Fourier Dimension Estimates for Sets of Exact Approximation Order: the Well-Approximable Case
FFourier Dimension Estimates for Sets of Exact ApproximationOrder: the Well-Approximable Case
Robert Fraser Reuben WheelerFebruary 4, 2021
Abstract
We obtain a Fourier dimension estimate for sets of exact approximation order introduced byBugeaud for certain approximation functions ψ . This Fourier dimension estimate implies that thesesets of exact approximation order contain normal numbers. Let E ⊂ R be a compact set. Frostman’s lemma [4] implies that the Hausdorff dimension dim H ( E ) of E is the supremum over all values of s < E supports a Borel probability measure µ satisfyingthe condition (cid:90) | (cid:98) µ ( ξ ) | | ξ | s − dξ < ∞ . This condition essentially says that | (cid:98) µ ( ξ ) | decays, in an L -average sense, like | ξ | − s/ .If, instead, we impose the condition that | (cid:98) µ ( ξ ) | decays pointwise like | ξ | − s/ for all s < s ≤
1, wesay that E has Fourier dimension at least s , written dim F ( E ) ≥ s . Clearly, we have that dim F ( E ) ≤ dim H ( E ) for every compact subset E ⊂ R . If a compact set E satisfies dim F ( E ) = dim H ( E ), then wesay that E is a compact Salem set .In fact, it is nontrivial to construct examples of Salem sets. The earliest constructions of Salem setsare random constructions, such as random cantor sets of Salem [14]. K¨orner established via a randomconstruction that the Fourier dimension of a set E ⊂ R can take any value from 0 to dim H ( E ) [10]. A classical result of Jarn`ık and Besicovitch [1] [7] concerns the Hausdorff dimension of the set of τ -approximable numbers. The τ -approximable numbers are the set E ( τ ) := { x : | x − r/q | ≤ q − τ for infinitely many pairs of integers ( r, q ). } For τ ≤
2, it is easy to see using the Dirichlet principle that E ( τ ) = R . Jarn`ık and Besicovitch showthat for τ >
2, dim H ( E ( τ )) = τ .Kaufman [9] established that, in fact, the set E ( τ ) has Fourier dimension equal to τ , implying that E ( τ ) is a Salem set. Notably, this is the first explicit non-random construction of a Salem set of Hausdorffdimension other than 0 or 1 in R .Fourier dimension calculations are of interest in metric Diophantine approximation because of acelebrated result of Davenport, Erd¨os, and Leveque [3]. This result concerns the presence of normalnumbers in subsets E ⊂ R .Specifically, Davenport, Erd¨os and Leveque show that if µ is a positive Borel probability measure,and a is a positive integer, then the sequence { a j x } ∞ n =1 is uniformly distributed modulo 1 if and only if ∞ (cid:88) N =1 N − N (cid:88) j =1 N (cid:88) k =1 (cid:98) µ ( m ( a j − a k )) < ∞ (1)for every nonzero integer m ∈ Z . If we crudely assume | (cid:98) µ ( ξ ) | ≤ | ξ | − s/ for every ξ ∈ R , the sum in k in(1) is essentially a geometric sum, so we have that the sum in j is bounded above by an m -dependent1 a r X i v : . [ m a t h . N T ] F e b onstant times N , and the sum certainly converges for all nonzero integers m , and all integers a ≥ µ is a Borel probability measure such that | (cid:98) µ ( ξ ) | ≤ | ξ | − s/ for some s >
0, then µ -almostevery point is a normal number. Note that a far weaker assumption on µ suffices to locate normalnumbers; see e.g. [11].In particular, any set E of positive Fourier dimension must contain normal numbers. It is thereforeof interest to find Fourier dimension estimates for subsets of R arising in Diophantine approximation. Ofcourse, the well-approximable numbers, being a Salem set of positive dimension, contain normal numbers.In fact, Kaufman also showed a Fourier dimension result for sets of badly -approximable numbers [8].The badly approximable numbers consist of those x ∈ R such that the partial quotients in thecontinued fraction expansion of x are bounded. Given a finite set S ⊂ N with at least two elements,we use the term S -badly-approximable numbers to refer to those real numbers x such that the partialquotients of the continued fraction expansion of x all lie in the finite set S . Kaufman [8] established that,if S is a finite set such that the Hausdorff dimension of the S -badly-approximable numbers is greaterthan 2 /
3, then the S -badly approximable numbers have positive Fourier dimension.The method used by Kaufman to estimate the Fourier dimension of the badly approximable numbersis very different from the method used to estimate the Fourier dimension of the well-approximablenumbers. For the well-approximable numbers, Kaufman’s argument relies on the cancellation of theexponential sum q (cid:88) r =0 e ( rs/q ) (2)for any integers s, q such that q does not divide s ; since s has a small number of divisors, the sum of (2)over all M/ ≤ q < M will also be small.In contrast, Kaufman’s Fourier dimension estimate [8] follows a rather different argument. This argu-ment relies on constructing a certain random measure on bounded integer sequences whose pushforwardunder the continued fraction map satisfies the relevant Fourier decay condition, which is established viaa van der Corput-type lemma.Queffelec and Ramar´e [12] improve the 2 / /
