Extremizers of the Alexandrov–Fenchel inequality within a new class of convex bodies

Lemma A.1

Let K ∈ 𝓚ⁿ be a convex body and let e, f be two linearly independent segments that are summands of K. Then e + f is also a summand of K.

Proof

To show that e + f slides freely in K (see [22, Section 3.2] and in particular Theorem 3.2.2 there), it suffices to consider two-dimensional slices of K parallel to e + f. Hence we can reduce to the case that K is two-dimensional.

Let ± u be the normals of e and ± v the normals of f. As F(e, ± v) are trivial, F(K ⊖ e, ± v) are translates of F(K, ± v). So translates of f are not only contained in F(K, ± v) but also in F(K ⊖ e, ± v). Then [22, Theorem 3.2.11] yields that f is a summand of K ⊖ e. This completes the proof. □

Lemma A.2

The function ζ : 𝓟ⁿ → 𝓟ⁿ that maps a polytope to its unique largest (i.e. inclusion-maximal) zonotope summand, centrally symmetric around the origin, is well-defined. Every zonotope summand of P ∈ 𝓟ⁿ is a summand of ζ(P).

Proof

We show that every polytope P has a unique largest zonotope summand. Let 𝓩(P) denote the nonempty set of origin centered zonotope summands of P. First note that summands of polytopes are polytopes (see [22, p. 157]) and polytopes that are zonoids are zonotopes (see [22, Corollary 3.5.7]). Hence the set of all origin centered zonotopes that are summands of P equals the set of all origin centered zonoids that are summands of P. The latter set is compact as the intersection of a compact set (the set of centered zonoids having mean width less or equal the mean width of P) and a closed set (the set of summands of P). It follows that there is a Z ∈ 𝓩(P) of maximum mean width. This Z is inclusion-maximal in 𝓩(P).

Let Y ∈ 𝓩(P). Then there exist pairwise linearly independent x₁, …, x_k ∈ 𝕊ⁿ⁻¹ and scalars y₁, …, y_k, z₁, …, z_k ≥ 0 such that

Assume for a contradiction that Y is not a summand of Z. Up to reordering of the indices, it follows that y₁ > z₁. Then y₁ [− x₁, x₁] is a summand of P, but, as Z is maximal, not a summand of

Let ℓ ∈ [k] be the largest index such that y₁ [−x₁, x₁] is a summand of P~:=P⊖∑i=2ℓzi[−xi,xi]. Then l < k and z_ℓ+1 [−x_ℓ+1, x_ℓ+1] is also a summand of P͠. Now Lemma A.1 shows that y₁ [−x₁, x₁] + z_ℓ+1 [−x_ℓ+1, x_ℓ+1] is a summand of P͠, but this contradicts the maximality of ℓ. Hence, every Y ∈ 𝓩(P) is a summand of Z, and there is only one maximal zonotope summand of P. □

Lemma A.3

Let X be a separable metric space and f : X → 𝓚ⁿ a function such that for any u ∈ 𝕊ⁿ⁻¹ and λ ∈ ℝ,

is closed. Then f is measurable.

Proof

Fix some countable and dense set Q ⊆ 𝕊ⁿ⁻¹. Let K ∈ 𝓚ⁿ and r > 0. By continuity of h_L for every L ∈ 𝓚ⁿ,

Taking the preimage under f, we get

By hypothesis, h(f(⋅), u) is Borel measurable for every u ∈ 𝕊ⁿ⁻¹. Since Q is countable, f^–1(B(K, r)) is a Borel set. Because balls like B(K, r) generate the Borel σ-algebra of 𝓚ⁿ, this shows that f is measurable. □

Lemma A.4

ζ is a measurable function.

Proof

We apply Lemma A.3. Let u ∈ 𝕊ⁿ⁻¹ and λ ∈ ℝ. It suffices to show that

is closed. Let (P_i) be a sequence in S_ζ(u, λ) that converges to P ∈ 𝓟ⁿ. Applying the Blaschke selection theorem to the bounded sequence of centered zonoids (ζ(P_i)), we find a subsequence (Q_i) that converges to a centered zonoid Z. Then Z is a summand of P. Because summands of polytopes are polytopes and polytopes that are zonoids are zonotopes, it follows that Z is a zonotope. So Z is a summand of ζ(P) and, in particular,

So P ∈ S_ζ(u, λ), proving that the latter set is closed. An application of Lemma A.3 concludes the proof. □

Definition A.5

Let (P_i) be a bounded sequence of (indecomposable) polytopes in ℝ³ with the following properties:

1. All facets are triangles.

2. For every i ∈ ℕ, P_i is a 3-dimensional (i + 3)-tope.

3. For every i ∈ ℕ, no two edges of P_i have the same direction.

4. If i, j ∈ ℕ are distinct and u is a facet normal of P_i, then F(P_j, u) is trivial.

5. If ℓ, i, j ∈ ℕ are distinct and e, f, g are edges of P_ℓ, P_i, P_j, then e + f + g is 3-dimensional. In particular, e + f is 2-dimensional.

