Antipodal Hadwiger numbers of finite-dimensional Banach spaces

TL;DR

This paper introduces and studies the antipodal Hadwiger numbers for finite-dimensional Banach spaces, providing exact values for Minkowski planes, estimates for classical spaces, and showing exponential growth with dimension.

Contribution

It defines the antipodal Hadwiger numbers, computes these for Minkowski planes, estimates them for ℓ_p^n spaces, and demonstrates exponential growth in dimension.

Findings

H'_ ext{α}(X)=4 for Minkowski planes

Estimates for H_ ext{α}(X) and H'_ ext{α}(X) in ℓ_p^n spaces

H'_ ext{α}(X) grows exponentially with dimension

Abstract

Let $X$ be a finite-dimensional Banach space; we introduce and investigate a natural generalization of the concepts of Hadwiger number $H(X)$ and strict Hadwiger number $H'(X)$. More precisely, we define the antipodal Hadwiger number $H_\alpha(X)$ as the largest cardinality of a subset $S \subseteq S_X$, such that $\forall x \neq y \in S \,\,\, \exists f \in B_{X^*}$ with \[1 \le f(x)-f(y) \,\,\, \textrm{and} \,\,\, f(y) \le f(z) \le f(x) \,\,\, \textrm{for} \,\,\, z \in S.\] The strict antipodal Hadwiger number $H'_\alpha(X)$ is defined analogously. We prove that $H'_\alpha(X)=4$ for every Minkowski plane and estimate (or in some cases compute) the numbers $H_\alpha(X)$ and $H'_\alpha(X)$, where $X=\ell_p^n, 1 < p \le +\infty$ and $n \ge 2$. We also show that the number $H'_\alpha(X)$ grows exponentially in $\dim X$.

Figures

Equations

$$1\le f(x)-f(y)\text{and}f(y)\le f(z)\le f(x)\text{for}z\in S.$$

$$1<f(x)-f(y)\text{and}f(y)\le f(z)\le f(x)\text{for}z\in S.$$

$$-\frac{1}{2}=f(e_j)\le f(\pm e_k)\le f(e_i)=\frac{1}{2}\text{for}k=1,2,\dots ,n$$

$$-1=f(-e_i)\le f(\pm e_k)\le f(e_i)=1\text{for}k=1,2,\dots ,n$$

$$-\frac{1}{2}=f(-e_j)\le f(\pm e_k)\le f(e_i)=\frac{1}{2}\text{for}k=1,2,\dots ,n$$

$$f(e_i-e_j)=\Vert e_i-e_j\Vert =\lambda -\mu \ge d=1+\epsilon .$$

$$p>p_0(n)\Rightarrow 1<n^{1/p}<2(1).$$

$$n^{1/p}<2\iff \log n^{1/p}<\log 2\iff \frac{1}{p}\log n<\log 2\iff p>{\alpha }_n=\frac{\log n}{\log 2}.$$

$$e_k^{\ast }(y)=-\frac{1}{n^{1/p}}\le e_k^{\ast }(z)\le e_k^{\ast }(x)=\frac{1}{n^{1/p}},\text{for}z\in S^{\prime }.$$

$$x_1=\left(\frac{1}{{\alpha }^{1/p}},0,\frac{1}{{\beta }^{1/p}}\right)x_2=\left(-\frac{1}{{\alpha }^{1/p}},0,\frac{1}{{\beta }^{1/p}}\right)$$

$$x_3=\left(0,\frac{1}{{\alpha }^{1/p}},\frac{1}{{\beta }^{1/p}}\right)x_4=\left(0,-\frac{1}{{\alpha }^{1/p}},\frac{1}{{\beta }^{1/p}}\right).$$

$$f_1=e_3^{\ast },f_2=\frac{e_1^{\ast }+e_2^{\ast }}{2^{1/q}}\text{and}{f}_3=\frac{e_1^{\ast }-e_2^{\ast }}{2^{1/q}},$$

$$f_1(x_1)=f_1(x_2)=f_1(x_3)=f_1(x_4)=\frac{1}{{\beta }^{1/p}}\text{and}$$

$$f_1(x_5)=f_1(x_6)=f_1(x_7)=f_1(x_8)=-\frac{1}{{\beta }^{1/p}}.$$

$$f_2(x_1)=f_2(x_3)=f_2(x_6)=f_2(x_8)=\frac{1}{2^{1/q}}\cdot \frac{1}{{\alpha }^{1/p}}>\frac{1}{2^{1/q}}\cdot \frac{1}{2^{1/p}}=\frac{1}{2^{1/p+1/q}}=\frac{1}{2}$$

$$\text{and}{f}_2(x_2)=f_2(x_4)=f_2(x_5)=f_2(x_7)=-\frac{1}{2^{1/q}}\cdot \frac{1}{{\alpha }^{1/p}}<-\frac{1}{2}.$$

$$f_3(x_1)=f_3(x_4)=f_3(x_6)=f_3(x_7)=\frac{1}{2^{1/q}}\cdot \frac{1}{{\alpha }^{1/p}}>\frac{1}{2}\text{and}$$

$$f_3(x_2)=f_3(x_3)=f_3(x_5)=f_3(x_8)=-\frac{1}{2^{1/q}}\cdot \frac{1}{{\alpha }^{1/p}}<-\frac{1}{2}.$$

$$d({\ell }_p^n,{\ell }_{\infty }^n)\le \frac{2}{d}<2,\text{since}d>1.$$

$$n^{1/p}=d({\ell }_p^n,{\ell }_{\infty }^n)\le \frac{2}{d}<2\Rightarrow p>\frac{\log n}{\log 2}={\alpha }_n$$

$$e_c(X)\le e(X)$$

$$H_{\alpha }^{\prime }(X)\ge e_c(X).$$

$$H_{\alpha }^{\prime }(X)\ge e_c(X)\ge \kappa (logn{)}^{\frac{1}{3}}.$$

$$\left\{\begin{array}[]{ll}<w_{i},w_{i}>=1,&i=1,2,\cdots,m\\ |<w_{i},w_{j}>|<\delta,&\textrm{for all distinct }i,j\in\{1,2,\cdots,m\}\\ \end{array}\right.$$

$$f(x)=<x,w_i-w_j>,x\in {\mathbb{R}}^n$$

$$f(w_j)\le f(w_k)\le f(w_i)(1).$$

$$L\subseteq \pi (B_Y)\subseteq (1+\delta )L(1)$$

$$f(z_i)-f(z_j)\ge d\text{ and }f(z_j)\le f(z_k)\le f(z_i),\text{ for }1\le k\le m.(2)$$

$$||z|{|}_{qt}\le ||z|{|}_L\le (1+\delta )||z|{|}_{qt},z\in Z(3)$$

Full text

[TABLE]

The strict antipodal Hadwiger number $H^{\prime}_{\alpha}(X)$ is defined analogously. We prove that $H^{\prime}_{\alpha}(X)=4$ for every Minkowski plane and estimate (or in some cases compute) the numbers $H_{\alpha}(X)$ and $H^{\prime}_{\alpha}(X)$, where $X=\ell_{p}^{n},1<p\leq+\infty$ and $n\geq 2$. We also show that the number $H^{\prime}_{\alpha}(X)$ grows exponentially in $\dim X$.

††2010 Mathematics Subject Classification: Primary 46B20;Secondary 52C99.

Key words and phrases: Hadwiger number, antipodal set, Banach-Mazur distance .

Introduction

If $X$ is any (real) Banach space, then $B_{X}$ and $S_{X}$ denote its closed unit ball and unit sphere respectively. A subset $S$ of a normed space $X$ is said to be $\delta$-separated, if $\|x-y\|\geq\delta$ for $x\neq y\in S$. Specifically $S$ is called equilateral, if there is a $\lambda>0$ such that for $x\neq y\in S$ we have $\|x-y\|=\lambda$; we also call $S$ a $\lambda$-equilateral set. Any equilateral set in an $n$-dimensional space is of cardinality at most $2^{n}$ and the maximum is attained only when $X=\ell_{\infty}^{n}$ (see [17]).

