John's Position is not good for approximation

TL;DR

The paper demonstrates limitations in approximating convex bodies in John's position with polytopes, showing that certain bodies cannot be closely approximated with polytopes having polynomial or subexponential facets.

Contribution

It establishes new lower bounds on the complexity of polytope approximations for convex bodies in John's position, highlighting fundamental limitations.

Findings

No polynomial-facet polytope can approximate certain convex bodies within a factor R_n

No subexponential-facet polytope can approximate some convex bodies within a factor of n

Results apply to high-dimensional convex geometry and approximation theory

Abstract

Recall that a convex body $K$ is in John's position if the unit Euclidean ball is the maximal volume ellipsoid contained in $K$. Approximating convex body in John's position by polytopes we obtain the following results. 1. Let $n>R_n\ge 1$ be a sequence such that $\lim_{n\rightarrow \infty} \frac{R_n}{n}=0$. For a sufficiently large $n$, we can construct a convex body $K\subset \mathbb{R}^n$ in John's position such that there is no $P$, polytope with a polynomial number of facets in $n$ such that $K\subset P\subset R_nK$; 2. For a sufficiently large $n$, there is a convex body $K\subset \mathbb{R}^n$ in John's position such that there is no $P$, polytope that has less than $\exp(cn)$ facets satisfies $K\subset P \subset \sqrt{n}K$.

Equations

$$d_{BM}(K,L):=\inf \{r\ge 1:\exists T\in G{L}_n(\mathbb{R})\text{ such that }K\subset TL\subset rK\}.$$

$$d_{BM}(K,L):=\inf\left\{\begin{array}[]{c}r\geq 1\,:\,\exists T\in GL_{n}(\mathbb{R})\text{ and }x,y\in\mathbb{R}^{n}\\ \text{ such that }K-x\subset T(L-y)\subset r(K-x)\end{array}\right\}.$$

$$B_2^n\subset K\subset P\subset \sqrt{n}{B}_2^n.$$

$$B_2^n\subset K\subset P\subset n{B}_2^n,$$

$$K\subset P\subset RK.$$

$$K\subset P\subset RK,$$

$$\text{number of facets of P}\ge \frac{c}{n}\cdot \left(\text{number of facets of }K\right).$$

$$K\subset P\subset R_{n}K.$$

$$K\subset P\subset \sqrt{n}K,$$

$$(1-\delta )P_{\delta }\subset K\subset P_{\delta }.$$

$$\frac{1}{2}P\subset K\subset P,$$

$$h_K\left(x\right):=\sup \left\{⟨x,y⟩:y\in K\right\}\forall x\in {\mathbb{R}}^n.$$

$${\rho }_K\left(x\right):=\sup \left\{r>0:rx\in K\right\}\forall x\in {\mathbb{R}}^n\backslash \left\{0\right\}.$$

$$K^{\circ }:=\{y\in {\mathbb{R}}^n:\forall x\in K⟨x,y⟩\le 1\}.$$

$${\Delta }_n=\left\{x\in {\mathbb{R}}^n:⟨x,u⟩\le 1\forall u\in {\left\{u_i\right\}}_{i=1}^{n+1}\right\},{\Delta }_n^{\circ }=\mathrm{conv}\left({\left\{u_i\right\}}_{i=1}^{n+1}\right).$$

$$\displaystyle\delta_{ij}:=\left\{\begin{array}[]{cc}1&\text{if $i=j$,}\\ 0&\text{otherwise.}\end{array}\right.$$

$$\langle x_{i},y\rangle\left\{\begin{array}[]{cc}=1&\text{ if $y=y_{i}$, }\\ \leq\frac{1}{2R}&\text{ if $y=y_{j}$ with $i\neq j$.}\\ \leq\frac{1}{2R}&\text{ if $y\in L^{\circ}$}\end{array}\right.$$

$$K\subset P\subset RK.$$

$$w_l=\sum _{i=1}^m{\lambda }_i{y}_i+{\lambda }_{m+1}y,$$

$$R⟨x_i,w_l⟩$$

$$\alpha$$

$$⟨{\alpha }^{\downarrow },{\theta }^{\uparrow }⟩$$

$$⟨{\theta }^{\downarrow },{\theta }^{\uparrow }⟩$$

$$=\frac{1}{7}\sqrt{1-{\left(\frac{1}{8}\right)}^2}\left(1-\sqrt{\frac{48}{63}}\right).$$