6. K:=∑i=1∞Pi is a well-defined convex body.

We call such a sequence admissible and K its associated body.

Remark A.6

If (P_i) is an admissible sequence of polytopes and K is its associated body, then K is a macroid. A suitable choice of a generating measure is given by

which is a probability measure on 𝓟ⁿ with bounded support. Recall that in general μ is not uniquely determined.

Example A.7

Let (P_i)_i∈ℕ be a bounded sequence of polytopes such that P_i is a 3-dimensional (i + 3)-tope having only triangular faces with no edge direction occurring twice. When we apply independent uniform random rotations to each of the P_i, we almost surely obtain an admissible sequence.

Remark A.8

Let K_i, i ∈ ℕ, and K be convex bodies in 𝓚ⁿ. Then K = ∑i=1∞ K_i holds (where the convergence of the partial sums is meant with respect to the Hausdorff metric) if and only if h_K = ∑i=1∞ h_Ki (where the convergence holds pointwise, but then also uniformly on the unit sphere). If u ∈ 𝕊^n–1, then the support sets of K and K_i, i ∈ ℕ, are related by F(K, u) = ∑i=1∞ F(K_i, u).

Remark A.9

Let (P_i)_i∈ℕ be an admissible sequence of polytopes and K its associated body. Items (iii) and (v) imply that if F(K, u) is an edge of K, then there is a unique i ∈ ℕ such that F(P_i, u) is an edge of P_i. In this situation, F(K, u) is a translate of F(P_i, u) and no other edge of any of the polytopes P_j, j ∈ ℕ, is parallel to F(K, u). From item (iv) we conclude that if F(K, u) is a triangular facet, then there is a unique i such that F(K, u) is a translate of F(P_i, u). See Lemma A.14 for further discussion.

Lemma A.10

Let Q ∈ 𝓟ⁿ be a polytope with macroid-generating measure ν on 𝓟ⁿ, that is,

Then supp ν ⊆ 𝓢(Q).

Proof

1. Let β > 0 be a lower bound on the lengths of the edges of Q, and let P ∈ 𝓢(Q) be nontrivial. Then βdiamP P is a summand of Q, as we show first.

   Let F(P, u) be an edge. Since F(P, u) is a scaled summand of F(Q, u), the latter must have an edge e (which is also an edge of Q) homothetic to F(P, u). The length of F(P, u) is at most diam P and the length of e is at least β, so e contains a translate of βdiamP F(P, u). Hence [22, Theorem 3.2.11] implies that βdiamP P is a summand of Q.

2. The set 𝓢(Q) is closed in 𝓟ⁿ, as we show next.

   Let (P_i) be a sequence in 𝓢(Q) converging to some P ∈ 𝓟ⁿ. If P is trivial, then P ∈ 𝓢(Q). Otherwise, there are a sequence of nontrivial polytopes (P_i) and a sequence of polytopes (R_i) such that Q=βdiamPiPi+Ri, and (R_i) must also converge to some R ∈ 𝓚ⁿ such that Q = βdiamP P + R. As R is a summand of P, it must be a polytope. So P ∈ 𝓢(Q).

3. Assume for a contradiction that there is some L ∈ supp ν ∖ 𝓢(Q). Then d := d(L, 𝓢(Q)) > 0 and λ := ν(B(L, d/2)) > 0. Define convex bodies L′ and R by

   hL′:=λ−1∫B(L,d/2)hPν(dP)andhR:=∫B(L,d/2)∁hPν(dP)

   so that Q = λL′ + R. It follows that R is a polytope and L′ ∈ 𝓢(Q), and hence

   d=d(L,S(Q))≤d(L,L′)≤d/2<d,

   which is a contradiction. □

Lemma A.11

For every edge e of a polytope P in ℝ³, there are a normal v ∈ 𝕊² of e and u ∈ ℚ³ ∖ span {e} such that u ⊥ v and e = F(P, v).

