⁿ. Proof Theorem 2.8 and Theorem 2.2 yields that Dⁿ1

k

is nowhere dense in Dⁿ, and the statement now follows from the fact that Dⁿ⁰ =S∞

k=1Dⁿ1 k

.

That Eⁿ⁰ is of the first category in Eⁿfollows from the facts that the mapping e : Dⁿ→ Eⁿ defined by e(C) = EC (where EC is the equivalence class of C) is continuous, and that Dⁿ is compact.

Theorem 2.10 Dⁿ is a compact metric space.

Proof Let Bⁿ be the metric space consisting of all distance functions dD(x) of non-empty compact subsets D of B(0, n), equipped with the distance d(f, g) given by d(f, g) = sup_x∈B(0,n){|f (x) − g(x)|}. Notice that there is a 1-1-correspondence between compact sets contained in B(0, n) and elements in Bⁿ.

First we show that Bⁿ is compact. Since every distance function dD is Lipschitz(1), the set Bⁿ is equi-continuous, and since dD(x) ≤ 2n for all x ∈ B(0, n), Arzela-Ascoli ’s theorem yields that every sequence {fk}^∞_k=1 contains a subsequence converging to a continuous function, say fkl → f as l → ∞.

So, what remains is to show that every such limit function f is the distance function for some non-empty compact subset D .

= {x ∈ B(0, n)|f (x) = 0}.

Assume that fkis a sequence of distance functions (to non-empty compact sets Dk) converging to a continuous function f . Since all compact sets Dk

in the convergent sequence are non-empty, every distance function fk in the sequence is zero at some point, say at xk. Since B(0, n) is compact, the sequence xk has a convergent subsequence, say xkl → x. Thus f is 0 at x, which means that D is non-empty. Being the pre-image of a closed set under a continuous map, D is closed, and hence compact. If g is the distance function of D, we need to show that f = g.

Assume that for some y ∈ B(0, n) the equality f (y) + = g(y) holds for some > 0. Then for all k large enough there exist points xk ∈ Dk such that

|xky| ≤ g(y)−₂. The sequence {xk} of points have a convergent subsequence, say xkm → x as m → ∞. Then fkm(x) ≤ |xkmx| < if m large. This means that f (x) = 0 and therefore x ∈ D. Then g(y) ≤ |xy| ≤ g(y) − ₂, which is a contradiction. Hence f (y) ≥ g(y).

Now, suppose that for some y ∈ B(0, n) the equality f (y) = g(y)+ holds for some > 0. This means that there is an x ∈ Bⁿ such that g(x) = 0 and

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with the property that the open ball (B(x,₂))^o does not contain points of any Dk for k sufficiently large. But g(x) = 0 implies f (x) = 0 which means that fk(x) < ₂ if k large enough. But this means that there are xk in Dk

such that |xkx| < ₂, so we have a contradiction.

Hence f (x) = g(x) for all x, and from this it follows that Bⁿ is compact.

Notice now that the subset of Bⁿ consisting of distance functions to compact sets containing B(0, 1) is closed. The same is true about the subset of Bⁿ consisting of distance functions to convex compact sets. Hence the subset of Bⁿ consisting of distance functions to convex compact sets containing B(0, 1) (i.e. Dⁿ) is closed, and since a closed subset of a compact space is compact, the theorem is proved.

Theorem 2.11 The map e : Dⁿ → Eⁿ is continuous.

Proof For every C, D ∈ Dⁿ with d(C, D) < r it is true that C ⊂ D + B(0, r) and B(0, r) ⊂ rD. Therefore C ⊂ (1 + r)D. Similarly, D ⊂ (1 + r)C. Thus C ⊂ (1 + r)D ⊂ (1 + r)²C, i.e. the distance dE(EC, ED) = ln dBM(EC, ED) ≤ 2 ln(1 + r) ≤ 2r is less than > 0 if r is small enough.

Hence e is continuous.

Let a Dⁿ-representative of an equivalence class EC mean a convex body C ∈ EC such that C ∈ Dⁿ. Notice that according to John’s theorem (Theo- rem 1.1), every equivalence class has a Dⁿ-representative.

Let EC be an equivalence class consisting of convex bodies with non-trivial critical sets, and let C1 and C2 be two Dⁿ-representatives of EC. Then there exists an affine map T such that T (C1) = C2. Assume that diam(C_i^∗) = ai. Since the quotient between two segments of a line is preserved by T , a2 ≤ na₁. Similarly, a1 ≤ na₂. Hence it is true that between two different Dⁿ- representatives in an equivalence class, the diameter of the critical set may differ at most by a factor n.

Therefore, if E_tⁿ .

= {EC ∈ Eⁿ|diam(C^∗) ≥ t for some Dⁿ-representative C of EC}, it is true that every equivalence class with non-trivial critical set is in Eⁿ1

k

for some sufficiently large k ∈ N. Hence Eⁿ⁰ =S∞ k=1Eⁿ1

k

. Theorem 2.12 Eⁿ⁰ is of the first category in Eⁿ.

Proof Since e is a continuous function from a compact space Dⁿ to a Haus- dorff space Eⁿ, it is closed. Since e(D_tⁿ) = E_tⁿ, Theorem 2.8 yields that E_tⁿ is closed in Eⁿ for every t > 0.

Since e is continuous, Theorem 2.2 yields that E_tⁿ is nowhere dense in Eⁿ. Now the statement follows from the fact that Eⁿ⁰ =S∞

k=1Eⁿ1 k.

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3 References

• B. Gr¨unbaum. Measures of symmetry for convex sets. Proceedings of Symposia in Pure mathematics Vol. III (1962) 233-270.

• P.C. Hammer. The centroid of a convex body. Proc. Amer. Math.

Soc. 2, (1951) 522-525.

• P.C. Hammer. Convex bodies associated with a convex body. Proc.

Amer. Math. Soc. 2, (1951) 781-793.

• P.C. Hammer and A. Sobczyk. Critical points of a convex body. Ab- stract 112, Bull. Amer. Math. Soc. 57, (1951) 127.

• F. John. Extremum problems with inequalities as subsidiary conditions. Courant Anniv. Volume, (1948) 187-204.

• S. Kaijser and Q. Guo. An estimate of the affine distance between convex sets. (1992)

• V.L. Klee. The critical set of a convex body. Amer. J. Math. 75, (1953) 178-188.

• B.H. Neumann. On some affine invariants of closed convex regions. J.

London. Math. Soc. 14, (1939) 262-272.

• W. S¨uss. ¨Uber eine Affininvariante von Eibereichen. Arch. Math 1, (1948) 127-128.