Shifted lattices and asymptotically optimal ellipses
R. S. Laugesen S. Liu
math.SP math.NT 35P15, 11P21, 52C05




]Richard S. Laugesen and Shiya Liu
Department of Mathematics, University of Illinois, Urbana,
IL 61801, U.S.A.
 Laugesen@illinois.edu   sliu63@illinois.edu



[2010]Primary 35P15. Secondary 11P21, 52C05


Translate the positive-integer lattice points in the first quadrant by some amount in the horizontal and vertical directions. Take a decreasing concave (or convex) curve in the first quadrant and construct a family of curves by rescaling in the coordinate directions while preserving area. Consider the curve in the family that encloses the greatest number of the shifted lattice points: we seek to identify the limiting shape of this maximizing curve as the area is scaled up towards infinity. 

The limiting shape is shown to depend explicitly on the lattice shift. The result holds for all positive shifts, and for negative shifts satisfying a certain condition. When the shift becomes too negative, the optimal curve no longer converges to a limiting shape, and instead we show it degenerates.

Our results handle the p-circle x^p+y^p=1 when p>1 (concave) and also when 0<p<1 (convex). Rescaling the p-circle generates the family of p-ellipses, and so in particular we identify the asymptotically optimal p-ellipses associated with shifted integer lattices. 

The circular case p=2 with shift -1/2 corresponds to minimizing high eigenvalues in a symmetry class for the Laplacian on rectangles, while the straight line case (p=1) generates an open problem about minimizing high eigenvalues of quantum harmonic oscillators with normalized parabolic potentials.  



Shifted lattices and asymptotically optimal ellipses
    [
    December 30, 2023
====================================================





§ INTRODUCTION


Among all ellipses centered at the origin with given area, consider the one enclosing the maximum number of positive integer lattice points. Does it approach a circular shape as the area tends to infinity? Antunes and Freitas <cit.>  showed the answer is yes. We tackle a variant of the problem in which the lattice is translated by some increments in the x- and y-directions, and show the asymptotically optimal ellipse is no longer a circle but an ellipse whose semi-axis ratio depends explicitly on the translation increments. This optimal ratio succeeds in “balancing” the horizontal and vertical empty strip areas created by the translation of the lattice; see <ref>. The precise statement is given in <ref>.




Generalized ellipses obtained by stretching a (concave or convex) smooth curve can be handled by our methods too, in <ref>. The results hold for all positive translations, and for small negative translations that satisfy a computable, curve-dependent criterion. 

When the curve is a straight line, one arrives at an open problem for right triangles that contain the most lattice points. The shape of these triangles exhibits a surprising clustering behavior as the area tends to infinity, as  revealed by our numerical investigations in  <ref>. This clustering conjecture has been investigated recently in the unshifted case by Marshall and Steinerberger <cit.>.

<ref> motivates this paper by connecting to spectral minimization problems for the Dirichlet Laplacian, and raises conjectures for the quantum harmonic oscillator and for a whole family of such Schrödinger eigenvalue problems. The recent advances on high eigenvalue minimization  began with work of Antunes and Freitas <cit.>, and continued with contributions from van den Berg, Bucur and Gittins <cit.>, van den Berg and Gittins <cit.>, Bucur and Freitas <cit.>, Gittins and Larson <cit.>, Larson <cit.>, and Marshall <cit.>. 

We show in <ref> that the original result of Antunes and Freitas does not extend to the subclass of symmetric eigenvalues. Instead, the optimal rectangle degenerates in the limit. 


The lattice point counting estimates in this paper are similar to those used for the Gauss circle problem, which aims for accurate asymptotics on the counting function inside the circle (and other closed curves) as the area grows to infinity. The best known error estimate on the Circle Problem is due to Huxley <cit.>. 

The lattice counting formulas in our paper differ somewhat from that work, because we consider only one quadrant of lattice points and our regions contain empty strips due to the translation of the lattice. Further, we focus on proving suitable inequalities (rather than asymptotics) for the counting function, in order to prevent the maximizing shape from degenerating. In essence, we develop inequalities that trade off the empty regions in the vertical and horizontal directions. After degeneration has been ruled out, we can invoke asymptotic formulas with error terms that need not be as good as Huxley's in order to prove convergence to a limiting shape.
 



§ RESULTS


Consider a strictly decreasing curve Γ lying in the first quadrant with x- and y-intercepts at L. Represent the curve as the graph of y= f(x) where f is strictly decreasing for x∈[0,L]. Denote the inverse function of f as g(y) for y ∈ [0,L]. Now compress the curve by a factor s>0 in the x-direction and stretch it by the same factor in the y-direction: 
 
    Γ(s) = graph of  sf(sx).
 
Next scale the curve Γ(s) by a factor r>0:

    rΓ(s)= graph of  rsf(sx/r) .
 


Given numbers σ, τ>-1, consider the translated or shifted positive-integer lattice

    (+σ) × (+τ) ,

which lies in the open first quadrant. Define the shifted-lattice counting function under the curve sΓ(s) to be


    N(r,s)    =number of shifted positive-integer lattice points lying inside or on rΓ(s)
       =#{ (j,k)∈ℕ×ℕ:k+τ≤ rsf ( (j+σ)s/r ) }.


The set S(r) consists of s-values that maximize N(r,s), that is, 

    S(r)= _s>0 N(r,s),      r>0 .
 

Write

    x^- = 
    
    0 ,     x ≥ 0, 
    
    |x| ,     x < 0
     .


Our first theorem will say that the maximizing set S(r) is bounded, under either of the following conditions on the shift parameters σ,τ>-1. 


Γ is concave and strictly decreasing, with

    max{f(1-σ^-/2-σ^-L),g(1-τ^-/2-τ^-L)} <2(1/2-σ^–τ^-)L.
 



Γ is convex and strictly decreasing, with 

    min{(1-σ^-)f(1-σ^-/2-σ^-L), (1-τ^-)g(1-τ^-/2-τ^-L)} > 2(σ^- + τ^-) L


and


    μ_f(σ)    def=min{ (1+σ)f(1+σ/2+σx)-f(x) : 1+σ/2+σL ≤ x ≤ L} > 0, 
    μ_g(τ)    def=min{ (1+τ)g(1+τ/2+τy)-g(y) : 1+τ/2+τL ≤ y ≤ L} > 0.



When σ,τ≥ 0,  
conditions <ref> and <ref> hold automatically (using that 0<f(x)<L and 0<g(y)<L when x,y ∈ (0,L)) and conditions <ref> and <ref> also hold (using that f and g are strictly decreasing and positive). Thus the Parameter Assumptions are significant only when σ<0 or τ<0. They are used to obtain an upper bound on the counting function: see the comments after <ref> and <ref>. 

 
If the curve Γ and the shift parameters σ,τ> -1/2 satisfy Parameter Assumption <ref> or <ref>, then for each >0 one has 

    S(r) ⊂[B(τ,σ)^-1-,B(σ,τ)+]     for all large r,

where 

    B(σ,τ)= 2+σ+τ + √( (2+σ + τ)^2 - 4(σ+1/2)τ ))/2(σ+1/2).



The bounding constant B(σ,τ) depends only on the shift parameters, not on the curve Γ. The bounding constant B(0,0)=4 in the unshifted case is consistent with our earlier work <cit.>. 

<ref> is proved in <ref>. Note it does not assume the curve is smooth. 

If the curve is smooth, then the optimal stretch set S(r) for maximizing the lattice count is not only bounded but converges asymptotically to a computable value, as stated in the next theorem. First we state the smoothness conditions to be used. 

 
Γ is concave, and for some (α ,β) ∈Γ with α,β>0 one has f∈ C^2[0,α], g ∈ C^2[0,β], with


    f'<0 on (0,α], f”<0 on [0,α], f” monotonic on [0, α],
       g'<0 on (0,β], g”<0 on [0,β], g” monotonic on [0, β].




