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On necessary multiplier conditions for Laguerre

arXiv:math/9307211v1 [math.CA] 9 Jul 1993On necessary multiplier conditions for Laguerreexpansions IIGeorge Gasper1 and Walter Trebels2Dedicated to Dick Askey and Frank Olver(March 5, 1992 version)Abstract. The necessary multiplier conditions for Laguerre expansions derivedin Gasper and Trebels [3] are supplemented and modified.

This allows us to placeMarkett’s Cohen type inequality [6] (up to the log–case) in the general framework ofnecessary conditions.Key words. Laguerre polynomials, necessary multiplier conditions, Cohen typeinequalities, fractional differences, weighted Lebesgue spacesAMS(MOS) subject classifications.

33C65, 42A45, 42C101IntroductionThe purpose of this sequel to [3] is to obtain a better insight into the structure ofLaguerre multipliers on Lp spaces from the point of view of necessary conditions.We recall that in [3] there occurs the annoying phenomenon that, e.g., the optimalnecessary conditions in the case p = 1 do not give the “right” unboundedness behaviorof the Ces`aro means. By slightly modifying these conditions we can not only remedythis defect but can also derive Markett’s Cohen type inequality [6] (up to the log–case) as an immediate consequence.For the convenience of the reader we briefly repeat the notation; we consider theLebesgue spacesLpw(γ) = {f : ∥f ∥Lpw(γ)= (Z ∞0|f(x)e−x/2|pxγ dx)1/p < ∞} ,1 ≤p < ∞,denote the classical Laguerre polynomials by Lαn(x), α > −1, n ∈N0 (see Szeg¨o [8,p.

100]), and setRαn(x) = Lαn(x)/Lαn(0),Lαn(0) = Aαn = n + αn=Γ(n + α + 1)Γ(n + 1)Γ(α + 1).1Department of Mathematics, Northwestern University, Evanston, IL 60208, USA. The work ofthis author was supported in part by the National Science Foundation under grant DMS-9103177.2Fachbereich Mathematik, TH Darmstadt, D–6100 Darmstadt, Germany.1

Associate to f its formal Laguerre seriesf(x) ∼(Γ(α + 1))−1∞Xk=0ˆfα(k)Lαk(x),where the Fourier Laguerre coefficients of f are defined byˆfα(n) =Z ∞0f(x)Rαn(x)xαe−x dx(1)(if the integrals exist). A sequence m = {mk} is called a (bounded) multiplier onLpw(γ), notation m ∈Mpw(γ), if∥∞Xk=0mk ˆfα(k)Lαk ∥Lpw(γ)≤C ∥f ∥Lpw(γ)for all polynomials f; the smallest constant C for which this holds is called the multi-plier norm ∥m ∥Mpw(γ) .

The necessary conditions will be given in certain “smoothness”properties of the multiplier sequence in question. To this end we introduce a fractionaldifference operator of order δ by∆δmk =∞Xj=0A−δ−1jmk+j(whenever the sum converges), the first order difference operator ∆2 with increment2 by∆2mk = mk −mk+2,and the notation∆2∆δmk = ∆δ+1mk + ∆δ+1mk+1.Generic positive constants that are independent of the functions (and sequences) willbe denoted by C. Within the setting of the Lpw(γ)-spaces our main results now read(with 1/p + 1/q = 1):Theorem 1.1 Let α, a > −1 and α + a > −1.

If f ∈Lpw(γ), 1 ≤p < 2, then ∞Xk=0|(k + 1)(γ+1)/p−1/2∆2∆a ˆfα(k)|q!1/q≤C ∥f ∥Lpw(γ),(2)providedγ + 1p≤α + ap+ 1if α + a ≤1/2,γ + 1p≤α + a2+ 1 + 12 1p −12!if α + a > 1/2.2

