A Perturbation of the Dunkl Harmonic Oscillator on the Line

Symmetry, Integrability and Geometry: Methods and Applications SIGMA 11 (2015), 059, 47 pages A Perturbation of the Dunkl Harmonic Oscillator on the Line Jes´us A. ´ ALVAREZ L ´ OPEZ †, Manuel CALAZA ‡and Carlos FRANCO † †Departamento de Xeometr´ıa e Topolox´ıa, Facultade de Matem´aticas, Universidade de Santiago de Compostela, 15782 Santiago de Compostela, Spain E-mail: jesus.alvar[email protected],carlosluis.franc[email protected] ‡Laboratorio de Investigaci´on 2 and Rheumatology Unit, Hospital Clinico Universitario de Santiago, Santiago de Compostela, Spain E-mail: manuel.c[email protected] Received February 19, 2015, in final form July 20, 2015; Published online July 25, 2015; Corrected June 28, 2017 https://doi.org/10.3842/SIGMA.2015.059 Abstract. Let Jσbe the Dunkl harmonic oscillator on R(σ > −1 2). For 0 <u<1 and ξ > 0, it is proved that, if σ > u −1 2, then the operator U=Jσ+ξ|x|−2u, with appropriate domain, is essentially self-adjoint in L2(R,|x|2σdx), the Schwartz space Sis a core of U1/2, and Uhas a discrete spectrum, which is estimated in terms of the spectrum of Jσ. A generalization Jσ,τ of Jσis also considered by taking different parameters σand τ on even and odd functions. Then extensions of the above result are proved for Jσ,τ , where the perturbation has an additional term involving, either the factor x−1on odd functions, or the factor xon even functions. Versions of these results on R+are derived. Key words: Dunkl harmonic oscillator; perturbation theory 2010 Mathematics Subject Classification: 47A55; 47B25; 33C45 1 Introduction The Dunkl operators on Rnwere introduced by Dunkl [8,9,10], and gave rise to what is now called the Dunkl theory [25]. They play an important role in physics and stochastic processes (see, e.g., [13,24,27]). In particular, the Dunkl harmonic oscillators on Rnwere studied in [11,19,20,23]. We will consider only this operator on R, where it is uniquely determined by one parameter. In this case, a conjugation of the Dunkl operator was previously introduced by Yang [28] (see also [21]). Let us fix some notation that is used in the whole paper. Let S=S(R) be the Schwartz space on R, with its Fr´echet topology. It decomposes as direct sum of subspaces of even and odd functions, S=Sev ⊕ Sodd. The even/odd component of a function in Sis denoted with the subindex ev/odd. Since Sodd =xSev, where xis the standard coordinate of R,x−1φ∈ Sev is defined for φ∈ Sodd. Let L2 σ=L2(R,|x|2σdx) (σ∈R), whose scalar product and norm are denoted by h,iσand k kσ. The above decomposition of Sextends to an orthogonal decomposition, L2 σ=L2 σ,ev ⊕L2 σ,odd, because the function |x|2σis even. Sis a dense subspace of L2 σif σ > −1 2, and Sodd is a dense subspace of L2 τ,odd if τ > −3 2. Unless otherwise stated, we assume σ > −1 2and τ > −3 2. The domain of a (densely defined) operator Pin a Hilbert space is denoted by D(P). If Pis closable, its closure is denoted by P. The domain of a (densely defined) sesquilinear form pin a Hilbert space is denoted by D(p). The quadratic form of pis also denoted by p. If pis closable, its closure is denoted by ¯ p. For an operator in L2 σpreserving the above decomposition, its restrictions to L2 σ,ev/odd will be indicated with the subindex ev/odd. The 2 J.A. ´ Alvarez L´opez, M. Calaza and C. Franco operator of multiplication by a continuous function hin L2 σis also denoted by h. The harmonic oscillator is the operator H=−d2 dx2+s2x2(s > 0) in L2 0with D(H) = S. The Dunkl operator on Ris the operator Tin L2 σ, with D(T) = S, determined by T=d dx on Sev and T=d dx + 2σx−1on Sodd, and the Dunkl harmonic oscillator on Ris the operator J=−T2+s2x2in L2 σwith D(J) = S. Thus Jpreserves the above decomposition of S, being Jev =H−2σx−1d dx and Jodd =H−2σd dxx−1. The subindex σis added to Jif needed. This J is essentially self-adjoint, and the spectrum of Jis well known [23]; in particular, J > 0. In fact, even for τ > −3 2, the operator Jτ,odd is defined in L2 τ,odd with D(Jτ,odd) = Sodd because it is a conjugation of Jτ+1,ev by a unitary operator (Section 2). Some operators of the form J+ξx−2 (ξ∈R) are conjugates of Jby powers |x|a(a∈R), and therefore their study can be reduced to the case of J[3]. Our first theorem analyzes a different perturbation of J. Theorem 1.1. Let 0< u < 1and ξ > 0. If σ > u −1 2, then there is a positive self-adjoint operator Uin L2 σsatisfying the following: (i)Sis a core of U1/2, and, for all φ, ψ ∈ S, U1/2φ, U1/2ψσ=hJφ, ψiσ+ξ|x|−uφ, |x|−uψσ.(1.1) (ii)Uhas a discrete spectrum. Let λ0≤λ1≤ ··· be its eigenvalues, repeated according to their multiplicity. There is some D=D(σ, u)>0, and, for each  > 0, there is some C=C(, σ, u)>0so that, for all k∈N, (2k+ 1 + 2σ)s+ξDsu(k+ 1)−u≤λk≤(2k+ 1 + 2σ)(s+ξsu) + ξCsu.(1.2) Remark 1.2. In Theorem 1.1, observe the following: (i) The second term of the right hand side of (1.1) makes sense because |x|−uS ⊂ L2 σsince σ > u −1 2. (ii) U=U, where U:= J+ξ|x|−2uwith D(U) = T∞ m=0 D(Um) (see [14, Chapter VI, Section 2.5]). The more explicit notation Uσwill be also used if necessary. (iii) The restrictions Uev/odd are self-adjoint in L2 σ,ev/odd and satisfy (1.1) with φ, ψ ∈ Sev/odd and (1.2) with keven/odd. In fact, by the comments before the statement, Uτ,odd is defined and satisfies these properties if τ > u −3 2. To prove Theorem 1.1, we consider the positive definite symmetric sesquilinear form udefined by the right hand side of (1.1). Perturbation theory [14] is used to show that uis closable and ¯ u induces a self-adjoint operator U, and to relate the spectra of Uand J. Most of the work is devoted to check the conditions to apply this theory so that (1.2) follows; indeed, (1.2) is stronger than a general eigenvalue estimate given by that theory (Remark 3.22). The following generalizations of Theorem 1.1 follow with a simple adaptation of the proof. If ξ < 0, we only have to reverse the inequalities of (1.2). In (1.1), we may use a finite sum Piξih|x|−uiφ, |x|−uiψiσ, where 0 < ui<1, σ > ui−1 2and ξi>0; then (1.2) would be modified by using maxiuiand miniξiin the left hand side, and maxiξiin the right hand side. In turn, this can be extended by taking Rp-valued functions (p∈Z+), and a finite sum Pih|x|−uiΞiφ, |x|−uiψiσin (1.1), where each Ξiis a positive definite self-adjoint endomorphism of Rp; then the minimum and maximum eigenvalues of all Ξiwould be used in (1.2). As an open problem, we may ask for a version of Theorem 1.1 using Dunkl operators on Rn, but we are interested in the following different type of extension. For σ > −1 2and τ > −3 2, let L2 σ,τ =L2 σ,ev ⊕L2 τ,odd, whose scalar product and norm are denoted by h,iσ,τ and k kσ,τ . Matrix expressions of operators refer to this decomposition. Let Jσ,τ =Jσ,ev ⊕Jτ,odd in L2 σ,τ , with D(Jσ,τ ) = S. The hypotheses of the generalization of Theorem 1.1 are rather involved to cover enough cases of certain application that will be indicated. A Perturbation of the Dunkl Harmonic Oscillator on the Line 3 Theorem 1.3. Let ξ > 0and η∈R, let 0< u < 1, σ > u −1 2, τ > u −3 2, θ > −1 2,(1.3) and set v=σ+τ−2θ. Suppose that the following conditions hold: (a)If σ=θ6=τand τ−σ6∈ −N, then σ−1< τ < σ + 1,2σ+1 2.