A Blichfeldt-type inequality for centrally symmetric convex bodies

Abstract

In this note, we derive an asymptotically sharp upper bound on the number of lattice points in terms of the volume of a centrally symmetric convex body. Our main tool is a generalization of a result of Davenport that bounds the number of lattice points in terms of volumes of suitable projections.

1 Introduction

Let $\mathcal K ^n$ be the set of all convex bodies in $\mathbb R ^n$, i.e., compact convex sets $K$ with nonempty interior $\text{ int}K$. Such a body $K$ is called centrally symmetric if $K=-K$. The family of $n$-dimensional lattices in $\mathbb R ^n$ is denoted by $\mathcal L ^n$ and the usual Lebesgue measure with respect to the $n$-dimensional Euclidean space by $\text{ vol}_n(\cdot )$. If the ambient space is clear from the context we omit the subscript and just write $\text{ vol}(\cdot )$. For a given bounded subset $S\subset \mathbb R ^n$, the number of lattice points in $S$ is denoted by $\text{ G}(S)=\#(S\cap \mathbb Z ^n)$. A lattice polytope is a polytope all of whose vertices are points in $\mathbb Z ^n$. Finally, for an $A\subseteq \mathbb R ^n$ the dimension of its affine hull will be denoted by $\dim A$.

We are interested in bounds on the volume in terms of the lattice point enumerator of a convex body. For $K\in \mathcal K ^n$ with $\dim (K\cap \mathbb Z ^n)=n$ a sharp lower bound on $\text{ vol}(K)$ was obtained by Blichfeldt [4]. It reads

$$\begin{aligned} \text{ vol}(K)\ge \frac{1}{n!}\left(\text{ G}(K)- n\right). \end{aligned}$$

We will call results of this kind Blichfeldt-type inequalities. On the other hand, the best known upper bound on $\text{ vol}(P)$ for a lattice polytope $P\in \mathcal K ^n$ is due to Pikhurko [15]

$$\begin{aligned} \text{ vol}(P)\le (8n)^n15^{n2^{2n+1}}\text{ G}(\text{ int}P) \end{aligned}$$

and holds under the condition that $\text{ G}(\text{ int}P)\ne 0$. On the class of centrally symmetric convex bodies, Blichfeldt [4] and van der Corput [6] obtained a sharp upper bound on the volume

$$\begin{aligned} \text{ vol}(K)\le 2^{n-1}\left(\text{ G}(\text{ int}K)+1\right),\quad K\in \mathcal K ^n_0. \end{aligned}$$

(1.1)

Bey, Henk and Wills [3] proposed the study of a reverse inequality also on the class of centrally symmetric convex bodies, and in [11] the authors derive, as a first step, Blichfeldt-type inequalities for lattice crosspolytopes, lattice zonotopes, and for centrally symmetric planar convex sets. Moreover, they conjecture that there is a constant $c\!>1$ such that $\text{ vol}(K)\!\ge \frac{c^n}{n!}\text{ G}(K)$, for every $K\in \mathcal K ^n_0$ with $\dim (K\cap \mathbb Z ^n)=n$.

In this work, we confirm this conjecture asymptotically by showing that for every $\varepsilon \in (0,1]$ and large enough $n\in \mathbb N$ a valid choice for this constant is $c=2-\varepsilon$. As the main ingredient to our argument we prove the following generalization of a classical result of Davenport [7]. Therein, we denote by $\mathcal Q (P)$ the set of all lattice parallelepipeds in $\mathcal K ^n$ whose edges are parallel to a given lattice parallelepiped $P\in \mathcal K ^n$. Moreover, $K|L$ denotes the orthogonal projection of $K$ onto the subspace $L$ and $\genfrac(){0.0pt}{}{[n]}{i}$ is the set of all $i$-element subsets of $[n]=\{1,\dots ,n\}$.

