Exact Kronecker Constants of Three Element Sets
aa r X i v : . [ m a t h . C A ] M a r EXACT KRONECKER CONSTANTS OFTHREE ELEMENT SETS
KATHRYN E. HARE and L. THOMAS RAMSEY Department of Pure Mathematics, University of Waterloo, Waterloo, Ont., Canada, N2L 3G1e-mail: [email protected] Department of Mathematics, University of Hawaii at Manoa, Honolulu, Hi, USA, 96822e-mail: [email protected]
Abstract.
For any three element set of positive integers, { a, b, n } , with a < b < n, n sufficiently large and gcd( a, b ) = 1, we find the least α suchthat given any real numbers t , t , t there is a real number x such thatmax {h ax − t i , h bx − t i , h nx − t i} ≤ α, where h·i denotes the distance to the nearest integer. The number α is knownas the angular Kronecker constant of { a, b, n } . We also find the least β suchthat the same inequality holds with upper bound β when we consider onlyapproximating t , t , t ∈ { , / } , the so-called binary Kronecker constant.The answers are complicated and depend on the congruence of n mod ( a + b ).Surprisingly, the angular and binary Kronecker constants agree except if n ≡ a mod ( a + b ).
1. Introduction
The classical Kronecker theorem states that if { r j } is any finite collec-tion of rationally independent real numbers, then given any sequence of realnumbers ( t j ) ⊆ [0 ,
1) and ε > x and ( k j ) such that | r j x − t j − k j | < ε for all j . This fails to be true if the { r j } are replaced by afinite collection of integers, { n j } , even allowing x to be any real number. Theangular Kronecker constant of the given set of integers S = { n j } , denoted α ( S ), is the infimum of the ε for which such an approximation can be madefor every sequence ( t j ). It is obvious that α ( S ) ≤ / S (finiteor infinite), and an application of the Baire category theorem shows that α ( S ) < / S . Without further knowledge about the set, Key words and phrases:
Kronecker constant, trigonometric approximation.
Mathematics Subject Classification:
Primary: 42A10; Secondary: 43A46, 11J71This work was partially supported by NSERC through application number 44597, andthe Edinburgh Math Soc. The authors would like to thank St. Andrews University fortheir hospitality when some of this research was done.
K.E. HARE and L.T. RAMSEY this is the best that can be said. Sets for which α ( S ) < / α ( S ) < / S -function is the restriction of the Fourier trans-form of a measure on the circle. In fact, this measure can be chosen to bediscrete and 1 / α < / S is Sidon.It could be helpful to know the Kronecker constants of finite sets in ad-dressing these questions as the Kronecker constant of an infinite set is thesupremum of the constants of its finite subsets. This is a difficult problem andcomplete answers are known only for special sets, such as two element sets[3], three element sets satisfying certain simple relations [5] and geometricsequences of the form { m j } dj =0 for an integer m [6].In this note we determine the Kronecker constants for all three elementsets of positive integers, { a, b, n } , where a and b are coprime and n is suffi-ciently large. The answers are surprisingly complicated, with different formu-las depending on the congruence of n mod ( a + b ). Our proof is algorithmic.Interestingly, we show that one can find the best approximate for a giventriple of real numbers ( t , t , t ) by either starting with the best or the ‘sec-ond best’ approximate for the pair ( t , t ) relative to the two element set { a, b } , and then, if necessary, making a slight modification. We call this the‘greedy algorithm’; see Section 2 for more details.A related problem is to determine how well the set of integers S canapproximate all { , / } -valued sequences ( t j ). The least ε for this approx-imation problem is known as the binary Kronecker constant, β ( S ). Likethe angular Kronecker constant, this constant is also known only in a fewspecial cases such as arithmetic progressions [7]. We calculate the binaryKronecker constants for three element sets, as well. An unexpected fact isthat α ( a, b, n ) = β ( a, b, n ) provided n a mod ( a + b ) . Here is our main theorem.
Theorem . Suppose a < b < n are positive integers with gcd( a, b ) = 1 and n suitably large. Assume aT ≡ a + b ) , n ≡ r mod ( a + b ) , and R ≡ rT mod ( a + b ) XACT KRONECKER CONSTANTS OF THREE ELEMENT SETS for r , R and T ∈ [0 , a + b ) . Then α ( a, b, n ) = n + a + ab − aR a + b )( a + n ) if ≤ R < an + ab an + bn + ab ) if R = an + bR a + b )( b + n ) if a < R ≤ an + 2 a + 2 ab − aR a + b )( a + n ) if R > a Moreover, α ( a, b, n ) = β ( a, b, n ) except if R = a , when α ( a, b, n ) > β ( a, b, n ) . We note that R = a if and only if n ≡ a mod ( a + b ). We also remarkthat if one checks the details of the proof, it can be seen how large n needsto be relative to the size of a and b .We conjecture that if gcd( a, b ) = m >
1, there will be ( a + b ) /m cases for α ( a, b, n ), determined by the congruence of n mod ( a/m + b/m ) . It seems reasonable to expect that with considerably more work the tech-niques of this paper could be generalized to sets S of size greater than 3,subject to the condition that the largest element is much larger than theothers. Remark . In discrete optimization, the closest vector problem is to find,given an additive subgroup L ⊆ Z n , the distance d ( v, L ) = min { ρ ( v − k ) : k ∈ L } for any v ∈ R n . This is known to be NP-hard for both the Euclideanand maximum norms ρ [9, p. 182].Finding Kronecker constants involves superimposing one additional layerof optimization. Indeed, if S = { n , ..., n d } ⊆ Z d , then α ( S ) = max { d ( v, Z d ) : v ∈ R d } , where ρ ( w ) = inf {k w − t ( n , ..., n d ) k : t ∈ R } . To date, the authorsdo not know the hardness level created by superimposing the additionallevel of optimization that is required to compute Kronecker constants. It isstriking that with S = { a, b, n } the Kronecker constant can now be computedinstantly for n large and gcd( a, b ) = 1, as proved in this paper, but that exactKronecker constants have eluded simplification for n relatively small. This isa strange kind of ‘hardness’, where smaller values of integers give the mostdifficult cases to analyze.Throughout the paper we will denote by E n the value on the right handside of the (claimed) formula for α ( a, b, n ). In Section 2 we prove that K.E. HARE and L.T. RAMSEY α ( a, b, n ) is dominated by the formulas specified above. In Section 3 wecompute β ( a, b, n ), showing that it agrees with these formulas when R = a .Since α ( a, b, n ) ≥ β ( a, b, n ), this proves the equality when R = a . The proofthat β ( a, b, n ) < α ( a, b, n ) = E n when R = a is handled directly in the finalsubsection. Notation and Definitions : Assume S = { n j } dj =1 is a set of d integerswith n < n < · · · < n d . Set n = ( n , . . . , n d ). We define the approximationcost for t = ( t , ..., t d ) ∈ R d , relative to S, as(1) µ S ( t ) = inf { k t − x · n + k k ∞ : x ∈ R , k ∈ Z d } = inf { kh t − x n ik ∞ : x ∈ R } . Here the symbol h u i denotes the distance to the nearest integer when u is areal number; for real vectors, u , it denotes the application of h·i component-wise. We omit the writing of the subscript S when the set S is clear.With this notation the angular Kronecker constant of S is α ( S ) = sup { µ ( t ) : t ∈ R d } and the binary Kronecker constant is β ( S ) = sup { µ ( t ) : t ∈ { , / } d } . As noted in [5], periodicity allows one to limit x to an interval of length 1and the vectors k to a finite set. Consequently, inf and sup can be replacedby min and max. A choice of x which minimizes µ S ( t ) is known as a bestapproximate for t relative to S .