(1.4) (b)If σ6=θ=τand σ−τ6∈ −N, then −τ, τ −1< σ < 3τ+ 1,11τ+ 2, τ + 1.(1.5) (c)If σ6=θ=τ+ 1 and σ−τ−16∈ −N, then τ+ 1 < σ < τ + 3,2τ+7 2.(1.6) (d)If σ6=θ6=τand σ−θ, τ −θ6∈ −N, then σ−τ 2−1,τ−σ 2,σ+τ−1 4,σ+3τ−2 14 ,3σ+τ−4 14 ,σ+τ−1 2< θ < σ+τ+1 2, τ−1< σ < τ + 3.(1.7) Then there is a positive self-adjoint operator Vin L2 σ,τ satisfying the following: (i)Sis a core of V1/2, and, for all φ, ψ ∈ S, V1/2φ, V1/2ψσ,τ =hJσ,τ φ, ψiσ,τ +ξ|x|−uφ, |x|−uψσ,τ +ηx−1φodd, ψevθ+φev, x−1ψoddθ.(1.8) (ii)Let ςk=σif kis even, and ςk=τif kis odd. Vhas a discrete spectrum. Its eigenvalues form two groups, λ0≤λ2≤ ··· and λ1≤λ3≤ ···, repeated according to their multiplicity, such that there is some D=D(σ, τ, u)>0and, for every  > 0, there are some C= C(, σ, τ, u)>0and E=E(, σ, τ, θ)>0so that, for all k∈N, λk≥(2k+ 1 + 2ςk)s−2|η|sv+1 2+ξDsu(k+ 1)−u−2|η|Esv+1 2,(1.9) λk≤(2k+ 1 + 2ςk)s+ξsu+ 2|η|sv+1 2+ξCsu+ 2|η|Esv+1 2.(1.10) (iii)Let ˜u∈Rsuch that 0, v, τ −2θ+1 2, σ −2θ−1 2<˜u < 1, v + 1, σ +1 2, τ +3 2,(1.11) and let ˆu= max{˜u, v + 1−˜u}. There is some D=D(σ, τ, u)>0and, for any  > 0, there is some e C=e C(, σ, τ, u)>0so that, for all k∈N, λk≥(2k+ 1 + 2ςk)s−|η|sˆu+ξDsu(k+ 1)−u−|η|e Csˆu.(1.12) (iv)If u=v+1 2and ξ≥ |η|, then there is some e D=e D(σ, τ, u)>0so that, for all k∈N, λk≥(2k+ 1 + 2ςk)s+ (ξ−|η|)e Dsu(k+ 1)−u.(1.13) 4 J.A. ´ Alvarez L´opez, M. Calaza and C. Franco (v)If we add the term ξ0hφev, ψeviσ+ξ00hφodd, ψoddiτto the right hand side of (1.8), for some ξ0, ξ00 ∈R, then the result holds as well with the additional term max{ξ0, ξ00}in the right hand side of (1.10), and the additional term, ξ0for k∈2Nand ξ00 for k∈2N+ 1, in the right hand sides of (1.9),(1.12)and (1.13). Remark 1.4. Note the following in Theorem 1.3: (i) Like in Remark 1.2(ii), we have V=V, where V=Uσ,ev η|x|2(θ−σ)x−1 η|x|2(θ−τ)x−1Uτ,odd , with D(V) = T∞ m=0 D(Vm). Note that the adjoint of |x|2(θ−σ)x−1:Sodd → |x|2(θ−σ)Sev, as a densely defined operator of L2 τ,odd to L2 σ,ev, is given by |x|2(θ−τ)x−1, with the appropriate domain. (ii) Taking θ0=θ−1>−3 2, since hxφ, ψiθ0=φ, x−1ψθ for all φ∈ Sev and ψ∈ Sodd, we can write (1.8) as V1/2φ, V1/2ψσ,τ =hJσ,τ φ, ψiσ,τ +ξ|x|−uφ, |x|−uψσ,τ +ηhφodd, xψeviθ0+hxφev, ψoddiθ0 for all φ, ψ ∈ S, and, correspondingly, V=Uσ,ev η|x|2(θ0−σ)x η|x|2(θ0−τ)x Uτ,odd . (iii) The conditions (1.4), (1.5) and (1.6) describe three convex open subsets of R2(Fig. 1). The condition (1.7) describes a convex open subset of R3(Fig. 2), which is symmetric with respect to the plane defined by σ=τ+ 1. It is a “semi-infinite bar” with 4 lateral faces, and 5 faces at the “bounded end.” (iv) In Theorem 1.3(iii), the condition (1.11) means that (1.3) also holds with ˜uand v+ 1 −˜u instead of u. There exists ˜usatisfying (1.11) just when 0, v, τ −2θ+1 2, σ −2θ−1 2<1, v + 1, σ +1 2, τ +3 2.(1.14) This property is satisfied in the cases (b) and (d) by (1.3), (1.5) and (1.7); in particular, we can take ˜u=v+1 2. In the case (a), if τ < 3σ, then (1.14) holds by (1.3) and (1.4). In the case (c), if σ < 3τ+ 4, then (1.14) holds by (1.3) and (1.6). The main arguments of the proofs of Theorems 1.1 and 1.3 are given in Sections 3–5. But some needed estimates are postponed to Sections 6and 7because they are of rather independent nature, and with rather long and tedious proofs. Versions of these results on R+are also derived in Section 8(Corollaries 8.1,8.2 and 8.3). In [4], these corollaries are used to study a version of the Witten’s perturbation ∆sof the Laplacian on strata with the general adapted metrics of [6,17,18]. This gives rise to an analytic proof of Morse inequalities in strata involving intersection homology of arbitrary perversity, which was our original motivation. The simplest case of adapted metrics, corresponding to the lower middle perversity, was treated in [2] using an operator induced by Jon R+. The perturbations of Jstudied here show up in the local models of ∆swhen general adapted metrics are considered. Some details of this application are given in Section 9. A Perturbation of the Dunkl Harmonic Oscillator on the Line 5 (a) Set defined by (1.4). (b) Set defined by (1.5). (c) Set defined by (1.6). Figure 1. Sets in Theorem 1.3(a),(b),(c). Figure 2. Set defined by (1.7) in Theorem 1.3(d). 2 Preliminaries The Dunkl annihilation and creation operators are B=sx+Tand B0=sx−T(s > 0). Like J, the operators Band B0are considered in L2 σwith domain S. They are perturbations of the usual annihilation and creation operators. The operators T,B,B0and Jare continuous on S. The following properties hold [3,23]: •B0is adjoint of B, and Jis essentially self-adjoint. •The spectrum of Jconsists of the eigenvalues1(2k+ 1 + 2σ)s,k∈N, of multiplicity one. •The corresponding normalized eigenfunctions φkare inductively defined by φ0=s(2σ+1)/4Γσ+1 2−1 2e−sx2/2,(2.1) φk=((2ks)−1 2B0φk−1if kis even, (2(k+ 2σ)s)−1 2B0φk−1if kis odd,k≥1.(2.2) •The eigenfunctions φkalso satisfy Bφ0= 0,(2.3) Bφk=((2ks)1 2φk−1if kis even, (2(k+ 2σ)s)1 2φk−1if kis odd,k≥1.(2.4) •T∞ m=0 DJm=S. 1It is assumed that 0 ∈N. 6 J.A. ´ Alvarez L´opez, M. Calaza and C. Franco By (2.1) and (2.2), we get φk=pke−sx2/2, where pkis the sequence of polynomials inductively given by p0=s(2σ+1)/4Γ(σ+1 2)−1 2and pk=((2ks)−1 2(2sxpk−1−Tpk−1) if kis even, (2(k+ 2σ)s)−1 2(2sxpk−1−Tpk−1) if kis odd,k≥1. Up to normalization, pkis the sequence of generalized Hermite polynomials [26, p. 380, Problem 25], and φkis the sequence of generalized Hermite functions. Each pkis of degree k, even/odd if kis even/odd, and with positive leading coefficient. They satisfy the recursion formula [3, equation (13)] pk=(k−1 2(2s)1 2xpk−1−(k−1+2σ)1 2pk−2if kis even, (k+ 2σ)−1 2(2s)1 2xpk−1−(k−1)1 2pk−2if kis odd.(2.5) When k= 2m+ 1 (m∈N), we have [3, equation (14)] x−1pk= m X i=0 (−1)m−ism!Γ(i+1 2+σ)s i!Γ(m+3 2+σ)p2i.