aa r X i v : . [ m a t h . N T ] J a n An Improved Construction of Progression-Free Sets
Michael Elkin ∗ November 1, 2018
Abstract
The problem of constructing dense subsets S of { , , . . . , n } that contain no arithmetictriple was introduced by Erd˝os and Tur´an in 1936. They have presented a constructionwith | S | = Ω( n log ) elements. Their construction was improved by Salem and Spencer, andfurther improved by Behrend in 1946. The lower bound of Behrend is | S | = Ω n √ √ log n · log / n ! . Since then the problem became one of the most central, most fundamental, and most in-tensively studied problems in additive number theory. Nevertheless, no improvement of thelower bound of Behrend was reported since 1946.In this paper we present a construction that improves the result of Behrend by a factorof Θ( √ log n ), and shows that | S | = Ω (cid:18) n √ √ log n · log / n (cid:19) . In particular, our result implies that the construction of Behrend is not optimal.Our construction is elementary and self-contained.
A subset S ⊆ { , , . . . , n } is called progression-free if it contains no three distinct elements i, j, ℓ ∈ S such that i is the arithmetic average of j and ℓ , i.e., i = j + ℓ . For a positive integer n ,let ν ( n ) denote the largest size of a progression-free subset S of { , , . . . , n } .Providing asymptotic estimates on ν ( n ) is a central and fundamental problem in additivenumber theory. This problem was introduced by Erd˝os and Turan [6] in 1936, and they haveshown that ν ( n ) = Ω( n log ). This estimate was improved by Salem and Spencer [12], and furtherimproved by Behrend [2] in 1946. Behrend have shown that ν ( n ) = Ω (cid:18) n √ √ log n · log / n (cid:19) , ∗ Department of Computer Science, Ben-Gurion University of the Negev, Beer-Sheva, Israel, [email protected]
This research has been supported by the Israeli Academy of Science, grant 483/06. ν ( n ) = O ( n log log n ) was proved in a seminal paper by Roth[11]. This bound was improved by Bourgain [3, 4], and the current state-of-the-art upper boundis ν ( n ) = O ( n · (log log n ) log / n ) [4]. The problem is also closely related to Szemer´edi theorem [13],and to the problem of finding arbitrarily long arithmetic progressions of prime numbers (see, e.g.,Green and Tao [8]), and to other central problems in the additive number theory. (See, e.g., theenlightening survey on Szemer´edi theorem in Scholarpedia.)In this paper we improve the lower bound of Behrend by a factor of Θ( √ log n ), and show that | S | = Ω (cid:18) n √ √ log n · log / n (cid:19) . Though the improvement is not large, our result demonstrates that the construction of Behrend isnot optimal. Also, despite very intensive research in this area, no improvement of Behrend lowerbound was achieved for more than sixty years.Our proof is elementary, and self-contained. The proof of Behrend is based on the observationthat a sphere in any dimension is convex, and thus cannot contain an arithmetic progression. Wereplace the sphere by a thin annulus, and demonstrate that this annulus contains a large convexlyindependent subset U of integer points. There is an inherent tradeoff between the width of theannulus and the size of U . In our construction we choose the largest width for which we are ableto show that U contains at least a constant fraction of all integer points of the annulus.The construction of Behrend was generalized by Rankin [10] to provide large subsets of { , , . . . , n } that contain no arithmetic progression of length k , for any fixed k . We believethat our technique will be useful for improving the lower bound of Rankin as well. Finally, like theconstruction of Behrend, our construction relies on the Pigeonhole Principle. Consequently, theresult of Moser [9] remains the best known lower bound achieved without relying on the Pigeon-hole Principle. However, we hope that our argument can be made independent of the PigeonholePrinciple. (See also Section 6.) For a pair a, b of real numbers, a ≤ b , we denote by [ a, b ] (respectively, ( a, b )) the closed (resp.,open) segment containing all numbers x , a ≤ x ≤ b (resp., a < x < b ). We also use the notation( a, b ] (respectively, [ a, b )) for denoting the segment containing all numbers x , a < x ≤ b (resp., a ≤ x < b ). For integer numbers n and m , n ≤ m , we denote by [ { n, m } ] the set of integernumbers { n, n + 1 , . . . , m } . If n = 1 then we use the notation [ { m } ] as a shortcut for [ { , m } ]. Fora real number x , we denote by ⌊ x ⌋ (respectively, ⌈ x ⌉ ) the largest (resp., smallest) integer numberthat is no greater (resp., no smaller) than x .A triple i, j, ℓ of distinct integer numbers is called an arithmetic triple if one of these numbersis the average of two other numbers, i.e., i = j + ℓ . A set S of integer numbers is called progression-free if it contains no arithmetic triple. For a positive integer number n , let ν ( n ) denote the largestsize of a progression-free subset S of [ { n } ].For a pair of integer functions f ( · ), g ( · ), we say that f ( n ) = O ( g ( n )) if there exists a positive(universal) constant c and a positive integer N such that for every n ≥ N , | f ( n ) | ≤ c · | g ( n ) | . In2his case we also say that g ( n ) = Ω( f ( n )). If both f ( n ) = O ( g ( n )) and g ( n ) = O ( f ( n )) hold, wesay that f ( n ) = Θ( g ( n )). If lim n →∞ f ( n ) g ( n ) = 0 we say that f ( n ) = o ( g ( n )). These definitions extendto positive real functions as well.Unless specified explicitly, log (respectively, ln) stands for the logarithm on base 2 (resp., e ).For a positive integer k and a vector v = ( v , v , . . . , v k ), let || v || = qP ki =1 v i denote the norm of the vector v . The expression || v || = P ki =1 v i will be referred to as the squared norm of thevector v .For three vectors v, u, w ∈ IR k , we say that v is a convex combination of u and w if there existsa real number p , 0 ≤ p ≤
