On the Graf s addition theorem for Hahn Exton q-bessel function

On the Graf s addition theorem for Hahn Exton q-bessel function Lazhar Dhaouadi, Ahmed Fitouhi To cite this version: Lazhar Dhaouadi, Ahmed Fitouhi. On the Graf s addition theorem for Hahn Exton q-bessel function. Constructive Approximation, Springer Verlag, 2, 34, pp.453-472. <.7/s365--932->. <hal-6528v2> HAL Id: hal-6528 https://hal.archives-ouvertes.fr/hal-6528v2 Submitted on 6 Sep 28 HAL is a multi-disciplinary open access archive for the deposit and dissemination of scientific research documents, whether they are published or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers. L archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

On the Graf s addition theorem for Hahn Exton q-bessel function Lazhar Dhaouadi Ahmed Fitouhi Abstract In this paper we study the positivity of the generalized q-translation associated with the q-bessel Hahn Exton function which is deduced by a new formulation of the Graf s addition formula related to this function. Introduction and Preliminaries It is well known that the generalized translation operator T associated with the Bessel function of the first kind is positive in the sense if f >, then Tf >. This property is easily seen when we write Tf as an integral representation with a kernel involving the area of some triangle (,9) and has several applications in many mathematical fields such that hypergroup structure, heat equation... In 22, the appropriated q-generalized translation for the q-bessel Hahn Exton function was founded 3 and the problem of its positivity asked. In literature we meet many attempts to show this property in particular case, nerveless they are no definitive response at to day. In 2 the authors prove that for the q-cosinus, the correspondent q-translation operator is positive if q,q for some q. In this work and owing a new formulation of the Graf s addition theorem 7 we give an affirmative answer about this theme by a technic involving some inclusion of sets. To make this work self containing, we begin by the following preliminaries. Throughout this paper we consider < q < and we adopt the standard conventional notations of 4. For complex a We put IPEB, 72 Zarzouna, Bizerte,Tunisia. E-mail lazhardhaouadi@yahoo.fr Faculté des sciences de Tunis, 6 Tunis, Tunisia. E-mail ahmed.fitouhi@fst.rnu.tn

n (a;q), (a;q) n ( aq i ), n.... Jackson s q-integral (see 5) over the interval, is defined by We denote by i f(x)d q x ( q) R + q {qn, n n Z}, q n f(q n ). and we consider L q,p,v the space of even functions f defined on R + q such that /p f q,p,v f(x) p x 2v+ d q x <. The q-bessel function of third kind and of order v, called also Hahn-Exton function,is defined by the q-hypergeometric function (see 8) J v (x,q) (qv+ ;q) (q;q) x v φ (,q v+,q,qx 2 ), R(v) >, and has a normalized form is given by j v (x,q) (q;q) (q v+ ;q) x v J v (x,q) φ (,q v+,q,qx 2 ) ( ) n q n(n+) 2 (q;q) n (q v+ x 2n. ;q) n It s an entire analytic function in z. In(2,6) the q-bessel Fourier transform F q,v and its properties were studied in great detail, it is defined as follow where n F q,v f(x) c q,v f(t)j v (xt,q 2 )t 2v+ d q t, c q,v (q 2v+2,q 2 ) q (q 2,q 2. ) There is many way to define the q-bessel translation operator 2,3. One of them can be enounced for suitable function f as follows: Tq,xf(y) v c q,v F q,v (f)(t)j v (xt,q 2 )j v (yt,q 2 )t 2v+ d q t, x,y R + q, f L q,,v. 2

Now we say that T v q,x is positive if T v q,xf for f. Let us putting the domain of positivity of T v q,x by Q v {q,, T v q,x is positive for all x R + q }. Trough the result of 2, Q v is not empty. So for q Q v the q convolution product of the two functions f and g L q,,v is defined by f q g(x) c q,v Tq,xf(y)g(y)y v 2v+ d q y. To close this section we present the following results proved in 2 which will be used in the remainder. Proposition j v (q n,q 2 ) ( q2 ;q 2 ) ( q 2v+2 ;q 2 ) (q 2v+2 ;q 2 ) Theorem The operator F q,v satisfies. For all functions f L q,,v, F 2 q,vf(x) f(x), x R + q. 2. If f L q,,v and F q,v f L q,,v then F q,v f q,v,2 f q,v,2. 2 The Graf s addition formula { if n q n2 (2v+)n if n <. The Graf s addition formula for Hahn-Exton q-bessel function proved by H.T.Koelink and F. Swarttouw 7 plays a central role here. It can be stated as follows. J v (Rq /2(y+z+v),q)J x v (q /2z,q) k Z J k (Rq /2(x+y+k),q)J v+k (Rq /2(y+k+v),q)J x (q /2(z k),q); and it is valid when z Z and R,x,y,v C satisfying R 2 q +R(x)+R(y) <, R(x) >, R. 3

