Inequalities and Asymptotics for some Moment Integrals
aa r X i v : . [ m a t h . C A ] A p r Inequalities and Asymptotics for some MomentIntegrals
Faruk Abi-Khuzam ∗† Department of MathematicsAmerican University of BeirutBeirut, LebanonOctober 16, 2018
Abstract
For α > β − >
0, we obtain two sided inequalities for the momentintegral I ( α, β ) = R R | x | − β | sin x | α dx .These are then used to give the ex-act asymptotic behavior of the integral as α → ∞ . The case I ( α, α )corresponds to the asymptotics of Ball’s inequality, and I ( α, [ α ] −
1) cor-responds to a kind of novel ”oscillatory” behavior.
Ball’s integral inequality [KB] , in connection with cube-slicing in R n , says thatfor all s ≥ , Z ∞−∞ | sin( πx ) πx | s dx ≤ r s , or Z ∞−∞ | sin xx | s dx ≤ π r s , with strict inequality except when s = 2. In particular, it suggests that theintegral decays like √ s as s → ∞ , and this is made precise by the followingasymptotic [NP] : lim s →∞ r s Z ∞−∞ | sin( πx ) πx | s dx = r π . Since q π <
1, the asymptotic result implies the inequality for large values of s . But there are no known ”easy” proofs of the inequality for the full rangeof values,the main difficulty being near small values of s . e.g. between 2 and ∗ Mathematics Subject Classification. Primary † Key words and phrases . Ball’s inequality,asymptotics,cube slicing, moments. I ( α, β ) = Z ∞−∞ | sin x | α | x | β dx, α > β − > . We shall obtain useful upper and lower bounds for this integral, and use them toobtain the asymptotic behavior of this integral. In addition, the inequalities ob-tained are indispensible in obtaining the asymptotic behavior, especially in theinteresting ”oscillatory” cases, I ( α, [ α ]) if α ≥
2, and I ([ α ] , α −
1) if α >
1, where[ α ] is the greatest integer in α . The oscillatory behavior makes it impossible toemploy the standard methods used in connection with Ball’s inequality.We place no restrictions on the indices α and β beyond those necessary toensure the convergence of the integral I ( α, β ). Indeed, the condition β > ∞ , and near 0, the inequality | sin x | α | x | β ≤ | x | α − β implies convergence, since α − β > − A natural way to deal with Ball’s inequality is to apply the sharp form of theHausdorff-Young inequality [WB]. This leads to two inequalities for the relevantintegral: the first works for all s ≥
2, but falls short of the required inequalityby supplying the larger constant √ e in place of √ . The second gives a constantsmaller than √ s ≥ . Proposition 1 (a) If s ≥ ,then Z ∞−∞ | sin( πx ) πx | s dx ≤ √ s · s(cid:18) s − (cid:19) s − < r es . (b) If s ≥ , q = s , and p is the index conjugate to q, then Z R (cid:12)(cid:12)(cid:12)(cid:12)(cid:18) sin πξπξ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) s ≤ r s (cid:18) √ pp + 1 (cid:19) q/p < r s . Proof.
For part (a), let χ = χ ( − / , / be the characteristic function of theinterval ( − / , / . Then its Fourier transform is given by ˆ χ ( ξ ) = sin πξπξ . Applying the sharp Hausdorff-Young inequality [WB], k ˆ χ k s ≤ C p k χ k p , where s ≥ , p = s ′ , the index conjugate to s , and C p is given by C p = p /p ( s ) − /s , we obtain Z R | sin πξπξ | s dξ ≤ (cid:16) p /p s − /s (cid:17) s/ . t now remains to compute (cid:16) p /p s − /s (cid:17) s/ = 1 √ s · (cid:18) ss − (cid:19) s − = 1 √ s · s(cid:18) s − (cid:19) s − < r es , s ≥ , and the inequality in (a) follows.To prove part (b), we employ the convolution g = χ ∗ χ of the same charac-teristic function. A simple computation gives g ( x ) = − x ≤ x ≤
11 + x − ≤ x ≤ | x | ≥ . Now ˆ g ( ξ ) = (cid:16) sin πξπξ (cid:17) , k g k pp = p +1 , and an application of the sharp-Hausdorff-Young inequality gives, for q ≥ , and the conjugate index p = q ′ , Z R (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:18) sin πξπξ (cid:19) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q ≤ (cid:16) p /p q − /q (cid:17) q/ (cid:18) p + 1 (cid:19) q/p = q − / (cid:18) √ pp + 1 (cid:19) q/p < √ pp + 1 · r q . Since √ pp +1 < , we obtain the inequality, R ∞−∞ | sin( πx ) πx | s dx < q s for all s ≥ . In this section we consider the question of obtaining upper and lower boundsfor the more general integral, namely R ∞−∞ | sin x | α | x | β dx . Those bounds are thenused to obtain the precise asymptotic behaviour of the integral as α → ∞ . Inaddition, the bounds make it possible to employ discontinuous functions suchas [ α ] in place of β , and then the asymptotic result also captures the preciseoscillations in the values of the integral, as α → ∞ . Theorem 2
Suppose α > β − > , and put I ( α, β ) = Z R | sin t | α | t | β dt = 2 Z ∞ | sin t | α | t | β dt, and φ ( α, β ) = α α − β +12 Γ( α + 1)Γ( α − β +12 + α + 1) , where Γ is the gamma-function.Then (cid:18) α (cid:19) α − β +12 Γ( α − β + 12 ) φ ( α, β ) ≤ I ( α, β ) ≤ (cid:18) α (cid:19) α − β +12 Γ (cid:18) α − β + 12 (cid:19) (cid:26) β − (cid:27) . n particular, if β = α , then r α √ α Γ( α + 1)Γ( α + ) √ π ≤ I ( α, α ) ≤ r α ! √ π (cid:26) α − (cid:27) . Proof.