2; inparticular, this condition holds if S = { , } . Hochman and Shmerkin [5] show that, without anyHausdorff dimension assumption, the set of S -badly-approximable numbers contains normal numbersfor any finite set S ⊂ N with at least two elements. In a recent work, Sahlsten and Stevens [13]improved on all of these results by showing, without any Hausdorff dimension assumption, that the S -badly-approximable numbers have positive Fourier dimension for any finite set S ⊂ N with at leasttwo elements.Of note is that, while Kaufman’s argument for the well-approximable numbers [9] works just as wellin the inhomogeneous setting, there does not seem to be an easy way to modify Kaufman’s argumentfor the badly approximable numbers, [8], to this case. Doing so would require a satisfactory analogue ofthe continued fraction expansion for the inhomogeneous version of the badly approximable numbers. Bugeaud [2] introduced sets of exact approximation order. We will now define an inhomogeneous ana-logue. Given an approximation function ψ and a real number θ ∈ [0 , ψ, θ ) to bethe set of real numbers x satisfying the pair of conditions: (cid:12)(cid:12)(cid:12)(cid:12) x − r − θq (cid:12)(cid:12)(cid:12)(cid:12) ≤ ψ ( q ) for infinitely many pairs ( r, q ) of relatively prime integers (cid:12)(cid:12)(cid:12)(cid:12) x − r − θq (cid:12)(cid:12)(cid:12)(cid:12) ≤ ψ ( q ) − cψ ( q ) for only finitely many pairs ( r, q ) of relatively prime integers and any c > ψ,
0) for certain functions ψ . Specifically,Bugeaud considers functions ψ such that the function x ψ ( x ) is nonincreasing. Bugeaud shows that theHausdorff dimension of Exact( ψ,
0) is λ + , where λ + ( ψ ) = − lim sup x →∞ log ψ ( q )log q ≥ . The upper Hausdorff dimension bound follows trivially from the Jarn`ık-Besicovitch theorem. For thelower bound, Bugeaud considers a subset of Exact( ψ,
0) consisting of numbers whose continued fractions2ave partial quotients that typically grow very slowly, except for some exceptional partial quotients thatare larger.The main result in this paper concerns the Fourier dimension of the sets Exact( ψ, θ ). Unlike Bugeaud,we will limit ourselves to approximation functions ψ of satisfying the property that λ ( ψ ) := − lim q →∞ log ψ ( q )log q (3)exists.Observe that the set Exact( ψ, θ ) is invariant under translations by integers. Therefore, we can viewExact( ψ, θ ) as a subset of the torus [0 ,
1) in a natural way. We will use the notation Exact [0 , ( ψ, θ ) torefer to this subset of the torus.Let θ ∈ [0 ,
1) be an irrational number. We define the
Diophantine approximation exponent of θ to be the infimum over values of γ such that the equation (cid:12)(cid:12)(cid:12)(cid:12) θ − pq (cid:12)(cid:12)(cid:12)(cid:12) ≤ q − γ has only finitely many solutions for integers p and q . Note that this implies that | qθ − p | ≤ q − γ +1 also has only finitely many solutions with q (cid:54) = 0.This means that if γ is the Diophantine approximation exponent of θ , η >
0, then we have that forany integers p and q , | qθ − p | ≥ q − γ − η +1 , provided q is sufficiently large depending on η and θ . Another way of saying this is that (cid:107) qθ (cid:107) ≥ q − γ − η +1 where (cid:107) qθ (cid:107) is the distance from qθ to the nearest integer. Theorem 1.1.
Let θ ∈ [0 , be either or an irrational number with finite Diophantine exponent γ ,taking γ = 1 if θ = 0 . Let ψ ( q ) be a positive, decreasing function of q . Suppose τ = λ ( ψ ) is such that τ > γ + (cid:112) (2 + γ ) − . (4) Then dim F Exact( ψ, θ ) is positive; moreover, we have the inequality dim F Exact( ψ, θ ) ≥ α := 2( β − τ ) τ ( β − , (5) where β := γ − ( τ − . (6)One can observe that the condition (4) implies that β > τ ; this implies that the right side of (5) ispositive. We will make some quick observations about Theorem 1.1. Remark 1.2. If θ = 0 , we are able to take γ = 1 . In this case, the inequality (4) reduces to τ > √ . (7) Remark 1.3.
For a fixed γ , observe that in the regime τ → ∞ , we have that β = γ − τ + O ( τ ) .Therefore, lim τ →∞ β − τ ) τ ( β − /τ = 1 . This means that, if τ is large, Theorem 1.1 “nearly” shows the set Exact( ψ, θ ) is a Salem set. In fact, our proof yields a slightly more general Fourier dimension estimate than the one in Theorem1.1. In order to state this estimate, we will introduce sets of tight approximation order.3 efinition 1.4 (Sets of tight approximation order) . Let ψ , ψ be a pair of approximation functionswith ψ ( q ) ≤ ψ ( q ) for all q . The set Tight( ψ , ψ , θ ) consists of those real numbers x satisfying theconditions (cid:12)(cid:12)(cid:12)(cid:12) x − r − θq (cid:12)(cid:12)(cid:12)(cid:12) ≤ ψ ( q ) for infinitely many pairs ( r, q ) of relatively prime integers, (cid:12)(cid:12)(cid:12)(cid:12) x − r − θq (cid:12)(cid:12)(cid:12)(cid:12) ≤ ψ ( q ) − cψ ( q ) for only finitely many ( r, q ) of relatively prime integers and any c > . The set Exact( ψ, θ ) is the same as the set Tight( ψ, ψ, θ ). Given appropriate conditions on ψ and ψ , we are able to estimate the Fourier dimension of the set Tight( ψ , ψ , θ ). Theorem 1.5.
Let θ ∈ [0 , be either or an irrational number with finite Diophantine exponent γ ,taking γ = 1 if θ = 0 . Let ψ ( q ) and ψ ( q ) be decreasing functions with ψ ( q ) ≤ ψ ( q ) for all q and suchthat ψ ( q ) − ψ ( q ) is decreasing.Let β := γ − ( τ − . (8) Suppose τ = λ ( ψ ) and τ = λ ( ψ ) are such that β > τ . Then dim F Tight( ψ , ψ , θ ) is positive;moreover, we have the inequality dim F Tight( ψ , ψ , θ ) ≥ α := 2( β − τ ) τ ( β − . (9) The key to adapting Kaufman’s argument to the set Tight( ψ , ψ , θ ) is an elementary lemma in Dio-phantine approximation. This lemma states that if a real number x is approximable by rationals at two“fairly close” scales, then x cannot be approximable at any intermediate scale.Let β (cid:15) = γ − ( τ − − (cid:15). (10) Lemma 2.1.