2. Upper bounds on the Kronecker constants2.1. Setting up a ‘greedy’ algorithm.
Throughout the remainder ofthe paper we assume a < b < n are fixed with gcd( a, b ) = 1 , and n is suitablylarge. When we write µ ( t , t , t ) we mean µ { a,b,n } ( t , t , t ). Recall that E n denotes the value claimed by Theorem 1 for α ( a, b, n ). It is known that forall positive integers a, b , α ( a, b, n ) → α ( a, b ) = 1 / (2 a + 2 b ) as n → ∞ . It is clear that lim n →∞ E n = 1 / (2 a + 2 b ), thus by taking n sufficiently largewe can assume without loss of generality that both α ( a, b, n ) and E n aresmaller than 1 / ( a + b ). XACT KRONECKER CONSTANTS OF THREE ELEMENT SETS t , t ), relativeto { a, b } , and then to modify it slightly to improve the approximation of t relative to n . In many cases, this suffices. In other cases a related, ‘secondbest’ approximate must be suitably modified. These ideas will be madeprecise in this section.In [5] it is shown that for any pair of real numbers, ( t , t ) , there arealways a real number x and integers k , k with the property(2) ax − ( t + k ) = − ( bx − ( t + k )) . In fact, there is always a triple, x, k , k , that not only satisfies (2), but isalso a best approximate to ( t , t ) , by which we mean that µ { a,b } ( t , t ) = k x ( a, b ) − ( t + k , t + k ) k ∞ . Given an x, k , k satisfying (2), we will put λ x = | ax − ( t + k ) | . If it is the case that λ x ≤ ( b − a ) / (2 n ), then choose z such that | nz − nx | ≤ nz ≡ t mod 1, say nz = t + k for integer k . An easy calculationgives | az − ( t + k ) | ≤ | a ( z − x ) | + | ax − ( t + k ) | ≤ an + λ x ≤ b + a n , | bz − ( t + k ) | ≤ | b ( z − x ) | + | bx − ( t + k ) | ≤ bn + λ x ≤ b − a n , and | nz − ( t + k ) | = 0 . This proves
Lemma . If λ x ≤ ( b − a ) / (2 n ) , then µ ( t , t , t ) ≤ k z ( a, b, n ) − ( t + k , t + k , t + k ) k ∞ ≤ b − a n . For sufficiently large n , (3 b − a ) / (2 n ) ≤ E n , thus our interest is (primarily)with those x such that λ x > ( b − a ) / (2 n ). K.E. HARE and L.T. RAMSEY
Next, we will explain what we mean by “modify it slightly to improvethe approximation of t relative to n ”. Take any a real number E ≥ λ x , andsuppose real number z and integer k have been chosen satisfying nz = t + k and | nz − nx | ≤ . If | nz − nx | ≤ λ x we will not modify x . Otherwise, our strategy will be toreplace x by a nearby point, x ± δ , to get a better approximate. This strategywill culminate with the bounds of Corollaries 6 and 8.Case 1: ax − ( t + k ) ≥
0. First, suppose z ≤ x . We will replace x by x − δ for a suitably small δ >
0. This will have the effect of bringing downthe size of the third component of k ( x − δ )( a, b, n ) − ( t + k , t + k , t + k ) k ∞ . However, we will pay a cost for this: the size of the second component willincrease. We will permit only enough of an increase to balance these approx-imations of t and t . Thus 0 < δ < x − z will be chosen so that | ( x − δ ) n − ( t + k ) | = | ( x − δ ) b − ( t + k ) | . Recalling that nz = t + k and xb − ( t + k ) = − λ x , this gives(3) δ = | nx − nz | − λ x b + n . Since ax − ( t + k ) = λ x >
0, it is clear that for small δ > , | ( x − δ ) a − ( t + k ) | ≤ ax − ( t + k ) = λ x = | xb − ( t + k ) | . As | ( x − δ ) a − ( t + k ) | changes more slowly than | ( x − δ ) b − ( t + k ) | , theinequality above actually holds for all δ >
0, consequently, k ( x − δ )( a, b, n ) − ( t + k , t + k , t + k ) k ∞ = | ( x − δ ) b − ( t + k ) | = | ( x − δ ) n − ( t + k ) | = λ x + b (cid:18) | nx − nz | − λ x b + n (cid:19) . This shows that if, in addition to the assumption z ≤ x , z satisfies therequirement λ x + b (cid:18) | nx − nz | − λ x b + n (cid:19) ≤ E, XACT KRONECKER CONSTANTS OF THREE ELEMENT SETS z ≥ x + nλ x − ( b + n ) Ebn , then(5) k ( x − δ )( a, b, n ) − ( t + k , t + k , t + k ) k ∞ ≤ E. Of course, (5) implies µ ( t , t , t ) ≤ E .Otherwise, z > x . In this case, we replace x by x + δ , where δ > t and t . This means we choose δ < z − x such that | ( x + δ ) n − ( t + k ) | = ( x + δ ) a − ( t + k ) , in other words, δ = | nx − nz | − λ x a + n . If ( x + δ ) b ≤ t + k , then clearly | ( x + δ ) b − ( t + k ) | ≤ | xb − ( t + k ) | ≤ ( x + δ ) a − ( t + k ) . But even when ( x + δ ) b > t + k , we will still have the bound | ( x + δ ) b − ( t + k ) | ≤ ( x + δ ) a − ( t + k ) , provided δ ≤ xa − ( t + k ) − ( xb − ( t + k )) b − a = 2 λ x b − a .But δ ≤ z − x ≤ /n , and by assumption 1 /n < λ x / ( b − a ), hence thiscondition is automatically satisfied. Therefore k ( x + δ )( a, b, n ) − ( t + k , t + k , t + k ) k ∞ = | ( x + δ ) a − ( t + k ) | = λ x + a (cid:18) | nx − nz | − λ x a + n (cid:19) . The same reasoning as in the first case shows that if, in addition to theassumption that z > x ,(6) z ≤ x + ( a + n ) E − nλ x an , K.E. HARE and L.T. RAMSEY then(7) k ( x − δ )( a, b, n ) − ( t + k , t + k , t + k ) k ∞ ≤ E. Thus again µ ( t , t , t ) ≤ E .Case 2: ax − ( t + k ) <
0. Then λ x = − ( ax − ( t + k )) = bx − ( t + k ).If z ≤ x, we choose 0 < δ < x − z so that | ( x − δ ) n − ( t + k ) | = | ( x − δ ) a − ( t + k ) | . The assumption λ x ≥ ( b − a ) / (2 n ) ensures that also | ( x − δ ) b − ( t + k ) | ≤ | ( x − δ ) a − ( t + k ) | , thus it follows that if(8) x ≥ z ≥ x + nλ x − ( a + n ) Ean , then(9) k ( x − δ )( a, b, n ) − ( t + k , t + k , t + k ) k ∞ ≤ E. If z ≥ x , we choose 0 < δ < z − x so that | ( x + δ ) n − ( t + k ) | = | ( x + δ ) b − ( t + k ) | , and again one can check that if(10) x ≤ z ≤ x + ( b + n ) E − nλ x bn , then(11) k ( x + δ )( a, b, n ) − ( t + k , t + k , t + k ) k ∞ ≤ E. These are the key ideas behind the next lemma.
