A REMARK ON THE MAHLER CONJECTURE: LOCAL MINIMALITY OF THE UNIT CUBE

FEDOR NAZAROV, FEDOR PETROV, DMITRY RYABOGIN, AND ARTEM ZVAVITCH

Abstract. We prove that the unit cube B_{∞}^{n} is a strict local minimizer for the Mahler volume product vol_{n}(K)vol_{n}(K^{∗}) in the class of origin symmetric convex bodies endowed with the Banach-Mazur distance.

1. Introduction

In 1939 Mahler [Ma] asked the following question. Let K ⊂ ℝ^{n}, n ⩾ 2, be a convex origin-symmetric body and let

K^{∗} := {ξ ∈ ℝ^{n} : x · ξ ⩽ 1 ∀x ∈ K}

be its polar body. Define P(K) = vol_{n}(K)vol_{n}(K^{∗}). Is it true that we always have

P(K) ⩾ P(B_{∞}^{n}),

where B_{∞}^{n} = {x ∈ ℝ^{n} : |x_{i}| ⩽ 1, 1 ⩽ i ⩽ n}?

Mahler himself proved in [Ma] that the answer is affirmative when n = 2. There are several other proofs of the two-dimensional result, see for example the proof of M. Meyer, [Me2], but the question is still open even in the three-dimensional case.

In the n-dimensional case, the conjecture has been verified for some special classes of bodies, namely, for bodies that are unit balls of Banach spaces with 1-unconditional bases, [SR], [R2], [Me1], and for zonoids, [R1], [GMR].

Bourgain and Milman [BM] (see also [Pi]) proved the inequality

P(K)^{1/n} ⩾ cP(B_{∞}^{n})^{1/n},

with some constant c > 0 independent of n. The best known constant c = π/4 is due to Kuperberg [Ku].

Note that the exact upper bound for P(K) is known:

P(K) ⩽ P(B_{2}^{n}),

where B_{2}^{n} is the n-dimensional Euclidean unit ball. This bound was proved by Santalo [Sa]. In [Pe] and [MeP] it was shown that the equality holds only if K is an ellipsoid.

Let d_{BM}(K, L) = inf{b/a : ∃T ∈ GL(n) such that aK ⊆ T L ⊆ bK} be the Banach-Mazur multiplicative distance between bodies K, L ⊂ ℝ^{n}. In this paper we prove the following result.

Date: January 13, 2011.

1991 Mathematics Subject Classification. Primary: 52A15, 52A21.

Key words and phrases. Convex body, Duality, Mahler Conjecture, Polytopes.

Supported in part by U.S. National Science Foundation grants DMS-0652684, DMS-0800243, DMS-0808908.

Theorem. Let K ⊂ ℝ^{n} be an origin-symmetric convex body. Then

P(K) ⩾ P(B_{∞}^{n}),

provided that d_{BM}(K, B_{∞}^{n}) ⩽ 1 + δ, and δ = δ(n) > 0 is small enough. Moreover, the equality holds only if d_{BM}(K, B_{∞}^{n}) = 1, i.e., if K is a parallelepiped.

Acknowledgment. We are indebted to Mathieu Meyer for valuable discussions.

Notation. Given a set F ⊂ ℝ^{n}, we define af(F) to be the affine subspace of the minimal dimension containing F, and l(F) to be the linear subspace parallel to af(F) of the same dimension. The boundary of a convex body K is denoted by ∂K. For a given set P ⊂ ℝ^{n}, we write P^{⊥} = {x ∈ ℝ^{n} : x · y = 0, ∀y ∈ P }. Let F be the set of all faces F of all dimensions of the cube B_{∞}^{n}. We denote by c_{F} the center of a face F ∈ F. We also denote B_{p}^{n} = {x ∈ ℝ^{n} : ∑_{i} |x_{i}|^{p} ⩽ 1}.

By C and c (with various indices and superscripts) we denote large and small positive constants respectively that may change from line to line and may depend on the dimension n, but on nothing else.

