Lattice points in stretched model domains of finite type in R d
aa r X i v : . [ m a t h . N T ] N ov LATTICE POINTS IN STRETCHED MODEL DOMAINS OFFINITE TYPE IN R d JINGWEI GUO AND WEIWEI WANG
Abstract.
We study an optimal stretching problem for certain convexdomain in R d ( d ≥
3) whose boundary has points of vanishing Gauss-ian curvature. We prove that the optimal domain which contains themost positive (or least nonnegative) lattice points is asymptotically bal-anced. This type of problem has its origin in the “eigenvalue optimiza-tion among rectangles” problem in spectral geometry. Our proof relieson two-term bounds for lattice counting for general convex domains in R d and an explicit estimate of the Fourier transform of the characteristicfunction associated with the specific domain under consideration. Introduction
Let Ω ⊂ R d be an Euclidean domain with sufficiently smooth boundary.A coordinate stretch of Ω is a domain of the form(1.1) A Ω := { ( a x , . . . , a d x d ) | ( x , . . . , x d ) ∈ Ω } , where a , . . . , a d are positive numbers, and A = diag( a , . . . , a d ) is the asso-ciated d × d diagonal matrix. Such a stretch is called a volume-preservingstretch if the stretching factors a , . . . , a d satisfy a · · · a d = 1, or in otherwords, det A = 1. In what follows we will always assume that the stretch A Ω is a volume-preserving stretch, and we will denote(1.2) a ∗ = k A − k ∞ = max { a − , . . . , a − d } . Since | A Ω | = | Ω | , by a simple geometric argument dated back to Gauss, itis easily seen that the number of lattice points in the enlarged domain, tA Ω,is equal to | Ω | t d + O ( t d − ) as t → ∞ . Note that although the leading termin this asymptotic formula is independent of the stretching factor A , theremainder term does depend on the shape of the domain and thus dependson the stretching factor A .The problem of estimating the growth rate of the remainder term Z d ∩ t Ω) − | Ω | t d as t → ∞ is now widely known as the lattice point problem , and has beenstudied extensively for over one hundred years. This is a highly nontrivial Mathematics Subject Classification.
Primary 11H06, 11P21, 52C07. Secondary42B10.
Key words and phrases.
Lattice points, optimal stretching, finite type domains.J.G. is partially supported by the NSFC Grant No. 11571131 and 11501535. problem. Even for the unit disk in the plane, in which case the problem isthe famous
Gauss circle problem , we are still far away from the conjecturedoptimal growth rate. For historical results on the lattice point problem, seefor example [10], [17], etc.Motivated by the “eigenvalue optimization among rectangles” problem inspectral geometry, in the paper [1] P. Antunes and P. Freitas studied thefollowing variation of the Gauss circle problem: find the “optimal stretchingfactor” that maximizes the remainder term N ∩ tA B ) − πt / , ( ∗ )where B = B (0 ,
1) is the unit disk in R , and A = diag( s, s − ). Notethat they only count positive lattice points, i.e. integer points in the firstquadrant. This is because they are considering corresponding Laplace eigen-values with Dirichlet boundary condition. Although there is no explicit wayto determine the “best” stretching factor s (not necessarily unique) for each t , they showed that as t → ∞ , the “best factor” s = s ( t ) →
1. In otherwords, among all ellipses of the same area, those that enclose the most lat-tice points in the first quadrant must be more and more “round”, as thearea goes to infinity.In the last couple of years, many people started to study the above typeof optimal stretching problem (for ellipses or ellipsoids) and the associatedshape optimization problem for eigenvalues. See for example [3], [4], [6], etc.Recently, in a pair of papers R. Laugesen and S. Liu [12] and S. Ariturkand R. Laugesen [2] extended P. Antunes and P. Freitas’ results to very gen-eral planar domains including p -ellipses for 0 < p < ∞ , p = 1. They showed,among others, that under mild assumptions on a given domain’s boundarythe most “balanced” domain will enclose the most positive (or least nonneg-ative) lattice points in the limit, where a domain in R d ( d ≥