We need first the following double inequality, − x ≤ sin xx ≤ e − x , ≤ x ≤ π. The left-hand inequality is easily proved by calculus. It will be used with ≤ x ≤√ . For the right-hand inequality, since ≤ x ≤ π , we may use the inequalitybetween the geometric and arithmetic mean of positive numbers to obtain n Y k =1 (cid:18) − x π k (cid:19) ≤ − x π n n X k =1 k ! n ≤ exp − x π n X k =1 k ! . Letting n → ∞ , and recalling the product representation of the sine function,and P ∞ k =1 1 k = π , we obtain the second inequality. The next step is to comparethe full integral in the theorem to an integral over the interval [0 , √ or over [0 , π ] . Z √ | sin t | α | t | β dt ≤ Z π | sin t | α | t | β dt ≤ Z ∞ | sin t | α | t | β dt = Z π | sin t | α | t | β dt + ∞ X k =1 Z ( k +1) πkπ | sin t | α | t | β dt = Z π | sin t | α | t | β dt + ∞ X k =1 Z π | sin t | α | t + kπ | β dt ≤ Z π | sin t | α | t | β dt ( ∞ X k =1 k + 1) β ) ≤ (cid:26) β − (cid:27) Z π sin α tt β dt. Using the above inequalities for sin xx , Z √ t α − β (cid:18) − t (cid:19) α dt ≤ Z π sin α tt β dt ≤ Z π t α − β exp (cid:18) − αt (cid:19) dt. Simple substitutions to change variables bring this double inequality to the form
12 6 α − β +12 Z x α − β − (1 − x ) α dx ≤ Z π sin α tt β dt ≤ (cid:18) α (cid:19) α − β +12 Z π α x α − β − e − x dx. f we extend the right most integral to [0 , ∞ ) , and then express both sides throughthe gamma function, we arrive at
12 6 α − β +12 Γ( α − β +12 )Γ( α + 1)Γ( α − β +12 + α + 1) ≤ Z π sin α tt β dt ≤ (cid:18) α (cid:19) α − β +12 Γ( α − β + 12 ) . This gives the first inequalities for I ( α, β ) , and so, the inequalities for I ( α, α ) . Corollary 3
Let I ( α, β ) be the integral in the theorem. (a) I f α − β = c is held constant, while α → ∞ , then lim α →∞ α c +12 I ( α, β ) = 6 c +12 Γ( c + 12 ) , c > − . In particular, the asymptotic for the integral in Ball’s inequality is lim α →∞ √ αI ( α, α ) = √ π. (b) If α − β = c , and c remains bounded as α → ∞ , then I ( α, β ) ∽ (cid:18) α (cid:19) α − β +12 Γ (cid:18) α − β + 12 (cid:19) , α → ∞ . In particular, I ( α, [ α ]) ∽ (cid:18) α (cid:19) α − [ α ]+12 Γ (cid:18) α − [ α ] + 12 (cid:19) , α → ∞ . Proof. (a) In the very special case where β = α , Stirling’s formula gives φ ( α, α ) = √ α Γ( α + 1)Γ( α + ) ∼ α α +1 e / ( α + 1 / α +1 = e / (1 + α ) α +1 ∼ , α → ∞ . From this, the case where α − β = c , a constant, is handled similarly : φ ( α, β ) = α c +12 Γ( α + 1)Γ( c +12 + α + 1) ∼ α α + c +12 + 12 e − α ( α + c +12 ) α + c +12 + e − ( α + c +12 ) = e c +12 (1 + c +12 α ) α + c +12 + ∼ , α → ∞ . (c) If α − β = c > −
1, and c is only bounded, then Stirling’s formula followedby the inequality (1 + c +12 α ) α ≤ e c +12 , giveslim inf α →∞ φ ( α, β ) = lim inf α →∞ α c +12 Γ( α + 1)Γ( c +12 + α + 1)5 lim inf α →∞ e c +12 (1 + c +12 α ) α + c +12 + ≥ lim inf α →∞ c +12 α ) c +12 + = 1 . So that lim inf α →∞ "(cid:18) α (cid:19) α − β +12 Γ( α − β + 12 ) − I ( α, β ) ≥ . The corresponding lim sup being clearly ≤ , we obtain I ( α, β ) ∽ (cid:18) α (cid:19) α − β +12 Γ (cid:18) α − β + 12 (cid:19) , α − β bounded , β → ∞ . We conclude by a generalization of the asymptotic result for a class of infiniteproducts. Let g be a function having an infinite product representation of theform g ( t ) = ∞ Y n =1 (cid:18) − t t n (cid:19) , where t n >