Lemma . Suppose there exist x ∈ R and integers k , k such that ax − ( t + k ) = − ( bx − ( t + k )) . Let | ax − ( t + k ) | = λ x and for E ≥ λ x , put z ( E, x ) = x + nλ x − ( b + n ) Ebn , z ( E, x ) = x + ( a + n ) E − nλ x an ,z ( E, x ) = x + nλ x − ( a + n ) Ean , z ( E, x ) = x + ( b + n ) E − nλ x bn . XACT KRONECKER CONSTANTS OF THREE ELEMENT SETS Assume λ x > ( b − a ) / (2 n ) . Then z ( E, x ) ≤ z ( E, x ) and z ( E, x ) ≤ z ( E, x ) .Furthermore,(i) If ax − ( t + k ) = λ x and there exists z ∈ [ z ( E, x ) , z ( E, x )] such that nz ≡ t mod 1 , then µ ( t , t , t ) ≤ E .(ii) If − ( ax − ( t + k )) = λ x and there exists z ∈ [ z ( E, x ) , z ( E, x )] such that nz ≡ t mod 1 , then µ ( t , t , t ) ≤ E . Proof. As E ≥ λ x , we have z ( E, x ) ≤ z ( E, x ) and z ( E, x ) ≤ z ( E, x ).We note that if there is some z ∈ [ z ( E, x ) , z ( E, x )] (or z ∈ [ z ( E, x ) , z ( E, x )])such that nz ≡ t mod 1, then there is a possibly different choice of z, be-longing to the same interval, still satisfying nz ≡ t mod 1 , and having theadditional property that | nx − nz | ≤
1. We will work with such a z .First, suppose ax − ( t + k ) = λ x . If z ( E, x ) ≤ z ≤ x , then it followsfrom (4) and (5) (with the δ described in (3)) that µ ( t , t , t ) ≤ k ( x − δ )( a, b, n ) − ( t + k , t + k , t + k ) k ∞ ≤ E. If, instead, x ≤ z ≤ z ( E, x ), we appeal to (6) and (7) to deduce that µ ( t , t , t ) ≤ k ( x + δ )( a, b, n ) − ( t + k , t + k , t + k ) k ∞ ≤ E. If − ( ax − ( t + k )) = λ x , we similarly call upon (8), (9), (10) and (11). Remark . Observe that the proof shows that if E = µ ( t , t , t ), thenone of x ± δ is a best approximate to ( t , t , t ) relative to { a, b, n } . Corollary . For all t , t , t we have µ ( t , t , t ) ≤ max (cid:18) n ( a + b ) µ { a,b } ( t , t ) + ab ab + an + bn , b − a n (cid:19) . Proof.
Choose x and integers k , k such that µ { a,b } ( t , t ) = | ax − ( t + k ) | = | bx − ( t + k ) | = λ x . As shown in Lemma 3, if λ x ≤ ( b − a ) / (2 n ), then µ ( t , t , t ) ≤ (3 b − a ) / (2 n ) , so assume otherwise. Set E = n ( a + b ) λ x + ab ab + an + bn = λ x + ab (1 − λ x )2 ab + an + bn . K.E. HARE and L.T. RAMSEY As µ { a,b } ( t , t ) < /
2, we have
E > λ x , so we may apply Lemma 4. Inparticular, the intervals [ z ( E, x ) , z ( E, x )] and [ z ( E, x ) , z ( E, x )] of Lemma4 are well defined and they have the same length:( a + n ) E − nλ x an + ( b + n ) E − nλ x bn = E (cid:18) ab + an + bnabn (cid:19) − λ x (cid:18) a + bab (cid:19) = 1 n .Hence both intervals contain some z with nz ≡ t mod 1 . Now apply theappropriate part of Lemma 4.
Remark . We point out that the calculations above show that if forsome λ x ∈ (0 , /
2) we let E = n ( a + b ) λ x + ab ab + an + bn and z j = z j ( E, x ) for j = 1 , , ,
4, then the intervals [ z , z ] and [ z , z ] bothhave length 1 /n .Another important quantity for us will be(12) L n = n + ab an + bn + ab ) . When n ≡ a mod ( a + b ), then L n = E n . For all n , L n ≤ E n and clearly L n > / (2 a + 2 b ). The significance of L n is that if λ = 1 / ( a + b ) − L n , then L n = n ( a + b ) λ + ab ab + an + bn . Thus a consequence of the previous corollary is
Corollary . For n large enough, if µ { a,b } ( t , t ) ≤ / ( a + b ) − L n , then µ ( t , t , t ) ≤ L n . Proof.
Observe that for large enough n , (3 b − a ) / n ≤ L n and applythe previous corollary. Since L n ≤ E n , the greedy algorthim,as described in the previous subsection, establishes that µ ( t , t , t ) ≤ E n for any pair ( t , t ) such that µ { a,b } ( t , t ) ≤ / ( a + b ) − L n . For other( t , t ) we will make use of a ‘second best’ point, which can also be naturallyconstructed, as explained in the next lemma. XACT KRONECKER CONSTANTS OF THREE ELEMENT SETS Lemma . There are a real number x ′ and integers k ′ , k ′ such that ax ′ − ( t + k ′ ) = − ( bx ′ − ( t + k ′ )) and k x ′ ( a, b ) − ( t + k ′ , t + k ′ ) k ∞ = 1 a + b − µ { a,b } ( t , t ) . Proof.
