An Analog of Titchmarsh’s Theorem of Jacobi Transform
Radouan Daher
Faculty of Sciences Ain Ckock, Casablanca, Morocco ra daher@yahoo.fr
Mohamed El Hamma
Faculty of Sciences Ain Ckock, Casablanca, Morocco m elhamma@yahoo.fr
Abstract
In this paper, we prove an analog of Titchmarsh’s theorem for the Jacobi transform for functions satisfying the Jacobi-Lipschitz condition in L 2 ( R + , Δ α,β ( t ) dt ), using a generalized translation operator.
Mathematics Subject Classification: 33C45; 43A90; 42C15
Keywords: Jacobi operator, Jacobi transform, generalized translation op- erator
1 Introduction and notations
In the present paper, we study a the set of functions in L 2 ( R + , Δ α,β ( t ) dt ) satis- fying the Cauchy Lipschitz condition, we use translation operator τ h associated of Jacobi operators, the operator has played a decisive role in the development of Euclidean harmonic analysis, as evidenced, for example, by landmark paper [3] by H¨ ormander.
Titchmarsh’s [7, Theorem 85] characterized the set of functions L 2 ( R ) satisfy- ing the Cauchy Lipschitz condition by means of asymptotic estimate growth of the norm of their Fourier transform, namely we have
Theorem 1.1 [7] Let α ∈ (0 , 1) and assume that f ∈ L 2 ( R ). Then the following are eguivalents:
(1) f ( t + h ) − f ( t ) L
2(
Ê= O ( h α ) as h −→ 0 (2)
|λ|≥r | f ( λ ) | 2 dλ = O ( r −2α ) r −→ ∞
where f stands for the Fourier transform of f
The main aim of this paper is to establish of Theorem 1.1 in Jacobi opera- tors setting by means the generalized translation operator τ h defined in Section 2. We point out that similar results have been established in the context of noncompact rank 1 Riemannian symmetric spaces [6].
In this section, we briefly collect the pertinnent definitions and facts relevant for Jacobi analysis, can be found in [4].
Let ( a ) 0 = 1 and ( a ) k = a ( a + 1) ... ( a + k − 1). The hypergeometric function
F ( a, b, c, z ) =
∞
k=0
( a ) k ( b ) k
( c ) k k ! z k , |z| < 1
the function z −→ F ( a, b, c, z ) is the unique solution of the differential equation z (1 − z ) u ”( z ) + ( c − ( a + b + 1) z ) u ( z ) − abu ( z ) = 0
which is regular in 0 and equals 1 there.
The Jacobi function with parametres ( α, β ) is defined by the formula ϕ (α,β) λ ( t ) = F ( 1
2 ( ρ − iλ ) , 1
2 ( ρ + iλ ) , α + 1 , − sinh 2 t )
For α ≥ −1 2 , α > β ≥ −1 2 , ρ = α + β + 1, the system {ϕ (α,β) λ } λ≥0 is a continuous orthonormal system in R + , with respect to the weight
Δ α,β ( t ) = (2 sinh t ) 2α+1 (2 cosh t ) 2β+1 The Jacobi operator
L = L α,β = d 2
dt 2 + ((2 α + 1) coth t + (2 β + 1) tanh t ) d dt
By means of which the Jacobi function ϕ (α,β) λ may alternatively be charac- terized as the unique solution to
L ϕ + ( λ 2 + ρ 2 ) ϕ = 0 (1)
on R + satisfying ϕ (α,β) λ (0) = 1, ϕ
λ (α,β) (0) = 0, and λ −→ ϕ (α,β) λ ( t ) is analytic for all t ≥ 0.
We adhere to the conventions and normalization used [2], the c-function c ( λ ) = 2 ρ Γ( iλ )Γ( 1 2 (1 + iλ ))
Γ( 1 2 ( ρ + iλ ))Γ( 1 2 ( ρ + iλ ) − β ) where α > 0 , α > β ≥ −1 2 .