Let $X$ be a finite-dimensional Banach  space. The Hadwiger number $H(X)$ of $X$ is the largest cardinality of a set $S\subseteq S_{X}$ such that $\|x-y\|\geq 1$, for $x\neq y\in S$. Also the strict Hadwiger number $H^{\prime}(X)$ of $X$ is the largest cardinality of a set $S\subseteq S_{X}$ such that $\|x-y\|>1$, for $x\neq y\in S$. It is clear that $H^{\prime}(X)\leq H(X)$. There exists an extensive bibliography on the above concepts (see [19]).

A subset $S$ of a normed space $X$ is said to be antipodal if for every $x,y\in S$ with $x\neq y$ there exists $f\in X^{*}$ such that $f(x)<f(y)$ and $f(x)\leq f(z)\leq f(y)\,\forall z\in S$. That is, for every $x,y\in S$ with $x\neq y$ there exist closed distinct parallel support hyperplanes $P(=\{z\in X:f(z)=f(x)\})$ and $Q(=\{z\in X:f(z)=f(y)\})$ with $x\in P$ and $y\in Q$. Every antipodal subset of an $n$-dimensional real vector space has cardinality at most $2^{n}$ by a result of Danzer and Grünbaum (see [2], also [17]) and this is attained only when the points of the antipodal set are the vertices of an $n$-dimensional parallelotope.

A bounded and separated antipodal subset of a normed space $X$ is a subset $S\subseteq B_{X}$, for which there is $d>0$ such that $\forall x\neq y\in S$ there is $f\in B_{X^{*}}$ with $d\leq f(x)-f(y)$ and $f(y)\leq f(z)\leq f(x)$ for $z\in S$. If X is a finite-dimensional real vector space then this concept of antipodality coincides with the classical one (see [2], [15]). The generalization of antipodality stated above was defined in [14] where the following Theorem (Th. 3 of [14]) was proved

Theorem**.**

Let $(X,\|\cdot\|)$ be a Banach space  and $S\subseteq X$ be a bounded and separated antipodal set with constant $d$. Then we have:

There is an equivalent norm $|||\cdot|||$ on $X$, such that S is an equilateral set in $(X,|||\cdot|||)$. 2. 2.

The Banach-Mazur distance between $(X,\|\cdot\|)$ and $(X,|||\cdot|||)$ satisfies the inequality $\,d((X,\|\cdot\|),(X,|||\cdot|||))\leq\frac{2}{d}$.

The Banach-Mazur distance between two isomorphic Banach spaces $X$ and $Y$ is $d(X,Y)=\inf\{\|T\|\cdot\|T^{-1}\|,\text{where }T:X\rightarrow Y\>\text{ is an isomorphism}\}$. It is easy to see that the dual spaces also have the same Banach-Mazur distance $d(X,Y)=d(X^{*},Y^{*})$. An equivalent (geometric) definition of the Banach-Mazur distance is (in finite dimensions), for $K,L\subseteq\mathbb{R}^{n}$ symmetric convex bodies $d(K,L)=\inf\{r>0:L\subseteq T(K)\subseteq r\cdot L,\text{where}\>T:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}\>\text{is a linear transformation}\}$.

Let $X$ be an $n$-dimensional Banach space. An Auerbach basis of $X$ is a biorthogonal system $\{(e_{i},e^{*}_{i}):i=1,2,\dots,n\}$ in $X\times X^{*}$ (i.e. $e^{*}_{i}(e_{j})=\delta_{ij},\,i,j=1,2,\dots,n$) such that $\{e_{i}:i=1,2,\dots,n\}$ is a basis of $X$ and $\|e_{i}\|=\|e^{*}_{i}\|=1$ for $i=1,2,\dots,n$. It is well known that any finite-dimensional Banach space admits an Auerbach basis (see [3]).

In the present paper we introduce and study some interesting analogues of the Hadwiger and the strict Hadwiger number for a finite-dimensional Banach space, which we call antipodal Hadwiger ($H_{\alpha}(X)$) and strict antipodal Hadwiger number ($H^{\prime}_{\alpha}(X)$). An analogue of the equilateral number $e(X)$, denoted by $e_{c}(X)$ is defined, which roughly is the largest cardinality of an equilateral subset of $X$ with center. The main results are the following:

We prove that $H^{\prime}_{\alpha}(X)=4$ for every Minkowski plane (Prop.4, Remarks 3(1)). 2. 2.

We estimate and in some cases find exact values of the numbers $H_{\alpha}(X)$ and $H^{\prime}_{\alpha}(X)$, when $X=\ell_{p}^{n},1<p\leq+\infty$ and $n\geq 2$ (Th.1). We also show that for an $n$-dimensional Banach space $X$ the number $H^{\prime}_{\alpha}(X)$ increases exponentially with $n$ (Th.4) and the number $e_{c}(X)$ is bounded below by an unbounded function $\varphi(n)$ (Th.3). 3. 3.

We compute the numbers $H_{\alpha}(X)$ and $H^{\prime}_{\alpha}(X)$ when $X=\ell_{1}^{3}$ or $X$ is the Petty space, i.e. $X=(\mathbb{R}^{3},\|\cdot\|)$, where $\|(x,y,z)\|=\sqrt{x^{2}+y^{2}}+|z|$ (Prop.10).

Acknowledgements. The authors wish to thank the anonymous referee for valuable suggestions and remarks, which helped us to complete our initial investigation and greatly contributed to the final form of this paper.

Antipodal Hadwiger number of a finite-dimensional Banach  space

We will always assume that $X$ is a finite-dimensional Banach  space. Given the definition of Hadwiger number and its variants (see [19]) it is natural to introduce the following definitions:

Definition 1**.**

The antipodal Hadwiger number $H_{\alpha}(X)$ is the largest cardinality of a set $S\subseteq S_{X}$ such that $\forall x\neq y\in S$ there is $f\in B_{X^{*}}$ with

[TABLE]

Clearly $S$ is a bounded and separated antipodal  set of unit vectors with $d=1$. In particular $S\subseteq S_{X}$ and $\|x-y\|\geq 1$ for $x\neq y\in S$, hence $H_{\alpha}(X)\leq H(X)$. 2. 2.

The strict antipodal Hadwiger number $H^{\prime}_{\alpha}(X)$ is the largest cardinality of a set $S\subseteq S_{X}$ such that $\forall x\neq y\in S$ there is $f\in B_{X^{*}}$ with

[TABLE]

As above, $S\subseteq S_{X}$ and $\|x-y\|>1$ for $x\neq y\in S$, hence $H^{\prime}_{\alpha}(X)\leq H^{\prime}(X)$.

**Remarks 1

**(1) Since every antipodal subset of an $n$-dimensional real vector space has cardinality at most $2^{n}$, we get that $H^{\prime}_{\alpha}(X)\leq H_{\alpha}(X)\leq 2^{n}$.

(2) If $Y$ is a subspace of $X$, then obviously $H_{\alpha}(Y)\leq H_{\alpha}(X)$ and $H^{\prime}_{\alpha}(Y)\leq H^{\prime}_{\alpha}(X)$.

(3) Let $S$ be an antipodal subset of $S_{X}$ such that $\|x-y\|\geq 1$ for $x\neq y\in S$. Since the space is finite-dimensional, $S$ is a bounded and separated antipodal  set, but the constant $d$ (see Definition 1) may be smaller than 1.

A simple example is a set of three consecutive vertices of a regular hexagon inscribed in the unit circle of $\ell_{2}^{2}$. These form an isosceles and obtuse triangle (with an angle of $120^{\circ}$) with 2 equal sides of length 1. Any functional $f\in B_{\ell_{2}^{2}}$ separating vertices $x,y$ which are at distance 1 gives an evaluation $|f(x)-f(y)|<1$.