$$\mathbb{P}\left(\left|I\cap J\right|\ge \frac{k}{2}\right)\le {\left(\frac{2k}{n}\right)}^{k/5}.$$

$$\mathbb{P}(|I\cap J|=l)=\frac{\left(\genfrac{}{}{0ex}{}{k}{l}\right)\left(\genfrac{}{}{0ex}{}{n-k}{k-l}\right)}{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}.$$

$$(\frac{a}{b}{)}^b\le (\genfrac{}{}{0ex}{}{a}{b})\le (\frac{ea}{b}{)}^b.$$

$$\displaystyle\begin{split}\mathbb{P}(|I\cap J|=l)&\leq&(\frac{ek}{l})^{l}(\frac{e(n-k)}{k-l})^{k-l}(\frac{k}{n})^{k}\\ &=&e^{k}(\frac{k^{2}}{ln})^{l}(\frac{n-k}{n})^{k-l}(\frac{k}{k-l})^{k-l}.\end{split}$$

$$(\frac{k^2}{ln}{)}^l\le (\frac{2k}{n}{)}^l\le (\frac{2k}{n}{)}^{\frac{k}{2}}.$$

$$(\frac{k}{k-l}{)}^{k-l}=(1+\frac{l}{k-l}{)}^{k-l}\le e^l\le e^k.$$

Full text

Department of Mathematics, University of Michigan, Ann Arbor, MI 48109-1043.

1 Introduction

One of the most natural questions in convex geometry is how well can a convex body in $\mathbb{R}^{n}$ (i.e., compact, convex set with non-empty interior) be approximated by polytopes with as few facets (or vertices) as possible. How closely a polytope approximates a convex body can be measured in different ways. In this paper, we are interested in the Banach-Mazur distance which, for (origin) symmetric convex bodies $K,L\subset\mathbb{R}^{n}$, is defined by

[TABLE]

Denote the Euclidean unit ball in $\mathbb{R}^{n}$ by $B_{2}^{n}$. The following lower bound was proven independently (and using different methods) in [2], [4], [6] and [7]. For any polytope $P$ with $m$ facets, one has $d_{BM}(B_{2}^{n},P)\geq c\sqrt{\frac{n}{\log(\frac{m}{n})}}$. Recently, Barvinok [3] showed that for any symmetric convex body $K\subset\mathbb{R}^{n}$, one can find a polytope $P$ with a number of facets, $m$, which is at least polynomial in $n$, such that $d_{BM}(K,P)=O(\sqrt{\frac{n\log(n)}{\log(m)}})$.

For the non-symmetric case, one has to modify the definition of the Banach-Mazur distance: For convex bodies $K,L\subset\mathbb{R}^{n}$, the Banach-Mazur distance $d_{BM}(K,L)$ is defined by

[TABLE]

The choice of the origin $x$ of $K$ in this definition is crucial. For a symmetric convex body, classical choices such as the center of mass, the center of the John ellipsoid, and the Santaló point, all coincide with the center of symmetry. However, this is not the case for a general convex body, a fact which makes the choice of origin an obstacle.

The following results make use of the center of mass as the origin. The first result, by Szarek [9], states that for any convex body in $\mathbb{R}^{n}$, there exists a polytope with either $m$ facets or $m$ vertices, such that $d_{BM}(K,P)\leq\frac{n}{\log(\frac{m}{n})}$. Using a random method, Brazitikos, Chasapis, and Hioni obtain an upper bound of the order of $\frac{n}{\sqrt{\log(\frac{m}{n})}}$, where $m$ is the number of vertices.

In this paper, we consider another natural choice of an origin. Namely, we fix the origin of a convex body $K\subset\mathbb{R}^{n}$ as the center of its John ellipsoid. The John ellipsoid associated to a convex body $K\subset\mathbb{R}^{n}$ is defined as the (unique) maximal volume ellipsoid contained in $K$. We say that a convex body is in John’s position if its maximal volume ellipsoid is the unit Euclidean ball $B_{2}^{n}$. For any convex body $K$, there exists an affine transformation $T$ such that $TK$ is in John’s position. If we consider convex bodies in John’s position, then the origin is the center of their John ellipsoids. John’s position has many consequences that have been useful for solving many problems, see e.g., [1] and the reference therein. For example, the only known proof showing that a symmetric convex body $K$ can be approximated by a polytope $P$ with polynomial number of facets in $n$ so that $d_{BM}(K,P)=O(\sqrt{n})$ relies on John’s position. More precisely, the proof uses the existence of $m=O\left(n^{2}\right)$ contact points $\{x_{i}\}_{i=1}^{m}$ (see definition below) that form an identity decomposition. The polytope is then defined by $P:=\{x\in\mathbb{R}^{n}\,:\,\left|\langle x,x_{i}\rangle\right|\leq 1\quad\forall i\leq m\}$. In particular, $P$ satisfies

[TABLE]

Thus, $d_{BM}(P,K)\leq\sqrt{n}$. We remark that Barvinok’s result [3] also relies on contact points. In the non-symmetric case, however, the above construction can only imply

[TABLE]

which shows that $d_{BM}(K,P)\leq n$.

