An analog of Titchmarsh’s theorem for the generalized Dunkl transform

DOI 10.1007/s11868-015-0130-z

An analog of Titchmarsh’s theorem for the generalized Dunkl transform

Radouan Daher¹ · Mohamed El Hamma¹

Received: 25 April 2015 / Accepted: 14 August 2015

© Springer Basel 2015

Abstract Using a generalized dual translation operator, we obtain an analog of Titch- marsh’s theorem for the generalized Dunkl transform for functions satisfying the Q-Dunkl Lipschitz condition in the space L²_Q.

Keywords Generalized Dunkl transform ·Differential-difference operator · Generalized dual translation operator

Mathematics Subject Classification 47B48

1 Introduction and preliminaries

Titchmarsh’s [5, Theorem 85] characterized the set of functions in L²(R)satisfying the Cauchy Lipschitz condition by means of an asymptotic estimate growth of the norm of their Fourier transform, namely we have

Theorem 1.1 [5] Letα∈(0,1)and assume that f ∈L²(R). Then the following are equivalent:

Dedicated to Professor François Rouvière for his 70’s birthday.

B

1 Department of Mathematics, Faculty of Sciences Aïn Chock, University of Hassan II, Casablanca, Morocco

1. f(.+h)− f(.)L²(R)=O(h^α)as h−→0 2.

|λ|≥r|f(λ)|²dλ=O(r^−2α)as r −→ ∞ where f stands for the Fourier transform of f .

In this paper, we prove an analog of this theorem for the generalized Dunkl transform in the space L²_Q.

Consider the first-order singular differential-difference operator on the real line Df(x)= d f

d x +

α+1 2

f(x)− f(−x)

x +q(x)f(x),

whereα >−¹₂ andq is aC^∞real-valued odd function onR. Forq =0, we obtain the classical Dunkl operator

D_αf(x)= d f d x +

α+1

2

f(x)− f(−x) x Put

Q(x)=exp

− x

0

q(t)dt

, (1)

withQis a even function.

We denote by:

1. L²_α(R)the class of measurable functions f onRwith the norm f2,α

R|f(x)|²|x|^2α+1d x 1/2

<+∞

2. L²_Q =L²_Q(R)the class of measurable functions f onRfor which f2,Q = Q f2,α <+∞,

whereQis given by formula (1).

The following statement is proved in [3]

Lemma 1.2 1. For eachλ∈C, the differential-difference equation Du =iλu, u(0)=1

admits a unique C^∞solution onR, denoted byψλgiven by ψ_λ(x)=Q(x)e_α(iλx), where e_αdenotes the Dunkl kernel onRdefined by

e_α(z)= j_α(i z)+ z

2α+2j_α+1(i z), z∈C

j_αbeing the normalized spherical Bessel function given by

j_α(z)=(α+1) ∞ n=0

(−1)ⁿ k!(n+p+1)

z 2

2n,z∈C,

2. For all x∈R,λ∈Cand n=0,1,2, . . . ,we have ∂ⁿ

∂λⁿψλ(x)

≤ Q(x)|x|ⁿe^{|I mλ||x} Lemma 1.3 For x∈Rthe following inequalities are fulfilled.

1. |j_α(x)| ≤1,

2. |1− j_α(x)| ≥c with|x| ≥1, where c >0is a certain constant which depends only onα.

Proof (Analog of Lemma 2.9 in [5]).

In the terms of j_α(x), we have (see [1])

1−j_α(x)=O(1), x≥1, (2)

1−j_α(x)=O(x²),0≤x≤1. (3)

Definition 1.4 The generalized Dunkl transform for a function f ∈L¹_Qis defined by FD(f)(λ)=

R f(x)ψ_−λ(x)|x|^2α+1d x.

From [2], we have two following theorems

Theorem 1.5 Let f ∈L¹_Qsuch thatFD(f)∈L¹_Q. Then f(x)(Q(x))²=m_α

RFD(f)(λ)ψ_λ(x)|λ|^2α+1dλ, where

m_α = 1

2^2α+2((α+1))²

Theorem 1.6 1. For every f ∈L²_Qwe have the Plancherel formula

R|f(x)|²(Q(x))²|x|^2α+1d x=m_α

R|FD(f)(λ)|²|λ|^2α+1dλ

2. The generalized Dunkl transformFDextends uniquely to an isometric isomorphism fromL²_QontoL_α².

From [2], we define the generalized dual translation operators are given by Thf(x)= Q(h)

Q(x)τ_α^−h(Q f)(x), where the Dunkl translation operators

τ_α^−hf(x)=

R f(z)dμ^α_h,x(z),

andμ^α_h,x is a finite signed measure onR, of total mass 1, with support [−|h| − |x|,||h| − |x||]∪[||h| − |x||,|h| + |x|]

and such thatμ^α_h,x ≤2.