This formula has originally been derived for v,x,y Z, R > by the interpretation of the Hahn-Exton q-bessel function as matrix elements of irreducible unitary representation of the quantum group of plane motions. If we replace q by q 2 and R by q r in the previous formula we get: J v (q y+z+v+r,q 2 )J x v (q z,q 2 ) k ZJ k (q x+y+k+r,q 2 )J v+k (q y+k+v+r,q 2 )J x (q z k,q 2 ), and put so and we have m y + z + v + r, x + y + k + r m + k + x z v, y + k + v + r m + k z, J v (q m,q 2 )J x v (q z,q 2 ) k ZJ k (q m+k+x z v,q 2 )J v+k (q m+k z,q 2 )J x (q z k,q 2 ). This last formula is valid for z Z and r,x,y,v C satisfying + 2R(r) + R(x) + R(y) +R(r)+R(m) R(z) R(v) >, R(x) >. In the above sum we replace z k by k we get J v (q m,q 2 )J x v (q z,q 2 ) J z k (q m+x v k,q 2 )J v+z k (q m k,q 2 )J x (q k,q 2 ), k Z The sum in the second member exists for z Z, m,v,x C, R(x) >. In fact there exist an infinity complex number r C for which + R(r) + R(m) R(z) R(v) >. Now using the definition of the normalized q-bessel function J x (q k,q 2 ) (q2x+2,q 2 ) (q 2,q 2 ) q vk j x (q k,q 2 ) ( q)c q,x q xk j x (q k,q 2 ), 4

we obtain q mv+z(x v) ( q) 2 c q,v c q,x v j v (q m,q 2 )j x v (q z,q 2 ) ( q)c q,x q xk J z k (q x v+m k,q 2 )J v+z k (q m k,q 2 )j x (q k,q 2 ). k Z Let and replace This implies λ q n, n Z, k k + n,m m + n,z z + n. q mv+z(x v) ( q) 2 c q,v c q,x v j v (q m λ,q 2 )j x v (q z λ,q 2 ) ( q)c q,x q xk J z k (q x v+m k,q 2 )J v+z k (q m k,q 2 )j x (q k λ,q 2 ) k Z ( q)c q,x q 2k(x+) q k(x+2) J z k (q x v+m k,q 2 )J v+z k (q m k,q 2 )j x (q k λ,q 2 ). We put k Z E v,x (q m,q z,q k ) ( q) 2 c q,v c q,x v q k(x+2) mv z(x v) J z k (q x v+m k,q 2 )J v+z k (q m k,q 2 ). The Proposition shows that the function: λ j v (q m λ,q 2 )j x v (q z λ,q 2 ), m,z Z, belongs to the space L q,,v. Theorem, part leads to the following statement: Proposition 2 For z,m Z and x,v C satisfying R(x v) >, R(v) > we have and j v (q m λ,q 2 )j x v (q z λ,q 2 ) c q,x E v,x (q m,q z,t)j x (λt,q 2 )t 2x+ d q t, E v,x (q m,q z,q k ) c q,x j v (q m λ,q 2 )j x v (q z λ,q 2 )j x (q k λ,q 2 )λ 2x+ d q λ. 5

3 Positivity of the q-translation operator This section is a direct application of the previous one but before any things we recall that the q-translation operator possess the q-integral representation (see 2) Proposition 3 Let f L q,,v then where T v q,xf(y) D v (x,y,z) c 2 q,v f(z)d v (x,y,z)z 2v+ d q z, j v (xt,q 2 )j v (yt,q 2 )j v (zt,q 2 )t 2v+ d q t. When q tends to we obtain at least formally the classical one and the q-kernel D v (x,y,z) tends to the classical one (,9) which involves the area of some triangle. Hence the positivity of T v q,x is subject of those of D v (x,y,z). A direct consequence of the above result is the fact that if the operator T v q,x is positive and f L q,,v then T v q,x f L q,,v ( in general it is not true without this hypothesis). Indeed Putting T v q,xf(y) y 2v+ d q t T v q,x f (y)y 2v+ d q t then we can write f(z) D v (x,y,z)y 2v+ d q y z 2v+ d q z. φ : t j v (xt,q 2 )j v (zt,q 2 ), D v (x,y,z) c q,v F q,v φ(y). This gives by the of the inversion formula in Theorem the important result as the classical; one Then we have and we obtain D v (x,y,z)y 2v+ d q y Fq,v 2 φ() φ(). T v q,xf(y) y 2v+ d q t f q,v,, f L q,,v T v q,x f L q,,v. 6