(2.6) The Pochhammer symbol could be used to simplify this expression, as well as many other expressions in Sections 3and 4. However there are quotients of gamma functions in Sections 4 and 5that can not be simplified in this way (see e.g. Proposition 4.7). Thus, for the sake of uniformity, we use gamma functions in all quotients of this type. Let jbe the positive definite symmetric sesquilinear form in L2 σ, with D(j) = S, given by j(φ, ψ) = hJφ, ψiσ. Like in the case of J, the subindex σwill be added to the notation T,B,B0 and φkand jif necessary. Observe that Bσ=(Bτon Sev, Bτ+ 2(σ−τ)x−1on Sodd,(2.7) B0 σ=(B0 τon Sev, B0 τ+ 2(τ−σ)x−1on Sodd.(2.8) The operator x:Sev → Sodd is a homeomorphism [3], which extends to a unitary operator x:L2 σ,ev →L2 σ−1,odd. We get xJσ,evx−1=Jσ−1,odd because xd2 dx2, x−1=−2d dx x−1. Thus, even for any τ > −3 2, the operator Jτ,odd is densely defined in L2 τ,odd, with D(Jτ,odd) = Sodd, and has the same spectral properties as Jτ+1,ev; in particular, the eigenvalues of Jτ,odd are (2k+1+2τ)s (k∈2N+ 1), and φτ,k =xφτ+1,k−1. To prove the results of the paper, alternative arguments could be given by using the expression of the generalized Hermite polynomials in terms of the Laguerre ones (see, e.g., [24, p. 525] or [25, p. 23]). In particular, certain asymptotic estimates of Laguerre functions [12,15] (see also [5,16]), yield the following asymptotic estimates of the generalized Hermite functions [1, Section 2.4]: there are some C, c > 0, depending only on σ, such that |φk(x)xσ| ≤                Cs ¯σ 2+1 4x¯σν¯σ 2−1 4if 0 < x ≤q1 sν , Cs1 4ν−1 4if q1 sν < x ≤pν 2s, Cs1 4(ν1 3+|sx2−ν|)−1 4if pν 2s< x ≤q3ν 2s, C(sx)1 2e−csx2if q3ν 2s< x, (2.9) where ¯σ= ¯σk=σ+1−(−1)k 2and ν=νk= 2k+1+2σ, with the proviso that we must take ν= 2 if k= 0 and σ < 1 2. A Perturbation of the Dunkl Harmonic Oscillator on the Line 7 3 The sesquilinear form t Let 0 < u < 1 such that σ > u −1 2. Then |x|−uS ⊂ L2 σ, and therefore a positive definite symmetric sesquilinear form tin L2 σ, with D(t) = S, is defined by t(φ, ψ) = |x|−uφ, |x|−uψσ=hφ, ψiσ−u. The notation tσmay be also used. The goal of this section is to study tand apply it to prove Theorem 1.1. Precisely, an estimation of the values t(φk, φ`) is needed. Lemma 3.1. For all φ∈ Sodd and ψ∈ Sev, t(B0φ, ψ)−t(φ, Bψ) = t(φ, B0ψ)−t(Bφ, ψ) = −2utx−1φ, ψ. Proof. By (2.7) and (2.8), for all φ∈ Sodd and ψ∈ Sev, t(B0 σφ, ψ)−t(φ, Bσψ) = hB0 σ−uφ, ψiσ−u−2ux−1φ, ψσ−u−hφ, Bσ−uψiσ−u =−2utx−1φ, ψ, t(φ, B0 σψ)−t(Bσφ, ψ) = hφ, B0 σ−uψiσ−u−hBσ−uφ, ψiσ−u−2ux−1φ, ψσ−u =−2utx−1φ, ψ. In the whole of this section, k,`,m,n,i,j,pand qwill be natural numbers. Let ck,` = t(φk, φ`) and dk,` =ck,`/c0,0. Thus dk,` =d`,k, and dk,` = 0 when k+`is odd. Since Z∞ −∞ e−sx2|x|2κdx =s−2κ+1 2Γκ+1 2(3.1) for κ > −1 2, we get c0,0= Γσ−u+1 2Γσ+1 2−1su.(3.2) Lemma 3.2. If k= 2m > 0, then dk,0=u √m m−1 X j=0 (−1)m−js(m−1)!Γ(j+1 2+σ) j!Γ(m+1 2+σ)d2j,0. Proof. By (2.2), (2.3), (2.6) and Lemma 3.1, ck,0=1 √2skt(B0φk−1, φ0) = 1 √2skt(φk−1, Bφ0)−2u √2sktx−1φk−1, φ0 =−2u √2sktx−1φk−1, φ0=u √m m−1 X j=0 (−1)m−js(m−1)!Γ(j+1 2+σ) j!Γ(m+1 2+σ)c2j,0. Lemma 3.3. If k= 2m > 0and `= 2n > 0, then dk,` =rm ndk−1,`−1+u √n n−1 X j=0 (−1)n−js(n−1)!Γ(j+1 2+σ) j!Γ(n+1 2+σ)dk,2j. Proof. By (2.2), (2.4), (2.6) and Lemma 3.1, ck,` =1 √2s`t(φk, B0φ`−1) = 1 √2`st(Bφk, φ`−1)−2u √2`stφk, x−1φ`−1 =rm nck−1,`−1+u √n n−1 X j=0 (−1)n−js(n−1)!Γ(j+1 2+σ) j!Γ(n+1 2+σ)ck,2j. 8 J.A. ´ Alvarez L´opez, M. Calaza and C. Franco Lemma 3.4. If k= 2m+ 1 and `= 2n+ 1, then dk,` =sn+1 2+σ m+1 2+σdk−1,`−1−u qm+1 2+σ n X j=0 (−1)n−jsn!Γ(j+1 2+σ) j!Γ(n+3 2+σ)dk−1,2j. Proof. By (2.2), (2.4), (2.6) and Lemma 3.1, ck,` =1 p2(k+ 2σ)st(B0φk−1, φ`) =1 p2(k+ 2σ)st(φk−1, Bφ`)−2u p2(k+ 2σ)stφk−1, x−1φ` =sn+1 2+σ m+1 2+σck−1,`−1−u qm+1 2+σ n X j=0 (−1)n−jsn!Γ(j+1 2+σ) j!Γ(n+3 2+σ)ck−1,2j. The following definitions are given for k≥`with k+`even. Let Πk,` =sm!Γ(n+1 2+σ) n!Γ(m+1 2+σ)(3.3) if k= 2m≥`= 2n, and Πk,` =sm!Γ(n+3 2+σ) n!Γ(m+3 2+σ)(3.4) if k= 2m+ 1 ≥`= 2n+ 1. Let Σk,` be inductively defined as follows2: Σk,0= m Y i=1 1−1−u i=Γ(m+u) m!Γ(u)(3.5) if k= 2m; Σk,` = Σk−1,`−1+u n−1 X j=0 (n−1)!Γ(j+1 2+σ) j!Γ(n+1 2+σ)Σk,2j(3.6) if k= 2m≥`= 2n > 0; and Σk,` = Σk−1,`−1−u n X j=0 n!Γ(j+1 2+σ) j!Γ(n+3 2+σ)Σk−1,2j(3.7) = 1−u n+1 2+σ!Σk−1,`−1−nu n+1 2+σ n−1 X j=0 (n−1)!Γ(j+1 2+σ) j!Γ(n+1 2+σ)Σk−1,2j(3.8) if k= 2m+ 1 ≥`= 2n+ 1. Thus Σ0,0= 1, Σ2,0=u, Σ4,0=1 2u(1 + u), and Σk,1= 1−u 1 2+σ!Σk−1,0(3.9) 2We use the convention that a product of an empty set of factors is 1. Such empty products are possible in (3.5) (when m= 0), in Lemma 3.10 and its proof, and in the proofs of Lemma 3.11 and Remark 3.19. Consistently, the sum of an empty set of terms is 0. Such empty sums are possible in Lemma 4.4 and its proof, and in the proof of Proposition 4.7. A Perturbation of the Dunkl Harmonic Oscillator on the Line 9 if kis odd. From (3.5) and using induction on m, it easily follows that Σk,0=u m m−1 X j=0 Σ2j,0(3.10) for k= 2m > 0. Combining (3.6) with (3.7), and (3.8) with (3.6), we get Σk,` = Σk−2,`−2−u n−1 X j=0 (n−1)!Γ(j+1 2+σ) j!Γ(n+1 2+σ)(Σk−2,2j−Σk,2j) (3.11) if k= 2m≥`= 2n > 0; and Σk,` = 1−u n+1 2+σ!Σk−2,`−2 + 1−u+n n+1 2+σ!u n−1 X j=0 (n−1)!Γ(j+1 2+σ) j!Γ(n+1 2+σ)Σk−1,2j(3.12) if k= 2m+ 1 ≥`= 2n+ 1 >1. Proposition 3.5. dk,` = (−1)m+nΠk,`Σk,` if k= 2m≥`= 2n, or if k= 2m+ 1 ≥`= 2n+ 1. Proof. We proceed by induction on kand l. The statement is obvious for k=`= 0 because d0,0= Π0,0= Σ0,0= 1. Let k= 2m > 0, and assume that the result is true for all d2j,0with j < m. Then, by Lemma 3.2, (3.3) and (3.10), dk,0=u √m m−1 X j=0 (−1)m−js(m−1)!Γ(j+1 2+σ) j!Γ(m+1 2+σ)(−1)jΠ2j,0Σ2j,0 = (−1)mu √m m−1 X j=0 s(m−1)!Γ(j+1 2+σ) j!Γ(m+1 2+σ)sj!Γ(1 2+σ) Γ(j+1 2+σ)Σ2j,0 = (−1)mΠk,0 u m m−1 X j=0 Σ2j,0= (−1)mΠk,0Σk,0. Now, take k= 2m≥`= 2n > 0 so that the equality of the statement holds for dk−1,`−1and all dk,2jwith j < n. Then, by Lemma 3.3, dk,` =rm n(−1)m+nΠk−1,`−1Σk−1,`−1 +u √n n−1 X j=0 (−1)n−js(n−1)!Γ(j+1 2+σ) j!Γ(n+1 2+σ)(−1)m+jΠk,2jΣk,2j. Here, by (3.3) and (3.4), pm/nΠk−1,`−1= Πk,`, and 1 √ns(n−1)!Γ(j+1 2+σ) j!Γ(n+1 2+σ)Πk,2j=1 √nsm!Γ(n+1 2+σ) (n−1)!Γ(m+1 2+σ) (n−1)!Γ(j+1 2+σ) j!Γ(n+1 2+σ) = Πk,` (n−1)!Γ(j+1 2+σ) j!Γ(n+1 2+σ). Thus, by (3.6), dk,` = (−1)m+nΠk,`Σk,`. 