In this paper we investigate the following problem.

Problem 1.1**.**

Let $R=o(n)$, and $K\subset\mathbb{R}^{n}$ be a convex body in John’s position. Is there a polytope $P$ whose number of facets is polynomial in $n$, such that

[TABLE]

We prove the following main theorem.

Theorem 1.2**.**

For a sufficiently large $n$ and for any $c_{0}\sqrt{n}\leq R\leq c_{1}n$, there exists a convex body $K\subset\mathbb{R}^{n}$ whose John’s ellipsoid is centered at the origin, and such that any polytope $P$ satisfying

[TABLE]

has at least $\exp(C\log(\frac{R^{2}}{n})\frac{n}{R^{2}}n)$ facets, where $c_{0},\,c_{1},\,C>0$ are some universal constants.

${\bf Remark:}$

Notice that the inclusion relations are invariant under invertible linear transformations. We may assume that the body $K$ is in John’s position. 2. 2.

For each $R\in\left[c_{0}\sqrt{n},\,c_{1}n\right]$, the body $K$ which is constructed in the Theorem 1.2 is a polytope. Moreover, for any polytope $P$ satisfying $K\subset P\subset RK$, we have:

[TABLE]

A direct consequence of Theorem 1.2 answers Problem 1.1 in the negative:

Corollary 1.3**.**

Let $R_{n}\rightarrow+\infty$ be a positive increasing sequence that satisfies $\lim_{n\rightarrow\infty}\frac{R_{n}}{n}\rightarrow 0$. For any constant $k>0$, there exists a convex body $K\subset\mathbb{R}^{n}$ in John’s position for a sufficiently large $n$ such that there is no polytope that has at most $n^{k}$ facets and satisfies

[TABLE]

In the other extreme, we have the following corollary:

Corollary 1.4**.**

For a sufficiently large $n$, there exists a convex body $K\subset\mathbb{R}^{n}$ in John’s position such that there is no polytope $P$ that has less than $\exp(cn)$ number of facets and satisfies

[TABLE]

where $c>0$ is a universal constant.

As we previously mentioned, for a symmetric convex body $K$, there exists a polytope $P$ with $O\left(n^{2}\right)$ facets such that $K\subset P\subset\sqrt{n}K$. Corollary 1.4 shows that approximating a non-symmetric body, in the same scale of $\sqrt{n}$ could be much more expensive.

The fact that Theorem 1.2 cannot provide a better result when $R=o(\sqrt{n})$ is not surprising. Using a net argument, one can derive the following:

Proposition 1.5**.**

Suppose $B_{2}^{n}\subset K\subset RB_{2}^{n}$. For a sufficiently small $\delta>0$, there exists a polytope $P_{\delta}$ with no more than $\exp(c\log(\frac{2R}{\delta})n)$ facets such that

[TABLE]

Applying the proposition to convex bodies in John’s position, we conclude the following.

Corollary 1.6**.**

Let $K$ be a convex body in $\mathbb{R}^{n}$ in John’s position, where $n$ is sufficiently large. Then, there exists a polytope $P$ with at most $\exp(c\log(n)n)$ facets such that

[TABLE]

where $c>0$ is a universal constant.

The paper is organized as follows. In Section 2 we introduce notation and recall some needed results. The proof of Theorem 1.2 is presented in Section 3. In Section 4 we prove Propostion 1.5.

