# What does the image of the integer lattice under a norm look like?

The question that I shall ask here has arisen in the context of Diophantine approximation. I find it rather interesting, and I have no idea how to answer it. Any help, advice, or suggestions for references would be much appreciated.

Let $$S$$ be a countably infinite subset of $$\mathbb{R}_{>0}$$ that satisfies the following conditions. Let $$n$$ be an arbitrary element of $$\mathbb{Z}_{\geq 2}.$$

1) For any integer $$k \geq 1,$$ we have $$k \cdot S \subseteq S.$$ (In particular, $$S$$ is unbounded.)

2) There exists a (necessarily unique) strictly increasing sequence $$t_\bullet = (t_k)_{k \geq 1}$$ such that $$\{t_k : k \geq 1\} = S.$$ Moreover, there exists a constant $$C > 0$$ such that for any $$k \geq 1,$$ we have $$0 < t_{k+1} - t_k \leq C.$$ (Notice that we clearly have $$\lim_{k \to +\infty} t_k = +\infty$$, in light of 1) above.)

3) There exists a constant $$D > 0$$ such that $$\limsup_{T \to +\infty} \frac{\mathrm{card}\{ k \in \mathbb{Z}_{\geq 1} : t_k \leq T\}}{T^n} \leq D.$$

Is it necessarily true that there exists a norm $$\eta$$ on $$\mathbb{R}^n$$ such that $$\eta(\mathbb{Z}^n) - \{0\} = S$$? (Clearly conditions 1), 2), and 3) are necessary in order for this to be true; I am wondering whether they are also, in fact, sufficient.)

• Probably you want to require $n \geq 2$, because for $n=1$ the only allowed $S$ are arithmetic progressions $-$ but there are non-AP's that satisfy (1,2,3), such as the positive integers not congruent to $1$ or $5 \bmod 6$. Nov 6, 2019 at 1:41
• Thank you; in fact, in my particular context, I also always assume $n \geq 2.$ Nov 6, 2019 at 1:52
• Also condition (2) is automatic from (1) once $S \neq \emptyset$: if $S \ni t_1$ then $S$ also contains $2t_1, 3t_1, 4t_1, \ldots$, so there is never a gap longer than $t_1$. Nov 6, 2019 at 5:03

I doubt that there is a simple characterization. In any case conditions (1,2,3) are not sufficient. For example, if $$n=2$$ then $$S$$ cannot be $${\bf Z}_{\geq 100}$$, and there are similar counterexamples for every $$n \geq 2$$, as a consequence of the following observation.
Proposition. If $$\eta$$ is a norm on $${\bf R}^n$$, and $$t_1$$ is the smallest element of $$\eta({\bf Z}^n)$$, then for any $$M>0$$ there are at most $$(2M+1)^n$$ integer vectors $$v$$ with $$\eta(v) \leq M t_1$$.
Proof: Consider the $$\eta$$-balls of radius $$t_1/2$$ centered on all such $$v$$. They have disjoint interiors, and are all contained in the $$\eta$$-ball of radius $$(2M+1)t_1/2$$ about $$0$$. Therefore their total volume is no larger than the volume of the ball of radius $$(2M+1)t_1/2$$. But the volume of a ball of radius $$r$$ is proportional to $$r^n$$. Hence the total number of radius $$t_1/2$$ balls is at most $$\left[ ((2M+1)t_1/2) \left/ \, (t_1/2) \right. \right]^n = (2M+1)^n$$, Q.E.D.
Equality is attained when $$\eta$$ is the sup norm and $$M$$ is an integer. (Then we get a perfect packing of cubes.)