A Khintchine-type theorem and solutions to linear equations in Piatetski-Shapiro sequences
aa r X i v : . [ m a t h . N T ] S e p A PERTURBED KHINTCHINE-TYPE THEOREM ANDSOLUTIONS TO LINEAR EQUATIONS IN PIATETSKI-SHAPIROSEQUENCES
DANIEL GLASSCOCK
Abstract.
Our main result concerns a perturbation of a classic theorem ofKhintchine in Diophantine approximation. We give sufficient conditions on asequence of positive real numbers ( ψ n ) n ∈ N and differentiable functions ( ϕ n : J → R ) n ∈ N so that for Lebesgue-a.e. θ ∈ J , the inequality k nθ + ϕ n ( θ ) k ≤ ψ n has infinitely many solutions. The main novelty is that the magnitude of theperturbation | ϕ n ( θ ) | is allowed to exceed ψ n , changing the usual “shrinkingtargets” problem into a “shifting targets” problem. As an application of themain result, we prove that if the linear equation y = ax + b , a, b ∈ R , hasinfinitely many solutions in N , then for Lebesgue-a.e. α >
1, it has infinitelymany or finitely many solutions of the form ⌊ n α ⌋ according as α < α > Introduction
Denote by { x } the fractional part of x ∈ R and by k x k the distance from x to theintegers. The main result in this paper concerns solutions in the positive integers n ∈ N to the system ((cid:13)(cid:13) θn + ϕ n ( θ ) (cid:13)(cid:13) ≤ ψ n { ρ n ( θ ) } ∈ I , (1)where ( ψ n ) n ∈ N is a sequence of positive real numbers, I is an interval in [0 , ϕ n ) n ∈ N and ( ρ n ) n ∈ N are sequences of differentiable functions on an interval J ⊆ R ,and θ ∈ J . More precisely, we show that under the right assumptions on thesequantities, the system in (1) has infinitely many solutions for Lebesgue-almostevery θ ∈ J .Because some of the assumptions are quite technical, we will state only thefollowing special case of the main theorem here in the introduction. The theoremis stated in its full generality in Section 3. Theorem 1 (Special case) . Let < σ < , and put ψ n = 1 /n σ . Let J ⊆ R be anon-empty, open interval. Let ( ϕ n ) n ∈ N and ( ρ n ) n ∈ N be sequences of C functionson J satisfying:(1) sup n ∈ N k ϕ n k J, ∞ < ∞ ;(2) (cid:0) k ϕ ′ n k J, ∞ (cid:1) n ∈ N is eventually non-increasing;(3) there exists ǫ > such that lim n →∞ k ϕ ′ n k J, ∞ n σ − ǫ = 0 ; and(4) lim n →∞ k ρ ′ n k J, ∞ n − (1+ σ ) = 0 . Mathematics Subject Classification.
Key words and phrases. metric Diophantine approximation, perturbed Khintchine-type theo-rem, Piatetski-Shapiro sequences, solutions to linear equations.
In addition, suppose that ρ n satisfies an equidistribution condition. Let I ⊆ [0 , be a non-empty interval. For Lebesgue-a.e. θ ∈ J , the system in (1) has infinitelymany solutions. Theorem 1 is best contextualized as a “perturbed” and “twisted” variant ofKhintchine’s classic result in Diophantine approximation [4]; see also [3, Theorem2.2]. Under the assumptions that ( nψ n ) n ∈ N is non-increasing and that P ∞ n =1 ψ n = ∞ , Khintchine proved that the inequality k θn k ≤ ψ n has infinitely many solutionsfor a.e. θ ∈ R . Interpreted in the language of dynamics, there are infinitely manytimes n ∈ N for which the point 0 under the rotation x x + θ on the torus R / Z lands into the “shrinking target” given by ( − ψ n , ψ n ).The quantity ϕ n ( θ ) signifies a perturbation of this rotation. In the case that themagnitude of the perturbation | ϕ n ( θ ) | is less than the accuracy of the approximation ψ n , a simple application of the triangle inequality eliminates the perturbation andreduces the inequality in (1) to one covered by Khintchine’s theorem. This is oneof the primary strengths of our main result: the magnitude of the perturbation isallowed to exceed ψ n . Dynamically, the targets we consider take the form ( − ϕ n ( θ ) − ψ n , − ϕ n ( θ ) + ψ n ); they are shrinking in length but no longer nested, leading to a“shifting targets” problem.Of course, solving such a shifting targets problem would be impossible withoutstrong restrictions on the nature of the perturbation. Indeed, with no restrictions,one could define ϕ n ( θ ) = − θn + ξ n ( θ ) for an arbitrary function ξ n : J → R andprove that there are infinitely many solutions to the inequality k ξ n ( θ ) k ≤ ψ n . Thereare few theorems to the author’s knowledge that supply sufficient conditions on asequence of functions ( ξ n ) n ∈ N to recover such a general result; for one such result,see [1]. In the case that ψ n = 1 /n σ , conditions (1), (2), and (3) are sufficientrestrictions on the perturbation to solve the shifting targets problem.The twist to this shifting targets problem is provided by the second condition in(1). The inclusion of this condition is primarily motivated by our main applicationand does not cause a great deal of added difficulty in the proof of Theorem 1.Setting I = [0 ,
1) in the statement of the theorem allows one to entirely disregardthis twist, and it is the author’s belief that the resulting theorem remains novel andpotentially useful outside the scope of this work.A illustrative example to which Theorem 1 applies is gotten by putting ψ n =1 /n σ and ϕ n ( θ ) = sin( n κ θ ) /n δ where 0 < δ < min( σ, κ ) and κ + 2 σ − δ <
1. (Take,for example, σ = κ = 1 / δ = 1 / ϕ n oscillates,and its magnitude is greater than ψ n . The proof of Theorem 1 makes use of themonotonicity of k ϕ ′ n k J, ∞ and its decay to find solutions to (1).At the heart of the proof is Lemma 8, a result on the number of solutions to aperturbed version of the inequality | qr − ps | ≤ L , one that arises frequently in thetheory. Key in counting solutions to this Diophantine inequality is the fact that thequantities involved are integers, making it amenable to techniques in basic numbertheory. The perturbed inequality no longer has this feature; counting its solutionsis the main technical difficulty overcome in this work.The primary motivation for Theorem 1 came from the desire to improve onthe main result in [2] concerning solutions to linear equations in Piatetski-Shapirosequences. A Piatetski-Shapiro sequence is a sequence of the form ( ⌊ n α ⌋ ) n ∈ N for Condition (C1) from the statement of Theorem 1 in its full generality in Section 3.