Lemma 1

Let $K\in \mathcal K ^n$ and let $P=\sum _{j=1}^n[0,z_j]$ be a lattice parallelepiped. Then

$$\begin{aligned} \text{ G}(K)\le \sum _{i=0}^n\sum _{J\in \genfrac(){0.0pt}{}{[n]}{i}}\text{ vol}_{n-i}(K|L_J^\perp )\text{ vol}_i(P_J), \end{aligned}$$

where $L_J=\text{ lin}\{z_j:j\in J\}$ and $P_J=\sum _{j\in J}[0,z_j]$, for $J\in \genfrac(){0.0pt}{}{[n]}{i}$. Equality holds if and only if $P$ is a fundamental cell of $\mathbb Z ^n$ and $K\in \mathcal Q (P)$.

As mentioned above, we use the preceding result to derive a Blichfeldt-type inequality for centrally symmetric convex bodies and thereby confirm conjectured bounds from [3, Conj. 1.1] and [11] asymptotically.

Theorem 1

For every $\varepsilon \in (0,1]$ there exists an $n(\varepsilon )\in \mathbb N$ such that for every $n\ge n(\varepsilon )$ and every $K\in \mathcal K ^n_0$ with $\dim (K\cap \mathbb Z ^n)=n$, we have

$$\begin{aligned} \text{ vol}(K)\ge \frac{(2-\varepsilon )^n}{n!}\text{ G}(K). \end{aligned}$$

The constant $2$ cannot be replaced by a bigger one.

As an application of this inequality, we bound the magnitude $\frac{\text{ G}(K)}{\text{ G}(K^{\star })\text{ vol}(K)}$ by constants that depend on the dimension $n$ but not on the body $K$. Herein, $K^{\star }=\{ x\in \mathbb R ^n:x^\mathrm{T} y\le 1,\quad \forall y\in K \}$ denotes the polar body of a given $K\in \mathcal K ^n_0$. Estimates of such kind were first studied and applied by Gillet and Soulé [9] who obtained $6^{-n}\le \frac{\text{ G}(K)}{\text{ G}(K^\star )\text{ vol}(K)}\le \frac{6^nn!}{c^n}$, for some $c\le 4$.

Theorem 2

For every $\varepsilon >0$ there exists an $n(\varepsilon )\in \mathbb N$ such that for every $n\ge n(\varepsilon )$ and every $K\in \mathcal K ^n_0$ with $\dim (K\cap \mathbb Z ^n)=n$, we have

$$\begin{aligned} (\pi +\varepsilon )^{-n}\le \frac{\text{ G}(K)}{\text{ G}(K^\star )\text{ vol}(K)}\le \frac{(\pi +\varepsilon )^nn!}{c^n}, \end{aligned}$$

where $c\le 4$ is an absolute constant.

The results of Lemma 1 and Theorem 1 are discussed in the subsequent section. In Sect. 3, we give the details of the proof of Theorem 2.

2 Proofs of Lemma 1 and Theorem 1

A convex body $T\in \mathcal K ^n$ is said to be a lattice tile with respect to the lattice $\Lambda \in \mathcal L ^n$ if $T$ tiles $\mathbb R ^n$ by vectors in $\Lambda$, that is, $\mathbb R ^n=\Lambda +T$ and $(x+\text{ int}T)\cap (y+\text{ int}T)=\emptyset$, for any different $x,y\in \Lambda$. It is well-known that every lattice tile is a polytope and thus we can assume that every lattice tile has the origin as a vertex. For a survey on tilings and references to the relevant literature the reader may consult [19].

In the case that $T$ is a fundamental cell of $\mathbb Z ^n$, the subsequent lemma was already mentioned by Betke and Wills [1] (see also [10, Sect. 3]).

Lemma 2

Let $K\in \mathcal K ^n$ and let $T$ be a lattice tile with respect to a sublattice $\Lambda$ of $\mathbb Z ^n$. Then

$$\begin{aligned} \text{ G}(K)\le \text{ vol}(K+T). \end{aligned}$$

If $T$ is a lattice parallelepiped $P$, then equality holds if and only if $\Lambda =\mathbb Z ^n$ and $K\in \mathcal Q (P)$.

proof

Since for any $x-y\in \Lambda$ we have $(x+\text{ int}T)\cap (y+\text{ int}T)=\emptyset$, unless $x=y$, every residue class modulo $\Lambda$ is a packing set of $T$. Let $\{z_1,\dots ,z_m\}\subseteq T\cap \mathbb Z ^n$ be a maximal subset of different representatives of residue classes modulo $\Lambda$. Writing $\Lambda _j=z_j+\Lambda$, we get that for all $j=1,\dots ,m$

$$\begin{aligned} \#(K\cap \Lambda _j)=\frac{\text{ vol}\left((K\cap \Lambda _j)+T\right)}{\text{ vol}(T)}. \end{aligned}$$