2) ( † ) is saidto be balanced if the ( d − ⊂ R d whose boundary ∂ Ω is C d +2 and has nowhere vanishing Gaussian curvature.(In Proposition 2.1 below we will prove two-term bounds for lattice count-ing for general strictly convex domains, by using which one can recover N.Marshall’s main theorem in [14].)The class of domains studied in [14] is quite general, though it does not in-clude some interesting domains having boundary points of vanishing Gauss-ian curvature, e.g. the super sphere { x ∈ R d : | x | ω + | x | ω + · · · + | x d | ω ≤ } , ω ≥ , ∗ In this paper we use N = { , , . . . } , Z + = { , , , . . . } , Z d ∗ = Z d \{ } , and R + = [0 , ∞ ). † For the sake of later usage, here we give the definition for any dimension d ≥ N OPTIMAL STRETCHING PROBLEM 3 a typical domain of finite type (in the sense of J. Bruna, A. Nagel, andS. Wainger [5]) in R d . In fact, even the classical lattice point problem fordomains having boundary points of vanishing Gaussian curvature is notwell understood, especially in high dimensions. For partial results we referinterested readers to [10], [17], [8], [9], and the references given there.We remark that if the boundary is “too flat” the optimal stretching prob-lem could be very subtle. For example, the case of triangles has been ana-lyzed by N. Marshall and S. Steinerberger in [15], where they showed thatthe optimal shape needs not to be asymptotically balanced.So a natural question is to study the optimal stretching problem for do-mains of finite type. In this paper we will explore this problem for thefollowing model domain of finite type in R d ( d ≥ D = { x ∈ R d : x ω + · · · + x ω d d ≤ } with ω , . . . , ω d ∈ N . For future purpose we will denote ω = max( ω , . . . , ω d ) . For each 1 ≤ j ≤ d , we let D j be the intersection of D with the coordinatehyperplane defined by x j = 0. For each t ≥ A ( t ) = argmax A N d ∩ tA D ) , where the argmax ranges over all positive definite diagonal matrices A ofdeterminant 1. We recall that the notation argmax x f ( x ) means the valuesof x for which f ( x ) attains the function’s largest value. So in our setting, A ( t ) means the stretching matrix A that makes the set tA D contain mostpositive lattice points. Note that in general such A is not unique, i.e. therecould be a set of “optimal factors”. So in what follows, when we write A ( t ) =diag( a ( t ) , a ( t ) , . . . , a d ( t )), we really mean that A ( t ) is any stretching factorthat maximizes the lattice counting function N d ∩ tA D ).Our main theorem is Theorem 1.1.
Let D be the domain defined by (1.3) . Then any maximalstretching factor A ( t ) = diag( a ( t ) , a ( t ) , . . . , a d ( t )) in (1.4) satisfies (1.5) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a j ( t ) − |D j | d p |D ||D | · · · |D d | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = O ( t − γ ) , ≤ j ≤ d, where |D j | is the ( d − -dimensional measure of D j whose formula is givenin Appendix A, and γ = min { ( d − / (2 ω ) , ( d − / (2 d + 2) } . Similarly we consider the optimal stretching problem for nonnegative lat-tice points. For each t ≥ e A ( t ) = argmin e A Z d + ∩ t e A D ) , where the argmin ranges over all positive definite diagonal matrices e A ofdeterminant 1. With the same understanding as above, when we write JINGWEI GUO AND WEIWEI WANG e A ( t ) = diag(˜ a ( t ) , ˜ a ( t ) , . . . , ˜ a d ( t )), we really mean that e A ( t ) is any stretch-ing factor that minimizes the lattice counting function Z d + ∩ t e A D ). Thenwe have Theorem 1.2.