 
Γ is convex, and for some (α ,β) ∈Γ with α,β>0 one has f∈ C^2[α,L], g ∈ C^2[β,L], with


    f'<0 on [α,L), f”>0 on [α,L], f” monotonic on [α,L],
       g'<0 on [β,L), g”>0 on [β,L], g” monotonic on [β,L].




 If the curve Γ and shift parameters σ, τ>-1/2 satisfy either Parameter Assumption <ref> and Concave Condition <ref>, or Parameter Assumption <ref> and Convex Condition <ref>, then the stretch factors maximizing N(r,s) approach

    s^*=√(τ + 1/2/σ+1/2)

as r →∞, with 

    S(r) ⊂ [s^* -O(r^-1/6), s^* + O(r^-1/6)],

and the maximal lattice count has asymptotic formula

    max_s>0 N(r, s) = r^2 (Γ)-2rL √((σ+1/2)(τ+1/2)) + O(r^2/3).

In particular, when the shift parameters σ and τ are equal, the optimal stretch factors for maximizing N(r,s) approach s^*=1 as r →∞.


The theorem follows from <ref> below, which has weaker hypotheses. 

We call the optimally stretched curve (s=s^*) “asymptotically balanced” in terms of the shift parameters, because the optimal shape balances the areas of the empty strips that are created by translation of the lattice: a horizontal rectangle of width rL/s^* and height τ+1/2 has the same area as a vertical rectangle of height rs^*L and width σ+1/2. (The “+1/2” arises from thinking of each lattice point as the center of a unit square.) Further, subtracting these two areas, each of which equals rL √((σ+1/2)(τ+1/2)), gives a heuristic derivation of the order-r correction term in the theorem. 



[Sufficient condition on shift parameters for the circle] When the curve Γ is the portion of the unit circle in the first quadrant, one takes L=1, f(x) = √(1-x^2), and α=β=1/√(2). Notice f is smooth and concave, with monotonic second derivative. By symmetry it suffices to consider σ≤τ. When σ≤τ≤ 0, Parameter Assumption <ref> says 

    √(1-(1+σ/2+σ)^2) < 2σ+2τ + 1.

When σ≤ 0≤τ, equality in Parameter Assumption <ref> would give a straight line. The resulting allowable region of (σ,τ)-shift parameters for <ref> is plotted on the left side of <ref>.



[Sufficient condition on shift parameters for p-circle with p=1/2]
Suppose Γ is the part of the 1/2-circle lying in the first quadrant, so that L=1, f(x) = (1-x^1/2)^2, and take α=β=1/4. Notice f is smooth and convex, with monotonic second derivative f”(x)=1/2x^-3/2. The region of allowable shift parameters for <ref> can be found numerically from Parameter Assumption <ref>, as shown on the right side of <ref>. 


Next we define weaker smoothness conditions. Let (α ,β) be a point on the curve Γ with α,β>0.

 Suppose Γ is concave, and:

 
  * f ∈ C^2(0,α], f'<0, f”<0, and a partition 0=α_0<α_1 < … < α_l = α exists such that f” is monotonic on (α_i-1,α_i) for each i=1,2,…,l; 
 
  * g ∈ C^2(0,β], g'<0, g”<0, and a partition 0=β_0<β_1 < … < β_m = β exists such that g” is monotonic on (β_i-1,β_i) for each i=1,2,…,m; 

  * positive functions δ(r) and ϵ(r) exist such that


    δ(r) = O(r^-2a_1) ,         f”(δ(r))^-1 = O(r^1-4a_2) , 
    ϵ(r) = O(r^-2b_1) ,         g”(ϵ(r))^-1 = O(r^1-4b_2) ,


as r →∞, for some numbers a_1,a_2,b_1,b_2>0;

  * and define a_3=1/2, b_3=1/2.

(The second condition in <ref> says that f”(x) cannot be too small as x → 0.)  


 Suppose Γ is convex, and:

 
  * f ∈ C^2[α,L), f'<0, f”>0, and a partition α=α_0<α_1 < … < α_l = L exists such that f” is monotonic on (α_i-1,α_i) for each i=1,2,…,l;  
 
  * g ∈ C^2[β,L), g'<0, g”>0, and a partition β=β_0<β_1 < … < β_m = L exists such that g” is monotonic on (β_i-1,β_i) for each i=1,2,…,m; 

  * positive functions δ(r) and ϵ(r) exist such that


    δ(r) = O(r^-2a_1) ,         f”(L-δ(r))^-1 = O(r^1-4a_2) , 
    ϵ(r) = O(r^-2b_1) ,         g”(L-ϵ(r))^-1 = O(r^1-4b_2) ,


as r →∞, for some numbers a_1,a_2,b_1,b_2>0;

  * and suppose f(x) = L+O(x^2a_3) as x → 0, and g(y) = L+O(y^2b_3) as y → 0, for some numbers a_3,b_3>0.

(The last condition says Γ cannot approach the axes too rapidly near the intercept points.)


Concave Condition <ref> implies Weaker Concave Condition <ref>, by choosing δ(r)=ϵ(r)=r^-1, a_1 = b_1 = 1/2 and a_2 = b_2 = 1/4, and noting that f”(0) ≠ 0, g”(0) ≠ 0. The same reasoning shows Convex Condition <ref> implies Weaker Convex Condition <ref> with a_3 = b_3 = 1/4, since 

    g(L)=0,  g'(L) ≤ 0,  g”(L)>0   ⟹   g(L-y) ≥ cy^2  for small y>0

where c>0, and substituting y=√(x/c) gives L-f(x) ≤√(x/c) for small x>0, and similarly for g. 

Thus <ref> follows immediately from the next result.


If the curve Γ and shift parameters σ, τ>-1/2 satisfy either Parameter Assumption <ref> and Weaker Concave Condition <ref>, or Parameter Assumption <ref> and Weaker Convex Condition <ref>, then the stretch factors maximizing N(r,s) approach

    s^*=√(τ + 1/2/σ+1/2)

as r →∞, with

    S(r) ⊂[ s^*-O(r^-), s^*+O(r^-) ]
 
where 

    =min{16,a_1,a_2,a_3,b_1,b_2,b_3 } .

Further, the maximal lattice count has asymptotic formula


    max_s>0 N(r, s) = r^2 (Γ)-2rL √((σ+1/2)(τ+1/2)) + O(r^1-2).




The proof in <ref> relies on lattice point counting propositions developed in <ref>, <ref> and <ref>. 


[p-circles]
Suppose Γ is the part of the p-circle |x|^p+|y|^p=1 lying in the first quadrant. When p>1 the curve is concave, and satisfies Weaker Concave Condition <ref> by <cit.>. When 0<p<1 it is convex and satisfies Weaker Convex Condition <ref> by <cit.>, noting that a_3=b_3=p/2 since f(x)=1+O(x^p) as x→ 0 and g(y)=1+O(y^p) as y → 0. 

Thus <ref> applies to each p-circle, p ≠ 1. The allowable shift parameters can be determined numerically from Parameter Assumption <ref> or <ref>, as in <ref> and <ref>. 



Next we show there can be no “universal” allowable region of negative shifts for <ref>. Specifically, for each choice of negative shifts σ, τ < 0, no matter how close to zero, a curve exists whose optimal stretch parameters grow to infinity or shrink to 0 as r →∞. That is, the optimal curve degenerates in the limit. 

 If -1<σ<0, τ > -1, then a concave C^2-smooth curve Γ exists, with intercepts at L=1, such that for each ϵ∈ (0,1) one has


    S(r) ⊂ (0,r^ϵ-1) ∪ (r^1-ϵ,∞)

for all large r.


The construction is given in <ref>. The point of the theorem is that as soon as one of the shift parameters is negative, a concave curve exists for which the maximizing stretch parameters approach either 0 or ∞ as r →∞. 