Since $T$ is a lattice tile, we have $\text{ vol}(T)=\det \Lambda =m$, and therefore

$$\begin{aligned} \text{ G}(K)=\sum _{j=1}^m\#(K\cap \Lambda _j)=\frac{1}{m}\sum _{j=1}^m \text{ vol}\left((K\cap \Lambda _j)+T\right)\le \text{ vol}(K+T). \end{aligned}$$

By compactness of the involved sets, equality holds if and only if $(K\cap \Lambda _j)+T=K+T$, for all $j=1,\dots ,m$. In particular, there can only be one residue class and thus $m=\det \Lambda =1$, i.e. $\Lambda =\mathbb Z ^n$. In the case that the lattice tile is a lattice parallelepiped $P=\sum _{i=1}^n[0,a_i],\,a_i\in \mathbb Z ^n$, every hyperplane supporting a facet of the convex polytope $K+P=(K\cap \mathbb Z ^n)+P$ is parallel to a hyperplane supporting a facet of $P$. Therefore, $K+P$ is a lattice translate of $\sum _{i=1}^n[0,t_ia_i]$ for some $t_i\in \mathbb N$, and so $K$ is a lattice translate of $\sum _{i=1}^n[0,(t_i-1)a_i]\in \mathcal Q (P)$.

Conversely, an easy calculation gives $\text{ G}(K)=\text{ vol}(K+P)$, for every $K\in \mathcal Q (P)$ and any fundamental cell $P$ of $\mathbb Z ^n$. $\square$

proof

(Lemma 1) The lattice parallelepiped $P=\sum _{j=1}^n[0,z_j]$ is clearly a lattice tile with respect to the sublattice of $\mathbb Z ^n$ which is spanned by $z_1,\dots ,z_n$. Based on an alternative proof by Ulrich Betke of an inequality of Davenport [7], we use Lemma 2 and develop the volume of $K+P$ into a sum of the mixed volumes $\text{ V}(K,n-i;P,i)$ of $K$ and $P$ (we refer to the books of Gardner [8] and Schneider [18] for details and properties of mixed volumes)

$$\begin{aligned} \text{ G}(K)\le \text{ vol}(K+P)=\sum \limits _{i=0}^n\genfrac(){0.0pt}{}{n}{i}\text{ V}(K,n-i;P,i). \end{aligned}$$

(2.1)

By the linearity and nonnegativity of the mixed volumes, we have

$$\begin{aligned} \text{ V}(K,n-i;P,i)&=\sum _{j_1=1}^n\dots \sum _{j_i=1}^n\text{ V}(K,n-i; [0,z_{j_1}],\dots ,[0,z_{j_i}])\nonumber \\&=\sum _{J\in \genfrac(){0.0pt}{}{[n]}{i}}i!\text{ V}(K,n-i;[0,z_j],j\in J), \end{aligned}$$

(2.2)

and by Eq. (A.41) in [8, App. A.5], it holds

$$\begin{aligned} \genfrac(){0.0pt}{}{n}{i}i!\text{ V}(K,n-i;[0,z_j],j\in J)&=\text{ vol}_{n-i}(K|L_J^\perp )\text{ vol}_i(P_J), \end{aligned}$$

(2.3)

for every $J\in \genfrac(){0.0pt}{}{[n]}{i}$. Finally, combining (2.1), (2.2) and (2.3) gives the desired result.

The equality characterization is inherited from Lemma 2 since (2.1) is the only step using an inequality. $\square$

For later reference, we state the following lemma that can be found for example in [16, Thm. 1] and [5, Lem. 3.1].