Let τ > , < c < , (cid:15) > be real numbers, and let x ∈ R . Suppose ψ ( q ) is a decreasingfunction with the property that λ ( ψ ) = τ and ψ ( q ) is chosen with ψ and ψ − ψ decreasing, suchthat λ ( ψ ) = τ , and λ ( ψ ) = τ . Let θ ∈ [0 , be either or an irrational number with Diophantineapproximation exponent γ , taking γ = 1 if θ = 0 . If x satisfies the pair of inequalities: ψ ( q ) − cψ ( q ) ≤ (cid:12)(cid:12)(cid:12)(cid:12) x − p − θq (cid:12)(cid:12)(cid:12)(cid:12) ≤ ψ ( q ) ψ ( q ) − cψ ( q ) ≤ (cid:12)(cid:12)(cid:12)(cid:12) x − p − θq (cid:12)(cid:12)(cid:12)(cid:12) ≤ ψ ( q ) where Q ( (cid:15) ) < q < q < q β (cid:15) and q is prime, then x does not satisfy any inequality of the form (cid:12)(cid:12)(cid:12)(cid:12) x − p − θq (cid:12)(cid:12)(cid:12)(cid:12) < ψ ( q ) − cψ ( q ) for q < q < q . Proof.
Suppose x satisfies the conditions of Lemma 2.1. Then we have the inequality ψ ( q ) − cψ ( q ) ≤ (cid:12)(cid:12)(cid:12)(cid:12) x − p − θq (cid:12)(cid:12)(cid:12)(cid:12) ≤ ψ ( q ) (11)We will now split into two cases depending on whether θ = 0.4 ase 1 Here we consider θ = 0. In this case, (11) reduces to ψ ( q ) − cψ ( q ) ≤ (cid:12)(cid:12)(cid:12)(cid:12) x − p q (cid:12)(cid:12)(cid:12)(cid:12) ≤ ψ ( q ) . (12)Observe that if there exist p, q such that (cid:12)(cid:12)(cid:12)(cid:12) x − pq (cid:12)(cid:12)(cid:12)(cid:12) < ψ ( q ) − cψ ( q ) , (13)with q > q , then we must have pq (cid:54) = p q ; this follows from (12) and the fact that ψ − cψ is decreasing.Thus, we have the inequality (cid:12)(cid:12)(cid:12)(cid:12) pq − p q (cid:12)(cid:12)(cid:12)(cid:12) = | pq − p q | qq ≥ qq since the numerator is a nonzero integer. On the other hand, equations (12) and (13) imply by thetriangle inequality that (cid:12)(cid:12)(cid:12)(cid:12) pq − p q (cid:12)(cid:12)(cid:12)(cid:12) ≤ ψ ( q ) + ψ ( q ) − cψ ( q ) ≤ ψ ( q ) , where the last inequality follows from the fact that ψ is decreasing.Combining these inequalities gives that 1 qq ≤ ψ ( q ) . (14)Take η >
0. At this stage, we select some Q η such that log(2 ψ ( q ))log q < − τ + η for all q ≥ Q η . Then, itfollows from (14), that if q ≥ Q η , q ≥ q τ − − η . A similar argument, using the fact that pq (cid:54) = p q by the primality of q , reveals that q ≥ q τ − − η and thus, we combine to get that q ≥ q ( τ − − η ) . So if we choose η sufficiently small that ( τ − − η ) > ( τ − − (cid:15) , we must have q ≥ q β (cid:15) , as desired. Case 2
Now, we will assume θ ∈ [0 ,
1) is an irrational number with Diophantine approximation expo-nent γ . In this case, we simply observe that if (11) and an analogue of (13) both hold, then we musthave, by the triangle inequality and the fact that ψ is decreasing, that (cid:12)(cid:12)(cid:12)(cid:12) p − θq − p − θq (cid:12)(cid:12)(cid:12)(cid:12) ≤ ψ ( q ) (15)and (cid:12)(cid:12)(cid:12)(cid:12) qp − pq − ( q − q ) θqq (cid:12)(cid:12)(cid:12)(cid:12) ≤ ψ ( q ) . (16)Now, qp − pq is an integer, as is ( q − q ). Therefore, if θ has Diophantine exponent γ , then we have || ( q − q ) θ || ≥ C η,θ ( q − q ) − γ +1 − η/ ≥ q − γ +1 − η , provided that q > Q η, , where Q η, is sufficiently large.Here, || · || denotes the distance to the nearest integer. Thus, for such q , we have (cid:12)(cid:12)(cid:12)(cid:12) qp − pq − ( q − q ) θqq (cid:12)(cid:12)(cid:12)(cid:12) ≥ q − γ +1 − η qq . (17)By combining inequalities (16) and (17), we get q − γ − η q − ≤ ψ ( q ) . q ≥ Q η, , then we have that 2 ψ ( q ) ≤ q − τ + η as in Case 1. Thus, for q > max( Q η, , Q η, ), we then have q − γ − η q − ≤ q − τ + η and solving for q yields q ≥ q ( γ + η ) − ( τ − − η )1 . By a similar argument, we observe q ≥ q ( γ + η ) − ( τ − − η ) . Combining these inequalities gives q ≥ q ( γ + η ) − ( τ − − η ) . If η is sufficiently small relative to (cid:15) , we then have the inequality q ≥ q β (cid:15) as desired.For the purposes of the rest of the argument, it will be important to have β (cid:15) > τ for sufficientlysmall (cid:15) . This leads to the restriction (4). In order to establish Theorem 1.5, we must construct, for any (cid:15) >
0, a finite Borel measure µ (cid:15) supportedon Tight( ψ , ψ , θ ) ∩ [0 ,
1] such that (cid:98) µ (cid:15) ( ξ ) ≤ C | ξ | − α/ (cid:15) for all ξ ∈ R . However, it is convenient to evaluate (cid:98) µ (cid:15) ( ξ ) at only integer values of ξ . For this purpose, it is convenient to introduce the set Tight [0 , ( ψ , ψ , θ ),a subset of the torus. To this end, we will construct a measure µ [0 , (cid:15) on the torus with support containedin Tight [0 , ( ψ , ψ , θ ). Observe that, as µ [0 , (cid:15) is a measure on the torus, it has a corresponding Fourier-Stieltjes series, the coefficients of which will be denoted (cid:91) µ [0 , (cid:15) ( ξ ). Such a measure µ [0 , (cid:15) can be associatedto a 1-periodic measure µ P(cid:15) supported on the real numbers.
