International Journal of Analysis and Applications
ISSN 2291-8639
Volume 8, Number 2 (2015), 123-129
http://www.etamaths.com

(δ,γ)-JACOBI-DUNKL LIPSCHITZ FUNCTIONS IN THE SPACE

L2(R,Aα,β(x)dx)

R. DAHER, S. EL OUADIH∗

Abstract. Using a generalized Jacobi-Dunkl translation, we obtain an ana-

log of Theorem 5.2 in Younis paper [7] for the Jacobi-Dunkl transform for

functions satisfying the (δ,γ)-Jacobi-Dunkl Lipschitz condition in the space

L2(R,Aα,β(x)dx),α ≥ β ≥ −12 ,α 6=
−1
2
.

1. Introduction and Preliminaries

Younis ([7], Theorem 5.2) characterized the set of functions in L2(R) satisfying
the Dini-Lipschitz condition by means of an asymptotic estimate growth of the
norm of their Fourier transforms, namely we have the following statement.
Theorem 1.1. [7] Let f ∈ L2(R). Then the following are equivalents:
(a) ‖f(x + h) −f(x)‖ = O

(
hη

(log 1
h

)γ

)
, as h → 0, 0 < η < 1,γ > 0

(b)

∫
|λ|≥r

|f̂(λ)|2dλ = O
(

r−2η

(log r)2γ

)
, as r →∞,

where f̂ stand for the Fourier transform of f.
In this paper, we obtain an analog of Theorem 1.1 for the Jacobi-Dunkl transform
on the real line. For this purpose, we use a generalized Jacobi-Dunkl translation
operator.
In this section , we recapitulate from [1,2,3,5] 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.

123



124 DAHER AND OUADIH

F is the Gausse hypergeometric function (see [1,6]).

ψ
α,β
λ is the unique C

∞-solution on R of the differential-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])

ψ
α,β
λ (x) = ϕ

α,β
µ (x) + i

λ

4(α + 1)
sinh(2x)ϕα+1,β+1µ (x), x ∈ R.

Denote L2α,β(R) = L
2
α,β(R,Aα,β(x)dx) the space of measurable functions f on R

such that

‖f‖L2
α,β

(R) =

(∫
R
|f(x)|2Aα,β(x)dx

)1/2
< +∞.

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

, we define the Jacobi-Dunkl transform of a function f ∈ L2α,β(R) by

Fα,βf(λ) =
∫
R
f(x)ψ

α,β
λ (x)Aα,β(x)dx, λ ∈ R,

and the inversion formula

f(x) =

∫
R
Fα,βf(λ)ψ

α,β
−λ (x)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‖ := ‖f‖L2
α,β

(R) = ‖Fα,β(f)‖L2(R,dσ(λ)).(1)



STRONG METRIZABILITY FOR CLOSED OPERATORS 125

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); λ,h ∈ R.(2)

For α ≥ −1
2

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

jα(x) = Γ(α + 1)

∞∑
n=0

(−1)n(x
2
)2n

n!Γ(n + α + 1)
.

Moreover, we see that

lim
x→0

jα(x) − 1
x2

6= 0,

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

|x| ≤ ε ⇒|jα(x) − 1| ≥ C1|x|2(3)

Lemma 1.2. (See[8],Lemma 3.1,Lemma 3.2) The following inequalities are valid
for Jacobi functions ϕα,βµ (x)

(c) |ϕα,βµ (x)| ≤ 1,
(d) |1 −ϕα,βµ (x)| ≤ x2(µ2 + ρ2).



126 DAHER AND OUADIH

Lemma 1.3. (See[4],Lemma 9) 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(x)| ≥ C2|1 − jα(µx)|.

2. Main Results

In this section we give the main results of this paper. We need first to define
(η,γ)-Jacobi-Dunkl Lipschitz class.

Definition 2.1. Let 0 < η < 1 and γ > 0. A function f ∈ L2α,β(R) is said to
be in the (η,γ)-Jacobi-Dunkl Lipschitz class, denoted by Lip(η,γ, 2), if

‖Thf(x) + T−hf(x) − 2f(x)‖ = O
(

hη

(log 1
h

)γ

)
, as h → 0.

Lemma 2.2. For f ∈ L2α,β(R), then

‖Thf(x) + T−hf(x) − 2f(x)‖2 = 4
∫
R
|ϕα,βµ (h) − 1|

2|Fα,βf(λ)|2dσ(λ).

Proof. We us formula (2), we conclude that

Fα,β(Thf + T−hf − 2f)(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(Thf + T−hf − 2f)(λ) = 2(ϕα,βµ (h) − 1)Fα,β(f)(λ).

By Parseval’s identity (formula (1)), we have the result.

Theorem 2.3. Let 0 < η < 1 , γ > 0 and f ∈ L2α,β(R). Then the following
conditions are equivalents
(i) f ∈ Lip(η,γ, 2)

(ii)

∫
|λ|≥r

|Fα,βf(λ)|2dσ(λ) = O
(

r−2η

(log r)2γ

)
, as r →∞.

