International Journal of Analysis and Applications
ISSN 2291-8639
Volume 8, Number 2 (2015), 104-109
http://www.etamaths.com

SOME REMARKS CONCERNING THE JACOBI-DUNKL

TRANSFORM IN THE SPACE Lp(R,Aα,β(t)dt)

R. DAHER, S. EL OUADIH∗ AND A. BELKHADIR

Abstract. In this paper, using a generalized Jacobi-Dunkl translation oper-

ator, we obtain a generalization of Titchmarsh’s theorem for the Dunkl trans-

form for functions satisfying the (φ,p)-Lipschitz Jacobi-Dunkl condition in the

space Lp(R,Aα,β(t)dt),α ≥ β ≥ −12 ,α 6=
−1
2
.

1. INTRODUCTION AND PRELIMINARIES

Titchmarsh’s [8,Theorem 85] characterized the set of functions in L2(R) satisfy-
ing the Cauchy-Lipschitz condition by means of an asymptotic estimate growth of
the norm of their Fourier transform, namely we have

Theorem 1.1. [8] Let α ∈ (0, 1) and assume that f ∈ L2(R). Then the fol-
lowing are equivalents
(a) ‖f(t + h) −f(t)‖ = 0(hα), as h → 0

(b)

∫
|λ|≥r

|f̂(λ)|2dλ = O(r−2α) as r →∞,

where f̂ stand for the Fourier transform of f.
In this paper, we prove a generalization of Theorem 1.1 for the Jacobi-Dunkl trans-
form for functions satisfying the (φ,p)-Lipschitz Jacobi-Dunkl condition in the space
Lp(R,Aα,β(t)dt), 1 < p ≤ 2. For this purpose, we use the generalized Jacobi-Dunkl
translation operator.
In this section, we recapitulate from [1,2,3,5,6] some results related to the harmonic
analysis associated with Jacobi-Dunkl operator Λα,β.
The Jacobi-Dunkl function with parameters (α,β),α ≥ β ≥ −1

2
,α 6= −1

2
, defined

by the formula:

∀x ∈ R,ψα,βλ (x) =




ϕα,βµ (x) −
i
λ
d
dx
ϕα,βµ (x) if λ ∈ C\{0}

1 if λ = 0

with λ2 = µ2 + ρ2, ρ = α + β + 1 and ϕα,βµ is the Jacobi function given by:

ϕα,βµ (x) = F

(
ρ + iµ

2
,
ρ− iµ

2
,α + 1,−(sinh(x))2

)
,

2010 Mathematics Subject Classification. 65R10.
Key words and phrases. Jacobi-Dunkl operator, Jacobi-Dunkl transform, generalized Jacobi-

Dunkl translation.

c©2015 Authors retain the copyrights of their papers, and all
open access articles are distributed under the terms of the Creative Commons Attribution License.

104



GELFAND TRIPLE ISOMORPHISMS 105

F is the Gausse hypergeometric function (see [1,7]).

ψ
α,β
λ is the unique C

∞-solution on R of the differentiel-difference equation


Λα,βU = iλU ,λ ∈ C

U(0) = 1
where Λα,β is the Jacobi-Dunkl operator given by:

Λα,βU(x) =
dU(x)
dx

+ [(2α + 1) coth x + (2β + 1) tanh x] ×
U(x) −U(−x)

2
].

The operator Λα,β is a particular case of the operator D given by

DU(x) =
dU(x)
dx

+
A′(x)

A(x)
×
(
U(x) −U(−x)

2

)
,

where A(x) = |x|2α+1B(x), and B a function of class C∞ on R, even and positive.
The operator Λα,β corresponds to the function

A(x) = Aα,β(x) = 2
ρ(sinh |x|)2α+1(cosh |x|)2β+1.

Using the relation

d

dx
ϕα,βµ (x) = −

µ2 + ρ2

4(α + 1)
sinh(2x)ϕα+1,β+1µ (x),

the function ψ
α,β
λ can be written in the form above (see [2])

∀x ∈ R,ψα,βλ (x) = ϕ
α,β
µ (x) + i

λ

4(α + 1)
sinh(2x)ϕα+1,β+1µ (x).

Denote L
p
α,β(R) = L

p
α,β(R,Aα,β(t)dt), 1 < p ≤ 2, the space of measurable functions

f on R such that

‖f‖p,α,β =
(∫

R
|f(t)|pAα,β(t)dt

)1/p
< +∞.