16 J.A. ´ Alvarez L´opez, M. Calaza and C. Franco (this also follows from (2.9)). For any  > 0, take some x0>0 and k0∈Nsuch that x−2u 0< /2 and Kk−1 6 0x1−2u 0< (1 −2u)/4. Then, for all odd natural k≥k0, t(φk)=2Zx0 0 φ2 k(x)x2(σ−u)dx + 2 Z∞ x0 φ2 k(x)x2(σ−u)dx ≤2Kk−1 6Zx0 0 x−2udx + 2x−2u 0Z∞ x0 φ2 k(x)x2σdx ≤2Kk−1 6x1−2u 0 1−2u+x−2u 0< , because 1 −2u > 0 and kφkkσ= 1. In the case where σ≥0, this argument is also valid when k is even. We do not know if infkt(φk)>0 when 1 2≤u < 1. Proof of Theorem 1.1.The positive definite sesquilinear form jof Section 2is closable by [14, Theorems VI-2.1 and VI-2.7]. Then, taking  > 0 so that ξsu−1<1, it follows from [14, Theorem VI-1.33] and Proposition 3.17 that the positive definite sesquilinear form u:= j+ξtis also closable, and D(¯ u) = D(j). By [14, Theorems VI-2.1, VI-2.6 and VI-2.7], there is a unique positive definite self-adjoint operator Usuch that D(U) is a core of D(¯ u), which consists of the elements φ∈D(¯ u) so that, for some χ∈L2 σ, we have ¯ u(φ, ψ) = hχ, ψiσfor all ψin some core of ¯ u(in this case, U(φ) = χ). By [14, Theorem VI-2.23], we have D(U1/2) = D(¯ u), Sis a core of U1/2(since it is a core of u), and (1.1) is satisfied. By Proposition 3.18, there is some D(σ, u) so that, for all s > 0 and k∈N, and every φis in the linear span of φ0, . . . , φk, we have t(φ)≥ Dsu(k+ 1)−ukφk2 σ. Moreover we can assume that the sequence (2k+ 1 + 2σ)s+ξDsu(k+ 1)−u is strictly increasing after reducing Dif necessary. So u(φ)≥(2k+ 1 + 2σ)s+ξDsu(k+ 1)−ukφk2 σ if φ∈ S is orthogonal in L2 σto the linear span of φ0, . . . , φk−1(assuming that this span is 0 when k= 0). Therefore Uhas a discrete spectrum satisfying the first inequality of (1.2) by the form version of the min-max principle [22, Theorem XIII.2]. The second inequality of (1.2) holds because ¯ u(φ)≤1 + ξsu−1¯ j(φ) + ξCsukφk2 σ for all φ∈D(¯ u) by Proposition 3.17 and [14, Theorem VI-1.18], since Sis a core of ¯ uand ¯ j. Remark 3.21. In the above proof, note that ¯ u=¯ j+ξ¯ tand D(¯ j) = DJ1/2. Thus (1.1) can be extended to φ, ψ ∈DU1/2using J1/2φ, J1/2ψσinstead of hJφ, ψiσ. Remark 3.22. Extend the definition of the above forms and operators to the case of ξ∈C. Then |¯ t(φ)| ≤ su−1<¯ j(φ)+Csukφk2 σfor all φ∈D(¯ j), like in the proof of Theorem 1.1. Thus the family ¯ u=¯ u(ξ) becomes holomorphic of type (a) by Remark 3.21 and [14, Theorem VII-4.8], and therefore U=U(ξ) is a self-adjoint holomorphic family of type (B). So the functions λk=λk(ξ) (ξ∈R) are continuous and piecewise holomorphic [14, Remark VII-4.22, Theorem VII-3.9, and VII-§3.4], with λk(0) = (2k+1+2σ)s. Moreover [14, Theorem VII-4.21] gives an exponential estimate of |λk(ξ)−λk(0)|in terms of ξ. But (1.2) is a better estimate. 4 Scalar products of mixed generalized Hermite functions Let σ, τ, θ > −1 2, and write v=σ+τ−2θ. This section is devoted to describe the scalar products ˆck,` = ˆcσ,τ,θ,k,` =hφσ,k, φτ,`iθ, A Perturbation of the Dunkl Harmonic Oscillator on the Line 17 which will be needed to prove Theorem 1.3. Note that ˆck,` = 0 if k+`is odd, and ˆcσ,τ,θ,k,` = ˆcτ,σ,θ,`,k (4.1) for all kand `. Of course, ˆck,` =δk,` if σ=τ=θ. According to Section 2, if kand `are odd, then ˆcσ,τ,θ,k,` is also defined when σ, τ, θ > −3 2, and we have ˆcσ,τ,θ,k,` =hxφσ+1,k−1, xφτ+1,`−1iθ= ˆcσ+1,τ+1,θ+1,k−1,`−1.(4.2) 4.1 Case where σ=θ6=τand τ−σ6∈ −N In this case, we have v=τ−σ. By (2.1) and (3.1), ˆc0,0=sv 2Γσ+1 21 2Γτ+1 2−1 2.(4.3) Lemma 4.1. If k > 0is even, then ˆck,0= 0. Proof. By (2.2), (2.3) and (2.7), ˆck,0=1 √2kshB0 σφσ,k−1, φτ,0iσ=1 √2kshφσ,k−1, Bτφτ,0iσ= 0. Lemma 4.2. If `= 2n > 0, then ˆc0,` =v √n n−1 X j=0 (−1)n−js(n−1)!Γ(j+1 2+τ) j!Γ(n+1 2+τ)ˆc0,2j. Proof. By (2.2), (2.3), (2.6) and (2.8), ˆc0,` =1 √2`shφσ,0, B0 τφτ,`−1iσ=1 √2`shφσ,0,(B0 σ−2vx−1)φτ,`−1iσ =1 √2`shBσφσ,0, φτ,`−1iσ−2v √2` n−1 X j=0 (−1)n−1−js(n−1)!Γ(j+1 2+τ) j!Γ(n+1 2+τ)ˆc0,2j =v √n n−1 X j=0 (−1)n−js(n−1)!Γ(j+1 2+τ) j!Γ(n+1 2+τ)ˆc0,2j. Lemma 4.3. If k= 2m > 0and `= 2n > 0, then ˆck,` =pn/mˆck−1,`−1. Proof. By (2.2), (2.4) and (2.7), ˆck,` =1 √2kshB0 σφσ,k−1, φτ,`iσ=1 √2kshφσ,k−1, Bτφτ,`iσ=rn mˆck−1,`−1. Lemma 4.4. If k= 2m+ 1 and `= 2n+ 1, then ˆck,` =n+1 2+σ q(m+1 2+σ)(n+1 2+τ) ˆck−1,`−1 −v qm+1 2+σ n−1 X j=0 (−1)n−jsn!Γ(j+1 2+τ) j!Γ(n+3 2+τ)ˆck−1,2j. 18 J.A. ´ Alvarez L´opez, M. Calaza and C. Franco Proof. By (2.2), (2.4), (2.6) and (2.7), ˆck,` =1 p2(k+ 2σ)shB0 σφσ,k−1, φτ,`iσ=1 p2(k+ 2σ)sφσ,k−1,Bτ−2vx−1φτ,`σ =sn+1 2+τ m+1 2+σˆck−1,`−1−v qm+1 2+σ n X j=0 (−1)n−jsn!Γ(j+1 2+τ) j!Γ(n+3 2+τ)ˆck−1,2j =n+1 2+σ q(m+1 2+σ)(n+1 2+τ) ˆck−1,`−1 −v qm+1 2+σ n−1 X j=0 (−1)n−jsn!Γ(j+1 2+τ) j!Γ(n+3 2+τ)ˆck−1,2j. Corollary 4.5. If k > `, then ˆck,` = 0. Proof. This follows by induction on `using Lemmas 4.1,4.3 and 4.4. Remark 4.6. By Corollary 4.5, in Lemma 4.4, it is enough to consider the sum with jrunning from mto n−1. Proposition 4.7. If k= 2m≤`= 2n, then ˆck,` = (−1)m+nsv 2sn!Γ(m+1 2+σ) m!Γ(n+1 2+τ) Γ(n−m+v) (n−m)!Γ(v), and, if k= 2m+ 1 ≤`= 2n+ 1, then ˆck,` = (−1)m+nsv 2sn!Γ(m+3 2+σ) m!Γ(n+3 2+τ) Γ(n−m+v) (n−m)!Γ(v). Proof. This is proved by induction on k. In turn, the case k= 0, ˆc0,` = (−1)nsv 2sΓ(1 2+σ) n!Γ(n+1 2+τ) Γ(n+v) Γ(v),(4.4) is proved by induction on `. If k=`= 0, (4.4) is (4.3). Given `= 2n > 0, assume that the result holds for k= 0 and all `0= 2n0< `. Then, by Lemma 4.2, ˆc0,` =v √n n−1 X j=0 (−1)n−js(n−1)!Γ(j+1 2+τ) j!Γ(n+1 2+τ)(−1)jsv 2sΓ(1 2+σ) j!Γ(j+1 2+τ) Γ(j+v) Γ(v) = (−1)nsv 2s(n−1)!Γ(1 2+σ) nΓ(n+1 2+τ) v Γ(v) n−1 X j=0 Γ(j+v) j!, obtaining (4.4) because Γ(p+1+t) p!=t p X i=0 Γ(i+t) i!(4.5) for all p∈Nand t∈R\(−N), as can be easily checked by induction on p. A Perturbation of the Dunkl Harmonic Oscillator on the Line 19 Given k > 0, assume that the result holds for all k0< k. If kis even, the statement follows directly from Lemma 4.3. If kis odd, by Lemma 4.4, Remark 4.6 and (4.5), ˆck,` =n+1 2+σ q(m+1 2+σ)(n+1 2+τ) (−1)m+nsv 2sn!Γ(m+1 2+σ) m!Γ(n+1 2+τ) Γ(n−m+v) (n−m)!Γ(v) −v qm+1 2+σ n−1 X j=m (−1)n−jsn!Γ(j+1 2+τ) j!Γ(n+3 2+τ) ×(−1)m+jsv 2sj!Γ(m+1 2+σ) m!Γ(j+1 2+τ) Γ(j−m+v) (j−m)!Γ(v) = (−1)m+nsv 2sn!Γ(m+1 2+σ) (m+1 2+σ)m!Γ(n+3 2+τ) 1 Γ(v) × Γ(n−m+v)(n+1 2+σ) (n−m)! −v n−m−1 X i=0 Γ(i+v) i!! = (−1)m+nsv 2sn!Γ(m+3 2+σ) m!Γ(n+3 2+τ) Γ(n−m+v) (n−m)!Γ(v). Remark 4.8. By (4.2), if kand `are odd, then Corollary 4.5 and Proposition 4.7 also hold when σ, τ > −3 2. 4.2 Case where σ6=θ6=τand σ−θ, τ −θ6∈ −N By (2.1) and (3.1), ˆc0,0=sv 2Γσ+1 2−1 2Γτ+1 2−1 2Γ(θ+1 2).(4.6) Lemma 4.9. If k= 2m > 0, then ˆck,0=σ−θ √m m−1 X i=0 (−1)m−is(m−1)!Γ(i+1 2+σ) i!Γ(m+1 2+σ)ˆc2i,0. Proof. By (2.2) and (2.8), ˆck,0=1 √2kshB0 σφσ,k−1, φτ,0iθ=1 2√mshB0 θφσ,k−1, φτ,0iθ+θ−σ √ms x−1φσ,k−1, φτ,0θ. Here, by (2.3), (2.6) and (2.7), hB0 θφσ,k−1, φτ,0iθ=hφσ,k−1, Bθφτ,0iθ=hφσ,k−1, Bτφτ,0iθ= 0, hx−1φσ,k−1, φτ,0iθ=− m−1 X i=0 (−1)m−is(m−1)!Γ(i+1 2+σ)s i!Γ(m+1 2+σ)ˆc2i,0. Lemma 4.10. If k= 2m > 0and `= 2n > 0, then ˆck,` =rn mˆck−1,`−1+σ−θ m m−1 X i=0 (−1)m−ism!Γ(i+1 2+σ) i!Γ(m+1 2+σ)ˆc2i,`. 