For convex curves, we do not know an analogue of <ref>: does a universal allowable region of (σ,τ) parameters exist in which <ref> holds for all C^2-smooth convex decreasing curves? 

The “bad” curve in <ref> can even be a quarter circle: 


If the curve Γ is the quarter unit circle, and σ , τ>-1 with either σ≤ -2/5 or τ≤ -2/5, then for each ϵ∈ (0,1) one has

    S(r) ⊂ (0,r^ϵ-1) ∪ (r^1-ϵ,∞)     for all large r.



The proof is in <ref>. And in <ref> we apply this result to Laplacian eigenvalue minimization on rectangles.




§ CONCAVE CURVES — COUNTING FUNCTION ESTIMATES
 
In order to prove <ref> we need to estimate the counting function. The curve Γ is taken  to be concave decreasing in the first quadrant, throughout this section. Denote the horizontal and vertical intercepts by x=L and y=M respectively, where L and M are positive but not necessarily equal. Allowing unequal intercepts is helpful for some of the results below. 

We start with a preliminary r-dependent bound on the maximizing set S(r). The proof of this bound also makes clear why N(r,s) attains its maximum as a function of s, for each fixed r, so that the set S(r) is well defined. 

If σ, τ > -1 then

    S(r)⊂[(1+τ)/rM, rL/(1+σ)]     whenever r ≥ (2+σ+τ)/√(LM).




The curve rΓ(s) with the particular choice s=√(L/M) has horizontal and vertical intercepts equal to
r√(LM). That intercept value is ≥ (2+σ+τ), by assumption on r in this lemma. Hence by
concavity, rΓ(s) encloses the  point (1+σ,1+τ) and so N(r,s)>0 for this particular value of s, which means the maximum of s ↦ N(r,s) is greater than 0. 

When s> rL/(1+σ), the x-intercept of rΓ(s) is less than 1+σ and so no shifted lattice points are enclosed, meaning N(r,s)=0. Thus the maximum is not attained for such s-values. Arguing similarly with the y-intercept shows the maximum is also not attained when s<(1+τ)/rM. The lemma follows. 



The last lemma required only that Γ be concave decreasing. Smoothness was not needed. Smoothness is not used in the next proposition either, which gives an upper bound on the counting function and so extends a result from the unshifted case <cit.>. 


Let σ, τ > -1. The number N(r, s) of shifted lattice points lying inside rΓ(s) satisfies


    N(r,s) ≤ r^2 (Γ)-C_1rs+σ^- τ^-


for all r ≥ (1-σ^-)s/L and s ≥ 1, where 


    C_1 = C_1(Γ, σ, τ) = 1/2(M-f(1-σ^-/2-σ^-L))-σ^-M-τ^-L.
 



The constant C_1 might or might not be positive. Parameter Assumption <ref> consists of the assumption C_1>0 along with the corresponding inequality for g, in the situation where L=M.

First suppose σ≤ 0, τ≤ 0. Write N for the number of shifted lattice points under Γ, and suppose L≥ 1+σ so that ⌊ L-σ⌋≥ 1. Extend the curve Γ horizontally from (0, M) to (σ, M), so that f(σ)=M. Construct triangles with vertices at (i-1+ σ, f(i-1+σ)), (i+σ, f(i+σ)), (i-1+σ, f(i+σ)) for i = 1,… ,⌊ L -σ⌋, as illustrated in <ref>. The rightmost vertex of the final triangle has horizontal coordinate ⌊ L-σ⌋ + σ, which is less than or equal to L. These triangles lie above the unit squares with upper right vertices at shifted lattice points, and lie below the curve Γ due to concavity. 



Hence 


    N + (triangles) ≤(Γ)-σ(M-τ)-τ(L-σ) -στ,


where the correction terms on the right side of the inequality represent the areas of the rectangular regions outside the first quadrant. 

Letting k = ⌊ L-σ⌋≥ 1, we compute

    (triangles)   =∑_i=1^k1/2(f(i-1+σ) - f(i+σ)) 
       =1/2(M-f(k+σ)) 
       ≥1/2(M- f(1+σ/2+σL))

because f is decreasing and k+σ≤ L< k+1+σ implies 

    k+σ > k+σ/k+1+σ L ≥1+σ/2+σ L .

Combining <ref> and <ref> proves 


    N ≤(Γ) - σ M -τ L -1/2(M- f(1+σ/2+σL)) +στ .


Now we replace Γ with the curve rΓ(s), meaning we replace N, L, M, f(x) with N(r,s), rs^-1L, rsM, rsf(sx/r) respectively, thereby obtaining the desired estimate <ref> (noting that L/s ≤ Ls since s ≥ 1). The restriction L≥ 1+σ becomes r ≥ (1+σ)s/L under the rescaling, and so we have proved the proposition in the case σ≤ 0, τ≤ 0. 

When σ> 0, τ> 0, the number of shifted lattice points inside rΓ(s) is less than or equal to the number when there is no shift (σ=τ=0), simply because the curve is decreasing. Thus this case of the proposition follows from the “σ,τ≤ 0” case above. 

When σ > 0, τ≤ 0, the number of shifted lattice points inside rΓ(s) is less than or equal to the number for σ = 0 with the same τ value, and so this case of the proposition follows also from the “σ,τ≤ 0” case above. A similar argument holds when σ≤ 0, τ > 0.  



Let σ, τ > -1. If s is bounded above and bounded below away from 0, as r →∞, then the number N(r, s) of shifted lattice points lying inside rΓ(s) satisfies


    N(r,s) ≤ r^2(Γ)-r(s^-1τ L + s(σ+1/2)M )+ o(r) .





Take c>1 and suppose c^-1<s<c throughout the rest of the proof. 

Suppose σ,τ≤ 0. Let K ≥ 1. Repeat the proof of <ref> except with the initial supposition L ≥ 1+σ replaced by L ≥ K + σ, and do not assume s ≥ 1. One finds 

    N(r,s) ≤ r^2 (Γ)-D_K rs-τ L r s^-1 + στ

for all r ≥ (K+σ)s/L, where 

    D_K = D_K(Γ, σ) = 1/2(M-f(K+σ/K+1+σL))+σ M.

We deduce


    lim sup_r →∞sup_s<c1/r( N(r,s) - r^2(Γ) + r(s(σ+1/2)M +s^-1τ L ) ) 
       ≤c/2 f(K+σ/K+1+σL) .


The last expression can be made arbitrarily small by choosing K sufficiently large (recall f(L)=0), and so the left side is ≤ 0. That proves the corollary when σ,τ≤ 0. 

Suppose σ >0, τ≤ 0. We will relate this case to the previous one. To emphasize the dependence of the counting function on the shift parameters, write N_σ,τ(r,s) for the counting function that was previously written N(r,s). Adding  columns of shifted lattice points at x=σ-⌈σ⌉ + 1, …, σ-1, σ gives the counting function N_σ,τ(r,s) where σ=σ-⌈σ⌉∈ (-1,0]. This counting function is related to the original one by


    N_σ,τ(r,s)   =N_σ,τ(r,s) + ∑_i=0^⌈σ⌉ - 1⌊ rsf(s(σ-i)/r)-τ⌋,
       =N_σ,τ(r,s) + ⌈σ⌉ rsM+o(r),


as r→∞, since s is bounded above and f is continuous with f(0)=M. Since σ,τ≤ 0, we may apply <ref> with σ replaced by σ to obtain

    N_σ,τ(r,s) ≤ r^2(Γ)-r(s^-1τ L + s(σ-⌈σ⌉ +1/2)M )+ o(r)   as r→∞.

Combining the above two formulas, we prove the corollary for σ>0, τ≤ 0.
 
When σ≤ 0, τ > 0, simply add the appropriate rows instead of columns and argue like above using ⌈τ⌉ instead of ⌈σ⌉, and using the boundedness of s^-1. Similarly, one can treat the case σ>0, τ > 0. 