1, such that v = p · u + (1 − p ) · w . A convex combination is called trivial if either p = 0 or p = 1. Otherwise, it is called non-trivial . For a set U ⊆ IR k of vectors,we say that U is a convexly independent set if it contains no three vectors v, u, w ∈ U such that v is a convex combination of u and w . For a set X ⊆ IR k of vectors, the exterior set of X , denoted Ext ( X ), is the subset of X that contains all vectors v ∈ X such that v cannot be expressed as anon-trivial convex combination of vectors from X .For a positive integer ℓ , let β ℓ denote the volume of an ℓ -dimensional ball of unit radius. It iswell-known (see, e.g, [7], p.3) that β ℓ = π ℓ/ Γ( ℓ + 1) , (1)where Γ( · ) is the (Euler) Gamma-function. We use the Gamma-function either with a positiveinteger parameter n or with a parameter n + for a positive integer n . In these cases the Gamma-function is given by Γ( n + 1) = n ! andΓ (cid:18) n + 12 (cid:19) = (2 n )! √ π n n ! . (2)(See [7], p.178.) Observe also thatΓ (cid:18) n + 12 (cid:19) = (cid:18) n − (cid:19) (cid:18) n − (cid:19) · . . . · · √ π ≥ ( n − √ π . (3)By definition, it is easy to verify that for an integer ℓ , ℓ ≥ β ℓ = Θ( β ℓ − √ ℓ ). The state-of-the-art lower bound for ν ( n ) due to Behrend [2] states that for every positive integer n , ν ( n ) = Ω (cid:18) n √ √ log n · log / n (cid:19) . (4)In this paper we improve this bound by a factor of Θ( √ log n ), and show that for every positiveinteger n , ν ( n ) = Ω (cid:18) n √ √ log n · log / n (cid:19) . (5)3ote that it is sufficient to prove this bound only for all sufficiently large values of n . The resultfor small values of n follows by using a sufficiently small universal constant c in the definition ofΩ-notation.We start with a short overview of the original construction of Behrend [2]. Fix a sufficientlylarge positive integer n . The construction involves a positive integer parameter k that will bedetermined later. Set y = n /k /
2. In what follows we assume that y is an integer. The case that y is not an integer is analyzed later in the sequel.Consider independent identically distributed random variables Y , Y , . . . , Y k , with each Y i dis-tributed uniformly over the set [ { , y − } ], for all i ∈ [ { k } ]. Set Z i = Y i , for all i ∈ [ { k } ], and Z = P ki =1 Z i . It follows that for all i ∈ [ { k } ],IE( Z i ) = y − X j =0 y · j = y y ) . Let µ Z = IE( Z ) denote the expectation of the random variable Z . It follows that µ Z = k y + Θ( k · y ) . (6)Also, for all i ∈ [ { k } ], Var ( Z i ) = IE( Z i ) − IE( Z i ) = IE( Y i ) − y + Θ( y ). Hence Var ( Z i ) = y y ) − y y ) = 445 · y + O ( y ) . Hence
Var ( Z ) = k · y ·
445 + O ( ky ) = k · y · · (1 + O ( 1 y )) , and the standard deviation of Z , σ Z , satisfies σ Z = √ k · y · · √ · (1 + O ( 1 y )) . (7)By Chebyshev inequality, for any a > | Z − µ Z | > a · σ Z ) ≤ a . Hence, for a fixed value of a , a >
0, at least (1 − a )-fraction of all vectors v from the set [ { , y − } ] k have squared norm that satisfies µ Z − a · σ Z ≤ || v || ≤ µ Z + a · σ Z . Note that each vector v ∈ [ { , y − } ] k has an integer squared norm. By Pigeonhole Principle, thereexists a value T such that µ z − a · σ Z ≤ T ≤ µ Z + a · σ Z that satisfies that at least (1 − a ) · a · σ Z · y k vectors from [ { , y − } ] k have squared norm T . Let S denote the set of these vectors. By (7), |S| ≥ (1 − a ) · a √ k · y · √ · (1 − O ( 1 y )) · y k = y k − √ k · c , c = c ( a ). Set a = 2. Now c = c (2) is a universal constant, andconsequently, |S| = Ω (cid:16) n k − k √ k (cid:17) . To maximize the right-hand-side, we set k = ⌈√ · log n ⌉ . Itfollows that |S| = Ω (cid:18) n √ √ log n · log / n (cid:19) . Observe that all vectors in S have the same norm √ T , and thus, for every three vectors v, u, w ∈ S , v = u + w . To obtain a progression-free set S ⊆ [ { n } ] we consider coordinates of vectors from S as digits of (2 y )-ary representation. Specifically, for every vector v = ( v , v , . . . , v k ) ∈ S , letˆ v = P k − i =0 v i +1 · (2 y ) i . The set S is now given by S = { ˆ v | v ∈ S} . Let f ( · ) : S → S denote thismapping.Note that for every v ∈ S , 0 < ˆ v ≤ (2 y ) k − n − . Observe also that since