20 J.A. ´ Alvarez L´opez, M. Calaza and C. Franco Proof. Like in the proof of Lemma 4.9, ˆck,` =1 2√mshB0 θφσ,k−1, φτ,`iθ+θ−σ √ms x−1φσ,k−1, φτ,`θ. Now, by (2.4), (2.6) and (2.7), hB0 θφσ,k−1, φτ,`iθ=hφσ,k−1, Bθφτ,`iθ=hφσ,k−1, Bτφτ,`iθ= 2√nsˆck−1,`−1, x−1φσ,k−1, φτ,`θ=− m−1 X i=0 (−1)m−is(m−1)!Γ(i+1 2+σ)s i!Γ(m+1 2+σ)ˆc2i,`. Lemma 4.11. If k= 2m+ 1 and `= 2n+ 1, then ˆck,` =m+1 2+θ q(m+1 2+σ)(n+1 2+τ) ˆck−1,`−1 −σ−θ qn+1 2+τ m−1 X i=0 (−1)m−ism!Γ(i+1 2+σ) i!Γ(m+3 2+σ)ˆc2i,`−1. Proof. By (2.2), ˆck,` =1 2q(n+1 2+τ)shφσ,k, B0 τφτ,`−1iθ, where, by (2.8), hφσ,k, B0 τφτ,`−1iθ=hφσ,k, B0 θφτ,`−1iθ=hBθφσ,k, φτ,`−1iθ =hBσφσ,k, φτ,`−1iθ+ 2(θ−σ)hx−1φσ,k, φτ,`−1iθ. Hence, by (2.4) and (2.6), ˆck,` =sm+1 2+σ n+1 2+τˆck−1,`−1−σ−θ qn+1 2+τ m X i=0 (−1)m−ism!Γ(i+1 2+σ) i!Γ(m+3 2+σ)ˆc2i,`−1 =m+1 2+θ q(n+1 2+τ)(m+1 2+σ) ˆck−1,`−1 −σ−θ qn+1 2+τ m−1 X i=0 (−1)m−ism!Γ(i+1 2+σ) i!Γ(m+3 2+σ)ˆc2i,`−1. Proposition 4.12. If k= 2mand `= 2n, then ˆck,` = (−1)m+nsv 2sm!n! Γ(m+1 2+σ)Γ(n+1 2+τ) × min{m,n} X p=0 Γ(p+1 2+θ)Γ(m−p+σ−θ)Γ(n−p+τ−θ) p!(m−p)!(n−p)!Γ(σ−θ)Γ(τ−θ), and, if k= 2m+ 1 and `= 2n+ 1, then ˆck,` = (−1)m+nsv 2sm!n! Γ(m+3 2+σ)Γ(n+3 2+τ) × min{m,n} X p=0 Γ(p+3 2+θ)Γ(m−p+σ−θ)Γ(n−p+τ−θ) p!(m−p)!(n−p)!Γ(σ−θ)Γ(τ−θ). A Perturbation of the Dunkl Harmonic Oscillator on the Line 21 Proof. The result is proved by induction on kand `. First, consider the case `= 0. When k=`= 0, the result is given by (4.6). Now, take any k= 2m > 0, and assume that the result holds for all ˆck0,0with k0= 2m0< k. Then, by Lemma 4.9 and (4.5), ˆck,0=σ−θ √m m−1 X i=0 (−1)m−is(m−1)!Γ(i+1 2+σ) i!Γ(m+1 2+σ) ×(−1)isv 2s1 i!Γ(i+1 2+σ)Γ(1 2+τ) Γ(1 2+θ)Γ(i+σ−θ) Γ(σ−θ) = (−1)msv 2sm! Γ(m+1 2+σ)Γ(1 2+τ) Γ(1 2+θ)(σ−θ) m m−1 X i=0 Γ(i+σ−θ) i!Γ(σ−θ) = (−1)msv 2s1 m!Γ(m+1 2+σ)Γ(1 2+τ) Γ(1 2+θ)Γ(m+σ−θ) Γ(σ−θ). From the case `= 0, the result also follows for the case k= 0 by (4.1). Now, take k= 2m > 0 and `= 2n > 0, and assume that the result holds for all ˆck0,`0with k0< k and `0≤`. By Lemma 4.10, ˆck,` =rn m(−1)m+n−2sv 2s(m−1)!(n−1)! Γ(m+1 2+σ)Γ(n+1 2+τ) × min{m−1,n−1} X q=0 Γ(q+3 2+θ)Γ(m−1−q+σ−θ)Γ(n−1−q+τ−θ) q!(m−1−q)!(n−1−q)!Γ(σ−θ)Γ(τ−θ) +σ−θ m m−1 X i=0 (−1)m−ism!Γ(i+1 2+σ) i!Γ(m+1 2+σ)(−1)i+nsv 2si!n! Γ(i+1 2+σ)Γ(n+1 2+τ) × min{i,n} X p=0 Γ(p+1 2+θ)Γ(i−p+σ−θ)Γ(n−p+τ−θ) p!(i−p)!(n−p)!Γ(σ−θ)Γ(τ−θ) = (−1)m+nsv 21 msm!n! Γ(m+1 2+σ)Γ(n+1 2+τ) × min{m−1,n−1} X q=0 Γ(q+3 2+θ)Γ(m−1−q+σ−θ)Γ(n−1−q+τ−θ) q!(m−1−q)!(n−1−q)!Γ(σ−θ)Γ(τ−θ) + (σ−θ) m−1 X i=0 min{i,n} X p=0 Γ(p+1 2+θ)Γ(i−p+σ−θ)Γ(n−p+τ−θ) p!(i−p)!(n−p)!Γ(σ−θ)Γ(τ−θ)!. Then the desired expression for ˆck,` follows because min{m−1,n−1} X q=0 Γ(q+3 2+θ)Γ(m−1−q+σ−θ)Γ(n−1−q+τ−θ) q!(m−1−q)!(n−1−q)!Γ(σ−θ)Γ(τ−θ) = min{m,n} X p=0 pΓ(p+1 2+θ)Γ(m−p+σ−θ)Γ(n−p+τ−θ) p!(m−p)!(n−p)!Γ(σ−θ)Γ(τ−θ), and, by (4.5), (σ−θ) m−1 X i=0 min{i,n} X p=0 Γ(p+1 2+θ)Γ(i−p+σ−θ)Γ(n−p+τ−θ) p!(i−p)!(n−p)!Γ(σ−θ)Γ(τ−θ) 22 J.A. ´ Alvarez L´opez, M. Calaza and C. Franco = (σ−θ) min{m−1,n} X p=0 m−1−p X j=0 Γ(p+1 2+θ)Γ(j+σ−θ)Γ(n−p+τ−θ) p!j!(n−p)!Γ(σ−θ)Γ(τ−θ) = min{m,n} X p=0 Γ(p+1 2+θ)(m−p)Γ(m−p+σ−θ)Γ(n−p+τ−θ) p!(m−p)!(n−p)!Γ(σ−θ)Γ(τ−θ).(4.7) Finally, take k= 2m+ 1 and `= 2n+ 1, and assume that the result holds for all ˆck0,`0with k0< k and `0< `. By Lemma 4.11, ˆck,` =(m+1 2+θ)(−1)m+nsv 2 q(m+1 2+σ)(n+1 2+τ)sm!n! Γ(m+1 2+σ)Γ(n+1 2+τ) × min{m,n} X p=0 Γ(p+1 2+θ)Γ(m−p+σ−θ)Γ(n−p+τ−θ) p!(m−p)!(n−p)!Γ(σ−θ)Γ(τ−θ) −σ−θ qn+1 2+τ m−1 X i=0 (−1)m−ism!Γ(i+1 2+σ) i!Γ(m+3 2+σ) ×(−1)i+nsv 2si!n! Γ(i+1 2+σ)Γ(n+1 2+τ) × min{i,n} X p=0 Γ(p+1 2+θ)Γ(i−p+σ−θ)Γ(n−p+τ−θ) p!(i−p)!(n−p)!Γ(σ−θ)Γ(τ−θ) = (−1)m+nsv 2sm!n! Γ(m+3 2+σ)Γ(n+3 2+τ) × min{m,n} X p=0 (m+1 2+θ)Γ(p+1 2+θ)Γ(m−p+σ−θ)Γ(n−p+τ−θ) p!(m−p)!(n−p)!Γ(σ−θ)Γ(τ−θ) −(σ−θ) m−1 X i=0 min{i,n} X p=0 Γ(p+1 2+θ)Γ(i−p+σ−θ)Γ(n−p+τ−θ) p!(i−p)!(n−p)!Γ(σ−θ)Γ(τ−θ)!. Then we get the stated expression for ˆck,` using (4.7) again.  Remark 4.13. By (4.2), if kand `are odd, then Proposition 4.12 also holds when σ, τ > −3 2. 5 The sesquilinear form t0 Consider the notation of Section 4. Since x−1Sodd =Sev, a sesquilinear form t0in L2 σ,τ , with D(t0) = S, is defined by t0(φ, ψ) = φev, x−1ψoddθ=hxφev, ψoddiθ−1. Note that t0is neither symmetric nor bounded from the left. The goal of this section is to study t0, and use it to prove Theorem 1.3. Let c0 k,` =t0(φσ,k, φτ,`). Clearly, c0 k,` = 0 if kis odd or `is even. A Perturbation of the Dunkl Harmonic Oscillator on the Line 23 5.1 Case where σ=θ=τ In this case, we have v= 0. Proposition 5.1. For k= 2mand `= 2n+ 1, if k > ` (m>n), then c0 k,` = 0, and, if k < ` (m≤n), then c0 k,` = (−1)n−ms1 2sn!Γ(m+1 2+σ) m!Γ(n+3 2+σ). Proof. This follows from (2.6) since ˆck,` =δk,` in this case.  Proposition 5.2. There is some ω=ω(σ)>0so that, for k= 2mand `= 2n+ 1, |c0 k,`|4s1 2(m+ 1)−ω(n+ 1)−ω. Proof. We can assume that m≤naccording to Proposition 5.1. Moreover |c0 k,`|4s1 2(m+ 1)σ 2−1 4(n+ 1)−σ 2−1 4 for all m≤nby Proposition 5.1 and Lemma 3.12. Therefore the result follows using Lemma 3.14, reversing the roles of mand n, because −σ 2−1 4<−u 2<0.  5.2 Case where σ=θ6=τand τ−σ6∈ −N Recall that v=τ−σin this case. Moreover c0 k,` = 0 if k > ` by (2.6) and Corollary 4.5. Proposition 5.3. For k= 2m<`= 2n+ 1 (m≤n), c0 k,` = (−1)m+ns1+v 2sn!Γ(m+1 2+σ) m!Γ(n+3 2+τ) Γ(n−m+1+v) (n−m)!Γ(1 + v). Proof. By (2.6), Corollary 4.5, Proposition 4.7 and (4.5), c0 k,` =s1 2 n X j=m (−1)n−jsn!Γ(j+1 2+τ) j!Γ(n+3 2+τ)(−1)m+jsv 2sj!Γ(m+1 2+σ) m!Γ(j+1 2+τ) Γ(j−m+v) (j−m)!Γ(v) = (−1)m+ns1+v 2sn!Γ(m+1 2+σ) m!Γ(n+3 2+τ) 1 Γ(v) n−m X i=0 Γ(i+v) i! = (−1)m+ns1+v 2sn!Γ(m+1 2+σ) m!Γ(n+3 2+τ) Γ(n−m+1+v) (n−m)!Γ(1 + v). Proposition 5.4. If (σ, τ)satisfies (1.4), then there is some ω=ω(σ, τ)>0so that, for k= 2m<`= 2n+ 1, |c0 k,`|4s1+v 2(m+ 1)−ω(n+ 1)−ω. Proof. By Proposition 5.3 and Lemma 3.12, |c0 k,`|4s1+v 2(m+ 1)σ 2−1 4(n+ 1)−τ 2−1 4(n−m+ 1)v. Then the result follows by Lemma 3.14, interchanging the roles of mand n, using the condition of Theorem 1.3(a).  24 J.A. ´ Alvarez L´opez, M. Calaza and C. Franco 5.3 Case where σ6=θ=τand σ−θ6∈ −N Recall that v=σ−τin this case. Proposition 5.5. For k= 2mand `= 2n+ 1, c0 k,` = (−1)m+ns1+v 2sm!n! Γ(m+1 2+σ)Γ(n+3 2+τ) min{m,n} X j=0 Γ(j+1 2+τ)Γ(m−j+v) j!(m−j)!Γ(v). Proof. Let jrun from 0 to min{m, n}. By (2.6), Corollary 4.5, Proposition 4.7 and (4.1), c0 k,` =s1 2X j (−1)n−jsn!Γ(j+1 2+τ) j!Γ(n+3 2+τ)(−1)j+msv 2sm!Γ(j+1 2+τ) j!Γ(m+1 2+σ) Γ(m−j+v) (m−j)!Γ(v) = (−1)m+ns1+v 2sm!n! Γ(m+1 2+σ)Γ(n+3 2+τ)X j Γ(j+1 2+τ)Γ(m−j+v) j!(m−j)!Γ(v). Proposition 5.6. If (σ, τ)satisfies (1.5), then there is some ω=ω(σ, τ)>0so that, for k= 2mand `= 2n+ 1, |c0 k,`|4s1+v 2(m+ 1)−ω(n+ 1)−ω. Proof. By Proposition 5.5 and Lemma 3.12, |c0 k,`|4s1+v 2(m+ 1)1 4−σ 2(n+ 1)−1 4−τ 2 min{m,n} X j=0 (m−j+ 1)v−1(j+ 1)τ−1 2. Then the result follows by Corollary 7.4, proved in Section 7, since (σ, τ) satisfies (1.5).  