How Can I Prove this Version of Minkowski’s Theorem: $vol(C)>k2^d$ with $2k$ Lattice Points?

Question

Question Prove that if $C\subseteq \mathbb{R}^d$ is convex, centrally symmetric and bounded, with $vol(C)>k2^d$, then $C$ contains at least $2k$ lattice points (of lattice $\mathbb{Z}^d$).

Note Minkowski’s theorem talks about $vol(C)>2^d$ and $C$ contains at least one point apart from the origin.

Attempt Let $C^\prime = \frac{1}{2}C$. Then, $vol(C^\prime)=(\frac{1}{2})^dvol(C)>k$. I don’t know how to go from here. Induction seems to be a bad choice since if we take object at $k=1$ away from object at $k=2$, the object we will have left will not be convex.

I will greatly appreciate your help.

Answer 1

You could try like in the proof of standard Minkowski. There are 2 steps.

1. If $D_1$ is such that $m(D) > m(\Gamma)$ (where $\Gamma$ is any fundamental parellelotope of the lattice $L$, then $D$ does not map injectively on $V/L$, that is, there exist $d_1$, $d_2$ in $D$ such that $0\ne d_1-d_2\in L$ ( the intuition is clear, the proof is a bit subtle, but standard nowadays).

1'. Generalization: if $D$ is such that $m(D) > k \cdot m(\Gamma)$, then there exists distinct $d_1$, $\ldots$, $d_{k+1} \in D$ such that $d_i-d_j \in L$ ( a similar proof)

2. Take $D = \frac{1}{2} C$. Then $m(D) > m(\Gamma)$, so there exist distinct $d_1$, $\ldots$, $d_{k+1} \in D$ with differences in $L$ Now $d_i = \frac{1}{2} c_i$. But $C$ is convex and symmetric so the $k$ distinct elements $\frac{c_i-c_{k+1}}{2}$ are in $C$ and in $L\backslash\{0\}$.

$\bf{Added:}$ We found $c_1$, $\ldots$, $c_{k+1}$ in $C$, distinct, such that $\frac{c_i-c_j}{2}\in L$ for all $i,j$. Now, this gets us $k$ distinct non-zero $\frac{c_i-c_{k+1}}{2}$ in $L \cap C$. Of course, since $C$ is symmetric, we also have $\frac{c_{k+1}-c_{i}}{2} \in C\cap L$. Now, we would like to make sure that

$$\frac{c_i-c_{k+1}}{2}\ne \frac{c_{k+1}-c_{j}}{2}$$

This will be guaranteed if we pick $c_{k+1}$ to be an extreme point of the set $\{c_1, \ldots, c_{k+1}\}$.

If $c_{k+1}$ is chosen as above, then we will have $2k$ elements in $C\cap L \backslash\{0\}$, done.