Using the eigenfunctions ψ
α,β
λ of the operator Λα,β called the Jacobi-Dunkl kernels,

we define the Jacobi-Dunkl transform by

Fα,βf(λ) =
∫
R
f(t)ψ

α,β
λ (t)Aα,β(t)dt, λ ∈ R,

and the inversion formula by

f(t) =

∫
R
Fα,βf(λ)ψ

α,β
−λ (t)dσ(λ),

where

dσ(λ) =
|λ|

8π
√
λ2 −ρ2|Cα,β(

√
λ2 −ρ2)|

IR\]−ρ,ρ[(λ)dλ.

Here,

Cα,β(µ) =
2ρ−iµΓ(α + 1)Γ(iµ)

Γ( 1
2
(ρ + iµ))Γ( 1

2
(α−β + 1 + iµ))

, µ ∈ C\(iN)

and IR\]−ρ,ρ[ is the characteristic function of R\] −ρ,ρ[.
The Jacobi-Dunkl transform is a unitary isomorphism from L2α,β(R) onto L

2(R,dσ(λ)),
i.e.

‖f‖2,α,β = ‖Fα,β(f)‖L2(R,dσ(λ)).(1)



106 DAHER, OUADIH AND BELKHADIR

Plancherel’s theorem (1) and the Marcinkiewics interpolation theorem (see [8]) we
get for f ∈ Lpα,β(R) with 1 < p ≤ 2 and q such that

1
p

+ 1
q

= 1,

‖Fα,β(f)‖Lq(R,dσ(λ)) ≤ K‖f‖p,α,β,(2)

where K is a positive constant (see [6]).
The operator of Jacobi-Dunkl translation is defined by

Txf(y) =

∫
R
f(z)dνα,βx,y (z), ∀x,y ∈ R

where να,βx,y (z),x,y ∈ R are the signed measures given by

dνα,βx,y (z) =




Kα,β(x,y,z)Aα,β(z)dz if x,y ∈ R∗

δx if y = 0
δy if x = 0

Here, δx is the Dirac measure at x. And,

Kα,β(x,y,z) = Mα,β(sinh(|x|) sinh(|y|) sinh(|z|))−2αIIx,y ×
∫ π

0

ρθ(x,y,z)

× (gθ(x,y,z))
α−β−1
+ sin

2β θdθ

Ix,y = [−|x|− |y|,−||x|− |y||] ∪ [||x|− |y||, |x| + |y|]
ρθ(x,y,z) = 1 −σθx,y,z + σ

θ
z,x,y + σ

θ
z,y,x

∀z ∈ R,θ ∈ [0,π],σθx,y,z =




cosh(x)+cosh(y)−cosh(z) cos(θ)
sinh(x) sinh(y)

,if xy 6= 0

0 ,if xy = 0

gθ(x,y,z) = 1 − cosh2(x) − cosh2(y) − cosh2(z) + 2 cosh(x) cosh(y) cosh(z) cos θ

t+ =




t ,if t > 0

0 ,if t ≤ 0
and,

Mα,β =




2−2ρΓ(α+1)√
πΓ(α−β)Γ(β+ 1

2
)

,if α > β

0 ,if α = β

In [2], we have

Fα,β(Thf)(λ) = ψ
α,β
λ (h)Fα,β(f)(λ).(3)

For α ≥ −1
2

, we introduce the Bessel normalized function of the first kind defined
by

jα(z) = Γ(α + 1)

∞∑
n=0

(−1)n(z
2
)2n

n!Γ(n + α + 1)
, z ∈ C.

Moreover, we see that

lim
z→0

jα(z) − 1
z2

6= 0,

by consequence, there exists C1 > 0 and η > 0 satisfying

|z| ≤ η ⇒|jα(z) − 1| ≥ C1|z|2.(4)



GELFAND TRIPLE ISOMORPHISMS 107

Lemma 1.1. Let α ≥ β ≥ −1
2
,α 6= −1

2
. Then for |ν| ≤ ρ, there exists a positive

constant C2 such that

|1 −ϕα,βµ+iν(t)| ≥ C2|1 − jα(µt)|.

Proof. (See[4],Lemma 9).

2. MAIN RESULT

In this section we give the main result of this paper. We need first to define
(φ,p)-Lipschitz Jacobi-Dunkl class.
Denote Nh by

Nh = Th + T−h − 2I
where I is the unit operator in the space L

p
α,β(R).

Definition 2.1. A function f ∈ Lpα,β(R) is said to be in (φ,p)-Lipschitz Jacobi-
Dunkl class, denoted by Lip(φ,p,α,β), if

‖Nhf‖p,α,β = O(φ(h)), as h → 0,

where φ(t) is a continuous increasing function on [0,∞), φ(0) = 0 and φ(ts) =
φ(t)φ(s) for all t,s ∈ [0,∞).