S ⊆ [ { , y − } ] k , the mapping f is one-to-one, i.e., if v = u , v, u ∈ S ,then ˆ v = ˆ u . Consequently, | S | = |S| = Ω (cid:18) n √ √ log n · log / n (cid:19) . Finally, we argue that S is a progression-free set. Suppose for contradiction that for three distinctnumbers ˆ v, ˆ u, ˆ w ∈ S , ˆ v = ˆ u + ˆ w . Let u, v, w be the corresponding vectors in S , v = ( v , v , . . . , v k ), u = ( u , u , . . . , u k ), w = ( w , w , . . . , w k ). Thenˆ v = k − X i =0 u i +1 + w i +1 · (2 y ) i = k − X i =0 v i +1 · (2 y ) i . However, since all the coordinates v , v , . . . , v k , u , u , . . . , u k , w , w , . . . , w k are in [ { , y − } ], itfollows that v i = u i + w i , for every index i ∈ [ { k } ]. Consequently, v = u + w , a contradiction to theassumption that || v || = || u || = || w || . Hence S is a progression-free set of size Ω( n √ √ log n · log / n ).Consider now the case that y = n /k is not an integer number. In this case the same constructionis built with ⌊ y ⌋ instead of y . Set n ′ = (2 ⌊ y ⌋ ) k . By previous argument, we obtain a progression-freeset S that satisfies | S | = Ω (cid:18) n ′ √ √ log n ′ · log / n ′ (cid:19) = Ω (cid:18) n ′ √ √ log n · log / n (cid:19) . Observe that nn ′ ≤ (cid:16) yy − (cid:17) k = 1 + Θ( ky ) = 1 + Θ (cid:16) √ log n (1 / √ ·√ log n (cid:17) .Hence | S | = Ω (cid:16) n √ √ log n · log / n (cid:17) , and we are done. In this section we present our construction of progression-free sets S ⊆ [ { n } ] with at leastΩ (cid:16) n √ √ log n · log / n (cid:17) elements. Fix k = ⌈√ n ⌉ , and y = n /k /
2. Observe that2 k/ √ √ · √ log n √ ≤ y ≤ · √ log n √ = 2 k/ . (8)5or convenience we assume that y is an integer. If this is not the case, the same analysis applieswith minor adjustments. (Specifically, we set y = ⌊ n /k / ⌋ . By the same argument as we used inSection 3, the resulting lower bound will be at most by a constant factor smaller than in the casewhen n /k / k -dimensional ball centered at the origin that has radius R ′ given by R ′ = µ Z = k y + Θ( ky ) . (9)(See (6).) By Chebyshev inequality, the annulus ˆ S of all vectors with squared norm in [ R ′ − · σ Z , R ′ + 2 · σ Z ] contains at least · y k integer points of the discrete cube C = [ { , y − } ] k .Fix a parameter g = ǫ · k , for a universal constant ǫ > S into ⌈ σ Z g ⌉ = ℓ annuli ˆ S , ˆ S , . . . , ˆ S ℓ , with the annulus ˆ S i containing all vectors withsquared norms in the range [ R ′ − σ Z + ( i − · g, R ′ − σ Z + i · g ), for i ∈ [ { ℓ − } ], and[ R ′ − σ Z + ( ℓ − σ Z , R ′ + 2 σ Z ] for i = ℓ .Observe that for distinct indices i, j ∈ [ { ℓ } ], the sets of integer points in ˆ S i and ˆ S j are disjoint.Thus, by the Pigeonhole Principle, there exists an index i ∈ [ { ℓ } ] such that the annulus ˆ S i containsat least 34 ℓ · y k = Ω( g · y k − √ k ) = Ω( ǫ √ k · y k − ) (10)integer points of C ∩ ˆ S . In other words, there exists a radius R , R ∈ [ R ′ − σ Z , R ′ + 2 σ Z ], suchthat the annulus S that contains all vectors with squared norm in the range [ R − g, R ] containsat least Ω( √ k · y k − ) integer points of C ∩ ˆ S .By (6), (7), and (9), R ≤ R ′ + 2 σ Z ≤ k · y + O ( k · y ) + O ( √ k · y ) ≤ k · y (cid:18) O (cid:18) √ k (cid:19)(cid:19) . (11)Let ˜ S be the set of integer points of C ∩ S . We will show that that ˜ S contains a convexlyindependent subset ˇ S with at least | ˇ S| ≥ | ˜ S| integer points. Consequently, | ˇ S| ≥ | ˜ S| √ k · y k − ) = Ω (cid:18) log / n · n √ √ log n (cid:19) . (12)Consider the set ˇ S = f ( ˇ S ) constructed from ˇ S by the mapping f described in Section 3. Since S is a convexly independent set, by the same argument as in Section 3, | ˇ S | = | ˇ S| , and moreover, ˇ S is a progression-free set. Hence | ˇ S | = Ω (cid:16) log / n · n √ √ log n (cid:17) , and our result follows.The following lemma is useful for showing an upper bound on the number of integer points in S that do not belong to the exterior set of ˜ S , Ext ( ˜ S ). This lemma is due to Coppersmith [5].Let B = B ( R,
0) denote the k -dimensional ball of radius R centered at the origin, and B = B ( R,
0) denote the set of integer points contained in this ball. Denote T = R . Lemma 4.1 [5] Let b ∈ B \ Ext ( B ) be an integer point that satisfies T − g ≤ || b || ≤ T . Thenthere exists a non-zero integer vector δ that satisfies ≤ h b, δ i ≤ g and < || δ || ≤ g . roof: Since b ∈ B \ Ext ( B ), there exist two integer points a and c in B and a constant p , 0 < p < b = p · a + (1 − p ) · c . Since a, c ∈ B , || a || , || c || ≤ T . Observe that either h a, b i or h c, b i isgreater or equal than || b || . (Otherwise, || b || = h pa + (1 − p ) c, b i = p · h a, b i + (1 − p ) · h c, b i < || b || ,contradiction.)Suppose without loss of generality that h a, b i ≥ || b || . Then h a − b, b i ≥
0. Set δ = a − b . Since a, b ∈ B are integer points, it follows that δ is an integer point as well. Moreover, since 0 < p < δ = 0. Moreover, T ≥ || a || = || b + δ || = || b || + 2 h b, δ i + || δ || . Recall that || b || ≥ T − g . Hence 2 h b, δ i + || δ || ≤ g . As h b, δ i = h a − b, b i ≥
0, it follows that h b, δ i , || δ || ≤ g , as required.Observe that δ ∈ IR k is an integer vector, and || δ || ≤ g = ǫ · k . Consequently, the vector δ may contain at most ǫ · k non-zero entries. This property will be helpful for our argument.Denote the number of integer vectors δ that have squared norm at most g by ˆ D ( g ). The nextlemma provides an upper bound on ˆ D ( g ). Lemma 4.2
For any ǫ > there exists η = η ( ǫ ) > such that lim ǫ → η ( ǫ ) = 0 , and ˆ D ( g ) = O (2 η · k ) . Proof:
Fix an integer value h , 1 ≤ h ≤ g . First, we count the number N ( h ) of k -tuples( q , q , . . . , q k ) of non-negative integer numbers that sum up to h .Consider permutations of ( k − h ) elements of two types, with h elements of the first typeand k − σ and σ ′ are said to be equivalent if they can be obtained one from another by permuting balls among themself, and permutingboundaries among themself.Let Π be the induced equivalence relation. Observe that there is a one-to-one mapping between k -tuples ( q , q , . . . , q k ) of non-negative integer numbers that sum up to h and the equivalenceclasses of the relation Π. Hence N ( h ) is equal to the number of equivalence classes of Π, i.e., N ( h ) = ( k − h )!( k − · h ! = (cid:18) k − hh (cid:19) . In a k -tuple ( δ , δ , . . . , δ k ) of integer numbers such that P ki =1 δ i = h , there can be at most h non-zero entries. Hence, for a fixed k -tuple of integers ( q , q , . . . , q k ) such that P ki =1 q i = h , theremay be at most 2 h k -tuples ( δ , δ , . . . , δ k ) of integers such that δ i = q i for every index i ∈ [ { k } ].Thus, the overall number D ( h ) of integer k -tuples ( δ , δ , . . . , δ k ) such that P ki =1 δ i = h satisfies D ( h ) ≤ h · N ( h ) = 2 h (cid:18) k − hh (cid:19) . Note that (cid:0) k − hh (cid:1) ≤ (cid:0) k − gg (cid:1) , for every integer h , 1 ≤ h ≤ g . Hence the number ˆ D ( g ) of integer k -tuples ( δ , δ , . . . , δ k ) with 1 ≤ P ki =1 δ i ≤ g satisfiesˆ D ( g ) = g X h =1 D ( h ) ≤ g X h =1 h · N ( h ) ≤ N ( g ) · g +1 ≤ g +1 · (cid:18) k + gg (cid:19) ≤ g +1 (cid:18) e ( k + g ) g (cid:19) g = 2 · (2 e ) g (cid:18) ǫ (cid:19) ǫ · k = 2 · (log 2 e +log(1+ ǫ )) ǫ · k . η = η ( ǫ ) = ǫ (log 2 e + log(1 + ǫ )). Then ˆ D ( g ) ≤ · η ( ǫ ) · k . Finally,lim ǫ → η ( ǫ ) = lim ǫ → log(1 + ǫ ) ǫ = 1ln 2 · lim y →∞ ln(1 + y ) y = 0 , completing the proof.Consider again the annulus S = { α ∈ IR k | T − g ≤ || α || ≤ T } , and the set ˜ S of integer pointsof S . For an integer vector δ that satisfies 0 < || δ || ≤ g , let ˆ Z ( δ ) denote the set of integer points b ∈ ˜ S that satisfy 0 ≤ h b, δ i ≤ g . Let ˆ W ( δ ) = ˆ Z ( δ ) ∩ C denote the intersection of ˆ Z ( δ ) with thediscrete cube C = [ { , y − } ] k , and let W ( δ ) = | ˆ W ( δ ) | . Also, let ˆ W = S { ˆ W ( δ ) | < || δ || ≤ g } ,and W = | ˆ W | .Let ˆ N denote the set of integer points of C ∩ S that do not belong to Ext ( B ), and N = | ˆ N | .By Lemma 4.1, ˆ N ⊆ ˆ W , and consequently, N ≤ W ≤ X { W ( δ ) | < || δ || ≤ g } . (13)Fix a vector δ , 0 < || δ || ≤ g . In the sequel we provide an upper bound for W ( δ ).Observe that since ˆ W ( δ ) is a set of integer points, it follows that for every b ∈ ˆ W ( δ ), h b, δ i ∈ [ { , g } ].For an integer number h ∈ [ { , g } ], let ˆ W ( δ, h ) denote the subset of ˆ W ( δ ) of integer points b that satisfy h b, δ i = h . Let W ( δ, h ) = | ˆ W ( δ, h ) | . Observe that for distinct values h = h ′ , h, h ′ ∈ [ { , g } ], the sets ˆ W ( δ, h ) and ˆ W ( δ, h ′ ) are disjoint. Consequently, W ( δ ) = g X h =0 W ( δ, h ) . (14)Next, we provide an upper bound for W ( δ, h ).Consider the hyperplane H = { α ∈ IR k | h α, δ i = h } . Observe that ˆ W ( δ, h ) = H ∩ S ∩ C is theintersection of the hyperplane H with the annulus S and with the discrete cube C.Let S denote the k -dimensional sphere with squared radius T centered at the origin, i.e., S = { α ∈ IR k | || α || = T } . Consider the intersection S ′ of S with the hyperplane H . Lemma 4.3 S ′ ⊆ H is a ( k − -dimensional sphere with squared radius ( T − h || δ || ) centered at h || δ || · δ . Proof:
For a vector α ∈ S ∩ H , || α − h || δ || · δ || = k X i =1 ( α i − h || δ || · δ i ) = || α || + h || δ || − h || δ || h α, δ i = || α || − h || δ || . (For the last equality, note that since α ∈ H , we have h α, δ i = h .)Recall that for a vector α ∈ S , T − g ≤ || α || ≤ T . Hence the intersection of the hyperplane H with the annulus S is the ( k − S ′ ⊆ H , centered at h || δ || · δ , containingvectors α such that T − g − h || δ || ≤ || α − h || δ || · δ || ≤ T − h || δ || . T ′ = T − h || δ || . Then S ′ is given by S ′ = { α ∈ H | T ′ − g ≤ || α − h || δ || · δ || ≤ T ′ } . Note that since h ≥ T ′ ≤ T for all h and δ .Recall that our goal at this stage is to provide an upper bound for the number W ( δ, h ) ofinteger points in ˆ W ( δ, h ) = H ∩ S ∩ C = S ′ ∩ C . Let C = [0 , y − k be the (continuous) cube. (Thediscrete cube C = [ { , y − } ] k is the set of integer points of C .) Let ˜ W = S ′ ∩ C . Since ˆ W ( δ, h ) isthe set of integer points in ˜ W , we are interested in providing an upper bound for the number ofinteger points in ˜ W . Our strategy is to show an upper bound for the ( k − Vol ( ˜ W ) of ˜ W , and to use standard estimates for the discrepancy between Vol ( ˜ W ) and the numberof integer points in ˜ W .Let H ′ = { α ∈ IR k | h α, δ i = 0 } be the parallel hyperplane to H that passes through the origin.Next, we construct an orthonormal basis Υ = { γ , γ , . . . , γ k − } for H ′ . This basis will be usefulfor estimating Vol ( ˜ W ).Recall that δ satisfies 0 < || δ || ≤ g = ǫ · k , and it is an integer vector. Consequently, δ = ( δ , δ , . . . , δ k ) contains at most g = ǫ · k non-zero entries. Let I ⊆ [ { k } ] be the subset ofindices such that δ i = 0. Let m = | I | . It follows that m ≤ g = ǫ · k .For every vector α = ( a , a , . . . , a k ) ∈ H ′ , it holds that X i ∈ I a i δ i = 0 . (15)Let γ (1) , γ (2) , . . . , γ ( m − be an arbitrary orthonormal basis for the solution space of the equation(15). These vectors are in IR m . For each index j ∈ [ { m − } ], we view the vector γ ( j ) as γ ( j ) = ( γ ( j ) i | i ∈ I ).We form orthonormal vectors ˆ γ (1) , ˆ γ (2) , . . . , ˆ γ ( m − ∈ IR k in the following way. For each index j ∈ [ { m − } ], and each index i ∈ I , the i th entry ˆ γ ( j ) i of ˆ γ ( j ) is set as γ ( j ) i , and for each index i ∈ [ { k } ] \ I , the entry ˆ γ ( j ) i is set as zero. Also, for each index i ∈ [ { k } ] \ I , we insert the vector ξ i = (0 , , . . . , , , , . . . , ξ i ∈ IR k , with 1 at the i th entry and zeros in all other entries into thebasis Υ. Observe that ξ i ∈ H ′ . The resulting basis Υ is { ˆ γ (1) , ˆ γ (2) , . . . , ˆ γ ( m − } ∪ { ξ i | i ∈ [ { k } ] \ I } .It is easy to verify that Υ is an orthonormal basis for H ′ .Order the vectors of Υ so that ˆ γ ( j ) = γ j for all j ∈ [ { , m − } ], and so that the vectors { ξ i | i ∈ [ { k } ] \ I } appear in an arbitrary order among γ m , γ m +1 , . . . , γ k − .Move the origin to the center h || δ || · δ of the annulus S ′ , and rotate the annulus so that newaxes become the colinear with vectors γ , γ , . . . , γ k − of the orthonormal basis Υ. Obviously, thismapping is volume-preserving.For a vector ζ ∈ H ′ , let ζ [Υ] , ζ [Υ] , . . . , ζ k − [Υ] denote the coordinates of ζ with respect tothe basis Υ, i.e., ζ i [Υ] = h ζ − h || δ || · δ, γ i i . Observe that since h δ, γ i i = 0 for all i ∈ [ { k − } ], itfollows that ζ i [Υ] = h ζ , γ i i , for all i ∈ [ { k − } ]. Lemma 4.4
For a vector ζ ∈ ˜ W = S ′ ∩ C , and an index i ∈ [ { m, . . . , k − } ] , we have ζ i [Υ] ≥ .In particular, ζ has at least (1 − ǫ ) · k non-negative coordinates with respect to the basis Υ . roof: Note that for every index i ∈ [ { m, k − } ], all entries of γ i are non-negative. (Becausethese are the vectors ξ j , j ∈ [ { k } ] \ I } of the standard Kronecker basis.) Since ζ ∈ C = [0 , y − k ,it follows that for all indices i ∈ [ { m, k − } ], the i th coordinate of ζ with respect to the basis Υis non-negative, that is, ζ i [Υ] = h ζ , γ i i ≥
0. Hence ζ has at least ( k − − ( m − ≥ (1 − ǫ ) · k non-negative coordinates with respect to the basis Υ.Recall that ˜ W ⊆ S ′ , and the annulus S ′ is given by (with respect to the basis Υ) S ′ = { α ∈ IR k − | T ′ − g ≤ || α || ≤ T ′ } . Let ˜ Q = { α ∈ IR k − | T ′ − g ≤ || α || ≤ T ′ , ∀ i ∈ [ { m, k − } ] , α i [Υ] ≥ } be the intersection of S ′ with the ( k − m ) half-spaces α i [Υ] ≥
0, for all i ∈ [ { m, k − } ]. Let S ′′ be the intersection of the annulus S ′ with the positive octant (with respect to Υ), i.e., S ′′ = { α ∈ (IR + ) k − | T ′ − g ≤ || α || ≤ T ′ } . It follows that ˜ W ⊆ ˜ Q , and Vol ( ˜ W ) ≤ Vol ( ˜ Q ) = 2 m · Vol ( S ′′ ) ≤ ǫ · k − · Vol ( S ′′ ).Next, we provide an upper bound for Vol ( S ′′ ). Lemma 4.5
For a sufficiently large integer k ,Vol ( S ′′ ) ≤ g · (cid:16) πe (cid:17) k/ · y k − · O ( √ k ) . Proof:
Let R ′ = √ T ′ . Observe that R ′ ≤ R = √ T . Let β k − be the volume of the ( k − Vol ( S ′′ ) = k − β k − (( T ′ ) k − − ( T ′ − g ) k − ). Note that( R ′ − g ) k − = (cid:16) − gR ′ (cid:17) k − · R ′ k − ≥ R ′ k − (cid:18) − g ( k − R ′ (cid:19) ≥ R ′ k − − R ′ k − g · k . (16)Hence by (1), Vol ( S ′′ ) ≤ k − · k · g · β k − · R ′ k − ≤ k − · k · g · π k − Γ (cid:0) k +12 (cid:1) · R k − . (17)By (11), T = R ≤ k · y (cid:16) O (cid:16) √ k (cid:17)(cid:17) , and so R ≤ q k · y (cid:16) O (cid:16) √ k (cid:17)(cid:17) . Hence Vol ( S ′′ ) ≤ k − · k · g · π k − Γ (cid:0) k +12 (cid:1) · (cid:18) k (cid:19) k − · y k − · O ( √ k ) . By Stirling formula, if k + 1 is even then for a sufficiently large k ,Γ (cid:18) k + 12 (cid:19) ≥ (cid:18) k − (cid:19) ! ≥ r k − · (cid:0) k − (cid:1) k − e k − = e / (cid:0) k (cid:1) k (cid:0) − k (cid:1) k e k/ ≥ · k k (2 e ) k .
10y (3), if k + 1 is odd then for a sufficiently large k ,Γ (cid:18) k + 12 (cid:19) = Γ (cid:18) k (cid:19) ≥ √ π (cid:18) k − (cid:19) ! ≥ πe √ · r k − · (cid:0) k − (cid:1) k − e k − ≥ πe √ k · (cid:0) k (cid:1) k · (cid:0) − k (cid:1) k e k ≥ √ k · k k/ (2 e ) k/ . Hence in both cases, for a sufficiently large k ,Γ (cid:18) k + 12 (cid:19) ≥ √ k · k k (2 e ) k . Consequently,
Vol ( S ′′ ) ≤ ( k · g ) · k − · π k √ k · (2 e ) k √ π · k k · k k − k − · y k − · O ( √ k ) = O (1) · ( k · g ) · k − √ k · (cid:16) πe (cid:17) k · y k − · O ( √ k ) ≤ g · (cid:16) πe (cid:17) k · y k − · O ( √ k ) . We conclude that