KHINTCHINE-TYPE THEOREM AND LINEAR EQUATIONS IN P-S SEQUENCES 3 non-integral α > ⌊ x ⌋ denotes the integer part (or floor) of x ∈ R . Theimage of n
7→ ⌊ n α ⌋ is denoted by PS( α ). We will say that the linear equation y = ax + b, a, b ∈ R (2)is solvable in PS( α ) if there are infinitely many distinct pairs ( x, y ) ∈ PS( α ) × PS( α )satisfying (2), and unsolvable otherwise. This terminology extends as expected tosolving equations and systems of equations in other subsets of N . Theorem 2.
Suppose a, b ∈ R , a
6∈ { , } , are such that (2) is solvable in N . ForLebesgue-a.e. α > , the equation (2) is solvable or unsolvable in PS ( α ) accordingas α < or α > . This theorem strengthens the main result in [2] by removing a restriction on a and b ; it is Theorem 1 that allows us to overcome that restriction. Theorem 2brings the the main result in [2] to its proper conclusion; the reader is encouragedto consult that paper for the motivation on finding solutions to linear equations andmore general combinatorial structure in Piatetski-Shapiro sequences. The followingcorollary demonstrates how Theorem 2 allows us to further that goal. Denote by Q + the set of positive rational numbers. Corollary 3.
For Lebesgue-a.e. α > , the limiting quotient set of PS ( α ) , ∞ \ N =1 n mn (cid:12)(cid:12)(cid:12) m, n ∈ PS ( α ) ∩ [ N, ∞ ) o , (3) is equal to Q + or { } according as α < or α > . In [2, Section 5], a number of questions were posed about further combinato-rial structure in Piatetski-Shapiro sequences. Our Theorem 2 resolves the first ofthose questions, but to the author’s knowledge, the other questions remain open.The interested reader is encouraged to consult that paper for further ideas andreferences.This paper is organized as follows. We begin by establishing notation and pre-liminary lemmas in Section 2, then we prove Theorem 1 in Section 3. The mainapplication to Piatetski-Shapiro sequences, Theorem 2, and Corollary 3 are provenin Section 4. 2.
Notation and auxillary lemmas
For x ∈ R , denote the distance to the nearest integer by k x k , the fractional partby { x } , the integer part (or floor) by ⌊ x ⌋ , and the ceiling by ⌈ x ⌉ := − ⌊− x ⌋ . Denotethe Lebesgue measure on R by λ , and denote the set of those points belonging toinfinitely many of the sets in the sequence ( E n ) n ∈ N by lim sup n →∞ E n . Given twopositive-valued functions f and g , we write f ≪ a ,...,a k g or g ≫ a ,...,a k f if thereexists a constant K > a , . . . , a k for which f ( x ) ≤ Kg ( x ) for all x in the domain common to both f and g . The supremumnorm of a real-valued continuous function ϕ on an interval J is denoted by k ϕ k J, ∞ .The following five lemmas play a key role in the proof of Theorem 1. The firstthree are standard. The fourth and fifth concern solutions to the fundamentalinequality | qr − ps | ≤ L and a perturbation of it. On a first reading, it would bepossible to skip to the statement and proof of the main theorem in the next sectionand use the remainder of this section simply as a reference. D. GLASSCOCK
Lemma 4 ([3, Lemma 1.6]) . Let I ⊆ R be an interval and A ⊆ I be measurable.If there exists a δ > such that for every sub-interval I ′ ⊆ I , λ ( A ∩ I ′ ) ≥ δλ ( I ′ ) ,then A is of full measure in I : λ ( I \ A ) = 0 . Lemma 5 ([3, Lemma 2.3]) . Let ( X, B , µ ) be a measure space with µ ( X ) < ∞ . If ( G n ) n ∈ N ⊆ B is a sequence of subsets of X for which P ∞ n =1 µ ( G n ) = ∞ , then µ (cid:18) lim sup n →∞ G n (cid:19) ≥ lim sup N →∞ N X n =1 µ ( G n ) ! N X n,m =1 µ ( G n ∩ G m ) ! − . In the following lemmas, an interval J ⊆ R reduced modulo 1 means the set (cid:8) { x } (cid:12)(cid:12) x ∈ J (cid:9) ⊆ [0 , Lemma 6.
Let J ⊆ R be a non-empty, open interval, and let J ′ ⊆ [0 , be J reduced modulo 1. Let α ∈ R . For all H ∈ N for which { α, α, . . . , Hα } ∩ Z = ∅ and all L ∈ N , L X ℓ =1 (cid:2) { αℓ } ∈ J ′ (cid:3) ≤ Lλ ( J ) + 2 LH + 1 + 6 H X h =1 h k hα k . Proof.
Note that J ′ is a union of at most two intervals in [0 , { αℓ } ) Lℓ =1 for each of these intervals, using the remarks onpages 130 and 131 in [3]. (cid:3) The following lemma is similar to, but does not follow from the statement of, [3,Lemma 6.2] and concerns the number of solutions to the inequality | qr − ps | ≤ L .Three aspects of this lemma will be most important in its application to the lemmafollowing it: 1) the implicit constant in the conclusion is independent of the lengthof the interval J ; 2) the interval J is permitted to contain 0; and 3) the lack ofrestriction on s leading to the asymmetry between the conditions on r/p and s/q . Lemma 7.
Let p, Q ∈ N , p an odd prime less than Q , L ≥ , Q ⊆ {
Q, . . . , Q − } \ p Z , and J ⊆ R be a non-empty, open interval with length λ ( J ) > p − . Thenumber of tuples ( q, r, s ) ∈ Z satisfying q ∈ Q , r (cid:14) p ∈ J, | qr − ps | ≤ L is ≪ λ ( J ) L |Q| + Q (log Q ) . Proof.
