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CHARACTERIZATION OF BESOV SPACES FOR THE DUNKL OPERATOR ON THE REAL LINE

CHOKRI ABDELKEFI AND MOHAMED SIFI DEPARTMENT OFMATHEMATICS

PREPARATORYINSTITUTE OFENGINEERSTUDIES OFTUNIS

1089 MONFLEURYTUNIS, TUNISIA

chokri.abdelkefi@ipeit.rnu.tn DEPARTMENT OFMATHEMATICS

FACULTY OFSCIENCES OFTUNIS

1060 TUNIS, TUNISIA

mohamed.sifi@fst.rnu.tn

Received 07 February, 2007; accepted 28 August, 2007 Communicated by S.S. Dragomir

ABSTRACT. In this paper, we define subspaces ofLpby differences using the Dunkl translation operators that we call Besov-Dunkl spaces. We provide characterization of these spaces by the Dunkl convolution.

Key words and phrases: Dunkl operators, Dunkl transform, Dunkl translation operators, Dunkl convolution, Besov-Dunkl spaces.

2000 Mathematics Subject Classification. 46E30, 44A15, 44A35.

1. INTRODUCTION

On the real line, Dunkl operators are differential-difference operators introduced in 1989, by C. Dunkl in [5] and are denoted byΛα,whereαis a real parameter>−12. These operators, are associated with the reflection groupZ2 onR. The Dunkl kernelEα is used to define the Dunkl transformFα which was introduced by C. Dunkl in [6]. Rösler in [13] showed that the Dunkl kernel satisfies a product formula. This allows us to define the Dunkl translationτx,x∈R. As a result, we have the Dunkl convolution∗α.

There are many ways to define Besov spaces (see [4, 12, 16]). This paper deals with Besov- Dunkl spaces (see [1, 2, 8]). Letβ > 0, 1 ≤ p < +∞ and1 ≤ q ≤ +∞, the Besov-Dunkl space denoted byBDβ,αp,q is the subspace of functionsf ∈Lpα)satisfying

Z +∞

0

x(f) +τ−x(f)−2fkp,α xβ

q

dx

x <+∞ if q <+∞

The authors thank the referee for his remarks and suggestions.

048-07

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and

sup

x∈(0,+∞)

x(f) +τ−x(f)−2fkp,α

xβ <+∞ if q = +∞,

whereµαis a weighted Lebesgue measure onR(see next section).

Put

A=

φ∈ S(R) : Z +∞

0

φ(x)dµα(x) = 0

,

whereS(R)is the space of even Schwartz functions onR. Givenφ ∈ A, we shall denote by Cφ,β,αp,q the subspace of functionsf ∈Lpα)satisfying

Z +∞

0

kf∗αφtkp,α

tβ

q

dt

t <+∞ if q <+∞

and

sup

t∈(0,+∞)

kf ∗αφtkp,α

tβ <+∞ if q= +∞, whereφt(x) = t2(α+1)1 φ(xt), for allt∈(0,+∞)andx∈R.

Our objective will be to prove thatBDp,qβ,α ⊂ Cφ,β,αp,q and when1< p < +∞,0 < β <1then BDβ,αp,q =Cφ,β,αp,q .

Observe that the Besov-Dunkl spaces are independent of the specific selection ofφinAand for 1 < p < +∞, 0 < β < 1, we have BDβ,αp,q ⊂ BDe p,qβ,α, where BDe β,αp,q is the subspace of functionsf ∈Lpα)satisfying

Z +∞

0

x(f)−fkp,α xβ

q

dx

x <+∞ if q <+∞

and

sup

x∈(0,+∞)

x(f)−fkp,α

xβ <+∞ if q = +∞, (see Remark 3.7 in Section 3, below).

Analogous results have been obtained for the weighted Besov spaces (see [3]).

