Let $v_1,\dots, v_d$ be a $\mathbb Z$-basis for the rank $d$ lattice $\lambda$ in $\mathbb R^d$. By definition, $v_1,\dots, v_d$ are also $\mathbb R$ linearly independent.
For $v=t_1v_1 +\cdots +t_d v_d \in \mathbb R$, with $t_i\in [0,1), \forall i$
Let $m(v)$ denote the minimal distance (usual Euclidean $l^2$ distance) from $\Lambda$ to $v$. I wonder if we can give $m(v)$ an (sharp, if possible) upper bound in terms of the norms of $v_1,\dots, v_d$?
This answer gives a bound $m(v)^2 \le \sum_{i=1}^d t_i(1-t_i) \|v_i\|^2$ but unfortunately its proof is very confusing to me. I couldn't understand how the "weight" they defined is applied to each factor and why there are cancellations for the crossed terms. The following part of that answer is most confusing:
Therefore in the expansion of the weighted sum $$ \sum_{x'} w(x') \|x'\|^2 = \sum_{x'} w(x') \sum_{i=1}^n \sum_{j=1}^n c'_i c'_j (l_i,l_j) $$ each $i=j$ term contributes $c_i (1-c_i) \|l_i\|^2$ and each $i \neq j$ cross-term contributes zero (why? Do you guys understand how the product $w(x')$ is applied to each term exactly?). Hence the weighted sum is $\sum_{i=1}^n c_i (1-c_i) \|l_i\|^2 = M$, and the inequaity is proved.
It has been a while and it might be hard to ask the author. I wonder if anyone could write here a new proof or give a clear explanation for the original proof there.