Let D be the domain defined by (1.3) . Then any minimalstretching factor e A ( t ) = diag(˜ a ( t ) , ˜ a ( t ) , . . . , ˜ a d ( t )) in (1.6) satisfies (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ˜ a j ( t ) − |D j | d p |D ||D | · · · |D d | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = O ( t − γ ) , ≤ j ≤ d, where γ = min { ( d − / (2 ω ) , ( d − / (2 d + 2) } . Proofs of these two theorems are essentially the same. We follow theframework of P. Antunes and P. Freitas [1], R. Laugesen and S. Liu [12],S. Ariturk and R. Laugesen [2], etc. We establish, in Section 2, bounds forlattice counting to control the stretching factor A . Then we prove, in Section3, asymptotic formulas for lattice counting to study the limiting behavior ofthe optimal stretching factor.One key ingredient in the proof is the following bound (given by [8, The-orem 2.1]) of the Fourier transform of the characteristic function of D . Let0 < c ≤ ξ ∈ S d − with | ξ d | ≥ c and λ > | b χ D ( λξ ) | . λ − d − Y l =1 min (cid:18) λ − ωl , | ξ l | − ωl − ωl − λ − (cid:19) , ( ‡ )where the implicit constant only depends on c and D .At last we remark that interesting results on the optimal stretching prob-lem in R with shifted lattices can be found in R. Laugesen and S. Liu [13].It would be interesting to investigate similar problems in high dimensions.2. Two-term bounds for lattice counting
In this section we prove high-dimensional analogues of [12, Proposition 6and 9] for strictly convex( § ) domains in R d . In the proof we use the parallelsection function of Ω (see A. Koldobsky [11, Chapter 2]) to transfer theproblem into a two-dimensional one (like N. Marshall did in [14]) and thencombine the methods used in [12] and [2] for planar curves that are concaveup or down.We notice that instead of proving two-term bounds N. Marshall [14]adopted a proof by contradiction using certain lower Riemann sums to provehis main theorem for convex domains with nonvanishing Gaussian curvature.By using Proposition 2.1 below it is not hard to recover his main theorem. ‡ For functions f and g with g taking nonnegative real values, f . g means | f | ≤ Cg for some constant C . The Landau notation f = O ( g ) is equivalent to f . g . § A domain Ω ⊂ R d is said to be strictly convex if the line segment connecting any twopoints x and y in Ω lies in the interior of Ω, except possibly for its endpoints. N OPTIMAL STRETCHING PROBLEM 5
One can also possibly improve the rate of convergence by using the methodin [7].We recall that for a positive definite diagonal matrix A with determinant1, the number a ∗ is defined as in (1.2). Proposition 2.1.
Let Ω ⊂ R d be strictly convex, compact, and symmetricwith respect to each coordinate hyperplane with C boundary. There existsa positive constant c depending only on Ω such that if t/a ∗ ≥ then (2.1) N d ∩ tA Ω) ≤ − d | Ω | t d − ca ∗ t d − and (2.2) Z d + ∩ tA Ω) ≥ − d | Ω | t d + ca ∗ t d − . Proof.