2. Description of the proof

The first difficulty in proving local minimality of the unit cube is that there are plenty of small perturbations with the same volume product, namely all close parallelepipeds. We overcome this difficulty by choosing a “canonical representative” in each class of affinely equivalent convex bodies. More precisely, we consider only the bodies K for which the unit cube is a parallelepiped of the least volume containing K. In addition to taking care of all close parallelepipeds, it allows us to fix 2n points on the boundary of K and K^{∗} (the centers of the (n − 1)-dimensional faces of B_{∞}^{n}). Our next step is to choose several additional points on the boundary of K and K^{∗} and to construct two (not necessarily convex) polytopes P ⊂ K and Q ⊂ K^{∗} such that

vol_{n}(P)vol_{n}(Q) ⩾ P(B_{∞}^{n}) − Cδ^{2},

where δ is the least positive number for which (1−δ)B_{∞}^{n} ⊂ K. We conclude that B_{∞}^{n} is a lower semi-stationary point for the volume product functional P. This means that the perturbation of B_{∞}^{n} by δ in the Banach-Mazur distance may result in decreasing the product volume only by δ^{2}, i.e., in the second order rather than in the first. Our last step is to show that either K contains a point outside (1 + cδ)P or K^{∗} contains a point outside (1 + cδ)Q for some small positive c. This allows us to conclude that P(K) exceeds vol_{n}(P)vol_{n}(Q) by at least cδ and get the final estimate

P(K) ⩾ P(B_{∞}^{n}) + cδ − Cδ^{2}

from which the strict local minimality follows immediately.

It is worth mentioning that the first part of the proof (lower semi-stationarity) works equally well for some other polytopes, for example, for the regular icosahedron and dodecahedron in ℝ^{3}. This indicates that the widely discussed idea to prove the Mahler conjecture by creating some kind of “gradient flow” on the class of convex bodies with respect to the volume product functional may be harder to realize than it seems.

3. Auxiliary results

Note that P(T K) = P(K) for all T ∈ GL(n). We will use this fact for choosing a canonical position for K.

Lemma 1. Let P be a parallelepiped of minimal volume containing a convex originsymmetric body K (note that every such parallelepiped is origin symmetric as well). Let T : ℝ^{n} → ℝ^{n} be a linear transformation such that P = T B_{∞}^{n}. Then T^{−1}K ⊂ B_{∞}^{n} and ±e_{j} ∈ ∂T^{−1}K, j = 1, ..., n.

Proof. Note that B_{∞}^{n} is a parallelepiped of minimal volume containing T^{−1}K. If e_{j} ∉ T^{−1}K, then there exists an affine hyperplane H ∋ e_{j} such that H ∩ T^{−1}K = ∅. Note that the volume of the parallelepiped bounded by H, −H, and the affine hyperplanes {x : x · e_{i} = ±1}, i ≠ j, equals vol_{n}(B_{∞}^{n}), and that this parallelepiped still contains K. But then we can shift H and −H towards K a little bit and a get a new parallelepiped of smaller volume containing K. □

We shall need the following simple technical lemma.

Lemma 2. Let P ⊂ ℝ^{n} be a star-shaped (with respect to the origin) polytope such that every (n−1)-dimensional face F of P has area at least A and satisfies dist(af(F), 0) ⩾ r, where dist denotes the Euclidean distance. Let x ∉ (1 + δ)P for some δ > 0. Then

vol_{n}(conv(P, x)) ⩾ vol_{n}(P) + \frac{δrA}{n}.

Proof. Let y = ∂P ∩ [0, x]. Let F be a face of P containing y. Then conv(P, x) \ P contains the pyramid with base F and apex x. The assumptions of the lemma imply that the height of this pyramid is at least δ dist(af(F), 0) ⩾ δr, so its volume is at least \frac{δrA}{n}. □

If K is sufficiently close to B_{∞}^{n}, then K is also close to any parallelepiped of minimal volume containing K.

Lemma 3. Let K be a convex body satisfying

(1 − δ)B_{∞}^{n} ⊂ K ⊂ B_{∞}^{n}.

Then there exists a constant C and a linear operator T such that

(1 − Cδ)B_{∞}^{n} ⊂ T^{−1}K ⊂ B_{∞}^{n},

and ±e_{i} ∈ T^{−1}K.

Proof. Let as before P = T B_{∞}^{n} be a parallelepiped of minimal volume containing K. Note that vol_{n}(P) ⩽ 2^{n}. On the other hand, if x ∈ P \ (1 + κ)(1 − δ)B_{∞}^{n}, then, by Lemma 2,

vol_{n}(P) ⩾ 2^{n}(1 − δ)^{n} + κ\frac{2^{n−1}}{n} (1 − δ)^{n}.