Vol ( ˜ W ) ≤ Vol ( ˜ Q ) ≤ ǫk − · Vol ( S ′′ ) ≤ · g · ǫk · (cid:16) πe (cid:17) k · y k − · O ( √ k ) . (18)Since ˜ W ⊆ ˜ Q , the number W ( δ, h ) of integer points in ˜ W is at most the number Q of integerpoints in ˜ Q . In Section 5 we will show that Q is not much larger than Vol ( ˜ Q ). Specifically, Q ≤ k O (1) · ǫk · (cid:16) πe (cid:17) k · y k − · O ( √ k ) . (19)We remark that this estimate is quite crude, as it says that the number Q of integer points in ˜ Q cannot be larger than by a factor of k O (1) than Vol ( ˜ Q ). However, it is sufficient for our argument.By (19), W ( δ, h ) ≤ Q ≤ k O (1) · ǫk · (cid:16) πe (cid:17) k · y k − · O ( √ k ) . (20)By (14), W ( δ ) = g X h =0 W ( δ, h ) ≤ ( g + 1) · k O (1) · ǫk · (cid:16) πe (cid:17) k · y k − · O ( √ k ) . Hence by (13), the overall number N of integer points in C ∩ S that do not belong to Ext ( B )(and thus, do not belong to Ext ( C ∩ S ), because S ⊆ B ) satisfies N ≤ X < || δ || ≤ g W ( δ ) ≤ k O (1) · ǫk · (cid:16) πe (cid:17) k · y k − · O ( √ k ) · ˆ D ( g ) . g ≤ k . By Lemma 4.2, and since for a sufficiently large k , k O (1) ≤ O ( √ k ) , it followsthat N ≤ k O (1) · ǫk · (cid:16) πe (cid:17) k · y k − · O ( √ k ) · O (2 η · k ) = 2 (( ǫ + η ( ǫ )+ O (1 / √ k ))+ log πe ) · k · y k − . By (10), the set ˜ S of integer points of C ∩ S contains | ˜ S| = Ω( ǫ √ k · y k − )integer points. By (8), y = 2 k/ > · O (cid:18) ǫ · √ k (cid:19) (( ǫ + η ( ǫ )+ O (1 / √ k ))+ log πe ) · k (21)whenever ǫ > k satisfy 1 > log πe ǫ + η ( ǫ )) + O (cid:18) √ k (cid:19) . By Lemma 4.2, lim ǫ → η ( ǫ ) = 0. Thus, for a sufficiently small universal constant ǫ >
0, andsufficiently large k , the inequality (21) holds, and thus | ˜ S| ≥ N . (More specifically, one needsto set ǫ so that 0 < ǫ + η ( ǫ ) < − log πe .) Hence the set ˜ S contains a subset ˇ S of integer pointsthat belong to Ext ( B ), and moreover, | ˇ S| ≥ | ˜ S| − N ≥ | ˜ S| = Ω( ǫ · √ k · y k − ) = Ω(log / n · n √ √ log n ) . Consider the annulus S ′ = { α = ( α , α , . . . , α k − ) ∈ IR k − | T ′ − g ≤ || α || ≤ T ′ } , and itsintersection ˜ Q with the half-spaces α i ≥ i ∈ [ { m, k − } ]. In this section we argue thatthe number Q of integer points in ˜ Q is not much larger than Vol ( ˜ Q ). Specifically, we show that Q = 2 O ( √ k ) · ǫk · (cid:16) πe (cid:17) k/ · y k − . (22)This proves (19), and hence completes the proof of our lower bound.Consider the ( k − B of squared radius t centered at the origin, for somesufficiently large t >
0. Let A ( B ) denote the number of integer points in B . For a positiveinteger j , let V j ( t ) denote the volume of the j -dimensional ball of squared radius t centered atthe origin. It is well-known (see, e.g., the survey of Adhikari [1]) that for a constant dimension k , | A ( B ) − V ( B ) | = O ( V k − ( t )). However, in our case the dimension k grows logarithmically with t .Fortunately, the following analogous inequality holds in this case: | A ( B ) − V ( B ) | = k O (1) · V k − ( t ) . (23)12e prove (23) in the sequel.Another subtle point is that we have rotated the vector space to move from the standardKronecker basis to the orthonormal basis Υ. (In fact, Υ is an orthonormal basis for the hyperplane H ′ , but it can be completed to an orthonormal basis for IR k by inserting the vector δ || δ || into it.)Consequently, the integer lattice was rotated as well, and so in our context A ( B ) is actually thenumber of points of the rotated integer lattice that are contained in B . These two quantitiesmay be slightly different. However, we argue below that the estimate (23) applies for the rotatedinteger lattice as well, for any rotation.Recall that m = | I | . Let˜ Q ext = { α = ( α , α , . . . , α k − ) ∈ IR k − : || α || ≤ T ′ , α i ≥ i ≥ m } (24)˜ Q int = { α = ( α , α , . . . , α k − ) ∈ IR k − : (25) || α || ≤ T ′ − ( g + 1) , α i ≥ i ≥ m } Observe that ˜ Q ⊆ ˜ Q ext \ ˜ Q int . Also, let ˜ Z denote˜ Z = { α = ( α , α , . . . , α k − ) ∈ IR k − : || α || ≤ T ′ , for all i ≥ m } . (26)The set ˜ Q ext (respectively, ˜ Q int ) is the intersection of the ( k − T ′ (resp., T ′ − ( g + 1)) centered at the origin with the half-spaces α i ≥ i ≥ m . The set ˜ Z is the intersection of the ( k − T ′ centered at the origin withthe half-spaces α i ≥ i ≥ m . The analogue of (23) that is required for our argument is | A ( ˜ Q ext ) − Vol ( ˜ Q ext ) | = k O (1) · Vol ( ˜ Z ) . (27)Given (27) we show (22) by the following argument. Lemma 5.1 A ( ˜ Q ) = 2 O ( √ k ) · ǫk · (cid:0) πe (cid:1) k/ · y k − . Proof:
By (27), A ( ˜ Q ) ≤ A ( ˜ Q ext ) − A ( ˜ Q int ) ≤ Vol ( ˜ Q ext ) + k O (1) · Vol ( ˜ Z ) − Vol ( ˜ Q int ) + k O (1) · Vol ( ˜ Z )= ( Vol ( ˜ Q ext ) − Vol ( ˜ Q int )) + k O (1) · Vol ( ˜ Z ) . Observe that
Vol ( ˜ Z ) = ǫk k − · β k − · ( T ′ ) k − . Also, since T ′ is much greater than g , Vol ( ˜ Q ext ) − Vol ( ˜ Q int ) = 2 ǫk k − · β k − (( T ′ ) k − − ( T ′ − ( g + 1)) k − ) ≤ O (1) · ǫk k − · β k − · k · ( g + 1) · ( T ′ ) k − . Hence A ( ˜ Q ) ≤ O (1) · ǫk k − · ( k O (1) · β k − + k · ( g + 1) · β k − ) · ( T ′ ) k − . Since β k − = Θ( k · β k − ) and g ≤ k , it follows that A ( ˜ Q ) ≤ k O (1) · ǫk k − · β k − · ( T ′ ) k − .
13y (11), T ′ ≤ k · y (1 + O ( √ k )). Also, β k − = π k − Γ( k +12 ) . Hence A ( ˜ Q ) = 2 O ( √ k ) · ǫk · (cid:16) πe (cid:17) k/ · y k − . Hence it remains to prove (27). Our proof is closely related to the argument in [7], pp. 94-97,and is provided for the sake of completeness. In addition, our argument is more general than theone in [7], as the latter argument applies only for balls, while our argument applies for intersectionsof balls with half-spaces.Fix m to be a positive integer number. (In our application m = | I | .) For positive integernumbers k and t , let Q k ( t ) denote the intersection of the k -dimensional ball B k ( t ) centered at theorigin with squared radius t with the half-spaces H ( i ) = { α = ( α , α , . . . , α k ) | α i ≥ } , for all i ≥ m . Let ¯ V k ( t ) denote the volume Vol ( Q k ( t )), and ¯ A k ( t ) denote the number of points of therotated integer lattice in Q k ( t ). Note that ¯ V k ( t ) = β k max { k − m +1 , } · t k/ . The next lemma provides anupper bound for the discrepancy between ¯ V k ( t ) and ¯ A k ( t ) in terms of ¯ V k − ( t ). Lemma 5.2