The contents of this paper are as follows. In Section 2, we collect some basic definitions and results about harmonic analysis associated with Dunkl operators .In Section 3, we prove the results about inclusion and coincidence between the spacesBDβ,αp,q andCφ,β,αp,q .

In what follows,crepresents a suitable positive constant which is not necessarily the same in each occurence.

2. PRELIMINARIES

On the real line, we consider the first-order differential-difference operator defined by Λα(f)(x) = df

dx(x) + 2α+ 1 x

f(x)−f(−x) 2

, f ∈ E(R), α >−1 2,

which is called the Dunkl operator. Forλ ∈ C, the Dunkl kernelEα(λ .)onRwas introduced by C. Dunkl in [5] and is given by

Eα(λx) =jα(iλx) + λx

2(α+ 1)jα+1(iλx), x∈R,

wherejα is the normalized Bessel function of the first kind of orderα (see [17]). The Dunkl kernelEα(λ .) is the unique solution on R of the initial problem for the Dunkl operator (see [5]).

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Letµαbe the weighted Lebesgue measure onRgiven by dµα(x) = |x|2α+1

2α+1Γ(α+ 1)dx.

For every 1 ≤ p ≤ +∞, we denote byLpα) the spaceLp(R, dµα)and we use k · kp,α as a shorthand fork · kLpα).

The Dunkl transformFαwhich was introduced by C. Dunkl in [6], is defined forf ∈L1α) by

Fα(f)(x) = Z

R

Eα(−ixy)f(y)dµα(y), x∈R. For allx, y, z∈R, consider

(2.1) Wα(x, y, z) = (Γ(α+ 1)2) 2α−1

πΓ(α+ 12)(1−bx,y,z+bz,x,y+bz,y,x)∆α(x, y, z), where

bx,y,z =

x2+y2−z2

2xy ifx, y ∈R\{0}, z ∈R

0 otherwise

and

α(x, y, z) =

([(|x|+|y|)2−z2][z2−(|x|−|y|)2])α−12

|xyz| if|z| ∈Sx,y

0 otherwise

where

Sx,y =h

||x| − |y||, |x|+|y|i . The kernelWα(see [13]) is even and we have

Wα(x, , y, z) =Wα(y, x, z) =Wα(−x, z, y) = Wα(−z, y,−x)

and Z

R

|Wα(x, y, z)|dµα(z)≤4.

In the sequel we consider the signed measureγx,y, onR, given by

(2.2) dγx,y(z) =





Wα(x, y, z)dµα(z) ifx, y ∈R\{0}

x(z) ify= 0

y(z) ifx= 0.

Forx, y ∈Randf a continuous function onR, the Dunkl translation operatorτxis given by τx(f)(y) =

Z

R

f(z)dγx,y(z).

According to [9], forx ∈R, τxis a continuous linear operator fromE(R)into itself and for all f ∈ E(R), we have

τx(f)(y) =τy(f)(x), τ0(f)(x) =f(x), forx, y ∈R, whereE(R)denotes the space ofC-functions onR.

According to [14, 15], the operator τx can be extended toLpα), 1 ≤ p ≤ +∞ and for f ∈Lpα)we have

(2.3) kτx(f)kp,α ≤4kfkp,α.

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Using the change of variablez = Ψ(x, y, θ) =p

x2+y2−2xycosθ, we have also (2.4) τx(f)(y) = cα

Z π 0

f(Ψ) +f(−Ψ) + x+y

Ψ (f(Ψ)−f(−Ψ))

α(θ), wheredνα(θ) = (1−cosθ) sinθdθandcα = 2Γ(α+1)πΓ(α+1

2).

From the generalized Taylor formula with integral remainder (see [11, Theorem 2, p. 349]), we have forf ∈ E(R)andx, y ∈R

x(f)−f)(y) = Z |x|

−|x|

sgn(x)

2|x|2α+1 + sgn(z) 2|z|2α+1

τzαf)(y)dµα(z).