The right hand side is greater than 2^{n} if κ > κ_{0} = 2n((1 − δ)^{−n} − 1). Thus, P ⊂ (1 + κ_{0})(1 − δ)B_{∞}^{n}, and thereby (1 − κ_{0})P ⊂ (1 − δ)B_{∞}^{n} ⊂ K. It remains to note that κ_{0} ⩽ 4n^{2}δ for sufficiently small δ > 0. □

Thus, replacing K by its suitable linear image we may assume everywhere below that K ⊂ B_{∞}^{n}, ±e_{j} ∈ ∂K, j = 1, …, n. Let δ > 0 be the minimal number such that (1 − δ)B_{∞}^{n} ⊂ K.

4. Computation of the kernel of the differential of the volume function

Choose some numbers a_{k} > 0, k = 0, …, n − 1, and define the polytope Q_{0} as the union of the simplices

S_{𝔽} = conv(0, a_{0}c_{F_{0}}, a_{1}c_{F_{1}}, …, a_{n−1}c_{F_{n−1}}),

where 𝔽 = {F_{0}, …, F_{n−1}} runs over all flags (F_{0} ⊂ F_{1} ⊂ F_{2} ⊂ ⋯ ⊂ F_{n−1}, dimF_{j} = j) of faces of the unit cube.

Choose now some points x_{F} close to x_{F}^{0} = a_{dimF} c_{F} and consider the polytope Q defined in the same way using the points x_{F}. Consider the function g({x_{F} }_{F ∈F}) = vol_{n}(Q). It is just a polynomial of degree n of the coordinates of x_{F}, so it is infinitely smooth.

Lemma 4. If Δx_{F} ∈ ℝ^{n}, Δx_{F} ⊥ c_{F} for all F, then {Δx_{F} } ∈ KerD_{{x_{F}^{0} }}g, where D_{X}g is the differential of g at the point X.

Proof. Since the kernel of the differential is a linear space, it suffices to check this for the vectors {Δx_{F} } in which only one ΔxF̃ ≠ 0. Due to symmetry, we may assume that cF̃ = (\underbrace{1, …, 1}_{k}, \underbrace{0, …, 0}_{n−k}). The space orthogonal to cF̃ is then generated by the

vectors e_{j}, j > k and e_{i} − e_{j}, 1 ⩽ i < j ⩽ k. Note now that the polytopes Q^{+} and Q^{−} built on the points x_{F}^{0} , F ≠ F̃, and x0 ; F̃ ± he_{j}, where j > k, are symmetric with respect to the symmetry e_{j} → −e_{j}, so their volumes are the same. On the other hand, the difference of their volumes in the first order is 2hD_{{x_{F}^{0} }}g({0, …, e_{j}, …, 0}), where e_{j} stands in the position corresponding to F̃ ∈ F. Thus,

D_{{x_{F}^{0} }}g({0, …, e_{j}, …, 0}) = 0.

To prove the equality D_{{x_{F}^{0} }}g({0, …, e_{i} − e_{j}, …, 0}) = 0, consider Q′ and Q′′ built using the points x_{F} = x_{F}^{0} , F ≠ F̃ and xF̃ = x0 ; F̃ + he_{i} or xF̃ = x0 ; F̃ + he_{j} respectively. They are also symmetric with respect to the symmetry e_{i} ↔ e_{j} and the difference of their volumes in the first order equals hD_{{x_{F}^{0} }}g({0, …, e_{i} − e_{j}, …, 0}). □

Below we shall also need the following elementary observation from real analysis.

Lemma 5. Let g(X) be a smooth function on ℝ^{N} and X_{0} ∈ ℝ^{N}. There exists a constant Const depending on g and X_{0} such that for all sufficiently small δ > 0 and all X_{1}, X_{2} ∈ ℝ^{N} satisfying

∥X_{1} − X_{0}∥, ∥X_{2} − X_{0}∥ ⩽ δ and X_{1} − X_{2} ∈ KerD_{X_{0}}g,

one has |g(X_{1}) − g(X_{2})| ⩽ Const δ^{2}.

Proof. Using the Taylor formula, we get

g(X_{j}) = g(X_{0}) + (D_{X_{0}}g) (X_{j} − X_{0}) + O(δ^{2}), where j = 1, 2.