5.4 Case where σ6=θ=τ+ 1 and σ−τ−16∈ −N Note that v=σ−τ−2 in this case. Moreover c0 k,` =φσ,k, x−1φτ,`τ+1 =hxφσ,k, φτ,`iτ=hφτ,`, xφσ,kiτ(5.1) for k= 2mand `= 2n+ 1 (Remark 1.4(ii)). Proposition 5.7. Let k= 2mand `= 2n+ 1. If k+ 1 < ` (m<n), then c0 k,` = 0. If k+ 1 ≥` (m≥n), then c0 k,` = (−1)m+nsv+1 2sm!Γ(n+3 2+τ) n!Γ(m+1 2+σ) Γ(m−n+v+ 1) (m−n)!Γ(v+ 1) . Proof. By (2.5) and (5.1), c0 k,` =sm+1 2+σ sˆcτ,σ,τ,`,k+1 +rm sˆcτ,σ,τ,`,k−1.(5.2) So c0 k,` = 0 if k+ 1 < ` by Corollary 4.5. When k+ 1 = `(m=n), by (5.2) and Proposition 4.7, c0 k,` =sm+1 2+σ ssv+2 2sΓ(n+3 2+τ) Γ(m+3 2+σ)=sv+1 2sΓ(n+3 2+τ) Γ(m+1 2+σ). A Perturbation of the Dunkl Harmonic Oscillator on the Line 25 When k−1≥`(m > n), by (5.2) and Proposition 4.7, c0 k,` =sm+1 2+σ s(−1)m+nsv+2 2sm!Γ(n+3 2+τ) n!Γ(m+3 2+σ) Γ(m−n+v+ 2) (m−n)!Γ(v+ 2) +rm s(−1)m+n−1sv+2 2s(m−1)!Γ(n+3 2+τ) n!Γ(m+1 2+σ) Γ(m−n+v+ 1) (m−1−n)!Γ(v+ 2) = (−1)m+nsv+1 2sm!Γ(n+3 2+τ) n!Γ(m+1 2+σ) Γ(m−n+v+ 1) (m−1−n)!Γ(v+ 2) m−n+v+ 1 m−n−1 = (−1)m+nsv+1 2sm!Γ(n+3 2+τ) n!Γ(m+1 2+σ) Γ(m−n+v+ 1) (m−n)!Γ(v+ 1) . Proposition 5.8. If (σ, τ)satisfies (1.6), then there is some ω=ω(σ, τ)>0so that, for k= 2mand `= 2n+ 1, |c0 k,`|4sv+1 2(m+ 1)−ω(n+ 1)−ω. Proof. By Proposition 5.7, we can assume that k+ 1 ≥`(m≥n), and, in this case, using also Lemma 3.12, we get |c0 k,`|4sv+1 2(m+ 1)1 4−σ 2(n+ 1)1 4+τ 2(m−n+ 1)v. Then the result follows using Lemma 3.14. 5.5 Case where σ6=θ6=τand σ−θ, τ −θ6∈ −N Proposition 5.9. For k= 2mand `= 2n+ 1, c0 k,` = (−1)m+ns1+v 2sm!n! Γ(m+1 2+σ)Γ(n+3 2+τ) × min{m,n} X p=0 Γ(p+1 2+θ)Γ(m−p+σ−θ)Γ(n−p+1+τ−θ) p!(m−p)!(n−p)!Γ(σ−θ)Γ(1 + τ−θ). Proof. By (2.6) and Proposition 4.12, c0 k,` =s1 2 n X j=0 (−1)n−jsn!Γ(j+1 2+τ) j!Γ(n+3 2+τ)(−1)m+jsv 2sm!j! Γ(m+1 2+σ)Γ(j+1 2+τ) × min{m,j} X p=0 Γ(p+1 2+θ)Γ(m−p+σ−θ)Γ(j−p+τ−θ) p!(m−p)!(j−p)!Γ(σ−θ)Γ(τ−θ) = (−1)m+ns1+v 2sm!n! Γ(m+1 2+σ)Γ(n+3 2+τ) × n X j=0 min{m,j} X p=0 Γ(p+1 2+θ)Γ(m−p+σ−θ)Γ(j−p+τ−θ) p!(m−p)!(j−p)!Γ(σ−θ)Γ(τ−θ). But, by (4.5), n X j=0 min{m,j} X p=0 Γ(m−p+σ−θ)Γ(j−p+τ−θ) (m−p)!(j−p)!Γ(σ−θ)Γ(τ−θ) 32 J.A. ´ Alvarez L´opez, M. Calaza and C. Franco where h(x) = −(γ+κ)x+κ(m+ 1) −γ. Observe that this expression is valid even when γ= 0 or κ= 0. Since f0and hhave the same zero set on [0, n], and they have the same sign on the complement of the zero set in [0, n], it is enough to analyze hto know where freaches its maximum on [0, n]. We consider several cases. Case where γ+κ= 0.Then h≡κ(m+ 2). If κ6= 0, then h6= 0 and sign h= sign κ. If κ= 0, then h≡0. Hence: max 0≤x≤nf(x) = (f(n)=(m−n+ 1)γ(n+ 1)κif κ=−γ≥0, f(0) = (m+ 1)γif κ=−γ≤0.(6.28) Case where γ+κ6= 0.Then hvanishes just at the point x0:= κ(m+ 1) −γ γ+κ. Case where γ+κ<0.We have h < 0 on (−∞, x0) and h > 0 on (x0,∞), yielding max 0≤x≤nf(x) =      f(n)=(m−n+ 1)γ(n+ 1)κif x0≤0, max{f(0), f(n)}if 0 ≤x0≤n, f(0) = (m+ 1)γif x0≥n. (6.29) Case where γ+κ<0and κ≥0; i.e., 0≤κ<−γ.Then x0<0, and therefore, by (6.29), 0≤κ<−γ⇒max 0≤x≤nf(x)=(m−n+ 1)γ(n+ 1)κ.(6.30) Case where γ+κ<0and γ≥0; i.e., 0≤γ < −κ.Then x0≥m+ 1, and therefore, by (6.29), 0≤γ < −κ⇒max 0≤x≤nf(x)=(m+ 1)γ.(6.31) Case where κ≤γ < 0.Then x0≥m 2, and we may have x0≤nor x0≥n. In any case, by (6.29), κ≤γ < 0⇒max 0≤x≤nf(x)=(m+ 1)γ.(6.32) Case where γ < κ<0.Then x0<m 2, and we may have x0≤0, 0 ≤x0≤nor x0≥n. In any case, by (6.29), γ < κ<0⇒max 0≤x≤nf(x) = ((m−n+ 1)γ(n+ 1)κor (m+ 1)γ.(6.33) Case where γ+κ>0.We have h > 0 on (−∞, x0) and h < 0 on (x0,∞), yielding max 0≤x≤nf(x) =      f(0) = (m+ 1)γif x0≤0, f(x0) if 0 ≤x0≤n, f(n)=(m−n+ 1)γ(n+ 1)κif x0≥n. (6.34) Case where γ+κ>0and κ≤0; i.e., −γ < κ≤0.Then x0<0, and therefore, by (6.34), −γ < κ≤0⇒max 0≤x≤nf(x)=(m+ 1)γ.(6.35) A Perturbation of the Dunkl Harmonic Oscillator on the Line 33 Case where γ+κ>0and γ≤0; i.e., −κ< γ ≤0.Then x0≥m+ 1, and therefore, by (6.34), −κ< γ ≤0⇒max 0≤x≤nf(x)=(m−n+ 1)γ(n+ 1)κ.(6.36) Case where γ+κ>0and γ, κ≥0.We may have x0≤0, 0 ≤x0≤nor n≤x0. Moreover f(x0) = γγκκ (γ+κ)γ+κ(m+ 2)γ+κ4(m+ 1)γ+κ. But, in this case, max 0≤x≤nf(x)≤max 0≤x≤n(m−x+ 1)γmax 0≤y≤n(y+ 1)κ= (m+ 1)γ(n+ 1)κ≤(m+ 1)γ+κ. Therefore, by (6.34), γ, κ≥0⇒max 0≤x≤nf(x)≤(m+ 1)γ(n+ 1)κ.(6.37) Gathering together (6.28), (6.30)–(6.33) and (6.35)–(6.37), we get (6.23). 6.2.3 Third list of conditions For all  > 0, n X p=0 (m−p+ 1)γ= m+1 X q=m−n+1 qγ≤       Zm+2 m−n+1 xγdx if γ≥0, (m−n+ 1)γ+Zm+1 m−n+1 xγdx if γ < 0 4     (m+ 1)γ+1 if γ > −1, 1 + ln(m+ 1) if γ=−1, (m−n+ 1)γ+1 if γ < −1 4     (m+ 1)γ+1 if γ > −1, (m+ 1)if γ=−1, (m−n+ 1)γ+1 if γ < −1. (6.38) The following gives a better estimate when γ≥0, and an alternative estimate when γ < 0: n X p=0 (m−p+ 1)γ≤((m+ 1)γ(n+ 1) if γ≥0, (m−n+ 1)γ(n+ 1) if γ < 0.(6.39) The estimate (6.39) is better than (6.38) when γ≥0 since m≥n, and (6.39) may be better or worse than (6.38) when γ < 0, depending on the values of mand n. Note that estimates of the type (6.39) for n P p=0 (p+1)κand n P p=0 (n−p+1)δare worse than (6.10) and (6.22). Thus it makes no sense to add this kind of estimate in Sections 6.2.1 and 6.2.2. On the other hand, we claim that (n−p+ 1)δ(p+ 1)κ≤     (n+ 1)κif δ≤κ,0, (n+ 1)δ+κif δ, κ≥0, (n+ 1)δif κ≤δ, 0 (6.40) for all p= 0, . . . , n. Combining (6.38)–(6.40), it follows that, for all  > 0, (m+ 1)α(n+ 1)β n X p=0 (m−p+ 1)γ(n−p+ 1)δ(p+ 1)κ4A3,(6.41) 34 J.A. ´ Alvarez L´opez, M. Calaza and C. Franco where A3=A3(m, n, α, β, γ, δ, κ, ) can be taken to be equal to (m+ 1)α+(n+ 1)β+δ+κand (m+ 1)α(n+ 1)β+δ+κ+1(m−n+ 1)γ)if γ=−1, 0 ≤δ, κ, (m+ 1)α(n+ 1)β+δ+κ(m−n+ 1)γ+1 and (m+ 1)α(n+ 1)β+δ+κ+1(m−n+ 1)γ)if γ < −1, 0 ≤δ, κ. By (6.41), applying Lemma 3.14 to the above list, we get the third list of conditions that guarantee (6.9) when m≥n: γ≤ −1,0≤δ, κ⇒(α+γ+ 1, α +β+δ+κ<0,or α+γ, α +β+δ+κ+ 1 <0.(6.42) To prove (6.40), it is enough to study the maximum of the C∞function f(x)=(n−x+ 1)δ(x+ 1)κ on [0, n] (the natural domain of fcontains (−1, n + 1)). We have f0(x)=(n−x+ 1)δ−1(x+ 1)κ−1h(x), where h(x) = −(δ+κ)x+κ(n+ 1) −δ. Observe that this expression is valid even when δ= 0 or κ= 0. Since f0and hhave the same zero set on [0, n], and they have the same sign on the complement of the zero set in [0, n], it is enough to analyze hto know where freaches its maximum on [0, n]. We consider several cases. Case where δ+κ= 0.Then h≡κ(n+ 2). If κ6= 0, then h6= 0 and sign h= sign κ. If κ= 0, then h≡0. Hence: max 0≤x≤nf(x) = (f(n)=(n+ 1)κif κ=−δ≥0, f(0) = (n+ 1)δif κ=−δ≤0.