By [2], we have the formula

FD(Thf)(λ)=ψ−λ(h)FD(f)(λ), f ∈L²_Q (4)

2 Main result

In this section we give the main result of this paper. We need first to define theQ-Dunkl Lipschitz class

Definition 2.1 Let β ∈ (0,1). A function f ∈ L²_Q is said to be in the Q-Dunkl Lipschitz class, denoted byLi p(Q,2, β), if

Thf(.)+T₋hf(.)−2Q(h)f(.)2,Q =O(Q(h)h^β)as h→0.

Theorem 2.2 Let f ∈L²_Q. Then the following are equivalents 1. f ∈ Li p(β,2,n),

2.

|λ|≥r|FD(f)(λ)|²|λ|^2α+1dλ=O(r^−2β)as r→ +∞.

Proof 1)⇒2)Assume that f ∈ Li p(Q,2, β). Then

Thf(.)+T₋hf(.)−Q(h)f(.)2,α,n =O(Q(h)h^β)as h →0.

From formula (4), we have

FD(Thf +T₋hf −2Q(h)f) (λ)=(ψ−λ(h)+ψ−λ(−h)−2Q(h))FD(f)(λ)

=(Q(h)e_α(−iλh)+Q(−h)e_α(iλh)−2Q(h))FD(f)(λ)

=((Q(h)(e_α(−iλh)+e_α(iλh))−2Q(h))FD(f)(λ)

=2Q(h)(j_α(λh)−1)FD(f)(λ)

Plancherel identity gives

Thf(.)+T₋hf(.)−2Q(h)f(.)²_2,Q=

R(2Q(h))²|1−j_α(λh)|²|FD(f)(λ)|²|λ|^2α+1dλ If|λ| ∈ [¹_h,_h²], then|λh| ≥1 and (2) of Lemma1.3implies that

1≤ 1

c²|1− j_α(λh)|² Then

1 h≤|λ|≤h²

|FD(f)(λ)|²|λ|^2α+1dλ≤ 1 c²

1 h≤|λ|≤²h

|1− j_α(λh)|²|FD(f)(λ)|²|λ|^2α+1dλ

≤ 1 c²

R|1−j_α(λh)|²|FD(f)(λ)|²|λ|^2α+1dλ

≤ 1 c²

1

4m_α(Q(h))²T_hf(.)+T_−hf(.)−2Q(h)f(.)²_2,Q

= O(h^2β) We obtain

r≤|λ|≤2r

|FD(f)(λ)|²|λ|^2α+1dλ=O(r^−2β)as r → +∞

Thus these existsC>0 such that

r≤|λ|≤2r

|FD(f)(λ)|²|λ|^2α+1dλ≤Cr^−2β

So that

|λ|≥r

|FD(f)(λ)|²|λ|^2α+1dλ=^+∞

j=0

2^jr≤|λ|≤2^j+1r

|FD(f)(λ)|²|λ|^2α+1dλ

≤C +∞

j=0

(2^jr)^−2β

≤Cr^−2β This proves that

|λ|≥r

|FD(f)(λ)|²|λ|^2α+1dλ=O(r^−2β)as r → +∞.

2)⇒1)Suppose now that

|λ|≥r|FD(f)(λ)|²|λ|^2α+1dλ=O(r^−2β)as r → +∞.

We write

R|1−j_α(λh)|²|FD(f)(λ)|²|λ|^2α+1dλ=I1+I2, where

I1=

|λ|<¹h

|1− j_α(λh)|²|FD(f)(λ)|²|λ|^2α+1dλ

and

I2=

|λ|≥¹h

|1−j_α(λh)|²|FD(f)(λ)|²|λ|^2α+1dλ

Estimate the summands I1and I2.

From inequality (1) of Lemma1.3, we have I2=

|λ|≥¹_h |1−j_α(λh)|²|FD(f)(λ)|²|λ|^2α+1dλ

≤4

|λ|≥¹_h |FD(f)(λ)|²|λ|^2α+1dλ

=O(h^2β).

Then

4(Q(h))²I2=O((Q(h))²h^2β).

To estimate I1, we use the inequality (3). Set φ(x)=

_+∞

x

|FD(f)(λ)|²|λ|^2α+1dλ.

An integration by parts, we obtain _x

0 |FD(f)(λ)|²|λ|^2α+1dλ= _x

0 −λ²φ(λ)dλ

= −x²φ(x)+2 _x

0

λφ(λ)dλ

≤2 _x

0 O(λ^1−2β)dλ

≤ O(x^2−β)

Then

R|1−j_α(λh)|²|FD(f)(λ)|²|λ|^2α+1dλ=O

h²

|λ|<¹_hλ²|FD(f)(λ)|²|λ|^2α+1dλ

+O(h^2β)

=O(h²h^−2+2β)+O(h^2β)

=O(h^2β)+O(h^2β)

=O(h^2β)

Therefore m_α

R(2Q(h))²|1−j_α(λh)|²|FD(f)(λ)|²|λ|^2α+1dλ=O((Q(h))²h^2β), and this ends the proof.

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