An Asymptotic Formula for the Sequence ||exp(i n h(t))||_A
aa r X i v : . [ m a t h . C V ] M a y AN ASYMPTOTIC FORMULA FOR THE SEQUENCE k exp( inh ( t )) k A BOGDAN M. BAISHANSKI AND JAN HLAVACEK
Abstract.
Given function f with an absolutely convergent Fourier series, f ( t ) = ∞ X ν = −∞ a ν e iνt , define the norm of f as k f k A = ∞ X ν = −∞ | a ν | . D. Girard has in the special case f ( t ) = g ` e it ´ , g ( z ) = z − α − αz , 0 < | α | < √ n k f n k A = 8 √ ` ´ (1 − β ) F „ ,
34 ; 32 ; 1 − β « + o (1) , n → ∞ , where F is the hypergeometric function and(2) β = 1 − | α | | α | . In our article we provide a generalization of Girard’s formula.We study the behavior of k f n k A as n → ∞ for f of the form e ih ( t ) , h ( t +2 π ) = h ( t )+2 kπ for some integer k , and h is a real, odd and twice continuouslydifferentiable. We show that if h ′′ has no zeros in (0 , π ), and if it satisfies anadditional condition near 0 and near π , then the following asymptotic formulaholds: 1 √ n k exp( inh ) k A = „ π « Z π p | h ′′ ( t ) | dt + o (1)as n → ∞ . As one of the corollaries to the result we obtain Girard’s formula.Moreover, we show that in the special case when h is 2 π -periodic, our formulaholds even if the parameter n tends to infinity through real values. Introduction
Throughout this article h will denote a real-valued function satisfying h ( t +2 π ) = h ( t ) + 2 kπ for some integer k , and such that the Fourier series ofexp ( ih ( t )) = ∞ X ν = −∞ a ν e iνt Mathematics Subject Classification.
Key words and phrases.
Absolutely convergent Fourier series, asymptotic formula, method ofstationary phase, modified equidistant Riemann sums, equidistribution, finite Blaschke products,Bessel functions, Van der Corput lemmas. is absolutely convergent. With usual notation, k exp ( ih ( t )) k A = ∞ X ν = −∞ | a ν | . The basic results on the asymptotic behavior of the sequence k exp ( inh ( t )) k A as n → ∞ were obtained in 1950’s.The most important of these results is due to A. Beurling and H. Helson: fromtheir theorem in [2] it follows that(1) If k exp ( inh ( t )) k A = O (1), n → ∞ , then h ( t ) = at + b where a is an integerand b is a real number.Next in importance are results obtained by J. P. Kahane [4], we quote the followingthree:(2) If h is continuous and piecewise linear, then k exp ( inh ( t )) k A = O (log n ), n → ∞ . (Note that here log n cannot be replaced by a sequence that wouldtend to infinity slower, since if h ( x ) = | x | on ( − π, π ] and h is 2 π -periodic,then k exp ( inh ( t )) k A = π log n + O (1))(3) If on an arbitrarily small interval h is twice differentiable with h ′′ ≥ µ > h ′′ ≤ − µ < λ > k exp ( inh ( t )) k A ≥ λ √ n .(4) If h is analytic and not a linear function, there exist positive constants λ and λ such that λ √ n ≤ k exp ( inh ( t )) k A ≤ λ √ n. Working in a different context B. Baishanski [1] has obtained the following result:(5) If f is analytic in the closed unit disk and (cid:12)(cid:12) f (cid:0) e it (cid:1)(cid:12)(cid:12) = 1 (i.e. if f is a finiteBlaschke product), then (cid:13)(cid:13) f n (cid:0) e it (cid:1)(cid:13)(cid:13) A → ∞ , n → ∞ , except in the trivial case f ( z ) = cz m , | c | = 1, m a non-negative integer.The results we quoted have important applications and (1) has been greatly gener-alized (for the generalization of (1) and applications of (1) – (4) see, for example,J. P. Kahane [5], for an application of (5) see P. Turan [7]).It may be not without interest to obtain more precise results on the asymptoticbehavior of (cid:13)(cid:13) e inh (cid:13)(cid:13) A . One such result was obtained by the late Denis Girard. Heproved [3]:(6) If f ( z ) = z − α − αz , < | α | < , then1 √ n (cid:13)(cid:13) f n (cid:0) e it (cid:1)(cid:13)(cid:13) A → √ (cid:18) Γ (cid:18) (cid:19)(cid:19) − p | α | | α | F (cid:18) ,
34 ; 32 ; 4 | α | (1 + | α | ) (cid:19) as n → ∞ . 2. Theorems
We generalize Girard’s result by the following
Theorem 1.
Let h be a real, twice continuously differentiable function, h ( t + 2 π ) = h ( t ) + 2 kπ for some integer k . If, in addition,(1) h ′′ has no zeros on (0 , π ) . N ASYMPTOTIC FORMULA FOR THE SEQUENCE k exp( inh ( t )) k A (2) h is odd(3) there exist functions m and m π , positive and increasing on (0 , c ) for some c > , and constant C > such that (3a) m (2 t ) m ( t ) ≤ C, m π (2 t ) m π ( t ) ≤ C, < t < c and m ( t ) ≤ h ′′ ( t ) ≤ Cm ( t ) , < t < c, (3b) m π ( t ) ≤ h ′′ ( π − t ) ≤ Cm π ( t ) , < t < c. (3c) then (4) 1 √ n (cid:13)(cid:13)(cid:13) e inh ( t ) (cid:13)(cid:13)(cid:13) A → (cid:18) π (cid:19) / Z π p h ′′ ( t ) dt, n → ∞ In the special case when in Theorem 1 the integer k is equal to 0, the result (4)can be strengthened and we have Theorem 2. If h is a π -periodic function satisfying the conditions of Theorem 1,then (5) 1 √ x (cid:13)(cid:13)(cid:13) e ixh ( t ) (cid:13)(cid:13)(cid:13) A → (cid:18) π (cid:19) / Z π p h ′′ ( t ) dt, as the real variable x tends to ∞ .Remark. If k = 0 in Theorem 1, then exp ( ixh ( t + 2 π )) = exp ( ixh ( t )) for every real t , unless x = mk for some integer m . Still we could try to interpret k exp( ixh ( t )) k A by restricting the function exp ( ixh ( t )) to an interval of length 2 π , and then ex-tending that restricted function periodically. But whatever restriction we choose,the extension will have jump discontinuities. Therefore its Fourier series will notbe absolutely convergent, and therefore for every x such that xk is not an integer, k exp( ixh ( t )) k A = ∞ . Remark.