Since L ≥
1, by replacing L with ⌈ L ⌉ , we may assume that L ∈ N . We wishto bound from above the cardinality of the set S = (cid:8) ( q, r, s, ℓ ) ∈ Z (cid:12)(cid:12) q ∈ Q , r (cid:14) p ∈ J, | ℓ | ≤ L, qr − ps = ℓ (cid:9) . For ( q, r, s, ℓ ) ∈ S , r ≡ qℓ (mod p ) , (4)where q denotes the positive integer less than p for which qq ≡ p ).If λ ( J ) ≥ (cid:14)
2, then for fixed q ∈ Q and | ℓ | ≤ L , the congruence (4) togetherwith r ∈ pJ ∩ Z implies that there are ≪ λ ( J ) choices for r for which ( q, r, s, ℓ ) ∈ S .Since q , r , and ℓ determine s , we have |S| ≪ λ ( J ) L |Q| .Suppose λ ( J ) < (cid:14)
2. The congruence (4) together with r (cid:14) p ∈ J gives (cid:26) qℓp (cid:27) = (cid:26) rp (cid:27) ∈ J ′ , KHINTCHINE-TYPE THEOREM AND LINEAR EQUATIONS IN P-S SEQUENCES 5 where J ′ = (cid:8) { x } (cid:12)(cid:12) x ∈ J (cid:9) ⊆ [0 ,
1) is J reduced modulo 1. It follows that if( q, r, s, ℓ ) ∈ S , then ( q, ℓ ) ∈ S ′ , where S ′ = (cid:8) ( q, ℓ ) ∈ Z (cid:12)(cid:12) q ∈ Q , | ℓ | ≤ L, (cid:8) qℓ (cid:14) p (cid:9) ∈ J ′ (cid:9) . Since λ ( J ) < (cid:14)
2, for fixed q ∈ Q and | ℓ | ≤ L , there is at most 1 choice for r forwhich there exists an s such that ( q, r, s, ℓ ) ∈ S . Since q, r, ℓ determine s , the map( q, r, s, ℓ ) ( q, ℓ ) from S into S ′ is injective, meaning |S| ≤ |S ′ | . We proceed bybounding |S ′ | from above.Setting [ expression ] to 1 if expression is true and 0 otherwise, |S ′ | = X q ∈Q L X ℓ = − L (cid:20)(cid:26) qℓp (cid:27) ∈ J ′ (cid:21) ≤ X q ∈Q L X ℓ =1 (cid:20)(cid:26) qℓp (cid:27) ∈ J ′ (cid:21) + L X ℓ =1 (cid:20)(cid:26) − qℓp (cid:27) ∈ J ′ (cid:21)! . Since |Q| ≪ Q (log Q ) and since the bound on the third term follows exactly asthe bound on the second, it suffices to show X q ∈Q L X ℓ =1 (cid:20)(cid:26) qℓp (cid:27) ∈ J ′ (cid:21) ≪ λ ( J ) L |Q| + Q (log Q ) . (5)Applying Lemma 6 with H = (cid:4) λ ( J ) − (cid:5) and α = q (cid:14) p , L X ℓ =1 (cid:20)(cid:26) qℓp (cid:27) ∈ J ′ (cid:21) ≪ λ ( J ) L + H X h =1 h (cid:13)(cid:13) hq (cid:14) p (cid:13)(cid:13) , where we used λ ( J ) ∈ ( p − , /
2) to give that 2 ≤ H < p and H − ≪ λ ( J ). Thischoice of H satisfies the conditions in Lemma 6: hq (cid:14) p / ∈ Z since q, h ∈ { , . . . , p − } .When we sum this expression over q ∈ Q , the first term on the right hand side of(5) follows immediately. Interchanging the order of summation in the second term,the second term on the right hand side of (5) would follow from X q ∈Q (cid:13)(cid:13) hq (cid:14) p (cid:13)(cid:13) ≪ Q log Q, and H X h =1 h − ≪ log Q. (6)The second inequality in (6) follows from the fact that λ ( J ) > Q − . For the firstinequality, note that for each q ∈ Q , there exists v q ∈ { , . . . ( p − / } for which (cid:13)(cid:13) hq (cid:14) p (cid:13)(cid:13) = v q (cid:14) p . Since v q = v q ′ only if q ≡ ± q ′ (mod p ), the map q v q is at most2(1 + Q (cid:14) p )-to-1. It follows that X q ∈Q (cid:13)(cid:13) hq (cid:14) p (cid:13)(cid:13) ≪ (cid:18) Qp (cid:19) p − X v =1 pv ≪ Q log Q, which completes the proof of (6) and the proof of the lemma. (cid:3) The following lemma is at the heart of Theorem 1 and concerns counting solutionsto a perturbation of the inequality | qr − ps | ≤ L . D. GLASSCOCK
Lemma 8.
Let p, Q ∈ N , p an odd prime less than Q , L ≥ , Q ⊆ {
Q, . . . , Q − } \ p Z , and J ⊆ R be a non-empty, open interval. Suppose (cid:8) ϕ n : J → R (cid:9) n ∈{ p }∪Q is a collection of C functions satisfying ω := 110 min n ∈{ p }∪Q λ ( J ) , LQ k ϕ ′ n k J, ∞ ! ≥
10 max n ∈{ p }∪Q p , LpQ , k ϕ n k J, ∞ n ! . (7) The number of tuples ( q, r, s ) ∈ Z satisfying q ∈ Q , rp , sq ∈ J, (cid:12)(cid:12) q (cid:0) r − ϕ p ( r/p ) (cid:1) − p (cid:0) s − ϕ q ( s/q ) (cid:1)(cid:12)(cid:12) ≤ L is ≪ λ ( J ) (cid:0) L |Q| + ω − Q (log Q ) (cid:1) . Proof.