In 2 and as a first preamble of this theme the authors proved that D (q m,q r,q k ) q2(r m)(k m) m (q 2(r m)+ ;q) φ (,q 2(r m)+,q;q 2(k m)+ ) 2 ( q)(q;q) I q q m J 2(r m) (q k m,q), which implies that the correspondent domain of T 2 q,x is given by Q 2,q, where q is the first zero of the q-hypergeometric function: q φ (,q,q,q); a second statement is easily given by the Proposition 2 with x v which gives and then E, (q m,q z,q k ) c q, j (q m λ,q 2 )j (q z λ,q 2 )j (q k λ,q 2 )λd q λ Hence we have c q, D (q m,q z,q k ), D (q m,q z,q k ) c q, E, (q m,q z,q k ) 2 ( q) q 2k J z k (q m k,q 2 ), Q,. Now we explicit the kernel in the production formula in terms of D v. Proposition 4 For n,m,k Z and < v we have E v,v (q m,q n,q k ) ( q) i (q 2v,q 2 ) i q i(2+2v) D v (q m,q i+n,q k ). Proof. From the following formula (see7, 5) which can be proved in a straightforward way by substitution of the defining series for the q-bessel 7

functions on both sides, by interchanging summations, and by evaluating the q-binomial series which occurs J x v (λ,q 2 ) λ v i where R(x v) > and R(x) > we obtain ( q)c q,x v j x v (λ,q 2 ) ( q)c q,x Put x v and change λ by q n λ we obtain j (q n λ,q 2 ) ( q)c q,v which gives from Proposition 2 i (q 2v,q 2 ) i q i(2+x) J x (λq i,q 2 ), i (q 2v,q 2 ) i q i(2+2x) j x (λq i,q 2 ). (q 2v,q 2 ) i q i(2+2v) j v (λq i+n,q 2 ), E v,v (q m,q n,q k ) c q,v j v (q m λ,q 2 )j (q n λ,q 2 )j v (q k λ,q 2 )λ 2v+ d q λ ( q) c 2 q,v i ( q) (q 2v,q 2 ) i q i(2+2v) j v (q m λ,q 2 )j v (q i+n λ,q 2 )j v (q k λ,q 2 )λ 2v+ d q λ i (q 2v,q 2 ) i q i(2+2v) D v (q m,q i+n,q k ). We justify the exchange of the signs sum and integral by i (q 2v,q 2 ) i q i(2+2v) j v (.,q 2 ) 2 q, j v(.,q 2 ) q,,v q 2(v+)m So we obtain the result. Proposition 5 Let < v < then j v (q m λ,q 2 )j v (q i+n λ,q 2 )j v (q k λ,q 2 )λ 2v+ dq λ i Q v,. (q 2v,q 2 ) i q i(2+2v) <. 8

Proof. For m,n,k Z and < v < we have E v,v (q m,q n,q k ) q k(v+2) mv J n k (q m k,q 2 )J v+n k (q m k,q 2 ) ( q)c q,v (q 2v,q 2 ) i ( q) (q 2,q 2 q i(2+2v) D v (q m,q i+n,q k ). ) i i In particular for m n k E v,v (,,) We introduce the following function ( q)c q,v J (,q 2 )J v (,q 2 ). φ v : q (q 2,q 2 ) 2 J v (,q 2 ). Let q.658 the first zero of φ and consider the graph of the function v φ v (q ), v,..8.6 x.4.2..2.3.4.5.6.7 We conclude that for < v < we have φ v (q ) <. Then there exist a very small ε > such that φ v (q ε) < and φ (q ε) >. Hence, this function q J (,q 2 )J v (,q 2 ) <, takes some negative values in the interval,, then for some entire i N As (q 2v,q 2 ) i q i(2+2v) D v (,q i,) <. (q 2v,q 2 ) i q i(2+2v) >, then D v (,q i,) < for some entire i N. Which prove that This finish the proof. Q v,. 9