Subtracting these two identities, we obtain

g(X_{1}) − g(X_{2}) = (D_{X_{0}}g) (X_{1} − X_{2}) + O(δ^{2}) = O(δ^{2}),

because (D_{X_{0}}g) (X_{1} − X_{2}) = 0. □

Let P ⊂ ℝ^{n} be a convex polytope. For a face F of P, we define its dual face F^{∗} of P^{∗} by F^{∗} = {y ∈ P^{∗} : x · y = 1 for all x ∈ F } (see Chapter 3.4 in [Gr]).

Lemma 6. Let P be a convex polytope such that 0 is in the interior of P. Let P^{∗} be its dual polytope. Choose some pair of dual faces F and F^{∗} of P and P^{∗} respectively and some points x ∈ F, x^{∗} ∈ F^{∗} in the relative interiors of F and F^{∗}. Assume that K is a convex body satisfying (1 − δ)P ⊂ K ⊂ P. Then there exists a pair of points y ∈ ∂K and y^{∗} ∈ ∂K^{∗} such that y · y^{∗} = 1 and ∥y − x∥, ∥y^{∗} − x^{∗}∥ ⩽ Cδ, where C > 0 does not depend on K or δ, but may depend on P, P^{∗}, F, F^{∗}, x and x^{∗}.

Proof. Since x · x^{∗} = 1 > 0, there exists a self-adjoint positive definite linear operator A such that Ax = x^{∗}. This operator can be chosen as follows: Let L be a 2dimensional plane through the origin containing both x and x^{∗}. A will act identically on L^{⊥}. To define its action on L, choose an orthogonal basis e_{1}, e_{2} in L such that e_{1} = x and put

A|_{L} = ( \begin{matrix}a & b \\ b & a′ \end{matrix} ) ,

where x^{∗} = ae_{1} + be_{2} and a′ > 0 is chosen so large that aa′ > b^{2}.

We will use below the following simple orthogonality relations:

(1) x ⊥ l(F^{∗}).

(2) x^{∗} ⊥ l(F).

(3) l(F) ⊥ l(F^{∗}).

(4) [A^{−1}l(F^{∗})]^{⊥} = span [x^{∗}, Al(F)] and [Al(F)]^{⊥} = span [x, A^{−1}l(F^{∗})].

(5) (x^{∗})^{⊥} ∩ span(x, A^{−1}l(F^{∗})) = A^{−1}l(F^{∗}). (1), (2) and (3) follow directly from the definition of F^{∗} (see Chapter 3.4 in [Gr]). Let us first prove (4). Since l(F) ⊥ l(F^{∗}) and A is self-adjoint, we also have Al(F) ⊥ A^{−1}l(F^{∗}). Also, since x ⊥ l(F^{∗}), we have x^{∗} = Ax ⊥ A^{−1}l(F^{∗}). Thus span(x^{∗}, Al(F)) ⊂ [A^{−1}l(F^{∗})]^{⊥}. On the other hand, x ∉ l(F), so x^{∗} = Ax ∉ Al(F) and

dim (span(x^{∗}, Al(F))) = 1 + dimF = n − dimF^{∗} = n − dimA^{−1}l(F^{∗}),

so A^{−1}l(F^{∗})^{⊥} can not be wider than span(x^{∗}, Al(F)). Similarly,

[Al(F)]^{⊥} = span [x, A^{−1}l(F^{∗})] .

To prove (5), we first note that A^{−1}l(F^{∗}) ⊥ x^{∗} (see (4)). Since x^{∗} · x = 1 ≠ 0, (x^{∗})^{⊥} ∩ span(x, A^{−1}l(F^{∗})) is a subspace of codimension 1 in span(x, A^{−1}l(F^{∗})), so it cannot be wider than A^{−1}l(F^{∗}).

Let K̃ = K ∩ span(x, A^{−1}l(F^{∗})) and let y ∈ K̃ maximize y · x^{∗}. Then y ∈ ∂K and a tangent plane to K at y contains an affine plane parallel to (x^{∗})^{⊥} ∩ span(x, A^{−1}l(F^{∗})) = A^{−1}l(F^{∗}). Therefore, there exists y^{∗} ∈ ∂K^{∗} ∩ [A^{−1}l(F^{∗})]^{⊥} = ∂K^{∗} ∩ span(x^{∗}, Al(F)) such that y · y^{∗} = 1.

Now let y = αx + h and y^{∗} = α^{∗}x^{∗} + h^{∗}, where h ∈ A^{−1}l(F^{∗}) and h^{∗} ∈ Al(F). Note that y · x^{∗} = α, so by maximality of y,

α = (y, x^{∗}) > (0, x^{∗}) = 0.