Since L ≥
1, by replacing L with ⌈ L ⌉ , we may assume that L ∈ N . We wishto bound the cardinality of the set T = (cid:26) ( q, r, s ) ∈ Z (cid:12)(cid:12) q ∈ Q , rp , sq ∈ J, (cid:12)(cid:12) q (cid:0) r − ϕ p ( r/p ) (cid:1) − p (cid:0) s − ϕ q ( s/q ) (cid:1)(cid:12)(cid:12) ≤ L (cid:27) . Let { J k } k ∈ K be a collection of | K | = (cid:4) λ ( J ) ω − (cid:5) open intervals in J , each oflength ω (cid:14)
5, covering J . For each k ∈ K , denote by T k the set (cid:26) ( q, r, s ) ∈ Z (cid:12)(cid:12) q ∈ Q , rp ∈ J k , sq ∈ J, (cid:12)(cid:12) q (cid:0) r − ϕ p ( r/p ) (cid:1) − p (cid:0) s − ϕ q ( s/q ) (cid:1)(cid:12)(cid:12) ≤ L (cid:27) , and note that since s (cid:14) q is required only to be in J , T ⊆ ∪ k ∈ K T k . Therefore, itsuffices to prove that for all k ∈ K , |T k | ≪ ωL |Q| + Q (log Q ) . (8)Fix k ∈ K , and suppose ( q, r , s ), ( q, r , s ) ∈ T k . Putting r = r − r and s = s − s , we will show that ( q, r , s ) ∈ T , where T = (cid:8) ( q, r , s ) ∈ Z (cid:12)(cid:12) q ∈ Q , r (cid:14) p ∈ ( − ω, ω ) , (cid:12)(cid:12) qr − ps (cid:12)(cid:12) ≤ L (cid:9) . To see that ( q, r , s ) ∈ T , note first that since J k is an open interval of length ω (cid:14) (cid:12)(cid:12) r /p − r /p (cid:12)(cid:12) < ω (cid:14)
5, meaning r (cid:14) p ∈ ( − ω, ω ). Moreover, (cid:12)(cid:12) s /q − s /q (cid:12)(cid:12) < ω :for i = 1 , (cid:12)(cid:12)(cid:12)(cid:12) r i − ϕ p ( r i /p ) p − s i − ϕ q ( s i /q ) q (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) r i p − s i q − (cid:18) ϕ p ( r i /p ) p − ϕ q ( s i /q ) q (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ LpQ , whereby it follows from several applications of the triangle inequality and (7) that (cid:12)(cid:12)(cid:12)(cid:12) s q − s q (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12) s q − r p (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) r p − r p (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) r p − s q (cid:12)(cid:12)(cid:12)(cid:12) < LpQ + k ϕ p k J, ∞ p + k ϕ q k J, ∞ q ! + ω ≤ ω
10 + ω < ω. KHINTCHINE-TYPE THEOREM AND LINEAR EQUATIONS IN P-S SEQUENCES 7
Finally, by the triangle inequality, the MVT, and (7), (cid:12)(cid:12) qr − ps (cid:12)(cid:12) = (cid:12)(cid:12) qr − ps − ( qr − ps ) (cid:12)(cid:12) ≤ (cid:12)(cid:12) qr − ps − (cid:0) qϕ p ( r /p ) − pϕ q ( s /q ) (cid:1)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:0) qϕ p ( r /p ) − pϕ q ( s /q ) (cid:1) − (cid:0) qϕ p ( r /p ) − pϕ q ( s /q ) (cid:1)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:0) qϕ p ( r /p ) − pϕ q ( s /q ) (cid:1) − ( qr − ps ) (cid:12)(cid:12) = (cid:12)(cid:12) q (cid:0) r − ϕ p ( r /p ) (cid:1) − p (cid:0) s − ϕ q ( s /q ) (cid:1)(cid:12)(cid:12) + (cid:12)(cid:12) q (cid:0) ϕ p ( r /p ) − ϕ p ( r /p ) (cid:1) − p (cid:0) ϕ q ( s /q ) − ϕ q ( s /q ) (cid:1)(cid:12)(cid:12) + (cid:12)(cid:12) q (cid:0) r − ϕ p ( r /p ) (cid:1) − p (cid:0) s − ϕ q ( s /q ) (cid:1)(cid:12)(cid:12) ≤ L + q (cid:12)(cid:12)(cid:12)(cid:12) r p − r p (cid:12)(cid:12)(cid:12)(cid:12) k ϕ ′ p k J, ∞ + p (cid:12)(cid:12)(cid:12)(cid:12) s q − s q (cid:12)(cid:12)(cid:12)(cid:12) k ϕ ′ q k J, ∞ + L ≤ L + 3 Qω max n ∈{ p }∪Q k ϕ ′ n k J, ∞ ≤ L. This shows that ( q, r , s ) ∈ T .In each (non-empty) fiber of the map T k → Z defined by ( q, r, s ) q , fix onepoint ( q, r q , s q ) ∈ T k . The map T k → Z defined by ( q, r, s ) ( q, r − r q , s − s q ) isinjective, and by the work above, its image lies in T . It follows that |T k | ≤ |T | ,meaning that in order to show (8), it suffices to show |T | ≪ ωL |Q| + Q (log Q ) . This inequality follows by applying Lemma 7 with p , Q , and Q as they are, 3 L as L , and ( − ω, ω ) as J , noting that 2 ω > p − follows from (7). (cid:3) Proof of main theorem
In this section, we state and prove the full version of Theorem 1. In order toproperly formulate the equidistribution condition, we need the following definition.
Definition 9.
For n ∈ N , let S n ⊆ Z be finite and F n : S n → R be a finite sequenceof elements of R indexed by S n . The sequence ( F n ) n ∈ N equidistributes modulo 1 iffor all intervals I ⊆ [0 , n →∞ | S n | (cid:12)(cid:12) { m ∈ S n | { F n ( m ) } ∈ I } (cid:12)(cid:12) = λ ( I ) . We will frequently denote a finite sequence F indexed by S ⊆ Z by (cid:0) F ( m ) (cid:1) m ∈ S .To ease notation in the statement and proof of the main theorem, any sumindexed over p or q will be understood to be a sum over prime numbers. For J ⊆ R and n ∈ N , the set nJ is { nj | j ∈ J } . Theorem 1.
Let J ⊆ R be a non-empty, open interval. Let ( ψ n ) n ∈ N be a sequencein (0 , / satisfying:(A1) sup n ∈ N ∞ X ℓ =1 ψ ℓ n ℓ ψ n ! < ∞ ;(A2) the sequence (cid:0) ψ n (cid:14) n (cid:1) n ∈ N is non-increasing;(A3) there exists c > such that for all n ∈ N , ψ n < cψ n ; and(A4) ∞ X p =2 ψ p = ∞ . D. GLASSCOCK
Let (cid:0) ϕ n : J → R (cid:1) n ∈ N be a sequence of C functions satisfying:(B1) sup n ∈ N k ϕ n k J, ∞ < ∞ ;(B2) the sequence (cid:0) k ϕ ′ n k J, ∞ (cid:1) n ∈ N is eventually non-increasing;(B3) lim N →∞ P Np =2 max( ψ p , k ϕ ′ p k J, ∞ )(log p ) (cid:16)P Np =2 ψ p (cid:17) = 0 ; and(B4) lim n →∞ k ϕ ′ n k J, ∞ nψ n = 0 .Let ( ρ n : J → R ) n ∈ N be a sequence of C functions satisfying: for all propersubintervals J ′ ⊆ J ,(C1) the sequence (cid:18)(cid:18) ρ n (cid:18) m − ϕ n ( m/n ) n (cid:19)(cid:19) m ∈ nJ ′ ∩ Z (cid:19) n ∈ N equidistributes modulo ; and(C2) lim n →∞ ψ n k ρ ′ n k J ′ , ∞ n = 0 .Let I ⊆ [0 , be a non-empty interval. For Lebesgue-a.e. θ ∈ J , the system ((cid:13)(cid:13) θn + ϕ n ( θ ) (cid:13)(cid:13) ≤ ψ n { ρ n ( θ ) } ∈ I (9) is solvable. Some discussion about the long list of assumptions is in order before the proof.The A conditions place restrictions on the accuracy of the approximation ψ n .Roughly speaking, conditions (A1) and (A2) say that ψ n decreases sufficientlyrapidly, while (A3) and (A4) say that ψ n does not decrease too rapidly. It isquick to check that (A2) is a weaker monotonicity assumption than the one thatappears in Khintchine’s theorem (that nψ n is non-increasing).It is easy to check that when 0 < σ <
1, the sequence ψ n = 1 /n σ satisfies allof the A conditions and (B3) with ψ p as the maximum. The main theorem in thisspecial case is reformulated in the introduction.As was discussed in the introduction, if k ϕ n k J, ∞ ≪ ψ n , then the perturbationin the inequality in (9) can be removed without any harm by rescaling ψ n . Thus,while it is not explicitly required, the theorem is most interesting when k ϕ n k J, ∞ exceeds ψ n .The theorem is such that if I = [0 , ρ n ’s or verify thatthe C conditions hold. Proof of Theorem 1.