The conditions (1) and (2) play an essential role in our proof, since theyimply that if νn is in the interior of the range of h ′ , then there are exactly twostationary points in the integral a n,ν = 12 π Z π − π exp ( i ( nh ( t ) − νt )) dt, and the contributions from these two points combine nicely.Obviously, h is odd if and only if the coefficients a n,ν are real. However, we neveruse directly the fact that a n,ν are real, neither would our proof be simplified by usingthe fact directly (indirectly it is reflected in the “nice” matching of contributionsfrom the two stationary points).Since h ′′ is odd and 2 π -periodic, we have that h ′′ (0) = h ′′ ( π ) = 0 and the condi-tion (3) is a condition on the behavior of h ′′ in a neighbourhood of its zeros. Thecondition (at x = 0) is satisfied for example if h ′′ ( t ) ∼ Ct α or h ′′ ( t ) ∼ Ct α | log t | β as t → + , where α > β ∈ R . (However, it is not difficult to give examples whenthe condition (3) is not satisfied. Two such examples are h ′′ ( t ) = exp (cid:0) − t (cid:1) and h ′′ ( t ) = t + t sin (cid:0) t (cid:1) on some interval (0 , c ).) BOGDAN M. BAISHANSKI AND JAN HLAVACEK
Notation.
We denote by C constants, dependent only on the function h , not alwaysthe same constant. Since arguments involving m and m π are similar, it will besufficient to consider only m , which from now on will be denoted m . We canassume that h ′′ > , π ), so h ′ is increasing there from α = h ′ (0) to β = h ′ ( π ).The inverse function of h ′ will be denoted by ψ . If α < νn < β , there exists one andonly one t n,ν , 0 < t n,ν < π such that h ′ ( t n,ν ) = νn or equivalently ψ (cid:0) νn (cid:1) = t n,ν .We set M ( x ) = m ( ψ ( x )) . Then M is defined and decreasing for α < x < γ ,where γ < β and ψ ( γ ) < c . The modulus of continuity of h ′′ will be denoted by ω : ω ( t ) = sup {| h ′′ ( x ) − h ′′ ( y ) | : x, y ∈ [0 , π ] , | x − y | < t } . Let Φ( x ) denote a function tending to infinity as x → ∞ , Φ( x ) = o ( √ x ). Additionalconditions will be imposed on Φ in the course of the proof. Finally, δ n = Φ( n ) √ n , and α n = max (cid:0) h ′ (2 δ n ) , α + n (cid:1) , β n = min (cid:0) h ′ ( π − δ n ) , β − n (cid:1) .The proof of Theorem 1 will be in five steps.First we show that X ν ∈E n | a n,ν | = O (log n ) , n → ∞ , so that the contribution of the external terms E n , i.e. the terms a n,ν such that ν ( αn + 1 , βn −
1) can be ignored in the proof of our formula.Second, we establish certain properties of h ′′ needed in the following steps.Third, we show that X ς ∈P n | a n,ν | = O (1) , n → ∞ where P n = { ν : αn + 1 < ν < α n n or β n n < ν < βn − } In the fourth step we use the method of stationary phase to approximate each ofthe central terms, that is terms a n,ν where νn ∈ ( α n , β n ). Finally, in the last step,we use Lemmas 3 and 4 to approximate the sum of the central terms and, usingthis approximation together with estimates obtained in previous steps, we provethe Theorem 1. 3. Lemmas
Lemma 1.
Let ϕ be a twice continuously differentiable function on the interval [ a, b ] such that ϕ ′ has no zeros in [ a, b ] . The following estimates hold for the integral I = Z ba e iϕ ( t ) dt (a) In the special case ϕ ( b ) = ϕ ( a )+ 2 kπ for some integer k , and ϕ ′ ( b ) = ϕ ′ ( a ) : | I | ≤ Var [ a,b ] (cid:18) ϕ ′ (cid:19) (b) In general case, provided ϕ ′ is monotone on [ a, b ] : | I | ≤ | ϕ ′ ( a ) | , | ϕ ′ ( b ) | ) N ASYMPTOTIC FORMULA FOR THE SEQUENCE k exp( inh ( t )) k A Proof.
A standard trick is used: the integral is multiplied and divided by iϕ ′ , thenthe integration by parts leads to the estimate | I | ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) e iϕ ( t ) iϕ ′ ( t ) (cid:12)(cid:12)(cid:12)(cid:12) bt = a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + Z ba (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ϕ ′′ ( t )( ϕ ′ ( t )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , from which both (a) and (b) follow. (cid:3) Lemma 2.
Let f be Riemann integrable on [ c, b ] for every c ∈ ( a, b ) , and let theimproper integral R ba + f exist. Let a j,n = a + j b − an , j = 0 , . . . , n , and let ξ j,n be suchthat a j − ,n ≤ ξ j,n ≤ a j,n for j = 1 , . . . , n .If there exists a monotone function M on [ a, b ] such that | f ( x ) | ≤ M ( x ) for x ∈ [ a, b ] and R ba + M < ∞ , then the equidistant Riemann sums of f on [ a, b ] modified by the omission of the initial summand (i.e. the sums b − an P f ( ξ j,n ) wherethe summation index does not run from j = 1 to n but from j = 2 to n ) convergeto R ba + f .In particular, (6) b − an X f (cid:18) kn (cid:19) −→ Z ba + f where the sum runs over all integers k such that an + 1 < k < bn .Remark. In the case when P denotes the full equidistant Riemann sum, taken overall k , an < k < bn , (6) may fail to hold. An example where it fails is provided bythe function f ( x ) = √ x − a in case a is irrational. Proof.
