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Stretching convex domains to capture
many lattice points

Nicholas F. Marshall Department of Mathematics, Yale University, New Haven, CT 06511, USA nicholas.marshall@yale.edu
Abstract.

We consider an optimal stretching problem for strictly convex domains in d\mathbb{R}^{d} that are symmetric with respect to each coordinate hyperplane, where stretching refers to transformation by a diagonal matrix of determinant 11. Specifically, we prove that the stretched convex domain which captures the most positive lattice points in the large volume limit is balanced: the (d1)(d-1)-dimensional measures of the intersections of the domain with each coordinate hyperplane are equal. Our results extend those of Antunes & Freitas, van den Berg, Bucur & Gittins, Ariturk & Laugesen, van den Berg & Gittins, and Gittins & Larson. The approach is motivated by the Fourier analysis techniques used to prove the classical #{(i,j)2:i2+j2r2}=πr2+𝒪(r2/3)\#\{(i,j)\in\mathbb{Z}^{2}:i^{2}+j^{2}\leq r^{2}\}=\pi r^{2}+\mathcal{O}(r^{2/3}) result for the Gauss circle problem.

Key words and phrases:
Lattice point counting, shape optimization, decay of Fourier transform
2010 Mathematics Subject Classification:
42B10, 52C07 (primary) and 11H06, 35P15, 90C27 (secondary)

1. Introduction

1.1. Introduction

In [1] Antunes & Freitas introduced a new type of lattice point problem; namely, among all ellipses that are symmetric about the coordinate axes and of fixed area, which captures the most positive lattice points? More precisely, for r1r\geq 1 what is:

a(r)=argmaxa>0#{(i,j)>02:(ia)2+(ja)2r2}.a(r)=\operatorname*{argmax}_{a>0}\,\#\left\{(i,j)\in\mathbb{Z}^{2}_{>0}:\left(\frac{i}{a}\right)^{2}+\left(ja\right)^{2}\leq r^{2}\right\}.

Determining a(r)a(r) for fixed r1r\geq 1 is challenging, even computationally [1]; however, Antunes & Freitas were able to prove that

limra(r)=1,\lim_{r\rightarrow\infty}a(r)=1,

i.e., the ellipse which captures the most positive lattice points for large areas approaches a circle. Moreover, a rate of convergence of at least

|a(r)1|=𝒪(r1/6)as r,|a(r)-1|=\mathcal{O}(r^{-1/6})\quad\text{as }r\rightarrow\infty,

was established in [10] (and implied by Section 5 in [1]).

Refer to caption
Figure 1. Eight ellipses which each capture the maximum possible number of positive lattice points for their given area.

This lattice point counting problem was originally motivated by a problem in high frequency shape optimization. Suppose that Ω2\Omega\subset\mathbb{R}^{2} is an a×1/aa\times 1/a rectangular domain. Then the eigenvalues of the Dirichlet Laplacian ΔΩ𝒟-\Delta_{\Omega}^{\mathcal{D}} are of the form

σ(ΔΩ𝒟)={π2((ia)2+(aj)2):i,j>0}.\sigma\left(-\Delta_{\Omega}^{\mathcal{D}}\right)=\left\{\pi^{2}\left(\left(\frac{i}{a}\right)^{2}+\left(aj\right)^{2}\right):i,j\in\mathbb{Z}_{>0}\right\}.

Thus, there is a bijection between the Dirichlet eigenvalues less than π2r2\pi^{2}r^{2}, and the positive lattice points in the ellipse (x/a)2+(ay)2r2(x/a)^{2}+(ay)^{2}\leq r^{2}. Hence, the statement limra(r)=1\lim_{r\rightarrow\infty}a(r)=1 can also be interpreted as the statement that the rectangle that minimizes the Dirichlet eigenvalues in the high frequency limit approaches the square.

1.2. Neumann eigenvalues

A dual result for eigenvalues of the Neumann Laplacian ΔΩ𝒩-\Delta_{\Omega}^{\mathcal{N}} was established by van den Berg, Bucur & Gittins [17], where as before Ω2\Omega\subset\mathbb{R}^{2} is an a×1/aa\times 1/a rectangular domain. The eigenvalues of ΔΩ𝒩-\Delta_{\Omega}^{\mathcal{N}} are of the form

σ(ΔΩ𝒩)={π2((ia)2+(aj)2):i,j0},\sigma\left(-\Delta_{\Omega}^{\mathcal{N}}\right)=\left\{\pi^{2}\left(\left(\frac{i}{a}\right)^{2}+\left(aj\right)^{2}\right):i,j\in\mathbb{Z}_{\geq 0}\right\},

and hence, the Neumann eigenvalues less than π2r2\pi^{2}r^{2} are in bijection with the nonnegative lattice points in the ellipse (x/a)2+(ay)2r2(x/a)^{2}+(ay)^{2}\leq r^{2}. In terms of a lattice point problem, van den Berg, Bucur & Gittins proved that if

a(r)=argmina>0#{(i,j)02:(ia)2+(ja)2r2},a(r)=\operatorname*{argmin}_{a>0}\,\#\left\{(i,j)\in\mathbb{Z}^{2}_{\geq 0}:\left(\frac{i}{a}\right)^{2}+\left(ja\right)^{2}\leq r^{2}\right\},

then limra(r)=1\lim_{r\rightarrow\infty}a(r)=1. That is, the ellipse which captures the least nonnegative lattice points in the large area limit approaches the circle; equivalently, the rectangle which maximizes the Neumann eigenvalues in the high frequency limit approaches the square.

1.3. Higher dimensions

Subsequently, the result for Dirichlet eigenvalues was generalized to three dimensions by Gittins & van den Berg [18], and recently to dd-dimensions for both the Dirichlet and Neumann cases by Gittins & Larson [5]. Specifically, Gittins & Larson show that in d\mathbb{R}^{d} the cuboid of unit measure which minimizes the Dirichlet eigenvalues in the high frequency limit approaches the cube, and similarly, that the cuboid of unit measure which maximizes the Neumann Laplacian eigenvalues in the high frequency limit approaches the cube. Both the Dirichlet and Neumann cases have corresponding lattice point problems analogous to the 22-dimensional case. The Dirichlet case corresponds to the following lattice point problem. Suppose A=diag(a1,a2,,ad)A=\operatorname{diag}(a_{1},a_{2},\ldots,a_{d}) is a positive diagonal matrix of determinant 11. For r1r\geq 1, let

A(r)=argmaxA#{(i1,i2,,id)>0d:(i1a1)2+(i2a2)2++(idad)2r2}.A(r)=\operatorname*{argmax}_{A}\,\#\left\{(i_{1},i_{2},\ldots,i_{d})\in\mathbb{Z}^{d}_{>0}:\left(\frac{i_{1}}{a_{1}}\right)^{2}+\left(\frac{i_{2}}{a_{2}}\right)^{2}+\cdots+\left(\frac{i_{d}}{a_{d}}\right)^{2}\leq r^{2}\right\}.

Then the result of [5] implies that limrAId=0\lim_{r\rightarrow\infty}\|A-\operatorname{Id}\|_{\infty}=0. Moreover, Gittins & Larson [5] show the following error estimate holds for dimensions d2d\geq 2:

AId=𝒪(rd12(d+1))as r,\left\|A-\operatorname{Id}\right\|_{\infty}=\mathcal{O}\left(r^{-\frac{d-1}{2(d+1)}}\right)\quad\text{as }r\rightarrow\infty,

and furthermore, they show a slightly improved error rate holds for d>5d>5 by applying a result of Götze [6]. In this paper, we establish a similar rate of convergence for general convex domains. We note that the Neumann case in dd-dimensions has an analogous dual formulation to the 22-dimensional case, and similar error estimates were established in [5]. We remark that results concerning the optimization of eigenvalues of the Laplacian with perimeter constraints have also been considered [2, 4, 16]. In particular, Bucur & Freitas [4] proved that any sequence of minimizers of the Dirichlet eigenvalues with a perimeter constraint converges to the unit disk in the high frequency limit, and that among kk-sided polygons, any sequence of minimizers converges to the regular kk-sided polygon in the high frequency limit.

1.4. Convex and concave curves

Laugesen & Liu [10] and Ariturk & Laugesen [3] generalized the two dimensional case of the above lattice point counting problems to certain classes of convex and concave curves. In particular, their results imply that among pp-ellipses |x/a|p+|ay|p=rp\left|x/a\right|^{p}+\left|ay\right|^{p}=r^{p} for p(0,){1}p\in(0,\infty)\setminus\{1\} the pp-ball captures the most positive lattice points. More generally, Laugesen & Liu and Ariturk & Laugesen show that the curves which capture the most positive integer lattice points in the large area limit are balanced: the distance from the origin to their points of intersection with the xx-axis and yy-axis are equal.

Refer to caption
Figure 2. Among all stretches of a concave curve, which captures the most positive lattice points?

1.5. Right triangles

The results of [3, 10] exclude the p=1p=1 case. In contrast to other values of pp, for the p=1p=1 case (where pp-ellipses are right triangles) Ariturk & Laugesen and Laugesen & Liu conjectured that the optimal triangle

“does not approach a 45-45-90 degree triangle as rr\rightarrow\infty. Instead one seems to get an infinite limit set of optimal triangles” (from [10]).

The author and Steinerberger [11] recently proved this conjecture and showed that the limit set is fractal of Minkowski dimension at most 3/43/4. Furthermore, all triangles which are optimal for infinitely many infinitely large areas have rational slopes, are contained in [1/3,3][1/3,3], and have 11 as a unique accumulation point.

Refer to caption Refer to caption
Figure 3. An illustration of the fractal set of infinitely many times optimal slopes.

1.6. Relation to spectral asymptotics

Suppose that RdR\subset\mathbb{R}^{d} is a cuboid, and let NR(λ)N_{R}(\lambda) denote the number of eigenvalues of the Dirichlet Laplacian ΔR𝒟-\Delta_{R}^{\mathcal{D}} that are less than λ\lambda

NR(λ)=#{μσ(ΔR𝒟):μλ}.N_{R}(\lambda)=\#\left\{\mu\in\sigma\left(-\Delta_{R}^{\mathcal{D}}\right):\mu\leq\lambda\right\}.

Then NR(λ)N_{R}(\lambda) has a two-term asymptotic formula in terms of the volume |R||R| and surface area |R||\partial R| of RR

NR(λ)=vd(2π)d|R|λd/2vd14(2π)d1|R|λd12+oR(λd12),asλ,N_{R}(\lambda)=\frac{v_{d}}{(2\pi)^{d}}|R|\lambda^{d/2}-\frac{v_{d-1}}{4(2\pi)^{d-1}}|\partial R|\lambda^{\frac{d-1}{2}}+o_{R}\left(\lambda^{\frac{d-1}{2}}\right),\quad\text{as}\quad\lambda\rightarrow\infty,

where vdv_{d} is the volume of the unit ball in d\mathbb{R}^{d}. A similar asymptotic formula holds for the eigenvalues of the Neumann Laplacian ΔR𝒩-\Delta_{R}^{\mathcal{N}}, see [8]. We remark that in 1913, Weyl, see page 199 of [21], speculated about the existence of such a two-term asymptotic formula for more general domains in 3\mathbb{R}^{3}, and the problem of establishing the above two-term asymptotic formula for general domains became known as Weyl’s Conjecture, see [8]. In 1980, Ivrii [9] proved that this two-term asymptotic formula indeed holds for more general domains under certain conditions. Suppose that RR is an a1×a2××ada_{1}\times a_{2}\times\cdots\times a_{d} cuboid. Then the eigenvalues of the Dirichlet Laplacian ΔR𝒟-\Delta^{\mathcal{D}}_{R} are of the form

σ(ΔR𝒟)={π2((i1a1)2+(i2a2)2++(idad)2):(i1,i2,,id)>0d},\sigma\left(-\Delta_{R}^{\mathcal{D}}\right)=\left\{\pi^{2}\left(\left(\frac{i_{1}}{a_{1}}\right)^{2}+\left(\frac{i_{2}}{a_{2}}\right)^{2}+\cdots+\left(\frac{i_{d}}{a_{d}}\right)^{2}\right):(i_{1},i_{2},\ldots,i_{d})\in\mathbb{Z}_{>0}^{d}\right\},

and thus, the eigenvalues of ΔR𝒟-\Delta_{R}^{\mathcal{D}} less than π2r2\pi^{2}r^{2} are in bijection with the positive lattice points in the ellipsoid (x1/a1)2+(x2/a2)2++(xd/ad)2r2(x_{1}/a_{1})^{2}+(x_{2}/a_{2})^{2}+\cdots+(x_{d}/a_{d})^{2}\leq r^{2}. If this cuboid RR has unit measure |R|=1|R|=1, then its surface area |R||\partial R| is given by

|R|=j=1d2aj1=2(trA1),whereA=diag(a1,a2,,ad).|\partial R|=\sum_{j=1}^{d}2a_{j}^{-1}=2\left(\operatorname{tr}A^{-1}\right),\quad\text{where}\quad A=\operatorname{diag}(a_{1},a_{2},\ldots,a_{d}).