Without loss of generality we assume a ∗ = a − . We will prove (2.1)below while the proof of (2.2) is essentially the same. We have that N d ∩ tA Ω) = X k ∈ N X k ′ ∈ N d − χ Ω ( k a t , . . . , k d a d t )and, by the convexity of Ω, that X k ′ ∈ N d − χ Ω ( k a t , . . . , k d a d t ) ≤ Z R d − χ Ω ( k a t , x a t , . . . , x d a d t ) d x ′ , where x ′ = ( x , . . . , x d ). Then, by change of variables and det( A ) = 1, weget that(2.3) N d ∩ tA Ω) ≤ a − t d − X k ∈ N f ( k a t ) , where(2.4) f ( x ) = Z R d − χ Ω ( x , x ′ ) d x ′ is (up to a constant multiple) the parallel section function of Ω in the direc-tion of x -axis, which is continuous from R + to R + , strictly decreasing andsupported on [0 , X ] for some constant X >
0, and bounded above by anabsolute constant f (0).We claim that there exists a positive constant c depending only on Ω suchthat(2.5) Z ∞ f ( x a t ) d x − X k ∈ N f ( k a t ) ≥ c. Since the integral in (2.5) is equal to 2 − d | Ω | a t , the desired estimate (2.1)follows easily from (2.3) and (2.5).If we denote F ( x ) = f ( x /a t ) , to prove the claim it suffices to show that the area between the curves of F and a step function P j ∈ N F ( j ) χ [ j − ,j ) ( x ) has a lower bound c (depending JINGWEI GUO AND WEIWEI WANG only on Ω), by comparing it with the total area of certain “inbetween”triangles.Notice that there must exist an interval [ p − δ, p + 2 δ ] ⊂ (0 , X ) with δ >
0, on which f ′′ is always nonnegative or nonpositive. Indeed, since f is two times continuously differentiable in some neighborhood of zero (see[11, Lemma 2.4]), if there exists a point p in that neighborhood such that f ′′ ( p ) = 0 then the continuity of f ′′ ensures the existence of such an interval.We first assume that a t ≥ /δ and study F mainly on the interval[( p − δ ) a t, ( p + δ ) a t ].If F is always nonpositive hence concave down we study the “inbetween”right triangles of width 1 with vertices ( i, F ( i + 1)), ( i + 1 , F ( i + 1)), and( i, F ( i )), with ⌊ ( p − δ ) a t ⌋ + 1 ≤ i ≤ ⌊ ( p + δ ) a t ⌋ −
1. The total area of thesetriangles is equal to ⌊ ( p + δ ) a t ⌋− X i = ⌊ ( p − δ ) a t ⌋ +1
12 ( F ( i ) − F ( i + 1))= 12 ( F ( ⌊ ( p − δ ) a t ⌋ + 1) − F ( ⌊ ( p + δ ) a t ⌋ )) ≥ (cid:18) f (cid:18) p − δ (cid:19) − f (cid:18) p + δ (cid:19)(cid:19) , (2.6)where in the last inequality we have used a t ≥ /δ .If F is always nonnegative hence concave up we study the “inbetween”right triangles of width 1 with vertices ( i, F ( i + 1)), ( i + 1 , F ( i + 1)), and( i, F ( i + 1) − F ′ ( i + 1)), where ⌊ ( p − δ ) a t ⌋ ≤ i ≤ ⌊ ( p + δ ) a t ⌋ −
2. The totalarea of these triangles is equal to ⌊ ( p + δ ) a t ⌋− X i = ⌊ ( p − δ ) a t ⌋ | F ′ ( i + 1) | ≥ ⌊ ( p + δ ) a t ⌋− X i = ⌊ ( p − δ ) a t ⌋ | F ( i + 1) − F ( i + 2) | = 12 ( F ( ⌊ ( p − δ ) a t ⌋ + 1) − F ( ⌊ ( p + δ ) a t ⌋ )) , which has the same lower bound as in (2.6).If 1 ≤ a t ≤ /δ , we observe that the left hand side of (2.5) is continuousin a t and always positive (since f is strictly decreasing), therefore it musthave a uniform lower bound c > c sufficiently small (say, smaller than c and the absoluteconstants in (2.6)) , we get (2.5). This finishes the proof of (2.1). (cid:3) Asymptotics for lattice counting
In this section we prove several asymptotic formulas for lattice counting.See N. Marshall [14] for the case when the domain’s boundary has nonvan-ishing Gaussian curvature.
N OPTIMAL STRETCHING PROBLEM 7
We first prove an asymptotic formula for the number of lattice points inthe domain tA D . Its proof is very standard except that we need to take careof the effect of the transformation A . Proposition 3.1.