Lemma 2.2. For f ∈ Lpα,β(R), then(∫
R

2q|ϕα,βµ (h) − 1|
q|Fα,βf(λ)|qdσ(λ)

)1
q

≤ K‖Nhf‖p,α,β

where 1
p

+ 1
q

= 1.

Proof. We us formula (3), we conclude that

Fα,β(Nhf)(λ) = (ψ
α,β
λ (h) + ψ

α,β
λ (−h) − 2)Fα,β(f)(λ),

Since

ψ
α,β
λ (h) = ϕ

α,β
µ (h) + i

λ

4(α + 1)
sinh(2h)ϕα+1,β+1µ (h),

ψ
α,β
λ (−h) = ϕ

α,β
µ (−h) − i

λ

4(α + 1)
sinh(2h)ϕα+1,β+1µ (−h),

and ϕα,βµ is even (see [2]), then

Fα,β(Nhf)(λ) = 2(ϕα,βµ (h) − 1)Fα,β(f)(λ).

By formula (2), we have the result.

Theorem 2.3. Let f(x) belong to Lip(φ,p,α,β). Then∫
|λ|≥r

|Fα,β(f)(λ)|qdσ(λ) = O(φ(r−q)), as r →∞,

where 1
p

+ 1
q

= 1.

Proof. Assume that f ∈ Lip(φ,p,α,β), then we have

‖Nhf‖p,α,β = O(φ(h)), as h → 0.



108 DAHER, OUADIH AND BELKHADIR

From Lemma 2.2, we have∫
R
|ϕα,βµ (h) − 1|

q|Fα,βf(λ)|qdσ(λ) ≤
Kq

2q
‖Nhf‖

q
p,α,β

By (4) and Lemma 1.1, we get∫
η
2h
≤|λ|≤η

h

|1−ϕα,βµ (h)|
q|Fα,β(f)(λ)|qdσ(λ) ≥ C

q
1C

q
2

∫
η
2h
≤|λ|≤η

h

|µh|2q|Fα,β(f)(λ)|qdσ(λ).

From η
2h
≤ |λ| ≤ η

h
we have( η

2h

)2
− ρ2 ≤ µ2 ≤

(η
h

)2
−ρ2

⇒ µ2h2 ≥
η2

4
−ρ2h2.

Take h ≤ η
3ρ

, then we have µ2h2 ≥ C3 = C3(η).
So,∫
η
2h
≤|λ|≤η

h

|1−ϕα,βµ (h)|
q|Fα,β(f)(λ)|qdσ(λ) ≥ C

q
1C

q
2C

q
3

∫
η
2h
≤|λ|≤η

h

|Fα,β(f)(λ)|qdσ(λ).

There exists then a positives constants C and K1 such that∫
η
2h
≤|λ|≤η

h

|Fα,β(f)(λ)|qdσ(λ) ≤ C
∫
R
|1 −ϕα,βµ (h)|

q|Fα,β(f)(λ)|qdσ(λ)

≤ K1φq(h) = K1φ(hq).
For all 0 < h < η

3ρ
. Then we have∫

r≤|λ|≤2r
|Fα,β(f)(λ)|qdσ(λ) ≤ K2φ(r−q), r →∞,

where K2 = K1φ(η
q2−q).

Furthermore, we obtain∫
|λ|≥r

|Fα,β(f)(λ)|qdσ(λ)

=

(∫
r≤|λ|≤2r

+

∫
2r≤|λ|≤4r

+

∫
4r≤|λ|≤8r

+ · ··

)
|Fα,β(f)(λ)|qdσ(λ)

≤ K2φ(r−q) + K2φ((2r)−q) + K2φ((4r)−q) + · · ·
≤ K2φ(r−q) + K2φ(2−q)φ(r−q) + K2φ((2−q)2)φ(r−q) + · · ·
≤ K2φ(r−q)(1 + φ(2−q) + φ((2−q)2) + · · ·).

We have φ(2−q) < 1, then∫
|λ|≥r

|Fα,β(f)(λ)|qdσ(λ) ≤ K3φ(r−q),

where K3 = K2(1 −φ(2−q))−1.
Finally, we get∫

|λ|≥r
|Fα,β(f)(λ)|qdσ(λ) = O(φ(r−q)), as r →∞.

Thus, the proof is finished.



GELFAND TRIPLE ISOMORPHISMS 109

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Departement of Mathematics, Faculty of Sciences Äın Chock, University Hassan II,

Casablanca, Morocco

∗Corresponding author