0, and c = P ∞ n =1 t − n < ∞ . We are interested in investigatinglim p →∞ R ∞ | g ( t ) | p t β dt , where 0 ≤ β <
1. Two examples of such a function are: f ( x ) = Z cos( xt ) h ( t ) dt, J ( x ) = 2 π Z cos xt √ − t dt. The first function f was considered in [KK] in connection with maximal mea-sures of sections of the n -cube. The second is the Bessel function of order 0.We first review the case where β = 0. If 0 ≤ t ≤ t , we need two inequalitiesanalogous to those obtained for the sinc function. If 0 < a i <
1, then we usethe double inequality1 − ( a + a + ... + a n ) ≤ n Y k =1 (1 − a k ) ≤ − n n X k =1 a k ! n , to obtain − t n X k =1 t − k ! ≤ n Y k =1 (cid:18) − t t k (cid:19) ≤ − t n n X k =1 t − k ! n ≤ exp − t n X k =1 t − k ! , and letting n → ∞ , get1 − ct ≤ | g ( t ) | ≤ e − ct , c = ∞ X k =1 t − k , ≤ t ≤ √ c , and the right inequality when0 ≤ t ≤ t .The left-hand inequality gives Z ∞ | g ( t ) | p dt ≥ Z √ c (1 − ct ) p dt = 12 √ c Z (1 − x ) p x − / dx = 12 √ c Γ( p + 1) √ π Γ( p + ) . By Stirling’s formula we obtainlim inf p →∞ √ p Z ∞−∞ | g ( t ) | p dt ≥ r πc , which suggests that the order of decay of the integral is √ p , and so leads natu-rally a consideration of √ p R ∞ | g ( t ) | p dt = R ∞ | g ( t √ p ) | p dt . Now use of two sidedinequalities gives (1 − c t p ) p ≤ | g ( t √ p ) | p ≤ e − ct , where the left-hand inequality holds true for 0 ≤ t ≤ √ p √ c , and the rigt-handinequality holds true for 0 ≤ t ≤ t √ p . It now becomes possible to use, exactlyas done in [NP], Lebesgue’s dominated convergence theorem to conclude thatactually lim p →∞ √ p R ∞−∞ | g ( t ) | p dt = p πc .In the general case where 0 < β <
1, if we were to try the same approach,we would need to know beforehand the expected rate of decay. Thus using oneof the inequalities above, we obtain Z ∞ | g ( t ) | p t β dt ≥ Z √ c (1 − ct ) p t − β dt = 12 √ c − β Z (1 − x ) p x − β dx = 12 √ c − β Γ( p + 1)Γ( − β )Γ( − β + p + 1) , leading to a sharp lower asymptotic, namelylim inf p →∞ p − β Z ∞ | g ( t ) | p t β dt ≥ Γ( − β )2 √ c − β .Once again this suggests that the expected decay is like p β − . So we make thesubstitution t = (1 − β ) − β p − / x − β , and find that p − β Z ∞ | g ( t ) | p t β = Z ∞ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) g (1 − β ) − β x − β √ p !(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p dx. The inequalities − c (1 − β ) − β x − β p ! p ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) g (1 − β ) − β x − β √ p !(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p ≤ exp (cid:16) − c (1 − β ) − β x − β (cid:17) p →∞ p − β Z ∞ | g ( t ) | p t β dt = Γ( − β )2 √ c − β . The author declares that there are no other contributors to this article, andthat he has no competing interests.
References [WB] W. Beckner, Inequalities in Fourier Analysis, Amer. Journal of Math., (2)102 (1975). no.1, 159-182.[KB] K. Ball,Cube Slicing in R n , Proc.Amer. Math. Soc.,Vol.97, 3(1986),465-473[NP] F.L.Nazarov and A.N.Podkorytov, Ball, Haagerup, and distribution func-tions, Complex analysis, operators, and related topics. Oper. Theory Adv.Appl., vol.113, Birkhauser, Basel, 2000, pp.247-267.[KK] H. Konig and A. Koldobsky, On the maximal measure of sections of the nn