(4) Any $\lambda$-equilateral set $S\subseteq B_{X}$ is a bounded and separated antipodal  set with $d=\lambda$ (see [14], Proposition 2). Since the usual basis $S=\{e_{1},e_{2},\dots,e_{n}\}$ of $\ell_{p}^{n},1<p<\infty$ is a $2^{\frac{1}{p}}$-equilateral set and $2^{\frac{1}{p}}>1$, it follows in particular that $H^{\prime}_{\alpha}(\ell_{p}^{n})\geq n$. If $p=1$, then the set $\{\pm e_{k}:k=1,2,\dots,n\}$ is 2-equilateral, hence $H^{\prime}_{\alpha}(\ell_{1}^{n})\geq 2n$.

In the sequel we will obtain lower estimates for the antipodal and the strict antipodal Hadwiger numbers of a finite-dimensional Banach  space.

Proposition 1**.**

Let X be an $n$-dimensional Banach space. Also let $\{(e_{i},e^{*}_{i}):i=1,2,\dots,n\}$ be an Auerbach basis of $X$. Then the set $A=\{\pm e_{i},i=1,2,\dots,n\}$ is a bounded and separated antipodal  subset of $X$ with constant $d=1$ and hence $H_{\alpha}(X)\geq 2n$.

Proof.

We check that $\forall x\neq y\in A$ there is $f\in B_{X^{*}}$ with $1\leq f(x)-f(y)$ and $f(y)\leq f(z)\leq f(x)$ for $z\in A$. We have the following cases:

(1) Let $x=e_{i}$ and $y=e_{j}$, $i\neq j$. Set $f=\frac{e^{*}_{i}-e^{*}_{j}}{2}$, then $\|f\|\leq 1$ and

[TABLE]

(2) Let $x=e_{i}$ and $y=-e_{i}$. Set $f=e^{*}_{i}$, then $\|f\|=1$ and

[TABLE]

(3) Let $x=e_{i}$ and $y=-e_{j}$, $i\neq j$. Set $f=\frac{e^{*}_{i}+e^{*}_{j}}{2}$, then $\|f\|\leq 1$ and

[TABLE]

(4) Let now $x=-e_{i}$ and $y=-e_{j}$, $i\neq j$. This case is similar to (1) and the proof is complete. ∎

In case when the space is smooth we have the following stronger result, which was proved in [7] (Prop. 3.9); see also [8] (Prop. 1.6):

Proposition 2**.**

Let X be an $n$-dimensional smooth Banach space and let $\{(e_{i},e^{*}_{i}):i=1,2,\dots,n\}$ be an Auerbach basis of $X$. Then the set $A=\{\pm e_{i},i=1,2,\dots,n\}$ is a bounded and separated antipodal  subset of $B_{X}$ with constant $d=1+\varepsilon$ for some $\varepsilon>0$. So when $X$ is smooth we have $H^{\prime}_{\alpha}(X)\geq 2n$.

Strengthening our assumption about the basis (supposing it is 1-suppression unconditional) we can prove the following:

Proposition 3**.**

Let X be an $n$-dimensional Banach space and let $\{e_{i},1\leq i\leq n\}$ be a 1-suppression unconditional normalized basis of $X$. If the norm of $X$ is strictly convex (or smooth), then the set $A=\{\pm e_{i},i=1,2,\dots,n\}$ is a bounded and separated antipodal  subset of $B_{X}$ with constant $d=1+\varepsilon$, hence $H^{\prime}_{\alpha}(X)\geq 2n$.

Proof.

Recall that the basis $\{e_{i},1\leq i\leq n\}$ of $X$ is 1-suppression unconditional, if for any $\alpha_{1},\alpha_{2},\dots,\alpha_{n}\in\mathbb{R}$ and $F\subseteq\{1,2,\dots,n\}$ we have $\|\sum_{k\in F}\alpha_{k}e_{k}\|\leq\|\sum_{k=1}^{n}\alpha_{k}e_{k}\|$. Since the basis is normalized, we get that the biorthogonal functionals $\{e^{*}_{i},1\leq i\leq n\}$ are also normalized. In particular $\{(e_{i},e^{*}_{i}):1\leq i\leq n\}$ is an Auerbach basis of $X$.

In both cases we use the fact that $\|e_{i}\pm e_{j}\|>1\,\,\,\text{for }1\leq i<j\leq n$. In any case we have $\|e_{i}\pm e_{j}\|\geq 1$, since $\{(e_{i},e^{*}_{i}):1\leq i\leq n\}$ is an Auerbach basis of $X$.

Let $X$ be strictly convex and assume that $\|e_{i}+e_{j}\|=1$ for some $1\leq i<j\leq n$. Then $e^{*}_{i}(e_{i}+e_{j})=e^{*}_{i}(e_{i})=1$, hence the normalized functional $e^{*}_{i}$ attains its maximum at two distinct points of the unit ball, a contradiction (see [12] §3.2). So $\|e_{i}+e_{j}\|>1$ and similarly $\|e_{i}-e_{j}\|>1$.

Let $d=\min\{\|e_{i}\pm e_{j}\|:1\leq i<j\leq n\}$. Then $d=1+\varepsilon$ for some $\varepsilon>0$. We prove the following:

(I) Given $1\leq i<j\leq n$ there are $\lambda,\mu\in\mathbb{R}$ such that the functional $f=\lambda e^{*}_{i}+\mu e^{*}_{j}$ satisfies $\|f\|\leq 1$, $f(e_{i}-e_{j})=\|e_{i}-e_{j}\|=\lambda-\mu\geq 1+\varepsilon$ and $\lambda=f(e_{i})\geq f(\pm e_{k})\geq f(e_{j})=\mu$, for $1\leq k\leq n$.

To prove this, note that the set $\{e^{*}_{i}:1\leq i\leq n\}$ is also 1-suppression unconditional normalized basis of $X^{*}$ and thus $\|e_{i}-e_{j}\|=\sup\{(xe^{*}_{i}+ye^{*}_{j})(e_{i}-e_{j}):\|xe^{*}_{i}+ye^{*}_{j}\|\leq 1\}$. So there are $\lambda,\mu\in\mathbb{R}$ such that the functional $f=\lambda e^{*}_{i}+\mu e^{*}_{j}$ gives

[TABLE]

It follows that

(a) $-1\leq\lambda,\mu\leq 1$ (since $f(e_{i})=\lambda,f(e_{j})=\mu$, $\|f\|\leq 1$ and $\|e_{i}\|=\|e_{j}\|=1$).

(b) $-1\leq\mu<0<\lambda\leq 1$ (since $\lambda-\mu\geq 1+\varepsilon$).

Consequently $\lambda=f(e_{i})\geq f(\pm e_{k})\geq f(e_{j})=\mu$, for $1\leq k\leq n$.

(II) Given $1\leq i<j\leq n$ there are $\lambda,\mu\in\mathbb{R}$ such that the functional $g=\lambda e^{*}_{i}+\mu e^{*}_{j}$ satisfies $\|g\|\leq 1$, $g(e_{i}+e_{j})=\|e_{i}+e_{j}\|=\lambda+\mu\geq 1+\varepsilon$ and $-\lambda=g(-e_{i})\leq g(\pm e_{k})\leq g(e_{j})=\mu$, for $1\leq k\leq n$.

The proof of this is similar.

(III) For $1\leq i\leq n$ set $f=e^{*}_{i}$. Then we have $-1=f(-e_{i})\leq f(\pm e_{k})\leq f(e_{i})=1$, for $1\leq k\leq n$, which completes the proof that $A$ is a bounded and separated antipodal  subset of $X$.

Let now $X$ be a smooth space. Assume that $\|e_{i}+e_{j}\|=1$ for some $1\leq i<j\leq n$. Then $e^{*}_{i}(e_{i}+e_{j})=e^{*}_{j}(e_{i}+e_{j})=1$, hence the normalized support functional of the vector $e_{i}+e_{j}$ is not unique, which contradicts the smoothness of $X$. So $\|e_{i}+e_{j}\|>1$ and similarly $\|e_{i}-e_{j}\|>1$. The rest of the proof proceeds as in the strictly convex case.

∎

**Remarks 2

**(1) It is clear that the assumption of strict convexity or smoothness in Prop.3 can be replaced by $\|e_{i}\pm e_{j}\|>1$ for $1\leq i<j\leq n$.