Also y · y^{∗} = αα^{∗} = 1, so α^{∗} > 0. Let ρ > 0 be such that B(x, ρ) ∩ af(F) ⊂ F and B(x^{∗}, ρ) ∩ af(F^{∗}) ⊂ F^{∗} where B(z, t) is the Euclidean ball of radius t centered at z. Since y ∈ ∂K and

K^{∗} ⊃ P^{∗} ⊃ F^{∗} ∋ x^{∗} + \frac{ρAh}{∥Ah∥},

we have

1 ⩾ y · (x^{∗} + \frac{ρAh}{∥Ah∥}) = α + \frac{ρAh · h}{∥Ah∥} ⩾ α + ρ′∥h∥, where ρ′ = \frac{ρ}{∥A∥∥A^{−1}∥}.

Since y^{∗} ∈ ∂K^{∗} and

K ⊃ (1 − δ)P ⊃ (1 − δ)F ∋ (1 − δ) [x + \frac{ρA^{−1}h^{∗}}{∥A^{−1}h^{∗}∥}] ,

we have

((1 − δ) [x + \frac{ρA^{−1}h^{∗}}{∥A^{−1}h^{∗}∥}]) · y^{∗} ⩽ 1

and

(1 − δ)^{−1} ⩾ [x + \frac{ρA^{−1}h^{∗}}{∥A^{−1}h^{∗}∥}] · y^{∗} = α^{∗} + \frac{ρA^{−1}h^{∗} · h^{∗}}{∥A^{−1}h^{∗}∥} ⩾ α^{∗} + ρ′∥h^{∗}∥.

Thus α ⩽ 1 and α^{∗} ⩽ 1/(1 − δ), which, together with αα^{∗} = 1, gives α ⩾ 1 − δ and α^{∗} ⩾ 1. Hence ρ′∥h∥ ⩽ δ, ρ′∥h^{∗}∥ ⩽ \frac{1}{1−δ} −1 and, thereby, ∥y −x∥, ∥y^{∗} −x^{∗}∥ ⩽ Cδ. □

Remark: Below (in Section 5) we will need Lemma 6 only for the case when x and x^{∗} are collinear. In this case we can choose A to be a pure homothety and get the points y = αx + h and y^{∗} = α^{∗}x^{∗} + h^{∗}, with h ∈ l(F^{∗}) and h^{∗} ∈ l(F).

Now define c_{F}^{∗} = \frac{1}{n−dimF} c_{F}. Choose positive numbers α_{F} and α_{F}^{∗} satisfying α_{F} α_{F}^{∗} = 1 and put y_{F} = α_{F} c_{F}, y_{F}^{∗} = α_{F}^{∗} c_{F}^{∗}.

Let Q = ∪_{𝔽}S_{𝔽}(Q) and Q′ = ∪_{𝔽}S_{𝔽}(Q′), where

S_{𝔽}(Q) = conv(0, y_{F_{0}}, y_{F_{1}}, …, y_{F_{n−1}}) and S_{𝔽}(Q′) = conv(0, y_{F_{0}}^{∗}, y_{F_{1}}^{∗}, …, y_{F_{n−1}}^{∗})

and 𝔽 runs over all flags 𝔽 = {F_{0}, …, F_{n−1}} of faces of B_{∞}^{n}.

Lemma 7.

vol_{n}(Q)vol_{n}(Q′) ⩾ P(B_{∞}^{n}).

Proof. For every flag 𝔽 = {F_{0}, …, F_{n−1}},

vol_{n}(S_{𝔽}(Q)) = vol_{n}(S_{𝔽}(B_{∞}^{n})) ∏_{j=0}^{n−1} α_{F_{j}}, where S_{𝔽}(B_{∞}^{n}) = conv(0, c_{F_{0}}, c_{F_{1}}, …, c_{F_{n−1}}),

and

vol_{n}(S_{𝔽}(Q′)) = vol_{n}(S_{𝔽}(B_{1}^{n})) ∏_{j=0}^{n−1} α_{F_{j}}^{∗}, where S_{𝔽}(B_{1}^{n}) = conv(0, c_{F_{0}}^{∗}, c_{F_{1}}^{∗}, …, c_{F_{n−1}}^{∗}).

Hence,

vol_{n}(S_{𝔽}(Q))vol_{n}(S_{𝔽}(Q′)) = vol_{n}(S_{𝔽}(B_{∞}^{n}))vol_{n}(S_{𝔽}(B_{1}^{n})).