Lemma 3

Let $K\in \mathcal K ^n_0$ and let $L$ be an $i$-dimensional linear subspace of $\mathbb R ^n$. Then

$$\begin{aligned} \text{ vol}(K)\le \text{ vol}_{n-i}(K|L^\perp )\text{ vol}_i(K\cap L)\le \genfrac(){0.0pt}{}{n}{i}\text{ vol}(K), \end{aligned}$$

and both inequalities are best possible.

proof

(Theorem 1) By assumption, we can find $n$ linearly independent lattice points $\{z_1,\dots ,z_n\}$ inside $K$. Applying Lemma 1 with respect to the lattice parallelepiped $P=\sum _{j=1}^n[0,z_j]$ gives

$$\begin{aligned} \text{ G}(K)\le \sum _{i=0}^n\sum _{J\in \genfrac(){0.0pt}{}{[n]}{i}}\text{ vol}_{n-i}(K|L_J^\perp )\text{ vol}_i(P_J), \end{aligned}$$

where $P_J=\sum _{j\in J}[0,z_j]$. By construction of the subspaces $L_J$, we have

$$\begin{aligned} \text{ vol}_i(K\cap L_J)\ge \text{ vol}_i\left(\text{ conv}\{\pm z_j:j\in J\}\right)=\frac{2^i}{i!}\text{ vol}_i(P_J) \end{aligned}$$

and together with Lemma 3, we get

$$\begin{aligned} \text{ G}(K)&\le \sum _{i=0}^n\sum _{J\in \genfrac(){0.0pt}{}{[n]}{i}}\frac{i!}{2^i} \text{ vol}_{n-i}(K|L_J^\perp )\text{ vol}_i(K\cap L_J)\nonumber \\&\le \text{ vol}(K)\sum _{i=0}^n\genfrac(){0.0pt}{}{n}{i}^2\frac{i!}{2^i}=\text{ vol}(K) \frac{n!}{2^n}L_n(2), \end{aligned}$$

(2.4)

where $L_n(x)=\sum _{k=0}^n\genfrac(){0.0pt}{}{n}{k}\frac{x^k}{k!}$ denotes the $n$th Laguerre polynomial. For two functions $f,g:\mathbb N \rightarrow \mathbb R$, we denote by $f(n)\approx g(n)$ that $\lim _{n\rightarrow \infty }\frac{f(n)}{g(n)}=1$. In Szegő’s book  [20, p. 199] one finds the approximation

$$\begin{aligned} L_n(x)\approx \frac{n^{-\frac{1}{4}}}{2\sqrt{\pi }}\frac{e^{-\frac{x}{2}}}{x^{\frac{1}{4}}}e^{2\sqrt{x(n+\frac{1}{2})}}\quad \text{ for} \text{ all} \text{ fixed}\, x>0. \end{aligned}$$

Therefore, by $\lim _{n\rightarrow \infty }e^{\frac{2\sqrt{2n+1}}{n}}=1$, we have

$$\begin{aligned} \frac{L_n(2)}{2^n}&\approx \frac{1}{2e\sqrt{\pi }\root 4 \of {2n}} \frac{e^{2\sqrt{2n+1}}}{2^n}<\frac{e^{2\sqrt{2n+1}}}{2^n}\le \frac{1}{(2-\varepsilon )^n} \end{aligned}$$

(2.5)

for every $\varepsilon \in (0,1]$ and large enough $n\in \mathbb N$. Hence, for large enough $n$, we arrive at

$$\begin{aligned} \text{ G}(K)\le \text{ vol}(K)\frac{n!}{2^n}L_n(2)\le \text{ vol}(K)\frac{n!}{(2-\varepsilon )^n}. \end{aligned}$$

In order to see that this inequality is asymptotically sharp, we consider the crosspolytope $C_{n,l}^\star =\text{ conv}\{\pm le_1,\pm e_2,\dots ,\pm e_n\}$. We have $\text{ G}(C_{n,l}^\star )=2(n+l)-1$ and $\text{ vol}(C_{n,l}^\star )=\frac{2^n}{n!}l$. Therefore, $\root n \of {\frac{n!\text{ vol}(C_{n,l}^\star )}{\text{ G}(C_{n,l}^\star )}}=\root n \of {\frac{2^nl}{2(n+l)-1}}$ tends to $2$ when $l$ and $n$ tend to infinity. On the other hand, the above inequality shows that for every $\varepsilon \in (0,1]$ we have $\root n \of {\frac{n!\text{ vol}(K)}{\text{ G}(K)}}\ge 2-\varepsilon$ for $n\rightarrow \infty$. $\square$

