Let $\Lambda$ be a lattice in $\mathbb{R}^n$. For $\bar{x} \in \mathbb{R}^n$, let 
$\| \bar{x} \| = max\_{1 \leq i \leq n} \{ |x\_i| \}$, i.e. the sup norm. Let $\lambda\_1, ..., \lambda\_n$
be a successive minima of $\Lambda$ with respect to the sup norm. 


I am interested in counting number of points 
$N(U) = \# \{ \bar{x} \in \Lambda : \| \bar{x} \| < U \}$ when say $\lambda\_{j} \leq U < \lambda\_{j+1}$. I am expecting some like
$$
\frac{U^j}{\lambda\_1 ... \lambda\_j} \ll N(U) \ll \frac{U^{j}}{\lambda\_1 ... \lambda\_{j}}
$$ 
when $\lambda\_{j} \leq U < \lambda\_{j+1}$, where the 
implicit constants may depend on $\det (\Lambda)$ and $n$, but not on $\Lambda$ itself.


I would appreciate any assistance, comments, or references. Thank you very much!


PS And I mean the inequalities as they are. Thank you!



Counting number of points in a lattice with bounded sup norm
Your expectation is correct. Let $\lambda\_{j} \leq U < \lambda\_{j+1}$. There are independent lattice vectors $v\_1,\dots,v\_j\in\Lambda$ such that $\|v\_i\|=\lambda\_i$ for $1\leq i\leq j$. 


Consider the linear combinations
$ c\_1v\_1+\dots +c\_j v\_j\in\Lambda$ with integral coefficients $c\_i\in\mathbb{Z}$ satisfying $|c\_i|\leq U/(2n\lambda\_i)$. Their sup-norms are $\leq U/2<U$, and their number is $\gg\_n U^j/(\lambda\_1\dots\lambda\_j)$. This proves the lower bound. 


Now consider the subspace $V\_j=\mathbb{R}v\_1+\dots+\mathbb{R}v\_j$ equipped with the $j$-volume coming from the euclidean norm in $\mathbb{R}^n$, i.e. an orthonormal basis in $V\_j$ spans a $j$-cube of $j$-volume $1$. Then, $\Lambda\_j=\Lambda\cap V\_j$ is a lattice in $V\_j$ with successive minima $\lambda\_1,\dots,\lambda\_j$, hence by [Minkowski's second theorem](http://en.wikipedia.org/wiki/Minkowski%27s_second_theorem) its fundamental parallelepiped $P\_j$ has $j$-volume
$$\text{vol}\_j(P\_j)\geq 2^{-j}\cdot\lambda\_1\dots\lambda\_j\cdot\text{vol}\_j(B\_j),$$
where $B\_j=\{ \bar{x} \in V\_j : \| \bar{x} \| \leq 1 \}.$
The sup-norm is majorized by the euclidean norm, hence $B\_j$ contains a euclidean $j$-ball of radius $1$, which shows that $\text{vol}\_j(P\_j)\gg\_j \lambda\_1\dots\lambda\_j$.
On the other hand,
$$\{ \bar{x} \in \Lambda : \| \bar{x} \| < U \}=\{ \bar{x} \in \Lambda\_j : \| \bar{x} \| < U \},$$ 
whence the translates of $P\_j$ by the elements of this set are pairwise disjoint and they all lie in $\{ \bar{x} \in V\_j : \| \bar{x} \| < (j+1)U \}$ whose $j$-volume is $\ll\_j U^j$. This proves the upper bound.


**Remark.** The above argument was enhanced and corrected by the OP's comments, for which I am grateful.