Let D be the domain defined by (1.3) . For any positivedefinite diagonal matrix A = diag( a , . . . , a d ) of determinant , if t/a ∗ ≥ then (3.1) Z d ∩ tA D ) = |D| t d + O (cid:18) a d − ω ∗ t ( d − − ω ) + a − d +1 ∗ t d − d +1 (cid:19) , where the implicit constant depends only on the domain D .Proof. Let 0 ≤ ρ ∈ C ∞ ( R d ) be a bump function such that supp ρ ⊂ B (0 , R R d ρ ( x ) d x = 1, ρ ε ( x ) = ε − d ρ ( ε − x ), 0 < ε <
1, and N A,ε ( t ) = X k ∈ Z d χ tA D ∗ ρ ε ( k ) , where χ tA D denotes the characteristic function of tA D . By the Poissonsummation formula, we have(3.2) N A,ε ( t ) = t d X k ∈ Z d b χ D ( tAk ) b ρ ( εk ) = |D| t d + R A,ε ( t ) , where R A,ε ( t ) = t d X k ∈ Z d ∗ b χ D ( tAk ) b ρ ( εk ) . By adding a partition of unity P dj =1 Ω j ≡ j supported inΓ j = { x ∈ R d : | x j | ≥ (2 d ) / | x |} and smooth away from the origin, we have R A,ε ( t ) = X ≤ i,j ≤ d S ij with S ij = t d X ( i ) Ω j ( Ak ) b χ D ( tAk ) b ρ ( εk ) , where the summation is over all lattice points k ’s with exactly 1 ≤ i ≤ d nonzero components such that Ak ∈ supp(Ω j ).We may assume j = d below since other cases are similar due to symmetry.If i = 1, by [8, Theorem 2.1] (see (1.7)) we have(3.3) | S d | . a − − P d − l =1 1 ωl d t d − − P d − l =1 1 ωl ≤ a P d − l =1 1 ωl ∗ t d − − P d − l =1 1 ωl , where in the last inequality we have used the definition of a ∗ .If 2 ≤ i ≤ d , by using [8, Theorem 2.1], | a l | ≥ a − ∗ , and | Ak | ≥ a − ∗ | k | ,and comparing the sums with integrals in polar coordinates, we get(3.4) | S id | . X S ∈P i ( N d ): d ∈ S a i +12 + α ( S ) ∗ t d − i +12 − α ( S ) (cid:16) ε − i − + α ( S ) (cid:17) , JINGWEI GUO AND WEIWEI WANG where N d = { , , . . . , d } , P i ( N d ) is the collection of all subsets of N d having i elements, and α ( S ) := X ≤ l ≤ d, l / ∈ S ω l . Note that the contribution coming from the term 1 on the right side of (3.4)is not larger than the right side of (3.3) if a ∗ < t .By (3.3), (3.4), and similar bounds for other j ’s we obtain that | R A,ε ( t ) | . X S ∈P ( N d ) a α ( S ) ∗ t d − − α ( S ) + X S ∈ P i ( N d ) , ≤ i ≤ d a i +12 + α ( S ) ∗ t d − i +12 − α ( S ) ε − i − + α ( S ) (3.5)whenever a ∗ < t .By M¨uller [16, Lemma 3] we have(3.6) N A,ε ( t − a ∗ ε ) ≤ Z d ∩ tA D ) ≤ N A,ε ( t + a ∗ ε ) . Let ε = ( t/a ∗ ) − d − d +1 . Note that t/ ≤ t ± a ∗ ε ≤ t/ t/a ∗ ≥ C for a sufficiently large constant C . Combining (3.2), (3.5), and (3.6) yields (cid:12)(cid:12)(cid:12) Z d ∩ tA D ) − |D| t d (cid:12)(cid:12)(cid:12) . a ∗ t d − ε + | R A,ε ( t ± a ∗ ε ) | . a d − ω ∗ t ( d − − ω ) + a − d +1 ∗ t d − d +1 , where the last inequality can be verified by a direct computation. Hence weget (3.1) when t/a ∗ ≥ C .If 1 ≤ t/a ∗ ≤ C applying the previous argument to Z d ∩ ( Ct ) A D ) yieldsthat Z d ∩ ( Ct ) A D ) . t d . Hence Z d ∩ tA D ) . t d , which trivially leads to the desired asymptotic formula (3.1). (cid:3) Recall that D j ⊂ R d is the intersection of D (defined by (1.3)) with thecoordinate hyperplane defined by x j = 0. As a consequence of Proposi-tion 3.1 we get the number of lattice points in the dilated and stretchedintersections. N OPTIMAL STRETCHING PROBLEM 9
Proposition 3.2.