Lemma For x R + q and R(v + t) > we have ( q)c q,v n Z q (+v+t)n j v (q n x,q 2 ) (q+v t ;q 2 ) (q +v+t ;q 2 ) x (+v+t). Proof. In 6 the following result was proved n Z Then we get q (+t)n J v (q n,q 2 ) (q+v t ;q 2 ) (q +v+t ;q 2 ), R(v + t) >. ( q)c q,v n Z Hence for x q k R + q we have ( q)c q,v This finish the proof. n Z Theorem 2 Let v x > then As consequence If v then Q v,. q (+v+t)n j v (q n,q 2 ) (q+v t ;q 2 ) (q +v+t ;q 2 ). q (+v+t)(n+k) j v (q n+k,q 2 ) (q+v t ;q 2 ) (q +v+t ;q 2 ). Q x Q v. If 2 v < then,q Q v,. If < v 2 then Q v,q. Proof. Let v > and < µ <. We will prove that Q v Q v+µ, which suffices to show that Q x Q v if x v. In the following we tack q Q v. We have c q,v+µ j v+µ (t,q 2 ) c q,v i (q 2µ,q 2 ) i q 2(v+)i j v (tq i,q 2 ),

and then which implies c q,v c 2 q,v+µ j v(tx,q 2 )j v+µ (ty,q 2 )j v (tz,q 2 ) c 3 (q 2µ,q 2 ) i (q 2µ,q 2 ) j q,v (q 2,q 2 q 2(+v)(i+j) ) j i,j,s j v (tx,q 2 )j v (tq i y,q 2 )j v (tq j z,q 2 ), where Note that c q,v T v,v+µ,v (x,y,z) c q,v i,j T v,w,α (x,y,z) c 2 q,w (q 2µ,q 2 ) i (q 2µ,q 2 ) j (q 2,q 2 ) j q 2(+v)(i+j) D v (x,q i y,q j z), j v (tx,q 2 )j w (ty,q 2 )j w (tz,q 2 )t 2α+ d q t. T v,v,v (x,y,z) D v (x,y,z). The exchange of the signs sum and integral is valid since: q 2(+v)(i+j) j v (q m t,q 2 ) j v (q n+i t,q 2 ) jv (q k+j t,q 2 ) t 2v+ d q t i,j + q 2(+v)(i+j) jv (q m t,q 2 ) jv (q n+i t,q 2 ) jv (q k+j t,q 2 ) t 2v+ d q t i,j q 2(+v)(i+j) j v (q m t,q 2 ) j v (q n+i t,q 2 ) jv (q k+j t,q 2 ) t 2v+ d q t. i,j Note that q 2(+v)(i+j) jv (q m t,q 2 ) jv (q n+i t,q 2 ) jv (q k+j t,q 2 ) t 2v+ d q t i,j jv (.,q 2 ) 3 q, i,j q 2(+v)(i+j) t 2v+ d q t <,

and q 2(+v)(i+j) j v (q m t,q 2 ) j v (q n+i t,q 2 ) jv (q k+j t,q 2 ) t 2v+ d q t i,j ( q) j,r i,j,r q 2(+v)(i+j r) jv (q m r,q 2 ) jv (q n+i r,q 2 ) jv (q k+j r,q 2 ) q 2(+v)j jv (q m r,q 2 ) jv (q k+j r,q 2 ) ( q) q 2(+v)(i r) jv (q n+i r,q 2 ), where r r. We write q 2(+v)(i r) jv (q n+i r,q 2 ) q 2(+v)n ( q) q 2(n+i r) jv (q n+i r,q 2 ) i Then j,r q 2(+v)n ( q) i im r < q 2(+v)n jv (.,q 2 ) q,,v. q 2(+v)j jv (q m r,q 2 ) jv (q k+j r,q 2 ) s,r j v (q m r,q 2 ) q 2(+v)j jv (q k+j r,q 2 ) j i q 2(+v)i j v (q i,q 2 ) jv (q m r,q 2 ) q 2(+v)(r k) q 2(k+j r) j (q k+j r,q 2 ) r s,r q 2(v+)k q q 2s j v (q m r,q 2 ) j q 2(v+)(r k) jv (.,q 2 ) q,,v jv (.,q 2 ) q, jk r r q 2(+v)j j (q j,q 2 ) q 2(v+)r <. Let < α < µ. We introduce the function A α,µ,v (x) as follows: A α,µ,v (x) c q,v t 2α j v+µ (t,q 2 )j v (xt,q 2 )t 2v+ d q t. 2