2 Preliminaries

For the standard Euclidean space $\mathbb{R}^{n}$, let $\langle\cdot,\cdot\rangle$ denote the usual inner product. For a vector $x\in\mathbb{R}^{n}$, let $|x|$ denote its Euclidean norm. Let $S^{n-1}$ be the unit sphere and $B_{2}^{n}$ be the unit Euclidean ball. Let $GL_{n}\left(\mathbb{R}\right)$ denote the group of invertible linear transformations. For $D_{1},\dots,D_{d}\subset\mathbb{R}^{n}$, let ${\rm conv}\left(\left\{D_{1},\cdots,D_{d}\right\}\right)$ denote the convex hull of the sets $D_{1},\dots,D_{d}$. Given a convex body $K\subset\mathbb{R}^{n},$ let $\partial K$ denote its boundary and $\text{vol}\left(K\right)$ denote the Lebesgue measure of $K$. We recall two standard functions associate with a convex body: Suppose $K$ is a convex body containing [math]. The support function $h_{K}$ of $K$ is defined by

[TABLE]

The radial function of $K$ is defined by

[TABLE]

In particular, $\frac{1}{\rho_{K}\left(x\right)}$ is also called the gauge function of $K$.

For a convex set $K\subset\mathbb{R}^{n}$ that contains [math], we define its polar $K^{\circ}$ by

[TABLE]

Suppose $K_{1},\dots,K_{d}\subset\mathbb{R}^{n}$ are convex sets containing the origin. One can verify that $\left(\cap_{i=1}^{d}K_{i}\right)^{\circ}={\rm conv}\left(\left\{K_{1}^{\circ},\cdots,K_{d}^{\circ}\right\}\right)$.

An ellipsoid $E\subset K$ is called the John ellipsoid of $K$ if $\text{vol}(E)\geq\text{vol}(E^{\prime})$ for any other ellipsoid $E^{\prime}\subset K$. It is well known that the John ellipsoid of a convex body exists and is unique. Furthermore, for any convex body $K$, there exists an affine transformation $T$ such that the John ellipsoid of $TK$ is $B_{2}^{n}$. As we mentioned before, we say that $K$ is in John’s position if its John ellipsoid is $B_{2}^{n}$.

Let $K\subset\mathbb{R}^{n}$ be a convex body in John’s position. A point $x\in\mathbb{R}^{n}$ is said to be a contact point of $K$ and $B_{2}^{n}$ if $x\in\partial K\cap\partial B_{2}^{n}$. A classical theorem of F. John provides a decomposition of the identity in terms of contact points (see e.g., Citation [1, p 52]):

Theorem 2.1**.**

Let $K$ be a convex body in $\mathbb{R}^{n}$ that contains $B_{2}^{n}$. Then, $K$ is in John’s position if and only if there exist contact points $x_{1},..,x_{m}$ and $a_{1},..,\,a_{m}>0$ such that

$\sum_{i=1}^{m}a_{i}x_{i}\otimes x_{i}=I_{n}$, and 2. 2.

$\sum_{i=1}^{m}a_{i}x_{i}=\vec{0}$.

Let $\Delta_{n}$ be the regular simplex in $\mathbb{R}^{n}$ that has an inner radius equal to $1$. Using the symmetry of $\Delta_{n}$ and uniqueness of the John ellipsoid, it is not difficult to check that $\Delta_{n}$ is in John’s position. Suppose $u_{1},\cdots,u_{n+1}$ are the contact points of $\Delta_{n}$. Then, $\langle u_{i},u_{j}\rangle=-\frac{1}{n}$ for $i\neq j$ and $\left\{-nu_{i}\right\}_{i=1}^{n+1}$ are the vertices of $\Delta_{n}$. We can express $\Delta_{n}$ and $\Delta_{n}^{\circ}$ in terms of $\left\{u_{i}\right\}_{i=1}^{n+1}$:

[TABLE]

For any integer $m\in\mathbb{N}$, we define $[m]:=\{1,\cdots,m\}$. For a subset $I\subset[m]$, let $|I|$ denote its cardinality. For $i,j\in\mathbb{N}$, we define

[TABLE]

3 Proof of the main result

In this section, we prove Theorem 1.2. The proof will be divided into three main propositions. The body $K$ is obtained by intersecting a simplex in John’s position with a large number of half spaces. As long as each half space contains the John ellipsoid of the simplex, the new body will be in John’s position as well. The construction of the body uses both certain structures and randomness.

3.1 Lower bound of facets

The first proposition below shows how to determine if a convex body cannot be approximated by polytopes with few facets.

Proposition 3.1**.**

Let $K:=\{x\in\mathbb{R}^{n}\,:\,\langle x,y_{i}\rangle\leq 1\quad\forall i\in[m]\}\cap L$, where $y_{1},...,y_{m}$ are vectors in $\mathbb{R}^{n}$ and $L$ is a convex body in $\mathbb{R}^{n}$ that has [math] as an interior point. Suppose there are points $x_{1},...,x_{m}\in K$ such that for some $R>1$, we have

[TABLE]

Then, there is no polytope $P$ that has less than $\frac{m}{2R}$ facets such that

[TABLE]

Proof.