For brevity, we will suppress the dependence on J , ( ψ n ) n ∈ N ,( ϕ n ) n ∈ N , ( ρ n ) n ∈ N , and I in the asymptotic notation appearing in the proof.Write J = ( j , j ), and let Θ ⊆ J be the set of those θ satisfying the conclusionof the theorem. To show that Θ is of full measure, it suffices by Lemma 4 to showthat there exists a δ > j < θ < θ < j , λ (Θ ∩ ( θ , θ )) ≥ δ ( θ − θ ) . (10)To this end, fix j < θ < θ < j . In what follows, the phrase “for all sufficientlylarge n ” means “for all n ≥ n ,” where n ∈ N may depend on any of the quantitiesand sequences introduced so far, including θ and θ . KHINTCHINE-TYPE THEOREM AND LINEAR EQUATIONS IN P-S SEQUENCES 9
For n ∈ N , define E n = { θ ∈ ( θ , θ ) | k θn + ϕ n ( θ ) k ≤ ψ n } ,F n = { θ ∈ ( θ , θ ) | { ρ n ( θ ) } ∈ I } . Put G n = E n ∩ F n , and note that lim sup n →∞ G n = Θ ∩ ( θ , θ ). Therefore, inorder to show (10), it suffices to prove that there exists a δ >
0, independent of θ , θ , for which λ lim sup p →∞ p prime G p ≥ δ ( θ − θ ) . (11)Passing to primes here makes parts of the later argument technically easier.To prove (11), it suffices by Lemma 5 to prove that ∞ X p =2 λ ( G p ) = ∞ (12)and that there exists a δ > θ , θ for whichlim sup N →∞ N X p =2 λ ( G p ) ! N X p,q =2 λ ( G p ∩ G q ) ! − ≥ δ ( θ − θ ) . (13)First we show (12) using Lemma 12. Fix 0 < η < min (cid:0) θ − j , j − θ , ( θ − θ ) / (cid:1) .For n ∈ N , let S n = (cid:8) m ∈ Z (cid:12)(cid:12) θ + η < m/n < θ − η (cid:9) ,T n = (cid:8) m ∈ Z (cid:12)(cid:12) θ − η < m/n < θ + η (cid:9) , and note that by the bounds on η , for n sufficiently large,( θ − θ ) n ≪ | S n | < | T n | ≪ ( θ − θ ) n. (14)To approximate the set E n by a union of intervals, define e n,m = m − ϕ n ( m (cid:14) n ) n ,E n,m = e n,m + 12 (cid:20) − ψ n n , ψ n n (cid:21) ,E ′ n,m = e n,m + 2 (cid:20) − ψ n n , ψ n n (cid:21) . It follows from the fact that ψ n < /
10, the assumptions in (B1) and (B4), thedefinition of E n , and estimates with the MVT that for n sufficiently large, the E ′ n,m ’s are disjoint and [ m ∈ S n E n,m ⊆ E n ⊆ [ m ∈ T n E ′ n,m . (15)This shows λ ( G n ) ≤ λ ( E n ) ≪ ( θ − θ ) ψ n .Let I ⊆ I be the middle third sub-interval of I . Claim 10.
For n sufficiently large and m ∈ S n , if { ρ n ( e n,m ) } ∈ I , then E n,m ⊆ F n . Proof.
Let θ ∈ E n,m . By the MVT, we see that for some ξ between θ and e n,m , (cid:12)(cid:12) ρ n ( θ ) − ρ n ( e n,m ) (cid:12)(cid:12) ≤ (cid:12)(cid:12) θ − e n,m (cid:12)(cid:12)(cid:12)(cid:12) ρ ′ n ( ξ ) (cid:12)(cid:12) ≪ ψ n n k ρ ′ n k J, ∞ . By the assumptions in (C2), the right hand side tends to zero as n → ∞ , so for n sufficiently large, (cid:12)(cid:12) ρ n ( θ ) − ρ n ( e n,m ) (cid:12)(cid:12) < λ ( I )3 . Therefore, if ρ n ( e n,m ) ∈ I , then for all θ ∈ E n,m , ρ n ( θ ) ∈ I . This implies that E n,m ⊆ F n . (cid:3) By the equidistribution assumption in (C2), for n sufficiently large, |{ m ∈ S n | { ρ n ( e n,m ) } ∈ I }|| S n | ≥ λ ( I )2 . Combining this with (14) and Claim 10, there are ≫ ( θ − θ ) n integers m ∈ S n for which E n,m ⊆ F n . It follows by the disjointness of the intervals E n,m that for n sufficiently large, λ ( G n ) ≫ ( θ − θ ) n ψ n n = ( θ − θ ) ψ n . (16)Now (12) follows by the assumption in (A4).Now we show (13) by estimating the “overlaps” between the G p ’s. First we showthat it suffices to prove that for all sufficiently large primes p (potentially dependingon θ , θ ) and for all N > p , N X q>p λ ( E p ∩ E q ) ≪ ( θ − θ ) N X q>p ψ p ψ q + max (cid:16) ψ p , k ϕ ′ p k J, ∞ (cid:17) (log p ) . (17)Indeed, suppose (17) holds for all primes p greater than some sufficiently large p ∈ N . Using the trivial bound λ ( G p ∩ G q ) ≤ λ ( E p ∩ E q ), it follows that N X p,q =2 λ ( G p ∩ G q ) ≤ N X p ≥ p q>p λ ( G p ∩ G q ) + N X p
p λ ( G q ) + N X q =2 λ ( G q ) ≪ ( θ − θ ) N X p ≥ p q>p ψ p ψ q + N X p ≥ p max (cid:16) ψ p , k ϕ ′ p k J, ∞ (cid:17) (log p ) + N X p
To show (17), note that by (15), the set E p is covered by a union of intervals E ′ p,r ,each of length 4 ψ p (cid:14) p . If p < q and E ′ p,r ∩ E ′ q,s = ∅ , then by estimating the distancebetween the midpoints of the intervals and using that ψ n /n is non-increasing (A2), | q ( r − ϕ p ( r/p )) − p ( s − ϕ q ( s/q )) | ≤ pq (cid:18) ψ p p , ψ q q (cid:19) = 4 qψ p , (18) λ (cid:0) E ′ p,r ∩ E ′ q,s (cid:1) ≤ (cid:18) ψ p p , ψ q q (cid:19) = 4 ψ q q . The left hand side of (17) is then N X q>p λ ( E p ∩ E q ) ≤ N X q>p X r ∈ T p s ∈ T q λ (cid:0) E ′ p,r ∩ E ′ q,s (cid:1) ≪ N X q>p ψ q q X r ∈ T p , s ∈ T q (18) holds . Now (17) will follow from Lemma 8 by partitioning { p, . . . , N } dyadically. Indeed,using (A2), the right hand side of the previous expression is ∞ X ℓ =0 X ℓ pp ψ p ψ q + max (cid:16) ψ p , k ϕ ′ p k J, ∞ (cid:17) (log p ) . This shows (17), completing the proof of the theorem. (cid:3) Solutions to linear equations in P-S sequences
In this section, we prove Theorem 2 and Corollary 3 with the help of Theorem1. The first step in the proof of Theorem 2 is a reduction of the problem to one inDiophantine approximation. This setting arises naturally when solving the equation a ⌊ n α ⌋ + b = ⌊ m α ⌋ for n . Theorem 11.