3 An application of Theorem 1

In this section, we give the details of the proof of Theorem 2. We proceed by showing that Theorem 1 can be used to derive an improvement of

$$\begin{aligned}\text{ G}(K)\text{ vol}(K^\star )\le 6^n.\end{aligned}$$

This inequality is due to Gillet and Soulé [9] and holds for every $K\in \mathcal K ^n_0$ with $\dim (K\cap \mathbb Z ^n)=n$. The volume of the Euclidean unit ball $B_n$ is denoted by

$$\begin{aligned} \kappa _n=\text{ vol}(B_n)=\frac{\pi ^{\frac{n}{2}}}{\Gamma (\frac{n}{2}+1)}, \end{aligned}$$

where $\varGamma (z)=\int _0^\infty e^{-t}t^{z-1}dt$ is the gamma function (cf. [8, p. 13]).

Corollary 1

For every $\varepsilon >0$ there exists an $n(\varepsilon )\in \mathbb N$ such that for every $n\ge n(\varepsilon )$ and every $K\in \mathcal K ^n_0$ with $\dim (K\cap \mathbb Z ^n)=n$, we have

$$\begin{aligned} \frac{c^n}{n!}\le \text{ G}(K)\text{ vol}(K^\star )\le (\pi +\varepsilon )^n. \end{aligned}$$

Here, $c\le 2$ is an absolute constant and the lower bound holds for every $n\in \mathbb N$ and arbitrary $K\in \mathcal K ^n_0$.

proof

For the lower bound, we combine Inequality (1.1) and the estimate

$$\begin{aligned} \text{ vol}(K)\text{ vol}(K^\star )&\ge \frac{C^n}{n!}, \end{aligned}$$

(3.1)

that is due to Bourgain and Milman [5] and holds for some universal constant $C\le 4$. Indeed, we have

$$\begin{aligned} \text{ G}(K)\text{ vol}(K^\star )\ge \frac{1}{2^n}\text{ vol}(K)\text{ vol}(K^\star )\ge \frac{(C/2)^n}{n!}. \end{aligned}$$

Now we restrict to $K\in \mathcal K ^n_0$ with $\dim (K\cap \mathbb Z ^n)=n$. The exact inequality (2.4) in the proof of Theorem 1 and the Blaschke-Santaló [17] inequality imply that

$$\begin{aligned} \text{ G}(K)\text{ vol}(K^\star )\le \frac{n!L_n(2)}{2^n}\text{ vol}(K)\text{ vol}(K^\star )\le \frac{n!\kappa _n^2L_n(2)}{2^n}. \end{aligned}$$

(3.2)

Hence, by (2.5), we have that for every $\varepsilon ^{\prime }\in (0,1]$ and large enough $n\in \mathbb N$

$$\begin{aligned} \text{ G}(K)\text{ vol}(K^\star )\le \frac{n!\kappa _n^2}{(2-\varepsilon ^{\prime })^n}. \end{aligned}$$

Stirling approximation and $\kappa _n=\frac{\pi ^\frac{n}{2}}{\varGamma (\frac{n}{2}+1)}$ give

$$\begin{aligned} \frac{n!\,\kappa _n^2}{(2-\varepsilon ^{\prime })^n}\approx \frac{\sqrt{2\pi n}(\frac{n}{e})^n\pi ^n}{(2-\varepsilon ^{\prime })^n\varGamma (\frac{n}{2}+1)^2}\approx \frac{\sqrt{2\pi n}(\frac{n}{e})^n\pi ^n}{(2-\varepsilon ^{\prime })^n\pi n(\frac{n}{2e})^n}\le \left(\frac{2\pi }{2-\varepsilon ^{\prime }}\right)^n. \end{aligned}$$

For every $\varepsilon \in (0,1)$ there exists an $\varepsilon ^{\prime }\in (0,1)$ such that $\frac{2\pi }{2-\varepsilon ^{\prime }}\le \pi +\varepsilon$, and we conclude that for large $n$, the inequality $\text{ G}(K)\text{ vol}(K^\star )\le (\pi +\varepsilon )^n$ holds. $\square$

In the planar case, we are able to give sharp bounds. Two sets are called unimodularly equivalent if there is a lattice preserving affine transformation that maps one onto the other. With $e_i$ we denote the $i$th unit vector in $\mathbb R ^n$.