Suppose there exists $\left\{w_{i}\right\}_{i=1}^{m_{1}}\subset\mathbb{R}^{n}$ such that $P:=\{x\in\mathbb{R}^{n}\,:\,\langle x,w_{l}\rangle\leq 1\,\,\forall l\in[m_{1}]\}$ satisfies

[TABLE]

The first inclusion indicates that $\left\{w_{i}\right\}_{i=1}^{m_{1}}\subset K^{\circ}$. The second inclusion is equivalent to the following: $\forall x\in\partial K$, $R\langle x,w_{l}\rangle\geq 1$ for some $l\in[m_{1}]$. Due to $\langle x_{i},\,y_{i}\rangle=1$, we also have $x_{i}\in\partial K$ for $i\in\left[m\right]$.

For $l\in[m_{1}]$, let $O_{l}$ be the sub-collection of $\{x_{i}\}_{i=1}^{m}$ such that $R\langle x_{i},w_{l}\rangle\geq 1$. Observe that $K^{\circ}=\text{conv}(\{y_{i}\}_{i=1}^{m},L^{\circ})$. Thus, $w_{l}$ can be expressed as a convex combination:

[TABLE]

where $y\in L^{\circ}$, $\lambda_{i}\geq 0$, and $\sum_{i=1}^{m+1}\lambda_{i}=1$.

This expression is not necessarily unique, but we fix one such expression. Taking inner product with $Rx_{i}$ we have

[TABLE]

If $x_{i}\in O_{l}$, then $\lambda_{i}\geq\frac{1}{2R}$. Due to $\sum_{i=1}^{m+1}\lambda_{i}=1$, we conclude that $|O_{l}|\leq 2R$.

Observe that $\cup_{l\in[m_{1}]}O_{l}=\{x_{i}\}_{i\in[m]}$; we conclude that $m_{1}\geq\frac{m}{2R}$. Therefore, $P$ has at least $\frac{m}{2R}$ facets. ∎

The example in the main theorem will be of the form $K:=\{x\,:\,\langle x,y_{i}\rangle\leq 1\,\,\forall i\in[m]\}\cap\Delta_{n}$, where $\{y_{i}\}_{i=1}^{m}\subset S^{n-1}$ and $\Delta_{n}$ is a regular simplex in John’s position. Then, we will find $\{x_{i}\}_{i=1}^{m}$, which satisfies the assumption of Proposition 3.1.

3.2 Structure

Here we have a deterministic proposition related to points in $S^{n-1}$.

Proposition 3.2**.**

Let $S:=S^{n-1}\cap\{x\,:\,\langle\beta,x\rangle=0\}$ for some $\beta\in S^{n-1}$. For $\theta\in S$, let $\theta^{\downarrow}:=-\frac{1}{8}\beta+\sqrt{1-\left(\frac{1}{8}\right)^{2}}\theta$ and $\theta^{\uparrow}:=\sqrt{1-(\frac{1}{7})^{2}}\beta+\frac{1}{7}\theta$. Then,

For $\alpha,\theta\in S$, $\langle\alpha^{\downarrow},\theta^{\uparrow}\rangle>0$ implies $\langle\alpha,\theta\rangle>\frac{3}{4}$, and 2. 2.

$\langle\theta^{\uparrow},\theta^{\downarrow}\rangle=\frac{1}{C_{0}}$ for $\theta\in S$ where $C_{0}:=\frac{1}{\frac{1}{7}\sqrt{1-\left(\frac{1}{8}\right)^{2}}\left(1-\sqrt{\frac{48}{63}}\right)}>1$.

In our construction of $K$, $y_{i}$ will be $\theta_{i}^{\uparrow}$ for some $\theta_{i}\in S$ and $x_{i}$ will be $C_{0}\theta_{i}^{\downarrow}$. In particular, the first statement of Proposition 3.2 implies that $\langle x_{i},y_{j}\rangle<0$ when $\langle\theta_{i},\,\theta_{j}\rangle<\frac{3}{4}$.

Proof.