To show that(7) D n = b − an n X j =2 f ( ξ j,n ) − Z ba + f tends to zero as n → ∞ , we start, given ε >
0, by choosing c = c ( ε ), d = d ( ε ), a < d < c < b such that(8) Z ca + M < ε. Since f is Riemann integrable on [ d, b ], | f | is bounded there by some constant K = K ( ε ). Let J = J ( n, ε ) be defined by the condition a J − ,n ≤ c < a J,n , and letthe intervals I = I ( ε ) and I n = I n ( ε ) be defined by I = [ c, b ] and I n = [ a J,n , b ]. Werewrite (7) as D n = b − an n X j = J +1 f ( ξ j,n ) − Z ba J,n f + b − an f ( ξ J,n ) − Z a J,n c f − Z ca + f + b − an J − X j =2 f ( ξ j,n ) BOGDAN M. BAISHANSKI AND JAN HLAVACEK so that(9) |D n | ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) b − an n X j = J +1 f ( ξ j,n ) − Z ba J,n f (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + b − an | f ( ξ J,n ) | + Z a J,n c | f | + Z ca + | f | + b − an J − X j =2 | f ( ξ j,n ) | We shall prove the lemma by showing that each of the five summands on the rightside of (9) is less than ε for n sufficiently large.To estimate the first summand, we shall make use of the following notation: If J is an interval, P = { x k , ≤ k ≤ m } a partition of J and f a function on J , thenΩ ( f, P , J ) = m X k =1 (Osc f ) ( x k − , x k ) ( x k − x k − )where (Osc f ) ( s, t ) = sup {| f ( x ) − f ( y ) | : s ≤ x, y ≤ t } . We can now write (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) b − an n X j = J +1 f ( ξ j,n ) − Z ba J,n f (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X j = J +1 Z a j,n a j − ,n ( f ( ξ j,n ) − f ( x )) dx (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ Ω( f, P n , I n ) , (10)where P n is the partition of the interval I n = [ a J,n , b ] consisting of points a j,n , J ≤ j ≤ n . Let f P n = P n ∪{ c } . Then f P n is a partition of I = [ c, b ] and (cid:13)(cid:13)(cid:13) f P n (cid:13)(cid:13)(cid:13) = b − an .Obviously(11) Ω ( f, P n , I n ) ≤ Ω (cid:16) f, f P n , I (cid:17) . Since f is Riemann integrable on I , Ω (cid:16) f, f P n , I (cid:17) → (cid:13)(cid:13)(cid:13) f P n (cid:13)(cid:13)(cid:13) →
0. It follows from(10) and (11) that the first summand in (9) is less than ε for large n .The second summand is less than b − an K ( ε ), the same holds for the third sum-mand. The fourth summand is less then ε by (8). The crucial step in the proof isthe estimate of the fifth summand.Since M is monotone decreasing on ( a, b ] we have, on one hand J − X j =2 | f ( ξ j,n ) | ≤ J − X j =2 M ( ξ j,n ) ≤ J − X j =2 M ( a j − ,n ) = J − X j =1 M ( a j,n )and on the other hand b − an M ( a j,n ) ≤ Z a j,n a j − ,n M ( x ) dx and so b − an J − X j =2 | f ( ξ j,n ) | ≤ Z a J − ,n a + M ( x ) dx ≤ Z ca + M ( x ) dx ≤ ε by (8). (cid:3) N ASYMPTOTIC FORMULA FOR THE SEQUENCE k exp( inh ( t )) k A Lemma 3.
Given an interval J and an array s n,k , ≤ s n,k ≤ , where n =1 , , . . . , and k runs through integers such that kn ∈ J , the following three conditionsare equivalent:(a) For any pair of intervals I , J , I ⊂ [0 , , J ⊂ J , we have, as n → ∞ , N n n → | I | · | J | , where N n is the number of indices k such that s n,k ∈ I and kn ∈ J , and | I | and | J | denote the lengths of the intervals I and J , respectively.(b) For any non-zero integer j and any subinterval J of J n X { k : kn ∈ J } exp (2 πijs n,k ) → , n → ∞ (c) If f is Riemann integrable on the interval J , and g is Riemann integrableon [0 , , then n X { k : kn ∈ J } f (cid:18) kn (cid:19) g ( s n,k ) → Z J f · Z g, n → ∞ Proof.
The proof is an adaptation of the well known method of Hermann Weyl.Let ( f, g ) n def = 1 n X k : kn ∈ J f (cid:18) kn (cid:19) g ( s n,k )( f, g ) def = Z J f · Z g and let F denotes the collection of pairs { f, g } such that ( f, g ) n → ( f, g ) as n → ∞ .We observe that(*) If f is Riemann integrable on J then { f, } ∈ F .It is easy to verify that condition (a) can be phrased as(a’) { χ J , χ I } ∈ F for any pair of intevals I , J , I ⊂ [0 , J ⊂ J and conditions (b) and (c) as(b’) { χ J , g j } ∈ F for g j ( x ) = exp (2 πijx ), j = ± , ± , . . . , and J any subinter-val of J ,(c’) { f, g } ∈ F for any complex-valued functions f and g , that are Riemann-integrable on J or [0 , { χ J , χ I } ∈ F for any pair of intervals I , J such that J ⊂ J and that theclosure of the interval I lies in the open interval (0 , { f, ϕ k } ∈ F and ϕ k converge to ϕ uniformly on [0 , { f, ϕ } ∈ F . BOGDAN M. BAISHANSKI AND JAN HLAVACEK (***)’ If f ≥ J and g is real-valued and if, for every ε >
0, there existfunctions ϕ and ϕ on [0 , ϕ ≤ g ≤ ϕ , such that { f, ϕ } ∈ F , { f, ϕ } ∈F and R ( ϕ − ϕ ) < ε , then { f, g } ∈ F .(***)” Similarly, if g ≥ f is real-valued, and if for every ε > ϕ and ϕ on J such that ϕ ≤ f ≤ ϕ , { ϕ , g } ∈ F , { ϕ , g } ∈ F and R J ( ϕ − ϕ ) < ε , then { f, g } ∈ F .We start the proof by observing that (*) and (b’) imply that { χ j , T } ∈ F forany trigonometric polynomial T ( x ) = P | ν |≤ N c ν e πiνx . From (**) we obtain that { χ J , ϕ } ∈ F for any continuous function ϕ on [0 ,
1] satisfying ϕ (0) = ϕ (1). Sincefor any ε > ϕ , ϕ with the property ϕ ≤ χ I ≤ ϕ and R ( ϕ − ϕ ) < ε (since the endpoints of I are inside (0 , { χ j , χ I } ∈ F , i.e (a”) is proved.From (*) we get { χ J , } ∈ F . Together with (a”) this implies that { χ J , ϕ } ∈ F for any step-function ϕ satisfying ϕ (0) = ϕ (1). If g is a real-valued Riemann-integrable function on [0 , ε > ϕ , ϕ that ϕ ≤ g ≤ ϕ and R ( ϕ − ϕ ) < ε . Therefore, using (***)’ again, we obtain { χ J , g } ∈ F , and { ϕ, g } ∈ F for ϕ a real-valued step-function on J . In particularthis holds for any g non-negative, Riemann-integrable. Using now (***)” we obtainthat { f, g } ∈ F for any f and g real-valued, Riemann-integrable, g non-negative.By linearity, the same holds if both f and g are complex-valued. This gives us (c’),which concludes the proof of the lemma. (cid:3) Remark.
If an array s n.k satisfies any of the three conditions in Lemma 3 (andtherefore all of these conditions), we say that the array is equidistributed on theinterval [0 , Lemma 4.