(2) An obvious example realizing Prop.3 is the set $\{\pm e_{i}:1\leq i\leq n\}$ in $\ell_{p}^{n},\,1<p<\infty,n\geq 2$.

(3) If $X$ is a Minkowski plane and $x,y\in S_{X}$ with $\|x-y\|=\|x+y\|=1$ then the points $\pm(x+y)$ and $\pm(x-y)$ lie on $S_{X}$ and are the vertices of a parallelogram inscribed in the unit circle. Since $\|\pm x\|=\|\pm y\|=1$, by Lemma 5, p.8 of [12] we get that all the segments joining neighbouring vertices must lie on the unit circle and hence the unit circle itself coincides with the parallelogram with vertices $\pm(x+y),\pm(x-y)$. It follows that $X$ is isometric to $\ell^{2}_{\infty}$. So if $X$ is not isometric to $\ell^{2}_{\infty}$ and we have $x,y\in S_{X}$ with $\|x-y\|=1$, then necessarily $\|x+y\|>1$. Similarly when $x,y\in S_{X}$ with $\|x-y\|=\|x+y\|=2$, it follows that $X$ is isometric to $\ell^{2}_{\infty}$.

Proposition 4**.**

Let $(X,\|\cdot\|)$ be a Minkowski plane. Then $H^{\prime}_{\alpha}(X)=4$.

Proof.

Let $\{(e_{i},e^{*}_{i}):i=1,2\}$ be an Auerbach basis of $X$. Since the vectors $e_{1}$ and $e_{2}$ are mutually orthogonal, the unit ball $B_{X}$ is supported at $e_{1}$ by a line $L_{1}=\{z\in X:e^{*}_{1}(z)=1\}$ parallel to $e_{2}$ and also supported at $e_{2}$ by a line $L_{2}=\{z\in X:e^{*}_{2}(z)=1\}$ parallel to $e_{1}$. The lines $L_{1},L_{2}$ are (non-parallel) sides of the parallelogram with vertices $\{\pm(e_{1}-e_{2}),\pm(e_{1}+e_{2})\}$ (and $\pm e_{1},\pm e_{2}$ are the midpoints of these sides). Since $\{e_{1},e_{2}\}$ is an Auerbach basis, we get that $1\leq\|e_{1}\pm e_{2}\|$ (and of course $\|e_{1}\pm e_{2}\|\leq 2$). If all the vertices of this parallelogram are of norm 2, then $X$ is isometric to $\ell^{2}_{\infty}$ (see Remark 2(3)) and the result follows.

Otherwise there is a pair of opposite vertices of norm less than 2 (say $\|\pm(e_{1}+e_{2})\|<2$). Now draw the diagonals of this parallelogram and take the points of intersection of these diagonals with the unit sphere $S_{X}$ (see Fig.1). We get 4 points (the points $\pm\frac{e_{1}-e_{2}}{\|e_{1}-e_{2}\|},\pm\frac{e_{1}+e_{2}}{\|e_{1}+e_{2}\|}$) which constitute a new parallelogram $ABCD$. This is a bounded and separated antipodal  set of 4 points lying on the unit sphere. Consider the norm one functionals $f$ and $g$ whose kernels are lines parallel to $AB$ and $BC$.

Let $A^{\prime}B^{\prime}C^{\prime}D^{\prime}$ be the homothetic copy of $ABCD$ of factor 2 with respect to the origin. Since $A^{\prime}B^{\prime}C^{\prime}D^{\prime}$ does not touch the unit sphere, $f$ and $g$ give an evaluation $>\frac{1}{2}$ and $<-\frac{1}{2}$ on opposite sides of $ABCD$. Indeed, let $x\in S_{X}$ such that $f(x)=1$, then $x\notin$ Ker$f$, so the half-line $L^{+}=\{\lambda x:\lambda\geq 0\}$ either intersects the line $A^{\prime}B^{\prime}$ or the line $D^{\prime}C^{\prime}$. Let for instance $L^{+}$ intersect $A^{\prime}B^{\prime}$ at $y=\lambda x$ and $AB$ at $z$. Then clearly $\lambda>1$ and $z=\frac{\lambda}{2}x$, thus $f(z)=\frac{\lambda}{2}f(x)=\frac{\lambda}{2}>\frac{1}{2}$. Similarly we get that also $g$ has the desired property. This way we obtain the constant $d>1$ in the definition of bounded and separated antipodal  subset of $X$. ∎

**Remarks 3

**(1) It is easy to see that in any Minkowski plane $(X,\|\cdot\|)$, there is an Auerbach basis $\{(e_{i},e^{*}_{i}):i=1,2\}$ of $X$ such that $\|e_{1}-e_{2}\|>1$ and $\|e_{1}+e_{2}\|>1$. Since any Auerbach basis of a 2-dimensional Banach space is 1-suppression unconditional, Prop.3 and Remark 2(1) imply that the set $A=\{\pm e_{1},\pm e_{2}\}$ is bounded and separated antipodal  with $d>1$, so we get still another proof of Prop.4. It is not known to us if in any $n$-dimensional Minkowski space ($n\geq 3$) there is an Auerbach basis satisfying $\|e_{i}\pm e_{j}\|>1$, $\forall i\neq j$.

In order to prove the 2-dimensional case, one can consider the inscribed centrally symmetric parallelogram of maximum area. If we call two non-opposite vertices $e_{1}$ and $e_{2}$, and if $e_{1}+e_{2}$ is on the unit circle, then, by area maximality, the line through $e_{2}$ parallel to $e_{1}$ supports the unit circle, so the segment from $e_{2}$ to $e_{1}+e_{2}$ lies on the boundary of the unit ball and then the parallelogram with vertices $\pm e_{1}$ and $\pm(e_{1}+e_{2})$ also has maximum area. It follows that the segment from $e_{1}$ to $-e_{2}$ is also on the unit circle (because of a supporting line parallel to $e_{1}+e_{2}$) and the unit ball is an affine regular hexagon. We may now pick any vertice $\alpha$ in the interior of the segment from $e_{1}$ to $-e_{2}$; then the parallelogram with vertices $\pm\alpha$ and $\pm(e_{1}+e_{2})$ (also of maximum area) yields a basis with the desired property.

(2) If $X$ is an $n$-dimensional space with $n\geq 3$ then $H^{\prime}_{\alpha}(X)\geq 4$, since any Minkowski plane admits a bounded and separated antipodal  set establishing this fact (see Prop.4). Moreover one can obtain an equilateral set of 4 points yielding the same result. By a result of V.V. Makeev, in any 3-dimensional space there are 4 equidistant points which are also equidistant from their common barycenter (see [11]). Assuming that the distance of a vertice from the barycenter is 1 (and that the barycenter coincides with the origin) we obtain a $\lambda$-equilateral set of 4 points lying on the unit sphere with $\lambda>1$ (note that the extreme points of the convex hull of a $\lambda$-equilateral set are exactly the points of the equilateral set and the common barycenter has distance $<\lambda$ from each of the extreme points). Taking into account Remarks 1(4) we have the result.

Antipodal Hadwiger number of $\ell_{p}$ spaces, $1<p\leq\infty$ and the exponential growth of $H^{\prime}_{\alpha}(X)$

In this chapter we evaluate the antipodal and strict antipodal Hadwiger numbers of many (finite dimensional) $\ell_{p}$ spaces. Let $\alpha_{n}=\frac{\log n}{\log 2}=\log_{2}n,\,n\geq 2$. This sequence is strictly increasing: $1=\frac{\log 2}{\log 2}=\alpha_{2}<\frac{\log 3}{\log 2}=\alpha_{3}<\frac{\log 4}{\log 2}=\alpha_{4}=2<\dots<\alpha_{n}<\dots$.

Proposition 5**.**

Let $n\geq 2$ and $1<p\leq\infty$, then the following hold:

When $p>\alpha_{n}$, then $H^{\prime}_{\alpha}(\ell_{p}^{n})=2^{n}$ (hence also $H_{\alpha}(\ell_{p}^{n})=2^{n}$). 2. 2.

When $p\geq\alpha_{n}$, then $H_{\alpha}(\ell_{p}^{n})=2^{n}$.

Proof.