We fix $\theta\in S$. For $\alpha\in S$, it can be expressed as

[TABLE]

where $s=\langle\alpha,\,\theta\rangle$ and $\alpha^{\prime}=\frac{\alpha-s\theta}{\left|\alpha-s\theta\right|}$. Notice that $\alpha^{\prime}\perp\theta$ and $\alpha^{\prime}\in S$. Thus,

[TABLE]

Suppose $\langle\alpha^{\downarrow},\,\theta^{\uparrow}\rangle>0$. Then, $s\geq\frac{7}{8}\sqrt{\frac{48}{49}\frac{64}{63}}=\sqrt{\frac{48}{63}}>\frac{3}{4}.$

Observe that $\langle\theta^{\downarrow},\theta^{\uparrow}\rangle$ is the same for all $\theta\in S$,

[TABLE]

Since $0<\frac{1}{7}\sqrt{1-\left(\frac{1}{8}\right)^{2}}\left(1-\sqrt{\frac{48}{63}}\right)<1$, we conclude that $C_{0}=\frac{1}{\frac{1}{7}\sqrt{1-\left(\frac{1}{8}\right)^{2}}\left(1-\sqrt{\frac{48}{63}}\right)}>1.$ ∎

3.3 Randomness

In the construction, we will choose $\left\{\theta_{i}\right\}$ independently and uniformly according to a probability distribution. In order to apply Proposition 3.1, we need to choose $\theta_{i}$ so that $\langle x_{i},\,y\rangle\leq\frac{1}{2R}$ for all $y\in\Delta_{n}^{\circ}$. Equivalently, $\rho_{\Delta_{n}}(\theta_{i}^{\downarrow})$ needs to be larger than $2RC_{0}$. The uniform randomness on $S$ does not work in this case. Thus, a probability that is compatible with the structure of $\Delta_{n}$ is required.

The following proposition is a tail bound for the hypergeometric distribution.

Proposition 3.3**.**

For a sufficiently large $n\in\mathbb{N}_{+}$, let $k$ be a positive integer satisfying $100<k<\frac{1}{2e^{8}}n$. Suppose $I,\,J$ are chosen independently and uniformly from $\left\{W\subset\left[n\right]\,:\,\left|W\right|=k\right\}$. Then,

[TABLE]

Proof.

We may assume that $J$ is fixed. For any positive integer $1\leq l\leq k$,

[TABLE]

For positive integers $a\geq b$, ${a\choose b}=\frac{a(a-1)\cdots(a-b+1)}{b(b-1)\cdots 1}$. A standard estimate of ${a\choose b}$ is the following:

[TABLE]

Applying these bounds to (2), we have

[TABLE]

Assuming that $l\geq\frac{k}{2}$ and $2k<n$,

[TABLE]

Also, using $(1+x)\leq e^{x}$ for $x\in\mathbb{R}$ we have

[TABLE]

Together with $(\frac{n-k}{n})^{k-l}\leq 1$,

[TABLE]

If $\frac{n}{2k}\geq e^{8}$, then $2k\leq\log(\frac{n}{2k})\frac{k}{4}$. We have the following:

[TABLE]

Using the union bound we get

[TABLE]

where the last inequality requires $k\geq 100$. ∎

3.4 The construction

Let $\Delta_{n}$ be the $n-$dimensional simplex in John’s position and $u_{1},\cdots,u_{n+1}$ be its contact points. We define $S:=S^{n-1}\cap\{x\in\mathbb{R}^{n}\,:\,\langle x,u_{1}\rangle=0\}$. Notice that $\Delta_{n}^{\prime}=\Delta_{n}\cap\{\langle x,u_{1}\rangle=0\}$ is a $\left(n-1\right)$-dimensional regular simplex. Let $v_{1},...,v_{n}\in S$ such that $\left\{c_{n}nv_{i}\right\}_{i=1}^{n}$ are vertices of $\Delta_{n}^{\prime}$. It is not hard to verify that $c_{n}\rightarrow 1$ as $n\rightarrow\infty$. Let $1\leq k\leq n$. For $I\subset[n]$ with $|I|=k$, we define

[TABLE]

Proposition 3.4**.**

For a sufficiently large $n$, let $1\leq k\leq\frac{n+3}{4}$. Suppose $I,J\subset\left\{W\subset\left[n\right]\,,\,\left|W\right|=k\right\}$ satisfy $|I\cap J|<\frac{k}{2}$. Then,

[TABLE]

Proof.

Because $\left\{v_{i}\right\}_{i=1}^{n}$ are in the vertex directions of a regular simplex, $\langle v_{i},v_{j}\rangle=-\frac{1}{n-1}$ if $i\neq j$. We have

[TABLE]

Thus,

[TABLE]

where $c_{n,k}=\frac{1}{\sqrt{1-\frac{k-1}{n-1}}}$. Suppose $1\leq k\leq\frac{n+3}{4}$, then $1\leq c_{n,k}\leq\sqrt{\frac{4}{3}}$.