As in [3] (see there the proof of Lemma 2.3) we immediately obtainTheorem 1.2 Let m = {mk} ∈Mpw(γ), 1 ≤p < 2, and let α and a be as in Theorem1.1. Thensupn 2nXk=n|(k + 1)(2γ+1)/p−(2α+1)/2∆2∆amk|q1k + 1!1/q≤C ∥m ∥Mpw(γ),(3)provided that in the case α + a ≤1/2 the conditionα + ap+ 1 ≥γ + 1p> (α + 1)/2 + 1/3pif 1 ≤p < 4/3(α + 1)/2 + 1/4if 4/3 ≤p < 2holds, and in the case α + a > 1/2 the conditionα + a2+ 1 + 12 1p −12!≥γ + 1p> (α + 1)/2 + 1/3pif 1 ≤p < 4/3(α + 1)/2 + 1/4if 4/3 ≤p < 2.In view of the results in [6], [3] and for an easy comparison we want to emphasize thecases γ = α and γ = αp/2.

Therefore, we stateCorollary 1a) Let m ∈Mpw(α), 1 ≤p < 2, and let α > −1 be such thatmax{1/(3p), 1/4} < (α + 1)(1/p −1/2). Then, with λ := (2α + 1)(1/p −1/2),supn 2nXk=n|(k + 1)λ∆2∆λ−1mk|q1k + 1!1/q≤C ∥m ∥Mpw(α) .b) Let m ∈Mpw(αp/2), 1 ≤p < 4/3, and (α −1)(1/p −1/2) ≥−1/2 .

Thensupn 2nXk=n|(k + 1)1/p−1/2∆2mk|q1k + 1!1/q≤C ∥m ∥Mpw(αp/2) .Remarks. 1) For polynomial f(x) =Pnk=0 ckLαk(x) Theorem 1.1 yields, by takingonly the (k = n)–term on the left hand side of (2),|cn|(n + 1)(γ+1)/p−1/2 ≤C ∥f ∥Lpw(γ),1 ≤p < 2(under the restrictions on γ of Theorem 1.1).

In particular, if we choose γ = α, thiscomprises formula (1.13) in Markett [6] for his basic case β = α. For γ = αp/2, iteven extends formula (1.14) in [6] to negative α’s as described in Corollary 1.3, b).The case 2 < p < ∞can be done by an application of a Nikolskii inequality, see [6].2) Analogously, Cohen type inequalities follow from Theorem 1.2; in particular, Corol-lary 1.3 yields3

Corollary 2 Let m = {mk}nk=0 be a finite sequence, 1 ≤p < 2, and α > −1.a) If m ∈Mpw(α) then(n + 1)(2α+2)(1/p−1/2)−1/2|mn| ≤C ∥m ∥Mpw(α),1 ≤p < 4α + 42α + 3,provided max{1/3p, 1/4} < (α + 1)(1/p −1/2).b) If m ∈Mpw(αp/2) and (α −1)(1/p −1/2) ≥−1/2, then(n + 1)2/p−3/2|mn| ≤C ∥m ∥Mpw(αp/2),1 ≤p < 4/3.With the exception of the crucial log–case, i.e. p0 = (4α + 4)/(2α + 3) or p0 = 4/3,resp., Corollary 1.4 contains Markett’s Theorem 1 in [6] and extends it to negativeα’s.

In particular we obtain for the Ces`aro means of order δ ≥0, represented by itsmultiplier sequence mδk,n = Aδn−k/Aδn, the “right” unboundedness behavior (see [4] )∥{mδk,n} ∥Mpw(α)≥C(n + 1)(2α+2)(1/p−1/2)−1/2−δ,1 ≤p <4α + 42α + 3 + 2δ.3) There arises the question, in how far the type of necessary conditions in [3] arecomparable with the present ones. Let λ > 1.

Since ∆2mk = ∆mk + ∆mk+1 weobviously havesupn 2nXk=n|(k + 1)(2γ+1)/p−(2α+1)/2∆2∆λ−1mk|q1k + 1!1/q(4)≤C supn 2nXk=n|(k + 1)(2γ+1)/p−(2α+1)/2∆λmk|q1k + 1!1/q.In general, a converse cannot hold as can be seen by the following example: chooseγ = α, λ = (2α + 1)(1/p −1/2) and mk = (−1)kk−ε, 0 < ε < 1. Thensupn 2nXk=n|(k + 1)∆mk|q1k + 1!1/q= ∞and hence by the embedding properties of the wbv–spaces, see [2], the right hand sideof (4) cannot be finite for all λ > 1.