Let < a < , I ⊆ [0 , be a non-empty interval, and κ, c, γ ∈ R with c > and γ = 0 . For Lebesgue-a.e. α > , the system (cid:13)(cid:13)(cid:13)(cid:13) na /α + κa /α αn α − (cid:13)(cid:13)(cid:13)(cid:13) ≤ cn α − { γn α } ∈ I (22) is solvable or unsolvable in N according as α < or α > .Proof of Theorem 2 assuming Theorem 11. Since a = 1, by interchanging x and y if necessary, we may assume that | a | <
1. By assumption, a = 0 and (2) is solvablein N , and this implies that a, b ∈ Q , a = a a > , a , a ∈ N , ( a , a ) = 1 , and a b ∈ Z . Let d ∈ { , . . . , a − } be such that da ≡ − ba (mod a ), and note that for all r ∈ R , a ⌊ r ⌋ + b ∈ Z ⇐⇒ ⌊ r ⌋ ≡ d (mod a ) ⇐⇒ (cid:26) ra (cid:27) ∈ (cid:20) da , d + 1 a (cid:19) . It follows that a ⌊ n α ⌋ + b ∈ PS( α ) ⇐⇒ ∃ k ∈ N , a ⌊ n α ⌋ + b = ⌊ k α ⌋⇐⇒ ( a ⌊ n α ⌋ + b ∈ Z and ∃ k ∈ N , a ⌊ n α ⌋ + b ≤ k α < a ⌊ n α ⌋ + b + 1 ⇐⇒ (cid:26) n α a (cid:27) ∈ (cid:20) da , d + 1 a (cid:19) and J n ∩ N = ∅ , (23) KHINTCHINE-TYPE THEOREM AND LINEAR EQUATIONS IN P-S SEQUENCES 13 where, by writing ⌊ n α ⌋ = n α −{ n α } and applying the Mean Value Theorem (MVT)twice, J n = h ( a ⌊ n α ⌋ + b ) /α , ( a ⌊ n α ⌋ + b + 1) /α (cid:17) = a /α n + U n + [ L n , R n ) ,U n = bα u − /αn , u n between an α and an α + b,L n = − aα { n α } l − /αn , l n between a ⌊ n α ⌋ + b and an α + b,R n = aα (cid:18) a − { n α } (cid:19) r − /αn , r n between an α + b and a ⌊ n α ⌋ + b + 1 . Note that J n , U n , u n , L n , l n , R n , and r n all depend on α . This shows so far that (2)is solvable in PS( α ) if and only if the system in (23) is solvable in N .We proceed by showing that solutions to (22) yield solutions to (23) and viceversa when I , κ , c , and γ are chosen appropriately. To this end, for i = 1 ,
2, let A = { α > | (23) is solvable in N } ,B i = { α > | (22) is solvable in N for a, I i , κ i , c i , γ i } . To prove Theorem 2, it suffices by Theorem 11 to find I i , κ i , c i , γ i , i = 1 ,
2, forwhich B ∩ (1 , ⊆ A ⊆ B . (24)We begin with the first containment in (24). Let I ′ = [1 (cid:14) , (cid:14) I = d/a + I ′ /a , γ = 1 /a , κ = b (cid:14) a , and c be a constant depending only on a to be specifiedmomentarily.Suppose α ∈ B ∩ (1 ,
2) and that n is a solution to (22); we will show that if n is sufficiently large, then it solves the system in (23). By (22), (cid:26) n α a (cid:27) = { γ n α } ∈ I ⊆ (cid:20) da , d + 1 a (cid:19) , which is the first condition in (23). This also implies that { n α } ∈ I ′ , which whencombined with the fact that 1 (cid:14) a >
1, means { n α } >
13 and 1 a − { n α } > . Combining these inequalities with the fact that α ∈ (1 ,
2) and, for n sufficientlylarge, an α + b + 1 ≤ an α , we get − L n = aα { n α } l − /αn ≫ a n α − ,R n = aα (cid:18) a − { n α } (cid:19) r − /αn ≫ a n α − . Let c be a third of the minimum of the constants implicit in the previous twoexpressions. With this choice, J n contains an open interval centered at a /α n + U n of length 6 c (cid:14) n α − . By the triangle inequality and an application of the MVT, for n sufficiently large, (cid:12)(cid:12)(cid:12)(cid:12)(cid:13)(cid:13)(cid:13) a /α n + U n (cid:13)(cid:13)(cid:13) − (cid:13)(cid:13)(cid:13)(cid:13) a /α n + κ a /α αn α − (cid:13)(cid:13)(cid:13)(cid:13)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12) U n − κ a /α αn α − (cid:12)(cid:12)(cid:12)(cid:12) ≤ c n α − , (25) from which it follows by (22) that (cid:13)(cid:13) a /α n + U n (cid:13)(cid:13) ≤ c (cid:14) n α − . This shows that J n contains the nearest integer to a /α n + U n ; in particular, J n ∩ N = ∅ , so n solves(23).The second containment in (24) is handled similarly. Let I = [0 , γ = 1, κ = b (cid:14) a , and c be a constant depending only on a to be specified momentarily.Suppose that α ∈ A and n solves (23); we will show that n satisfies (22). The secondcondition in (22) is satisfied automatically by our choice of I . For n sufficientlylarge, a ⌊ n α ⌋ + b ≥ an α (cid:14)
2, whereby | L n | , | R n | ≤ c (cid:14) n α − , where c is chosen(depending only on a ) to satisfy both inequalities. Since J n contains an integer, itmust be that (cid:13)(cid:13) a /α n + U n (cid:13)(cid:13) ≤ c (cid:14) n α − . It follows by the triangle inequality andthe MVT just as in (25) with an upper bound of c (cid:14) n α − that n satisfies the firstcondition in (22). (cid:3) To prove Theorem 11, we first change variables under t a ( x ) = (log a x ) − toarrive at the equivalent Theorem 11 ′ . Note that when 0 < a <
1, the function t a isincreasing and infinitely differentiable on ( a, t a on ( a,
1) is non-singularwith respect to the Lebesgue measure.