Lemma 3.1.
Suppose that µ [0 , (cid:15) is a measure on the torus with support contained in Tight [0 , ( ψ , ψ , θ ) ,with the property that | (cid:91) µ [0 , (cid:15) ( ξ ) | ≤ C (1 + | ξ | ) − α/ (cid:15) for all ξ ∈ Z \{ } . Let φ ∈ C ∞ c be any smoothfunction supported in [0 , such that φµ P(cid:15) is not the zero measure. Then there exists a C not dependingon ξ such that (cid:100) φµ P(cid:15) ( ξ ) ≤ C (1 + | ξ | ) − α/ (cid:15) for all ξ ∈ R .Proof. Observe that the Fourier transform of µ P(cid:15) , viewed as a tempered distribution on R , is given by ∞ (cid:88) s = −∞ (cid:98) µ [0 , ( s ) δ s , where δ s is the Dirac mass centered at s .Therefore, we can make sense of (cid:100) φµ P(cid:15) as the convolution of (cid:99) µ P(cid:15) and (cid:98) φ . This convolution is equal to (cid:90) (cid:98) φ ( ξ − s ) d (cid:99) µ P(cid:15) ( s ) = ∞ (cid:88) s = −∞ (cid:98) φ ( ξ − s ) (cid:91) µ [0 , (cid:15) ( s ) . We now apply our assumption on (cid:91) µ [0 , (cid:15) ( s ), as well as the Schwartz bound on (cid:98) φ , to conclude | (cid:100) φµ P(cid:15) ( ξ ) | (cid:46) ∞ (cid:88) s = −∞ (1 + | ξ − s | ) − (1 + | s | ) − α/ (cid:15) . (18)We will now write the sum in (18) as S + S , where S = (cid:88) | s − ξ |≤| ξ | / (1 + | ξ − s | ) − (1 + | s | ) − α/ (cid:15) .S = (cid:88) | s − ξ | > | ξ | / (1 + | ξ − s | ) − (1 + | s | ) − α/ (cid:15) .
6e will first estimate S . Observe that if | s − ξ | ≤ | ξ | /
2, we must have 1 + | s | ≥ | ξ | /
3. Therefore, wehave S (cid:46) (1 + | ξ | / − α/ (cid:15) (cid:88) | u |≤| ξ | / (1 + | u | ) − (cid:46) (1 + | ξ | ) − α/ (cid:15) . This gives the desired estimate for S . It remains to estimate S . In order to estimate S , observe thatthe inequality | s − ξ | ≥ | ξ | / | s − ξ | ≥ | s | /
10. Applying this estimate gives S ≤ (cid:88) | s − ξ |≥| ξ | / (1 + | ξ | / − (1 + | s | / − (1 + | s | ) − α/ (cid:15) (cid:46) (1 + | ξ | ) − (cid:88) | s − ξ |≥| ξ | / (1 + | s | ) − − α/ (cid:15) ≤ (1 + | ξ | ) − ∞ (cid:88) s = −∞ (1 + | s | ) − − α/ (cid:15) (cid:46) (1 + | ξ | ) − . Adding S and S gives the result.We will need one more result that goes in the other direction—a result that allows us to lift compactlysupported, bounded, measurable functions f on R to bounded, measurable functions f [0 , the torus.We emphasize that the following lemma allows us to control the Fourier coefficients of f [0 , by knowing (cid:98) f ( s ) for integer values s .Let f ∈ L ∞ ( R ) be compactly supported. Define f P by f P ( x ) = (cid:88) j ∈ Z f ( x + j ) . The assumptions on f guarantee that f P ( x ) converges a.e. to a 1-periodic function in L ∞ ( R ). Thisfunction can naturally be associated to a function f [0 , on the torus. Lemma 3.2.
Let f ∈ L ∞ ( R ) be a compactly supported function, and define f [0 , as above. Then (cid:91) f [0 , ( s ) = (cid:98) f ( s ) for all integers s .Proof. We have (cid:91) f [0 , ( s ) = (cid:90) e − πisx f [0 , ( x ) dx = (cid:90) e − πisx f P ( x ) dx = (cid:90) e − πisx (cid:88) j ∈ Z f ( x + j ) dx = (cid:88) j ∈ Z (cid:90) j +1 j e − πis ( x − j ) f ( x ) dx = (cid:88) j ∈ Z (cid:90) j +1 j e − πisx f ( x ) dx = (cid:90) R e − πisx f ( x ) dx = (cid:98) f ( s ) . A single-scale estimate
Lemma 3.1 reduces the proof of Theorem 1.5 to finding a measure µ supported on the torus such that (cid:98) µ ( s ) ≤ | s | − α/ (cid:15) for integers s . This measure will be constructed as a weak-limit of products of functions,each of which is a sum of smoothed indicator functions of balls of an appropriate scale.For now, we consider functions supported on the interval R . We will later lift this function to thetorus. We define a function g M at scale M which we use to construct our measure supported in the exactorder set. The function we consider is supported in the set (cid:26) ψ ( q ) − c M ψ ( q ) ≤ (cid:12)(cid:12)(cid:12)(cid:12) x − r − θq (cid:12)(cid:12)(cid:12)(cid:12) ≤ ψ ( q ); q prime , M ≤ q < M, ≤ r < q (cid:27) ⊂ R , (19)where we take c M = M − (cid:15)/ . (20)Observe that 0 < c M <
1, with c M close to 0 if M is chosen sufficiently large. For the remainder of thissection, we will suppress the dependence on M and write c for c M .For a given prime q ∈ [ M, M ), the interval I q,r = { x : ψ ( q ) − cψ ( q ) ≤ x − r − θq ≤ ψ ( q ) } can beexpressed as I q,r = x q,r + cψ ( q )[ − / , / , where x q,r = r − θq + ψ ( q ) − ( c/ ψ ( q ).Let φ be a smooth function with supp φ ⊂ [ − / , /
2] for which | (cid:98) φ ( ξ ) | ≤ C exp (cid:0) −| ξ | / (cid:1) for large | ξ | .Such a function φ is provided by Ingham [6], who in fact constructs a real-valued function whose squaresatisfies the desired properties.We define g M ( x ) = (cid:88) M ≤ q< M q prime (cid:88) ≤ r Let g M , f M be defined as above. Then (cid:98) f M (0) = 1 , (22) (cid:98) f M ( s ) = 0 if ≤ | s | < M , (23) | (cid:98) f M ( s ) | ≤ C (cid:15) M − (cid:15) if M ≤ | s | ≤ M τ (1+ (cid:15)/ , (24) | (cid:98) f M ( s ) | ≤ exp (cid:16) −| sM τ | / (cid:17) if | s | ≥ M τ (1+ (cid:15)/ . (25)Proof. Equation (22) follows directly from our normalization of f M .For the proof of (23), (24), and (25), we will begin by computing (cid:98) g M ( s ) explicitly. By equation (21),we have (cid:98) g M ( s ) = (cid:88) M ≤ q< Mq prime (cid:88) ≤ r 200 is positive, it follows that for M sufficiently large, we have thebound | (cid:98) f M ( s ) | ≤ exp (cid:16) −| s | / M − τ / (cid:17) , establishing the desired bound (25). 