A Blichfeldt-type inequality for the surface area

TL;DR

This paper establishes a new inequality relating the number of lattice points in a convex body to its volume and surface area, extending Blichfeldt's classical volume-based bound with novel surface area considerations.

Contribution

It introduces a Blichfeldt-type inequality involving surface area, providing bounds on lattice points that improve upon classical volume-based estimates, especially for non-lattice translates.

Findings

New inequality linking lattice points, volume, and surface area

Improved bounds for non-lattice translates of lattice polytopes

Stronger 3D inequality with a factor of 2 on surface area

Abstract

In 1921 Blichfeldt gave an upper bound on the number of integral points contained in a convex body in terms of the volume of the body. More precisely, he showed that $#(K\cap\Z^n)\leq n! \vol(K)+n$, whenever $K\subset\R^n$ is a convex body containing $n+1$ affinely independent integral points. Here we prove an analogous inequality with respect to the surface area $\F(K)$, namely $ #(K\cap\Z^n) < \vol(K) + ((\sqrt{n}+1)/2) (n-1)! \F(K)$. The proof is based on a slight improvement of Blichfeldt's bound in the case when $K$ is a non-lattice translate of a lattice polytope, i.e., $K=t+P$, where $t\in\R^n\setminus\Z^n$ and $P$ is an $n$-dimensional polytope with integral vertices. Then we have $#((t+P)\cap\Z^n)\leq n! \vol(P)$. Moreover, in the 3-dimensional case we prove a stronger inequality, namely $#(K\cap\Z^n) < \vol(K) + 2 \F(K)$.