But since ∆2∆λ−1mk = ∆λ−1∆2mk ∼(k+1)−ε−λ,the left hand side of (4) is finite for all λ > 1.Theorem 1.1 will be proved in Section 2 by interpolating between (L1, l∞)– and(L2, l2)–estimates. The a ̸= 0 case is an easy consequence of the case a = 0 when4

one uses the basic formula (see formula (3) in [3] and Remark 3 preceding Section 3there)∆aRαk(x) =Γ(α + 1)Γ(α + a + 1)xaRα+ak(x),x > 0, a > −1 −min{α, α/2 −1/4},(5)where in the case a > −(2α + 1)/4 the series for the fractional difference convergesabsolutely. In Section 3, a necessary (L1, l1)–estimate is derived and it is comparedwith a corresponding sufficient (l1, L1)–estimate.2Proof of Theorem 1.1Let us first handle the (L2, l2)–estimate.

Since∆2∆a ˆfα(k) = ∆1+a ˆfα(k) + ∆1+a ˆfα(k + 1)it follows from the Parseval formula preceding Corollary 2.5 in [3] that ∞Xk=0|qAα+1+ak∆2∆a ˆfα(k)|2!1/2≤CZ ∞0|f(t)e−t/2t(α+1+a)/2|2 dt1/2. (6)Concerning the (L1, l∞)–estimate we first restrict ourselves to the case a = 0.

Defineµ ∈R by2 1p −12!µ = γp −α + 12;with the notation Lαk(t) = (Aαk/Γ(α + 1))1/2Rαk(t)e−t/2tα/2 it follows that|∆2 ˆfα(k)| = C|Z ∞0f(t){Lαk(t)/qAαk −Lαk+2(t)/qAαk+2}e−t/2tα/2 dt|≤C(k + 1)−1−α/2Z ∞0|f(t)||t−µ−1/2Lαk(t)|e−t/2t(α+1)/2+µ dt+C(k + 1)−α/2Z ∞0|f(t)||t−µ−1/2{Lαk(t) −Lαk+2(t)}|e−t/2t(α+1)/2+µ dt = I + II.We distinguish the two cases α ≤1/2 and α > 1/2:First consider the case α ≤1/2 . By the asymptotic estimates for Lαk(t) −Lαk+2(t) inAskey and Wainger [1, p.699], see formula (2.12) in [6], it follows for γ ≤α + p −1that∥t−µ−1/2{Lαk(t) −Lαk+2(t)} ∥∞≤C(k + 1)−1−µ5

so thatII ≤C(k + 1)−1−µ−α/2Z ∞0|f(t)|e−t/2t(α+1)/2+µ dt,γ ≤α + p −1. (7)By Lemma 1, 4th case, in [5]∥t−µ−1/2Lαk(t) ∥∞≤C(k + 1)−µ−5/6so that triviallyI ≤C(k + 1)−1−µ−α/2Z ∞0|f(t)|e−t/2t(α+1)/2+µ dt,γ + 1p≤α + 12−13p + 23.By Lemma 1, 5th case, in [5]∥t−µ−1/2Lαk(t) ∥∞≤C(k + 1)µ+1/2so thatI ≤C(k + 1)µ−1/2−α/2Z ∞0|f(t)|e−t/2t(α+1)/2+µ dt,≤C(k + 1)−1−µ−α/2Z ∞0|f(t)|e−t/2t(α+1)/2+µ dt,γ + 1p> α + 12−13p + 23,provided that µ −(α + 1)/2 ≤−1 −µ −α/2 which is equivalent to µ ≤−1/4 orγ ≤3p/4 −1/2 + αp/2.