The Dunkl convolutionf ∗αg, of two continuous functionsf andgonRwith compact support, is defined by

(f ∗α g)(x) = Z

R

τx(f)(−y)g(y)dµα(y), x∈R. The convolution∗α is associative and commutative (see [13]).

We have the following results (see [14]).

i) Assume that p, q, r ∈ [1,+∞[satisfying 1p + 1q = 1 + 1r (the Young condition). Then the map(f, g)→f ∗α gdefined onCc(R)×Cc(R), extends to a continuous map from Lpα)×Lqα)toLrα)and we have

(2.5) kf ∗α gkr,α ≤4kfkp,αkgkq,α. ii) For allf ∈L1α)andg ∈L2α), we have

(2.6) Fα(f ∗αg) = Fα(f)Fα(g)

and forf ∈L1α),g ∈Lpα)and1≤p < ∞, we get

(2.7) τt(f ∗α g) =τt(f) ∗αg =f ∗ατt(g), t∈R. 3. CHARACTERIZATION OFBESOV-DUNKLSPACES

Letβ >0,1≤p < +∞and1 ≤q ≤+∞. We say that a measurable functionf onRis in the Besov-Dunkl spaceBDp,qβ,αiff ∈Lpα)with

Z +∞

0

x(f) +τ−x(f)−2fkp,α xβ

q

dx

x <+∞ if q <+∞

and

sup

x∈(0,+∞)

x(f) +τ−x(f)−2fkp,α

xβ <+∞ if q = +∞.

Remark 3.1. Note that forf ∈Lpα)the functionR→R+,x7→ kτx(f) +τ−x(f)−2fkp,α is measurable (see [10, Lemma 1, (ii)]).

Lemma 3.2. Let0< β < 1,1≤p <+∞,1≤q≤+∞andf ∈Lpα). IfΛα(f)∈Lpα) thenf ∈ BDβ,αp,q.

Proof. Using the generalized Taylor formula, Minkowski’s inequality for integrals and (2.3), we can write

x(f) +τ−x(f)−2fkp,α ≤ kτx(f)−fkp,α+kτ−x(f)−fkp,α

≤ckΛα(f)kp,α Z x

−x

1

2|x|2α+1 + 1 2|z|2α+1

α(z),

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hence we obtain forx >0,

x(f) +τ−x(f)−2fkp,α ≤c xkΛα(f)kp,α. Then it follows that forA >0

Z +∞

0

x(f) +τ−x(f)−2fkp,α

xβ

q

dx x ≤c

Z A 0

xkΛα(f)kp,α

xβ

q

dx x +c

Z +∞

A

kfkp,α xβ

q

dx

x <+∞.

Here whenq= +∞, we make the usual modification. This completes the proof.

Example 3.1. Let0< β < 1,1≤ p <+∞and1 ≤q ≤ +∞. By Lemma 3.2, we can assert that

(1) S(R), Cc1(R)⊂ BDp,qβ,α.

(2) The functionsg, hn defined on R, by g(x) = e−|x| andhn(x) = coshxnx, n ∈ N are in BDβ,αp,q.

Now, in order to establish that for all φ ∈ A, BDp,qβ,α ⊂ Cφ,β,αp,q and for 1 < p < +∞, 0< β <1,BDβ,αp,q =Cφ,β,αp,q , we need to show some useful lemmas.

Lemma 3.3. Letφ ∈ A, 1≤p < +∞andr >0, then there exists a constantc >0such that for allf ∈Lpα)andt >0, we have

(3.1) kφtαfkp,α ≤c Z +∞

0

min x

t

2(α+1)

, t

x r

x(f) +τ−x(f)−2fkp,αdx x . Proof. Lett >0, we have

Z +∞

0

φt(x)dµα(x) = Z +∞

0

φ(x)dµα(x) = 0 and

tαf)(y) = Z

R

φt(x)τy(f)(−x)dµα(x)

= Z

R

φt(x)τy(f)(x)dµα(x), then we can write fory∈R

2(φtαf)(y) = Z

R

φt(x) [τy(f)(x) +τy(f)(−x)−2f(y)]dµα(x)

= 2 Z +∞

0

φt(x) [τx(f)(y) +τ−x(f)(y)−2f(y)]dµα(x).