Let ϕ be a real-valued twice differentiable function on an interval J and let ϕ ′′ ( x ) ≥ ρ > for x ∈ J . Then the array (cid:10) nϕ (cid:0) kn (cid:1)(cid:11) , where h s i denotes thefractional part of s , for n = 1 , , . . . and kn ∈ J , is equidistributed on [0 , .Proof. It is suficient to show that the condition (b) of Lemma 3 is satisfied. Let j denote any integer different from zero. Since exp (2 πijx ) has period 1, we get forany subinterval J = [ c, d ] of J (12) U n,j def = X k | kn ∈ J exp (2 πijs n,k ) = X nc ≤ k ≤ nd exp (cid:18) πijnϕ (cid:18) kn (cid:19)(cid:19) To estimate the last expression, we use one of Van der Corput’s lemmas [8, ch. 5,lemma 4.6]:(*) If f ′′ ( x ) ≥ µ > f ′′ ( x ) ≤ − µ < a, b ], then(13) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X a In this section we will provide an estimate for the sum of the external terms,namely(14) X ν ∈E n | a n,ν | = O (log n ) , n → ∞ where the external terms E n are the terms where νn ≤ α + n or νn > β − n . Wewill provide an estimate for the sum of the terms where νn ≤ α + n , the estimatefor the other case can be obtained in a similar way.We start by splitting the sum into three parts:(15) X νn ≤ α + n | a n,ν | = X α − n < νn ≤ α + n | a n,ν | + X α − < νn ≤ α − n | a n,ν | + X νn ≤ α − | a n,ν | We can easily dismiss the first sum, as it has at most three terms, each less thanor equal to one.For the second sum, we will use the following estimate of | a n,ν | :(16) | a n,ν | = 12 π (cid:12)(cid:12) b n,ν + b n,ν (cid:12)(cid:12) ≤ π | b n,ν | where(17) b n,ν = Z π e i ( nh ( t ) − νt ) dt. Applying Lemma 1 part (b) with ϕ ( t ) = nh ( t ) − νt we obtain(18) | b n,ν | ≤ | nα − ν | , | nβ − ν | )This gives us estimate(19) X α − ≤ νn ≤ α − n | b n,ν | ≤ X α − ≤ νn ≤ α − n αn − ν ≤ C log( n )For the third summand we apply Lemma 1 part (a), again with ϕ ( t ) = nh ( t ) − νt ,but with the interval ( a, b ) = ( − π, π ). It is easy to verify that if νn ≤ α − ϕ satisfiesthe hypothesis of Lemma 1 part (a), and we obtain(20) | a n,ν | = 12 π (cid:12)(cid:12)(cid:12)(cid:12)Z π − π e i ( nh ( t ) − νt ) dt (cid:12)(cid:12)(cid:12)(cid:12) ≤ π Var ( − π,π ) (cid:18) ϕ ′ (cid:19) . Since ϕ ′ is even and monotone on (0 , π ), we haveVar ( − π,π ) (cid:18) ϕ ′ (cid:19) = 2 (cid:18) ϕ ′ (0) − ϕ ′ ( π ) (cid:19) ≤ n β − α (cid:0) α − νn (cid:1) which together with (20) gives(21) X νn <α − | a n,ν | ≤ C X νn <α − n β − α (cid:0) α − νn (cid:1) ≤ C Z α − −∞ β − α ( α − x ) dx which is a constant independent of n .Then (15) together with (16), (19) and (21) will give us (14).5. Some Auxiliary Results on h ′′ and √ h ′′ The function h ′′ ( ψ ( t )) is integrable on [ α, β ] = [ h ′ (0) , h ′ ( π )] (and therefore √ h ′′ ( ψ ( t )) is also integrable on [ α, β ]).To prove this fact, we consider the integral I ( τ , τ ) = Z h ′ ( τ ) h ′ ( τ ) h ′′ ( ψ ( t )) dt, < τ < τ < π. Making the change of variable u = ψ ( t ) we get t = h ′ ( u ), dt = h ′′ ( u ) du and so I ( τ , τ ) = R τ τ du = τ − τ , which converges as τ → τ → π − .For further reference, we note(22) Z h ′ ( d ) h ′ (0) h ′′ ( ψ ( t )) dt = d, ≤ d ≤ π If d > h ′′ ( x ) ≤ < x ≤ d , then we deducefrom (22) that(23) Z h ′ ( d ) h ′ (0) p h ′′ ( ψ ( t )) dt ≤ d. Finally, we note that the same change of variable as above ( u = ψ ( t )) bringsanother simplification:(24) Z h ′ ( τ ) h ′ ( τ ) p h ′′ ( ψ ( t )) dt = Z τ τ p h ′′ ( u ) du. The function h ′′ ( ψ ( t )) has on [ h ′ (0) , h ′ (4 c )] a monotone majorant M ( t ) whichis integrable. (Therefore √ h ′′ ( ψ ( t )) has also such a majorant: p M ( t )). (Bothstatements are valid also on the interval [ h ′ ( π − c ) , h ′ ( π )], possibly with differentmajorant M ( t )).Clearly, since h ′′ ( x ) ≥ m ( x ) on [0 , c ], we have that M ( t ) = m ( ψ ( t )) is a mono-tone majorant of h ′′ ( ψ ( t )) . The function M is integrable because it is itself majorizedby the integrable function Ch ′′ ( ψ ( t )) : this follows from the inequality h ′′ ( x ) ≤ Cm ( x ), x ∈ [0 , c ]. N ASYMPTOTIC FORMULA FOR THE SEQUENCE k exp( inh ( t )) k A According to the paragraphs 5.1 and 5.2, conditions of Lemma 2 are satisfiedby the functions h ′′ ( ψ ( t )) and √ h ′′ ( ψ ( x )) on the interval [0 , c ]. By Lemma 2 wehave then(25) X nh ′ (0)+1 ≤ k ≤ nh ′ ( d ) n h ′′ (cid:0) ψ (cid:0) kn (cid:1)(cid:1) ≤ C Z d M ( x ) dx, < d ≤ h ′ (4 c ) . Since M ( x ) ≤ Ch ′′ ( ψ ( x )) for x ∈ ( h ′ (0) , h ′ (4 c )], we deduce from (22) and (25) that(26) 0 ≤ X nh ′ (0)+1 ≤ k ≤ nh ′ ( d ) n h ′′ (cid:0) ψ (cid:0) kn (cid:1)(cid:1) ≤ Cd if 0 < d ≤ c .Similarly, using (23) we obtain(27) 0 ≤ X nh ′ (0)+1 ≤ k ≤ nh ′ ( d ) n q h ′′ (cid:0) ψ (cid:0) kn (cid:1)(cid:1) ≤ Cd if 0 ≤ d ≤ min (4 c, d ) where h ′′ ( x ) ≤ < x ≤ d . In this paragraph we prove the following inequality to which we shall referrepeatedly:(28) h ′′ ( t ) ≤ Ch ′′ ( ξ )in each of the following cases:(a) 0 < t < ξ < c (b) π − c < ξ < t < π (c) if δ is any positive number less than c and if 2 δ < t < π − δ , | ξ − t | < δ .Note that the constant C is (28) does not depend on δ .To prove the inequality, we will consider each of the three cases.