Proposition 1

Let $K\in \mathcal K ^2_0$ be such that $\dim (K\cap \mathbb Z ^2)=2$. Then

$$\begin{aligned} 2\le \text{ G}(K)\text{ vol}(K^\star )\le 21. \end{aligned}$$

In the upper bound equality holds if and only if $K$ is unimodularly equivalent to the hexagon $H=\text{ conv}\{\pm e_1,\pm e_2,\pm (e_1+e_2)\}$. The lower bound is also best possible and holds for arbitrary $K\in \mathcal K ^2_0$.

Proof

For the upper bound, we can restrict to lattice polygons $P\in \mathcal{P }^2_0$ since for $P_K=\text{ conv}\{K\cap \mathbb Z ^2\}$ we clearly have $\text{ G}(K)\text{ vol}(K^\star )\le \text{ G}(P_K)\text{ vol}(P_K^\star )$. By the well-known formula of Pick [14], we get

$$\begin{aligned} \text{ G}(P)\text{ vol}(P^\star )&=\left(\text{ vol}(P)+\frac{1}{2}\text{ G}(\partial P)+1\right)\text{ vol}(P^\star )\\&=\left(\text{ vol}(P)+\frac{1}{2}\text{ G}(P)-\frac{1}{2}\text{ G}(\text{ int}P)+1\right)\text{ vol}(P^\star ) \end{aligned}$$

and therefore

$$\begin{aligned} \text{ G}(P)\text{ vol}(P^\star )=2\text{ vol}(P)\text{ vol}(P^\star )+\text{ vol}(P^\star )\left(2-\text{ G}(\text{ int}P)\right). \end{aligned}$$

Using the Blaschke-Santaló inequality [17] in the plane gives $\text{ G}(P)\text{ vol}(P^\star )\le 2\text{ vol}(P)\text{ vol}(P^\star )\le 2\pi ^2<21$, whenever $\text{ G}(\text{ int}P)>1$. Up to unimodular equivalence there are only three centrally symmetric lattice polygons with exactly one interior lattice point: the square $[-1,1]^2$, the diamond $\text{ conv}\{\pm e_1,\pm e_2\}$ and the hexagon $H$ (see for example [13, Prop. 2.1]). Among these three the hexagon is the only maximizer of $\text{ G}(P)\text{ vol}(P^\star )$.

For the lower bound, we use Mahler’s inequality $\text{ vol}(K)\text{ vol}(K^\star )\ge 8$, for $K\in \mathcal K ^2_0$ (see [12]). Following the lines of the proof of Corollary 1, this yields $\text{ G}(K)\text{ vol}(K^\star )\ge 2$. For small $\varepsilon >0$, the square $[-1+\varepsilon ,1-\varepsilon ]^2$ shows that this is best possible. $\square$

Remark 1

Based on computer experiments in small dimensions, we conjecture that the hexagon in the plane is an exception and that for $n\ge 3$ the maximizing example is the standard crosspolytope $C_n^\star =\text{ conv}\{\pm e_1,\ldots ,\pm e_n\}$, i.e. $\text{ G}(K)\text{ vol}(K^\star )\le (2n+1)2^n$, for every $K\in \mathcal K ^n_0$ with $\dim (K\cap \mathbb Z ^n)=n$.

Finally, we are ready to prove Theorem 2. The arguments are similar to those given by Gillet and Soulé [9, Sect. 1.6]. We repeat them here with the necessary adjustments to our bounds and for the sake of completeness.

Proof

(Theorem 2) The upper bound can be derived by applying the lower bound to $K^\star$ and by using Inequality (3.1) of Bourgain and Milman. In fact, there is an absolute constant $c\le 4$ with

$$\begin{aligned} \frac{\text{ G}(K)}{\text{ G}(K^\star )\text{ vol}(K)}=\frac{\text{ G}(K)\text{ vol}(K^\star )}{\text{ G}(K^\star ) \text{ vol}(K)\text{ vol}(K^\star )}\le \frac{(\pi +\varepsilon )^n}{\text{ vol}(K)\text{ vol}(K^\star )} \le \frac{(\pi +\varepsilon )^nn!}{c^n}. \end{aligned}$$

Here, $\varepsilon >0$ and $n\in \mathbb N$ is large enough.