For a sufficiently large real t > and an integer k ≥ , | ¯ A k ( t ) − ¯ V k ( t ) | = O ( k / · ¯ V k − ( t )) . Remark:
This lemma applies even if k = k ( t ) is a function of t .Before proving Lemma 5.2, we first provide a number of auxiliary lemmas that will be usefulfor its proof. We start with Euler Sum-formula ([7], Satz 29.1, p.185). Lemma 5.3
For a real-valued function f ( u ) differentiable in a segment [ a, b ] , X a<ℓ ≤ b f ( ℓ ) = Z ba f ( u ) du + ψ ( a ) · f ( a ) − ψ ( b ) · f ( b ) + Z ba ψ ( u ) · f ′ ( u ) du , where ψ ( u ) = u − ⌊ u ⌋ − . In addition, we will use the following property of the function ψ ( · ). Lemma 5.4
For any two real numbers κ and λ , κ ≤ λ , − ≤ R λκ ψ ( u ) du ≤ . Proof:
Observe that Z ψ ( u ) du = Z (cid:18) u − ⌊ u ⌋ − (cid:19) du = Z (cid:18) u − (cid:19) du = Z − tdt = 0 . Moreover, for any integer j , R j +1 j ψ ( u ) du = R j +1 j (cid:0) u − ⌊ u ⌋ − (cid:1) du = R − tdt = 0. Hence R j ψ ( u ) du = R − j ψ ( u ) du = 0, for any positive integer j .It follows that Z λκ ψ ( u ) du = Z ⌈ κ ⌉ κ ψ ( u ) du + Z ⌊ λ ⌋⌈ κ ⌉ ψ ( u ) du + Z λ ⌊ λ ⌋ ψ ( u ) du = Z ⌈ κ ⌉ κ ψ ( u ) du + Z λ ⌊ λ ⌋ ψ ( u ) du . κ ′ = κ − ⌊ κ ⌋ , λ ′ = λ − ⌊ λ ⌋ . The integral R ⌈ κ ⌉ κ ψ ( u ) du is equal to 0 if κ = ⌈ κ ⌉ , and is equal to R κ ′ ψ ( u ) du , otherwise. The latter integral satisfies Z κ ′ ψ ( u ) du = Z κ ′ − tdt = 12 (1 − κ ′ ) κ ′ ≤ . Also, R κ ′ ψ ( u ) du = (1 − κ ′ ) κ ′ ≥ R λ ⌊ λ ⌋ ψ ( u ) du is equal to 0 if λ = ⌊ λ ⌋ , and it is equal to R λ ′ ψ ( u ) du , otherwise. Thelatter integral satisfies Z λ ′ ψ ( u ) du = Z λ ′ − − tdt = 12 ( λ ′ − λ . Hence − ≤ R λ ′ ψ ( u ) du ≤ , and thus, − ≤ R λκ ψ ( u ) du ≤ ψ ( · ). Lemma 5.5
For a positive real number t and a positive integer p ≥ , | Z √ t u · ψ ( u )( t − u ) p − du | ≤ t p − . Proof:
Since f ( u ) = u is a monotone increasing function, there exists ξ ∈ [0 , √ t ] such that Z √ t u · ψ ( u )( t − u ) p − du = √ t Z √ tξ ψ ( u )( t − u ) p − du . Since g ( u ) = ( t − u ) p − is a monotone decreasing function in [ ξ, √ t ], there exists η ∈ [ ξ, √ t ] suchthat the right-hand-side is equal to √ t · ( t − ξ ) p − R ηξ ψ ( u ) du . By Lemma 5.4, |√ t ( t − ξ ) p − Z ηξ ψ ( u ) du | ≤ √ t · t p − = t p − . Note also that | Z − u · ψ ( u )( t − u ) p − du | ≤ | Z − ( t − u ) p − du | ≤ · t p − . Hence for any t and p as above, | Z √ t − u · ψ ( u )( t − u ) p − du | ≤ t p − + 12 · t p − ≤ t p − (cid:18) √ t (cid:19) . (28)We are now ready to prove Lemma 5.2. Proof of Lemma 5.2:
15e prove by induction on k that there exists a universal constant c > | ¯ A k ( t ) − ¯ V k ( t ) | ≤ c · k − X j =1 p j ! (cid:18) √ t (cid:19) · ¯ V k − ( t ) . (29)The constant c will be determined later.The induction base is k = 5. It is well-known (see, e.g., [1]) that | ¯ A ( t ) − ¯ V ( t ) | = O ( ¯ V ( t )) = O ( t / ).Next, we prove the induction step.In all summations below, ℓ is an integer index. The analysis splits into two cases. The firstcase is k + 1 < m , and the second is k + 1 ≥ m . In the first case¯ A k +1 ( t ) = X | ℓ |≤√ t ¯ A k ( t − ℓ ) = X | ℓ |≤√ t ¯ V k ( t − ℓ ) + X | ℓ |≤√ t ( ¯ A k ( t − ℓ ) − ¯ V k ( t − ℓ )) , and so | ¯ A k +1 ( t ) − X | ℓ |≤√ t ¯ V k ( t − ℓ ) | = | X | ℓ |≤√ t ( ¯ A k ( t − ℓ ) − ¯ V k ( t − ℓ )) | ≤ X | ℓ |≤√ t | ¯ A k ( t − ℓ ) − ¯ V k ( t − ℓ ) | . In the second case the same inequalities apply, but the index ℓ runs in the range 0 ≤ ℓ ≤ √ t in allsummations. It turns out to be more convenient to have the index ℓ vary in the range − ≤ ℓ ≤ √ t rather than 0 ≤ ℓ ≤ √ t in these summations.By the induction hypothesis (that is, by (29)), | ¯ A k ( t − ℓ ) − ¯ V k ( t − ℓ ) | ≤ c · k − X j =1 p j ! (cid:18) √ t (cid:19) · ¯ V k − ( t − ℓ ) . Hence | ¯ A k +1 ( t ) − X | ℓ |≤√ t ¯ V k ( t − ℓ ) | ≤ c · k − X j =1 p j ! (cid:18) √ t (cid:19) · X | ℓ |≤√ t ¯ V k − ( t − ℓ ) . (30)Next, we estimate P | ℓ |≤√ t ¯ V k ( t − ℓ ) via Euler Sum-formula (Lemma 5.3). In the first case, wesubstitute a = −√ t , b = √ t , and f ( u ) = ¯ V k ( t − u ). Then f ( a ) = f ( b ) = ¯ V k (0) = 0, and dfdu ( u ) = ddu ¯ V k ( t − u ) = β k ddu ( t − u ) k = − β k · k · ( t − u ) k − u . By Lemma 5.3 it follows that X | ℓ |≤√ t ¯ V k ( t − ℓ ) = Z √ t √ t ¯ V k ( t − u ) du − k · β k Z √ t √ t ψ ( u )( t − u ) k − udu . (31)In the second case ( k ≥ m − a = − , b = √ t , and again f ( a ) = f ( b ) = 0. Also, dfdu ( u ) = − β k · k − m +1 · k · ( t − u ) k − u , X − <ℓ ≤√ t ¯ V k ( t − ℓ ) = Z √ t − ¯ V k ( t − u ) du − k · β k k − m +1 Z √ t − ψ ( u )( t − u ) k − udu . (32)In the first case, since h ( u ) = uψ ( u ) is an even function on IR \ ZZ , the right-hand-side in (31) isequal to Z √ t √ t ¯ V k ( t − u ) du − k · β k Z √ t ψ ( u )( t − u ) k − udu . Let J denote | R √ t u · ψ ( u )( t − u ) k − du | . By Lemma 5.5, J ≤ t k − . Hence X | ℓ |≤√ t ¯ V k ( t − ℓ ) = Z √ t √ t ¯ V k ( t − u ) du − kβ k · J = ¯ V k +1 ( t ) − kβ k · J .