(6.43) Case where δ+κ6= 0.Then hvanishes just at the point x0:= κ(n+ 1) −δ δ+κ. Case where δ+κ>0.We have h > 0 on (−∞, x0) and h < 0 on (x0,∞), yielding max 0≤x≤nf(x) =      f(0) = (n+ 1)δif x0≤0, f(x0) if 0 ≤x0≤n, f(n)=(n+ 1)κif x0≥n. (6.44) Case where δ+κ>0and δ, κ≥0.We may have x0≤0, 0 ≤x0≤nor n≤x0. Moreover f(x0) = δδκκ (δ+κ)δ+κ(n+ 2)δ+κ4(n+ 1)δ+κ. But, in this case, max 0≤x≤nf(x)≤max 0≤x≤n(n−x+ 1)δmax 0≤y≤n(y+ 1)κ= (n+ 1)δ+κ. Therefore, by (6.44), δ, κ≥0⇒max 0≤x≤nf(x)≤(n+ 1)δ+κ.(6.45) Gathering together (6.43) and (6.45), we get the second case of (6.40). The other cases will not be used, and they follow with similar arguments. A Perturbation of the Dunkl Harmonic Oscillator on the Line 35 6.2.4 Fourth list of conditions We have (p+ 1)κ≤((n+ 1)κif κ≥0, 1 if κ≤0(6.46) for p= 0, . . . , n. Moreover, by (6.10), (6.38) and (6.39), for all  > 0, n X p=0 (n−p+ 1)2δ= n+1 X q=1 q2δ4     (n+ 1)2δ+1 if δ > −1 2, (n+ 1)if δ=−1 2, 1 if δ < −1 2, (6.47) n X p=0 (m−p+ 1)2γ4     (m+ 1)2γ+1 if γ > −1 2, (m+ 1)if γ=−1 2, (m−n+ 1)2γ+1 if γ < −1 2, (6.48) n X p=0 (m−p+ 1)2γ≤((m+ 1)2γ(n+ 1) if γ≥0, (m−n+ 1)2γ(n+ 1) if γ < 0.(6.49) The estimate (6.49) is better than (6.48) when γ≥0, and it may be better or worse than (6.48) when γ < 0, depending on the values of mand n. By the Cauchy–Schwartz inequality, n X p=0 (m−p+ 1)γ(n−p+ 1)δ≤  n X p=0 (m−p+ 1)2γ  1 2  n X p=0 (n−p+ 1)2δ  1 2 . Therefore, by (6.46)–(6.49), (m+ 1)α(n+ 1)β n X p=0 (m−p+ 1)γ(n−p+ 1)δ(p+ 1)κ4A4,(6.50) where A4=A4(m, n, α, β, γ, δ, κ) can be taken to be equal to (m+ 1)α(n+ 1)β+δ+κ+1 2(m−n+ 1)γ+1 2and (m+ 1)α(n+ 1)β+δ+κ+1(m−n+ 1)γ)if γ < −1 2< δ, 0 ≤κ. By (6.50), applying Lemma 3.14 to the above list, we get the fourth list of conditions that guarantee (6.9) when m≥n: γ < −1 2< δ, 0≤κ⇒(α+γ+1 2, α +β+δ+κ+1 2<0,or α+γ, α +β+δ+κ+ 1 <0.(6.51) 6.2.5 Fifth list of conditions This is analogous to the estimates of Section 6.2.4, interchanging the roles of δand κ. We have (n−p+ 1)δ≤((n+ 1)δif δ≥0, 1 if δ≤0(6.52) for p= 0, . . . , n. Moreover, by (6.10), for all  > 0, n X p=0 (p+ 1)2κ4     (n+ 1)2κ+1 if κ>−1 2, (n+ 1)if κ=−1 2, 1 if κ<−1 2. (6.53) 36 J.A. ´ Alvarez L´opez, M. Calaza and C. Franco Applying the Cauchy–Schwartz inequality, we get n X p=0 (m−p+ 1)γ(p+ 1)κ≤  n X p=0 (m−p+ 1)2γ  1 2  n X p=0 (p+ 1)2κ  1 2 . Therefore, by (6.52), (6.53), (6.48) and (6.49), for all  > 0, (m+ 1)α(n+ 1)β n X p=0 (m−p+ 1)γ(n−p+ 1)δ(p+ 1)κ4A5,(6.54) where A5=A5(m, n, α, β, γ, δ, κ, ) can be taken to be equal to (m+ 1)α+(n+ 1)β+δ+and (m+ 1)α(n+ 1)β+δ+1 2+(m−n+ 1)γ)if γ=κ=−1 2, 0 ≤δ, (m+ 1)α+(n+ 1)β+δand (m+ 1)α(n+ 1)β+δ+1 2(m−n+ 1)γ)if κ< γ =−1 2, 0 ≤δ, (m+ 1)α(n+ 1)β+δ+(m−n+ 1)γ+1 2and (m+ 1)α(n+ 1)β+δ+1 2+(m−n+ 1)γ)if γ < κ=−1 2, 0 ≤δ, (m+ 1)α(n+ 1)β+δ(m−n+ 1)γ+1 2and (m+ 1)α(n+ 1)β+δ+1 2(m−n+ 1)γ)if γ, κ<−1 2, 0 ≤δ. By (6.54), applying Lemma 3.14 to the above list, we get the fifth list of conditions that guarantee (6.9) when m≥n: γ, κ≤ −1 2,0≤δ⇒(α+γ+1 2, α +β+δ < 0,or α+γ, α +β+δ+1 2<0.(6.55) 6.2.6 Sixth list of conditions We have (m−p+ 1)γ≤((m+ 1)γif γ≥0, (m−n+ 1)γif γ≤0(6.56) for p= 0, . . . , n. Moreover, by (6.10), for all  > 0, n X p=0 (n−p+ 1)2δ= n+1 X q=1 q2δ4     (n+ 1)2δ+1 if δ > −1 2, (n+ 1)if δ=−1 2, 1 if δ < −1 2. (6.57) Applying the Cauchy–Schwartz inequality, we get n X p=0 (n−p+ 1)δ(p+ 1)κ≤  n X p=0 (n−p+ 1)2δ  1 2  n X p=0 (p+ 1)2κ  1 2 . Therefore, by (6.53), (6.56) and (6.57), for all  > 0, (m+ 1)α(n+ 1)β n X p=0 (m−p+ 1)γ(n−p+ 1)δ(p+ 1)κ4A6,(6.58) A Perturbation of the Dunkl Harmonic Oscillator on the Line 37 where A6=A6(m, n, α, β, γ, δ, κ, ) can be taken to be equal to (m+ 1)α+γ(n+ 1)β+if (κ≤δ=−1 2,0≤γ, or δ≤κ=−1 2,0≤γ, (m+ 1)α+γ(n+ 1)βif δ, κ<−1 2, 0 ≤γ. By (6.58), applying Lemma 3.14 to the above list, we get the sixth list of conditions that guarantee (6.9) when m≥n: δ, κ≤ −1 2,0≤γ⇒α+γ, α +β+γ < 0.(6.59) 6.2.7 Seventh list of conditions By (6.10), (6.38) and (6.39), for all  > 0, n X p=0 (n−p+ 1)3δ= n+1 X q=1 q3δ4     (n+ 1)3δ+1 if δ > −1 3, (n+ 1)if δ=−1 3, 1 if δ < −1 3, (6.60) n X p=0 (p+ 1)3κ4     (n+ 1)3κ+1 if κ>−1 3, (n+ 1)if κ=−1 3, 1 if κ<−1 3, (6.61) n X p=0 (m−p+ 1)3γ4     (m+ 1)3γ+1 if γ > −1 3, (m+ 1)if γ=−1 3, (m−n+ 1)3γ+1 if γ < −1 3, (6.62) n X p=0 (m−p+ 1)3γ≤((m+ 1)3γ(n+ 1) if γ≥0, (m−n+ 1)3γ(n+ 1) if γ < 0.(6.63) Note that (6.63) is better than (6.62) for γ≥0, and it is an alternative estimate for γ < 0. Applying the generalized H¨older inequality [7], we get n X p=0 (m−p+ 1)γ(n−p+ 1)δ(p+ 1)κ ≤  n X p=0 (m−p+ 1)3γ  1 3  n X p=0 (n−p+ 1)3δ  1 3  n X p=0 (p+ 1)3κ  1 3 . Therefore, by (6.60)–(6.63), (m+ 1)α(n+ 1)β n X p=0 (m−p+ 1)γ(n−p+ 1)δ(p+ 1)κ4A7,(6.64) where A7=A7(m, n, α, β, γ, δ, κ) can be taken to be equal to (m+ 1)α(n+ 1)β+δ+κ+2 3(m−n+ 1)γ+1 3and (m+ 1)α(n+ 1)β+δ+κ+1(m−n+ 1)γ)if γ < −1 3< δ, κ, (m+ 1)α(n+ 1)β+δ+1 3(m−n+ 1)γ+1 3and (m+ 1)α(n+ 1)β+δ+2 3(m−n+ 1)γ)if γ, κ<−1 3< δ. 38 J.A. ´ Alvarez L´opez, M. Calaza and C. Franco By (6.64), applying Lemma 3.14 to the above list, we get the seventh list of conditions that guarantee (6.9) when m≥n: γ < −1 3< δ, κ⇒(α+γ+1 3, α +β+δ+κ+2 3<0,or α+γ, α +β+δ+κ+ 1 <0,(6.65) γ, κ<−1 3< δ ⇒(α+γ+1 3, α +β+δ+1 3<0,or α+γ, α +β+δ+2 3<0.(6.66) 6.2.8 Obtaining the sets Sijk from the lists of conditions The left hand side of the conditions from the lists of Sections 6.2.1–6.2.7 involve only (γ, δ, κ). Now, we indicate which of them define sets covering every Qijk, for the chosen subindices ijk equal to 515, 522, 252, 155, 212. Those conditions will produce the definition of Sijk ⊂R2×Qijk so that (6.9) holds for m≥n. The set Q515 is given by the left hand side of (6.25), whose right hand side is (6.1), defining S515. Any (γ, δ, κ)∈Q522 satisfies the left hand side of (6.59), whose right hand side is (6.2), defining S522. Any (γ, δ, κ)∈Q252 satisfies the left hand side of (6.13) or (6.14), and satisfies the left hand side of (6.55) and (6.66). So, when m≥n, the estimate (6.9) is guaranteed for any (α, β, γ, δ, κ)∈R2×Q252 satisfying both (6.13) and (6.14), or any of (6.55) or (6.66). On R2×Q252, these conditions mean that (α, β, γ, δ, κ) satisfies (6.3) and (6.4), defining S252. Any (γ, δ, κ)∈Q155 satisfies the left hand side of (6.13) or (6.14), and satisfies the left hand side of (6.51), (6.42) and (6.65). So, when m≥n, the estimate (6.9) is guaranteed for any (α, β, γ, δ, κ)∈R2×Q155 satisfying both (6.13) and (6.14), or any of (6.51), (6.42) or (6.65). On R2×Q155, these conditions mean that (α, β, γ, δ, κ) satisfies (6.3) and (6.5), defining S155. Any (γ, δ, κ)∈Q212 satisfies the left hand side of (6.26) or (6.27). So, when m≥n, the estimate (6.9) is guaranteed for any (α, β, γ, δ, κ)∈R2×Q212 satisfying both (6.26) and (6.27). On R2×Q212, these conditions become (6.6) and (6.7), defining S212. 