Proof

Let U be the relatively open normal cone of the edge e. Then span U is two-dimensional, and U is open in span U. Choose w ∈ 𝕊² ∩ U^⊥, so that span U = w^⊥. Let × denote the cross product. The continuous and surjective map ℝ³ → span U, x ↦ w × x, maps the dense set S := ℚ³ ∖ span {e} ⊆ ℝ³ to the dense set {w} × S ⊆ span U. Because U ∖ {0} ⊆ span U is nonempty and open, there must be some ṽ ∈ (U ∖ {0}) ∩ ({w} × S). By construction, there is u ∈ S such that ṽ = w × u.

Now, v := ∥ṽ∥^–1 ṽ ∈ 𝕊² is an inner normal of e, i.e. e = F(P, v), and is orthogonal to u ∈ S = ℚ³ ∖ span {e}. □

Lemma A.12

There is a set 𝓜 ⊆ 𝓟³ of full μ-measure that satisfies the following property: If some P ∈ 𝓜 has an edge with direction v ∈ 𝕊², then one of the P_i also has an edge in direction v.

Proof

We intend to use Lemma A.11. If u ∈ ℝ³ ∖ {0}, then

and hence the weak continuity and the Minkowski linearity of the area measures imply that

So S1′ (π_u^⊥ K, ⋅) is a discrete Borel measure (i.e., has countable support) in u^⊥ ∩ 𝕊². Denoting by

the set of its atoms, we obtain from special cases of [12, Theorem 2.23, Lemma 3.4] that

So the set

has full μ-measure.

Since for each u ∈ ℚ³ ∖ {0} the set ω_u is countable, the set of pairs

is countable. Using the notation from Remark 2.18, we define

The set 𝓜₂ has full μ-measure. To see this, it is sufficient to consider v ∈ 𝕊² and P ∈ 𝓟³ with F(P, v) ∉ supp F_v(μ) and to show that P ∉ supp μ. By assumption, there is a neighbourhood U′ of F(P, v) with F_v(μ)(U′) = 0, hence μ({P′ : F(P′, v) ∈ U′}) = 0. Since {P′ : F(P′, v) ∈ U′} is a neighbourhood of P, it follows that P ∉ supp μ.

Furthermore, for all P ∈ 𝓜₂ and (u, v) ∈ C, Lemma A.10 shows that F(P, v) is a scaled summand of F(K, v).

Now let 𝓜 := 𝓜₁ ∩ 𝓜₂. Assume P ∈ 𝓜 and that e is an edge of P. By Lemma A.11, there are u ∈ ℚ³ ∖ span e and v ∈ u^⊥ ∩ 𝕊² such that e = F(P, v). In particular, F(π_u^⊥ P, v) is nontrivial and S1′ (π_u^⊥ P, {v}) > 0. Since P ∈ 𝓜₁, it follows that v ∈ ω_u and so (u, v) ∈ C. Since P ∈ 𝓜₂, this implies that e = F(P, v) is a scaled summand of the nontrivial support set F(K, v), which is then either an edge or a parallelogram. If it is an edge, it has the same direction as e and also is an edge of one of the P_i and we are done. If it is a parallelogram, then one of the sides of the parallelogram must have the same orientation as e, and one of the P_i has an edge with this orientation. This concludes the proof. □

Lemma A.13

There is a set 𝓜′ of full μ-measure such that for each P ∈ 𝓜′ and for each u ∈ 𝕊², F(P, u) is a scaled summand of F(K, u).

Proof

1. Let P ∈ 𝓜, where 𝓜 is as in the statement of Lemma A.12. If F(P, u) is trivial, so is the claim. If F(P, u) is an edge, then by Lemma A.12, there is i ∈ ℕ such that F(P, u) is homothetic to the edge F(P_i, u) and hence a scaled summand of F(K, u).

2. Now consider the case that P ∈ 𝓜 and F(P, u) is a facet. The edges of the polytopes P_i, i ∈ ℕ, together have only countably many directions. Denote the countable set of these directions by A ⊂ 𝕊². The facet F(P, u) is incident to (at least) two edges with linearly independent directions v, w ∈ A that determine the facet normal u up to sign σ ∈ {−1, 1} via

   ϕ:{−1,1}×{(a,b)∈A2:a≠±b}→R3,(σ,u,v)↦σu×v∥u×v∥.

   So the facet normals of P are contained in the countable image of ϕ, which is independent of the choice of P.