Let $J\subset[n]$ with $|J|=k$. Then,

[TABLE]

Suppose $|I\cap J|<\frac{k}{2}$ and $n$ is large enough. Then,

[TABLE]

∎

We are now ready to prove the main theorem.

Proof of Theorem 1.2 .

For a sufficiently large $n$, we fix $100\leq k\leq\frac{n}{2e^{8}}$. Then, $n$ and $k$ satisfy the assumptions in Proposition 3.3 and Proposition 3.4. Let $m\subset\mathbb{N}$ be an integer that we will specify later. Let $I_{1},\,I_{2},\dots,\,I_{m}$ be chosen independently and uniformly from $\{W\subset[n]\,:\,|W|=k\}$. Let $u_{1}$ be the vector $\beta$ described in Proposition 3.2. In particular, $S:=S^{n-1}\cap\{x\,:\,\langle u_{1},\,x\rangle=0\}$. We adapt the definition of $\theta^{\uparrow}$ and $\theta^{\downarrow}$ for $\theta\in S$. Let

[TABLE]

By Proposition 3.3, we have $\mathbb{P}(|I_{i}\cap I_{j}|\geq\frac{k}{2})\leq(\frac{2k}{n})^{\frac{k}{5}}$. A union bound argument shows that

[TABLE]

By setting $m=(\frac{n}{2k})^{k/20}$, we have

[TABLE]

Since the probability is strictly smaller than $1$, there exists a sample such that $|I_{i}\cap I_{j}|<\frac{k}{2}$ for all $1\leq i<j\leq m$. From now on, we fix such a sample.

We want to apply Proposition 3.1 with $L=\Delta_{n}$, $y_{i}=v_{I_{i}}^{\uparrow}$, and $x_{i}=\frac{1}{\langle v_{I_{i}}^{\downarrow},\,v_{I_{i}}^{\uparrow}\rangle}v_{I_{i}}^{\downarrow}=C_{0}v_{I_{i}}^{\downarrow}$, where $C_{0}$ is the constant defined in Proposition 3.2. We start verifying the assumptions that are described in Proposition 3.1.

First, $\Delta_{n}^{\circ}=\text{conv}\{u_{1},\cdots,u_{n+1}\}$. Because $\{c_{n}nv_{i}\}_{i=1}^{m}\subset\Delta_{n}^{{}^{\prime}}\subset\Delta_{n}$, $\langle c_{n}nv_{i},u_{j}\rangle\leq 1$ for $i\in[n]$ and $j\in[n+1]$. Thus, for any $I\subset[n]$ with $|I|=k$ and $j\in[n+1]$,

[TABLE]

Since $\langle-u_{i},u_{j}\rangle=-(1+\frac{1}{n})\delta_{ij}+\frac{1}{n}\leq\frac{1}{n}$,

[TABLE]

Since $\frac{n}{2e^{8}}\geq k\geq 1$ , $1\leq c_{n,k}\leq\sqrt{\frac{4}{3}}$ and $c_{n}\rightarrow 1$ as $n\rightarrow+\infty$,

[TABLE]

We conclude that

[TABLE]

Every $y\in\Delta_{n}^{\circ}$ can be written as a convex combination of $\left\{u_{i}\right\}_{i=1}^{n+1}$. Thus, the same inequality holds:

[TABLE]

Let $i,j\in[m+1]$ with $i\neq j$. Applying Proposition 3.4, we obtain $\langle v_{I_{i}},v_{I_{j}}\rangle<\frac{3}{4}$ since $\left|I_{i}\cap I_{j}\right|<\frac{3}{4}$. According to Proposition 3.2, we obtain $\langle C_{0}v_{I_{i}}^{\downarrow},v_{I_{j}}^{\uparrow}\rangle<0$. In the case $i=j$, by definition we have $\langle C_{0}v_{I_{i}}^{\downarrow},v_{I_{j}}^{\uparrow}\rangle=1$. To summarize,

[TABLE]

Now, we can apply Proposition 3.1 with $m=(\frac{n}{2k})^{k/10}$, $y_{i}=u_{I_{i}}^{\uparrow}$, $x_{i}=C_{0}u_{I_{i}}^{\downarrow}$, $L=\Delta_{n}$ and $R=\frac{n}{6C_{0}\sqrt{k}}$ with the condition that $100\leq k\leq\frac{n}{2e^{8}}$. Expresssing these relations in terms of $R$ and $n$, we have