It follows that | ¯ V k +1 ( t ) − X | ℓ |≤√ t ¯ V k ( t − ℓ ) | = 2 kβ k · J ≤ kβ k · t k − . (33)In the second case by (32) and since R √ t ¯ V k ( t − u ) du = ¯ V k +1 ( t ), it follows that X − <ℓ ≤√ t ¯ V k ( t − ℓ ) = Z − ¯ V k ( t − u ) du + ¯ V k +1 ( t ) − k · β k k − m +1 Z √ t − ψ ( u ) u ( t − u ) k − du . Let J ′ denote | R √ t − u · ψ ( u )( t − u ) k − du | . By (28), J ′ ≤ t k − + 12 · t k − ≤ t k − (cid:18) √ t (cid:19) . (34)Since ¯ V k ( t − u ) ≥ u , − ≤ u ≤
0, the integral R − ¯ V k ( t − u ) du is non-negative as well.Thus, | ¯ V k +1 ( t ) − X − ≤ ℓ ≤√ t ¯ V k ( t − ℓ ) | ≤ k · β k k − m +1 · J ′ ≤ k · β k k − m +1 · t k − (cid:18) √ t (cid:19) . In the first case, by the triangle inequality, by (30), and by (33), | ¯ A k +1 ( t ) − ¯ V k +1 ( t ) | ≤ | ¯ A k +1 ( t ) − X | ℓ |≤√ t ¯ V k ( t − ℓ ) | (35)+ | X | ℓ |≤√ t ¯ V k ( t − ℓ ) − ¯ V k +1 ( t ) |≤ c · k − X j =1 p j ! X | ℓ |≤√ t ¯ V k − ( t − ℓ ) + | X | ℓ |≤√ t ¯ V k ( t − ℓ ) − ¯ V k +1 ( t ) |≤ c · k − X j =1 p j ! X | ℓ |≤√ t ¯ V k − ( t − ℓ ) + 2 kβ k · t k − . (36)17nalogously, in the second case, | ¯ A k +1 ( t ) − ¯ V k +1 ( t ) | ≤ c · k − X j =1 p j ! (cid:18) √ t (cid:19) X − ≤ ℓ ≤√ t ¯ V k − ( t − ℓ ) (37)+ 2 k · β k k +2 − m · t k − · (cid:18) √ t (cid:19) . However, in the first case P | ℓ |≤√ t ¯ V k − ( t − ℓ ) ≤ R √ t −√ t ¯ V k − ( u ) du = ¯ V k − ( t ). In the second case, X − ≤ ℓ ≤√ t ¯ V k − ( t − ℓ ) = X ≤ ℓ ≤√ t ¯ V k − ( t − ℓ ) ≤ Z √ t ¯ V k − ( u ) du = ¯ V k − ( t ) . Hence in both cases the first terms in (36) and in the right-hand-side of (37) are at most c · k − X j =1 p j ! (cid:18) √ t (cid:19) ¯ V k − ( t ) . Consequently, in both cases, | ¯ A k +1 ( t ) − ¯ V k +1 ( t ) | ≤ c · k − X j =1 p j ! (cid:18) √ t (cid:19) · ¯ V k − ( t )+ 2 k · β k max { k +2 − m, } · t k − · (cid:18) √ t (cid:19) . By (1), β k = Θ (cid:16) β k − √ k (cid:17) . Set c to be a universal constant such that c ≥ √ k · β k β k − , for all integer k ≥ | ¯ A k +1 ( t ) − ¯ V k +1 ( t ) | ≤ c · k − X j =1 p j ! (cid:18) √ t (cid:19) · ¯ V k − ( t )+ c · √ k · (cid:18) √ t (cid:19) · β k − max { ( k − − m +1 , } · t k − = c · k − X j =1 p j ! (cid:18) √ t (cid:19) ¯ V k − ( t ) . Finally, P kj =1 √ j ≤ k / , completing the proof. In this paper we improved the lower bound of Behrend by a factor of Θ( √ log n ). As was alreadymentioned, both Behrend’s and our proof arguments rely on the Pigeonhole Principle. It isreasonable to believe that by choosing T = R = µ Z (see (9)) one can get an annulus withat least as many integer points as in the annulus S chosen via the Pigeonhole Principle. To provethat this is the case one should probably use normal approximation of the discrete random variable18 (see Sections 3 and 4), and employ probablistic estimates to argue that the probability that Z isbetween ( µ Z − ǫk ) and ( µ Z + ǫk ) is at least as large as the probability that it is between ( µ Z − σ Z )and ( µ Z + 2 σ Z ), divided by ǫk σ Z . Although this appears to be quite clear intuitively, so far we werenot able to find sufficiently precise probabilistic estimates to prove this statement formally. Oncethis intuition is formalized, our construction will become independent of the Pigeonhole Principle.This, in turn, would be a significant improvement of the lower bound of Moser [9]. Acknowledgements
The author is indebted to Don Coppersmith, who was offered a coauthorship on this paper. Inparticular, Lemma 4.1 is due to Don. In addition, fingerprints of Don can be found in numerousother places in this paper.The author is grateful to Benny Sudakov for introducing him to the problem. The authorthanks also Noga Alon, Eitan Bachmat, B´ela Bollob´as, Danny Berend, Alexander Razborov,Oded Regev, Alex Samorodnitsky, and Shakhar Smorodinsky, for encouragement and for helpfuldiscussions.
References [1] S. D. Adhikari. Lattice points in spheres.
Bulletin of the Allahabad Mathematical Society ,8-9:1–13, 1993-1994.[2] F. Behrend. On sets of integers which contain no three terms in arithmetic progression.
Proc.Nat. Acad. Sci. , 32:331–332, 1946.[3] J. Bourgain. On triples in arithmetic progression.
GAFA , 9:968–984, 1999.[4] J. Bourgain. Roth’s theorem in progressions revisited. manuscript , 2007.[5] D. Coppersmith. personal communication, 2003.[6] P. Erd˝os and P. Tur´an. On some sequences of integers.
J. London Math. Society , 11:261–264,1936.[7] F. Fricker.
Einfuhrung in die Gitterpunktlehre . Birkhauser, 1982.[8] B. Green and T. Tao. New bounds for Szemer´edi’s theorem, II: A new bound for r ( n ). manuscript , 2008.[9] L. Moser. On non-averaging sets of integers. Canadian J. Math. , 5:245–253, 1953.[10] R. Rankin. Sets not containing more than a given number of terms in arithmetic progression.
Proc. Roy. Soc. Edinburgh Section A , 65:332–344, 1960.[11] K. Roth. On certain sets of integers.
J. London Math. Society , 28:245–252, 1953.1912] R. Salem and D. Spencer. On sets of integers which contain no three in arithmetic progression.