   For each u ∈ 𝕊², the set F(K, u) is a polytope. Consider the set of full μ-measure

   M3:=⋂u∈imϕF(⋅,u)−1(suppFu(μ)).

   If P is also in 𝓜₃, then by Lemma A.10 and u ∈ imϕ, the support set F(P, u) is a scaled summand of F(K, u).

   Now the assertion follows from (I) and (II) with 𝓜′ := 𝓜 ∩ 𝓜₃. □

Lemma A.14

Let e be an edge of F(K, u) for some u ∈ 𝕊². Then there is a unique i ∈ ℕ such that F(P_i, u) has an edge homothetic to e, e is in fact a translate of F(P_i, u), and this edge is unique among the edges of P_i.

Proof

The uniqueness statements immediately follow from the properties of admissible sequences (P_i). Note that an edge of F(K, u) need not be an edge of K as defined here.

If F(K, u) is a singleton, it does not have any edges.

If F(K, u) is an edge, then there is i ∈ ℕ such that F(P_i, u) is a translate of F(K, u).

If F(K, u) is a triangle, then there is i ∈ ℕ such that F(P_i, u) is a translate of F(K, u). So a translate of e is an edge of F(P_i, u).

If F(K, u) is a parallelogram, then there are unique i, j ∈ ℕ with i ≠ j such that F(P_i, u) is a translate of an edge of P_i, F(P_j, u) is a translate of an edge of P_j and F(P_i, u) + F(P_j, u) is a translate of F(K, u). So e is either a translate of F(P_i, u) or of F(P_j, u). □

Lemma A.15

The polytopes with a nontrivial zonotope summand are contained in a μ-zero set 𝓝.

Proof

Recall the measurable function ζ from Lemmas A.2 and A.4. The macroid Z generated by ζ(μ), the image measure of μ under the map ζ, is a zonoid and summand of K, implying

Because the right-hand side is a discrete measure, so is S₂(Z, ⋅). If we can show that Z has no facets, then S₂(Z, ⋅) is a discrete measure without atoms, hence zero. Then Z is at most one-dimensional. If we can also show that Z has no edges, then Z must be trivial, and the set 𝓝 of polytopes P with nontrivial ζ(P) is a μ-zero set, proving the claim.

It remains to show that Z has no facets or edges. We aim at a contradiction and assume that F(Z, u) is a facet or an edge.

Since F(Z, u) is a summand of the polytope F(K, u), it has an edge e that is homothetic to an edge of F(K, u). Because Z is centrally symmetric around the origin, F(Z, -u) = -F(Z, u) also has an edge that is a translate of e, and therefore F(K, − u) has an edge that is homothetic to e. Hence, F(K, u) and F(K, − u) both contain an edge homothetic to e. By Lemma A.14, and especially the uniqueness statement, there is i ∈ ℕ such that F(P_i, u) and F(P_i, − u) intersect in the very same edge homothetic to e. But this contradicts P_i being 3-dimensional, and Z cannot have edges or facets. This completes the proof. □

Lemma A.16

The generating measure μ is supported in translates of pos ((P_i)_i), the set of translates of finite positive (Minkowski) combinations of polytopes from the sequence (P_i)_i∈ℕ.

Proof

By Lemmas A.13 and A.15, μ is supported in the polytopes P that have no nontrivial zonotope summand and such that for all u ∈ 𝕊², the support set F(P, u) is a scaled summand of F(K, u). Let 𝓜₄ be a set of such polytopes of full μ-measure, and let P ∈ 𝓜₄. For the proof, we may assume that P is nontrivial.

The strategy of proof is to show that there exist nonnegative numbers α_i(P), for i ∈ ℕ, at most finitely many of which are non-zero, such that P͠ := ∑i=1∞ α_i(P)P_i has the following property: All facets of P and P͠ are triangles or parallelograms, the triangular facets of P are translates of the triangular facets of P͠, and vice versa, and a similar fact holds for the facets that are parallelograms. Once this is proved, it follows that P and P͠ are translates of each other.

1. All facets of K are triangles and parallelograms. The only scaled summands of a triangle are homothets of that triangle; the only scaled summands of a parallelogram are (possibly degenerate) parallelograms with the same edge directions but possibly different proportions. So all facets of P are of this kind.

2. Let u ∈ 𝕊² be such that F(P, u) is a triangular facet. Then F(K, u) is homothetic to F(P, u), that is, there are unique α_u > 0 and t_u ∈ ℝⁿ such that F(P, u) = α_u F(K, u) + t_u. Moreover, there are unique i = i(u) ∈ ℕ and tu′ ∈ ℝⁿ such that F(K, u) is a translate of F(P_i, u) and F(P, u) = α_u F(P_i, u) + tu′ .