Choose a best approximate x and integers k , k such that k x ( a, b ) − ( t + k , t + k ) k ∞ = µ { a,b } ( t , t ) . In particular, ax − ( t + k ) = − ( bx − ( t + k )), consequently, x = t + k + t + k a + b and(13) | ax − ( t + k ) | = (cid:12)(cid:12)(cid:12)(cid:12) at − bt − ( bk − ak ) a + b (cid:12)(cid:12)(cid:12)(cid:12) . This quantity is minimized when we make either the choice bk − ak = ⌊ at − bt ⌋ or the choice bk − ak = ⌊ at − bt ⌋ +1, depending on whichgives the lesser value. (Of course, we can find such integers k , k in eithercase because a, b are coprime.) Without loss of generality, assume the choice bk − ak = ⌊ at − bt ⌋ gives the minimal answer (the other case is symmetric).That means µ { a,b } ( t , t ) = ax − ( t + k ) = at − bt − [ at − bt ] a + b . Choose integers g, h such that ag − bh = 1 and put(14) k ′ = k − h and k ′ = k − g. Then bk ′ − ak ′ = ⌊ at − bt ⌋ +1 and if we take(15) x ′ = t + k ′ + t + k ′ a + b then ax ′ − ( t + k ′ ) = − ( bx ′ − ( t + k ′ )) < k x ′ ( a, b ) − ( t + k ′ , t + k ′ ) k ∞ = − ( ax ′ − ( t + k ′ ))= − (cid:18) at − bt − ( ⌊ at − bt ⌋ + 1) a + b (cid:19) = 1 a + b − µ { a,b } ( t , t ) . (16)2 K.E. HARE and L.T. RAMSEY
Remark . By construction x − x ′ = ( g + h ) / ( a + b ). The reader shouldobserve that Lemma 4 applies to this x ′ , as well as the best approximate x . Lemma . Choose n so large that / ( a + b ) − L n > ( b − a ) / n . Suppose µ { a,b } ( t , t ) = k x ( a, b ) − ( t + k , t + k ) k ∞ := λ x . Choose integers g, h such that ag − bh = 1 and define x ′ , k ′ , k ′ as in (14) and (15). Assume λ x = ax − ( t + k ) > a + b − L n . Fix E ≥ L n and define z = z ( E, x ) and z = z ( E, x ) as in Lemma 4. Put z = x ′ + n (cid:0) a + b − λ x (cid:1) − ( a + n ) Ean , z = x ′ + ( b + n ) E − n (cid:0) a + b − λ x (cid:1) bn . If there is some z with nz ≡ t mod 1 , satisfying z ≤ z ≤ z or z ≤ z ≤ z ,then µ ( t , t , t ) ≤ E. Proof.
Put λ x ′ := | ax ′ − ( t + k ′ ) | = − ( ax ′ − ( t + k ′ )). As shown in(16), λ x ′ = 1 / ( a + b ) − λ x , and since λ x = µ { a,b } ( t , t ) ≤ / (2( a + b )), itfollows that b − a n < λ x ≤ λ x ′ < L n ≤ E. In the notation of Lemma 4, z = z ( E, x ′ ) and z = z ( E, x ′ ), hence a directapplication of that lemma yields the result. We remind the reader that the numbers E n are defined to be the right handside of the formulas given in Theorem 1. Consider any ( t , t , t ) and choose x ∈ R and integers k , k such that µ { a,b } ( t , t ) = k x ( a, b ) − ( t + k , t + k ) k ∞ . Without loss of generality we can assume µ { a,b } ( t , t ) = ax − ( t + k ) = λ x (rather than − ( ax − ( t + k ))), for otherwise replace t = ( t , t , t ) by( − t , − t , − t ) , noting that µ ( t ) = µ ( − t ).If λ x ≤ / ( a + b ) − L n , then by Corollary 8, µ ( t , t , t ) ≤ L n ≤ E n .Hence we can assume λ x > / ( a + b ) − L n and we define x ′ , z , z , z and z as in Lemma 11, with E n taking the role of E .When R = a , the strategy of the proof is to check that z − z + z − z ≥ /n and that either z − z or z − z is equal to j/n for some integer XACT KRONECKER CONSTANTS OF THREE ELEMENT SETS j . Thus there is always a choice of z with nz ≡ t mod 1 and satisfying z ∈ [ z , z ] ∪ [ z , z ]. Appealing to Lemma 11 shows that µ ( t , t , t ) ≤ E n .We will outline the details for R < a and leave the other R = a cases forthe reader. First, note that, regardless of the choice of R, (17) z − z = x − x ′ + 2( a + n ) E n an − a ( a + b ) . As observed in Remark 10, x − x ′ = ( g + h ) / ( a + b ) , where g and h are integerssuch that ag − bh = 1. As a ( g + h ) = 1 + h ( a + b ) ≡ a + b ), there isan integer v such that T = g + h + v ( a + b ).Choosing integers M and M ′ such that n = M ( a + b ) + r and R = M ′ ( a + b ) + rT, and substituting in the value of E n for R < a gives theidentity z − z − n = g + ha + b − a ( a + b ) + n + a ( a + b ) − aRan ( a + b ) − n = M ( g + h ) − M ′ − rvn . This shows there is an integer j = M ( g + h ) − M ′ − rv such that z = z + j/n .A straight forward calculation gives z − z + z − z = 2( an + bn + 2 ab ) E n − nabn . It follows that z − z + z − z ≥ /n if and only if E n ≥ ( n + ab ) / (2( an + bn + 2 ab )) and this latter condition is certainly true.Now suppose that R = a . Again, using (17), taking the value of E n for R = a, and simplifying gives z − z − aE n n ( a + b ) = g + ha + b + n + aban ( a + b ) − a ( a + b ) = n ( g + h ) + bn ( a + b ) . Since a T ≡ a = R ≡ rT and T is relatively prime to a + b , we have n ≡ r ≡ a mod ( a + b ) . Hence there is an integer M ′′ such that n = M ′′ ( a + b ) + a . Using againthe identity a ( g + h ) = 1 + h ( a + b ) gives n ( g + h ) + bn ( a + b ) = M ′′ ( g + h ) n + a ( g + h ) + bn ( a + b ) = M ′′ g + M ′′ h + ah + 1 n . K.E. HARE and L.T. RAMSEY
Thus z = z + 2 aE n n ( a + b ) + sn for some integer s . In particular, z > z + s/n .We argue next that z ≤ z + s/n , which is equivalent to proving that z − z ≥ aE n / ( na + nb ). To see this, note that the definition of z and z gives z − z = ( b + n ) E n bn + ( a + n ) E n an − a + b − λ x a − a + b − λ x b . Because 1 / ( a + b ) − λ x < L n = E n we have z − z − aE n n ( a + b ) > E n · (cid:18) b + nbn + a + nan − an ( a + b ) (cid:19) − E n · (cid:18) a + 1 b (cid:19) = 2 bE n ( a + b ) n > . Therefore [ z , z + s/n ] ⊂ [ z , z ] ∪ [ z + s/n, z + s/n ]. The length, V, of [ z , z + s/n ] is the sum of the lengths of the intervals [ z , z ] and [ z + s/n, z + s/n ],but with the length of the overlap, the subinterval [ z + s/n, z ], subtracted: V = z − z + z − z − aE n n ( a + b )= 2( an + bn + 2 ab ) E n − nabn − aE n n ( a + b )= 1 n + b ( ab + n )( a + b ) n ( ab + an + bn ) > n . We can again conclude that there is an integer z ∈ [ z , z ] ∪ [ z , z ], with nz ≡ t mod 1.