The Jacobi transform of a function f ∈ L 2 ( R + , Δ α,β ( t ) dt ) is defined by
f ( λ ) =
∞
0 f ( t ) ϕ (α,β) λ ( t )Δ α,β ( t ) dt (2) and the inversion formula is statement thet (cf.[4])
f ( t ) = 1 2 π
∞
0
f ( λ ) ϕ (α,β) λ ( t ) dμ ( λ ) (3) where dμ ( λ ) = |c ( λ ) | −2 dλ .
The Jacobi transform a unitary isomorphism from L 2 ( R + , Δ α,β ( t ) dt ) onto L 2 ( R + , 2π 1 dμ ( λ )), i.e
f = f L
2(
Ê+,Δ
α,β(t)dt) = f L
2(
Ê+,
12π
dμ(λ)) (4)
The limiting case α = β = −1 2 is the Fourier-cosine transform, which we will not study.
We have
L f ( λ ) = − ( λ 2 + ρ 2 ) f ( λ ) (5) The generalized translation operator was defined by Flensted-Jensen and Koornwinder [2, Formula (5.1)] given by
τ y f ( x ) =
∞
0 f ( z ) K ( x, y, z )Δ α,β ( z ) dz with kernel
K ( x, y, z ) = 2 −2ρ Γ( α + 1)(cosh x cosh cosh z ) −α−β−1
Γ( 1 2 )Γ( α + 1 2 )(sinh x sinh y sinh z ) 2α (1 − B 2 ) α−
12× F (( α + β, α − β, α + 1 2 , 1
2 (1 − B )) for |x − y| < z < x + y and K ( x, y, z ) = 0 elsewhere and
B = cosh 2 x + cosh 2 y + cosh 2 z − 1 2 cosh x cosh y cosh z in [1], we have
τ h f ( λ ) = ϕ (α,β) λ ( h ) f ( λ ) (6)
For α ≥ −1 2 , we introduce the Bessel normalized function of the first kind
j α defined by
j α ( z ) = Γ( α + 1)
∞
n=0
( − 1) n ( z 2 ) 2n
n !Γ( n + α + 1) , z ∈ C (7) Moreover, from (7) we see that
z−→0 lim
j α ( z ) − 1
z 2 = 0 (8)
by consequence, there exist c > 0 and η > 0 satisfying
|z| ≤ η = ⇒ |j α ( z ) − 1 | ≥ c|z| 2 (9) Lemma 1.2 The following inequalities are valid for Jacobi functions ϕ (α,β) λ ( t )
(1) |ϕ (α,β) λ ( t ) | ≤ 1 (10)
(2) 1 − ϕ (α,β) λ ( t ) ≤ t 2 ( λ 2 + ρ 2 ) (11) Proof: analog (see [lemmas 3.1-3.2, 3])
Lemma 1.3 Let α > −1 2 , α ≥ β ≥ −1 2 . Then for |η| ≤ ρ , there existes a positive constant c 1 such that
| 1 − ϕ (α,β) μ+iη ( t ) | ≥ c 1 | 1 − j α ( μt ) | (12) Proof: (see[Lemma 9, 1])
2 An analog of Titchmarsh’s Theorem
In this section we give the main resultat of this paper, we need first to define Jacobi-Lipschitz class.
Definition 2.1 Let δ ∈ (0 , 1). A function f ∈ L 2 ( R + , Δ α,β ( t ) dt ) is said to be in the Jacobi-Lipschitz class, denote by Lip ( δ, 2), if
τ h f ( x ) − f ( x ) = O ( h δ ) , as h −→ 0
Theorem 2.2 Let f ∈ L 2 ( R + , Δ α,β ( t ) dt ). Then the following are equiva- lents
1. f ∈ Lip ( δ, 2).
2. ∞
r | f ( λ ) | 2 dμ ( λ ) = O ( r −2δ ) as r −→ + ∞ .