Using Minkowski’s inequality for integrals, we obtain kφtαfkp,α

Z +∞

0

t(x)| kτx(f) +τ−x(f)−2fkp,αα(x)

≤c Z +∞

0

x t

2(α+1) φx

t

x(f) +τ−x(f)−2fkp,αdx (3.2) x

≤c Z +∞

0

x t

2(α+1)

x(f) +τ−x(f)−2fkp,α

dx x . (3.3)

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On the other hand, since the functionφbelongs toS(R), then forr >0there exists a constant csuch that

x t

2(α+1)+r φx

t

≤c.

By (3.2), we obtain

(3.4) kφtαfkp,α ≤c Z +∞

0

t x

r

x(f) +τ−x(f)−2fkp,αdx x .

From (3.3) and (3.4), we deduce (3.1).

Lemma 3.4. Letφ ∈ Aand1 < p < +∞, then there exists a constantc > 0such that for all f ∈Lpα)andx >0, we have

(3.5) kτx(f) +τ−x(f)−2fkp,α ≤c Z +∞

0

minn 1,x

t

okφtαfkp,αdt t . Proof. Put for0< ε < δ <+∞

fε,δ(y) = Z δ

ε

tαφtαf)(y)dt

t , y∈R. By interchanging the orders of integration and (2.7), we obtain

τx(fε,δ)(y) = Z δ

ε

τxtαφtαf)(y)dt t

= Z δ

ε

xt)∗αφtαf)(y)dt

t , y ∈R, x∈(0,+∞), so we can write forx∈(0,+∞)andy∈R,

x(fε,δ) +τ−x(fε,δ)−2fε,δ)(y) = Z δ

ε

[(τxt) +τ−xt)−2φt)∗αφtαf](y)dt t . Using Minkowski’s inequality for integrals and (2.5), we get

k(τx(fε,δ) +τ−x(fε,δ)−2fε,δkp,α ≤ Z δ

ε

k(τxt) +τ−xt)−2φt)∗αφtαfkp,αdt (3.6) t

≤c Z δ

ε

xt) +τ−xt)−2φtk1,αtαfkp,αdt t . Forx, t∈(0,+∞), we have

xt) +τ−xt)−2φtk1,α

= Z

R

Z

R

φt(z)(dγx,y(z) +dγ−x,y(z))

−2φt(y)

α(y)

= Z

R

Z

R

φz t

(dγx,y(z) +dγ−x,y(z))

−2φy t

t−2(α+1)α(y).

By (2.1) and the change of variablez0 = zt , we have Wα(x, y, z0t)t2(α+1)=Wαx

t,y t, z0

, then from (2.2), we get

x,y(z) =dγx

t,yt(z0) and dγ−x,y(z) = dγ−x

t ,yt(z0),

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hence

xt) +τ−xt)−2φtk1,α (3.7)

= Z

R

Z

R

φ(z0) dγx

t,yt(z0) +dγ−x

t ,yt(z0)

−2φy t

t−2(α+1)α(y)

= Z

R

h

τx

t(φ)y t

−x

t (φ)y t

i

t−2(α+1)−2φt(y)

α(y)

=

τx

t(φ) +τ−x

t (φ)−2φ

t

1,α

= τx

t(φ) +τ−x

t (φ)−2φ 1,α.

Sinceφ ∈ S(R), then using (2.4) and [7, Theorem 2.1] (see also [11, Theorem 2, p. 349]), we can assert that

τx

t(φ) +τ−x

t (φ)−2φ

1,α≤cx

tkφ0k1,α ≤cx t. On the other hand, by (2.3) we have

τx

t(φ) +τ−x

t (φ)−2φ 1,α

≤10kφk1,α ≤c, then we get,

(3.8)

τx

t(φ) +τ−x

t (φ)−2φ 1,α

≤cminn 1,x

t o

.