[TABLE]

The lower bound of the facets of the polytope $P$ in Proposition 3.1 is $\frac{m}{2R}$. To simplify $m$, we further restrict $R>\sqrt{en}$ so that $\frac{R^{2}}{n}>e$. Since $C_{0}>1$, we have $\log\left(18C_{0}^{2}\right)>0$. Thus,

[TABLE]

Then,

[TABLE]

when $C^{\prime}=\frac{1}{720C_{0}^{2}}>0$. In order to take care of the $\log(2n)$ term we need to check the last term carefully. First,

[TABLE]

for $R>\sqrt{en}$. Let $c_{1}=\min\left\{1,\,\sqrt{\frac{C^{\prime}}{8}},\,\frac{1}{60C_{0}}\right\}$. Suppose $R=c_{1}n$, we have

[TABLE]

where the last inequality holds for large $n$. Together with $2\log(n)\geq\log(2n)$,

[TABLE]

when $R=c_{1}n$. Since $\log(\frac{R^{2}}{n})\frac{n}{R^{2}}$ is a decreasing function for $R>\sqrt{en}$, $\frac{1}{2}C^{\prime}\log(\frac{R^{2}}{n})\frac{n}{R^{2}}\geq\log(2n)$ for $\sqrt{en}<R<c_{1}n$. Therefore, we conclude that for $c_{0}\sqrt{n}<R<c_{1}n$,

[TABLE]

where $C>0$ is an universal constants. Therefore, for $c_{0}\sqrt{n}\leq R\leq c_{1}n$, there exists a convex body $K\subset\mathbb{R}^{n}$ in John’s position such that no polytope $P$ that has less than $\exp(C\log(\frac{R^{2}}{n})\frac{n^{2}}{R^{2}})$ facets satisfies

[TABLE]

∎

4 Upper bound for small $R$

Proposition 4.1**.**

Suppose $B_{2}^{n}\subset K\subset RB_{2}^{n}$. For $0<\delta<1$, there exists a polytope $P_{\delta}$ with no more than $\exp(c\log(\frac{2R}{\delta})n)$ facets such that $(1-\delta)P_{\delta}\subset K\subset P_{\delta}$. Here $c>0$ is a universal constant.

Proof.

Let $B_{2}^{n}\subset K\subset RB_{2}^{n}$ be a convex body. Let $h:S^{n-1}\rightarrow\left[1,\,R\right]$ be the support function of $K$. Observe that $h$ is also the gauge function of $\frac{1}{R}B_{2}^{n}\subset K^{\circ}\subset B_{2}^{n}$ . Thus, $h$ is a $R-$Lipschitz continuous function.

Let $\mathscr{N}$ be a $\frac{\delta}{2R}$-net of $S^{n-1}$. We define

[TABLE]

Thus, $P_{\delta}$ is a polytope with at most $\left|\mathscr{N}\right|$ facets. Recall that by a volumetric argument, the size of a $\varepsilon-$net on $S^{n-1}$ is bounded by $\exp\left(c\log\left(\frac{1}{\varepsilon}\right)n\right)$ for an universal constant $c>0$. Hence, $P_{\delta}$ has no more than $\exp\left(c\log\left(\frac{2R}{\delta}\right)n\right)$ number of facets.

Since

[TABLE]

we have $K\subset P_{\delta}$. Observe that

[TABLE]

For $x\in\partial K$, there exists $\theta$ such that \langle x,\,\theta\rangle=h\text{\left(\theta\right)}. We pick $\alpha\in\mathscr{N}$ such that $\left\|\alpha-\theta\right\|<\frac{\delta}{2R}$. Then,

[TABLE]

Since $h$ is a $R-$Lipschitz continuous function, we have

[TABLE]

Together with $h\left(\alpha\right)\geq 1$, the equation (10) becomes

[TABLE]

Thus, $x\notin\left(1-\delta\right)P_{\delta}$. In paricular, we conclude the radial function of $K$ is always greater than the radial funciton of $\left(1-\delta\right)P_{\delta}$. Therefore, we have $\left(1-\delta\right)P_{\delta}\subset K$. ∎

Acknowledgement

I am grateful to my Advisor, Mark Rudelson for fruitful discussions. I am also grateful to the anonymous reviewer for his/her suggestions.