3. Lower bounds on the Kronecker constants
Continue with the standard assumptions: a , b and n are positive integerswith a < b < n , gcd( a, b ) = 1 and n sufficiently large. Recall that E n is thevalue claimed by Theorem 1 for α ( a, b, n ).For R = a , the proof that α ( a, b, n ) ≥ E n will follow from showing that E n is equal to the binary Kronecker constant, β ( a, b, n ) , since it is obvious XACT KRONECKER CONSTANTS OF THREE ELEMENT SETS α ( a, b, n ) ≥ β ( a, b, n ) for all a, b, n . Unfortunately, when R = a , we willsee that β ( a, b, n ) < E n . Thus a different and more direct argument will begiven to establish the specified lower bound in that case. These argumentsare very technical.The key to obtaining the needed lower bounds is that, depending on thesize of µ { a,b } ( t , t ), we can restrict the search for the best approximate x forthe ( t , t , t )-approximation problem to a small range of real numbers. Thefirst step towards this is to describe the uniqueness modulo 1 of some of theintervals used in the argument. Lemma . Suppose that for some real t and t there are real numbers x and y and integers j , j , k and k such that ax − t − j = ay − t − k = − ( bx − t − j ) = − ( by − t − k ) . Then there is an integer s such that ( y, j , k ) = ( s + x, as + j , bs + k ) . Proof.
By arguments similar to those used in the proof of Lemma 9, at − bt + aj − bj a + b = at − bt + ak − bk a + b . Consequently, a ( k − j ) = b ( k − j ). Because gcd( a, b ) = 1 there is aninteger s such that k − j = as and thus k − j = bs . It follows that y − x = k − j + k − j a + b = s . As noted in[7], there is a “toggling trick” for { , / } - valued functions. Suppose θ is defined on { n j } ⊆ Z by θ ( n j ) = θ j ∈ { , / } for all j . Then we have µ { n j } ( θ ) = µ { n j } ( e θ ) where e θ j = θ j if n j is even and e θ j = 1 / − θ j if n j isodd. With { n , n } = { a, b } , since the assumption that gcd( a, b ) = 1 impliesat least one of a or b is odd, the four binary possibilities break into pairs,which are equivalent under toggling. One of these pairs includes (0 ,
0) with µ { a,b } (0 ,
0) = 0. So computing β ( a, b ) reduces to computing just one of thefour binary possibilities.6 K.E. HARE and L.T. RAMSEY
Lemma . We have β ( a, b ) = α ( a, b ) = 1 / (2 a + 2 b ) . To be more precise, β ( a, b ) = µ { a,b } (0 , /
2) = µ { a,b } (1 / , if a, b are both odd µ { a,b } (1 / ,
0) = µ { a,b } (1 / , / if a is even, b odd µ { a,b } (0 , /
2) = µ { a,b } (1 / , / if a is odd, b even . Furthermore, if µ { a,b } ( θ , θ ) = 0 , then for all real θ , µ ( θ , θ , θ ) ≤ (3 b − a ) / (2 n ) . Proof.
Suppose that a is odd and that θ ( a ) = 0 = θ , θ ( b ) = 1 / θ .From [5], we know there are a real number x and integers k , k such that µ { a,b } ( θ , θ ) = k ( θ , θ ) − x ( a, b ) + ( k , k ) k ∞ . As in (13), µ { a,b } (0 , /
2) = | aθ − bθ + ak − bk | a + b = | (2 k + 1) a − k b | a + 2 b . Because a is odd, it follows that β ( a, b ) ≥ µ { a,b } (0 , / ≥ / (2 a + 2 b ).But µ { a,b } (0 ,
0) = µ { a,b } (1 / , /
2) = 0 , and we always have, µ { a,b } (0 , / ≤ α ( a, b ) = 1 / (2 a + 2 b ) ([5]), hence β ( a, b ) = µ { a,b } (0 , /
2) = α ( a, b ) = 1 / (2 a +2 b ).The case a is even is similar.The last claim of the lemma follows from Lemma 3. Notation and Elementary Observations:
For the remainder of thissubsection we will set( t , t ) = (cid:26) (1 / ,
0) if b is odd(0 , /
2) if b is even . It is easy to see from the toggling trick and the previous lemma that β ( a, b, n ) = max { µ ( t , t , t ) : t = 0 , / } . To calculate µ ( t , t , t ) , we will first exhibit a specific interval that containsa best approximate to t = ( t , t , t ). This is done in Lemma 14. In Prop.15 we use ideas from the proof of Lemma 4 to calculate µ ( t , t , t ) by mini-mizing kh t − y ( a, b, n ) ik ∞ over y in this interval. The final step is to find themaximum value of µ ( t , t , t ) over t ∈ { , / } . XACT KRONECKER CONSTANTS OF THREE ELEMENT SETS b is odd, then gcd(2 a, b ) = 1 , so there are integers G, H such that2 aG − bH = 1. Notice that H is odd. Set g = 2 G and h = H . Put k = ( H − / k = G. If, instead, b is even there are integers H and (odd) G such that aG − b H = 1.Set g = G and h = 2 H , k = H and k = ( G − / . In either case, ag − bh = 1. Put(18) x = g + h a + b ) . As usual, set λ x := ax − t − k . The reader should check that λ x = − ( bx − t − k ) = 1 / (2 a + 2 b ). For z ∈ R , define(19) y z = ( x + | nz − nx |− λ x a + n if z ≥ xx − | nz − nx |− λ x b + n if z < x Observe that the expression, y z , has the form x ± δ that appears in the leadup to the proof of Lemma 4.Since λ x = 1 / (2( a + b )),(20) n ( a + b ) λ x + ab ab + an + bn = n + 2 ab an + bn + 2 ab ) := E .Recall that E was an important number in the upper bound argument.Let z = z ( E, x ) and z = z ( E, x ) be as defined in Lemma 4. Our nextstep is to show that we can find a best approximate to ( t , t , t ) in [ y z , y z ]. Lemma . Assume the notation is as above. For each choice of t ∈{ , / } there is some y ∈ [ y z , y z ] and integer k such that µ ( t , t , t ) = k ( t , t , t ) − y ( a, b, n ) + ( k , k , k ) k ∞ . Proof.