From (3.6), (3.7) and (3.8), we obtain

(3.9) kτx(fε,δ) +τ−x(fε,δ)−2fε,δkp,α ≤c Z δ

ε

minn 1,x

t

okφtαfkp,αdt t . Using (2.6), observe that

Z

R

(φ∗αφ)(x)|x|2α+1dx= 2α+1Γ(α+ 1)Fα(φ∗αφ)(0)

= 2α+1Γ(α+ 1)(Fα(φ)(0))2

= 2α+1Γ(α+ 1) Z

R

φ(z)dµα(z) 2

= 0, and sinceφ∗αφis in the Schwarz spaceS(R), we have

Z

R

|log|x|| |φ∗αφ(x)| |x|2α+1dx <+∞.

Then, by the Calderón reproducing formula related to the Dunkl operators (see [10, Theorem 3]), we have

ε→0, δ→+∞lim fε,δ =c f, inLpα).

From (2.3) and (3.9), we deduce (3.5).

Lemma 3.5. Let0 ≤ ε, r < +∞andr > β >0, then there exists constantsc1, c2 > 0such that we have

(3.10)

Z +∞

0

y z

β

min y

z ε

, z

y r

dy

y ≤c1, z ∈(0,+∞)

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and

(3.11)

Z +∞

0

y z

β

min y

z ε

, z

y r

dz

z ≤c2, y∈(0,+∞).

Proof. We can write

Z +∞

0

y z

β

min y

z ε

, z

y r

dy y

=z−(β+ε) Z z

0

yβ+ε−1dy+zr−β Z +∞

z

yβ−r−1dy≤c1, z ∈(0,+∞) and

Z +∞

0

y z

β

min y

z ε

, z

y r

dz z

=yβ−r Z y

0

z−β+r−1dz+yβ+ε Z +∞

y

z−β−ε−1dz ≤c2, y ∈(0,+∞),

which proves the results.

Theorem 3.6.

(1) Let1≤p < +∞,1≤q≤+∞andβ >0, then we have for allφ∈ A

(3.12) BDβ,αp,q ⊂ Cφ,β,αp,q .

(2) Let1< p < +∞,1≤q≤+∞and0< β <1, then we have for allφ ∈ A

(3.13) BDp,qβ,α =Cφ,β,αp,q .

Proof. Putωpα(f)(x) = kτx(f) +τ−x(f)−2fkp,α forf ∈ Lpα)andq0 = q−1q the conjugate ofqwhen1< q <+∞.

• We start with the proof of the inclusion (3.12). Suppose that 1 ≤ p < +∞, 1 ≤ q ≤ +∞,φ∈ A,r > β andf ∈ BDp,qβ,α.

Case whenq= 1. By (3.1) and Fubini’s theorem, we have Z +∞

0

kf ∗αφtkp,α tβ

dt t ≤c

Z +∞

0

Z +∞

0

min x

t

2(α+1)

, t

x r

ωpα(f)(x)t−β−1dtdx x

≤c Z +∞

0

ωαp(f)(x)

Z +∞

0

min x

t

2(α+1)

, t

x r

t−β−1dt dx

x

≤c Z +∞

0

ωαp(f)(x)

x−r Z x

0

tr−β−1dt+x2(α+1) Z +∞

x

t−β−2α−3dt dx

x

≤c Z +∞

0

ωpα(f)(x) xβ

dx

x <+∞, hencef ∈ Cφ,β,αp,1 .