   Also note that there are at most two triangular facets of P that have an edge parallel to a fixed direction; otherwise, there would be an i ∈ ℕ such that P_i also had more than two such facets, contradicting the hypothesis that P_i has at most one edge parallel to a given direction.

3. We observe that dim P = 3. Recall that P is nontrivial. If dim P = 1, then P is a segment, which is a zonotope, a contradiction. If dim P = 2, then P is a triangle or a non-degenerate parallelogram, which is a zonotope. The latter is excluded. Hence P is a triangle with P = F(P, v) = F(P, − v) for a unit vector v. Let e be an edge of P. Then P_i(v) and P_i(−v) both contain an edge parallel to e. Hence, i := i(v) = i(v) and F(P_i, v) = F(P_i, − v) and thus dim P_i = 2, a contradiction.

4. Let G be the graph with the edges of P as G-vertices, where two edges, i.e. G-vertices, are connected if and only if they are opposite edges in a parallelogram facet of P. Since every edge is only part of two facets, the maximum degree of a G-vertex, i.e. an edge of P, is two. The connected components of G are cycles or chains.

   Let us first make sure that no cycles can occur. Assume that the edge e of P with direction u is part of a cycle. Then π_u^⊥ P is a convex polygon and π_u^⊥ e is one of its vertices. The two edges incident to π_u^⊥ e are projections of parallelograms that connect e to the two neighbors of e in G, and an induction shows that all support sets of π_u^⊥ P either are projections of edges parallel to e or parallelograms connecting two such edges. For the sake of applying [22, Theorem 3.2.22], let F(e, v) be an edge of e. In this case, v ∈ u^⊥, and so e is a summand of F(P, v). Then [22, Theorem 3.2.22] guarantees that e is a summand of P, in contradiction to P having no nontrivial zonotope summand. Therefore, the connected component of any edge e of P is a chain e₁ − … − e_k of e-translates.

   The endpoints of this chain must be edges of two triangular facets of P. By (ii), there can be no other chain with edges parallel to e₁, …, e_k. So if f is an edge parallel to e, then f = e_j for some j ∈ [k] and f is a translate of e. Moreover, for any edge e of P there are exactly two triangular facets of P with an edge parallel to e.

5. Let u, v ∈ 𝕊², and i := i(u) as in (ii), such that F(P, u) is a triangle and F(P_i, v) is a facet adjacent to F(P_i, u) via an edge e. By (iii), there is exactly one w ∈ 𝕊² besides u such that F(P, w) is a triangle with an edge parallel to e. By (ii), F(P_i, w) ≠ F(P_i, u) is then also a triangle with an edge parallel to e. Because P_i contains no other edge parallel to e, it follows that v = w. So we have

   F(P,u)=αuF(Pi,u)+tu,F(P,v)=αvF(Pi,v)+tv.

   By (iii), all edges of P parallel to e are translates of each other, hence it follows that α_u = α_v.

6. Let u, v ∈ 𝕊², and i := i(u) as in (ii), such that F(P, u) and F(P_i, v) are triangles. Then the triangles F(P_i, u) and F(P_i, v) are connected via a chain of neighboring facets. Iteration of (v) shows that F(P, v) is a triangle and α_u = α_v. So α_u only depends on P and i(u), and we set α_i(u)}(P) := α_u > 0. If i ∈ ℕ and P contains no triangular facet F(P, w) with i(w) = i, then we set α_i(P) := 0.

   For each i ∈ ℕ, there are uncountably many edge normals of P_i but only countably many facet normals of K. Let u ∈ 𝕊² be such that F(P_i, u) is an edge and F(K, u) is not a facet and in fact a translate of F(P_i, u). Then for each P ∈ 𝓜₄, F(P, u) is a summand of F(K, u) and satisfies V(F(P, u)) = α_i(P) V(F(P_i, u)). This shows that 𝓜₄ ∋ P ↦ α_i(P) is measurable,

   V(F(K,u))=∫M4V(P,u)μ(dP)=∫M4αi(P)μ(dP)V(F(Pi,u))

   and thus we get

   ∫M4αi(P)μ(dP)=1. (16)

7. Let P͠ := ∑i=1∞ α_i(P)P_i, involving only finitely many nonzero summands. Note that dimP͠ = 3, since α_i(P) > 0 for some i ∈ ℕ. Every facet of P or P͠ is either triangular or a parallelogram. The preceding items show that the triangular facets of P are translates of the triangular facets of P͠, and vice versa.

