This is an idle curiosity related to Guass's circle problem. Consider any sequence of bounded convex sets $V_1, V_2, \ldots \subseteq \mathbb{R}^2$. Let $L_n$ be the set of lattice points of $V_n$, i.e. $L_n = V_n \cap \mathbb{Z}^2$.

Additionally, suppose we have the following condition (motivated below in Notes): for any $\alpha \in \mathbb{Q}$, for the projection $\pi_\alpha$ from $\mathbb{R}^2$ to the line $y = \alpha x$, $$ \left| \pi_\alpha (L_n) \right| \to \infty. $$ Then must it be the case that $$ \lim_{n \to \infty} \frac{\operatorname{area}(V_n)}{|L_n|} = 1? $$


If we do not require the condition involving $\pi_\alpha$, then we may take $V_n$ to be a long rectangle along the line $y = \alpha x$ for a counterexample. (Specifically, if the line is $y = 0$ we take the rectangle $[-1.9, 1.9] \times [-n,n]$; if the line is $y = x$ we take the rectangle from $(-n,-n)$ to $(n,n)$ of width almost $\sqrt{2}$, etc.)

I have not specified any condition relating the different convex sets $V_n$; for instance, we don't assume $V_1 \subseteq V_2 \subseteq V_3 \subseteq \ldots$. Thus, I am essentially asking for a sort of uniform convergence of the lattice-point approximation to the area of a convex set, i.e. convergence across all convex sets at once.

I believe my condition is sufficient, but I do not know a proof approach. Perhaps Pick's theorem is useful if we begin by constructing a convex polygon with integer-coordinate vertices around $L_n$.

Del points out in the comments that my conditions do not exclude $V_n$ from being a thin rectangle of some irrational slope. This is true, and requiring the projection condition for $\alpha \in \mathbb{R}$ as well won't help. So maybe this can be used to make a counterexample.

  • $\begingroup$ I think it should be $\alpha\in \mathbb{R}$. Otherwise a family of thin rectangles with irrational slope should still provide a counterexample right? Or maybe they won't work and that's also part of the point? $\endgroup$ – Del Sep 16 '16 at 23:16
  • $\begingroup$ @Del Well, the reason I didn't allow $\alpha$ irrational is that it doesn't actually add anything -- any two distinct integer-coordinate points project to different points, so the condition is just saying that $\pi_{\alpha} (L_n) \to \infty$ which is already implied by the condition when $\alpha$ is rational. $\endgroup$ – 6005 Sep 16 '16 at 23:24
  • $\begingroup$ @Del However, maybe your idea will give a counterexample to my conjecture. The thin rectangles with irrational slopes are, to be sure, a nasty example that I didn't have in mind and that my conditions don't exclude. $\endgroup$ – 6005 Sep 16 '16 at 23:26
  • $\begingroup$ Your condition implies that the width of the convex sets along rational directions has to go to infinity, but not necessarily along irrational ones. However this doesn't make an explicit counterexample. it could well be the case that any constant-width irrational-slope family of rectangles nevertheless satisfies the limit about the area... $\endgroup$ – Del Sep 16 '16 at 23:36

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