Consider the set $S=\{-1,1\}^{n}$ and note that it is an antipodal subset of $\mathbb{R}^{n}$.

Assume first that $1<p<\infty$. Then $\|x\|_{p}=n^{1/p}$, for every $x\in S$. Since $\lim_{p\rightarrow\infty}n^{1/p}=1$ and $1<n^{1/p},\,n\geq 2$ and $p>1$, we get that there is $p_{0}(n)$ such that

[TABLE]

We will show that the least number $p_{0}(n)$ such that inequality (1) holds is $\alpha_{n}$. Indeed, for $n\geq 2$ and $p>1$ we have

[TABLE]

Set $S^{\prime}=\frac{1}{n^{1/p}}\cdot S$, for $p\geq\alpha_{n}$ and $n\geq 2$ and observe that $S^{\prime}$ is a bounded and separated antipodal  subset of $S_{\ell_{p}^{n}}$ with constant $d=\frac{2}{n^{1/p}}\geq 1$. This is true because, given $x\neq y\in S^{\prime}$, $x=(x_{1},x_{2},\dots,x_{n}),\,y=(y_{1},y_{2},\dots,y_{n})$, there is $1\leq k\leq n$ with $x_{k}\neq y_{k}$ such that the numbers $x_{k},y_{k}$ have different signs and $|x_{k}|=|y_{k}|=\frac{1}{n^{1/p}}$. Without loss of generality, let $x_{k}>0$ and $y_{k}<0$. Then

[TABLE]

It is clear from these inequalities that $H^{\prime}_{\alpha}(\ell_{p}^{n})=2^{n}$ for $p\in(\alpha_{n},\infty)$ (since then $d=\frac{2}{n^{1/p}}>1$) and $H_{\alpha}(\ell_{p}^{n})=2^{n}$ for $p=\alpha_{n}$ (since then $n^{1/p}=2$ and thus $d=1$).

In case when $p=\infty$, it is easily verified that the set $S$ itself is bounded and separated antipodal  subset of $S_{\ell_{\infty}^{n}}$ with constant $d=2$ (actually $S$ is a 2-equilateral set, cf. Remark 1(4)), hence $H^{\prime}_{\alpha}(\ell_{\infty}^{n})=2^{n}$.

The proof of the Proposition is now complete. ∎

**Remarks 4

**(1) For $n=2$ we already have the stronger result $H^{\prime}_{\alpha}(X)=4$ for any Minkowski plane, by Proposition 4.

(2) For $n=3$ we get that, when $p>\alpha_{3}=\frac{\log 3}{\log 2}\simeq 1.58$, then $H^{\prime}_{\alpha}(\ell_{p}^{3})=2^{3}=8$. In particular $H^{\prime}_{\alpha}(\ell_{2}^{3})=2^{3}=8$.

The following Proposition settles the situation in case of a 3-dimensional space for the remaining $1<p\leq\alpha_{3}\simeq 1.58$:

Proposition 6**.**

When $1<p<\infty$, then $H^{\prime}_{\alpha}(\ell_{p}^{3})=2^{3}=8$.

Proof.

Let $2<\beta<2^{p}$, it then holds that $\frac{2}{\beta^{1/p}}>1$. Take $\alpha>0$ (and $\alpha<2$) such that $\frac{1}{\alpha}+\frac{1}{\beta}=1\Leftrightarrow\alpha=\frac{\beta}{\beta-1}$. We consider the points

[TABLE]

[TABLE]

Also set $x_{5}=-x_{1},\,x_{6}=-x_{2},\,x_{7}=-x_{3}$ and $x_{8}=-x_{4}$. Since $x_{4}-x_{1}=\left(-\frac{1}{\alpha^{1/p}},-\frac{1}{\alpha^{1/p}},0\right)=x_{2}-x_{3}$, the points $x_{1},x_{2},x_{3},x_{4}$ are vertices of a parallelogram (actually an orthogonal parallelogram). Thus, the points $\pm x_{1},\pm x_{2},\pm x_{3},\pm x_{4}$ are vertices of an orthogonal parallelepiped. Observe that $\|x_{k}\|_{p}=1$, $k=1,2,3,4$, so $\pm x_{k}\in S_{\ell_{p}^{3}}$, $k=1,2,3,4$ and also $\|x_{k}-x_{l}\|_{p}>1$, $1\leq k<l\leq 8$.

Obviously $S=\{x_{k}:k=1,2,\dots,8\}$ is an antipodal subset of $S_{\ell_{p}^{3}}$. We will show that it is bounded and separated antipodal  with constant $d>1$. Consider the functionals

[TABLE]

where $q$ satisfies $\frac{1}{p}+\frac{1}{q}=1$ and $\{e_{1},e_{2},e_{3}\}$ is the usual basis of $\ell_{p}^{3}$, $1<p<\infty$. Note that $f_{1},f_{2},f_{3}\in S_{\ell_{q}^{3}}$ (the unit sphere of the dual space $\ell_{q}=\ell_{p}^{*}$). Geometrically, the kernels of $f_{1},f_{2},f_{3}$ are the planes $z=0,y=-x$ and $y=x$ of $\mathbb{R}^{3}$ respectively.

The following are easy to check:

[TABLE]

[TABLE]

Therefore the points $\{x_{1},x_{2},x_{3},x_{4}\}$ and $\{x_{5},x_{6},x_{7},x_{8}\}$ are separated by planes parallel to $z=0$ and the difference is $\frac{1}{\beta^{1/p}}-\left(-\frac{1}{\beta^{1/p}}\right)=\frac{2}{\beta^{1/p}}>1$. Also

[TABLE]

[TABLE]

So the separation now is achieved by planes parallel to $y=-x$ and the difference is $\frac{1}{2^{1/q}}\cdot\frac{1}{\alpha^{1/p}}-\left(-\frac{1}{2^{1/q}}\cdot\frac{1}{\alpha^{1/p}}\right)=\frac{2}{2^{1/q}\cdot\alpha^{1/p}}>1$. We also have

[TABLE]

[TABLE]

The separation of the points is now achieved by planes parallel to $y=x$ and the corresponding difference is $\frac{1}{2^{1/q}}\cdot\frac{1}{\alpha^{1/p}}-\left(-\frac{1}{2^{1/q}}\cdot\frac{1}{\alpha^{1/p}}\right)=\frac{2}{2^{1/q}\cdot\alpha^{1/p}}>1$.

From the above calculations, we conclude that $d=\min\{\frac{2}{\beta^{1/p}},\frac{2}{2^{1/q}\cdot\alpha^{1/p}}\}>1$ and thus $H^{\prime}_{\alpha}(\ell_{p}^{3})=2^{3}=8,\,\,\forall p>1$.

∎

Using the same method of proof, one can prove the following generalization:

Proposition 7**.**

When $n\geq 3$ and $1<p<\infty$, then $H^{\prime}_{\alpha}(\ell_{p}^{n})\geq 4n-4$.

Proof.

Assigning to the last coordinate the value $\pm\frac{1}{\beta^{1/p}}$ and placing $\pm\frac{1}{\alpha^{1/p}}$ each time in one of the first $n-1$ coordinates with 0 in every other place, one obtains the $4n-4$ required vectors. ∎

**Remark 5

**It is clear from Remark 4(1) that the above result also holds true for $n=2$. Since the Banach space $\ell_{p}^{n},\,1<p<\infty,\,n\geq 2$ is smooth, we get from Prop. 2 that $H^{\prime}_{\alpha}(\ell_{p}^{n})\geq 2n$. Proposition 7 though provides us with a better lower bound, $H^{\prime}_{\alpha}(\ell_{p}^{n})\geq 4n-4$.

Proposition 8**.**

Let $n\geq 4$ ($\alpha_{n}\geq 2$), then the following hold:

When $2\leq p\leq\alpha_{n}$, then $4n-4\leq H^{\prime}_{\alpha}(\ell_{p}^{n})<2^{n}$. 2. 2.

When $2\leq p<\alpha_{n}$, then $4n-4\leq H^{\prime}_{\alpha}(\ell_{p}^{n})\leq H_{\alpha}(\ell_{p}^{n})<2^{n}$. 3. 3.