Substituting λ=π2r2\lambda=\pi^{2}r^{2} into the above two-term asymptotic formula for NR(λ)N_{R}(\lambda) for this cuboid gives

NR(π2r2)=12dvdrd12dvd1(trA1)rd1+oR(rd1),asr.N_{R}(\pi^{2}r^{2})=\frac{1}{2^{d}}v_{d}\,r^{d}-\frac{1}{2^{d}}v_{d-1}\left(\operatorname{tr}A^{-1}\right)r^{d-1}+o_{R}\left(r^{d-1}\right),\quad\text{as}\quad r\rightarrow\infty.

We emphasize that the error term in this asymptotic formula depends implicitly on the cuboid RR, and therefore, this asymptotic formula by itself is insufficient to determine which cuboid RR of unit measure maximizes NR(λ)N_{R}(\lambda) as λ\lambda\rightarrow\infty. Addressing this issue is the main challenge in [1, 5, 17, 18]. If we were to ignore the error term, then maximizing NR(λ)N_{R}(\lambda) among cuboids of unit measure would be equivalent to minimizing trA1\operatorname{tr}A^{-1}. By the arithmetic mean geometric mean inequality

trA1=da11+a21++ad1dd(a11a21ad1)1/d=d,\operatorname{tr}A^{-1}=d\cdot\frac{a_{1}^{-1}+a_{2}^{-1}+\cdots+a_{d}^{-1}}{d}\geq d\cdot\left(a_{1}^{-1}\cdot a_{2}^{-1}\cdot\cdots\cdot a_{d}^{-1}\right)^{1/d}=d,

with equality if and only if A=IdA=\operatorname{Id}. Since 2trA1=|R|2\operatorname{tr}A^{-1}=|\partial R|, this inequality can be interpreted as an isoperimetric inequality for cuboids: the surface area of a cuboid of unit measure is greater than or equal to the surface area of the unit cube with equality if and only if the cuboid is the unit cube. Ultimately, after the main challenge of dealing with the error term has been appropriately handled, this isoperimetric inequality for cuboids is the reason that the cube is asymptotically optimal in [1, 5, 17, 18]. In this paper, we consider a positive lattice point counting problem in more general domains; specifically, we study the number of positive lattice points in a fixed convex domain Ω\Omega that has been scaled by r1r\geq 1 and stretched by a linear transformation AA represented by a positive diagonal matrix of determinant 11. In this generalized setting, the term trA1\operatorname{tr}A^{-1} similarly arises, and we use Fourier analysis to develop lattice point counting results which lead to uniform error estimates for optimal stretching problems.

1.7. Motivation

In this paper, our motivation is twofold: first, the results of Laugesen & Liu and Ariturk & Laugesen show that the asymptotic balancing observed in lattice point problems for ellipses [1, 5, 17, 18] extends from ellipses to a more general class of convex and concave curves, at least in two dimensions, and second, the analysis of the p=1p=1 case in [11] shows that the convergence breaks down when the curves become flat. This phenomenon is common in harmonic analysis, where the decay of the Fourier transform is dependent on non-vanishing curvature. To briefly review the relation of Fourier analysis to lattice point problems, suppose ff is a CC^{\infty} function on d\mathbb{R}^{d} of compact support. Then the Poisson summation formula states that

ndf(n)=ndf^(n),\sum_{n\in\mathbb{Z}^{d}}f(n)=\sum_{n\in\mathbb{Z}^{d}}\widehat{f}(n),

where f^\widehat{f} denotes the Fourier transform of ff

f^(ξ)=df(x)e2πixξ𝑑x.\widehat{f}(\xi)=\int_{\mathbb{R}^{d}}f(x)e^{-2\pi ix\cdot\xi}dx.

Suppose that χ(x)\chi(x) is the indicator function for a domain Ωd\Omega\subset\mathbb{R}^{d}. Then the indicator function for the scaled domain rΩr\Omega is χr(x)=χ(x/r)\chi_{r}(x)=\chi(x/r), whose Fourier transform, by a change of variables is

χ^r(ξ)=rdχ^(rξ).\widehat{\chi}_{r}(\xi)=r^{d}\widehat{\chi}(r\xi).

Moreover, since χ^r(0)=|Ω|rd\widehat{\chi}_{r}(0)=|\Omega|r^{d}, if the Poisson summation formula could be applied to χr\chi_{r}, it would express the number of lattice points inside rΩr\Omega as |Ω|rd|\Omega|r^{d} plus the sum of χ^r\widehat{\chi}_{r} over the nonzero lattice points. Unfortunately, the Fourier transforms of indicator functions do not decay rapidly enough for the Poisson summation formula to be applied (these functions lack sufficient smoothness). However, this issue can be resolved by smoothing the indicator function by convolution with a bump function, and useful estimates can be obtained. This approach can be used to establish the classical

#{(i,j)2:i2+j2r2}=πr2+𝒪(r2/3),\#\{(i,j)\in\mathbb{Z}^{2}:i^{2}+j^{2}\leq r^{2}\}=\pi r^{2}+\mathcal{O}(r^{2/3}),

result for the Gauss circle problem accredited to Sierpiński [12], van der Corput [19], and Voronoi [20], which was the first non-trivial step towards the conjectured result:

#{(i,j)2:i2+j2r2}=πr2+𝒪(r1/2+ε),\#\{(i,j)\in\mathbb{Z}^{2}:i^{2}+j^{2}\leq r^{2}\}=\pi r^{2}+\mathcal{O}\left(r^{1/2+\varepsilon}\right),

for all ε>0\varepsilon>0. Currently the best known result is 𝒪(r131/208)\mathcal{O}(r^{131/208}) due to Huxley [7]. The argument for the 𝒪(r2/3)\mathcal{O}(r^{2/3}) error term for the circle in 2\mathbb{R}^{2} can be generalized for convex domains Ωd\Omega\subset\mathbb{R}^{d} whose boundary Ω\partial\Omega has nowhere vanishing Gauss curvature [13]. This generalization results in the following bound on the number of enclosed lattice points:

#{ndrΩ}=|Ω|rd+𝒪(rd2dd+1).\#\{n\in\mathbb{Z}^{d}\cap r\Omega\}=|\Omega|r^{d}+\mathcal{O}\left(r^{d-\frac{2d}{d+1}}\right).

In this paper, we follow a similar approach to the proof of this result, but additionally handle the effects of stretching the domain. More specifically, given a strictly convex domain Ω\Omega which is symmetric with respect to each coordinate hyperplane, we consider the positive lattice points contained in the domain A(rΩ)={A(rx):xΩ}A(r\Omega)=\{A(rx):x\in\Omega\}, where AA is a positive diagonal matrix of determinant 11, and r1r\geq 1 is a scaling factor.

Refer to caption
Figure 4. The circle under the group action of positive diagonal matrices of determinant 11.

In the following we compute an asymptotic expansion for #{ndA(rΩ)}\#\{n\in\mathbb{Z}^{d}\cap A(r\Omega)\} which includes both the effects of the diagonal transformation AA and scaling factor rr; we use this expansion to derive uniform error estimates for an optimal stretching problem. The resulting theorem extends the results of [1, 3, 5, 17, 18]. We note that we do not completely recover the results of [3, 10], since our approach can only represent curves which can be realized as the boundary of a convex domain whose boundary has nowhere vanishing Gauss curvature. However, there is some hope that the presented framework could be used to fully generalize the results of Laugesen & Liu [10] and Ariturk & Laugesen [3] by using more delicate bounds on the decay of the Fourier transform. Thus the presented approach may be useful for proving further generalizations.

2. Main Result

2.1. Main Result

Suppose that Ωd\Omega\subset\mathbb{R}^{d} is a bounded convex domain whose boundary Ω\partial\Omega is Cd+2C^{d+2} and has nowhere vanishing Gauss curvature. Furthermore, suppose that Ω\Omega is symmetric with respect to each coordinate hyperplane and balanced in the following sense.

Definition 2.1.

We say that a bounded convex domain Ωd\Omega\subset\mathbb{R}^{d} is balanced if

|{(x1,x2,,xd)Ω:xj=0}|=1for j=1,2,,d,\left|\left\{(x_{1},x_{2},\ldots,x_{d})\in\Omega:x_{j}=0\right\}\right|=1\quad\text{for }j=1,2,\ldots,d,

where |||\cdot| denotes the (d1)(d-1)-dimensional Lebesgue measure.

Note that the balanced assumption does not restrict the domains for which the below Theorem applies. Rather, the assumption that Ω\Omega is balanced is equivalent to choosing the unique balanced representative BΩB\Omega for Ω\Omega, where BB is a positive diagonal matrix. Suppose that A=diag(a1,a2,,ad)A=\operatorname{diag}(a_{1},a_{2},\ldots,a_{d}) is a positive definite diagonal matrix of determinant 11. To reiterate, we define

A(rΩ)={A(rx):xΩ},A(r\Omega)=\{A(rx):x\in\Omega\},

for scaling factor r1r\geq 1. That is to say, A(rΩ)A(r\Omega) is the domain Ω\Omega scaled by rr and transformed by AA.

Refer to caption Refer to caption
Figure 5. Each domain Ωd\Omega\subset\mathbb{R}^{d} which contains the origin has a unique balanced representative; the domains illustrated on the left share the balanced representative illustrated on the right.
Theorem 2.1.

Suppose Ωd\Omega\subset\mathbb{R}^{d} and A:ddA:\mathbb{R}^{d}\rightarrow\mathbb{R}^{d} are as described above. Then for d2d\geq 2 and r1r\geq 1,

#{n>0dA(rΩ)}=12d|Ω|rd12d(trA1)rd1+𝒪(A12dd+1rd2dd+1),\#\left\{n\in\mathbb{Z}^{d}_{>0}\cap A(r\Omega)\right\}=\frac{1}{2^{d}}|\Omega|r^{d}-\frac{1}{2^{d}}\left(\operatorname{tr}A^{-1}\right)r^{d-1}+\mathcal{O}\left(\|A^{-1}\|^{\frac{2d}{d+1}}_{\infty}r^{d-\frac{2d}{d+1}}\right),

where the implicit constant is independent of AA and rr. Moreover, if

A(r)=argmaxA#{n>0dA(rΩ)},A(r)=\operatorname*{argmax}_{A}\,\#\left\{n\in\mathbb{Z}^{d}_{>0}\cap A(r\Omega)\right\},

then

A(r)Id=𝒪(rd12(d+1))as r,\|A(r)-\operatorname{Id}\|_{\infty}=\mathcal{O}\left(r^{-\frac{d-1}{2(d+1)}}\right)\quad\text{as }r\rightarrow\infty,

where the implicit constant is independent of AA and rr.

In the following remark, we describe specifically how the implicit constants in Theorem 2.1 depend on Ωd\Omega\subset\mathbb{R}^{d}, and provide conditions under which the expansion and convergence rate results in Theorem 2.1 hold uniformly over a family of domains.

Remark 2.1.

The implicit constants in Theorem 2.1 which depend on the domain Ωd\Omega\subset\mathbb{R}^{d} can be chosen in terms of the following three quantities: a lower bound on the Gauss curvature of Ω\Omega, an upper bound on the diameter of Ω\Omega, and a lower bound on the inradius of Ω\Omega. Therefore, the result of Theorem 2.1 holds uniformly over any family of admissible domains, which have uniform bounds for these three quantities. Indeed, constants depending on Ωd\Omega\subset\mathbb{R}^{d} enter the proof of Theorem 2.1 from three sources. First, we use a constant C>0C>0 such that Ω[C,C]d\Omega\subset[-C,C]^{d}; since Ω\Omega contains the origin it suffices to choose CC equal to the diameter of Ω\Omega. Second, we use a constant c>0c>0 from Lemma 3.1; by the proof of Lemma 3.1, this constant can be chosen as one divided by the inradius of Ω\Omega. Third, we implicitly use a constant when using the bound in Lemma 3.3 for the decay of the Fourier transform of the indicator function for Ω\Omega; this implicit constant can be chosen in terms of a lower bound on the Gauss curvature of the domain, cf. [13].