The next proposition gives an asymptotic approximation to N(r,s), assuming the curve is concave decreasing and has suitably monotonic second derivative.  



Let σ, τ> -1 and 0 ≤ q < 1. If Weaker Concave Condition <ref> holds and s+s^-1 = O(r^q) then 


    N(r,s)=r^2 (Γ)-r( s^-1(τ + 1/2)L+s(σ+1/2)M)+ O(r^Q)


as r →∞, where

    Q = max{23, 12+32q , 1-2a_1+q,1-2a_2+32q, 1-2b_1+q, 1-2b_2+32q }.

Special cases: (i) If q=0 then Q = 1-2e where e = min{16, a_1, a_2,b_1, b_2 }. 

(ii) If Concave Condition <ref> holds then Q=max{23, 12+32q }. 


The numbers a_1,a_2,b_1,b_2 come from Weaker Concave Condition <ref>. That Condition also involves a point (α,β) ∈Γ with α,β>0, which we use in the following proof. 


The idea is to translate and truncate the curve rΓ(s) as in <ref>, in order to reduce to an unshifted lattice problem.  Then we invoke known results from our earlier paper <cit.> (which builds on work of Krätzel <cit.> and a theorem of van der Corput). 

Step 1 — Translating and truncating.
Notice rs →∞ and rs^-1→∞ as r →∞, since s=O(r^q) and s^-1=O(r^q) with q<1. Thus by taking r large enough, we insure

    rs^-1 g ( s^-11+τ/r) > rs^-1α > 1+σ ,     
    rs f ( s 1+σ/r) > rs β > 1+τ .

For all large r one also has δ(r)<α and ϵ(r)<β, by Weaker Concave Condition <ref>. 

Given a large r satisfying the above conditions, and a corresponding s>0, we let

    α = rs^-1α - (1+σ) ,     β = rs β - (1+τ) ,

and

    L = rs^-1 g ( s^-11+τ/r) - (1+σ) ,     M = rs f ( s 1+σ/r) - (1+τ) ,

so that 

    0 < α < L,      0 < β < M .



Consider the point O=(1+σ,1+τ) in the first quadrant. Regard this point as the new origin, and let Γ be the portion of rΓ(s) lying in the new first quadrant — see <ref>. That is, Γ is the graph of 

    f(x) = rs f ( s x+1+σ/r) - (1+τ) ,      0 ≤ x ≤L ,

and also of its inverse function 

    g(y) = rs^-1 g ( s^-1y+1+τ/r) - (1+σ) ,      0 ≤ y ≤M .

Notice (α,β) ∈Γ, since f(α)=β. Write N for the number of positive-integer lattice points under the curve Γ. That is, 

    N = #{ (j,k) ∈× : k ≤f(j) } .

This N does not count the lattice points in the first column or row, which arise from j=0 or k=0.

Weaker Concave Condition <ref> guarantees that f is C^2-smooth on the interval [0,α], with f^' < 0 and f^''<0 there, and similarly g is C^2-smooth on [0,β] with g^' < 0 and g^''<0 there. 

Next, we partition the interval [0,α] as 0=α_0 < α_1 < … < α_l = α where the interior partition points are chosen to be the elements of

    { rs^-1α_i - (1+σ) : i=1,…,l-1 }

that happen to lie between 0 and α. Observe f^'' is monotonic on each subinterval of the partition, by Weaker Concave Condition <ref>. Similarly, g^'' is monotonic on each subinterval of the corresponding partition 0=β_0 < β_1 < … < β_m = β of the interval [0,β].

Let 

    δ = [ rs^-1δ(r) - (1+σ) ]^+ ,     ϵ = [ rs ϵ(r) - (1+τ) ]^+ ,

so that 0 ≤δ<α and 0 ≤ϵ<β. 

To relate some of these old and new quantities, we denote antiderivatives of f, g by

    F(x) = ∫_0^x f(t)  t,       G(y) = ∫_0^y g(t)  t ,

and observe that


    ()    =r^2(Γ) - r^2 (F((1+σ) s/r)+ G((1+τ) s^-1/r))+ (1+σ)(1+ τ),
     '(x)   = s^2f'(sx+1+σ/r),    ”(x)= s^3/rf”(sx+1+σ/r),
    ∫_0^ | ”(x)|^1/3 x    = r^2/3∫_(1+σ)s/r^α | f ”(x)|^1/3 x ≤ r^2/3∫_0^α | f
    ”(x)|^1/3 x,
    ∑_i=1^l1/|”(_i)|^1/2   ≤∑_i=1^lr^1/2s^-3/2/|f”(α_i)|^1/2,


and similarly for  except with s replaced by s^-1.

Step 2 — Estimating the counting function.
Applying part (a) of <cit.> to the curve  and using the preceding relationships, we get


    |-r^2 (Γ)+ r^2 (F((1+σ) s/r)+ G((1+τ) s^-1/r)) 
       + r/2(s f((1+σ)s/r) + s^-1g((1+τ)s^-1/r) ) |
       ≤ 6r^2/3(∫_0^α |f”(x)|^1/3 x+∫_0^β |g”(y)|^1/3 y)  +175r^1/2(s^-3/2/|f”(δ(r))|^1/2+s^3/2/|g”(ϵ(r))|^1/2)
       +525r^1/2(∑_i=1^ls^-3/2/|f”(α_i)|^1/2+∑_j=1^ms^3/2/|g”(β_j)|^1/2) +1/4(∑_i=1^l s^2|f'(α_i)|+∑_j=1^m s^-2|g'(β_j)|)
       +r/2(s^-1δ(r)+sϵ(r))+l+m
    + 1/2(1+σ)+1/2(1+τ)+(1+σ)(1+τ)+1 ,

where we dealt with the term involving |”(δ)|^-1/2 in <cit.> as follows. One has ”(δ)=r^-1s^3 f”(z) where z = r^-1s(δ+1+σ) ≥δ(r), and so by monotonicity of f” on each subinterval of the partition (as assumed in Weaker Concave Condition <ref>) one concludes

    |”(δ)|≥ r^-1s^3 min{ |f”(δ(r))|,  |f”(α_1)| , …,  |f”(α_l )| }.

Thus the term involving |”(δ)|^-1/2 can be estimated by the sum of terms involving |f”(δ(r))|^-1/2 and |f”(α_i)|^-1/2. 

The right side of <ref> already has the desired order O(r^Q), by direct estimation and using that s+s^-1=O(r^q) and 2q < 12+32q since q<1. 

Step 3 — Understanding the left side of inequality <ref>. 
It remains to deal with the terms on the left of <ref>. Clearly N(r,s) and  count the same lattice points, except that N(r,s) also counts the points in the first row and column. That is, 


    = N(r,s) - ⌊ rsf((1+σ)s/r)-τ⌋ - ⌊ rs^-1 g((1+τ)s^-1/r)-σ⌋ +1 
        = N(r,s) -  rsf((1+σ)s/r) -τ - rs^-1 g((1+τ)s^-1/r)-σ + ρ(r,s)


for some number ρ(r,s) ∈ [1,3]. Substitute this formula into the left side of <ref>. Substitute also the following expressions, which are obtained from <ref>: 


    rsf((1+σ)s /r)    =rsM +O(s^2),
    
    r^2F((1+σ) s/r)    =rs (1+σ) M +O(s^2),


and similarly for g and G.

The proposition now follows straightforwardly, since O(s^2)=O(r^2q). 





If f is decreasing and concave on [0,L] then 

    f(x)=f(0) +O(x),      F(x)=f(0)x + O(x^2),     as x → 0,

where F(x) = ∫_0^x f(t)  t is the antiderivative of f(x). 



The difference quotient (f(x)-f(0))/x is a decreasing function of x since f is concave, and it is less than or equal to 0 since f is decreasing. Hence the difference quotient is bounded, and so f(x)=f(0) +O(x). Integrating completes the proof. 