When $p=\alpha_{n}$, then $4n-4\leq H^{\prime}_{\alpha}(\ell_{p}^{n})<H_{\alpha}(\ell_{p}^{n})=2^{n}$

Proof.

To prove (1), given $p,n$ satisfying $2\leq p\leq\alpha_{n}$ assume that the contrary holds, i.e. $H^{\prime}_{\alpha}(\ell_{p}^{n})=2^{n}$. Then $B_{\ell_{p}^{n}}$ contains a bounded and separated antipodal  subset $S$ with constant $d>1$ and cardinality $|S|=2^{n}$. By Theorem 3 of [14] stated in the Introduction, there is an equivalent norm $|||\cdot|||$ on $\mathbb{R}^{n}$ which admits an equilateral set of cardinality $2^{n}$, hence $(\mathbb{R}^{n},|||\cdot|||)$ is isometric to $\ell_{\infty}^{n}$. Moreover the same Theorem yields for the Banach-Mazur distance of the norms $\|\cdot\|_{p}$ and $|||\cdot|||$ that

[TABLE]

But for $p\geq 2$ we know that $d(\ell_{p}^{n},\ell_{\infty}^{n})=n^{1/p}$, see [4]. Hence

[TABLE]

which contradicts our assumption. Taking into account Proposition 7, the proof of (1) is complete.

The proof of (2) is similar. Concerning (3), the inequalities follow from (1) and the equality follows from Proposition 5(2). ∎

The following Theorem summarizes all the previous results about $\ell_{p}^{n}$, $1<p\leq+\infty$ spaces:

Theorem 1**.**

Let $1<p\leq+\infty$, then the following hold:

When $n=2$ or $3$ , then $H^{\prime}_{\alpha}(\ell_{p}^{n})=4n-4=2^{n}=H_{\alpha}(\ell_{p}^{n})$. 2. 2.

When $n\geq 4$, then we have:

1. (a)

$H^{\prime}_{\alpha}(\ell_{p}^{n})\geq 4n-4$** 2. (b)

when $2\leq p<\alpha_{n}$, then $4n-4\leq H^{\prime}_{\alpha}(\ell_{p}^{n})\leq H_{\alpha}(\ell_{p}^{n})<2^{n}$ 3. (c)

when $p=\alpha_{n}$, then $4n-4\leq H^{\prime}_{\alpha}(\ell_{p}^{n})<2^{n}=H_{\alpha}(\ell_{p}^{n})$, in particular for $n=4$ we get that $p=\alpha_{4}=2$ and $12\leq H^{\prime}_{\alpha}(\ell_{2}^{4})<16=H_{\alpha}(\ell_{2}^{4})$ and 4. (d)

when $p>\alpha_{n}$, then $H^{\prime}_{\alpha}(\ell_{p}^{n})=2^{n}=H_{\alpha}(\ell_{p}^{n})$.

We now introduce the number $e_{c}(X)$ for a finite-dimensional Banach space $X$ and show that it is bounded below by an unbounded function of the dimension of $X$. Since $H^{\prime}_{\alpha}(X)\geq e_{c}(X)$, we have that $H^{\prime}_{\alpha}(X)$ also has this property.

Recall that $e(X)$ (=the equilateral number of $X$) denotes the largest size of an equilateral subset of $X$. We define analogously the number $e_{c}(X)$ to be the largest size of a $\lambda$-equilateral subset of $S_{X}$, where $1<\lambda\leq 2$. It is clear that

[TABLE]

and also by Remark 1(4) that

[TABLE]

Now, one can prove by similar arguments (we omit the details) the following variant of a significant Theorem of Brass and Dekster (see Theorem 8 of [17]):

Theorem 2**.**

Let $X$ be an $n$-dimensional Banach space ($n\geq 2$) with Banach-Mazur distance $d(X,\ell_{2}^{n})\leq 1+\frac{1}{3(n+1)}$. Then any $\lambda$-equilateral set in $S_{X}$, where $\lambda\in\left(1,1+\frac{1}{3(n+1)}\right)$, of at most $n-1$ points can be extended to a $\lambda$-equilateral set in $S_{X}$ of $n$ points.

We can prove using Dvoretzky’s Theorem and Theorem 2, in the same way as Theorem 7 of [17] is proved, that if dim$X=n$ then $e_{c}(X)\geq\kappa(logn)^{\frac{1}{3}}$ for some constant $\kappa>0$ and $n$ sufficiently large.

So from the above observations we get the following:

Theorem 3**.**

Let $X$ be an $n$-dimensional Banach space. Then

[TABLE]

Note. Using the techniques of Swanepoel and Villa in [18] (see also Th.4.3 of [9]) one can show that $H^{\prime}_{\alpha}(X)\geq e_{c}(X)\geq e^{\kappa_{1}\sqrt{logn}}$, for some constant $\kappa_{1}>0$. We also note that for $n=2$ or $3$ the inequality $e_{c}(X)\geq n+1$ holds; for the case $n=2$ we refer the reader to [10], Prop. 1.2 and for $n=3$ to Remarks 3(2). It is not difficult to check that $e_{c}(\ell_{p}^{n})\geq n+1$ for $1<p<+\infty$, $n\geq 1$ and in particular that $e_{c}(\ell_{2}^{n})=n+1$. Also (obviously) $e_{c}(\ell_{1}^{n})\geq 2n$ and $e_{c}(\ell_{\infty}^{n})=2^{n}$.

In the sequel we will prove that $H^{\prime}_{\alpha}(X)$ actually increases exponentially in the dimension of $X$. We need the following Lemma (see [17], Lemma 2), which according to [17] is a special case of the Johnson-Lindenstrauss flattening Lemma and also results from the Gilbert-Varshamov lower bound for binary codes, see also [16] Th.13.

Lemma 1**.**

For each $\delta>0$ there exist $\varepsilon=\varepsilon(\delta)$ and $n_{0}=n_{0}(\delta)\geq 1$ such that for all $n\geq n_{0}$ there exist $m>(1+\varepsilon)^{n}$ vectors $w_{1},\cdots,w_{m}\in\mathbb{R}^{n}$ ($w_{i}=\frac{x_{i}}{\sqrt{n}},\,x_{i}\in\{-1,1\}^{n}$) satisfying

[TABLE]

We may take $\varepsilon=\frac{\delta^{2}}{2}$ and $n_{0}\geq\frac{120log2}{25\delta^{4}-\delta^{6}}$.

We first prove the result for the spaces $\ell_{2}^{n}$. The essential part of this proof is contained in the proof of Th.2 in [17].

Proposition 9**.**

Let $X=\ell_{2}^{n}$, then the number $H^{\prime}_{\alpha}(X)$ increases exponentially in the dimension of $X$.

Proof.

Let $0<\delta\leq\frac{1}{3}$, then by Lemma 1 there exist $\varepsilon=\varepsilon(\delta)>0$ and $n_{0}=n_{0}(\delta)\geq 1$ such that for all $n\geq n_{0}$ there exist $m>(1+\varepsilon)^{n}$ vectors $w_{1},\cdots,w_{m}\in\mathbb{R}^{n}$ satisfying the two conclusions. We will prove that the set $\{w_{1},\cdots,w_{m}\}$ is bounded and separated antipodal  subset of $S_{\ell_{2}^{n}}$ with constant $d>\sqrt{2-2\delta}\,(\geq\sqrt{\frac{4}{3}}>1)$.

By the first conclusion of Lemma 1 we have $||w_{i}||_{2}=1$ for $i=1,\cdots,m$. For any $i\neq j\in\{1,2,\cdots,m\}$ we define the linear functional

[TABLE]

with $||f||_{2}=||w_{i}-w_{j}||_{2}$.