When Ωd\Omega\subset\mathbb{R}^{d} is a dd-dimensional ellipsoid, the Theorem implies the Dirichlet Laplacian results of Gittins & Larson [5]. Applying the Theorem for dimension d=2d=2 recovers the original result of Antunes & Freitas [1] and agrees with the error estimate in [10]. Specifically, in dimension d=2d=2, the set of all positive diagonal matrices of determinant 11 is the 11-parameter family A=diag(1/a,a)A=\operatorname{diag}(1/a,a) for a>0a>0 and the result of the Theorem can be stated as:

Corollary 2.1.

In the case Ω2\Omega\subset\mathbb{R}^{2}, the Theorem gives

#{n>02A(rΩ)}=14|Ω|r214(a+1a)r+𝒪(a4/3r2/3),\#\left\{n\in\mathbb{Z}^{2}_{>0}\cap A(r\Omega)\right\}=\frac{1}{4}|\Omega|r^{2}-\frac{1}{4}\left(a+\frac{1}{a}\right)r+\mathcal{O}\left(a^{4/3}r^{2/3}\right),

and

|a(r)1|=𝒪(r1/6).|a(r)-1|=\mathcal{O}\left(r^{-1/6}\right).

As a second Corollary we can establish dual results for the nonnegative lattice point problem, which corresponds to the Neumann Laplacian results of Gittins & Larson [5].

Corollary 2.2.

Suppose Ωd\Omega\subset\mathbb{R}^{d} and A:ddA:\mathbb{R}^{d}\rightarrow\mathbb{R}^{d} satisfy the hypotheses of the Theorem, and additionally suppose 1A1Cr1\leq\|A^{-1}\|_{\infty}\leq Cr for a fixed constant C>0C>0. Then

#{n0dA(rΩ)}=12d|Ω|rd+12d(trA1)rd1+𝒪(A12dd+1rd2dd+1),\#\left\{n\in\mathbb{Z}^{d}_{\geq 0}\cap A(r\Omega)\right\}=\frac{1}{2^{d}}|\Omega|r^{d}+\frac{1}{2^{d}}\left(\operatorname{tr}A^{-1}\right)r^{d-1}+\mathcal{O}\left(\|A^{-1}\|^{\frac{2d}{d+1}}_{\infty}r^{d-\frac{2d}{d+1}}\right),

where the implicit constant is independent of AA and rr. Moreover, if

A(r)=argminA#{n0dA(rΩ)},A(r)=\operatorname*{argmin}_{A}\,\#\{n\in\mathbb{Z}^{d}_{\geq 0}\cap A(r\Omega)\},

then

A(r)Id=𝒪(rd12(d+1)),\|A(r)-\operatorname{Id}\|_{\infty}=\mathcal{O}\left(r^{-\frac{d-1}{2(d+1)}}\right),

where the implicit constant is independent of AA and rr.

The assumption 1A1Cr1\leq\|A^{-1}\|_{\infty}\leq Cr is needed for the expansion to hold, but not necessary for the convergence results as it serves to avoid the case where A(rΩ)A(r\Omega) contains more than order rdr^{d} lattice points on the coordinate hyperplanes which is clearly non-optimal for the argmin\operatorname*{argmin}. The proof of this Corollary follows from the arguments in Step 3.3 of the proof of Theorem 2.1.

2.2. Generalization

In the following we describe a generalization of Theorem 2.1. In particular, we remark how Theorem 2.1 can be generalized to domains Ω\Omega which are not necessarily symmetric with respect to each coordinate hyperplane by considering the lattice points {nd:nj0,j=1,,d}\{n\in\mathbb{Z}^{d}:n_{j}\not=0,\forall j=1,\ldots,d\} rather than the positive lattice points >0d\mathbb{Z}^{d}_{>0}.

Remark 2.2.

Suppose that Ωd\Omega\subset\mathbb{R}^{d} is a bounded convex domain whose boundary Ω\partial\Omega is Cd+2C^{d+2} and has nowhere vanishing Gauss curvature. Moreover, suppose Ω\Omega is balanced in the sense of Definition 2.1, and contains the origin. Define 0d:={nd:nj0,j=1,,d}\mathbb{Z}^{d}_{\not=0}:=\{n\in\mathbb{Z}^{d}:n_{j}\not=0,\,\forall j=1,\ldots,d\}. Let A=diag(a1,a2,,ad)A=\operatorname{diag}(a_{1},a_{2},\ldots,a_{d}) be a positive diagonal matrix of determinant 11. Then

#{n0dA(rΩ)}=|Ω|rd(trA1)rd1+𝒪(A12dd+1rd2dd+1),\#\left\{n\in\mathbb{Z}^{d}_{\not=0}\cap A\left(r\Omega\right)\right\}=|\Omega|r^{d}-\left(\operatorname{tr}A^{-1}\right)r^{d-1}+\mathcal{O}\left(\|A^{-1}\|^{\frac{2d}{d+1}}_{\infty}r^{d-\frac{2d}{d+1}}\right),

where the implicit constant is independent of AA and rr. Moreover, if

A(r)=argmaxA#{n0dA(rΩ)},A(r)=\operatorname*{argmax}_{A}\,\#\left\{n\in\mathbb{Z}^{d}_{\not=0}\cap A\left(r\Omega\right)\right\},

then

A(r)Id=𝒪(rd12(d+1)),\|A(r)-\operatorname{Id}\|_{\infty}=\mathcal{O}\left(r^{-\frac{d-1}{2(d+1)}}\right),

where the implicit constant is independent of AA and rr. The proof of this statement is immediate from Step 3.3 of the proof of Theorem 2.1.

3. Proof of main result

3.1. Proof strategy

Before discussing the technical details, we describe the proof strategy. The proof is divided into five steps. First, we establish a new lattice point counting result which holds uniformly over a family of positive diagonal matrices AA with determinant 11. Specifically, we show that

#{ndA(rΩ)}=|Ω|rd+𝒪(A12dd+1rd2dd+1),\#\{n\in\mathbb{Z}^{d}\cap A(r\Omega)\}=|\Omega|r^{d}+\mathcal{O}\left(\|A^{-1}\|_{\infty}^{\frac{2d}{d+1}}r^{d-\frac{2d}{d+1}}\right),

when 1A1Cr1\leq\|A^{-1}\|_{\infty}\leq Cr where C>0C>0 is a fixed constant. The key observation in the proof of this expansion is that the classical Fourier analysis techniques used to study the Gauss circle problem can be applied if the indicator function of the domain A(rΩ)A(r\Omega) is mollified by a bump function which has been appropriately stretched and scaled. Second, as a consequence of this result we show that the number of lattice points on the coordinate hyperplanes is, as expected, equal to (trA1)rd1\left(\operatorname{tr}A^{-1}\right)r^{d-1} plus an appropriate error term; specifically,

#{ndA(rj=1dΩj)}=(trA1)rd1+𝒪(A12dd+1rd2dd+1),\#\left\{n\in\mathbb{Z}^{d}\cap A\left(r\bigcup_{j=1}^{d}\Omega_{j}\right)\right\}=(\operatorname{tr}A^{-1})r^{d-1}+\mathcal{O}\left(\|A^{-1}\|_{\infty}^{\frac{2d}{d+1}}r^{d-\frac{2d}{d+1}}\right),

where Ωj\Omega_{j} denotes the intersection of Ω\Omega with the hyperplane orthogonal to the jj-th coordinate vector. Third, we combine these two results to produce a positive lattice point counting result for A(rΩ)A(r\Omega):

#{n>0dA(rΩ)}=12d|Ω|rd12d(trA1)rd1+𝒪(A12dd+1rd2dd+1).\#\left\{n\in\mathbb{Z}_{>0}^{d}\cap A(r\Omega)\right\}=\frac{1}{2^{d}}|\Omega|r^{d}-\frac{1}{2^{d}}\left(\operatorname{tr}A^{-1}\right)r^{d-1}+\mathcal{O}\left(\|A^{-1}\|_{\infty}^{\frac{2d}{d+1}}r^{d-\frac{2d}{d+1}}\right).

Fourth, we show that this positive lattice point counting result implies that

A(r)Id0,as r,whereA(r)=argmaxA#{n>0dA(rΩ)},\|A(r)-\operatorname{Id}\|\rightarrow 0,\quad\text{as }r\rightarrow\infty,\quad\text{where}\quad A(r)=\operatorname*{argmax}_{A}\,\#\left\{n\in\mathbb{Z}^{d}_{>0}\cap A(r\Omega)\right\},

where the argmax\operatorname*{argmax} is taken over positive diagonal matrices AA of determinant 11. We note that clearly A(r)1=𝒪(r)\|A(r)^{-1}\|_{\infty}=\mathcal{O}(r) since otherwise A(rΩ)A(r\Omega) will not contain any positive lattice points; we proceed by considering two cases. First, we consider the case where

A1is on the order ofr,\|A^{-1}\|_{\infty}\quad\text{is on the order of}\quad r,

where our positive lattice point counting result for A(rΩ)A(r\Omega) provides no information because the error term is order rdr^{d}. However, in such an extreme situation we are able to independently show non-optimality. Second, we assume that

A1=o(r),\|A^{-1}\|_{\infty}=o(r),

and use the positive lattice point counting result to show convergence. Specifically, we assume

A1=ψ(r)r,whereψ(r)0,asr.\|A^{-1}\|_{\infty}=\psi(r)r,\quad\text{where}\quad\psi(r)\rightarrow 0,\quad\text{as}\quad r\rightarrow\infty.

Substituting A1=ψ(r)r\|A^{-1}\|_{\infty}=\psi(r)r into the positive lattice point counting result for A(rΩ)A(r\Omega) gives

#{n>0dA(rΩ)}12d|Ω|rd(12dCψ(r)d1d+1)ψ(r)rd,\#\left\{n\in\mathbb{Z}_{>0}^{d}\cap A(r\Omega)\right\}\leq\frac{1}{2^{d}}|\Omega|r^{d}-\left(\frac{1}{2^{d}}-C\psi(r)^{\frac{d-1}{d+1}}\right)\psi(r)r^{d},

where CC is a fixed constant. Since ψ0\psi\rightarrow 0 the term in the parentheses we eventually be positive, and hence, comparison to the case where A=IdA=\operatorname{Id} implies convergence. Fifth, and finally, given the fact that A(r)A(r) converges to Id\operatorname{Id} we establish the convergence rate

A(r)Id=𝒪(rd12(d+1)),\|A(r)-\operatorname{Id}\|_{\infty}=\mathcal{O}\left(r^{-\frac{d-1}{2(d+1)}}\right),

using a standard arithmetic mean geometric mean argument.

3.2. Useful lemmata

The proof of the main result relies on three lemmata: two geometric in nature, and one related to the decay of the Fourier transform of indicator functions of convex domains with nowhere vanishing Gauss curvature. Lemmata 3.1 and 3.2 are proved in this section, and are motivated by similar technical results in [14], while Lemma 3.3 appears in standard references, cf., [13, 14, 15].

Lemma 3.1.

Suppose that Ωd\Omega\subset\mathbb{R}^{d} is a bounded convex open domain which contains the origin. Then there exists a constant c>0c>0 such that for all 0<r<0<r<\infty, all 0<δ<0<\delta<\infty, and all 0|y|δ0\leq|y|\leq\delta

xrΩx+y(r+cδ)Ω.x\in r\Omega\implies x+y\in(r+c\delta)\Omega.
Proof.

Since Ω\Omega is open, bounded, and contains the origin 0d\vec{0}\in\mathbb{R}^{d}, there exists a constant ε>0\varepsilon>0 such that

{xd:|x0|ε}Ω.\{x\in\mathbb{R}^{d}:|x-\vec{0}|\leq\varepsilon\}\subset\Omega.

Set c=1/εc=1/\varepsilon, and suppose that 0<r<0<r<\infty, xrΩx\in r\Omega, and 0<|y|δ0<|y|\leq\delta are given. Then

y=|y|εεy|y|{xd:|x0|cδε}cδΩ,y=\frac{|y|}{\varepsilon}\cdot\varepsilon\frac{y}{|y|}\in\{x\in\mathbb{R}^{d}:|x-\vec{0}|\leq c\delta\varepsilon\}\subseteq c\delta\Omega,

and therefore,

x+yrΩ+cδΩ(r+cδ)Ω,x+y\in r\Omega+c\delta\Omega\subseteq(r+c\delta)\Omega,

where the final inclusion follows from the convexity of Ω\Omega. ∎

Refer to caption
Figure 6. Geometrically Lemma 3.1 implies that {xd:d(x,rΩ)δ}(r+cδ)Ω\{x\in\mathbb{R}^{d}:d(x,r\Omega)\leq\delta\}\subseteq(r+c\delta)\Omega.