But this is no further restriction since for α ≤1/2 thereholds α + p −1 ≤3p/4 −1/2 + αp/2. Summarizing, for −1 < α ≤1/2, γ ≤α + p −1and µ = (γ/p −(α + 1)/2)/2(1/p −1/2) we have thatsupk|(k + 1)1+µ+α/2∆2 ˆfα(k)| ≤CZ ∞0|f(t)|e−t/2t(α+1)/2+µ dt.

(8)Now consider the case α > 1/2. Then, by formula (2.12) in [6], (7) is obviously truewhen (γ + 1)/p ≤α/2 + 1 + (1/p −1/2)/2.

Again, the application of Lemma 1 in[5] requires γ ≤α + p −1, which for α > 1/2 is less restrictive than (γ + 1)/p ≤α/2 + 1 + (1/p −1/2)/2. Its 4th case now leads toI ≤C(k + 1)−11/6−µ−α/2Z ∞0|f(t)|e−t/2t(α+1)/2+µ dt,γ + 1p≤α + 12−13p + 23,and its 5th case toI ≤C(k + 1)µ−1/2−α/2Z ∞0|f(t)|e−t/2t(α+1)/2+µ dt,(γ + 1)/p > α + 12−13p + 23.6

But µ −1/2 −α/2 ≤−µ −1 −α/2 if (γ + 1)/p ≤α/2 + 1 + (1/p −1/2)/2; so that,summarizing, (8) also holds under this restriction for α > 1/2.Now an application of the Stein and Weiss interpolation theorem (see [7]) with Tf ={Tf(k)}and Tf(k) =qAα+1k∆2 ˆfα(k) gives the assertion of Theorem 1.1 in the casea = 0.If a ̸= 0 then by (1), the definition of ∆2∆a, and by (5)∆2∆a ˆfα(k) = C{∆ˆfα+a(k) + ∆ˆfα+a(k + 1)} = C∆2 ˆfα+a(k),since already the condition γ < α + a + 1 (which implies no new restriction) givesabsolute convergence of the infinite sum and integral involved (see the formula follow-ing (9) in [3]) and Fubini’s Theorem can be applied. Hence all the previous estimatesremain valid when α is replaced by α + a.3A variant for integrable functionsTheorem 1.1 gives a necessary condition for a sequence {fk} to generate with respectto Lαk an L1w(γ)–function.

But this condition is hardly comparable with the followingsufficient one which is a slight modification of Lemma 2.2 in [3].Theorem 3.1 Let α > −1 and δ > 2γ −α + 1/2 ≥0. If {fk} is a bounded sequencewith limk→∞fk = 0 and∞Xk=0(k + 1)δ+α−γ|∆δ+1fk| ≤K{fk},then there exists a function f ∈L1w(γ) with ˆfα(k) = fk for all k ∈N0 and∥f ∥L1w(γ)≤C K{fk}for some constant C independent of the sequence {fk}.The proof follows along the lines of Lemma 2.2 in [3] since the norm of the Ces`arokernelχα,δn (x) = (AδnΓ(α + 1))−1nXk=0Aδn−kLαk(x) = (AδnΓ(α + 1))−1Lα+δ+1n(x)can be estimated with the aid of Lemma 1 in [5] by∥χα,δk∥L1w(γ)≤C(k + 1)α−γ,δ > 2γ −α + 1/2The variant of Theorem 1.1 in the case p = 1 is7

Theorem 3.2 If α > −1 and γ > max{−1/3, α/2 −1/6}, then∞Xk=0(k + 1)γ−2/3|∆2γ−α+1/3 ˆfα(k)| ≤C ∥f ∥L1w(γ) .A comparison of the sufficient condition and the necessary one nicely shows where theL1w(γ)–functions live; in particular we see that the “smoothness” gap (the differenceof the orders of the difference operators) is just greater than 7/6. It is clear thatTheorem 3.2 can be modified by using the ∆2–operator.