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Case whenq= +∞. By (3.1), we have kφtαfkp,α ≤c

Z t 0

x t

2(α+1)

ωpα(f)(x)dx x +

Z +∞

t

t x

r

ωαp(f)(x)dx x

≤c sup

x∈(0,+∞)

ωpα(f)(x) xβ

t−2(α+1) Z t

0

x2α+1+βdx+tr Z +∞

t

x−β−r−1dx

≤ctβ sup

x∈(0,+∞)

ωαp(f)(x) xβ , then we deduce thatf ∈ Cφ,β,αp,+∞.

Case when1< q <+∞. By (3.1) again, we have fort >0 kφtαfkp,α

tβ ≤c

Z +∞

0

x t

β

min x

t

2(α+1)

, t

x

rωpα(f)(x) xβ

dx x . Put

K(x, t) =x t

βmin x

t

2(α+1), t

x

r

.

Using Hölder’s inequality and (3.10), we can write kφtαfkp,α

tβ ≤c

Z +∞

0

(K(x, t))q10

(K(x, t))1qωpα(f)(x) xβ

dx x

≤c

Z +∞

0

K(x, t)

ωpα(f)(x) xβ

q dx

x 1q

.

Then by Fubini’s theorem and (3.11), we have Z +∞

0

tαfkp,α tβ

q

dt t ≤c

Z +∞

0

ωαp(f)(x) xβ

qZ +∞

0

K(x, t)dt t

dx x

≤c Z +∞

0

ωαp(f)(x) xβ

q

dx

x <+∞, which proves the result.

• Let us now prove the equality (3.13). Assumef ∈ Cφ,β,αp,q , φ ∈ A and0 < β < 1. For 1< p <+∞and1≤q ≤+∞, we have to show only thatf ∈ BDp,qβ,α.

Case whenq= 1. By (3.5) and Fubini’s theorem, we have Z +∞

0

ωpα(f)(x) xβ

dx x ≤c

Z +∞

0

Z +∞

0

minn 1,x

t

okφtαfkp,αx−β−1dt t dx

≤c Z +∞

0

tαfkp,α

Z +∞

0

minn 1,x

t o

x−β−1dx dt

t

≤c Z +∞

0

tαfkp,α 1

t Z t

0

x−βdx+ Z +∞

t

x−β−1dx dt

t

≤c Z +∞

0

tαfkp,α

tβ

dt

t <+∞, then we obtain the result.

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Case whenq= +∞. By (3.5), we get ωpα(f)(x)≤c

Z x 0

tαfkp,αdt t +

Z +∞

x

x

tkφtαfkp,αdt t

≤c sup

t∈(0,+∞)

tαfkp,α tβ

Z x 0

tβ−1dt+x Z +∞

x

tβ−2dt

≤cxβ sup

t∈(0,+∞)

tαfkp,α tβ , so, we deduce thatf ∈ BDp,+∞β,α .

Case when1< q <+∞. By (3.5) again, we have forx >0 ωpα(f)(x)

xβ ≤c Z +∞

0

t x

β

minn 1,x

t

okφtαfkp,α tβ

dt t . Put

R(x, t) = t

x

βminn 1,x

t o

. Using Hölder’s inequality and (3.10), we can write

ωpα(f)(x) xβ ≤c

Z +∞

0

(R(x, t))q10

(R(x, t))1qtαfkp,α tβ

dt t

≤c

Z +∞

0

R(x, t)

tαfkp,α tβ

q

dt t

1q , then by Fubini’s theorem and (3.11), we have

Z +∞

0

ωαp(f)(x) xβ

q dx

x ≤c Z +∞

0

tαfkp,α

tβ

qZ +∞

0

R(x, t)dx x

dt t

≤c Z +∞

0

tαfkp,α tβ

q

dt

t <+∞,

thus the result is established.

Remark 3.7. By proceeding in the same manner as in Lemma 3.4 and (2) of Theorem 3.6, we can assert that for1< p < +∞and0< β < 1, we haveCφ,β,αp,q ⊂BDe p,qβ,α , hence from (3.13) we conclude thatBDβ,αp,q ⊂BDe β,αp,q.

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