§ CONVEX CURVES — COUNTING FUNCTION ESTIMATES
 
Assume the curve Γ is convex decreasing, throughout this section. We will prove estimates for convex curves analogous to the work in <ref> for concave curves. 

<ref> below is an improved  r-dependent bound on the optimal stretch factors, generalizing Ariturk and Laugesen's lemma from the unshifted situation <cit.>. By “improved” we refer to the upper and lower bounds: for instance, when σ =0 the upper bound in <ref> improves on the bound in <ref> by a factor of 2. This tighter bound on the optimal stretch factor gives us more flexibility when deriving the two-term counting estimate in <ref>. 

In the next lemma we assume for simplicity that the x- and y-intercepts are both L, so that we need not change the definitions of μ_f(σ) and μ_g(τ) in <ref>.


If σ, τ > -1 with μ_f(σ)> 0 and μ_g(τ) > 0, then

    S(r)⊂[ 2+τ/rL , rL/2+σ]

whenever 


    r ≥max((2+σ)√(2(1+τ)/Lμ_f(σ)), (2+τ)√(2(1+σ)/Lμ_g(τ))) .




 

Claim 1: N(r,s)=0 if s ∈( 0,(1+τ)/rL ] or s ∈[ rL/(1+σ),∞). Indeed, the curve rΓ(s) has x- and y-intercepts at rL/s and rsL, respectively, and so if rL/s ≤ 1+σ or rsL ≤ 1+τ then the point (1+σ,1+τ) is not enclosed by the curve and so the lattice count N(r,s) is zero. 

Claim 2: if <ref> holds and s∈( rL/(2+σ),rL/(1+σ) ) then 

    N(r,s) < N(r,1+σ/2+σ s) .

To prove this claim, notice the x-intercept satisfies 

    1+σ < rL/s < 2+σ ,

and so only the first column of shifted lattice points (the points with x-coordinate at 1+σ) can contribute to the count inside rΓ(s). Hence N(r,s)= ⌊ rsf((1+σ)s/r) - τ⌋. Meanwhile, if we count shifted lattice points in the first two columns (where x=1+σ and x=2+σ) we find 


    N(r,1+σ/2+σ s) 
       ≥⌊ rs1+σ/2+σf( (1+σ)^2s/(2+σ )r)- τ⌋ + ⌊ rs1+σ/2+σf( (1+σ)s/r) - τ⌋
       > rs1+σ/2+σf( (1+σ)^2s/(2+σ )r) + rs1+σ/2+σf( (1+σ)s/r) -2τ -2 
       = rsf((1+σ)s/r) + rs/2+σ( (1+σ)f( (1+σ)^2s/(2+σ )r) -f( (1+σ)s/r) ) -2(1+τ)  
       ≥ rsf((1+σ)s/r) + rs/2+σμ_f(σ) -2(1+τ) 
       >  rsf((1+σ)s/r) ≥ N(r,s) ,


where to get the final line we use that rs/2+σμ_f(σ)>2(1+τ), which follows from s>rL/(2+σ) and the lower bound on r in <ref>. The proof of Claim 2 is complete. 

Claim 3: if <ref> holds and s ∈( (1+τ)/rL , (2+τ)/rL ) then 

    N(r,s) < N(r,2+τ/1+τ s ) .

The proof is analogous to Claim 2, except counting in rows instead of columns.

Claim 4: if <ref> holds then the maximizing s-values for N(r,s) lie in the interval [ (2+τ)/rL, rL/(2+σ) ]. To see this, note that N(r,s^')>0 for some s^'>0, by the strict inequality in Claim 2, and so the maximum does not occur in the intervals considered in Claim 1. The maximum does not occur in the interval considered in Claim 2, as that claim itself shows, and similarly for Claim 3. Thus the maximum must occur in the remaining interval, which proves Claim 4 and thus finishes the proof of the lemma. 



The next bound generalizes work of Ariturk and Laugesen <cit.> from the unshifted situation (σ=τ=0) to the shifted case.


Let σ,τ> -1. The number N(r, s) of shifted lattice points lying inside rΓ(s) satisfies


    N(r,s) ≤ r^2 (Γ)-C_2rs + σ^- τ^-


for all r ≥ (2-σ^-)s/L and s≥ 1, where 

    C_2 = C_2(Γ, σ, τ) = 1/2(1-σ^-)f(1-σ^-/2-σ^-L)-σ^-M-τ^-L .



The constant C_2 need not be positive. That is why hypothesis <ref> in Parameter Assumption <ref> includes (for L=M) the assertion that C_2>0.  



First consider σ≤0, τ≤0. Write N for the number of shifted lattice points under Γ. Suppose L≥ 2+σ. Extend the curve horizontally from (0, M) to (σ, M), so that f(σ)=M. Construct a trapezoid (see <ref>) with vertices at ( σ,f(1+σ) ), ( 1+σ, f(1+σ) ), (0,h), (σ,h) where h=f(1+σ)-(1+σ) f'(1+σ). Also construct triangles with vertices (i-1+σ, f(i+σ)), (i+σ, f(i+σ)), (i-1+ σ, f(i+σ)-f'(i+σ)), where i = 2,… ,⌊ L -σ⌋. These triangles lie above the squares with upper right vertices at the shifted lattice points, and like  below the curve by convexity, as <ref> illustrates. Hence 

    N + (trapezoid and triangles) ≤(Γ)-σ(M-τ)-τ(L-σ) -στ



Let k = ⌊ L-σ⌋≥ 2, so that k+σ≤ L < k+σ+1. Then


    (trapezoid)
       = 1/2(base+top) · (height) 
        = - 1/2 (1-σ) · (1+σ) f'(1+σ) 
       ≥1/2 (1+σ) ( f(1+σ) - f(2+σ) )


by convexity, and using that 1-σ≥ 1. Further, convexity implies


    (triangles)   = - 1/2∑_i=2^k  f'(i+σ) 
       ≥1/2∑_i=2^k-1(f(i+σ)-f(i+1+σ) ) + 1/2(f(k+σ)-f(L) )
       =1/2f(2+σ) .


Hence


    (trapezoid) + (triangles) 
       ≥1/2(1+σ)f(1+σ) - 1/2σ f(2+σ) 
       ≥1/2(1+σ)f(1+σ/2+σL) - 1/2σ f(2+σ/2+σL)


since f is decreasing and L/(2+σ) ≥ 1. Combining <ref> and <ref> and using f(L)=0 proves 


    N ≤(Γ) - σ M -τ L -1/2(1+σ)f(1+σ/2+σL) +στ .



Now replace Γ with the curve rΓ(s), meaning replace N, L, M, f(x) with N(r,s), rs^-1L, rsM, rsf(sx/r) respectively. Using s ≥ 1, we know L/s ≤ Ls; the assumption L ≥ 2+σ becomes r ≥ (2+σ)s/L. Thus we obtain <ref> in the case σ≤ 0, τ≤ 0. 

One may now deduce the remaining cases as was done in the proof of <ref>. 



Let σ, τ > -1. If s is bounded above and bounded below away from 0, as r →∞, then the number N(r, s) of shifted lattice points lying inside rΓ(s) satisfies


    N(r,s) ≤ r^2(Γ)-r(s^-1τ L + s(σ+1/2)M )+ o(r) .





Fix c>1 and assume c^-1<s<c in the rest of the proof. 

Suppose σ,τ≤ 0, and let K ≥ 2. Repeat the proof of <ref> except with the initial requirement L ≥ 2+σ replaced by L ≥ K + σ, and do not assume s ≥ 1. The argument gives 

    N(r,s) ≤ r^2 (Γ) -E_K rs - τ L rs^-1 +στ .

for all r ≥ (K+σ)s/L, where 

    E_K = E_K(Γ, σ) = 1/2(1+σ)f(1+σ/K+σL) - 1/2σ f(2+σ/K+σL) + σ M .