Theorem 11 ′ . Let < a < , I ⊆ [0 , be a non-empty interval, and κ, c, γ ∈ R with c > and γ = 0 . For Lebesgue-a.e. a < θ < , the system (cid:13)(cid:13)(cid:13)(cid:13) nθ + κθt a ( θ ) n t a ( θ ) − (cid:13)(cid:13)(cid:13)(cid:13) ≤ cn t a ( θ ) − n γn t a ( θ ) o ∈ I (26) is solvable or unsolvable in N according as θ < √ a or θ > √ a . This change of variables reveals the form of Theorem 11 as a perturbation of therotation considered in Khintchine’s theorem with a twist. We cannot directly applyTheorem 1 because, as it stands now, the accuracy of the approximation ψ n in (26)is a function of the variable θ . We are able to eliminate this unpleasant feature bybreaking the interval ( a, √ a ) up into suitable subintervals and replacing the upperbound with a worst case bound on that subinterval.The following equidistribution lemma will help us verify condition (C1) whenapplying Theorem 1 to prove Theorem 11 ′ . Lemma 12.
Let < a < , γ ∈ R \ { } , and J = ( j , j ) be a non-empty, openinterval with closure J ⊆ ( a, √ a ) . Let (cid:0) ϕ n : J → R (cid:1) n ∈ N be a sequence of C functions such that for all i ∈ { , , , } , sup n ∈ N k ϕ ( i ) n k J, ∞ < ∞ . The sequence (cid:18)(cid:16) γn t a ( m/n − ϕ n ( m/n ) /n ) (cid:17) m ∈ nJ ∩ Z (cid:19) n ∈ N equidistributes modulo .Proof. Let N n = (cid:12)(cid:12) { m ∈ Z | m/n ∈ J } (cid:12)(cid:12) , and note that N n (cid:14)(cid:0) nλ ( J ) (cid:1) → n → ∞ .For n, h ∈ N , let f n ( x ) = 1 n (cid:18) x + ⌊ j n ⌋ − ϕ n (cid:18) x + ⌊ j n ⌋ n (cid:19)(cid:19) ,g n ( x ) = γn t a ( f n ( x )) , g n,h ( x ) = g n ( x + h ) − g n ( x ) . KHINTCHINE-TYPE THEOREM AND LINEAR EQUATIONS IN P-S SEQUENCES 15
Note that f n and g n are C functions on [1 , N n ] and that g n,h is a C functionon [1 , N n − h ]. Because the sequences of supremum norms of the derivatives ofthe ϕ n ’s are bounded, we see that for n sufficiently large, for all x ∈ [1 , N n ], | f ′ n ( x ) − n − | ≪ n − while | f ( i ) n ( x ) | ≪ n − ( i +1) for i ∈ { , } .Using this notation, we must show1 N n N n X i =1 δ { g n ( i ) } −→ λ | [0 , as n → ∞ , (27)where δ x denotes the unit point mass at the point x ∈ [0 ,
1) and convergence ofprobability measures is meant in the weak- ∗ topology.Since t a is strictly increasing on ( a, √ a ) and J ⊆ ( a, √ a ), we can fix σ , σ suchthat for all x ∈ J , 1 < σ < t a ( x ) < σ < . To handle the exponential sum estimates that follow, we will show that there existpositive constants C and C (depending on a and γ ) such that for all h ∈ N , allsufficiently large n ∈ N , and all x ∈ [1 , N n − h ], C h n − (3 − σ ) ≤ (cid:12)(cid:12) g ′′ n,h ( x ) (cid:12)(cid:12) ≤ C h n − (3 − σ ) . (28)By the MVT, g ′′ n,h ( x ) = hg ′′′ n ( ξ x ) for some ξ x ∈ ( x, x + h ), so it suffices to show thatfor all h ∈ N , all sufficiently large n ∈ N , and all x ∈ [1 , N n ], C n − (3 − σ ) ≤ | g ′′′ n ( x ) | ≤ C n − (3 − σ ) . (29)Writing g ′′′ n ( x ) explicitly reveals that g ′′′ n ( x ) = g n ( x )(log n ) (cid:18) f ′ n ( x ) f n ( x ) (cid:19) t a ( f n ( x )) − (log a ) − r n ( x )log n ! , where r ( x ) is a sum of nine terms, five of which are of the form c (log a ) i (log n ) − (3 − i ) (log f n ( x )) − j where c ∈ { , , } , i ∈ { , } and j ∈ { , , , } , and four of whichare of the form c (log a ) i (log n ) − (3 − i ) f n ( x ) j f ( j +1) n ( x ) f ′ n ( x ) − ( j +1) (log f n ( x )) − k where c ∈ { , − , − } , i, j ∈ { , } , and k ∈ { , , } . By the bounds on the derivates of f n , for sufficiently large n , | r n ( x ) | is bounded from above uniformly in x ∈ [1 , N n ].The inequality in (29) follows for n sufficiently large since for all x ∈ [1 , N n ], theterms f n ( x ) and t a ( f n ( x )) / − (log a ) are bounded from above and away from 0, n σ ≤ (cid:12)(cid:12) g n ( x )(log n ) (cid:12)(cid:12) ≤ n σ and (2 n ) − ≤ | f ′ n ( x ) | ≤ n − .To prove (27), it suffices by Weyl’s Criterion ([5], Chapter 1, Theorem 2.1) toshow that for all b ∈ Z \ { } ,1 N n N n X i =1 e (cid:0) bg n ( i ) (cid:1) −→ n → ∞ , where e ( x ) = e πix . By the van der Corput Difference Theorem ([5], Chapter 1,Theorem 3.1) and another application of Weyl’s Criterion, it suffices to prove thatfor all h ∈ N and for all b ∈ Z \ { } ,1 N n − h N n − h X i =1 e (cid:0) bg n,h ( i ) (cid:1) −→ n → ∞ . (30) An exponential sum estimate ([5], Chapter 1, Theorem 2.7) gives us that1 N n − h (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N n − h X i =1 e (cid:0) bg n,h ( i ) (cid:1)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:18) | b | | g ′ n,h ( N n − h ) − g ′ n,h (1) | + 2 N n − h (cid:19) p | b | δ + 3 ! , where δ = C hn − (3 − σ ) from (28). By the MVT and the upper bound from (28),we see the right hand side is bounded from above for sufficiently large n by (cid:18) | b | C h n − (3 − σ ) + 2 N n − h (cid:19) n (3 − σ ) / p | b | C h + 3 ! ≪ n (3 − σ ) / n p | b | h , where the implicit constant depends on a , γ , η , and η . The limit in (30) followssince (3 − σ ) / < (cid:3) Now we can deduce Theorem 11 ′ from Theorem 1. Proof of Theorem 11 ′ . Let Θ ⊆ ( a,