From the inversion formula in Theorem (t t 2α j v+µ (t,q 2 ) L q,,v ) we get c q,v A α,µ,v (x)j v (xt,q 2 )x 2v+ d q x t 2α j v+µ (t,q 2 ). Let x. From Lemma we obtain A α,µ,v (x) c q,v ( q) n Zq 2(α+v+)n j v (xq n,q 2 )j v+µ (q n,q 2 ) c q,v ( q) ( ) i q i(i+) (q 2+2v,q 2 ) n Z i i (q 2,q 2 q 2(α+v++i)n j v+µ (q n,q 2 ) ) i c q,v ( q) ( ) i q i(i+) (q 2+2v,q 2 ) i (q 2,q 2 x 2i q 2(α+v++i)n j v+µ (q n,q 2 ) ) i c q,v c q,v+µ c q,v c q,v+µ Note that i n Z ( ) i q i(i+) (q 2+2w,q 2 ) i (q 2,q 2 x 2i(q2(+v+µ) 2(α+v++i),q 2 ) ) i (q 2(α+v++i),q 2 ) ( ) i q i(i+) (q 2+2v,q 2 ) i (q 2,q 2 x 2i(q2(µ α) q 2i,q 2 ) ) i (q 2(α+v++i),q 2. ) i i < µ α <. The exchange of the signs sum is valid since. Indeed, let q k x then we have q i(i+) q 2ik q 2(α+v++i)n jv+µ (q n,q 2 ) n Z i + + q i(i+) q 2ik q 2(α+v++i)n jv+µ (q n,q 2 ) i n q i(i+) q 2ik q 2(α+v++i)n jv+µ (q n,q 2 ) i n q i(i+) q 2ik q 2(α+v+)n i n q i(i+) q 2ik i n q 2(α+v++i)n+n2 +(2v+2µ+)n. 3

One has to observe that the first sum exist. The second sum also: q i(i+) q 2ik q 2(α+v++i)n+n2 +(2v+2µ+)n i n n q 2(α+v++i)n+n2 +(2v+2µ+)n q i(i+) q 2ik q 2(α+v+)n+(2v+2µ+)n+(2k+)n q (i n)2 q (2k+)(i n) n n i n i i q 2(µ α)n+(2k+)n q i2 q (2k+)i If x > then we obtain A α,µ,v (x) q 2(µ α)n+(2k+)n q i2 q (2k+)i <. n c q,v ( q) n Zq 2(α+v+)n j v (xq n,q 2 )j v+µ (q n,q 2 ) i c q,v ( q) ( ) i q i(i+) (q 2+2v,q 2 ) n Z i i (q 2,q 2 q 2(α+v++i)n j v (q n x,q 2 ) ) i c q,v ( q) ( ) i q i(i+) (q 2+2v+2µ,q 2 ) i (q 2,q 2 q 2(α+v++i)n j v (q n x,q 2 ) ) i i x 2(α+v+) ( ) i q i(i+) (q 2+2v+2µ,q 2 ) i (q 2,q 2 x 2i(q2(+v) 2(α+v++i),q 2 ) ) i (q 2(α+v++i),q 2 ) i x 2(α+v+) ( ) i q i(i+) (q 2+2v+2µ,q 2 ) i (q 2,q 2 x 2i (q 2α q 2i,q 2 ) ) i (q 2(α+v++i),q 2 <. ) Now, we write i c q,v A α,µ,v (x)t v,v+µ,v (x,y,z)x 2v+ d q x c q,v A α,µ,v (x) c 2 q,v j w (xt,q 2 )j v (yt,q 2 )j v (zt,q 2 )t 2v+ d q t x 2v+ d q x c 2 q,v+µ c q,v A α,µ,v (x)j v (xt,q 2 )x 2v+ d q x j v+µ (yt,q 2 )j v+µ (zt,q 2 )t 2v+ d q t c 2 q,v+µ n Z j v+µ (t,q 2 )j v+µ (yt,q 2 )j v+µ (zt,q 2 )t 2(v+α)+ d q t T v+µ,v+µ,v+α (,y,z). 4