Here
E > / (2 a + 2 b ) = λ x = ax − t − k >
0. For n sufficientlylarge we have λ x ≥ ( b − a ) / (2 n ), hence we may use Lemma 4.The choice of E ensures that z − z = 1 /n (see Remark 7). Thus for eachreal t there is an integer k such that z = ( t + k ) /n belongs to the interval8 K.E. HARE and L.T. RAMSEY [ z , z ]. Consequently, an application of Lemma 4 implies µ ( t , t , t ) ≤ E .Furthermore, the proof of that lemma shows that if ( t + k ) /n is z or z ,then E = k ( t , t , t ) − y z ( a, b, n ) + ( k , k , k ) k ∞ . As noted in [5], there is a real number u and an integer vector k such that µ ( t , t , t ) = k ( t , t , t ) − u ( a, b, n )+ k k ∞ . Because µ ( t , t , t ) ≤ α ( a, b, n ) < /
2, the vector k = ( j, k, m ) where j is the unique integer nearest to au − t ,etc.Case 1: Suppose that ( t + j ) /a ≤ ( t + k ) /b . One can easily check thatthere is an integer H ≥ t + kb − t + ja = λ x ( a + b ) + Hab .
Set y = ( t + t + j + k ) / ( a + b ) so that ay − t − j = − ( by − t − k ) = at − bt + ak − bja + b = λ x + Ha + b . First, suppose that H ≥
1. Since λ x >
0, if u ≥ y then µ ( t , t , t ) ≥ h au − t i = au − t − j ≥ ay − t − j ≥ a + b , while if u ≤ y , then µ ( t , t , t ) ≥ h bu − t i = t + k − bu ≥ t + k − by ≥ a + b . However, for n large enough, µ ( t , t , t ) ≤ α ( a, b, n ) < / ( a + b ). Thiscontradiction forces H = 0.With H = 0, the real number y meets the hypotheses of Lemma 12, hencethere is an integer s such that ( y, j, k ) = ( x + s, as + k , bs + k ).If u > y z + s , then because y z > x we have u > x + s > s + ( t + j ) /a =( t + j ) /a . From the proof of Lemma 4 we see that h au − t i = au − t − j = a ( u − s ) − t − k > ay z − t − k = E .
But, as observed in the second paragraph, µ ( t , t , t ) ≤ E , so this gives acontradiction. XACT KRONECKER CONSTANTS OF THREE ELEMENT SETS u < y z + s , then u < ( t + k ) /b and h bu − t i > k + t − by z = E . Again, this contradicts the first paragraph.Therefore u − s ∈ [ y z , y z ] and it has the same approximation propertiesfor ( t , t , t ) as does u , hence the lemma holds.Case 2: Otherwise, ( t + j ) /a > ( t + k ) /b . In this case, there is an integer H ≥ t + ja − t + kb = H − ( a + b ) λ x ab = 2 H − ab . Let y = ( t + t + j + k ) / ( a + b ). The reader can check that by − t − k = − ( ay − t − j ) = 2 H − a + 2 b . If H ≥ u ≥ y, then u > ( t + k ) /b and thus h bu − t i = | bu − t − k | = bu − t − k ≥ by − t − k ≥ a + 2 b . If, instead, u ≤ y , then u < ( t + j ) /a and thus h au − t i = | au − t − j | = j + t − au ≥ j + t − ay ≥ a + 2 b . However, for n large enough, µ ( t , t , t ) ≤ α ( a, b, n ) < / (2 a + 2 b ), thus H = 1.Now we use the fact that t ∈ { , / } . Clearly µ ( t , t , t ) = k ( − t , − t , − t ) − ( − u )( a, b, n ) + ( − j, − k, − m ) k ∞ . When t = 0 let j ′ = − j , otherwise let j ′ = − j −
1. In either case, − t − j = t + j ′ , so − t − ( − u ) a − j = t − ( − u ) a + j ′ . Define k ′ and m ′ similarly.Again, in either case − t − k = t + k ′ , so ( t + j ′ ) /a < ( t + k ′ ) /b . Moreover, µ ( t , t , t ) = k ( t , t , t ) − ( − u )( a, b, n ) + ( j ′ , k ′ , m ′ ) k ∞ . Therefore − u, with the integers j ′ , k ′ and m ′ , meets the conditions of thefirst part of this proof, where the conclusion of the lemma has already beenestablished.With these technical results, we can now calculate µ ( t , t , t ).0 K.E. HARE and L.T. RAMSEY
Proposition . Let n ≡ r mod (2 a + 2 b ) , with r ∈ [0 , a + 2 b ) . Let ( g + h ) r ≡ S mod (2 a + 2 b ) with S ∈ [0 , a + 2 b ) . Then µ ( t , t ,
0) = a + b ) if S = 0 , , a + 2 b − n + bS a + b )( b + n ) if ≤ S ≤ an + 2 a + 2 ab − aS a + b )( a + n ) if a < S ≤ a + 2 b − and µ ( t , t , /
2) = a + b ) if S = a + b, a + b ± n + a + ab − aS a + b )( a + n ) if ≤ S < a + b − n − ab − b + bS a + b )( b + n ) if a + b + 1 < S ≤ a + bn + 3 a + 3 ab − aS a + b )( a + n ) if a + b < S < a + 2 b . Proof.