9 A Convolution Stability Lemma In this section, we establish a convolution stability lemma. This lemma will later be used in Section 6, incombination with Lemma 4.1, applied at different scales as part of an induction argument, to completethe proof of Theorem 1.5.To this end, we will consider a sequence { M j } ∞ j =1 of positive numbers whose growth rate is dictatedby Lemma 2.1.This convolution stability lemma provides an estimate for F ∗ G , where functions F and G satisfycertain bounds following Lemma 4.1. In practice, the function G will be (cid:98) f M ∗ (cid:98) f M ∗ · · · ∗ (cid:98) f M j for someappropriate j , and F will be taken to be (cid:98) f M j +1 .In this section, we will assume that τ and τ satisfy the condition β > τ . Recall we defined β (cid:15) = γ − ( τ − − (cid:15) = β − (cid:15) in the equation (10). We consider only those (cid:15) small enough so that β (cid:15) > τ . Lemma 5.1 (Convolution Stability Lemma) . Let M j +1 = M β (cid:15) j , where β (cid:15) > τ ≥ . Let (cid:15) > besufficiently small that the quantity δ := β (cid:15) (1 − (cid:15) ) − τ (1 + (cid:15) ) τ ( β (cid:15) − (cid:15) ) , (31) satisfies δ − (cid:15) > , and let F, G : Z → C be functions satisfying the following estimates: F (0) = 1 , (32) F ( s ) = 0 if ≤ | s | < M j +1 , (33) | F ( s ) | ≤ C (cid:15) M − (cid:15)j +1 if M j +1 ≤ | s | ≤ M τ (1+ (cid:15)/ j +1 , (34) | F ( s ) | ≤ exp − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) sM τ j +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) / if | s | ≥ M τ (1+ (cid:15)/ j +1 . (35)and G (0) ≤ , (36) | G ( s ) | ≤ | s | − δ + (cid:15) if | s | ≤ M τ (1+ (cid:15) ) j , (37) | G ( s ) | ≤ exp − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) sM τ j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) / if | s | ≥ M τ (1+ (cid:15) ) j . (38)Then, provided that M is sufficiently large depending on (cid:15) , we have the following three conclusions:(a) | F ∗ G ( s ) − G ( s ) | ≤ M − δj +1 if | s | ≤ M τ (1+ (cid:15) ) j (b) | F ∗ G ( s ) | ≤ | s | − δ + (cid:15) if M τ (1+ (cid:15) ) j < | s | ≤ M τ (1+ (cid:15) ) j +1 (c) | F ∗ G ( s ) | ≤ exp − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) sM τ j +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) / if | s | > M τ (1+ (cid:15) ) j +1 . For reference, we here give a sketch of F and G corresponding to some index j . The usage of ≈ inthis figure indicates an (cid:15) -loss in the exponent on M j or M j +1 .Note that in region A , we have that | G ( s ) | (cid:46) | s | − δ . In region B , G ( s ) decays rapidly. In regions A and B (not including 0), F vanishes. In region C , F decays rapidly.10igure 1: Sketch of F and G Remark 5.2. Lemma 5.1 is called a convolution stability lemma because of the bound (a), whichshows that for small values of s , the convolution F ∗ G will be very close to G .The bound (b) will dictate the Fourier decay of the infinite product measure supported on our set.Note that the bound | s | − δ + (cid:15) is significantly worse than the bound M − (cid:15)j +1 available for F in this region-this is the reason for the loss in Fourier dimension compared to the set of well-approximable numbersfrom Kaufman’s argument.The bound (c) will allow for the convolution stability lemma to be applied inductively. Although thisbound gets slightly worse at each stage of the induction, it will always be good enough to match theconditions required for G at the next stage of the induction.Proof. We will first prove (a). To this end, we assume | s | ≤ M τ (1+ (cid:15) ) j . We write F ∗ G ( s ) = (cid:88) t ∈ Z F ( t ) G ( s − t ) . The main contribution to this sum will come from the t = 0 term, which is precisely G ( s ). Additionally,there is no contribution for 1 ≤ | t | < M j +1 because F ( t ) = 0 there. Thus, we see that | F ∗ G ( s ) − G ( s ) | ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:88) M j +1 ≤| t |≤ M τ (cid:15) ) j +1 F ( t ) G ( s − t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:88) M τ (cid:15) ) j +1 < | t | F ( t ) G ( s − t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . (39)We now estimate the size of the first term on the right hand side of (39) and consider the corresponding M j +1 ≤ | t | ≤ M τ (1+ (cid:15) ) j +1 . We have | F ( t ) | ≤ C (cid:15) M − (cid:15)j +1 . Furthermore, because | s | ≤ M τ (1+ (cid:15) ) j , we have | s − t | ≥ M j +1 , provided M is chosen large enough. Thus, we see | G ( s − t ) | ≤ exp − (cid:32) M j +1 M τ j (cid:33) / ≤ exp − M β(cid:15) − τ j √ . Recall that β (cid:15) − τ > 0. Combining the bounds on F and G and counting the number of terms in the11um, the first sum of (39) is bounded by C (cid:15) exp − M β(cid:15) − τ j √ M − (cid:15) + τ (1+ (cid:15) ) j +1 = C (cid:15) exp − M β(cid:15) − τ β(cid:15) j +1 √ M − (cid:15) + τ (1+ (cid:15) ) j +1 ≤ M − δj +1 , provided M is chosen large enough.The final step in proving (a) is to bound the second term on the right hand side of (39). For | t | > M τ (1+ (cid:15) ) j +1 , we still have | s − t | ≥ M τ (1+ (cid:15) ) j , so we can apply the tail estimates for both F and G .Thus, F ( t ) ≤ exp − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) tM τ j +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) / . For simplicity, we observe that | s − t | ≥ | t | / | G ( s − t ) | ≤ 1. So we estimate (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:88) M τ (cid:15) ) j +1 < | t | F ( t ) G ( s − t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:88) | t |≥ M τ (cid:15) ) j +1 exp − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) tM τ j +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) / ≤ M − δj +1 , by comparing to the corresponding integral, if M is sufficiently large. Summing the two terms completesthe proof of (a).We now prove (b). We consider those s such that M τ (1+ (cid:15) ) j < | s | ≤ M τ (1+ (cid:15) ) j +1 . We bound F ∗ G , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:88) t ∈ Z F ( s − t ) G ( t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:88) | t |≤ M τ (cid:15) ) j F ( s − t ) G ( t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:88) M τ (cid:15) ) j ≤| t |≤ M τ (cid:15) ) j +1 F ( s − t ) G ( t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:88) M τ (cid:15) ) j ≤| t | F ( s − t ) G ( t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . (40)The main contribution to this bound will be the first term. For such values of | t | ≤ M τ (1+ (cid:15) ) j , we havethe estimate | G ( t ) | ≤ | t | − δ + (cid:15) for t (cid:54) = 0 and | G ( t ) | ≤ t = 0.Continuing our analysis of the first term of (40), where | t | ≤ M τ (1+ (cid:15) ) j , we have that | F ( s − t ) | ≤ C (cid:15) M − (cid:15)j +1 . As a result, for an appropriate constant K , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:88) | t |≤ M τ (cid:15) ) j F ( s − t ) G ( t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ M − (cid:15)j +1 C (cid:15) + 4 C (cid:15) M τ (cid:15) ) j (cid:88) t =1 | t | − δ + (cid:15) . ≤ KM − (cid:15) +( − δ + (cid:15) +1) τ β(cid:15) (1+ (cid:15) ) j +1 ≤ K | s | τ (cid:15) ) ( − (cid:15) +( − δ + (cid:15) +1) τ β(cid:15) (1+ (cid:15) )) ≤ | s | − δ + (cid:15) , since our choice of δ guarantees that the penultimate exponent on | s | above is − δ and we can absorb theconstant K in to the | s | (cid:15) term, which is possible if M is sufficiently large.We now consider the second term in the bound (40). The relevant t are those with M τ (1+ (cid:15) ) j ≤ | t | ≤ M τ (1+ (cid:15) ) j +1 , including the case in which t = s . For such t , we have a bound of 1 on | F ( s − t ) | and a boundof exp (cid:18) − (cid:12)(cid:12)(cid:12) tM τ j (cid:12)(cid:12)(cid:12) / (cid:19) for | G ( t ) | . As there are at most 4 M τ (1+ (cid:15) ) j +1 such values of t , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:88) M τ (cid:15) ) j ≤| t |≤ M τ (cid:15) ) j +1 F ( s − t ) G ( t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ M τ (1+ (cid:15) ) j +1 exp − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) M τ (1+ (cid:15) ) j M τ j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) / ≤ | s | − δ + (cid:15) , if M is taken large enough depending on the choice of (cid:15) .Finally, we consider the final term in (40). Here | t | ≥ M τ (1+ (cid:15) ) j +1 , and we are well within the regionon which the tail bounds can be applied for both F and G . We will not need both tail bounds, however;12e will simply bound | F ( s − t ) | by 1 on this region, and use the tail bound exp (cid:18) − (cid:12)(cid:12)(cid:12) tM τ j (cid:12)(cid:12)(cid:12) / (cid:19) for G ( t ).Then, by comparing to the integral and ensuring that M is sufficiently large, we find (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:88) M τ (cid:15) ) j ≤| t | F ( s − t ) G ( t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ | s | − δ + (cid:15) . The established bounds on each of the terms in (40) combine to show | F ∗ G ( s ) | ≤ | s | − δ + (cid:15) , completingthe proof of (b).It remains to prove (c). Let s be such that | s | > M τ (1+ (cid:15) ) j +1 . Writing the convolution as for (b), webound F ∗ G ( s ) as follows: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:88) t ∈ Z F ( s − t ) G ( t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:88) | t |≤ | s || s − t |≥ | s | F ( s − t ) G ( t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:88) | t |≤ | s || s − t | < | s | F ( s − t ) G ( t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:88) | t | > | s | F ( s − t ) G ( t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . (41)The thrust of the proof is that, because | s | is so large, we are always in a situation for which the tailbounds on either F or G will apply.We consider the first term of (41), where | t | ≤ | s | and | s − t | ≥ | s | . For such t , including t = 0, we havethe bound | G ( t ) | ≤ 2. On the other hand, for such t , we certainly have | s − t | ≥ M τ (1+ (cid:15) ) j +1 ≥ M τ (1+ (cid:15)/ j +1 ,provided M is taken sufficiently large. Thus, we have | F ( s − t ) | ≤ exp − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) s M τ j +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) / . Summing over | t | ≤ | s | , we see (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:88) | t |≤ | s || s − t |≥ | s | F ( s − t ) G ( t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ | s | exp − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) s M τ j +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) / ≤ 13 exp − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) sM τ j +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) / , provided M is large enough depending on (cid:15) .Next, we will bound the second term of (41). The relevant t are such that | s − t | < | s | (includingthe t = s term). Note that for such t , we certainly have | t | ≥ | s | ≥ M τ (1+ (cid:15) ) j , if M is large enough. Wealso have | F ( s − t ) | ≤ 1, and | G ( t ) | ≤ exp (cid:18) − (cid:12)(cid:12)(cid:12) s M τ j (cid:12)(cid:12)(cid:12) / (cid:19) . Observe that the total number of values of t summed is at most 4 | s | . If M is sufficiently large, keeping in mind that | s | ≥ M τ (1+ (cid:15) ) j +1 , we observe that4 | s | is much less than exp( | s | / ((2 M j ) − τ / − M − τ / j +1 )). Thus, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:88) | t |≤ | s || s − t | < | s | F ( s − t ) G ( t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ | s | exp − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) s M τ j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) / ≤ 13 exp − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) sM τ j +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) / . It remains to bound the third term in (41). Here, | t | ≥ | s | , and we have | s − t | ≥ | t | . We will usean estimate of 1 for | G ( t ) | and a bound of exp (cid:18) − (cid:12)(cid:12)(cid:12) t M τ j +1 (cid:12)(cid:12)(cid:12) / (cid:19) for | F ( s − t ) | , as we can apply the tailbound on F . Hence, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:88) | t | > | s | F ( s − t ) G ( t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:88) | t |≥ | s | exp − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t M τ j +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) / ≤ 13 exp − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) sM τ j +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) / , by comparison with the corresponding integral.We arrive at the desired bound, (c), by summing the three terms.13 Construction of the Measure In this section, we complete the proof of Theorem 1.5. Recall that δ = β (cid:15) (1 − (cid:15) ) − τ (1 + (cid:15) ) τ ( β (cid:15) − (cid:15) ) ;this was defined at (31). The stated lower bound on the Fourier dimension, α = β − τ ) τ ( β − , was givenat (9). To prove Theorem 1.5, it suffices to construct a measure µ (cid:15) on the torus [0 , | (cid:98) µ (cid:15) ( s ) | (cid:46) | s | − δ + (cid:15) .Let (cid:15) > 0, and let M be a number so large that Lemma 5.1 applies (with M j +1 = M β (cid:15) j for all j > (cid:80) ∞ j =1 M − δj = (cid:80) ∞ j =1 M − jβ (cid:15) δ < . For each j , define f M j as in Lemma 4.1,and f [0 , M j as in Lemma 3.2. We define the function µ ( k ) (cid:15) = (cid:81) kj =1 f [0 , M j . We claim that the measures µ ( k ) (cid:15) have a subsequence with a weak limit µ (cid:15) with the desired properties. In the proof, we conflate thefunction µ ( k ) (cid:15) with the absolutely continuous measure whose Radon-Nikodym derivative is µ ( k ) (cid:15) .The proof of this will require us to estimate (cid:98) µ ( k ) (cid:15) ( s ) for integer values s . We will obtain the followingestimate by applying Lemma 5.1 inductively. Lemma 6.1. Let M = 0 for convenience. We have the following estimates on (cid:98) µ ( k ) (cid:15) ( s ) for any integers k ≥ and s ∈ Z : − k (cid:88) j =1 M − δj < (cid:98) µ ( k ) (cid:15) (0) < k (cid:88) j =1 M − δj , (42) | (cid:98) µ ( k ) (cid:15) ( s ) | < | s | − δ + (cid:15) + k (cid:88) j = J M − δj < | s | − δ + (cid:15) if M τ (1+ (cid:15) ) J − < | s | ≤ M τ (1+ (cid:15) ) J for ≤ J ≤ k, (43) | (cid:98) µ ( k ) (cid:15) ( s ) | ≤ exp (cid:32) − (cid:12)(cid:12)(cid:12)(cid:12) sM τ k (cid:12)(cid:12)(cid:12)(cid:12) / (cid:33) if | s | > M τ (1+ (cid:15) ) k . (44) Proof. We prove this lemma by induction. The base case of this lemma is implied by Lemma 4.1 appliedto f [0 , M . So we need only show the inductive step.Suppose, for some k > 1, we have that µ ( k − (cid:15) satisfies the estimates in Lemma 6.1. We must showthat µ ( k ) (cid:15) also satisfies these estimates. Our tool for this is Lemma 5.1. Observe that, by definition, wehave that µ ( k ) (cid:15) = µ ( k − (cid:15) f [0 , M k . Therefore, by the convolution rule for the Fourier transform, we have (cid:98) µ ( k ) (cid:15) = (cid:98) µ ( k − (cid:15) ∗ (cid:91) f [0 , M k . The estimates (42), (43), and (44) imply that (cid:98) µ ( k − (cid:15) is able to serve as the function G in Lemma 5.1.Note that (42) for µ ( k ) (cid:15) follows immediately by combining (a) of Lemma 5.1 and (42) for µ ( k − (cid:15) . Similarly,for 1 ≤ J ≤ k − 1, we have that (43) holds for µ ( k ) (cid:15) by combining (a) of Lemma 5.1 and (43) for µ ( k − (cid:15) .The J = k case of the estimate (43) for µ ( k ) (cid:15) is an immediate consequence of (b) of Lemma 5.1. Finally,the estimate (44) is given by (c) of Lemma 5.1.We are now in a position to define our measure µ (cid:15) . It is clear from the Banach-Alaoglu theorem thatsome subsequence of the µ ( k ) (cid:15) converges weakly to some measure µ (cid:15) . The estimate (42) shows that theweak-limit of this subsequence is a nonzero finite measure, and it is clear from the fact that µ ( k ) (cid:15) ≥ k that µ (cid:15) ≥ 0. The estimate (43) implies that | (cid:98) µ (cid:15) ( s ) | ≤ | s | − δ + (cid:15) for all s . Therefore, in orderto establish the Fourier dimension bound, the only statement it remains to prove about µ (cid:15) is that itssupport is contained in Tight [0 , ( ψ , ψ , θ ). Lemma 6.2. The measure µ (cid:15) is supported on Tight [0 , ( ψ , ψ , θ ) . roof. The measures µ ( k ) (cid:15) have nested, decreasing support, so we must havesupp µ (cid:15) ⊂ (cid:92) k supp µ ( k ) (cid:15) = (cid:92) k supp f [0 , M k . So it is sufficient to prove that (cid:92) k supp f [0 , M k ⊂ Tight [0 , ( ψ , ψ , θ ) . By construction, supp f M k is contained in the set given at (19), with c M = c M k = M − (cid:15)/ k . Let c < K = K ( (cid:15), c ) such that M − (cid:15)/ K < c .Suppose x ∈ (cid:84) k supp f [0 , M k . Let x ∗ in R be the element of the interval [0 , 1) that is congruent modulo1 to x . Then for any k , there exists an integer z k such that x ∗ + z k ∈ supp f M k . This means that, forany k > K , there exists a pair ( r k , q k ) with 0 ≤ r k < q k and M k ≤ q k < M k and q k prime such that ψ ( q ) − cψ ( q ) < (cid:12)(cid:12)(cid:12) x ∗ − r k − z k q k − θq k (cid:12)(cid:12)(cid:12) < ψ ( q ). Letting r (cid:48) k = r k − z k q k , this gives, for every k > K , a pair( r (cid:48) k , q k ) such that ψ ( q k ) − cψ ( q k ) < (cid:12)(cid:12)(cid:12) x ∗ − r (cid:48) k − θq k (cid:12)(cid:12)(cid:12) < ψ ( q k ). Since q k +1 (cid:46) q β (cid:15) k for every k , Lemma 2.1shows that, provided k > K (cid:48) for an appropriate value K (cid:48) ( (cid:15), c ), there no pair ( r, q ) such that q k < q < q k +1 and (cid:12)(cid:12)(cid:12) x ∗ − r − θq (cid:12)(cid:12)(cid:12) ≤ ψ ( q ) − cψ ( q ). Because this works for all k > K (cid:48)(cid:48) := max( K, K (cid:48) ), this shows that (cid:12)(cid:12)(cid:12) x ∗ − r − θq (cid:12)(cid:12)(cid:12) > ψ ( q ) − cψ ( q ) for all pairs ( r, q ) with q > q K (cid:48)(cid:48) , establishing the result.We have shown the support of µ ( k ) (cid:15) is contained in Tight [0 , ( ψ , ψ , θ ). Applying Lemma 3.1 givesthe desired measure on Tight( ψ , ψ , θ ). Acknowledgements The authors would like to thank Sanju Velani and Evgeniy Zorin, without whom this project would nothave been possible.This material is based upon work supported by the National Science Foundation under Award No.1803086.The second author was supported by The Maxwell Institute Graduate School in Analysis and itsApplications, a Centre for Doctoral Training funded by the UK Engineering and Physical SciencesResearch Council (Grant EP/L016508/01), the Scottish Funding Council, Heriot-Watt University andthe University of Edinburgh. References [1] A. S. Besicovitch. Sets of Fractional Dimensions (IV): On Rational Approximation to Real Numbers. J. London Math. Soc. , S1-9(2):126, 1934.[2] Y. Bugeaud. Sets of exact approximation order by rational numbers. Math. Ann. , 327(1):171–190,2003.[3] H. Davenport, P. Erd˝os, and W. J. LeVeque. On Weyl’s criterion for uniform distribution. MichiganMath. J. , 10:311–314, 1963.[4] O. Frostman. Potentiel d´equilibre et capacit´e des ensembles avec quelques applications `a la th´eoriedes fonctions. Lunds Univ. Math Sem. , 3:1–118, 1935.[5] M. Hochman and P. Shmerkin. Equidistribution from fractal measures. Invent. Math. , 202(1):427–479, 2015.[6] A. E. Ingham. A Note on Fourier Transforms. J. London Math. Soc. , 9(1):29–32, 1934.[7] V. Jarn´ık. Diophantischen Approximationen und Hausdorffsches Mass. Mat. Sborjnik , 36:371–382,1929. 158] R. Kaufman. Continued fractions and Fourier transforms. Mathematika , 27(2):262–267 (1981),1980.[9] R. Kaufman. On the theorem of Jarn´ık and Besicovitch. Acta Arith. , 39(3):265–267, 1981.[10] T. W. K¨orner. Hausdorff and Fourier dimension. Studia Math. , 206(1):37–50, 2011.[11] A. D. Pollington, S. Velani, A. Zafeiropoulos, and E. Zorin. Inhomogeneous Diophantine Approxi-mation on M -sets with restricted denominators. arXiv e-prints , page arXiv:1906.01151, June 2019.[12] M. Queff´elec and O. Ramar´e. Analyse de Fourier des fractions continues `a quotients restreints. Enseign. Math. (2) , 49(3-4):335–356, 2003.[13] T. Sahlsten and C. Stevens. Fourier transform and expanding maps on Cantor sets. arXiv e-prints ,page arXiv:2009.01703, September 2020.[14] R. Salem. On singular monotonic functions whose spectrum has a given Hausdorff dimension.
M τ (1+ (cid:15)/ , we take advantage of the choice of φ . First, observe that, because ψ isdecreasing, we can bound cψ ( q ) s from below by cψ (2 M ) s . By (20), this is M − (cid:15)/ ψ (2 M ) s. Providedthat M is sufficiently large depending on (cid:15) , we can use (3) to conclude that | cψ ( q ) s | ≥ − τ − (cid:15)/ M − τ − (cid:15)/ | s | . Therefore, applying our assumption on φ , we have that (cid:98) φ ( cψ ( q ) s ) ≤ C exp (cid:16) − − / τ + (cid:15)/ M − / τ + (cid:15)/ | s | / (cid:17) . (30)Now, we estimate the series (29) using the bound (30). There are no more than M terms in this sum,so, for an appropriate constant C , | (cid:98) f M ( s ) | ≤ C log M exp (cid:16) − − / τ + (cid:15)/ M − / τ + (cid:15)/ | s | / (cid:17) . Since we are in the regime | s | > M τ (1+ (cid:15)/ , | (cid:98) f M ( s ) | ≤ C log M exp (cid:16) − (cid:16) | s | / M − τ / (cid:17) (cid:16) − / τ + (cid:15)/ M (cid:15)τ / − (cid:15)/ (cid:17)(cid:17) . Because the exponent (cid:15)τ / − (cid:15)/