For the lower bound, let $L=\text{ lin}(K^\star \cap \mathbb Z ^n)$ and $k=\dim L$. Let us abbreviate $L_n=L_n(2)$. We use the exact inequality (3.2) from the proof of Corollary 1 in the sublattice $\mathbb Z ^n\cap L$ and obtain

$$\begin{aligned} \text{ G}(K^\star )&=\text{ G}(K^\star \cap L)\le \frac{k!\kappa _k^2L_k}{2^k}\frac{\det ((\mathbb Z ^n\cap L)^\star )}{\text{ vol}_k((K^\star \cap L)^\star )}\nonumber \\&=\frac{k!\kappa _k^2L_k}{2^k}\frac{\det (\mathbb Z ^n|L)}{\text{ vol}_k(K|L)}, \end{aligned}$$

(3.3)

where $\det (\cdot )$ denotes the determinant of a lattice. By $K=(K^\star )^{\star }$, we have $\text{ int}K=\{ x\in K:|x^\mathrm{T} y|<1,\quad \forall y\in K^{\star } \}$ and thus $x^\mathrm{T} y=0$, for all $x\in \text{ int}K\cap \mathbb Z ^n$ and $y\in K^\star \cap \mathbb Z ^n$. Therefore, using Inequality (1.1) in the sublattice $\mathbb Z ^n\cap L^\perp$, we get

$$\begin{aligned} \text{ G}(K)&\ge \text{ G}(\text{ int}K)=\text{ G}(\text{ int}K\cap L^\perp )\nonumber \\&\ge \frac{1}{2^{n-k}}\frac{\text{ vol}_{n-k}(\text{ int}K\cap L^\perp )}{\det (\mathbb Z ^n\cap L^\perp )}=\frac{1}{2^{n-k}}\frac{\text{ vol}_{n-k}(K\cap L^\perp )}{\det (\mathbb Z ^n\cap L^\perp )}. \end{aligned}$$

(3.4)

Combining (3.3), (3.4) and the lower bound in Lemma 3 gives

$$\begin{aligned} \frac{\text{ G}(K)}{\text{ G}(K^\star )}\ge \frac{4^k}{2^nk!\kappa _k^2L_k}\frac{\text{ vol}_{n-k}(K\cap L^\perp )\text{ vol}_k(K|L)}{\det (\mathbb Z ^n\cap L^\perp )\det (\mathbb Z ^n|L)}\ge \frac{4^k\text{ vol}(K)}{2^nk!\kappa _k^2L_k}. \end{aligned}$$

What remains is to show that the function $g(k)=\frac{4^k}{k!\,\kappa _k^2\,L_k}$ is nonincreasing for $k\ge 0$, because together with the previous inequality we then arrive at

$$\begin{aligned} \frac{\text{ G}(K)}{\text{ G}(K^\star )\text{ vol}(K)}\ge \frac{4^k}{2^nk!\kappa _k^2L_k} \ge \frac{2^n}{n!\kappa _n^2L_n}. \end{aligned}$$

In the proof of Corollary 1, we have seen that for every $\varepsilon >0$ the right hand side of the above inequality is at least $(\pi +\varepsilon )^{-n}$ for large enough $n\in \mathbb N$.

Now $g(k)\ge g(k+1)$ if and only if $\frac{k+1}{4}\ge \frac{\kappa _k^2}{\kappa _{k+1}^2}\frac{L_k}{L_{k+1}}$. By the estimate $\frac{\kappa _k^2}{\kappa _{k+1}^2}\le \frac{k+2}{2\pi }$ (see [2, Lem. 1]), it is therefore enough to prove $\frac{k+1}{4}\ge \frac{k+2}{2\pi }\frac{L_k}{L_{k+1}}$. After an elementary calculation, we see that this follows from the recurrence relation $L_{k+1}=\sum _{i=0}^{k+1}\genfrac(){0.0pt}{}{k+1}{i}\frac{2^i}{i!}=2L_k-\frac{2^k(k-1)}{(k+1)!}$. $\square$