To justify the exchange of the signs integrals we write A α,µ,v (x) j v (xt,q 2 ) x 2v+ d q x j v (yt,q 2 ) j v (zt,q 2 ) td q t j v (.,q 2 ) 2 q, j v(.,q 2 ) q,,v A α,µ,v q,,v z. Not that x A α,µ,v (x) is continued at and A α,µ,v (x) j v+µ (.,q 2 ) q, j v (.,q 2 ) q,,α+v x 2(α+v+), as x. On the other hand then we obtain for all δ > T v+µ,v+µ,v+α (,y,z) A α,µ,v (x)x 2v+ d q x, In the following we assume that < y,z. Let δ A α,µ,v (x) T v,v+µ,v (x,y,z) δ x 2v+ d q x. inf T v,v+µ,v(,y,z) y,z Not that δ exist and strictly positive, indeed the following function (y,z) T v,v+µ,v (,y,z) is continuous on the compact,,. Hence, there exist such that δ T v,v+µ,v (,y,z ) (y,z ),, If we assume that T v,v+µ,v (,y,z ) then i,j inf T v,v+µ,v(,y,z). y,z (q 2µ,q 2 ) i (q 2µ,q 2 ) j (q 2,q 2 ) j q 2(+v)(i+j) D v (,q i y,q j z ), which implies that c q,v F q,v t j v (t,q 2 )j v (z t,q 2 ) (q i y ) D v (,q i y,z ), i N. 5

From Proposition and the fact that there exist σ > such that jv (z,q 2 ) σ e z, z C we see that this function z F q,v t j v (t,q 2 )j v (z t,q 2 ) (z) is analytic. Then for all x R q F q,v t j v (t,q 2 )j v (z t,q 2 ) (x) j v (t,q 2 )j v (z t,q 2 ), t R q, but this is absurd. Then δ >. Now we have If then T v+µ,v+µ,v+α (,y,z) δ T v,v+µ,v (x,y,z), < x. δ > T v,v+µ,v (x,y,z), x >. A α,µ,v (x) T v,v+µ,v (x,y,z) δ x 2v+ d q x Otherwise, there exist s > such that + A α,µ,v (x) T v,v+µ,v (x,y,z) δ x 2v+ d q x. δ > T v,v+µ,v (x,y,z), x > s. because T v,v+µ,v (x,y,z) < c 2 q,v+µ jv+µ (.q 2 ) 2 q, jv (.q 2 ) q,v, x 2(v+), as x. For δ δ we obtain T v+µ,v+µ,v+α (,y,z) + + s s A α,µ,v (x) T v,v+µ,v (x,y,z) δ x 2v+ d q x A α,µ,v (x) T v,v+µ,v (x,y,z) δ x 2v+ d q x A α,µ,v (x) T v,v+µ,v (x,y,z) δ x 2v+ d q x I (δ) + I 2 (δ) + I 3 (δ). 6

Note that δ I (δ) is a decreesing function tends towards and I (δ ) > δ I 2 (δ) is an increasing function tends towards + and I 2 (δ ) < δ I 3 (δ) is an increasing function tends towards + and I 3 (δ ) > then there exist δ > δ such that I (δ) + I 2 (δ) which implies In the end T v+µ,v+µ,v+α (,y,z) I 3 (δ) >. lim T v+µ,v+µ,v+α(,y,z) α µ lim c 2 α µ q,v+µ c 2 q,v+µ j v+µ (t,q 2 )j v+µ (yt,q 2 )j v+µ (zt,q 2 )t 2(v+α)+ d q t j v+µ (t,q 2 )j v+µ (yt,q 2 )j v+µ (zt,q 2 ) ( lim α µ t 2(v+α)+) d q t T v+µ,v+µ,v+α (,y,z) D v+µ (,y,z). It is not hard to justify the exchange of the signs integral and limit, in fact j v+µ (t,q 2 )j v+µ (yt,q 2 )j v+µ (zt,q 2 )t 2(v+α)+ t j v+µ (yt,q 2 )j v+µ (zt,q 2 )t 2(v+α)+ q, jv+µ (t,q 2 ). From the following identity ( D v+µ (x,y,z) x 2(v+µ+) D v+µ, y x, z ), x we deduce that D v+µ (x,y,z), x,y,z R + q. The fact that leads to the result. Q 2,q and Q, 7

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