Let $k\in\{1,2,\cdots,m\}$ with $k\neq i,j$, we will prove that

[TABLE]

It suffices to prove that

$\max\{|f(w_{i}-w_{k})|,|f(w_{j}-w_{k})|:k=1,2,\cdots,m\text{ and }k\neq i,j\}\leq f(w_{i}-w_{j})=||w_{i}-w_{j}||_{2}^{2}$. Setting $\alpha=<w_{i},w_{j}>$ (and taking into account Lemma 1 and the fact that $0<\delta\leq\frac{1}{3}$) we have $f(w_{i}-w_{k})=<w_{i}-w_{k},w_{i}-w_{j}>=<w_{i},w_{i}>-<w_{i},w_{j}>-<w_{k},w_{i}>+<w_{k},w_{j}>\,<1-\alpha+\frac{2}{3}=\frac{5}{3}-\alpha$.

Also $f(w_{i}-w_{j})=<w_{i}-w_{j},w_{i}-w_{j}>=<w_{i},w_{i}>+<w_{j},w_{j}>$ $-2<w_{i},w_{j}>=2-2\alpha$. Since $|\alpha|=|<w_{i},w_{j}>|<\delta\leq\frac{1}{3}$, it holds that $\frac{5}{3}-\alpha<2-2\alpha$. Similarly $-f(w_{i}-w_{k})=\alpha-\frac{5}{3}<2-2\alpha$, hence $|f(w_{i}-w_{k})|<f(w_{i}-w_{j})$. The other inequality required can be proved analogously and so (1) holds true.

Now setting $g=\frac{f}{||f||_{2}}$ we have $||g||_{2}=1$ and

$g(w_{i}-w_{j})=\frac{1}{||f||_{2}}f(w_{i}-w_{j})=\frac{1}{||w_{i}-w_{j}||_{2}}||w_{i}-w_{j}||_{2}^{2}=||w_{i}-w_{j}||_{2}=\sqrt{2-2<w_{i},w_{j}>}>\sqrt{2-2\delta}\geq\sqrt{\frac{4}{3}}>1$ and the proof of the Proposition is complete. ∎

**Remark 6

**The proof of Proposition 9 implies that, choosing $0<\delta\leq\frac{1}{3}$ small enough, the distance $||w_{i}-w_{j}||_{2}=\sqrt{2-2<w_{i},w_{j}>}>\sqrt{2-2\delta}$ (for $i\neq j$) becomes arbitrarily close to $\sqrt{2}$. Also note that the set $\{w_{1},\cdots,w_{m}\}$ is strictly antipodal, since the inequalities (1) of Prop.9 are strict.

Using the previous Proposition and an important result of Milman (the Quotient of Subspace Theorem, see [13], also [6]) the way it was used by Bourgain in [5], Th. 4.3, we are now in a position to prove the general result:

Theorem 4**.**

Let $X$ be a finite-dimensional Banach space, then the number $H^{\prime}_{\alpha}(X)$ increases exponentially in the dimension of $X$.

Proof.

From Milman’s result there is a function $\psi:(0,+\infty)\rightarrow(0,+\infty)$, such that for all $\delta>0$ and for every Banach space $X$ with dim$X=n$, there exist subspaces $Z\subseteq Y\subseteq X$ satisfying dim$Z\geq\psi(\delta)n$ and there is an ellipsoid $L\subseteq Z$, with

[TABLE]

where $\pi:Y\rightarrow Z$ is the orthogonal projection of $Y$ onto $Z$. (There exists a proper euclidean structure on $X$ used in the proof of Milman’s Theorem. We consider the orthogonal projection $P_{Z}:Y\rightarrow Z$ with respect to this structure, which algebraically coincides with the quotient map $\pi:Y\rightarrow Y/Z^{\perp}$; hence $\pi(B_{Y})$ induces the quotient norm $||z||_{qt}=\inf\{||z-z_{0}||:z_{0}\in Z^{\perp}\}$ on $Z$ which is isomorphic with $Y/Z^{\perp}$, see [6] Ths. 5.2.1, 5.3.1 and [20], Prop. 2.3).

Assume that $\delta\in\left(0,\frac{1}{3}\right)$. Then by the previous Proposition there is $\varepsilon=\varepsilon(\delta)>0$, such that for $n\in\mathbb{N}$ large enough there exist $m\geq(1+\varepsilon)^{\psi(\delta)n}$ vectors $z_{1},z_{2},\cdots,z_{m}$ on the boundary of the euclidean unit ball $L$ in $Z$ with the set $S=\{z_{1},z_{2},\cdots,z_{m}\}$ being a bounded and separated antipodal  set with constant $d>\sqrt{2-2\delta}$. This means that for $i\neq j\in\{1,2,\cdots,m\}$ there is a linear functional $f\in Z^{*}$ with $||f||_{L}=1$ and such that

[TABLE]

Now (1) is equivalent to

[TABLE]

which implies for the dual norms that

[TABLE]

It is clear that for $i\leq m$ there is $y_{i}\in B_{Y}$ such that $\pi(y_{i})=z_{i}$. We note that we can select $y_{i}\in S_{Y}$ so that $\pi(y_{i})=z_{i}$ holds (if $z_{i}\in S_{Y}$ then set $y_{i}=z_{i}$, otherwise for any $z_{0}\in Z^{\perp}\setminus\{0\}$, there is $\lambda>0$ so that $y_{i}=z_{i}+\lambda z_{0}\in S_{Y}$ and $\pi(y_{i})=z_{i}$).

Next, we will prove that the set $S^{\prime}=\{y_{1},y_{2},\cdots,y_{m}\}$ is a bounded and separated antipodal  subset of $S_{Y}$ (hence also of $S_{X}$) with constant $\geq\frac{\sqrt{2-2\delta}}{1+\delta}$. Let $1\leq i\neq j\leq m$. Take an $f\in Z^{*}$ with $||f||_{L}=1$ such that the inequalities in (2) hold and set $g=f\circ\pi$. Then (using (4) and $||\pi||=1$) we have

[TABLE]

Observe that, since $g(y_{k})=f(\pi(y_{k}))=f(z_{k}),\,1\leq k\leq m$, from (2) we have

[TABLE]

Also $g(y_{i})-g(y_{j})=f(\pi(y_{i}))-f(\pi(y_{j}))=f(z_{i})-f(z_{j})\geq d$. Relation (5) implies that

[TABLE]

For $0<\delta<\sqrt{5}-2$ we have that $\frac{\sqrt{2-2\delta}}{1+\delta}>1$ (and $\frac{\sqrt{2-2\delta}}{1+\delta}\rightarrow\sqrt{2}$ when $\delta\rightarrow 0,\,\delta>0$). The proof is now complete.

∎

We also investigated the spaces $\ell_{1}^{3}$ and the Petty space on $\mathbb{R}^{3}$ (see [17], [19]) with respect to their (strict) antipodal Hadwiger numbers. In both spaces, the strict antipodal Hadwiger numbers are as big as possible:

Proposition 10**.**

$H_{\alpha}(\ell_{1}^{3})=H^{\prime}_{\alpha}(\ell_{1}^{3})=2^{3}=8$. 2. 2.

If $(X,\|\cdot\|)$ is the Petty space on $\mathbb{R}^{3}$, where $\|(x,y,z)\|=\sqrt{x^{2}+y^{2}}+|z|$, then $H_{\alpha}(X)=H^{\prime}_{\alpha}(X)=2^{3}=8$.

Proof.

To prove (1), set $x_{1}=\left(1,1,-\frac{1}{3}\right),\,x_{2}=\left(1,-\frac{1}{3},1\right),\,x_{3}=\left(-\frac{1}{3},1,1\right)$ and $O=$conv $\{\pm x_{1},\pm x_{2},\pm x_{3}\}$. The Minkowski functional of $O$ defines a norm and the corresponding space is isometric to $\ell_{1}^{3}$ through the linear isometry designated by $T(e_{i})=x_{i},\,i=1,2,3$ ($\{e_{1},e_{2},e_{3}\}$ is the usual basis of $\ell_{1}^{3}$). Let also $C_{3}=B_{\ell_{\infty}^{3}}=[-1,1]^{3}$. In [21] Fei Xue observed that the octahedron $O$ and the cube $C_{3}$ satisfy

[TABLE]

and obtained for the Banach-Mazur distance of the spaces $\ell_{1}^{3}$ and $\ell_{\infty}^{3}$ the upper bound $d(\ell_{1}^{3},\ell_{\infty}^{3})\leq\frac{9}{5}<2$.