Hence


    lim sup_r →∞sup_s<c1/r( N(r,s) - r^2(Γ) + r(s(σ+1/2)M +s^-1τ L ) ) 
       ≤c/2| M - (1+σ)f(1+σ/K+σL) + σ f(2+σ/K+σL) | .


The last expression can be made arbitrarily small by choosing K sufficiently large (recall f(0)=M), and so the left side is ≤ 0, which proves the corollary when σ,τ≤ 0. 

By arguing as in the proof of <ref>, one handles the other three cases for σ and τ.




In the next proposition we state a two-term asymptotic for lattice point counting under convex curves.




Let σ, τ> -1. If Weaker Convex Condition <ref> holds and s+s^-1 = O(1) then 


    N(r,s)=r^2 (Γ)-r( s^-1(τ + 1/2)L+s(σ+1/2)M)+ O(r^1-2)


as r →∞, where = min{16, a_1, a_2, a_3,b_1, b_2, b_3 }. In particular, if Convex Condition <ref> holds  then <ref> holds with =16. 


<ref> does not assume the intercepts L and M are equal, and so we modify Weaker Convex Condition <ref> by taking each occurrence of “L” that relates to the function g and changing it to “M”, and changing the a_3-condition to f(x) = M+O(x^2a_3).


We use the idea from <ref>: translate and truncate the curve rΓ(s) to reduce to an unshifted lattice problem, and then use results from Ariturk and Laugesen's paper <cit.>. 

Assume rΓ(s) does not pass through any point in the shifted lattice. This assumption will be removed in the final step of the proof. 

Step 1 — Translating and truncating.
Keep the notation from the proof of <ref>, except redefine the quantities δ and ϵ to be

    δ = [ L + 1+σ - rs^-1 (L - δ(r))  ]^+ ,     ϵ = [ M + 1+τ - rs (M - ϵ(r))  ]^+ .

Arguing as in Step 1 of that proof, we have


    0 < α < ⌊L⌋,      0 < β < ⌊M⌋ ,


by taking r large enough, and also

    0 ≤δ< ⌊L⌋-α,     0 ≤ϵ< ⌊M⌋-β .


Step 2 — Estimating the counting function.
Recall F represents the antiderivative of f, defined in <ref>. Applying part (a) of <cit.> to the curve  and using the relationships between the unshifted and shifted quantities as in the proof of <ref>, we get


    |-r^2 (Γ) + r^2(F((1+σ)s/r) + G((1+τ)s^-1/r))
       + r/2(sf((1+σ)s/r) +s^-1g((1+τ)s^-1/r)))|
       ≤ 6r^2/3(∫_α^L f”(x)^1/3 x+∫_β^M g”(y)^1/3 y)
    +175r^1/2(s^-3/2/f”(L- δ(r))^1/2
       +
    s^3/2/g”(M-ϵ(r))^1/2)+700r^1/2(∑_i=0^l-1s^-3/2/f”(α_i)^1/2+∑_j=0^m-1
    s^3/2/g”(β_j)^1/2)
       +1/4(
    ∑_i=0^l-1s^2|f'(α_i)|+∑_j=0^m-1s^-2|g'(β_j)| ) +1/2r(s^-1δ(r)+sϵ(r))+
    l+m
        +1/2(1+σ)+1/2(1+τ)+(1+σ)(1+τ)+5
     
       +rs^-1g((1+τ)/rs)-(1+σ)/rsf((1+σ)s/r)-(1+τ)+rsf((1+σ)s/r)-(1+τ)/rs^-1g((1+τ)/rs)-(1+σ),

where we estimated the term involving ”(-δ)^-1/2 as follows. One has ”(-δ) =r^-1s^3 f”(z) where 

    z=r^-1s (L-δ+1+σ) ≤ L-δ(r) ,

and so by monotonicity of f” on each subinterval of the partition (as assumed in Weaker Convex Condition <ref>) one concludes

    ”(-δ)≥ r^-1s^3 min{ f”(L-δ(r)),  f”(α_0),…,f”(α_l-1)}.

Thus the term involving ”(-δ)^-1/2 can be estimated by the sum of terms involving f”(L-δ(r))^-1/2 and f”(α_i)^-1/2.

The right side of <ref> has the form O(r^1-2e), by arguing directly with s+s^-1=O(1) and the assumptions in Weaker Convex Condition <ref>, and estimating the last two terms in <ref> by

    rs^-1g((1+τ)/rs)-(1+σ)/rsf((1+σ)s/r)-(1+τ)=s^-1L-o(1)/sM-o(1) = O(1)

and similarly with f and g interchanged. 

Step 3 — Understanding the left side of inequality <ref>. The terms on the left of <ref> are dealt with in the same manner as in Step 3 of <ref>, except replacing <ref> with the last assumption in Weaker Convex Condition <ref>, as follows. Substituting x=(1+σ)s/r into f(x)=M+O(x^2a_3) and into F(x)=Mx+O(x^1+2a_3) gives

    rsf((1+σ)s /r)    =rsM +O(r^1-2a_3),
    
    r^2F((1+σ) s/r)    =rs (1+σ) M +O(r^1-2a_3),

since s+s^-1=O(1). One argues similarly for g and G. Thus we have finished the proof under the assumption that rΓ(s) passes through no lattice points. 

Step 4 — Finishing the proof. Now drop the assumption that rΓ(s) passes through no lattice points. Notice the counting function N(r,s) is increasing in the r-variable. Fix the r and s values, and modify the functions δ(·) and ϵ(·) to be continuous at r. For sufficiently small η>0 we have N(r+η,s)=N(r,s), because the r-variable would have to increase by some positive amount for the curve rΓ(s) to reach any new lattice points. Since no lattice points lie on the curve (r + η)Γ(s), Steps 1–3 above apply to that curve. Hence by continuity as η→ 0, the conclusion of the proposition holds also for rΓ(s).





§ LOWER BOUND ON THE COUNTING FUNCTION FOR DECREASING Γ


We need a rough lower bound on the counting function, in order to prove boundedness of the maximizing set in <ref>. Assume the curve Γ is strictly decreasing in the first quadrant, and has x- and y-intercepts at L and M. The intercepts need not be equal, in the next lemma. 


The number N(r, s) of shifted lattice points lying inside rΓ(s) satisfies

    N(r,s)≥ r^2(Γ)-r(s^-1(1+τ)L+s(1+σ)M),      r,s>0.




We split the proof into two cases, and later rescale to handle the general curve. Write N for the number of shifted lattice points under Γ. 

Case I: The point (1+σ,1+τ) lies outside the curve Γ, and so N=0. Then the rectangles with vertices (0,0),(L,0),(L,1+τ),(0,1+τ) and (0,0),(1+σ,0),(1+σ,M),(0,M) cover Γ since the curve is decreasing, and so by comparing areas one has 


    N +(1+τ)L + (1+σ)M≥(Γ).



Case II: The point (1+σ,1+τ) lies inside the curve. We shift the origin to =(1+σ, 1+τ) and draw new axes, denoting the x- and y-intercepts on the new axes by  and ; see <ref>. The part of Γ lying in the new first quadrant is .



Each lattice point corresponds to a square whose lower left vertex sits at that point. These squares cover  since the curve is strictly decreasing. The remaining area under Γ is covered by the two rectangles described in Case I. The sum of the areas of the squares and rectangles must exceed the area under Γ, and so <ref> holds once again. 

To complete the proof, simply replace the curve Γ with rΓ(s), meaning that in <ref> we replace N, L, M with N(r,s), rs^-1L, rsM respectively. The lemma follows.





§ PROOF OF <REF>


We prove the theorem in two parts: first for concave curves, and then for convex curves. When Γ is concave, we will utilize the bound on S(r) in <ref> and the two-term upper bound on the counting function in <ref>, along with the improved upper bound in <ref> and the rough lower bound on the counting function in <ref>. 

Recall the intercepts are assumed equal (L=M) in this theorem. 