The application of this Lemma to lattice point counting problems occurs when developing bounds for indicator functions in terms of mollified indicator functions. Specifically, Lemma 3.1 is typically applied as follows. Suppose χ\chi is the indicator function for the set Ωd\Omega\subset\mathbb{R}^{d}, and define

χr(x)=χ(x/r),\chi_{r}(x)=\chi(x/r),

which is the indicator function for rΩr\Omega. Let φ\varphi denote a CC^{\infty} bump function supported on the unit ball which integrates to 11. Set

φδ(x)=δdφ(x/δ).\varphi_{\delta}(x)=\delta^{-d}\varphi(x/\delta).

Suppose c>0c>0 is chosen in accordance to Lemma 3.1. Then we claim that for all xdx\in\mathbb{R}^{d}

χr(x)[χr+cδφδ](x),\chi_{r}(x)\leq[\chi_{r+c\delta}*\varphi_{\delta}](x),

where * denotes convolution. Indeed, by the choice of c>0c>0, for all |y|δ|y|\leq\delta

χr(x)=1χr+cδ(xy)=1.\chi_{r}(x)=1\implies\chi_{r+c\delta}(x-y)=1.

Therefore, if χr(x)=1\chi_{r}(x)=1, then

[χr+cδφδ](x)=|y|δχr+cδ(xy)φδ(y)𝑑y=|y|δφδ(y)𝑑y=1.[\chi_{r+c\delta}*\varphi_{\delta}](x)=\int_{|y|\leq\delta}\chi_{r+c\delta}(x-y)\varphi_{\delta}(y)dy=\int_{|y|\leq\delta}\varphi_{\delta}(y)dy=1.

In the proof of the Theorem in the following section a similar result is required. However, in this case the indicator functions and bump functions under consideration are transformed by a positive diagonal linear transformation AA of determinant 11. More precisely, we consider

χA,r(x)=χ(A1x/r)andφA,δ=δdφ(A1x/δ),\chi_{A,r}(x)=\chi(A^{-1}x/r)\quad\text{and}\quad\varphi_{A,\delta}=\delta^{-d}\varphi(A^{-1}x/\delta),

where AA is a positive diagonal matrix of determinant 11. Let

χA,r,δ=χA,rφA,δ.\chi_{A,r,\delta}=\chi_{A,r}*\varphi_{A,\delta}.

In the following Lemma, we repeat the above analysis to develop upper and lower bounds for χA,r\chi_{A,r} in terms of the smoothed version χA,r,δ\chi_{A,r,\delta}.

Lemma 3.2.

Suppose χA,r\chi_{A,r} and χA,r,δ\chi_{A,r,\delta} are as defined above. Then there exists a constant c>0c>0 such that for all r>0r>0, all δr/(1+c)\delta\leq r/(1+c), and all positive diagonal matrices AA of determinant 11

χA,rcδ,δ(x)χA,r(x)χA,r+cδ,δ(x),\chi_{A,r-c\delta,\delta}(x)\leq\chi_{A,r}(x)\leq\chi_{A,r+c\delta,\delta}(x),

for all xdx\in\mathbb{R}^{d}.

Proof.

By Lemma 3.1 we may choose a constant c>0c>0 such that for all r>0r>0 and all δ>0\delta>0

xrΩ|y|δx+y(r+cδ)Ω.x\in r\Omega\wedge|y|\leq\delta\implies x+y\in(r+c\delta)\Omega.

To establish the upper bound it suffices to show that χA,r(x)=1χA,r+cδ,δ=1\chi_{A,r}(x)=1\implies\chi_{A,r+c\delta,\delta}=1. By definition

χA,r+cδ,δ(x)=|A1y|δχr+cδ(A1(xy))φδ(A1y)𝑑y.\chi_{A,r+c\delta,\delta}(x)=\int_{|A^{-1}y|\leq\delta}\chi_{r+c\delta}(A^{-1}(x-y))\varphi_{\delta}(A^{-1}y)dy.

Since AA is a diagonal matrix of determinant 11, by a change of variables of integration we conclude

|A1y|δχr+cδ(A1(xy))φδ(A1y)𝑑y=|y|δχr+cδ(A1xy)φδ(y)𝑑y.\int_{|A^{-1}y|\leq\delta}\chi_{r+c\delta}(A^{-1}(x-y))\varphi_{\delta}(A^{-1}y)dy=\int_{|y|\leq\delta}\chi_{r+c\delta}(A^{-1}x-y)\varphi_{\delta}(y)dy.

If χA,r(x)=1\chi_{A,r}(x)=1, then by the choice of c>0c>0 we have χr+cδ(A1xy)=1\chi_{r+c\delta}(A^{-1}x-y)=1, when |y|δ|y|\leq\delta. Thus

|y|δχr+cδ(A1xy)φδ(y)𝑑y=|y|δφδ(y)𝑑y=1,\int_{|y|\leq\delta}\chi_{r+c\delta}(A^{-1}x-y)\varphi_{\delta}(y)dy=\int_{|y|\leq\delta}\varphi_{\delta}(y)dy=1,

which establishes the upper bound. Next, to establish the lower bound we show that χA,r(x)=0χA,rcδ,δ(x)=0\chi_{A,r}(x)=0\implies\chi_{A,r-c\delta,\delta}(x)=0. Let x~=xy\tilde{x}=x-y and r~=rcδ\tilde{r}=r-c\delta. By Lemma 3.1

x~r~Ω|y|δx~+y(r~+cδ)Ω.\tilde{x}\in\tilde{r}\Omega\wedge|y|\leq\delta\implies\tilde{x}+y\in(\tilde{r}+c\delta)\Omega.

Writing the contrapositive of this statement in terms of xx and rr gives

xrΩ|y|δxy(rcδ)Ω.x\not\in r\Omega\wedge|y|\leq\delta\implies x-y\not\in(r-c\delta)\Omega.

Using the same change of variables of integration as above, we have

χA,rcδ,δ(x)=|y|δχrcδ(A1xy)φδ(y)𝑑y.\chi_{A,r-c\delta,\delta}(x)=\int_{|y|\leq\delta}\chi_{r-c\delta}(A^{-1}x-y)\varphi_{\delta}(y)dy.

If χA,r(x)=0\chi_{A,r}(x)=0, then the derived implication implies that χrcδ(A1xy)=0\chi_{r-c\delta}(A^{-1}x-y)=0 for all |y|δ|y|\leq\delta; thus

|y|δχrcδ(A1xy)φδ(y)𝑑y=0,\int_{|y|\leq\delta}\chi_{r-c\delta}(A^{-1}x-y)\varphi_{\delta}(y)dy=0,

which completes the proof. ∎

Lemma 3.3.

Suppose Ωd\Omega\subset\mathbb{R}^{d} is a bounded convex domain with Cd+2C^{d+2} boundary Ω\partial\Omega with nowhere vanishing Gauss curvature. If χ\chi is the indicator function of Ω\Omega, then

|χ^(ξ)|=𝒪(|ξ|d+12),as |ξ|.|\widehat{\chi}(\xi)|=\mathcal{O}\left(|\xi|^{-\frac{d+1}{2}}\right),\quad\text{as }|\xi|\rightarrow\infty.

That is to say, the Fourier transform of an indicator function decays one order better than the Fourier transform of the corresponding surface carried measure dμd\mu, which decays like

|dμ^|=𝒪(|ξ|d12).|\widehat{d\mu}|=\mathcal{O}\left(|\xi|^{-\frac{d-1}{2}}\right).

The proof of this lemma involves an integration by parts of an expression for the Fourier transform of the corresponding surface carried measure, which is where the extra order of convergence arises.

3.3. Proof of Theorem 2.1

For clarity, we have divided the proof of the Theorem into five steps. First, we fix notation. We say

f(x)hg(x)if and only iff(x)Chg(x),f(x)\lesssim_{h}g(x)\quad\text{if and only if}\quad f(x)\leq C_{h}g(x),

for a fixed constant Ch>0C_{h}>0 only depending on hh. Throughout the proof, we assume that r1r\geq 1, d2d\geq 2, and that Ωd\Omega\subset\mathbb{R}^{d} is a bounded convex domain with a Cd+2C^{d+2} boundary Ω\partial\Omega with nowhere vanishing Gauss curvature. In particular, we assume that

Ω[C,C]dwhereC>0is a fixed constant.\Omega\subset[-C,C]^{d}\quad\text{where}\quad C>0\quad\text{is a fixed constant.}

Let A=diag(a1,a2,,ad)A=\operatorname{diag}(a_{1},a_{2},\ldots,a_{d}) denote a positive diagonal matrix of determinant 11. Without loss of generality we may assume that a1a2ada_{1}\leq a_{2}\leq\ldots\leq a_{d}, and set a=1/a1=A1a=1/a_{1}=\|A^{-1}\|_{\infty}.

Step 3.1.

Suppose 1aCr1\leq a\leq Cr. Then

#{ndA(rΩ)}=|Ω|rd+𝒪(a2dd+1rd2dd+1),\#\{n\in\mathbb{Z}^{d}\cap A(r\Omega)\}=|\Omega|r^{d}+\mathcal{O}\left(a^{\frac{2d}{d+1}}r^{d-\frac{2d}{d+1}}\right),

where the implicit constant is independent of AA and rr.

Proof.

Let χ\chi denote the indicator function for the given domain Ω\Omega, and set

χA,r,δ(x)=[χA,rφA,δ](x),\chi_{A,r,\delta}(x)=[\chi_{A,r}*\varphi_{A,\delta}](x),

where χA,r=χ(A1x/r)\chi_{A,r}=\chi(A^{-1}x/r) and φA,δ=δdφ(A1x/δ)\varphi_{A,\delta}=\delta^{-d}\varphi(A^{-1}x/\delta) for some CC^{\infty} bump function φ\varphi supported on the unit ball in d\mathbb{R}^{d} and such that

dφ(x)𝑑x=1.\int_{\mathbb{R}^{d}}\varphi(x)dx=1.

We denote our “smoothed” approximation of the number of lattice points enclosed by A(rΩ)A(r\Omega) by

NA,r,δ=ndχA,r,δ(n).N_{A,r,\delta}=\sum_{n\in\mathbb{Z}^{d}}\chi_{A,r,\delta}(n).

Since χA,r,δ\chi_{A,r,\delta} is CC^{\infty} and of compact support, by the Poisson summation formula

NA,r,δ=ndχ^A,r,δ(n)=ndrdχ^(Anr)φ^(Anδ).N_{A,r,\delta}=\sum_{n\in\mathbb{Z}^{d}}\widehat{\chi}_{A,r,\delta}(n)=\sum_{n\in\mathbb{Z}^{d}}r^{d}\widehat{\chi}(Anr)\widehat{\varphi}(An\delta).

Since χ^(0)=|Ω|\widehat{\chi}(0)=|\Omega| and φ^(0)=1\widehat{\varphi}(0)=1, we can break the sum up into three parts:

NA,r,δ=|Ω|rd+0<|n|<a/δrdχ^(Anr)φ^(Anδ)+|n|a/δrdχ^(Anr)φ^(Anδ).N_{A,r,\delta}=|\Omega|r^{d}+\sum_{0<|n|<a/\delta}r^{d}\widehat{\chi}(Anr)\widehat{\varphi}(An\delta)+\sum_{|n|\geq a/\delta}r^{d}\widehat{\chi}(Anr)\widehat{\varphi}(An\delta).

Since χ^\widehat{\chi} is the Fourier transform of an indicator function of a convex domain with nowhere vanishing Gauss curvature, it follows from Lemma 3.3 that

|χ^(ξ)|Ω|ξ|d+12.|\widehat{\chi}(\xi)|\lesssim_{\Omega}|\xi|^{-\frac{d+1}{2}}.

Furthermore, since φ^\widehat{\varphi} is the Fourier transform of a CC^{\infty} function with compact support if follows that

|φ^(ξ)|φ|ξ|N,|\widehat{\varphi}(\xi)|\lesssim_{\varphi}|\xi|^{-N},

for all N0N\geq 0. To evaluate the first sum we use the fact that |χ^|Ω|ξ|d+12|\widehat{\chi}|\lesssim_{\Omega}|\xi|^{-\frac{d+1}{2}}, and |φ^|φ1|\widehat{\varphi}|\lesssim_{\varphi}1

0<|n|a/δrdχ^(Anr)φ^(Anδ)Ω,φ0<|n|a/δrdrd+12ad+12|n|d+12.\sum_{0<|n|\leq a/\delta}r^{d}\widehat{\chi}(Anr)\widehat{\varphi}(An\delta)\lesssim_{\Omega,\varphi}\sum_{0<|n|\leq a/\delta}r^{d}r^{-\frac{d+1}{2}}a^{\frac{d+1}{2}}|n|^{-\frac{d+1}{2}}.