The set $S=\frac{5}{9}\{-1,1\}^{3}$ is the set of vertices of a parallelepiped and all of its points belong to the boundary of the octahedron $O$ (hence are vectors of $\ell_{1}$-norm 1). One can readily check that

[TABLE]

[TABLE]

and the other points are the symmetric of these 4. We will show that $S$ is a bounded and separated antipodal  subset of $S_{\ell_{1}^{3}}$ with constant $d=\frac{10}{9}>1$.

The functionals separating opposite faces of the parallelepiped are the $e_{i}^{*},\,i=1,2,3$ (since $O\subseteq C_{3}$, we have that $|e_{i}^{*}(x,y,z)|\leq 1$ for any $(x,y,z)\in O$, hence $e_{i}^{*}\in B_{(\ell_{1}^{3})^{*}}$). For instance

[TABLE]

[TABLE]

and the evaluations for the other faces are similar (see also the proof of Proposition 6), which implies that $d=\frac{5}{9}-(-\frac{5}{9})=\frac{10}{9}>1$.

Now for the proof of (2) one may consider the points $A(-0.18,0,0.82)$, $B(0.82,0,-0.18)$, $C(0.32,0.6,0.32)$ and $D(0.32,-0.6,0.32)$. These points lie on the unit sphere of the Petty space and form a parallelogram, as $\overrightarrow{AC}=\overrightarrow{DB}=(0.5,0.6,-0.5)$. So these points together with the symmetric points $A^{\prime},B^{\prime},C^{\prime},D^{\prime}$ with respect to the origin form a parallelepiped with all vertices on the unit sphere. To find the three functionals separating opposite faces, we first calculate the equations of three planes, each one defined by three vertices of the parallelepiped. We have:

[TABLE]

[TABLE]

[TABLE]

Taking into account the dual norm $\|(x,y,z)\|_{*}=\max\{\sqrt{x^{2}+y^{2}},|z|\}$ and normalizing, each of these planes yields a functional which, of course, evaluates all points on a face the same way. So set $f_{1}=(1,0,1),\,f_{2}=\frac{1}{\sqrt{1.36}}(0.6,1,-0.6)$ and $f_{3}=\frac{1}{\sqrt{1.36}}(0.6,-1,-0.6)$ which are all of norm 1. The separation of the faces goes as follows:

[TABLE]

[TABLE]

with $f_{1}(A)-f_{1}(A^{\prime})=1.28>1$.

[TABLE]

[TABLE]

also

[TABLE]

[TABLE]

with $f_{2}(A^{\prime})-f_{2}(A)=f_{3}(A^{\prime})-f_{3}(A)=\frac{1.2}{\sqrt{1.36}}\simeq 1.02899>1$. Hence $d\simeq 1.02899>1$ and the conclusion follows.

∎

**Remarks 7

**(1) Using the Theorem stated in the Introduction (Th. 3 of [14]) we obtain an upper bound for the Banach-Mazur distance of the Petty space $X$ from $\ell_{\infty}^{3}$. We apply the Theorem as in the proof of Proposition 8 and conclude that $d(X,\ell_{\infty}^{3})<2$. One can find a better upper bound by direct calculation. Since $d(X,\ell_{\infty}^{3})=d(X^{*},\ell_{1}^{3})$, it suffices to evaluate the distance between the dual spaces, which is easier. The ball of $X^{*}$ is the right cylinder $B_{X^{*}}=\{(x,y,z):\sqrt{x^{2}+y^{2}}\leq 1\;\;\textrm{and}\;\;|z|\leq 1\}$. Consider the points $A(0,-1,1),\,B(0.8,0.6,1)$ and $C(-0.8,0.6,1)$ of $S_{X^{*}}$ along with the symmetric points $A^{\prime},B^{\prime},C^{\prime}$ with respect to the origin. Let $O=$ conv $\{\pm\overrightarrow{OA},\pm\overrightarrow{OB},\pm\overrightarrow{OC}\}$ (an octahedron producing a space linearly isometric to $\ell_{1}^{3}$). We calculate the largest $\alpha>\frac{1}{2}$ such that

[TABLE]

The right-hand inclusion is obvious. To find the optimal value of $\alpha$ one has to check which homothetic copy of $B_{X^{*}}$ touches some of the faces of the octahedron in a single point (it suffices to check the upper 4 faces due to symmetry). The upper 4 faces define the planes

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Note that if the upper half of the octahedron touches a homothetic copy $\alpha\cdot B_{X^{*}}$ of the cylinder, then the common point must lie on the circle $x^{2}+y^{2}=\alpha^{2},\,z=\alpha$. Solving the systems of equations of each of the above 4 planes together with $x^{2}+y^{2}=\alpha^{2}$ and $z=\alpha$ we find that for $\alpha\simeq 0.56416$ there is at most one common point of a face with the corresponding homothetic copy of $B_{X^{*}}$, hence $0.56416\cdot B_{X^{*}}\subseteq O\subseteq B_{X^{*}}$ and thus $d(X,\ell_{\infty}^{3})=d(X^{*},\ell_{1}^{3})\leq\frac{1}{0.56416}\simeq 1.77254$.

(2) Concerning the original Hadwiger number of the Petty space $X$, we obtain a lower bound of 14 by using the 1-separated set of 14 points on the unit sphere of $X$:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Observe that the points $\alpha_{1},\alpha_{2}$ are the symmetric of $\alpha_{5},\alpha_{4}$ with respect to the YZ-plane, the points $\alpha_{8},\alpha_{7},\alpha_{6}$ are the symmetric of $\alpha_{2},\alpha_{3},\alpha_{4}$ with respect to the XZ-plane and the points $b_{1},b_{2},b_{3}$ are the symmetric of $b_{4},b_{5},b_{6}$ with respect to the XY-plane.

Also the set of points

[TABLE]

[TABLE]

[TABLE]

[TABLE]

shows that $H(X)\geq 14$. Both of the above pointsets are maximal 1-separated subsets of $S_{X}$. Furthermore the subset $\{c_{1},c_{2},\cdots,c_{8}\}\cup\{b_{1},b_{4}\}$ of the second pointset yields the lower bound of 10 for the strict Hadwiger number of $X$, i.e. $H^{\prime}(X)\geq 10$. It seems likely that $H(X)=14$, but the details of such a proof are not yet clear.

(3) About the Hadwiger number of the euclidean spaces the exact values are known in case when $n=2,3,4,8$ and $24$ (see [1], also [19]). When $2\leq n\leq 6$, the best-known lower bounds are larger than $2^{n}$ and for $n=7$ we have $126\leq H(\ell_{2}^{7})\leq 134$. Unlike that, when $8\leq n\leq 24$, the best-known upper bounds are smaller than $2^{n}$.

We conclude with some open problems and questions:

Find better upper and lower bounds for the numbers:

1. (a)

$H^{\prime}_{\alpha}(\ell_{p}^{n})$ and $H_{\alpha}(\ell_{p}^{n})$, for $n\geq 4$ and $1\leq p<2$, 2. (b)

$H^{\prime}(\ell_{\alpha_{n}}^{n})$, for $n\geq 4$ and 3. (c)

$H^{\prime}_{\alpha}(\ell_{2}^{n})$ and $H_{\alpha}(\ell_{2}^{n})$, for $5\leq n\leq 24$

Of particular interest is the case $n=4$ (then $\alpha_{4}=2$) and $n=8,\,24$, where the exact values $H(\ell_{2}^{n})$ are known. 2. 2.

Is there a 3-dimensional Banach space $X$, with $H_{\alpha}(X)<8$? Such a space would be a candidate to have Banach-Mazur distance $d(X,\ell_{\infty}^{3})\geq 2$ (see the proof of Prop.8). 3. 3.

Let $X$ be an $n$-dimensional Banach space with $n\geq 4$. Does the inequality $e_{c}(X)\geq n+1$ (or at least $H^{\prime}_{\alpha}(X)\geq n+1$) hold? (See Theorem 3 and the Note following it).