The factors on the right hand side do not depend on the flag 𝔽. Thus,

vol_{n}(Q)vol_{n}(Q′) = ∑_{𝔽}vol_{n}(S_{𝔽}(Q)) ∑_{𝔽}vol_{n}(S_{𝔽}(Q′))

⩾ (∑_{𝔽} \sqrt{vol_{n}(S_{𝔽}(Q))vol_{n}(S_{𝔽}(Q′))})^{2} = (∑_{𝔽} \sqrt{vol_{n}(S_{𝔽}(B_{∞}^{n}))vol_{n}(S_{𝔽}(B_{1}^{n}))})^{2}

= ∑_{𝔽}vol_{n}(S_{𝔽}(B_{∞}^{n})) ∑_{𝔽}vol_{n}(S_{𝔽}(B_{1}^{n})) = vol_{n}(B_{∞}^{n})vol_{n}(B_{1}^{n}) = P(B_{∞}^{n}).

□

5. Lower stationarity of B_{∞}^{n}

Now apply Lemma 6 to B_{∞}^{n} and B_{1}^{n} and the points c_{F} ∈ F and c_{F}^{∗} = \frac{1}{n−dimF} c_{F} ∈ F^{∗}, where F^{∗} is the face of B_{1}^{n} dual to F. Since c_{F} and c_{F}^{∗} are collinear, we get points

x_{F} = α_{F} c_{F} + h_{F} ∈ ∂K and x_{F}^{∗} = α_{F}^{∗} c_{F}^{∗} + h_{F}^{∗} ∈ ∂K^{∗},

where α_{F} α_{F}^{∗} = 1, h_{F} ∈ l(F^{∗}), h_{F}^{∗} ∈ l(F) and |α_{F} − 1|, |α_{F}^{∗} − 1|, ∥h_{F} ∥, ∥h_{F}^{∗} ∥ ⩽ Cδ.

Since ±e_{j} ∈ ∂K and ±e_{j} ∈ ∂K^{∗}, we can choose x_{F} = y_{F} = x_{F}^{∗} = y_{F}^{∗} = c_{F} = c_{F}^{∗} when dimF = n − 1.

Put y_{F} = α_{F} c_{F} and y_{F}^{∗} = α_{F}^{∗} c_{F}^{∗}, and consider the polytopes

P = ∪_{𝔽}conv(0, x_{F_{0}}, …, x_{F_{n−1}}) and P′ = ∪_{𝔽}conv(0, x_{F_{0}}^{∗}, …, x_{F_{n−1}}^{∗}),
Q = ∪_{𝔽}conv(0, y_{F_{0}}, …, y_{F_{n−1}}) and Q′ = ∪_{𝔽}conv(0, y_{F_{0}}^{∗}, …, y_{F_{n−1}}^{∗}).

Note that x_{F} − y_{F} = h_{F}, x_{F}^{∗} − y_{F}^{∗} = h_{F}^{∗} and h_{F} , h_{F}^{∗} ⊥ c_{F}.

Thus by Lemmata 4, 5.

|vol_{n}(P) − vol_{n}(Q)| ⩽ Cδ^{2} and |vol_{n}(P′) − vol_{n}(Q′)| ⩽ Cδ^{2},

whence

vol_{n}(P)vol_{n}(P′) ⩾ vol_{n}(Q)vol_{n}(Q′) − Cδ^{2} ⩾ P(B_{∞}^{n}) − Cδ^{2},

where the last inequality follows from Lemma 7.

Since K ⊃ P and K^{∗} ⊃ P′, it remains to show that for some c > 0, either K ⊄ (1+cδ)P, or K^{∗} ⊄ (1+cδ)P′. Then, by Lemma 2, either vol_{n}(K) ⩾ vol_{n}(P)+c′δ, or vol_{n}(K^{∗}) ⩾ vol_{n}(P′) + c′δ. This yields

P(K) ⩾ P(B_{∞}^{n}) + c′′δ − Cδ^{2} > P(B_{∞}^{n}),

provided that δ > 0 is small enough.