Therefore,

0<|n|a/δrdχ^(Anr)φ^(Anδ)Ω,φrd12ad+120<|n|a/δ|n|d+12.\sum_{0<|n|\leq a/\delta}r^{d}\widehat{\chi}(Anr)\widehat{\varphi}(An\delta)\lesssim_{\Omega,\varphi}r^{\frac{d-1}{2}}a^{\frac{d+1}{2}}\sum_{0<|n|\leq a/\delta}|n|^{-\frac{d+1}{2}}.

The sum on the right hand side can be compared to the integral

|x|a/δ|x|d+12𝑑x=Cd0a/δtd+12td1𝑑t=Cdad12δd12.\int_{|x|\leq a/\delta}|x|^{-\frac{d+1}{2}}dx=C_{d}\int_{0}^{a/\delta}t^{-\frac{d+1}{2}}t^{d-1}dt=C_{d}a^{\frac{d-1}{2}}\delta^{-\frac{d-1}{2}}.

Thus we conclude that

0<|n|a/δrdχ^(Anr)φ^(Anδ)Ω,φrd12adδd12.\sum_{0<|n|\leq a/\delta}r^{d}\widehat{\chi}(Anr)\widehat{\varphi}(An\delta)\lesssim_{\Omega,\varphi}r^{\frac{d-1}{2}}a^{d}\delta^{-\frac{d-1}{2}}.

For the second sum, we use the fact that

|φ^(ξ)|φ|ξ|d2,|\widehat{\varphi}(\xi)|\lesssim_{\varphi}|\xi|^{-\frac{d}{2}},

and use the same decay of χ^\widehat{\chi} as before. This yields

|n|a/δrdχ^(Anr)φ^(Anδ)Ω,φ|n|a/δrdrd+12ad+12|n|d+12ad2|n|d2δd2.\sum_{|n|\geq a/\delta}r^{d}\widehat{\chi}(Anr)\widehat{\varphi}(An\delta)\lesssim_{\Omega,\varphi}\sum_{|n|\geq a/\delta}r^{d}r^{-\frac{d+1}{2}}a^{\frac{d+1}{2}}|n|^{-\frac{d+1}{2}}a^{\frac{d}{2}}|n|^{-\frac{d}{2}}\delta^{-\frac{d}{2}}.

Hence

|n|a/δrdχ^(Anr)φ^(Anδ)Ω,φrd12δd2ad+12|n|a/δ|n|d12.\sum_{|n|\geq a/\delta}r^{d}\widehat{\chi}(Anr)\widehat{\varphi}(An\delta)\lesssim_{\Omega,\varphi}r^{\frac{d-1}{2}}\delta^{-\frac{d}{2}}a^{d+\frac{1}{2}}\sum_{|n|\geq a/\delta}|n|^{-d-\frac{1}{2}}.

We can bound the sum on the right hand side by comparison to the integral

|x|a/δ|x|d1/2𝑑x=Cda/δtd12td1𝑑t=Cda12δ12.\int_{|x|\geq a/\delta}|x|^{-d-1/2}dx=C_{d}\int_{a/\delta}^{\infty}t^{-d-\frac{1}{2}}t^{d-1}dt=C_{d}a^{-\frac{1}{2}}\delta^{\frac{1}{2}}.

So

|n|a/δrdχ^(Anr)φ^(Anδ)Ω,φrd12adδd12.\sum_{|n|\geq a/\delta}r^{d}\widehat{\chi}(Anr)\widehat{\varphi}(An\delta)\lesssim_{\Omega,\varphi}r^{\frac{d-1}{2}}a^{d}\delta^{-\frac{d-1}{2}}.

Moreover, since the bound for both sums is the same

|NA,r,δ|Ω|rd|Ω,φrd12adδd12.\left|N_{A,r,\delta}-|\Omega|r^{d}\right|\lesssim_{\Omega,\varphi}r^{\frac{d-1}{2}}a^{d}\delta^{-\frac{d-1}{2}}.

Set

δ=a2dd+1rd1d+1.\delta=a^{\frac{2d}{d+1}}r^{-\frac{d-1}{d+1}}.

Substituting this value of δ\delta into the last inequality yields

|NA,r,δ|Ω|rd|Ω,φa2dd+1rd2dd+1.\left|N_{A,r,\delta}-|\Omega|r^{d}\right|\lesssim_{\Omega,\varphi}a^{\frac{2d}{d+1}}r^{d-\frac{2d}{d+1}}.

By Lemma 3.2 there exists a constant c>0c>0 such that for all r>0r>0, all δr/(1+c)\delta\leq r/(1+c), and all positive diagonal matrices AA of determinant 11

χA,rcδ,δ(x)χA,r(x)χA,r+cδ(x),\chi_{A,r-c\delta,\delta}(x)\leq\chi_{A,r}(x)\leq\chi_{A,r+c\delta}(x),

for all xdx\in\mathbb{R}^{d}. However, we do not in general know that δr/(1+c)\delta\leq r/(1+c). Therefore, in the following we consider two cases:

1a(1+c)d+12dr,and(1+c)d+12dr<aCr.1\leq a\leq(1+c)^{-\frac{d+1}{2d}}r,\quad\text{and}\quad(1+c)^{-\frac{d+1}{2d}}r<a\leq Cr.

Case 1

If 1a(1+c)d+12dr1\leq a\leq(1+c)^{-\frac{d+1}{2d}}r, then

δ=a2dd+1rd1d+1δr/(1+c).\delta=a^{\frac{2d}{d+1}}r^{-\frac{d-1}{d+1}}\implies\delta\leq r/(1+c).

Thus, we may apply Lemma 3.2 to conclude

χA,rcδ,δ(x)χA,r(x)χA,r+cδ,δ(x).\chi_{A,r-c\delta,\delta}(x)\leq\chi_{A,r}(x)\leq\chi_{A,r+c\delta,\delta}(x).

Therefore, by the definition of NA,r,δN_{A,r,\delta}

NA,rcδ,δ#{ndA(rΩ)}NA,r+cδ,δ.N_{A,r-c\delta,\delta}\leq\#\{n\in\mathbb{Z}^{d}\cap A(r\Omega)\}\leq N_{A,r+c\delta,\delta}.

Moreover, since δr/(1+c)\delta\leq r/(1+c) applying the bound derived above gives

|NA,r+cδ,δ|Ω|rd|Ω,φa2dd+1rd2dd+1,\left|N_{A,r+c\delta,\delta}-|\Omega|r^{d}\right|\lesssim_{\Omega,\varphi}a^{\frac{2d}{d+1}}r^{d-\frac{2d}{d+1}},

and similarly,

|NA,rcδ,δ|Ω|rd|Ω,φa2dd+1rd2dd+1.\left|N_{A,r-c\delta,\delta}-|\Omega|r^{d}\right|\lesssim_{\Omega,\varphi}a^{\frac{2d}{d+1}}r^{d-\frac{2d}{d+1}}.

Therefore, we conclude

|#{ndA(rΩ)}|Ω|rd|Ω,φa2dd+1rd2dd+1.\left|\#\{n\in\mathbb{Z}^{d}\cap A(r\Omega)\}-|\Omega|r^{d}\right|\lesssim_{\Omega,\varphi}a^{\frac{2d}{d+1}}r^{d-\frac{2d}{d+1}}.

Note φ\varphi can be fixed such that it only depends on Ω\Omega (more specifically, dependent only on the dimension of Ω\Omega), so the proof is complete for this case.

Case 2

If (1+c)d+12dr<aCr(1+c)^{-\frac{d+1}{2d}}r<a\leq Cr, then we define

c~:=a(1+c)d+12dr1.\tilde{c}:=a(1+c)^{\frac{d+1}{2d}}r^{-1}.

Since the determinant of AA is equal to 11, there exists 1<k<d1<k<d such that

a1ak1ak+1ad.a_{1}\leq\ldots\leq a_{k}\leq 1\leq a_{k+1}\leq\ldots\leq a_{d}.

Define

a~j={ajc~1for 1jkajc~kdkfor k<jd.\tilde{a}_{j}=\left\{\begin{array}[]{cc}a_{j}\cdot\tilde{c}^{-1}&\text{for }1\leq j\leq k\\ a_{j}\cdot\tilde{c}^{\frac{k}{d-k}}&\text{for }k<j\leq d\end{array}\right..

Suppose A~=diag(a~1,a~2,,a~d)\tilde{A}=\operatorname{diag}(\tilde{a}_{1},\tilde{a}_{2},\ldots,\tilde{a}_{d}). Then by construction

det(A~)=1andA~1=(1+c)d+12dr.\det(\tilde{A})=1\quad\text{and}\quad\left\|\tilde{A}^{-1}\right\|_{\infty}=(1+c)^{-\frac{d+1}{2d}}r.

By the domain monotonicity of lattice point counting and the result from Case 1

#{ndA(rΩ)}#{ndA~(c~rΩ)}=𝒪((c~r)d).\#\{n\in\mathbb{Z}^{d}\cap A(r\Omega)\}\leq\#\{n\in\mathbb{Z}^{d}\cap\tilde{A}(\tilde{c}r\Omega)\}=\mathcal{O}\left((\tilde{c}r)^{d}\right).

Since 1c~cd+12dC1\leq\tilde{c}\leq c^{\frac{d+1}{2d}}C and the constants cc and CC only depend on Ω\Omega we conclude that

#{ndA(rΩ)}=𝒪(rd),\#\{n\in\mathbb{Z}^{d}\cap A(r\Omega)\}=\mathcal{O}\left(r^{d}\right),

where the implicit constant only depends on Ω\Omega. This completes the proof of Step 3.1.

Step 3.2.

Suppose that 1aCr1\leq a\leq Cr, and assume that Ω\Omega is balanced in the sense of Definition 2.1. Let Ωj\Omega_{j} denote the intersection of Ω\Omega with the coordinate hyperplane orthogonal to the jj-th coordinate vector. Then

#{ndA(rj=1dΩj)}=(trA1)rd1+𝒪(a2dd+1rd2dd+1),\#\left\{n\in\mathbb{Z}^{d}\cap A\left(r\bigcup_{j=1}^{d}\Omega_{j}\right)\right\}=(\operatorname{tr}A^{-1})r^{d-1}+\mathcal{O}\left(a^{\frac{2d}{d+1}}r^{d-\frac{2d}{d+1}}\right),

where the implicit constant is independent of AA and rr.

Proof.

If d=1d=1, then the statement is trivial. We consider two cases: d=2d=2 and d>2d>2.

Case 1

If the dimension d=2d=2, then the set of positive diagonal matrices of determinant 11 is the 11-parameter family A=diag(1/a,a)A=\operatorname{diag}(1/a,a). Therefore, it suffices to show that

#{n2A(rj=12Ωj)}=(a+1a)r+𝒪(a43r23),\#\left\{n\in\mathbb{Z}^{2}\cap A\left(r\bigcup_{j=1}^{2}\Omega_{j}\right)\right\}=\left(a+\frac{1}{a}\right)r+\mathcal{O}\left(a^{\frac{4}{3}}r^{\frac{2}{3}}\right),

for A=diag(1/a,a)A=\operatorname{diag}(1/a,a). In this case, Ω1\Omega_{1} and Ω2\Omega_{2} are the intersection of Ω\Omega with the yy-axis and xx-axis, respectively. Moreover, since we have assumed Ω\Omega is balanced |Ωj|=1|\Omega_{j}|=1 for j=1,2j=1,2. Therefore, when Ω\Omega is scaled by rr and transformed by AA, the total number of points on the axes will be ar+r/a+𝒪(1)ar+r/a+\mathcal{O}(1), and thus the above statement holds.

Case 2

Suppose the dimension d>2d>2. Let AjA_{j} denote the (d1)×(d1)(d-1)\times(d-1) positive diagonal matrix formed by removing the jj-th row and jj-th column from AA. Write:

#{ndAj(rΩj)}=#{ndaj1d1Aj(raj1d1Ωj)}.\#\left\{n\in\mathbb{Z}^{d}\cap A_{j}(r\Omega_{j})\right\}=\#\left\{n\in\mathbb{Z}^{d}\cap a_{j}^{\frac{1}{d-1}}A_{j}\left(ra_{j}^{-\frac{1}{d-1}}\Omega_{j}\right)\right\}.