Proof. (i) ⇒ (ii). Assume that f ∈ Lip(η,γ, 2) . Then we have

‖Thf(x) + T−hf(x) − 2f(x)‖ = O
(

hη

(log 1
h

)γ

)
, as h → 0

From Lemma 2.2, we have

‖Thf(x) + T−hf(x) − 2f(x)‖2 = 4
∫
R
|ϕα,βµ (h) − 1|

2|Fα,βf(λ)|2dσ(λ).

By (3) and Lemma 1.3, we get∫
ε
2h
≤|λ|≤ε

h

|1−ϕα,βµ (h)|
2|Fα,β(f)(λ)|2dσ(λ) ≥ C21C

2
2

∫
ε
2h
≤|λ|≤ε

h

|µh|4|Fα,β(f)(λ)|2dσ(λ).



STRONG METRIZABILITY FOR CLOSED OPERATORS 127

From ε
2h
≤ |λ| ≤ ε

h
we have( ε

2h

)2
− ρ2 ≤ µ2 ≤

(ε
h

)2
−ρ2

⇒ µ2h2 ≥
ε2

4
−ρ2h2.

Take h ≤ ε
3ρ

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

h

|1−ϕα,βµ (h)|
2|Fα,β(f)(λ)|2dσ(λ) ≥ C21C

2
2C

2
3

∫
ε
2h
≤|λ|≤ε

h

|Fα,β(f)(λ)|2dσ(λ).

There exists then a positive constant C4 such that∫
ε
2h
≤|λ|≤ε

h

|Fα,β(f)(λ)|2dσ(λ) ≤ C4
∫
R
|1 −ϕα,βµ (h)|

2|Fα,β(f)(λ)|2dσ(λ)

= O

(
h2η

(log 1
h

)2γ

)
.

We obtain ∫
r≤|λ|≤2r

|Fα,β(f)(λ)|2dσ(λ) ≤ C
r−2η

(log r)2γ
,

where C is a positive constant. Now,∫
|λ|≥r

|Fα,β(f)(λ)|2dσ(λ) =
∞∑
i=0

∫
2ir≤|λ|≤2i+1r

|Fα,β(f)(λ)|2dσ(λ)

≤ C
(

r−2η

(log r)2γ
+

(2r)−2η

(log 2r)2γ
+

(4r)−2η

(log 4r)2γ
+ · · ·

)
≤ C

r−2η

(log r)2γ
(
1 + 2−2η + (2−2η)2 + (2−2η)3 + · · ·

)
≤ Kη

r−2η

(log r)2γ
,

where Kη = C(1 − 2−2η)−1 since 2−2η < 1.
Consequently∫

|λ|≥r
|Fα,βf(λ)|2dσ(λ) = O

(
r−2η

(log r)2γ

)
, as r →∞.

(ii) ⇒ (i). Suppose now that∫
|λ|≥r

|Fα,βf(λ)|2dλ = O
(

r−2η

(log r)2γ

)
, as r →∞,

and write ∫
R
|1 −ϕα,βµ (h)|

2|Fα,β(f)(λ)|2dσ(λ) = I1 + I2

where

I1 =

∫
|λ|< 1

h

|1 −ϕα,βµ (h)|
2|Fα,β(f)(λ)|2dσ(λ),

I2 =

∫
|λ|≥1

h

|1 −ϕα,βµ (h)|
2|Fα,β(f)(λ)|2dσ(λ).



128 DAHER AND OUADIH

Firstly, we use the formula |ϕα,βµ (h)| ≤ 1 and

I2 ≤ 4
∫
|λ|≥1

h

|Fα,β(f)(λ)|2dσ(λ) = O
(

h2η

(log 1
h

)2γ

)
.(4)

Set

φ(λ) =

∫ ∞
λ

|Fα,β(f)(x)|2dσ(x).

An integration by parts gives∫ x
0

λ2|Fα,β(f)(λ)|2dσ(λ) =
∫ x

0

−λ2φ′(λ)dλ

= −x2φ(x) + 2
∫ x

0

λφ(λ)dλ

≤ 2
∫ x

0

λ1−2δ(log λ)−2γdλ

= O(x2−2δ(log x)−2γ).

We use the formula (d) of Lemma 1.2

I1 ≤
∫
|λ|< 1

h

|1 −ϕα,βµ (h)||Fα,β(f)(λ)|
2dσ(λ)

≤
∫
|λ|< 1

h

(µ2 + ρ2)h2|Fα,β(f)(λ)|2dσ(λ)

≤ h2
∫
|λ|< 1

h

λ2|Fα,β(f)(λ)|2dσ(λ)

= O

(
h2h−2+2η

(
log

1

h

)−2γ)
..

Hence,

I1 = O

(
h2η

(log 1
h

)2γ

)
.(5)

Finally, we conclude from (4) and (5) that∫
R
|1 −ϕα,βµ (h)|

2|Fα,β(f)(λ)|2dσ(λ) = O
(

h2η

(log 1
h

)2γ

)
.

And this ends the proof.

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STRONG METRIZABILITY FOR CLOSED OPERATORS 129

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

Casablanca, Morocco

∗Corresponding author