6. The conclusion of the proof

Note that at least one of the coordinates of one of the xF̃ with dimF̃ = 0 is at most 1 − δ. Indeed, assume that all coordinates are greater than (1 − δ′) in absolute value with some δ′ < δ. Define D = conv{x_{F} : F ∈ F, dimF = 0} ⊂ K. Let z ∈ D^{∗}. Choose F so that (x_{F})_{j}z_{j} ⩾ 0 for all j = 1, …, n. Then

1 ⩾ x_{F} · z ⩾ (1 − δ′) ∑_{j} |z_{j}|.

Thus, D^{∗} ⊂ (1 − δ′)^{−1}B_{1}^{n} and D ⊃ (1 − δ′)B_{∞}^{n}, contradicting the minimality of δ.

Due to symmetry, we may assume without loss of generality that F̃ = {(1, …, 1)} and that (xF̃)_{1} ⩽ 1 − δ. Assume that K ⊂ (1 + cδ)P. Consider the point x̃ = (1−δ, c′δ, …, c′δ), where c′ = 1/(n− \frac{5}{4}). Then x̃ ∈ (1−c′′δ)P^{∗}, where c′′ = 1/(4n−5). Indeed, it is enough to check that x̃ · x_{F} ⩽ 1 − c′′δ for all vertices x_{F} of P. If F ≠ {(1, …, 1)}, then all coordinates of x_{F} do not exceed 1 and at least one does not exceed 1/2. Thus, if δ is small enough, we get

x̃ · x_{F} ⩽ (1 − δ) + (n − 2)c′δ + \frac{c′δ}{2} = 1 − δ + (n − \frac{3}{2})c′δ = 1 − c′′δ.

If F = {(1, …, 1)}, then

x̃ · x_{F} ⩽ (1 − δ)^{2} + (n − 1)c′δ = 1 − 2δ + \frac{n − 1}{n − \frac{5}{4}} δ + δ^{2} ⩽ 1 − 2δ + \frac{4}{3}δ + δ^{2} ⩽ 1 − c′′δ,

provided that δ > 0 is small enough. Therefore if c < c′′, we get x̃ ∈ \frac{1}{1+cδ}P^{∗} ⊂ K^{∗}. Now note that for every x ∈ P′, we have

|x_{1}| + (1 − C′δ) ∑_{j⩾2}|x_{j}| ⩽ 1,

provided C′ is chosen large enough. Indeed, again it is enough to check this for the vertices x_{F}^{∗} of P′. If c_{F} ≠ (±1, 0, …, 0) we have ∑_{j⩾2}|(x_{F}^{∗})_{j}| ⩾ 1/3, so

|(x_{F}^{∗})_{1}| + (1 − C′δ) ∑_{j⩾2}|(x_{F}^{∗})_{j}| ⩽ ∑_{j⩾1}|(x_{F}^{∗})_{j}| − C′δ ∑_{j⩾2}|(x_{F}^{∗})_{j}| ⩽ 1 + nCδ − \frac{C′δ}{3} ⩽ 1,

provided that C′ ⩾ 3nC, where C is the constant such that ∥x_{F}^{∗} − c_{F}^{∗} ∥ ⩽ Cδ. If c_{F} = (±1, 0 …, 0), then x_{F} = ±e_{1} and the inequality is trivial.

Now it remains to note that

|x̃_{1}|+(1−C′δ) ∑_{j⩾2}|x̃_{j}| = 1−δ +(1−C′δ)(n−1)c′δ = 1+c′′δ −C′(n−1)c′δ^{2} > 1+cδ,

provided that c < c′′/2 and δ is small enough, whence x̃ ∉ (1 + cδ)P′.

References

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[R1] S. Reisner, Zonoids with minimal volume–product. Math. Zeitschrift 192 (1986), 339–346.

[R2] S. Reisner, Minimal volume product in Banach spaces with a 1-unconditional basis. J. London Math. Soc. 36 (1987), 126–136.

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[SR] J. Saint Raymond, Sur le volume des corps convexes sym etriques. Seminaire d’initiation ‘a l’Analyse, 1980/1981, Publ. Math. Univ. Pierre et Marie Curie, Paris, 1981.

Department of Mathematics, University of Wisconsin, Madison 480 Lincoln Drive Madison, WI 53706, USA

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St. Petersburg Department of Steklov Institute of Mathematics, Fontanka 27, St.Petersburg, 191023, Russia

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Department of Mathematics, Kent State University, Kent, OH 44242, USA

  E-mail address: ryabogin@math.kent.edu

Department of Mathematics, Kent State University, Kent, OH 44242, USA

  E-mail address: zvavitch@math.kent.edu