Observe that

detaj1d1Aj=1and(aj1d1Aj)1(aj1d1)a.\det a_{j}^{\frac{1}{d-1}}A_{j}=1\quad\text{and}\quad\left\|\left(a_{j}^{\frac{1}{d-1}}A_{j}\right)^{-1}\right\|_{\infty}\leq\left(a_{j}^{-\frac{1}{d-1}}\right)a.

Therefore, by the result from Step 3.1:

|#{ndaj1d1Aj(raj1d1Ωj)}|Ωj|aj1rd1|Ω(aj1d1a)2(d1)d(aj1d1r)(d1)2(d1)d.\left|\#\left\{n\in\mathbb{Z}^{d}\cap a_{j}^{\frac{1}{d-1}}A_{j}\left(ra_{j}^{-\frac{1}{d-1}}\Omega_{j}\right)\right\}-|\Omega_{j}|a_{j}^{-1}r^{d-1}\right|\lesssim_{\Omega}\left(a_{j}^{-\frac{1}{d-1}}a\right)^{\frac{2(d-1)}{d}}\left(a_{j}^{-\frac{1}{d-1}}r\right)^{(d-1)-\frac{2(d-1)}{d}}.

Observe that the total contribution of aja_{j} to the right hand side of this inequality is

aj(1d1)(2(d1)d+(d1)2(d1)d)=aj1.a_{j}^{\left(-\frac{1}{d-1}\right)\left(\frac{2(d-1)}{d}+(d-1)-\frac{2(d-1)}{d}\right)}=a_{j}^{-1}.

Therefore, bounding aj1a_{j}^{-1} by a=A1a=\|A^{-1}\|_{\infty} gives

|#{ndaj1d1Aj(raj1d1Ωj)}|Ωj|aj1rd1|Ωa(a2(d1)dr(d1)2(d1)d).\left|\#\left\{n\in\mathbb{Z}^{d}\cap a_{j}^{\frac{1}{d-1}}A_{j}\left(ra_{j}^{-\frac{1}{d-1}}\Omega_{j}\right)\right\}-|\Omega_{j}|a_{j}^{-1}r^{d-1}\right|\lesssim_{\Omega}a\cdot\left(a^{\frac{2(d-1)}{d}}r^{(d-1)-\frac{2(d-1)}{d}}\right).

A direct computation shows that

(a2dd+1rd2dd+1)(a2(d1)dr(d1)2(d1)d)1=a2d2+drd2+d2d2+d.\left(a^{\frac{2d}{d+1}}r^{d-\frac{2d}{d+1}}\right)\left(a^{\frac{2(d-1)}{d}}r^{(d-1)-\frac{2(d-1)}{d}}\right)^{-1}=a^{\frac{2}{d^{2}+d}}r^{\frac{d^{2}+d-2}{d^{2}+d}}.

Thus, since we have assumed that aCra\leq Cr, it follows that

a(a2(d1)dr(d1)2(d1)d)Ωa2dd+1rd2dd+1.a\cdot\left(a^{\frac{2(d-1)}{d}}r^{(d-1)-\frac{2(d-1)}{d}}\right)\lesssim_{\Omega}a^{\frac{2d}{d+1}}r^{d-\frac{2d}{d+1}}.

Therefore, we obtain the bound

|#{ndAj(rΩj)}|Ωj|aj1rd1|Ωa2dd+1rd2dd+1.\left|\#\{n\in\mathbb{Z}^{d}\cap A_{j}(r\Omega_{j})\}-|\Omega_{j}|a_{j}^{-1}r^{d-1}\right|\lesssim_{\Omega}a^{\frac{2d}{d+1}}r^{d-\frac{2d}{d+1}}.

By the balanced assumption |Ωj|=1|\Omega_{j}|=1. Therefore, summing over j=1,,dj=1,\ldots,d yields

|j=1d#{ndAj(rΩj)}(trA1)rd1|Ωa2dd+1rd2dd+1.\left|\sum_{j=1}^{d}\#\{n\in\mathbb{Z}^{d}\cap A_{j}(r\Omega_{j})\}-(\operatorname{tr}A^{-1})r^{d-1}\right|\lesssim_{\Omega}a^{\frac{2d}{d+1}}r^{d-\frac{2d}{d+1}}.

Next, we will use a similar argument to analyze dAj,k(rΩj,k)\mathbb{Z}^{d}\cap A_{j,k}(r\Omega_{j,k}), where Ωj,k=ΩjΩk\Omega_{j,k}=\Omega_{j}\cap\Omega_{k} and Aj,kA_{j,k} denotes the (d2)×(d2)(d-2)\times(d-2) matrix formed by removing the jj-th and kk-th rows, and jj-th and kk-th columns from AA. Write:

#{ndAj,k(rΩj,k)}=#{nd(ajak)1d2Aj,k(r(ajak)1d2Ωj)}.\#\left\{n\in\mathbb{Z}^{d}\cap A_{j,k}(r\Omega_{j,k})\right\}=\#\left\{n\in\mathbb{Z}^{d}\cap(a_{j}a_{k})^{\frac{1}{d-2}}A_{j,k}\left(r(a_{j}a_{k})^{-\frac{1}{d-2}}\Omega_{j}\right)\right\}.

In this case, applying Step 3.1 yields:

|#{nd(ajak)1d2Aj,k(r(ajak)1d2Ωj)}|Ωj,k|aj1ak1rd2|\left|\#\left\{n\in\mathbb{Z}^{d}\cap(a_{j}a_{k})^{\frac{1}{d-2}}A_{j,k}\left(r(a_{j}a_{k})^{-\frac{1}{d-2}}\Omega_{j}\right)\right\}-|\Omega_{j,k}|a_{j}^{-1}a_{k}^{-1}r^{d-2}\right|
Ω((ajak)1d2a)2(d2)d1((ajak)1d2r)(d2)2(d2)d1.\lesssim_{\Omega}\left((a_{j}a_{k})^{-\frac{1}{d-2}}a\right)^{\frac{2(d-2)}{d-1}}\left((a_{j}a_{k})^{\frac{1}{d-2}}r\right)^{(d-2)-\frac{2(d-2)}{d-1}}.

Observe that the total contribution of aja_{j} and aka_{k} to the right hand side is

(ajak)(1d2)(2(d2)d1+(d2)2(d2)d1)=aj1ak1.(a_{j}a_{k})^{\left(-\frac{1}{d-2}\right)\left(\frac{2(d-2)}{d-1}+(d-2)-\frac{2(d-2)}{d-1}\right)}=a_{j}^{-1}a_{k}^{-1}.

Therefore, bounding aj1ak1a_{j}^{-1}a_{k}^{-1} by a2a^{2} yields the bound

|#{ndAj,k(rΩj,k)}|Ωj,k|aj1ak1rd2|Ωa2(a2(d2)d1r(d2)2(d2)d1).\left|\#\left\{n\in\mathbb{Z}^{d}\cap A_{j,k}(r\Omega_{j,k})\right\}-|\Omega_{j,k}|a_{j}^{-1}a_{k}^{-1}r^{d-2}\right|\lesssim_{\Omega}a^{2}\left(a^{\frac{2(d-2)}{d-1}}r^{(d-2)-\frac{2(d-2)}{d-1}}\right).

A direct computation yields:

(a2dd+1rd2dd+1)(a2(d2)d1r(d2)2(d2)d1)1=a4d21r2d26d21.\left(a^{\frac{2d}{d+1}}r^{d-\frac{2d}{d+1}}\right)\left(a^{\frac{2(d-2)}{d-1}}r^{(d-2)-\frac{2(d-2)}{d-1}}\right)^{-1}=a^{\frac{4}{d^{2}-1}}r^{\frac{2d^{2}-6}{d^{2}-1}}.

Therefore, since we have assumed aCra\leq Cr, it follows that

a2(a2(d2)d1r(d2)2(d2)d1)Ωa2dd+1rd2dd+1.a^{2}\left(a^{\frac{2(d-2)}{d-1}}r^{(d-2)-\frac{2(d-2)}{d-1}}\right)\lesssim_{\Omega}a^{\frac{2d}{d+1}}r^{d-\frac{2d}{d+1}}.

Similarly, it follows that

|Ωj,k|aj1ak1rd2Ωa2dd+1rd2dd+1.|\Omega_{j,k}|a_{j}^{-1}a_{k}^{-1}r^{d-2}\lesssim_{\Omega}a^{\frac{2d}{d+1}}r^{d-\frac{2d}{d+1}}.

Therefore, we conclude that

#{ndAj,k(rΩj,k)}Ωa2dd+1rd2dd+1.\#\left\{n\in\mathbb{Z}^{d}\cap A_{j,k}(r\Omega_{j,k})\right\}\lesssim_{\Omega}a^{\frac{2d}{d+1}}r^{d-\frac{2d}{d+1}}.

Summing over 1j<kd1\leq j<k\leq d gives

1j<kd#{ndAj,k(rΩj,k)}Ωa2dd+1rd2dd+1.\sum_{1\leq j<k\leq d}\#\left\{n\in\mathbb{Z}^{d}\cap A_{j,k}(r\Omega_{j,k})\right\}\lesssim_{\Omega}a^{\frac{2d}{d+1}}r^{d-\frac{2d}{d+1}}.

Recall that we previously established the inequality

|j=1d#{ndAj(rΩj)}(trA1)rd1|Ωa2dd+1rd2dd+1.\left|\sum_{j=1}^{d}\#\left\{n\in\mathbb{Z}^{d}\cap A_{j}(r\Omega_{j})\right\}-(\operatorname{tr}A^{-1})r^{d-1}\right|\lesssim_{\Omega}a^{\frac{2d}{d+1}}r^{d-\frac{2d}{d+1}}.

These last two inequalities can be used to deduce the result in combination with the observation that:

j=1d#{ndAj(rΩj)}1j<kd#{ndAj,k(rΩj,k)}#{ndA(rj=1dΩj)},\sum_{j=1}^{d}\#\left\{n\in\mathbb{Z}^{d}\cap A_{j}(r\Omega_{j})\right\}-\sum_{1\leq j<k\leq d}\#\left\{n\in\mathbb{Z}^{d}\cap A_{j,k}(r\Omega_{j,k})\right\}\leq\#\left\{n\in\mathbb{Z}^{d}\cap A\left(r\bigcup_{j=1}^{d}\Omega_{j}\right)\right\},

and

#{ndA(rj=1dΩj)}j=1d#{ndAj(rΩj)}.\#\left\{n\in\mathbb{Z}^{d}\cap A\left(r\bigcup_{j=1}^{d}\Omega_{j}\right)\right\}\leq\sum_{j=1}^{d}\#\{n\in\mathbb{Z}^{d}\cap A_{j}(r\Omega_{j})\}.

Step 3.3.

Assume that Ω\Omega is balanced in the sense of Definition 2.1 and is symmetric with respect to each coordinate hyperplane. Then

#{n>0dA(rΩ)}=12d|Ω|rd12d(trA1)rd1+𝒪(a2dd+1rd2dd+1).\#\left\{n\in\mathbb{Z}_{>0}^{d}\cap A(r\Omega)\right\}=\frac{1}{2^{d}}|\Omega|r^{d}-\frac{1}{2^{d}}\left(\operatorname{tr}A^{-1}\right)r^{d-1}+\mathcal{O}\left(a^{\frac{2d}{d+1}}r^{d-\frac{2d}{d+1}}\right).
Proof.

We partition the proof into two cases:

1aCr,andCr<a<.1\leq a\leq Cr,\quad\text{and}\quad Cr<a<\infty.

Case 1

If 1aCr1\leq a\leq Cr, then the results from Steps 3.1 and 3.2 hold. By the assumed symmetry of the domain, the number of positive lattice points contained in Ω\Omega is equal to the number of lattice points in Ω\Omega minus those in j=1dΩj\bigcup_{j=1}^{d}\Omega_{j} divided by 2d2^{d}:

#{n>0dA(rΩ)}=12d(#{ndA(rΩ)}#{ndA(rj=1dΩj)}).\#\left\{n\in\mathbb{Z}^{d}_{>0}\cap A(r\Omega)\right\}=\frac{1}{2^{d}}\left(\#\left\{n\in\mathbb{Z}^{d}\cap A(r\Omega)\right\}-\#\left\{n\in\mathbb{Z}^{d}\cap A\left(r\bigcup_{j=1}^{d}\Omega_{j}\right)\right\}\right).

Combing the results of Steps 3.1 and 3.2 yields the result.