 §.§ Part 1: Γ is concave and Parameter Assumption <ref> holds
 
The proof has two steps. Step 1 shows S(r) is bounded above and below away from 0, for large r. Step 2 uses this boundedness to improve the asymptotic bound on S(r), revealing that it depends only on σ and τ and not the curve Γ. 

Step 1. Take s∈ S(r) and suppose r ≥ (2+σ+τ)/L. Then <ref> says s≤ rL/(1+σ), so that 

    r ≥(1+σ)s/L≥(1-σ^-)s/L.

If s≥ 1 then <ref> implies


    N(r,s)≤ r^2 (Γ)-C_1rs + σ^- τ^- .


Parameter Assumption <ref> guarantees here that C_1>0.

The lower bound in <ref> with “s = 1” says

    N(r,1) ≥ r^2 (Γ)-(2+σ+τ)Lr.


Since s∈ S(r) is a maximizing value, one has N(r,s) ≥ N(r,1), and so the preceding two inequalities give 

    s ≤(2+σ+τ)L/C_1 + σ^- τ^- L/(2+σ+τ)C_1

when r ≥ (2+σ+τ)/L and s ≥ 1. Thus S(r) is bounded above for all large r. 

Similarly if s∈ S(r) then s^-1 is bounded above, by interchanging the roles of the horizontal and vertical axes in the argument above. Thus the set S(r) is bounded below away from 0, for large r.

Step 2. The number 

    s = lim sup_s ∈ S(r), r →∞ s

is finite and positive by Step 1. Combining the inequality N(r,s) ≥ N(r,1) with estimate <ref> and <ref> (which relies on the boundedness of S(r)) we obtain  

    (σ+1/2) s^2 - (2+σ+τ) s + τ≤ 0

after letting r →∞. Notice σ+1/2>0 by hypothesis in <ref>. Hence s is bounded above by the larger root of the quadratic; that is,

    s≤ B(σ,τ)= 2+σ+τ + √( (2+σ + τ)^2 - 4(σ+1/2)τ)/2(σ+1/2) .

Similarly lim sup_r→∞ s^-1≤ B(τ,σ), by interchanging the roles of the axes. The proof of <ref> is complete, in the concave case. 
 


 §.§ Part 2: Γ is convex and Parameter Assumption <ref> holds
 

Take s ∈ S(r) and suppose r satisfies (<ref>), recalling there that μ_f(σ) and μ_g(τ) are positive by Parameter Assumption <ref>. Now proceed as in Part 1 of the proof, simply replacing <ref>, <ref> and <ref> with <ref>, <ref> and <ref>, respectively.




§ PROOF OF <REF>
 

Recall the intercepts are equal, L=M, in this theorem. 

The optimal stretch parameters are bounded above and bounded below away from 0 as r →∞, by <ref>. (It suffices to use the curve-dependent bound from Step 1 of that proof; we do not need the curve-independent bound B(σ,τ) from Step 2.) 

Hence by <ref> (if Γ is concave) or <ref> (if Γ is convex), 


    N(r,s) =r^2 (Γ)-rL(s^-1(τ + 1/2)+s(σ+1/2))+ O(r^1-2)


when s ∈ S(r); this estimate holds also when s>0 is any fixed value. Thus for s ∈ S(r) and s^*=√((τ + 1/2)/(σ+1/2)) we have


    N(r,s)   ≤ r^2 (Γ)-rL(s^-1(τ+ 1/2)+s(σ+1/2))+ O(r^1-2), 
    
    N(r,s^*)    ≥ r^2 (Γ)-2rL√((τ+1/2)(σ + 1/2)) + O(r^1-2),


as r →∞. Notice N(r, s^*)≤ N(r,s) because s ∈ S(r) is a maximizing value, and so


    s^-1(τ + 1/2)+s(σ+1/2) ≤ 2√((τ + 1/2)(σ+1/2))+  O(r^-2) .


Therefore s =s^*+ O(r^-), by <ref> below with a = τ + 1/2, b = σ+1/2. 

For the final statement of the theorem, when s∈ S(r) one has 

    2√((τ + 1/2)(σ+1/2))≤ s^-1(τ + 1/2)+s(σ+1/2)≤ 2√((τ + 1/2)(σ+1/2))+  O(r^-2)

by the arithmetic–geometric mean inequality and <ref>. Multiplying by rL and substituting into <ref> gives the asymptotic formula <ref>. 


 
When a, b, s >0 and 0 ≤ t ≤√(ab), 

    s^-1a + sb ≤ 2√(ab) +t   ⟹  | s - √(a/b)| ≤ 3(ab)^1/4/ b√(t) .



 By taking the square root on both sides of the inequality 

    ( (s^-1a)^1/2-(sb)^1/2)^2 = s^-1a + sb - 2√(ab)≤ t

and then using that the number (ab)^1/4 lies between (s^-1a)^1/2 and (sb)^1/2 (because it is their geometric mean), we find

    | (ab)^1/4 - (sb)^1/2 | ≤ t^1/2 .

Hence (ab)^1/4 - t^1/2≤ (sb)^1/2≤ (ab)^1/4 + t^1/2. Squaring and using that
t ≤ (ab)^1/4t^1/2 (when t ≤√(ab)) proves the lemma.  





§ PROOF OF <REF> AND <REF>




 §.§ Proof of <ref>



Fix σ∈ (-1,0) and τ>-1. Since 0<1+σ<1, we may choose m ∈ large enough that


    (1+σ)^2m < 1/2m+1 .


Defining ϕ(x) = 1-x^2m for 0 ≤ x ≤ 1, one checks ϕ(1+σ) > area under graph of ϕ. Thus one may choose 0<δ<1 small enough that the function

    f(x) = 1-δ x^2 - (1-δ) x^2m ,      0 ≤ x ≤ 1 ,

satisfies

    f(1+σ) > area under graph of f .

Observe f is smooth and strictly decreasing, with f” < 0 on [0,1], so that its graph Γ is concave. The inverse function g satisfies the same conditions. 

The curve rΓ(r) is the graph of r^2f(x) for 0 ≤ x ≤ 1. This curve contains only the first column of shifted lattice points (the points with x-coordinate 1+σ), and so 


    N(r,r) 
        = ⌊ r^2 f(1+σ) - τ⌋
       ≥ r^2 f(1+σ) - τ - 1.



Now fix 0<ϵ<1. If s ∈ [r^ϵ-1,r^1-ϵ] then s + s^-1 = O(r^1-ϵ), and so <ref> with q=1-ϵ and L=M=1 gives that


    N(r,s) 
        = r^2 (Γ) - r(s(σ+1/2) + s^-1(τ + 1/2)) + O(r^2-3ϵ/2) 
        = r^2 (Γ) + o(r^2) .


Since (Γ) < f(1+σ), we conclude that for all large r,

    N(r,s) < N(r,r)

and so s ∉ S(r), which proves the theorem. 



 §.§ Proof of <ref>



By symmetry, we may suppose σ≤ -2/5.

The argument is the same as for <ref>, except now the curve is a quarter circle, described by f(x) = √(1-x^2). The only point to check in the proof is that 

    f(1+σ) > (Γ)

when -1<σ≤-2/5, which reduces to the fact that 4/5 > π/4. 




§ NUMERICAL EXAMPLES, AND CONJECTURES FOR TRIANGLES (P=1) 

 
<ref>(a) illustrates the convergence of s ∈ S(r) to s^*, when Γ is a quarter circle and the shifts are positive. The convergence is erratic, while still obeying the decay rate O(r^-1/6) as promised by <ref>. <ref>(b) shows the degeneration that can occur when the shifts are negative, as explained in <ref>. 