8. It remains to consider the parallelogram facets. Let u ∈ 𝕊². If F(K, u) is not a parallelogram, it is a singleton, an edge or a triangle. For all P ∈ 𝓜₄, neither F(P, u) nor F(P͠, u) is then a parallelogram.

9. We consider the situation from (viii). From now on, we assume that F(K, u) is a parallelogram. We choose v, w ∈ 𝕊² ∩ u^⊥ such that the edges of F(K, u) are F(F(K, u), ± v) and F(F(K, u), ± w). There are unique distinct i, j ∈ ℕ such that F(F(K, u), ± v) are translates of F(P_i, u) and F(F(K, u), ± w) are translates of F(P_j, u). Let P ∈ 𝓜₄. Then F(P͠, u) is a translate of

   αi(P)F(Pi,u)+αj(P)F(Pj,u),

   which might be a singleton, an edge or a parallelogram. On the other hand, F(P, u) is a translate of

   F(F(P,u),v)+F(F(P,u),w).

10. We consider the situation from (ix) and aim to show that F(P, u) and F(P͠, u) are translates of each other. In the current item, we show that the conclusion holds at least if P is taken from a subset of 𝓜₄ of full measure. The argument will be completed in (xi).

    Let P ∈ 𝓜₄. If F(F(P, u), v) is not a singleton, then it is an edge parallel to F(P_i, u). Hence it must be a translate of α_i(P)F(P_i, u). In either case,

    V(F(F(P,u),v))≤αi(P)V(F(Pi,u)). (17)

    Relation (17) can be used to bound the integrand in

    V(F(Pi,u))=V(F(F(K,u),v))=∫M4V(F(F(P,u),v))μ(dP).

    But then (16) from (vi) implies that equality must hold in (17), for μ-almost all polytopes P ∈ 𝓜₄. Hence, F(F(P, u), v) is a translate of α_i(P) F(P_i, u), and a similar argument shows that F(F(P, u), w) is a translate of α_j(P) F(P_j, u), for μ-almost all P ∈ 𝓜₄. So there is a measurable set 𝓜₅(u) ⊆ 𝓜₄ of full μ-measure such that for all P ∈ 𝓜₅(u), a translate of F(P, u) is

    αi(P)F(Pi,u)+αj(P)F(Pj,u),

    and hence also a translate of F(P͠, u).

11. Finally, set 𝓜₅ := ⋂_u 𝓜₅(u), where we take the countable intersection over all normals of parallelogram facet normals of K. Then 𝓜₅ is a measurable set of full μ-measure and for all P ∈ 𝓜₅ and u ∈ 𝕊², F(P, u) is a parallelogram if and only if F(P͠, u) is, and in this case both are translates of each other: When F(K, u) is a parallelogram, it follows from (x) that F(P, u) and F(P̃, u) are translates, and if it is not, neither of them is a parallelogram due to (viii).

The proof is concluded by an application of Minkowski’s uniqueness theorem for area measures of convex polytopes. □

Lemma A.17

Let (P_i) be an admissible sequence and K its associated body together with a macroid-generating measure μ. Then for all i ∈ ℕ there is P ∈ supp μ such that P_i is a scaled summand of P.

Proof

Let u be the normal of a (necessarily triangular) facet of P_i. Then F(K, u) is a translate of this triangular facet. Clearly, μ is concentrated on supp μ and, according to Lemma A.16, on the set of translates of all finite positive combinations of the P_j, j ∈ ℕ. By (10) we have

hence there is P ∈ pos((P_j)_j) ∩ supp μ such that F(P, u) is nontrivial. This can only be the case if P_i is a scaled summand in the finite positive combination defining P, as F(P_j, u) is trivial for all j ≠ i. □

Theorem A.18

Let (P_i) be an admissible sequence and K its associated body. Then the macroid K is not a polyoid.

Proof

Assume for contradiction that K is a k-polyoid with a generating measure μ supported in the space of k-topes. Lemma A.17 shows that P_k+1 is a scaled summand of some P ∈ supp μ. But then P is not a k-tope (see, e.g., [6, Lemma 2.3]), in contradiction to the property supp μ ⊆ Pk3 of μ. So K is not a k-polyoid for any k ∈ ℕ. □

Remark A.19

Recall that Example A.7 shows that such K really exists.