(a) By monotonicity of m and the inequality m ≤ h ′′ ≤ Cm on [0 , c ] we get h ′′ ( t ) ≤ Cm ( t ) ≤ Cm ( ξ ) ≤ Ch ′′ ( ξ ).(b) is analogous to (a).(c) This case we need to subdivide into three subcases:(c1) 2 c ≤ t ≤ π − c , | ξ − t | ≤ δ < c (c2) 2 δ ≤ t ≤ c , | ξ − t | ≤ δ < c (c3) π − c ≤ t ≤ π − δ , | ξ − t | ≤ δ < c The case (c1) is simple: If I = [2 c, π − c ] and J = [ c, π − c ], it is sufficientto choose C so that sup I h ′′ < C inf J h ′′ .It remains only to consider the case (c2), because (c3) can be treatedthe same way. This is the only delicate case, and this is the only placein the proof of the Theorem 1 where we use the assumption (3a), that is m (2 t ) m ( t ) ≤ C for 0 ≤ t ≤ c .We observe that 0 ≤ t − δ < ξ < t + δ < c , and so(29) m ( t − δ ) ≤ m ( ξ ) ≤ h ′′ ( ξ ) . Since t − δ < t < c , the assumption (3a) gives us m (2( t − δ )) ≤ Cm ( t − δ )and from (29) it follows that m (2( t − δ )) ≤ Ch ′′ ( ξ ). On the other hand, t ≥ δ implies t ≤ t − δ ), and so m ( t ) ≤ m (2( t − δ )). Now h ′′ ( t ) ≤ Cm ( t ) ≤ Cm (2( t − δ )) ≤ Ch ′′ ( ξ ), so (28) holds. Total contribution of periphery terms We will now prove that(30) X ν ∈P n | a n,ν | = O (1) , n → ∞ . Here P n = P ln ∪ P rn , P ln = { ν : αn + 1 < ν < α n n } , P rn = { ν : β n n < ν < βn − } ,α n = max (cid:18) h ′ (2 δ n ) , h ′ (0) + 1 n (cid:19) ,β n = min (cid:18) h ′ ( π − δ n ) , h ′ ( π ) − n (cid:19) . (We should mention that one of the sets P ln or P rn , or both of them, may be empty).We shall consider only the left peripheral terms, the right periphery terms can betreated similarly.It will be sufficient to consider the integrals b n,ν = Z π exp ( i ( nh ( t ) − νt )) dt, where αn + 1 < ν < α n n , or equivalently h ′ (0) + n < νn < h ′ (2 δ n ). We write b n,ν = Z δ n exp ( i ( nh ( t ) − νt )) dt + Z π δ n exp ( i ( nh ( t ) − νt )) dt. For the first summand we use the trivial estimate(31) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z δ n exp ( i ( nh ( t ) − νt )) dt (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ δ n . To estimate the second summand, we shall apply Lemma 1, part (b). We set ϕ ( t ) = nh ( t ) − νt and observe that since h ′′ > , π ], the function ϕ ′ isstrictly increasing. Observing that ν ∈ P ln implies h ′ ( t n,ν ) = νn < h ′ (2 δ n ), we have t n,ν < δ n . Since ϕ ′ ( t n,ν ) = 0, we get that ϕ ′ ( t ) ≥ ϕ ′ (3 δ n ) > δ n , π ]. So, byLemma 1, part (b), we get that(32) (cid:12)(cid:12)(cid:12)(cid:12)Z π δ n exp ( i ( nh ( t ) − νt )) dt (cid:12)(cid:12)(cid:12)(cid:12) ≤ ϕ ′ (3 δ n ) = 2 n (cid:0) h ′ (3 δ n ) − νn (cid:1) . But νn = h ′ ( t n,ν ) and(33) h ′ (3 δ n ) − νn = h ′ (3 δ n ) − h ′ ( t n.ν ) = h ′′ ( ξ ) (3 δ n − t n,ν )for some ξ ∈ ( t n,ν , δ n ). Since t n,ν < δ n , we have 3 δ n − t n,ν ≥ δ n . On the otherhand, by the inequality (28), case (a), since 0 < t n,ν < ξ < δ n < c , we have h ′′ ( ξ ) ≥ C h ′′ ( t n,ν ). Thus we derive from (33) that h ′ (3 δ n ) − νn ≥ δ n h ′′ ( t n,ν ) C . N ASYMPTOTIC FORMULA FOR THE SEQUENCE k exp( inh ( t )) k A From this estimate and (32) we obtain (cid:12)(cid:12)(cid:12)(cid:12)Z π δ n exp ( i ( nh ( t ) − νt )) dt (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cnδ n h ′′ ( t n,ν ) . The last estimate and (31) imply(34) X ν ∈P ln | b n,ν | ≤ δ n X ν ∈P ln Cδ n X ν ∈P ln nh ′′ ( t n,ν ) . We have δ n X ν ∈P ln δ n · card { ν : αn + 1 < ν < α n n }≤ δ n · ( α n − α ) · n = ( h ′ (2 δ n ) − h ′ (0)) · nδ n = h ′′ ( ξ ) · nδ n for some ξ ∈ (0 , δ n ). Since h ′′ ( x ) ≤ Cm ( x ) and m is increasing, we get h ′′ ( ξ ) · nδ n ≤ Cm ( ξ ) nδ n ≤ Cm (2 δ n ) nδ n Since, on the other hand, m ( x ) ≤ h ′′ ( x ) = h ′′ ( x ) − h ′′ (0) ≤ ω ( x ), and ω (2 x ) ≤ ω ( x ), we have Cm (2 δ n ) nδ n ≤ Cω (2 δ n ) nδ n ≤ Cω ( δ n ) nδ n This, together with δ n = Φ( n ) √ n , and our choice of Φ( n ) which will be made in (55),namely ω ( δ n )Φ( n ) = 1, gives us(35) δ n X ν ∈P ln ≤ Cω ( δ n )Φ( n ) = C Φ( n ) = o (1) , n → ∞ . We will now show that the second term on the right hand side of (34) is bounded.Using (3b) we get X ν ∈P ln nh ′′ (cid:0) ψ (cid:0) νn (cid:1)(cid:1) ≤ n X αn +1 <ν<α n n m (cid:0) ψ (cid:0) νn (cid:1)(cid:1) That can then be written as 1 n X M (cid:16) νn (cid:17) where the sum runs over ν satisfying h ′ (0) + n ≤ νn ≤ h ′ (2 δ n ). Since M is decreas-ing, this sum is less than or equal to Z h ′ (2 δ n ) h ′ (0) M ( x ) dx = Z h ′ (2 δ n ) h ′ (0) m ( ψ ( x )) dx ≤ C Z h ′ (2 δ n ) h ′ (0) h ′′ ( ψ ( x )) dx = 2 δ n C where the last step follows from (22). Thus(36) 1 δ n X ν ∈P ln nh ′′ ( t n,ν ) ≤ C. From the previous estimates (34), (35) and (36), and the observation (16) that | a n,ν | ≤ | b n,ν | we get X ν ∈P ln | a n,ν | = O (1) , n → ∞ . Since similar estimate holds for P rn , we obtain (30).7. Central Terms In this section we show that, for ν ∈ C n , i.e. nα n ≤ ν ≤ nβ n ,(37) a n,ν = r π p nh ′′ ( t n,ν ) cos (cid:16) ρ n,ν + π (cid:17) + e R n,ν where ρ n,ν = n ( h ( t n,ν ) − h ′ ( t n,ν ) t n,ν ) and(38) (cid:12)(cid:12)(cid:12) e R n,ν (cid:12)(cid:12)(cid:12) ≤ Cnh ′′ ( t n,ν ) δ n + nω ( δ n ) δ n . We note, in passing, that ρ n,ν = − nh ∗ (cid:0) νn (cid:1) , where h ∗ is the Legendre transformof the function h , i.e. h ∗ ( x ) = xψ ( x ) − h ( ψ ( x )), ψ being the inverse function of thederivative of h .We write(39) b n,ν = B n,ν + R (1) n,ν where(40) B n,ν = Z t n,ν + δ n t n,ν − δ n exp ( i ( nh ( t ) − νt )) dt and(41) R (1) n,ν = Z t n,ν − δ n exp ( i ( nh ( t ) − νt )) dt + Z πt n,ν + δ n exp ( i ( nh ( t ) − νt )) dt. Let ϕ ( t ) = nh ( t ) − νt , then since h ′ is increasing we have for t ∈ [0 , t n,ν − δ n ] that ϕ ′ ( t ) = nh ′ ( t ) − ν = n ( h ′ ( t ) − h ′ ( t n,ν )) ≤ n ( h ′ ( t n,ν − δ n ) − h ′ ( t n,ν )) = − nδ n h ′′ ( ξ )for some ξ between t n,ν − δ n and t n,ν . So by part (b) of Lemma 1, and then bycase (c) of inequality (28) we obtain (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z t n,ν − δ n exp ( i ( nh ( t ) − νt )) dt (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ nh ′′ ( ξ ) δ n ≤ Cnδ n h ′′ ( t n,ν ) . Same type of estimate holds for R πt n,ν + δ n exp ( i ( nh ( t ) − νt )) dt , so we obtain(42) (cid:12)(cid:12)(cid:12) R (1) n,ν (cid:12)(cid:12)(cid:12) ≤ Cnh ′′ ( t n,ν ) δ n . Now we consider the integral B n,ν defined in (40). Writing h ( t ) = h ( t n,ν ) + h ′ ( t n,ν )( t − t n,ν ) + h ′′ ( ξ )2 ( t − t n,ν ) for | t − t n,ν | < δ n and some ξ between t and t n,ν ,and using h ′ ( t n,ν ) = νn , we can easily check that(43) exp ( i ( nh ( t ) − νt )) = e iρ n,ν e inh ′′ ( t n,ν )( t − t n,ν ) / + R (2) n,ν ( t ) N ASYMPTOTIC FORMULA FOR THE SEQUENCE k exp( inh ( t )) k A where (cid:12)(cid:12)(cid:12) R (2) n,ν ( t ) (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) e i n ( h ′′ ( ξ ) − h ′′ ( t n,ν ) ) ( t − t n,ν ) − (cid:12)(cid:12)(cid:12) ≤ n | h ′′ ( ξ ) − h ′′ ( t n,ν ) | ( t − t n,ν ) ≤ n ω ( δ n ) δ n . From the last estimate and (43) we obtain that(44) B n,ν = g B n,ν + R (2) n,ν where(45) g B n,ν = e iρ n,ν Z t n,ν + δ n t n,ν − δ n exp (cid:18) i h ′′ ( t n,ν )2 n ( t − t n,ν ) (cid:19) dt and(46) (cid:12)(cid:12)(cid:12) R (2) n,ν (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z t n,ν + δ n t n,ν − δ n R (2) n,ν ( t ) dt (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ nω ( δ n ) δ n . A change of variable u = q nh ′′ ( t n,ν )2 ( t − t n,ν ) transforms (45) into g B n,ν = √ e iρ n,ν p nh ′′ ( t n,ν ) Z δ n q nh ′′ ( tn,ν )2 − δ n q nh ′′ ( tn,ν )2 e iu du = 2 √ e iρ n,ν p nh ′′ ( t n,ν ) Z ∞ e iu du − Z ∞ δ n q nh ′′ ( tn,ν )2 e iu du ! . (47)Since R ∞ e iu du = √ π e iπ/ and (cid:12)(cid:12)(cid:12)(cid:12)Z ∞ x e iu du (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)Z ∞ x e it √ t dt (cid:12)(cid:12)(cid:12)(cid:12) ≤ x , we get from (47) that(48) g B n,ν = √ πe i ( ρ n,ν + π ) p nh ′′ ( t n,ν ) + R (3) n,ν where(49) (cid:12)(cid:12)(cid:12) R (3) n,ν (cid:12)(cid:12)(cid:12) ≤ nh ′′ ( t n,ν ) 1 δ n . From (39), (44) and (48) we obtain(50) b n,ν = √ π p nh ′′ ( t n,ν ) e i ( ρ n,ν + π ) + gg R n,ν , where gg R n,ν = R (1) n,ν + R (2) n,ν + R (3) n,ν , and, by (42), (46) and (49) we get(51) (cid:12)(cid:12)(cid:12)(cid:12) gg R n,ν (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cnh ′′ ( t n,ν ) 1 δ n + nω ( δ n ) δ n . Since h is odd, a n,ν = 2 Re ( b n,ν )2 π and, by (50)(52) a n,ν = r π p nh ′′ ( t n,ν ) cos (cid:16) ρ n,ν + π (cid:17) + g R n,ν , where (cid:12)(cid:12)(cid:12) g R n,ν (cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12) gg R n,ν (cid:12)(cid:12)(cid:12)(cid:12) . By summing over the central terms, we obtain(53) X nα n ≤ ν ≤ nβ n | a n,ν | = r π X nα n ≤ ν ≤ nβ n p nh ′′ ( t n,ν ) (cid:12)(cid:12)(cid:12) cos (cid:16) ρ n,ν + π (cid:17)(cid:12)(cid:12)(cid:12) + R n where(54) | R n | ≤ X nα n ≤ ν ≤ nβ n (cid:12)(cid:12)(cid:12)(cid:12) gg R n,ν (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cδ n X nα n ≤ ν ≤ nβ n nh ′′ ( t n,ν ) + Cn ω ( δ n ) δ n (Here we have used the fact that the number of central terms is ( β n − α n ) n < Cn .)We observe that X nα n ≤ ν ≤ nβ n nh ′′ ( t n,ν ) = X nα n ≤ ν ≤ nβ n nh ′′ (cid:0) ψ (cid:0) νn (cid:1)(cid:1) is bounded, since it is positive and smaller than X nα +1 ≤ ν ≤ nβ − nh ′′ (cid:0) ψ (cid:0) νn (cid:1)(cid:1) = X nα +1 ≤ ν ≤ nh ′ ( π ) 1 nh ′′ (cid:0) ψ (cid:0) νn (cid:1)(cid:1) + X nh ′ ( π ) <ν ≤ nβ − nh ′′ (cid:0) ψ (cid:0) νn (cid:1)(cid:1) and the last two summands converge to a finite limit. (Here we appealed toLemma 2, the conditions in that lemma being satisfied by observations