6.2.9 The preliminary estimate is satisfied on a convex set Let us show the convexity of the set of elements x= (α, β, γ, δ, κ)∈R5satisfying (6.9) for m≥n, with ω=ω(x)>0. For i= 0,1, suppose that xi= (αi, βi, γi, δi,κi) satisfies (6.9) for m≥nwith ωi=ω(xi)>0. Recall that the case where m≤nfollows from the case where m≥nby using the mapping (6.8). For 0 <t<1, let xt= (αt, βt, γt, δt,κt) = (1 −t)x0+tx1, ωt= (1 −t)ω0+tω1>0. By H¨older inequality, for all m≥n, n X p=0 (m−p+ 1)γt(n−p+ 1)δt(p+ 1)κt = n X p=0(m−p+ 1)γ0(n−p+ 1)δ0(p+ 1)κ01−t(m−p+ 1)γ1(n−p+ 1)δ1(p+ 1)κ1t ≤  n X p=0 (m−p+ 1)γ0(n−p+ 1)δ0(p+ 1)κ0  1−t A Perturbation of the Dunkl Harmonic Oscillator on the Line 39 ×  n X p=0 (m−p+ 1)γ1(n−p+ 1)δ1(p+ 1)κ1  t . So (m+ 1)αt(n+ 1)βt n X p=0 (m−p+ 1)γt(n−p+ 1)δt(p+ 1)κt 4(m+ 1)−ω0(n+ 1)−ω01−t(m+ 1)−ω1(n+ 1)−ω1t= (m+ 1)−ωt(n+ 1)−ωt. Thus xtsatisfies (6.9) for m≥nwith ωt. This completes the proof of Lemma 6.1. 7 The main estimates Here, we show the estimates used in the proofs of Propositions 5.6 and 5.10. We continue with the notation of Section 6. Moreover let (σ, τ, θ) denote the standard coordinates of R3. Consider the affine injection R3→R5and the affine isomorphism of R3defined by (σ, τ, θ)7→ 1 4−σ 2,−1 4−τ 2, σ −θ−1, τ −θ, θ −1 2,(7.1) (σ, τ, θ)7→ (τ+ 1, σ −1, θ).(7.2) The mapping (7.2) is the reflection with respect to the plane defined by σ=τ+ 1, and it corresponds to the mapping (6.8) via (7.1). Let ˇ K,ˇ K0⊂R3be the inverse images of ˇ S,ˇ S0 by (7.1), and let ˇ Kconv,ˇ K0 conv be their convex hulls. So ˇ Kconv,ˇ K0 conv are contained in the inverse images of ˇ Sconv,ˇ S0 conv by (7.1), and ˇ K0,ˇ K0 conv are the images of ˇ K,ˇ Kconv by (7.2). Thus ˇ Kconv ∩ˇ K0 conv is symmetric with respect to the plane σ=τ+ 1. We will show the following. Lemma 7.1. ˇ Kconv ∩ˇ K0 conv consists of the elements (σ, τ, θ)∈R3that satisfy (1.7). The following is a direct consequence of Lemmas 6.1 and 7.1. Corollary 7.2. If (σ, τ, θ)∈R3satisfies (1.7), then there is some ω > 0such that (6.9)holds with the image (α, β, γ, δ, κ)of (σ, τ, θ)by (7.1). Let ˇ J⊂R2be the inverse image of ˇ Kconv ∩ˇ K0 conv by the affine injection R2→R3, (σ, τ)7→ (σ, τ, τ). The following is a direct consequence of Lemma 7.1. Lemma 7.3. ˇ Jconsists of the elements (σ, τ)∈R2that satisfy (1.5). Lemma 7.3 and Corollary 7.2 have the following direct consequence. Corollary 7.4. If (σ, τ)∈R2satisfies (1.5), then there is some ω > 0such that (6.9)holds with the image (α, β, γ, δ, κ)of (σ, τ, τ)by (7.1). Let us prove Lemma 7.1. For the subindices ijk equal to 515, 522, 252, 155, 212, let Kijk and Rijk be the inverse images of Sijk and R2×Qijk by the mapping (7.1). Thus ˇ K=K515 ∪K522 ∪ K252 ∪K155 ∪K212. Moreover, for every θ∈R, let I1 1(θ)=(−∞, θ], I2 1(θ)=(−∞, θ −1], I3 1=−∞,−1 2, I1 2(θ) = θ, θ +1 2, I2 2(θ) = θ−1, θ −1 2, I3 2=−1 2,0, I1 3(θ) = θ+1 2, θ +2 3, I2 3(θ) = θ−1 2, θ −1 3, I3 3=0,1 6, I1 4(θ) = θ+2 3, θ + 1, I2 4(θ) = θ−1 3, θ, I3 4=1 6,1 2, 40 J.A. ´ Alvarez L´opez, M. Calaza and C. Franco I1 5(θ)=[θ+ 1,∞), I2 5(θ) = [θ, ∞), I3 5=1 2,∞. It can be directly checked that Rijk =(σ, τ, θ)∈R3|(σ, τ)∈I1 i(θ)×I2 j(θ), θ ∈I3 k. Simple computations show that, via (7.1), the conditions defining the sets Sijk (Section 6.1) become the following descriptions of the sets Kijk (Fig. 3(a)): K515:This is the subset of R515 defined by σ 2−3 4< θ, σ −τ−3<0. K522:This is the subset of R522 defined by σ 2−3 4,σ−τ 2−1< θ. K252:This is the subset of R252 defined by σ+τ−1 2< θ, (7.3)        σ 2−1 4,τ−σ 2< θ, or σ 2−5 12,τ−σ 2+1 3< θ, or σ 2−3 4< θ, τ −σ+ 1 <0. K155:This is the subset of R155 defined by (7.3) and              σ 2+1 4< θ, τ −σ−1<0,or σ 2−1 4< θ, τ −σ < 0,or σ 2−5 12 < θ, τ −σ+1 3<0,or σ 2−3 4< θ, τ −σ+ 1 <0. K212:This is the subset of R212 defined by σ 2−1 4≤θ⇒θ < σ+τ+1 2, θ≤σ 2−1 4⇒σ 2−3 4,σ−τ 2−1< θ. With tedious computations assisted by graphics produced with Mathematica, it follows that ˇ Kconv is the open subset of R3defined by (Fig. 3(b)) σ−τ 2−1,τ−σ 2,σ+τ−1 4,σ+3τ−2 14 ,σ+τ−1 2< θ < σ+τ+1 2, τ −1< σ < τ + 3.(7.4) This is a “semi-infinite bar” with 4 lateral faces, and 4 faces at the “bounded end”. Applying the affine transformation (7.2) to this description, we get that ˇ K0 conv consists of the triples (σ, τ, θ)∈R3satisfying the following conditions: σ−τ 2−1,τ−σ 2,σ+τ−1 4,3σ+τ−4 14 ,σ+τ−1 2< θ < σ+τ+1 2, τ −1< σ < τ + 3.(7.5) Combining (7.4) and (7.5), it follows that ˇ Kconv ∩ˇ K0 conv is given by (1.7) (Fig. 2), completing the proof of Lemma 7.1. A Perturbation of the Dunkl Harmonic Oscillator on the Line 41 (a) ˇ K(b) ˇ Kconv Figure 3. The sets ˇ Kand ˇ Kconv. Remark 7.5. In Sections 6.2.1–6.2.7, we have only written the cases that provide the most general conditions to define S515,S522,S252,S155,S212. But indeed much more hidden work was needed to produce this shorter proof: •We have computed all cases in Sections 6.2.1–6.2.7, giving rise to seven long lists of conditions that guarantee (6.9) when m≥n. •We have studied which of those conditions are the most general ones on every R2×Qijk, for all ijk = 1,...,5. This produces 125 sets Sijk, whose inverse images by (7.1) give 125 sets Kijk. The corresponding unions are denoted by Sand K, and their convex hulls by Sconv and Kconv. •We got that 41 sets Kijk are empty, including the 25 sets of the form Kij1, and the remaining 84 sets Kijk fit together forming a “semi-infinite bar” (Fig. 4). •With tedious computations, we have shown that Kconv is given by (7.4). •We have chosen the most simple family, K515,K522,K252,K155,K212, defining the same convex hull (ˇ Kconv =Kconv). •Finally, we have made some attempts to improve the estimates of Section 6.2.7 by using more general versions of the H¨older inequality [7]. Some better estimates were obtained in this way, but they produce the same set Kconv after taking the convex hull. Remark 7.6. The set Sconv may have a simple expression, like Kconv, but its computation became too involved. This is the reason we have used Kconv, obtaining the conditions of Theorem 1.3, which are general enough for our applications in [4]. But, of course, the inverse image of Sconv by (7.1) is possibly larger than Kconv. Therefore a simple expression of Sconv would possibly give a better version of Theorem 1.3. Even a simple expression of ˇ Sconv would possibly give a better version of Theorem 1.3. 8 Operators induced on R+ Let Sev/odd,+={φ|R+|φ∈ Sev/odd}. For c, d > −1 2, let L2 c,+=L2(R+, x2cdx) and L2 c,d,+= L2 c,+⊕L2 d,+, whose scalar products are denoted by h,icand h,ic,d, respectively. For c1, c2, d1, d2∈R, let P0=H−2c1x−1d dx +c2x−2, Q0=H−2d1 d dxx−1+d2x−2. Morever let ξ > 0 and η, θ ∈R.