1) be the set of those θ satisfying the conclusionof Theorem 11 ′ . We will show that Θ is of full Lebesgue measure by showingseparately that it has full measure in the intervals ( a, √ a ) and ( √ a, ∩ ( a, √ a ) is of full measure, we will cover ( a, √ a ) by short intervalsand apply Theorem 1 to each one. We will define ( ψ n ) n ∈ N , ( ϕ n ) n ∈ N , and ( ρ n ) n ∈ N so that every θ ∈ ( a, √ a ) for which (9) is solvable is a θ for which (26) is solvable.Note that ( a, √ a ) can be covered by open intervals of the form ( j , j ) withclosure contained in ( a, √ a ) with the property that t a ( j ) − t a ( j ) < − t a ( j ) . (31)Let J = ( j , j ) be such an interval. We will define ( ψ n ) n ∈ N , ( ϕ n ) n ∈ N , and ( ρ n ) n ∈ N and verify that the conditions of Theorem 1 hold.For n ∈ N , put ψ n = min( c, − ) n t a ( j ) − . Because 0 < t a ( j ) − <
1, it follows from the discussion just after the statementof Theorem 1 that conditions (A1) through (A4) and condition (B3) (with ψ p asthe maximum) hold for this choice of ψ n .Define ϕ n : J → R to be ϕ n ( θ ) = κθt a ( θ ) n t a ( θ ) − . Since t a is infinitely differentiable on J with derivatives bounded uniformly fromabove and away from 0, the function ϕ n is C on J and for i ∈ { , , , } , k ϕ ( i ) n k J, ∞ ≪ (log n ) i n t a ( θ ) − . (32)Condition (B1) in Theorem 1 follows from (32) with i = 0. To verify (B2), notethat ϕ ′ n ( θ ) = κn t a ( θ ) − (cid:18) log n log θ + 1 + log θ log a (cid:19) . It follows that for n sufficiently large, k ϕ ′ n k J, ∞ = | ϕ ′ n ( j ) | . This sequence is even-tually decreasing, verifying (B2). Condition (B3) with k ϕ ′ p k J, ∞ as the max followsfrom (32), a calculation, and the inequality in (31). Condition (B4) follows from acalculation and the inequality t a ( j ) − t a ( j ) <
1, which follows from the fact that1 < t a ( j ) < t a ( j ) < KHINTCHINE-TYPE THEOREM AND LINEAR EQUATIONS IN P-S SEQUENCES 17
Define ρ n : J → R to be ρ n ( θ ) = γn t a ( θ ) . Let J ′ = ( j ′ , j ′ ) be a proper subintervalof J , and note that k ρ n k J ′ , ∞ ≪ n t a ( j ′ ) log n . In order to verify (C1), we have only tonote that the conditions of Lemma 12 are met by the inequalities in (32). Condition(C2) follows from a calculation and the fact that t a ( j ′ ) < t a ( j ).It is simple to check that every θ ∈ ( a, √ a ) for which (9) is solvable is a θ forwhich (26) is solvable. Since the conditions of Theorem 1 hold, the set Θ ∩ J is offull measure. Since ( a, √ a ) was covered by such intervals J , the set Θ ∩ ( a, √ a ) isof full measure.To show that the set Θ ∩ ( √ a,
1) is of full measure, we will show that for all θ > √ a , the set ( θ , \ Θ has zero measure. Define ϕ n as above, and let σ ∈ (1 , t a ( θ ) − H n = (cid:26) θ ∈ ( θ , (cid:12)(cid:12)(cid:12)(cid:12) (cid:13)(cid:13) θn + ϕ n ( θ ) (cid:13)(cid:13) ≤ n σ (cid:27) . If θ ∈ ( θ , \ Θ, then (26) is solvable, meaning that for infinitely many n ∈ N , k θn + ϕ n ( θ ) k ≤ (cid:14) n σ . It follows that( θ , \ Θ ⊆ lim sup n →∞ H n . (33)Just as in (15), H n is covered by a union of ≪ (1 − θ ) n intervals, each of length ≪ n − ( σ +1) . Since σ > P ∞ n =1 λ ( H n ) < ∞ . By the first Borel-Cantelli Lemma,lim sup n →∞ H n has zero measure, so (33) implies that ( θ , \ Θ has zero measure. (cid:3)
Finally, we deduce Corollary 3 from Theorem 2.
Proof of Corollary 3.
Denote by Q ( α ) the limit quotient set in (3). Note that a ∈ Q ( α ) if and only if the linear equation y = ax is solvable in PS( α ). ByTheorem 2, the set \ a ∈ Q + \{ } (cid:8) α ∈ (1 , (cid:12)(cid:12) the equation y = ax is solvable in PS( α ) (cid:9) is of full measure in the interval (1 , α < Q ( α ) = Q + . On the other hand, Theorem 2 gives that the set [ a ∈ Q + \{ } (cid:8) α ∈ (2 , ∞ ) (cid:12)(cid:12) the equation y = ax is solvable in PS( α ) (cid:9) is of zero measure in (2 , ∞ ), proving that for Lebesgue-a.e. α > Q ( α ) = { } . (cid:3) References [1] J. W. S. Cassels. Some metrical theorems of Diophantine approximation. II.
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