Let A = diag( a , . . . , a d ) be a positive definite diagonalmatrix of determinant , if t/a ∗ ≥ then Z d ∩ tA ( ∪ dj =1 D j )) − d X j =1 a − j |D j | t d − = O (cid:18) a d − ω ∗ t ( d − − ω ) + a − d +1 ∗ t d − d +1 (cid:19) , (3.7) where the implicit constant depends only on the domain D .Proof. For 1 ≤ j ≤ d we let A j = diag( a , . . . , a j − , a j +1 , . . . , a d ) be a( d − × ( d −
1) matrix and use the notation D j to represent the set { ( x , . . . , x j − , x j +1 , . . . , x d ) ∈ R d − : X ≤ l ≤ d,l = j x ω l l ≤ } as well. The precise meaning of D j will be clear from the context and thisabuse of notation will not cause any problem. Note that Z d − ∩ tA j D j ) = (cid:18) Z d − ∩ (cid:18) ta − d − j (cid:19) (cid:18) a d − j A j (cid:19) D j (cid:19) . Applying Proposition 3.1 to the right hand side of the equation above yields (cid:12)(cid:12)(cid:12) Z d − ∩ tA j D j ) − a − j |D j | t d − (cid:12)(cid:12)(cid:12) . a − j (cid:18) a d − ω ∗ t ( d − − ω ) + a − d ∗ t d − d (cid:19) . (3.8)For 1 ≤ j = k ≤ d let A j,k be the ( d − × ( d −
2) diagonal matrix byremoving the j -th and k -th columns and rows from A , and D j,k = D j ∩ D k ⊂ R d . As above we abuse the notation D j,k for a subset of R d − . Since Z d − ∩ tA j,k D j,k ) = (cid:16) Z d − ∩ (cid:16) t ( a j a k ) − d − (cid:17) (cid:16) ( a j a k ) d − A j (cid:17) D j,k (cid:17) , applying Proposition 3.1 again gives (cid:12)(cid:12)(cid:12) Z d − ∩ tA j,k D j,k ) − ( a j a k ) − |D j,k | t d − (cid:12)(cid:12)(cid:12) . ( a j a k ) − (cid:18) a d − ω ∗ t ( d − − ω ) + a − d − ∗ t d − d − (cid:19) , (3.9)where |D j,k | denotes the ( d − D j,k .Since d X j =1 Z d − ∩ tA j D j ) − X ≤ j Let A = diag( a , . . . , a d ) be a positive definite diagonal ma-trix of determinant , if t/a ∗ ≥ then N d ∩ tA D ) = 2 − d |D| t d − − d d X j =1 a − j |D j | t d − + O (cid:18) a d − ω ∗ t ( d − − ω ) + a − d +1 ∗ t d − d +1 (cid:19) (3.10) and Z d + ∩ tA D ) = 2 − d |D| t d + 2 − d d X j =1 a − j |D j | t d − + O (cid:18) a d − ω ∗ t ( d − − ω ) + a − d +1 ∗ t d − d +1 (cid:19) , (3.11) where the implicit constants depend only on the domain D .Proof. By symmetry of the domain D we have that N d ∩ tA D ) = 2 − d (cid:16) Z d ∩ tA D ) − Z d ∩ tA ( ∪ dj =1 D j )) (cid:17) . Then (3.10) follows from (3.1) and (3.7).For nonnegative lattice points, we know Z d + ∩ tA D ) = 2 − d (cid:16) Z d ∩ tA D ) + Z d ∩ tA ( ∪ dj =1 D j )) + R D ( A, t ) (cid:17) , where(3.12) R D ( A, t ) . X ≤ j In this section we prove Theorem 1.1. The nonnegative case, i.e. Theorem1.2, can be handled essentially in the same way.Let B = diag |D | d p |D ||D | · · · |D d | , . . . , |D d | d p |D ||D | · · · |D d | ! be a diagonal d × d matrix. Using (3.10) with A = B , we get(4.1) N d ∩ tB D ) ≥ − d | D | t d − Ct d − for sufficiently large t , where C = 2 − d d ( Q dj =1 |D j | ) /d . N OPTIMAL STRETCHING PROBLEM 11 For every sufficiently large t let A ( t ) be a fixed optimal stretching matrix(see (1.4)). Thus(4.2) N d ∩ tB D ) ≤ N d ∩ tA ( t ) D ) . Denote a ∗ ( t ) = k A ( t ) − k ∞ . Then t/a ∗ ( t ) ≥ , otherwise tA ( t ) D does not contain any positive lattice point. ThereforeProposition 2.1 gives(4.3) N d ∩ tA ( t ) D ) ≤ − d |D| t d − ca ∗ ( t ) t d − for some constant c > D . Combining (4.1), (4.2), and(4.3) yields a ∗ ( t ) ≤ C/c, which means that for sufficiently large t any optimal stretching factor A ( t )has its a ∗ bounded uniformly from above.Then (3.10) gives N d ∩ tA ( t ) D ) = 2 − d |D| t d − − d d X j =1 a j ( t ) − |D j | t d − + O ( t ( d − − ω ) + t d − d +1 )and N d ∩ tB D ) = 2 − d |D| t d − − d d d p |D ||D | · · · |D d | t d − + O ( t ( d − − ω ) + t d − d +1 ) . Combining these two asymptotics with (4.2) yields d X j =1 a j ( t ) − |D j | d p |D ||D | · · · |D d | ≤ d + O (cid:16) t − d − ω + t − d − d +1 (cid:17) . Then (1.5) follows easily from Lemma B.1. This finishes the proof. Appendix A. Computation of the volume of D ∩ R d + In this part we show that the volume of D ∩ R d + is(A.1) V ( ω , . . . , ω d ) := d X l =1 ω ω · · · ω d ω l ! − Γ( ω )Γ( ω ) · · · Γ( ω d )Γ( P dl =1 1 ω l ) . As a consequence, the |D j | ’s in Theorem 1.1 are explicitly given by |D j | = 2 d − V ( ω , . . . , ω j − , ω j +1 , . . . , ω d ) . To prove (A.1), we denote V D = |D ∩ R d + | and we change coordinates from x l to y l = x k l l , where k l = ω l / 2. Then we have V D = Z { x ∈ R d : x ω + ··· + x ωdd ≤ }∩ R d + d x · · · d x d = Z { y ∈ R d : y + ··· + y d ≤ }∩ R d + d Y l =1 k l y kl − l d y · · · d y d . The last integral can be computed in polar coordinates ( r, θ , . . . , θ d − ),where r = | y | = q y + y + · · · + y d such that y m = r cos θ m Q m − l =1 sin θ l for 1 ≤ m ≤ d − y d = r Q d − l =1 sin θ l . We then get V D = Z Z π · · · Z π k ( r cos θ ) k − k ( r sin θ cos θ ) k − · · · k m ( r cos θ m m − Y l =1 sin θ l ) km − · · · k d ( r d − Y l =1 sin θ l ) kd − r d − sin d − θ sin d − θ · · · sin θ d − d r d θ · · · d θ d − . A direct computation gives V D =( d X l =1 ω ω · · · ω d ω l ) − B ( 12 k , d X l =2 k l ) B ( 12 k , d X l =3 k l ) · · · B ( 12 k d − , k d )=( d X l =1 ω ω · · · ω d ω l ) − Γ( ω )Γ( ω ) · · · Γ( ω d )Γ( P dl =1 1 ω l ) , where Γ( x ) is the gamma function and B ( x, y ) = Γ( x )Γ( y ) / Γ( x + y ) is thebeta function. Appendix B. An elementary lemma The following lemma is a generalization of [12, Lemma 13]. Lemma B.1. If < ε < and s , . . . , s d are positive numbers such that Q dj =1 s j = 1 and P dj =1 s j ≤ d + ε , then (B.1) s j = 1 + O ( ε / ) , ≤ j ≤ d. Proof. The assumptions imply that ( d + 1) − d ≤ s j ≤ d + 1 for 1 ≤ j ≤ d ,namely, all s j ’s are uniformly bounded from above and below. Denote s j =exp( u j ). Then P dj =1 u j = 0. Hence, by Taylor’s formula, we have d + ε ≥ d X j =1 s j = d + 12 d X j =1 e c j u j , N OPTIMAL STRETCHING PROBLEM 13 where each c j is between 0 and u j (which means each exp( c j ) is between1 and s j , thus uniformly bounded from below). 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