Quite different behavior occurs when Γ is a straight line with slope -1, in other words, when the curve is the 1-ellipse described by f(x)=1-x, which is not covered by our results in <ref>. Here N(r,s) counts the shifted lattice points inside the right triangle with vertices at (r/s,0), (0,rs) and the origin. <ref> insures the maximizing set S(r) is bounded above and below, being contained in [B(τ,σ)^-1-,B(σ,τ)+] for all large r. This boundedness depends on Parameter Assumption <ref> holding, which in this case says

    (2-max(σ^-,τ^-))(1-2σ^- - 2τ^-) > 1 .

In particular, S(r) is bounded for the 1-ellipse if σ=τ>-0.117. <ref> for convergence of S(r) does not apply, though, to the 1-ellipse. 

The numerical plots in <ref> suggest S(r) might not converge, and might instead cluster at many different heights. Are those heights determined by a number theoretic property of some kind? (Such behavior would be particularly interesting when the shifts are σ=τ=-1/2, since those shifted lattice points correspond to energy levels of harmonic oscillators in 2-dimensions, as explained in the next section.) For a more detailed discussion and precise conjecture on this open problem for p=1 in the unshifted case, see our work in <cit.> and the partial results of Marshall and Steinerberger <cit.>.

The numerical method that generated the figures is described in <cit.> for p=1. It adapts easily to handle other values of p, in particular p=2 (the circle), and the code is available in <cit.>. 






§ FUTURE DIRECTIONS — OPTIMAL QUANTUM OSCILLATORS




 §.§ Literature on spectral minimization

Antunes and Freitas <cit.> investigated the problem of maximizing the number of first-quadrant lattice points in ellipses with fixed area, and showed that optimal ellipses must approach a circle as the radius approaches infinity. In terms of eigenvalues of the Dirichlet Laplacian on rectangles having fixed area, their result says that the rectangle minimizing the n-th eigenvalue must approach a square as n →∞. Their intuition is that high eigenvalues should be asymptotically minimal for the “most symmetrical” domain. 

Besides Antunes and Freitas's work on eigenvalue minimization <cit.>, we mention that van den Berg and Gittins <cit.> showed the cube is asymptotically minimal in 3-dimensions as n →∞, while Gittins and Larson <cit.> handle all dimensions ≥ 2. Van den Berg, Bucur and Gittins <cit.> proved an analogous asymptotic maximization result for eigenvalues of the Neumann Laplacian on rectangles. Bucur and Freitas <cit.> showed for general domains in 2 dimensions that eigenvalue minimizing regions become circular in the limit, under a perimeter normalization. Larson <cit.> shows among convex domains that the disk asymptotically minimizes the Riesz means of the  eigenvalues, for Riesz exponents ≥ 3/2. Eigenvalue minimizing domains have been studied numerically by Oudet <cit.>, Antunes and Freitas <cit.>, and Antunes and Oudet <cit.>, <cit.>. Incidentally, Colbois and El Soufi <cit.> proved subadditivity of n ↦λ_n^* (the minimal value of the n-eigenvalue), from which it follows that the the famous Pólya conjecture λ_n ≥ 4π n / in 2-dimensions would be a corollary of the conjecture that the eigenvalue minimizing domain approaches a disk as n →∞. 

A different way of extending the work of Antunes and Freitas is to investigate lattice point counting inside more general curves, not just ellipses. Laugesen and Liu <cit.> and Laugesen and Ariturk <cit.> showed this can be done for p-ellipses with p ≠ 1, and for more general concave and convex curves in the first quadrant too. Marshall <cit.> has extended the results to strongly convex domains in all dimensions, using somewhat different methods. For more on the literature see <cit.>.  




 §.§ Spectral application of <ref>

This paper sheds new light on the rectangular result of Antunes and Freitas. Consider the family of rectangles defined by

    [0,2π s^-1] × [0,2π s]

for s>0. The “even–even” eigenfunctions (that are symmetrical with respect to the two axes through the center) have the form

    u = sin( s(j-1/2)x ) sin( s^-1(k-1/2)y )

with corresponding eigenvalues

    λ = ( s(j-1/2) )^ 2 + ( s^-1(k-1/2) )^ 2 ,

for j,k ≥ 1. These even–even eigenvalues have counting function 


    #{λ≤ r^2 }
        = number of points in the shifted lattice ( - 1/2) × (-1/2) lying
       inside or on the ellipse (sx)^2+(s^-1y)^2 ≤ r^2
        = N(r,s)


where the shift parameters are σ=τ=-1/2 and the curve Γ is the quarter-circle.

<ref> says the set S(r) of s-values that maximize the counting function does not approach 1 as r →∞. Instead, the maximizing s-values approach 0 or ∞. Thus the even–even symmetry class of eigenvalues on the rectangle behaves quite differently from the full collection of eigenvalues studied by Antunes and Freitas. The asymptotically optimal rectangle for maximizing the counting function as r →∞ (or equivalently, minimizing the n-th eigenvalue as n →∞) is not the square but rather the degenerate rectangle. 



 §.§ Open problem for harmonic oscillators

A quantum analogue of the Antunes–Freitas theorem for rectangles would be to minimize the n-th energy level among the following family of harmonic oscillators. For each s>0, consider 


    -Δ u + 1/4( (sx)^2 + (s^-1y)^2 ) u = E u ,      x,y ∈ ,


with boundary condition u → 0 as |(x,y)| →∞. Write s_n for an s-value that minimizes the n-th eigenvalue E_n. By analogy with Antunes and Freitas's theorem for Dirichlet rectangles, one might conjecture that s_n → 1 as n →∞. In fact, the behavior is quite different, as we now explain. 

Let us translate the harmonic oscillator problem into a shifted lattice point counting problem. The 1-dimensional oscillator equation -u^'' + 1/4 x^2 u = E u has eigenvalues E=j-1/2 for j=1,2,3,.…. By separating variables and rescaling, one finds that equation <ref> has spectrum 

    { E_n } = { s(j-1/2) + s^-1(k-1/2) : j,k = 1,2,3,…} .
 
Hence the number of harmonic oscillator eigenvalues less than or equal to r equals the number of points in the shifted lattice ( - 1/2) × ( - 1/2) lying below the straight line sx+s^-1y=r, which is given by our counting function N(r,s) where Γ is the straight line y=1-x (the 1-ellipse) and the shift parameters are σ=τ=-1/2. To minimize the eigenvalues we should maximize the counting function. 

The numerical evidence in the left part of <ref> suggests that the s-values maximizing the counting function N(r,s) do not converge to 1 as r →∞. Rather, the optimal s-values seem to cluster at various heights. (For a precise such clustering conjecture in the unshifted case, see <cit.>.) Thus the family of harmonic oscillators exhibits strikingly different spectral behavior from the family of Dirichlet rectangles. 



 §.§ Interpolating family of Schrödinger potentials

The family of Schrödinger potentials |sx|^q + |s^-1y|^q, where 2<q<∞ and s>0, interpolates between the harmonic oscillator (q=2) and the infinite potential well (q=∞) that corresponds to the Dirichlet Laplacian on a rectangular domain. We conjecture that when 2<q<∞, the set S(r) of values maximizing the eigenvalue counting function will converge to 1 as r →∞. This conjecture would provide a 1-parameter family of quantum oscillators for which the analogue of the Antunes–Freitas theorem holds true, with the family terminating in an exceptional endpoint case: the harmonic oscillator. 

The difficulty is that the eigenvalues of the 1-dimensional oscillator with potential |x|^q do not grow at a precisely regular rate. Hence to tackle the conjecture, one will need to extend the current paper from shifted lattices, where each row and column of the lattice is translated by the same amount, and find a way to handle deformed lattices, where the amount of translation varies with the rows and columns. This challenge remains for the future. 




§ ACKNOWLEDGMENTS

This research was supported by grants from the Simons 
Foundation (#429422 to Richard Laugesen) and the University 
of Illinois Research Board (award RB17002). 
The material in this paper forms part of Shiya Liu's Ph.D. dissertation at the University of Illinois, Urbana–Champaign <cit.>. 



99

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