made insections 5.1 and 5.2) Therefore (54) gives | R n | ≤ C (cid:18) δ n + n ω ( δ n ) δ n (cid:19) . Choosing δ n so that δ n = n ω ( δ n ) δ n , i.e. choosing Φ n so that(55) ω (cid:18) Φ n √ n (cid:19) Φ n = 1 , the past estimate simplifies to(56) | R n | ≤ Cδ n = C √ n Φ n . Remark. Since it is a property of a modulus of continuity that there exists C > ω ( x ) > Cx , we obtain from (55) that P hi n = O ( n / ), n → ∞ .8. Final Step Taking into account contributions of external (14) and periphery (30) terms,together with (53) and (56) and the last remark, we obtain(57)1 √ n ∞ X ν = −∞ | a n,ν | = r π nβ n X ν = nα n n p h ′′ ( t n,ν ) (cid:12)(cid:12)(cid:12) cos (cid:16) ρ n,ν + π (cid:17)(cid:12)(cid:12)(cid:12) + O (cid:18) n (cid:19) , n → ∞ . N ASYMPTOTIC FORMULA FOR THE SEQUENCE k exp( inh ( t )) k A We shall prove that(58) D n = nβ n X ν = nα n n p h ′′ ( t n,ν ) (cid:12)(cid:12)(cid:12) cos (cid:16) ρ n,ν + π (cid:17)(cid:12)(cid:12)(cid:12) − π Z βα dt p h ′′ ( ψ ( t )) → , n → ∞ . Using the simplification of the last integral given in (24), i.e. making change ofvariable x = ψ ( t ), we obtain from (57) and (58) the conclusion of Theorem 1To prove (58), for each sufficiently small ε > D n = A n,ε + B n,ε + C n,ε − E ε where A n,ε = nh ′ ( ε ) X ν = nα n n p h ′′ ( t n,ν ) (cid:12)(cid:12)(cid:12) cos (cid:16) ρ n,ν + π (cid:17)(cid:12)(cid:12)(cid:12) ,C n,ε = nβ n X ν = nh ′ ( π − ε ) n p h ′′ ( t n,ν ) (cid:12)(cid:12)(cid:12) cos (cid:16) ρ n,ν + π (cid:17)(cid:12)(cid:12)(cid:12) ,E ε = 2 π Z h ′ ( ε ) h ′ (0) dt p h ′′ ( ψ ( t )) + 2 π Z h ′ ( π ) h ′ ( π − ε ) dt p h ′′ ( ψ ( t )) , and B n,ε = nh ′ ( π − ε ) X ν = nh ′ ( ε ) n p h ′′ ( t n,ν ) (cid:12)(cid:12)(cid:12) cos (cid:16) ρ n,ν + π (cid:17)(cid:12)(cid:12)(cid:12) − π Z h ′ ( π − ε ) h ′ ( ε ) dt p h ′′ ( ψ ( t )) . Since all the summands in A n,ε are non-negative, we get0 ≤ A n,ε ≤ X nα +1 ≤ ν ≤ nh ′ ( ε ) n p h ′′ ( t n,ν ) · | A n,ε | ≤ Cε , and, in the same way, | C n,ε | ≤ Cε .We obtain similarly from (23) that 0 ≤ E ε ≤ ε . It follows then that(59) |D n | ≤ | B n,ε | + 2( C + 1) ε. Now we let ϕ ( x ) = π ( h ( ψ ( x )) − xψ ( x )) (= − π h ∗ ( x ) where h ∗ denotes the Le-gendre transform of h ), and observe that ϕ ′ ( x ) = − π ψ ( x ), and ϕ ′′ ( x ) = − π ψ ′ ( x ) = − π h ′′ ( ψ ( x )) . Let J ε denote the interval [ h ′ ( ε ) , h ′ ( π − ε )]. For x ∈ J ε we have ψ ( x ) ∈ [ ε, π − ε ], and it follows that ϕ is twice continuously differentiable on J ε ,and that its second derivative is bounded away from zero on J ε . Therefore, byLemma 4, the array of fractional parts of nϕ (cid:0) νn (cid:1) = ρ n,ν π , n = 1 , , , . . . , νn ∈ J ε isequidistributed (in the sense of Lemma 3).Now we shall apply Lemma 3 to the interval J ε , the array of fractional partsof ρ n,ν π , and the functions f ( x ) = √ h ′′ ( ψ ( x )) and g ( x ) = (cid:12)(cid:12) cos π (cid:0) x + (cid:1)(cid:12)(cid:12) , which areRiemann integrable on J ε and [0 , g being periodic with period 1.Since1 n X { ν | νn ∈ J ε } f (cid:16) νn (cid:17) g (cid:16) ρ n,ν π (cid:17) = 1 n X { ν | νn ∈ J ε } q h ′′ (cid:0) ψ (cid:0) νn (cid:1)(cid:1) (cid:12)(cid:12)(cid:12) cos (cid:16) ρ n,ν + π (cid:17)(cid:12)(cid:12)(cid:12) , we obtain by Lemma 3(c) that the last expression tends to Z Jε p h ′′ ( ψ ( x )) dx · Z (cid:12)(cid:12)(cid:12)(cid:12) cos π (cid:18) x + 14 (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) dx = 2 π Z J ε p h ′′ ( ψ ( x )) dx. Therefore B n,ε → n → ∞ for any ε > 0, so from (59) we get D n → n → ∞ . This proves (58) and therefore Theorem 1.9. Proof of Theorem 2 Proof. Let S ( x ) = S ( x, h ) = 1 √ x ∞ X ν = −∞ | a x,ν | = 1 √ x (cid:13)(cid:13)(cid:13) e ixh ( t ) (cid:13)(cid:13)(cid:13) A where a x,ν = 12 π Z π − π exp ( i ( xh ( t ) − νt ) dt, and let L = L ( h ) = (cid:18) π (cid:19) Z π p | h ′′ ( t ) | dt. We shall prove that(1) S ( x ) is continuous on (1 , ∞ ), and(2) S ( nα ) → L as n → ∞ for any α real, α = 0.It is a well known fact, obtained by a category argument, that (1) and (2) implythat S ( x ) → L as the real variable x tends to ∞ . Therefore, to prove Theorem 2,it is sufficient to prove (1) and (2).Applying Lemma 1 part (a) to the case ϕ ( t ) = xh ( t ) − νt on [ − π, π ], we see thatfor any A > C > ≤ | x | ≤ A , | ν | ≥ A max | h ′ | , then | a x,ν | ≤ Cν . By the Weierstrass M-test the series P | a x,ν | is uniformly convergent on[ − A, A ], and since a x,ν are continuous function of x for all ν , S ( x, h ) is continuouson (1 , ∞ ).If h is 2 π -periodic, so is αh , and αh satisfies the conditions of Theorem 1 forall α > 0. So, by Theorem 1, S ( n, αh ) → L ( αh ) and we obtain S ( nα, h ) = √ α S ( n, αh ) → √ α L ( αh ) = L ( h ), which proves (2). (cid:3) Corollaries Corollary 1. (Oral communication of George Stey) Let J ν denotes the Besselfunction of integer order ν . Then √ x ∞ X ν = −∞ | J ν ( x ) | → (cid:0) (cid:1) as x tends to ∞ .Proof. From the well known fact e ix sin( t ) = ∞ X ν = −∞ J ν ( x ) e iνt we obtain that ∞ X ν = −∞ | J ν ( x ) | = (cid:13)(cid:13)(cid:13) e ix sin( t ) (cid:13)(cid:13)(cid:13) A N ASYMPTOTIC FORMULA FOR THE SEQUENCE k exp( inh ( t )) k A Since h ( t ) = sin( t ) satisfies the conditions of Theorem 2, and since Z π p | h ′′ ( t ) | dt = Z π p sin( t ) dt = B (cid:18) , (cid:19) = 4 π √ π Γ (cid:0) (cid:1) we obtain from the Theorem 2 that1 √ x ∞ X ν = −∞ | J ν ( x ) | → (cid:0) (cid:1) , x → ∞ . (cid:3) Corollary 2. Let (60) f ( t ) = J Y j =1 e it − α j − α j e it where α j ∈ (0 , for all j = 1 , . . . , J .Then (61) k f n k A √ n = (cid:18) π (cid:19) Z − vuut J X j =1 α j (cid:0) − α j (cid:1)(cid:0) α j − α j u (cid:1) (cid:0) − u (cid:1) − du + o (1) as n → ∞ Proof. It is obvious that f is continuous 2 π -periodic infinitely differentiable func-tion, and it is easy to see that f ( t ) = e − ih ( t ) , where(62) h ( t ) = J X j =1 i log (cid:18) e it − α j − α j e it (cid:19) . The first three derivatives of h are h ′ ( t ) = − J X j =1 − α j α j − α j cos t (63a) h ′′ ( t ) = J X j =1 α j (cid:0) − α j (cid:1) sin t (cid:0) α j − α j cos t (cid:1) (63b) h ′′′ ( t ) = J X j =1 α j (cid:0) − α j (cid:1) (cid:0) − α j + cos( t ) + α j cos( t ) + α j cos(2 t ) (cid:1)(cid:0) α j − α j cos( t ) (cid:1) (63c)As each of the terms of the sum in (63b) is positive, it is obvious that h ′′ ( t ) > t ∈ (0 , π ). Finally, h ′′′ (0) = J X j =1 α j (1 + α j )(1 − α j ) = 0(64) h ′′′ ( π ) = J X j =1 α j (1 − α j )(1 + α j ) = 0 . (65)Therefore the function h satisfies all the conditions of the Theorem 1, and since k f n k A = (cid:13)(cid:13) e − inh ( t ) (cid:13)(cid:13) A = P ∞ ν = −∞ | a n,ν | , where f n ( t ) = P ∞ ν = −∞ a n,ν e iνt , we have(66) 1 √ n ∞ X ν = −∞ | a n,ν | = (cid:18) π (cid:19) Z π vuut J X j =1 α j (cid:0) − α j (cid:1) sin t (cid:0) α j − α j cos t (cid:1) dt + o (1) as n → ∞ Substituting u = cos( t ) and simplifying will result in (61). (cid:3) Corollary 3. (Girard’s formula) If g ( z ) = z − α − αz , < | α | < then (67) (cid:13)(cid:13) g n (cid:0) e it (cid:1)(cid:13)(cid:13) A √ n = 16 √ (cid:18) Γ (cid:18) (cid:19)(cid:19) − p | α | | α | F (cid:18) , 34 ; 32 ; 4 | α | (1 + | α | ) (cid:19) + o (1) as n → ∞ .Proof. Without loss of generality we may assume that α is real and positive. Settingin Corollary2 J = 1, f ( t ) = g (cid:0) e it (cid:1) and α = α , we obtain(68) k f n k A √ n = (cid:18) π (cid:19) Z − s α (1 − α )(1 + α − αu ) (cid:0) − u (cid:1) − du + o (1) as n → ∞ .Substituting s = u the integral will become(69) 2 p α (1 − α )(1 + α ) Z (cid:18) − α (1 + α ) s (cid:19) − (1 − s ) − s − d s which can be written as(70) 8 π (cid:18) Γ (cid:18) (cid:19)(cid:19) − √ α α F (cid:18) , 34 ; 32 ; 4 α (1 + α ) (cid:19) (see for example theorems 16 and 21 in [6]) Substituting this for the integral in (68)will give us (67). (cid:3) Two Open Questions Theorem 1 and Corollary 2 make us wonder to what extent the convergence ofthe sequence √ n k exp( inh ( t )) k A is an exceptional phenomenon, and the followingtwo questions arise:(1) If the set of conditions in Theorem 1 is weakened so that the condition (2): h is odd is replaced by the condition (2’): h ′′ has no zeroes in ( − π, √ n k exp( inh ( t )) k A converges?(2) If B ( z ) is a finite Blaschke product, is it true that √ n (cid:13)(cid:13) B n (cid:0) e it (cid:1)(cid:13)(cid:13) A con-verges? References [1] Bogdan M. Bajsanski. Sur une classe g´en´erale de proc´ed´es de sommations du type d’Euler–Borel. Acad. Serbe Sci. Publ. Inst. Math. , X:131–152, 1956.[2] A. Beurling and H. Helson. Fourier–Stieltjes transforms with bounded powers. Math. Scand. ,1:120–126, 1953.[3] Dennis. M. Girard. The behavior of the norm of an automorphism of the unit disk. Pacific J.Math. , 47(2):443–456, 1973.[4] J. P. Kahane. Sur certaines classes de s´eries de Fourier absolument convergentes. J. Math.Pures. Appl. , 35(9):249–259, 1956.[5] J. P. Kahane. Transformees de fourier des fonctions sommables. In Proceedings of the In-ternational Congress of Mathematicians , Djursholm, 1962. Institut Mittag–Leffler, InstitutMittag–Leffler.[6] Earl D. Rainville. Special Functions . The Macmillan Company, 1960. N ASYMPTOTIC FORMULA FOR THE SEQUENCE k exp( inh ( t )) k A [7] P. Turan. A remark concerning the behaviour of a power series on the periphery of its conver-gence circle. Publ. Inst. Math. Acad. Serbe Sci. , 12:19–26, 1958.[8] A. Zygmund. Trigonometric Series , volume 1. Cambridge University Press, second edition,1956. Bogdan M. Baishanski, Department of Mathematics, The Ohio State University, 231W 18th Avenue, Columbus, OH 43210 E-mail address : [email protected] Jan Hlavacek, Department of Mathematical Sciences, Saginaw Valley State Univer-sity, 7400 Bay Road, University Center, MI 48710 E-mail address ::