Proof: 1 = ⇒ 2:Assume that f ∈ Lip ( δ, 2). Then we have τ h f ( x ) − f ( x ) = O ( h δ ) , as h −→ 0 formulas (4) and (6) gives
τ h f ( x ) − f ( x ) 2 =
∞
0 | 1 − ϕ (α,β) λ ( h ) | 2 | f ( λ ) | 2 dμ ( λ ) Since (12) and λ ∈ R + , we have
ηh 2hη
| 1 − ϕ (α,β) λ ( h ) | 2 | f ( λ ) | 2 dμ ( λ ) ≥ c 1
ηh 2hη
| 1 − j α ( λh ) | 2 | f ( λ ) | 2 dμ ( λ ) From (9), we obtain
ηh 2hη
| 1 − ϕ (α,β) λ ( h ) | 2 | f ( λ ) | 2 dμ ( λ ) ≥ c 1 c 2 η 4 16
ηh 2hη
| f ( λ ) | 2 dμ ( λ ) There exists then a positive constant K such that
ηh 2hη
| f ( λ ) | 2 dμ ( λ ) ≤ K
∞
0 | 1 − ϕ (α,β) λ ( h ) | 2 | f ( λ ) | 2 dμ ( λ )
≤ Kh 2δ For all h > 0. Then
2r
r
| f ( λ ) | 2 dμ ( λ ) ≤ Kr −2δ for all r > 0. Furthermore, we have
∞
r
| f ( λ ) | 2 dμ ( λ ) =
∞
i=0
2
i+1r
2
ir
| f ( λ ) | 2 dμ ( λ )
≤ K
∞
i=0
(2 i r ) −2δ
≤ Kr −2δ This prove
∞
r
| f ( λ ) | 2 dμ ( λ ) = O ( r −2δ ) as r −→ + ∞
2 = ⇒ 1: Suppose now that
∞
r
| f ( λ ) | 2 dμ ( λ ) = O ( r −2δ ) as r −→ + ∞ we write
τ h f ( x ) − f ( x ) 2 = I 1 + I 2 where
I 1 =
1h
0 | 1 − ϕ (α,β) λ ( h ) | 2 | f ( λ ) | 2 dμ ( λ ) and
I 2 =
∞
h1
| 1 − ϕ (α,β) λ ( h ) | 2 | f ( λ ) | 2 dμ ( λ ) Estimate the summands I 1 and I 2 .
we have from (10)
I 2 ≤ 4
∞
1h
| f ( λ ) | 2 dμ ( λ ) = O ( h 2δ ) To estimate I 1 , we use the inequality (10).
I 1 =
1h
0 | 1 − ϕ (α,β) λ ( h ) | 2 | f ( λ ) | 2 dμ ( λ ) ≤ 2
1h
0 | 1 − ϕ (α,β) λ ( h ) || f ( λ ) | 2 dμ ( λ ) From the inequality (11), we have
I 1 ≤ 2 h 2
1h
0
( λ 2 + ρ 2 ) | f ( λ ) | 2 dμ ( λ )
= 2 ρ 2 h 2
1h
0 | f ( λ ) | 2 dμ ( λ ) + 2 h 2
1h
0 λ 2 | f ( λ ) | 2 dμ ( λ ) Note that
2 ρ 2 h 2
1h
0 | f ( λ ) | 2 dμ ( λ ) ≤ 2 ρ 2 h 2
∞
0 | f ( λ ) | 2 dμ ( λ )
= 2 ρ 2 f 2
= O ( h 2δ ) since 2 δ < 2
We put
ψ ( r ) =
∞
r
| f ( λ ) | 2 dμ ( λ ) Integrating by parts, we obtain
2 h 2
1h
0 λ 2 | f ( λ ) | 2 dμ ( λ ) = 2 h 2
1h
0
( −r 2 ψ
( r ) dr
= 2 h 2 ( − 1 h 2 ψ ( 1
h ) + 2
1h
0 rψ ( r ) dr )
= − 2 ψ ( 1
h ) + 4 h 2
1h
0 rψ ( r ) dr
≤ 4 Ch 2
1h