To begin, we observe that there are integers M and N such that n = M (2 a + 2 b ) + r and r ( g + h ) = N (2 a + 2 b ) + S . As x = ( g + h ) / (2 a + 2 b ),we have nx = M ( g + h ) + N + S/ (2 a + 2 b ) . Let E , z , z , k , k and y z be as defined in the preamble to Lemma14. Recall that z < x < z and the choice of E ensures that z − z =1 /n . Also, ( t + k ) /a < x < ( t + k ) /b . We note that if z ≤ x , then z ≤ y z ≤ x and, further, that the mapping z y z is strictly decreasing for z ≤ x − / (2 n ( a + b )). Conversely, if z ≥ x , then x ≤ y z ≤ z and the mapping z y z is strictly increasing for z ≥ x + 1 / (2 n ( a + b )).For n sufficiently large, we have ( t + k ) /a ≤ z and z ≤ ( t + k ) /b .Consequently, for all u ∈ [ z , z ] (and hence for all u ∈ [ y z , y z ]), au − t − k ≥ t + k − bu ≥ . Moreover, for u < x , au − t − k < ax − t − k = 12 a + 2 b = t + k − bx < t + k − bu. XACT KRONECKER CONSTANTS OF THREE ELEMENT SETS u > x we have au − t − k > / (2 a + 2 b ) > t + k − bu .We will assume t = 1 /
2. The case t = 0 is similar and will be discussedbriefly at the end of the proof.By Lemma 14, there is a real number u ∈ [ y z , y z ] and an integer k suchthat µ ( t , t , t ) = k ( t , t , / − u ( a, b, n ) + ( k , k , k ) k ∞ . As proven in [5], at least two of t − ua + k , t − ub + k and 1 / − un + k have opposite signs and absolute values equal to µ ( t , t , / au − t − k = t + k − ub and µ ( t , t , t ) = au − t − k ;Type 2 : au − t − k = 1 / k − un , µ ( t , t , t ) = au − t − k and nottype 1;Type 3 : un − / − k = t + k − ub , µ ( t , t , t ) = t − ub + k and nottype 1 or 2.We begin with Type 1. The remarks above imply that u = x , µ ( t , t , t ) =1 / (2 a + 2 b ), and | / k − nx | ≤ / (2 a + 2 b ). Also, k is the nearest integerto nx − /
2. As(21) nx − / M ( g + h ) + N + S − a − b a + 2 b , this forces S to be a + b − a + b , or a + b + 1.Conversely, if S is one of these three values, then letting u = x and k = M ( g + h ) + N gives us a Type 1 case.Next, suppose we are in the Type 2 situation. If u < x , then we wouldhave t + k − bu > au − t − k = µ ( t , t , / u ≥ x .Because au − t − k > , we have nu − t − k <
0. Hence u < ( t + k ) /n and k > nu − t ≥ nx − /
2, so that k ≥ ⌈ nx − / ⌉ .Let z = ( t + k ) /n . If z ≤ x + 1 / (2 n ( a + b )), then | nz − nx | = | k + 12 − nx | ≤ a + b ) . That would imply S = a + b ± a + b and thus we would be in Type 1.So z > x + 1 / (2 n ( a + b )) and because z > x we have u = x + δ = y z , asseen in the proof of Lemma 4 (see especially Remark 5). Since u ≤ y z , we2 K.E. HARE and L.T. RAMSEY know that z ≤ z < x + 1 /n . Also, from the proof of Lemma 4, we have(22) µ ( t , t , t ) = au − t − k = nλ x + a | nz − nx | a + n . Subcase 1: S ≤ a + b . If k > ⌈ nx − / ⌉ , then being an integer, k ≥ ⌈ nx − / ⌉ + 1 and thus z = ( k + t ) /n ≥ ( nx − / / /n = x + 1 /n > z . As this is impossible, k = ⌈ nx − / ⌉ , consequently (21) implies k = M ( g + h ) + N . Hence nz − nx = M ( g + h ) + N + 12 − [ M ( g + h ) + N + S a + b ) ] = a + b − S a + b ) . In this case, applying (22) and simplifying yields, µ ( t , t , t ) = n − aS + a ( a + b )2( a + b )( a + n ) . Using the formula nz − nx = (( a + n ) E − nλ x ) /a , allows one to show that2 b − a + b ) < n ( z − x ) < b a + b ) . Thus, having z ≤ z is equivalent to a + b − S < b and hence a − b < S .Subcase 2: S > a + b . Again, if k > ⌈ nx − / ⌉ , then z = k + t n ≥ nx − / / n = x + 1 n > z . So k = ⌈ nx − / ⌉ = M ( g + h ) + N + 1 and therefore nz − nx = M ( g + h )+ N +1+ 12 − (cid:20) M ( g + h ) + N + S a + 2 b (cid:21) = 3 a + 3 b − S a + 2 b . This gives us µ ( t , t , t ) = n a +2 b + a (3 a +3 b − S )2 a +2 b a + n = n − aS + 3 a ( a + b )2( a + b )( a + n ) . XACT KRONECKER CONSTANTS OF THREE ELEMENT SETS z ≤ z is equivalent to 3 a + 3 b − S < b or 3 a + b < S .Thus Type 2 implies 0 ≤ S < a + b − a + b < S < a + 2 b .Finally, assume we are in Type 3. In this case u ≤ x for otherwise au − t − k > t + k − bu = µ ( t , t , / t + k − bu > t + k − nu <
0, so k ≤ nu − / ≤ nx − / z = ( t + k ) /n gives z < u ≤ x . If z ≥ x − / (2 n ( a + b )) , then | nx − k − t | ≤ / (2 a + 2 b ). But this can happen only in Type 1.So z < x − / (2 n ( a + b )). Again, the proof of Lemma 4 implies u = y z and µ ( t , t , t ) = | y z b − ( t + k ) | = | y z n − ( t + k ) | = λ x + b ( | nx − nz | − λ x ) b + n . Subcase 1: S ≥ a + b . If k < ⌊ nx − / ⌋ , then z = k + t n ≤ nx − / / − n = x − n < z .But this is impossible since u ≥ y z implies z ≥ z > x − /n . Hence nx − nz = M ( g + h ) + N + S a + 2 b − (cid:20) M ( g + h ) + N + 12 (cid:21) = S − a − b a + 2 b and therefore µ ( t , t , t ) = n a +2 b + b ( S − a − b )2 a +2 b b + n = n + bS − b ( a + b )2( a + b )( b + n ) . The condition that z ≥ z is equivalent to nx − nz ≤ nx − nz , which in turnis equivalent to S − a − b ≤ a , and thus to S ≤ a + b . Excluding Type 1,we have a + b + 1 < S ≤ a + b .Subcase 2: S < a + b . Similar arguments to above show that k = ⌊ nx − / ⌋ = M ( g + h ) + N −
1. Thus nx − nz = M ( g + h ) + N + S a + 2 b − (cid:20) M ( g + h ) + N − (cid:21) = S + a + b a + 2 b . Here, having z ≥ z is equivalent to S + a + b ≤ a and thus S ≤ a − b , whichis not possible because S ≥ S in the appropriate categories implies the desired result.The arguments are similar when t = 0: Type 1 arises when S = 0 , a + 2 b −
1, Type 2 when 2 a < S ≤ a + 2 b − ≤ S ≤ a. K.E. HARE and L.T. RAMSEY
Corollary . (i) If n a mod ( a + b ) , then β ( a, b, n ) = E n .(ii) If n ≡ a mod ( a + b ) , then β ( a, b, n ) = n + ab a + b )( a + n ) < E n . Proof.
This is just a matter of checking which is greater, µ ( t , t ,
0) or µ ( t , t , / n a mod ( a + b ) (i.e. R = a )now follows immediately. Corollary . If n a mod ( a + b ) , then α ( a, b, n ) = β ( a, b, n ) = E n . Proof.