Theorem 3.2 does not followfrom the p = 1 case of Lemma 2.1 in [3] since that estimate would lead to the divergentsumP∞k=0(k + 1)−1 ∥f ∥L1w(γ).ProofBy formula (5) we have∆2γ−α+1/3 ˆfα(k) = CZ ∞0f(t)R2γ+1/3k(t)t2γ+1/3e−t dt= C(k + 1)−γ−1/6Z ∞0f(t)L2γ+1/3k(t)tγ+1/6e−t/2 dtand hence∞Xk=0(k + 1)γ−2/3|∆2γ−α+1/3 ˆfα(k)| ≤CZ ∞0|f(t)|∞Xk=0(k + 1)−5/6|t1/6L2γ+1/3k(t)|tγe−t/2 dtif the right hand side converges. To show this we discuss for j ∈Zsup2j≤t≤2j+1∞Xk=0(k + 1)−5/6|t1/6L2γ+1/3k(t)|and prove that this quantity is uniformly bounded in j, whence the assertion.First consider those j ≥0 for which there exists a nonnegative integer n such that0 ≤k ≤2n implies 3ν/2 := 3(2k + 2γ + 4/3) ≤2j but such that this inequalityfails to hold for k ≥2n+1; the latter assumption in particular implies that essentiallyν/2 ≥2j+1 for k ≥2n+4.

Since ∥t1/6L2γ+1/3k(t) ∥∞≤C(k + 1)−1/6 by Lemma 1 in [5],we obviously have∞Xk=0(k + 1)−5/6|t1/6L2γ+1/3k(t)| ≤2nXk=0+∞Xk=2n+4. .

. + O(1).

(9)For k = 0, . .

. , 2n we can now apply the fourth case of formula (2.5) in [5] to obtain|t1/6L2γ+1/3k(t)| ≤Ce−µ2j for some positive constant µ and the first sum on the right8

hand side of (9) is bounded uniformly in j. In consequence of the choice of n thesecond case of formula (2.5) in [5] can be used for k ≥2n+4, giving∞Xk=2n+4(k + 1)−5/6|t1/6L2γ+1/3k(t)| ≤Ct−1/12∞Xk=2n+4(k + 1)−13/12 = O(1)since 2j ≤t ≤2j+1 and j and n are comparable.Now consider the remaining j’s: We have to split up the sumP∞k=0 .

. .

into two parts,one where k is such that 2jν ≥1 (this contribution has just been seen to be uniformlybounded in j), the other where k is such that 2jν ≤1. To deal with the last casechoose again n to be the greatest integer such that 2n+2 + 4γ + 8/3 ≤2−j; this time,n and −j are comparable and we obtain by the first case of (2.5) in [5]2nXk=0(k + 1)−5/6|t1/6L2γ+1/3k(t)| ≤Ctγ+1/32nXk=0(k + 1)γ−2/3 = O(1)if 2j ≤t ≤2j+1, γ > −1/3, which completes the proof.9

References[1] R. Askey and S. Wainger, Mean convergence of expansions in Laguerre andHermite series, Amer. J.

Math., 87 (1965), pp. 695 – 708.

[2] G. Gasper and W. Trebels, A characterization of localized Bessel poten-tial spaces and applications to Jacobi and Hankel multipliers, Studia Math., 65(1979), pp. 243 – 278.

[3] G. Gasper and W. Trebels, Necessary multiplier conditions for Laguerreexpansions, Canad. J.

Math., 43 (1991), pp. 1228 – 1242.

[4] E. G¨orlich and C. Markett, A convolution structure for Laguerre series,Indag. Math., 44 (1982), pp.

161 – 171. [5] C. Markett, Mean Ces`aro summability of Laguerre expansions and normestimates with shifted parameter, Anal.

Math., 8 (1982), pp. 19 – 37.

[6] C. Markett, Cohen type inequalities for Jacobi, Laguerre and Hermite ex-pansions, SIAM J. Math.

Anal., 14 (1983), pp. 819 – 833.

[7] E.M. Stein and G. Weiss, Interpolation of operators with change of mea-sures,Trans. Amer.

Math. Soc., 87 (1958), pp.

159 – 172. [8] G. Szeg¨o, Orthogonal Polynomials, 4th ed., Amer.

Math. Soc.

Colloq. Publ.23, Providence, R.I., 1975.10

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