Case 2

If Cr<a<Cr<a<\infty, then we argue as follows. Recall that C>0C>0 is a constant such that Ω[C,C]d\Omega\subset[-C,C]^{d}. Therefore, if a>Cra>Cr, then a1<1/(Cr)a_{1}<1/(Cr) and hence

A(rΩ)(1,1)×d1.A(r\Omega)\subseteq(-1,1)\times\mathbb{R}^{d-1}.

In particular, it follows that

#{n>0dA(rΩ)}=0.\#\{n\in\mathbb{Z}^{d}_{>0}\cap A(r\Omega)\}=0.

Therefore, the statement to prove reduces to

0=12d|Ω|rd12d(trA1)rd1+𝒪(a2dd+1rd2dd+1),0=\frac{1}{2^{d}}|\Omega|r^{d}-\frac{1}{2^{d}}\left(\operatorname{tr}A^{-1}\right)r^{d-1}+\mathcal{O}\left(a^{\frac{2d}{d+1}}r^{d-\frac{2d}{d+1}}\right),

which trivially holds because the error term dominates the right hand side since Cr<a<Cr<a<\infty. ∎

Step 3.4.

Suppose that Ω\Omega is balanced in the sense of Definition 2.1 and is symmetric with respect to each coordinate hyperplane. Let

A(r)=argmaxA#{n>0dA(rΩ)},A(r)=\operatorname*{argmax}_{A}\,\#\left\{n\in\mathbb{Z}^{d}_{>0}\cap A(r\Omega)\right\},

where the argmax\operatorname*{argmax} ranges over all positive diagonal matrices AA of determinant 11. Write A(r)=diag(a1(r),a2(r),,ad(r))A(r)=\operatorname{diag}(a_{1}(r),a_{2}(r),\ldots,a_{d}(r)). Without loss of generality suppose a1(r)a2(r)ad(r)a_{1}(r)\leq a_{2}(r)\leq\cdots\leq a_{d}(r) and let a(r)=1/a1(r)=A1(r)a(r)=1/a_{1}(r)=\|A^{-1}(r)\|_{\infty}. Then

limra(r)=1.\lim_{r\rightarrow\infty}a(r)=1.
Remark 3.1.

Two technical remarks are in order about the definition of A(r)A(r). First, we argue why the argmax\operatorname*{argmax} exists. As noted in Step 3.3, if Cr<a<Cr<a<\infty, then A(rΩ)A(r\Omega) will not contain any positive lattice points; therefore, the admissible AA for the argmax\operatorname*{argmax} can be restricted to those satisfying

1A1CrA(rΩ)[(Cr)d,(Cr)2]d,1\leq\left\|A^{-1}\right\|_{\infty}\leq Cr\implies A(r\Omega)\subset\left[-(Cr)^{d},(Cr)^{2}\right]^{d},

where C>0C>0 is a constant such that Ω[C,C]d\Omega\subset[-C,C]^{d}. Since the set [(Cr)2,(Cr)2]d[-(Cr)^{2},(Cr)^{2}]^{d} contains finitely many positive lattice points, we conclude that the argmax\operatorname*{argmax} exists. Second, when the argmax\operatorname*{argmax} is not unique, A(r)A(r) should be interpreted as being equal to an arbitrary element from the maximal set; all convergence results are independent of this choice.

Proof of Step 3.4.

By the argument in Case 2 of Step 3.3, we may assume 1a(r)Cr1\leq a(r)\leq Cr. First, we will suppose that there exists a constant c>0c>0 such that

1cra(r)Cr,\frac{1}{c}r\leq a(r)\leq Cr,

for arbitrarily large values rr. That is to say, we suppose that there exists a sequence rnr_{n} tending towards infinity such that rn/ca(rn)Crnr_{n}/c\leq a(r_{n})\leq Cr_{n}. We show that this assumption leads to a contradiction, which implies that

a(r)/r0,as r.a(r)/r\rightarrow 0,\quad\text{as }r\rightarrow\infty.

While producing this contradiction, we write A=A(r)A=A(r) and a=a(r)a=a(r) to simplify notation. Let χ\chi denote the indicator function for Ω\Omega. Then

#{n>0dA(rΩ)}=n>0dχ(A1n/r).\#\left\{n\in\mathbb{Z}^{d}_{>0}\cap A(r\Omega)\right\}=\sum_{n\in\mathbb{Z}^{d}_{>0}}\chi(A^{-1}n/r).

Let (n1,n)>0×>0d1(n_{1},n^{\prime})\in\mathbb{Z}_{>0}\times\mathbb{Z}_{>0}^{d-1}, and let A1A_{1} denote the (d1)×(d1)(d-1)\times(d-1) matrix formed by removing the 11-st row and 11-st column from AA. With this notation

#{n>0dA(rΩ)}=n1>0n>0d1χ(n1a1r,A11nr).\#\left\{n\in\mathbb{Z}^{d}_{>0}\cap A(r\Omega)\right\}=\sum_{n_{1}\in\mathbb{Z}_{>0}}\sum_{n^{\prime}\in\mathbb{Z}^{d-1}_{>0}}\chi\left(\frac{n_{1}}{a_{1}r},\frac{A_{1}^{-1}n^{\prime}}{r}\right).

Since by assumption r/caCrr/c\leq a\leq Cr, and a=1/a1a=1/a_{1}, we have

1Cra1c.\frac{1}{C}\leq ra_{1}\leq c.

For simplicity, assume ra1=cra_{1}=c. In the course of the proof, we will only use the fact that ra1ra_{1} is bounded above by a fixed constant. We remark that the method of estimating the sum in this part of the proof is similar to the arguments in [5, 18]. With the notation ra1=cra_{1}=c

#{n>0dA(rΩ)}=n1>0n>0d1χ(nc,A11nr).\#\{n\in\mathbb{Z}^{d}_{>0}\cap A(r\Omega)\}=\sum_{n_{1}\in\mathbb{Z}_{>0}}\sum_{n^{\prime}\in\mathbb{Z}^{d-1}_{>0}}\chi\left(\frac{n}{c},\frac{A_{1}^{-1}n^{\prime}}{r}\right).

We assert that:

n>0d1χ(nc,A11nr)0d1χ(n1c,A11xr)𝑑x.\sum_{n^{\prime}\in\mathbb{Z}^{d-1}_{>0}}\chi\left(\frac{n}{c},\frac{A_{1}^{-1}n^{\prime}}{r}\right)\leq\int_{\mathbb{R}^{d-1}_{\geq 0}}\chi\left(\frac{n_{1}}{c},\frac{A_{1}^{-1}x^{\prime}}{r}\right)dx^{\prime}.

Indeed, each lattice point (n1,n)(n_{1},n^{\prime}) can be identified with the (d1)(d-1)-dimensional cube with sides parallel to the coordinates axes and vertices (n1,n)(n_{1},n^{\prime}) and (n1,n1)(n_{1},n^{\prime}-\vec{1}), where 1\vec{1} denotes a (d1)(d-1)-dimensional vector of ones. Since the set A(rΩ)A(r\Omega) is convex, each of these cubes is contained in A(rΩ)A(r\Omega) and their first coordinate is equal to n1n_{1}. Moreover, this collection of cubes is disjoint since each pair of vertices is unique. Therefore, the sum is a lower approximation of the integral. Thus we have

#{n>0dA(rΩ)}n1>00d1χ(n1c,A11xr)𝑑x.\#\left\{n\in\mathbb{Z}^{d}_{>0}\cap A(r\Omega)\right\}\leq\sum_{n_{1}\in\mathbb{Z}_{>0}}\int_{\mathbb{R}^{d-1}_{\geq 0}}\chi\left(\frac{n_{1}}{c},\frac{A_{1}^{-1}x^{\prime}}{r}\right)dx^{\prime}.

Define the function fA1,r:f_{A_{1},r}:\mathbb{R}\rightarrow\mathbb{R} by

fA1,r(x1)=0d1χ(x1c,A11xr)𝑑x.f_{A_{1},r}(x_{1})=\int_{\mathbb{R}^{d-1}_{\geq 0}}\chi\left(\frac{x_{1}}{c},\frac{A_{1}^{-1}x^{\prime}}{r}\right)dx^{\prime}.

The function fA1,r(x1)f_{A_{1},r}(x_{1}) is supported on some interval [0,b][0,b] such that 0<b<cC0<b<cC, where C>0C>0 is a constant such that Ω[C,C]d\Omega\subset[-C,C]^{d}. Define

Ωx1/c:=Ω{x1c}×d1such thatfA1,r(x1)=12d1|A1(rΩx1/c)|,\Omega_{x_{1}/c}:=\Omega\cap\left\{\frac{x_{1}}{c}\right\}\times\mathbb{R}^{d-1}\quad\text{such that}\quad f_{A_{1},r}(x_{1})=\frac{1}{2^{d-1}}|A_{1}(r\Omega_{x_{1}/c})|,

where |||\cdot| denotes the (d1)(d-1)-dimensional Lebesgue measure, and where the factor 1/2d11/2^{d-1} arises since the integral is taken over 0d1\mathbb{R}^{d-1}_{\geq 0}. Since det(A1)a1=1\det(A_{1})\cdot a_{1}=1 and a1=c/ra_{1}=c/r it follows that

fA1,r(x1)=12d1rc|rΩx1/c|=12d1rd1c|Ωx1/c|.f_{A_{1},r}(x_{1})=\frac{1}{2^{d-1}}\frac{r}{c}|r\Omega_{x_{1}/c}|=\frac{1}{2^{d-1}}r^{d}\frac{1}{c}|\Omega_{x_{1}/c}|.

Since Ω\Omega is symmetric about the coordinate axes and is strictly convex, the maximum value of fA1,r(x1)f_{A_{1},r}(x_{1}) occurs at x1=0x_{1}=0, fA1,r(x1)f_{A_{1},r}(x_{1}) is strictly decreasing on [0,b][0,b], and fA1,r(b)=0f_{A_{1},r}(b)=0. Moreover, since we assume the boundary of Ω\Omega is Cd+2C^{d+2} the function fA1,rf_{A_{1},r} is certainly C1C^{1}, which is sufficient for our purposes. Integrating fA1,rf_{A_{1},r} on [0,b][0,b] yields

0bfA1,r(x1)𝑑x1=12d1rd0b1c|Ωx1/c|𝑑x1=12d|Ω|rd.\int_{0}^{b}f_{A_{1},r}(x_{1})dx_{1}=\frac{1}{2^{d-1}}r^{d}\int_{0}^{b}\frac{1}{c}|\Omega_{x_{1}/c}|dx_{1}=\frac{1}{2^{d}}|\Omega|r^{d}.

However, our previous arguments show that

#{n>0dA(rΩ)}n1>0fA1,r(x1)=12d1rdn1=1b1c|Ωx1/c|.\#\left\{n\in\mathbb{Z}^{d}_{>0}\cap A(r\Omega)\right\}\leq\sum_{n_{1}\in\mathbb{Z}_{>0}}f_{A_{1},r}(x_{1})=\frac{1}{2^{d-1}}r^{d}\sum_{n_{1}=1}^{\lfloor b\rfloor}\frac{1}{c}|\Omega_{x_{1}/c}|.

We assert that there exists ε>0\varepsilon>0 such that

n1=1b1c|Ωn1/c|(1ε)0b1c|Ωx1/c|𝑑x1.\sum_{n_{1}=1}^{\lfloor b\rfloor}\frac{1}{c}|\Omega_{n_{1}/c}|\leq(1-\varepsilon)\int_{0}^{b}\frac{1}{c}|\Omega_{x_{1}/c}|dx_{1}.

Indeed, the sum on the left hand side is a lower Riemann sum for the integral of the strictly decreasing C1C^{1} function (1/c)|Ωx1/c|(1/c)|\Omega_{x_{1}/c}|. Moreover, the constant ε>0\varepsilon>0 can be chosen to hold uniformly for any lower Riemann sum of the integral which discretizes the integral into at most cC\lfloor cC\rfloor pieces.

Refer to caption
Figure 7. An approximation of the area enclosed by an ellipse by squares [i11,i1]×[i21,i2][i_{1}-1,i_{1}]\times[i_{2}-1,i_{2}].

Therefore, we conclude that

#{n>0dA(rΩ)}n1=1bfA1,r(n1)(1ε)0bfA1,r(x1)𝑑x1=(1ε)12d|Ω|rd.\#\{n\in\mathbb{Z}^{d}_{>0}\cap A(r\Omega)\}\leq\sum_{n_{1}=1}^{\lfloor b\rfloor}f_{A_{1},r}(n_{1})\leq(1-\varepsilon)\int_{0}^{b}f_{A_{1},r}(x_{1})dx_{1}=(1-\varepsilon)\frac{1}{2^{d}}|\Omega|r^{d}.