We have already seen that α ( a, b, n ) ≤ E n . Obviously, α ( a, b, n ) ≥ β ( a, b, n ) = E n , hence we have the equalities if R = a. n ≡ a mod ( a + b ) . As with n a mod ( a + b ), we show that there is (up to mod 1) oneinterval in which to search for the optimal approximation point. However,in this case it will not suffice to consider only t ∈ { , / } as the angularKronecker constant is greater than the binary Kronecker constant.Recall that for such n , E n = L n where L n was defined in (12). Lemma . Assume n ≡ a mod ( a + b ) ( R = a ) and that the real num-bers x , t , t and integers k and k satisfy λ x := ax − t − k = − ( bx − t − k ) = 1 a + b − L n . Define z = z ( L n , x ) and z = z ( L n , x ) as in Lemma 4. For z ∈ [ z , z ] ,define y z as in (19). Then, for all real t , there is some y ∈ [ y z , y z ] andinteger k such that µ ( t , t , t ) = k ( t + k , t + k , t + k ) − y ( a, b, n ) k ∞ . Proof.
Since λ x = 1 / ( a + b ) − L n , we have z − z = 1 /n (see Remark 7and the following discussion). Thus, given any t , there is an integer k suchthat z = ( t + k ) /n ∈ [ z , z ]. Since lim n L n = 1 / (2 a + 2 b ) , for large enough n we have λ x ≥ ( b − a ) / (2 n ), thus the proof of Lemma 4 shows that(23) µ ( t , t , t ) ≤ k ( t + k , t + k , t + k ) − y z ( a, b, n ) k ∞ ≤ L n . XACT KRONECKER CONSTANTS OF THREE ELEMENT SETS u ∈ R and integers j, k, m such that µ ( t , t , t ) = k ( t , t , t ) − u ( a, b, n ) + ( j, k, m ) k ∞ .One can check that λ x = at − bt + ak − bk a + b .Consequently, if ( t + j ) /a ≤ ( t + k ) /b, then there is a non-negative integer H such that t + kb − t + ja = H + ( a + b ) λ x ab . If H ≥
1, then arguments similar to those used as in Lemma 14 show that µ ( t , t , t ) ≥ / ( a + b ). Hence H = 0. But then one can argue, as inLemma 14, that there is an integer s such that if u − s / ∈ [ y z , y z ] , theneither h au − t i > L n or h bu − t i > L n . Since we know from (23) that µ ( t , t , t ) ≤ L n , this is a contradiction.If, instead, ( t + k ) /b < ( t + j ) /a , then there is an integer H ≥ t + ja − t + kb = H − ( a + b ) λ x ab . Let y = ( t + t + j + k ) / ( a + b ). As by − t − k = − ( ay − t − j ) = bt − at + bj − aka + b = Ha + b − λ x , it follows that if u ≥ y , then h bu − t i ≥ H/ ( a + b ) − λ x , while if u ≤ y then h au − t i ≥ H/ ( a + b ) − λ x . Hence µ ( t , t , t ) ≥ a + b − λ x = L n But we already know that µ ( t , t , t ) ≤ L n and this approximation accuracycan be achieved by some y ∈ [ y z , y z ] . Proposition . α ( a, b, n ) ≥ L n for all n sufficiently large. Proof.
Because a and b are positive integers, the function ( t , t ) ( at − bt ) / ( a + b ) is onto R from R . In particular, there are real numberssuch that ( at − bt ) / ( a + b ) = 1 / ( a + b ) − L n . K.E. HARE and L.T. RAMSEY
Set k = k = 0 and x = ( t + t + k + k ) / ( a + b ). Then(24) λ x := ax − t − k = − ( bx − t − k ) = at − bt + ak − bk a + b = 1 a + b − L n , so x , t , t , k and k satisfy the hypotheses of the Lemma 18. Let z and z be provided by that lemma. Let t = nz . By Lemma 18 there is a real u ∈ [ y z , y z ] and integer k such that µ ( t , t , t ) = k ( t + k , t + k , t + k ) − u ( a, b, n ) k ∞ . Because µ ( t , t , t ) ≤ α ( a, b, n ) < /
2, we know that k is the unique nearestinteger to t − ua , the same for k and t − bu , and for k and t − nu . Bythe proof of Lemma 4, there is an integer k such that the real number y z satisfies L n = ay z − t − k = − ( ny z − t − k ) and z = ( t + k ) /n. By our choice of t as nz , we know that k = 0.Suppose z − / (2 n ) ≤ u < y z . Then 1 / ≥ t − nu ≥ h t − nu i = t − nu . Consequently, h t − nu i = t − nu > t − ny z = L n ≥ µ ( t , t , t ) . This contradiction excludes u from [ z − / (2 n ) , y z ).Suppose y z < u ≤ z − / (2 n ). As noted in the proof of Lemma 4,we have y z ≥ z . Furthermore, we have 1 / ≤ t − nu < h t − nu i = 1 − t + nu . Using the fact that z = z − /n and some detailsfrom the proof of Lemma 4 gives the inequalities h t − nu i = 1 − t + nu > − nz + ny z = 1 − ( nz + 1) + ny z ≥ L n ≥ µ ( t , t , t ) . This contradiction excludes u from ( y z , z − / (2 n )].Thus u = y z or u = y z . If u = y z , then k = k = 0 and µ ( t , t , t ) = L n .Suppose u = y z . Here t = nz = nz + 1 and thus z = ( t − /n . By theproof of Lemma 4, L n = t + k − by z = ny z − t − ( −
1) = ny z − t + 1 ≥ | t + k − ay z | . For n large enough, L n < / ( a + b ) < /
2, thus k = − L n = k ( t + k , t + k , t + k ) − y z ( a, b, n ) k ∞ = µ ( t , t , t ) . XACT KRONECKER CONSTANTS OF THREE ELEMENT SETS Corollary . If n ≡ a mod ( a + b ) , then α ( a, b, n ) = L n . This completes the proof of Theorem 1.
Remark . ¿From the proof of Prop. 19 one can find rational t , t , t ,depending only on a, b, n , so that the bound µ ( t , t , t ) is optimal. Indeed,the following choice will work: Let t = 0, t = a + ba (cid:18) a + b − L n (cid:19) and t = n ( a + n )( n + ab )2 an ( an + bn + ab ) .We note that when x = t / ( a + b ), then λ x satisfies the identities in (24) andsimplifying gives z := x + ( a + n ) L n − nλ x an = ( a + n )( n + ab )2 an ( an + bn + ab ) . Hence t = nz , so this choice of t , t , t satisfies all the requirements of theproof of the proposition. References [1] J. Galindo and S. Hernandez, The concept of boundedness and the Bohrcompactification of a MAP abelian group,
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