However, applying Step 3.3 with A=IdA=\operatorname{Id} yields

#{n>0dId(rΩ)}=12d|Ω|rd12ddrd1+𝒪(rd2dd+1).\#\{n\in\mathbb{Z}_{>0}^{d}\cap\operatorname{Id}(r\Omega)\}=\frac{1}{2^{d}}|\Omega|r^{d}-\frac{1}{2^{d}}dr^{d-1}+\mathcal{O}\left(r^{d-\frac{2d}{d+1}}\right).

When rr is sufficiently large

εrd<12ddrd1+𝒪(rd2dd+1),-\varepsilon r^{d}<-\frac{1}{2^{d}}dr^{d-1}+\mathcal{O}\left(r^{d-\frac{2d}{d+1}}\right),

which contradicts the optimality of such ar/ca\geq r/c. Therefore, we conclude that

a(r)=ψ(r)r,whereψ(r)0,asr.a(r)=\psi(r)r,\quad\text{where}\quad\psi(r)\rightarrow 0,\quad\text{as}\quad r\rightarrow\infty.

By the Step 3.3 of the proof:

#{n>0dA(r)(rΩ)}=12d|Ω|rd12d(trA(r)1)rd1+𝒪(a(r)2dd+1rd2dd+1).\#\{n\in\mathbb{Z}_{>0}^{d}\cap A(r)(r\Omega)\}=\frac{1}{2^{d}}|\Omega|r^{d}-\frac{1}{2^{d}}\left(\operatorname{tr}A(r)^{-1}\right)r^{d-1}+\mathcal{O}\left(a(r)^{\frac{2d}{d+1}}r^{d-\frac{2d}{d+1}}\right).

Substituting a(r)=ψ(r)ra(r)=\psi(r)r yields

#{n>0dA(r)(rΩ)}12d|Ω|rd(12dC1ψ(r)d1d+1)ψ(r)rd,\#\left\{n\in\mathbb{Z}_{>0}^{d}\cap A(r)(r\Omega)\right\}\leq\frac{1}{2^{d}}|\Omega|r^{d}-\left(\frac{1}{2^{d}}-C_{1}\psi(r)^{\frac{d-1}{d+1}}\right)\psi(r)r^{d},

for some constant C1C_{1}. Since ψ(r)0\psi(r)\rightarrow 0, we may choose rr large enough such that

12dC1ψ(r)d1d+112d+1.\frac{1}{2^{d}}-C_{1}\psi(r)^{\frac{d-1}{d+1}}\geq\frac{1}{2^{d+1}}.

Thus for large enough rr,

#{n>0dA(r)(rΩ)}12d|Ω|rd12d+1ψ(r)rd.\#\left\{n\in\mathbb{Z}_{>0}^{d}\cap A(r)(r\Omega)\right\}\leq\frac{1}{2^{d}}|\Omega|r^{d}-\frac{1}{2^{d+1}}\psi(r)r^{d}.

However, if such a situation is optimal, it must be competitive with the situation A=IdA=\operatorname{Id}, where

#{n>0dId(rΩ)}=12d|Ω|rdd2drd1+𝒪(rd2dd+1).\#\left\{n\in\mathbb{Z}_{>0}^{d}\cap\operatorname{Id}(r\Omega)\right\}=\frac{1}{2^{d}}|\Omega|r^{d}-\frac{d}{2^{d}}r^{d-1}+\mathcal{O}\left(r^{d-\frac{2d}{d+1}}\right).

Therefore, we conclude that

ψ(r)rd=𝒪(rd1),\psi(r)r^{d}=\mathcal{O}\left(r^{d-1}\right),

which implies

ψ(r)=𝒪(r1).\psi(r)=\mathcal{O}\left(r^{-1}\right).

Therefore, the set of optimal aa is uniformly bounded. Convergence then immediately follows from the result from Step 3.3, i.e., from the equation:

#{n>0dA(rΩ)}=12d|Ω|rd12d(trA1)rd1+𝒪(a2dd+1rd2dd+1).\#\left\{n\in\mathbb{Z}_{>0}^{d}\cap A(r\Omega)\right\}=\frac{1}{2^{d}}|\Omega|r^{d}-\frac{1}{2^{d}}\left(\operatorname{tr}A^{-1}\right)r^{d-1}+\mathcal{O}\left(a^{\frac{2d}{d+1}}r^{d-\frac{2d}{d+1}}\right).

Indeed, since aa is uniformly bounded the second term determines the effect of AA when rr is large. Therefore, in order to maximize #{n>0dA(rΩ)}\#\{n\in\mathbb{Z}_{>0}^{d}\cap A(r\Omega)\} the coefficient of rd1r^{d-1} must be minimized. More precisely, A(r)A(r) can be characterized in terms of the following minimization problem

A(r)=argminA(trA1)+𝒪(rd1d+1)A(r)=\operatorname*{argmin}_{A}\left(\operatorname{tr}A^{-1}\right)+\mathcal{O}\left(r^{-\frac{d-1}{d+1}}\right)

where the argmin\operatorname*{argmin} is taken over positive diagonal matrices AA of determinant 11 such that 1A1c01\leq\|A^{-1}\|_{\infty}\leq c_{0}, where c01c_{0}\geq 1 is a fixed constant. By the arithmetic mean geometric mean inequality, any positive diagonal matrix AA of determinant 11 satisfies

trA1d(detA1)1/d=d,\operatorname{tr}A^{-1}\geq d\left(\det A^{-1}\right)^{1/d}=d,

with equality if and only if A=IdA=\operatorname{Id}. Therefore, we conclude that

trA(r)1d,as r.\operatorname{tr}A(r)^{-1}\rightarrow d,\quad\text{as }r\rightarrow\infty.

Since trA1\operatorname{tr}A^{-1} is a continuous function, and equality holds in the arithmetic mean geometric mean inequality if and only if A=IdA=\operatorname{Id} we conclude

A(r)Id0,as r,\|A(r)-\operatorname{Id}\|_{\infty}\rightarrow 0,\quad\text{as }r\rightarrow\infty,

as was to be shown. ∎

In the fifth step, we establish a rate of convergence.

Step 3.5.

Suppose that Ω\Omega is balanced in the sense of Definition 2.1 and is symmetric with respect to each coordinate hyperplane. Let

A(r)=argmaxA#{n>0dA(rΩ)},A(r)=\operatorname*{argmax}_{A}\,\#\{n\in\mathbb{Z}^{d}_{>0}\cap A(r\Omega)\},

where the argmax\operatorname*{argmax} ranges over all positive diagonal matrices AA of determinant 11. Then

A(r)Id=𝒪(rd12(d+1)).\|A(r)-\operatorname{Id}\|_{\infty}=\mathcal{O}\left(r^{-\frac{d-1}{2(d+1)}}\right).
Proof.

Applying the result of Step 3.3 for A=IdA=\operatorname{Id} gives

#{n>0dId(rΩ)}=12d|Ω|rd12ddrd1+𝒪(rd2dd+1).\#\{n\in\mathbb{Z}^{d}_{>0}\cap\operatorname{Id}(r\Omega)\}=\frac{1}{2^{d}}|\Omega|r^{d}-\frac{1}{2^{d}}dr^{d-1}+\mathcal{O}(r^{d-\frac{2d}{d+1}}).

By the arithmetic mean geometric mean inequality

trA1d(detA1)1/d=d.\operatorname{tr}A^{-1}\geq d\left(\det A^{-1}\right)^{1/d}=d.

Therefore, since A(r)A(r) maximizes #{n>0dA(rΩ)}\#\{n\in\mathbb{Z}^{d}_{>0}\cap A(r\Omega)\} over all AA, we must have

12dtrA1(r)rd112ddrd1Ωrd2dd+1,\frac{1}{2^{d}}\operatorname{tr}A^{-1}(r)r^{d-1}-\frac{1}{2^{d}}dr^{d-1}\lesssim_{\Omega}r^{d-\frac{2d}{d+1}},

which simplifies to

trA1(r)dΩrd1d+1.\operatorname{tr}A^{-1}(r)-d\lesssim_{\Omega}r^{-\frac{d-1}{d+1}}.

To complete the proof it suffices to show that:

18AId2trA1d.\frac{1}{8}\|A-\operatorname{Id}\|^{2}_{\infty}\leq\operatorname{tr}A^{-1}-d.

Indeed then,

18AId2trA1dΩrd1d+1AIdΩrd12(d+1).\frac{1}{8}\|A-\operatorname{Id}\|^{2}_{\infty}\leq\operatorname{tr}A^{-1}-d\lesssim_{\Omega}r^{-\frac{d-1}{d+1}}\quad\implies\quad\|A-\operatorname{Id}\|_{\infty}\lesssim_{\Omega}r^{-\frac{d-1}{2(d+1)}}.

Without loss of generality, suppose that

A(r)=diag(a1,a2,,ad)wherea1a2ad.A(r)=\operatorname{diag}\left(a_{1},a_{2},\ldots,a_{d}\right)\quad\text{where}\quad a_{1}\leq a_{2}\leq\ldots\leq a_{d}.

There are two cases to consider

Case 1

A(r)Id=|ad1|\|A(r)-\operatorname{Id}\|_{\infty}=|a_{d}-1|. In this case, suppose that ad=1+εa_{d}=1+\varepsilon where ε>0\varepsilon>0. Then

A(r)Id=ε.\|A(r)-\operatorname{Id}\|_{\infty}=\varepsilon.

Since the determinant of AA is equal 11, it follows that the product a1a2ad1=11+εa_{1}\cdot a_{2}\cdot\cdots\cdot a_{d-1}=\frac{1}{1+\varepsilon}. Therefore, by the arithmetic mean geometric mean inequality,

(d1)(1+ε)1d1+(11+ε)1a1+1a2+1ad1+(11+ε)=trA1.(d-1)(1+\varepsilon)^{\frac{1}{d-1}}+\left(\frac{1}{1+\varepsilon}\right)\leq\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots\frac{1}{a_{d-1}}+\left(\frac{1}{1+\varepsilon}\right)=\operatorname{tr}A^{-1}.

Expanding the left hand side in a Taylor series yields

d+dε22(d1)+𝒪(ε3)=trA1.d+\frac{d\varepsilon^{2}}{2(d-1)}+\mathcal{O}(\varepsilon^{3})=\operatorname{tr}A^{-1}.

And therefore, when ε\varepsilon is sufficiently small,

12A(r)Id2=12ε2trA1d.\frac{1}{2}\|A(r)-\operatorname{Id}\|_{\infty}^{2}=\frac{1}{2}\varepsilon^{2}\leq\operatorname{tr}A^{-1}-d.

Case 2

A(r)Id=|a11|\|A(r)-\operatorname{Id}\|_{\infty}=|a_{1}-1|. Suppose that a1=11+εa_{1}=\frac{1}{1+\varepsilon}. When ε\varepsilon is sufficiently small

A(r)Id=|a11|2ε.\|A(r)-\operatorname{Id}\|_{\infty}=|a_{1}-1|\leq 2\varepsilon.

Furthermore, since the determinant of AA is equal to 11

a2a3ad=1+ε.a_{2}\cdot a_{3}\cdot\cdots\cdot a_{d}=1+\varepsilon.

Therefore, by the arithmetic mean geometric mean inequality

(1+ε)+(d1)(11+ε)1d1(1+ε)+1a2+1a3+1ad=trA1.(1+\varepsilon)+(d-1)\left(\frac{1}{1+\varepsilon}\right)^{\frac{1}{d-1}}\leq(1+\varepsilon)+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\cdots\frac{1}{a_{d}}=\operatorname{tr}A^{-1}.

Expanding the left hand side in a Taylor series yields

d+dε22(d1)+𝒪(ε3)trA1.d+\frac{d\varepsilon^{2}}{2(d-1)}+\mathcal{O}(\varepsilon^{3})\leq\operatorname{tr}A^{-1}.

Therefore, when ε\varepsilon is sufficiently small,

18AId|212ε2trA1d.\frac{1}{8}\|A-\operatorname{Id}|^{2}\leq\frac{1}{2}\varepsilon^{2}\leq\operatorname{tr}A^{-1}-d.

Moreover, in either case

18AId2trA1d,\frac{1}{8}\|A-\operatorname{Id}\|^{2}\leq\operatorname{tr}A^{-1}-d,

and the proof is complete. ∎

Acknowledgment. We are grateful to Stefan Steinerberger for many useful discussions, and Andrei Deneanu for insightful comments. Additionally, we would like to thank Richard Laugesen for valuable feedback leading to corrections in the proof and a much improved exposition, and the referees for their helpful comments.

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