Volume 89 No. 3 2013, 335-341
ISSN: 1311-8080 (printed version); ISSN: 1314-3395 (on-line version) url:http://www.ijpam.eu
doi:http://dx.doi.org/10.12732/ijpam.v89i3.4
A P
ijpam.eu
TITCHMARSH’S THEOREM FOR THE JACOBI TRANSFORM IN THE SPACE L
2(α,β)(
R+)
R. Daher
1, H. Lahlali
2, M. El Hamma
3§1,2,3
Department of Mathematics Faculty of Sciences A¨ın Chock
University of Hassan II Casablanca, MOROCCO
Abstract: Using a generalized translation operator, we obtain a generalization of Titchmarsh’s theorem of the Jacobi transform for functions satisfying the ψ- Jacobi Lipschitz condition.
AMS Subject Classification: 33C45, 43A90, 42C15, 44A20
Key Words: Jacobi operator, Jacobi transform, generalized translation op- erator
1. Introduction and Preliminaries
In [3], we proved an analog of Titchmarsh’s theorem for the Jacobi transform in the space L
2(α,β)(
R+) of functions satisfying the Jacobi-Lipschitz condition. In this paper we prove the generalization of this theorem of functions satisfying the ψ-Jacobi Lipschitz condition. For this purpose, we use a generalized translation operator.
Now, we collect some basic facts on the Jacobi transform, and more details about the Jacobi transform can be found in [1] and [5].
Received: April 12, 2013
c 2013 Academic Publications, Ltd.url: www.acadpubl.eu
§Correspondence author
The Jacobi function ϕ
λ(t) = ϕ
(α,β)λ(t) of order (α, β) (α ≥ −
12, α >
β ≥ −
12) is the unique C
∞function on
Rwith equals 1 at 0 and satisfies the differential equation
(D
α,β+ λ
2+ ρ
2)ϕ
λ(t) = 0, where ρ = α + β + 1 and
D
α,β= d
2dt
2+ ((2α + 1) coth t + (2β + 1) tanh t) d dt .
Lemma 1.1. The following inequalities are valid for a Jacobi function ϕ
λ(t) (λ, t ∈
R+).
1. |ϕ
λ(t)| ≤ 1,
2. 1 − ϕ
λ(t) ≤ t
2(λ
2+ ρ
2),
3. there is a constant c > 0 such that
1 − ϕ
λ(t) ≥ c for λt ≥ 1.
Proof. See [6, Lemmas 3.1-3.2]
Consider the Hilbert space L
2(α,β)(
R+) = L
2(
R+, ∆
(α,β)(t)dt) with the norm kf k
2,(α,β)=
Z
∞ 0|f (x)|
2∆
(α,β)(x)dx
1/2, where
∆
(α,β)(t) = (2 sinh t)
2α+1(2 cosh t)
2β+1. The c-function is defined by (see [4])
c(λ) = 2
ρΓ(iλ)Γ(
12(1 + iλ)) Γ(
12(ρ + iλ))Γ(
12(ρ + iλ) − β)) , where α ≥ −
12and α > β ≥ −
12.
The Jacobi transform of a function f ∈ L
2(α,β)(
R+) is defined by f b (λ) =
Z
∞ 0f(t)ϕ
λ(t)∆
(α,β)(t)dt
The inverse formula [5] is defined by f (t) = 1
2π Z
∞0
f b (λ)ϕ
λ(t)dµ(λ), where dµ(λ) = |c(λ)|
−2dλ.
The Jacobi transform a unitary isomorphism from L
2(α,β)(
R+) onto L
2(
R+,
1
2π
dµ(λ)), i.e.
kf k
2,(α,β)= k f b k
L2(R+,12πdµ(λ))
(1)
Recall from [4] the generalized translation operator τ
hof a suitable function f on
R+, is defined by
τ
hf (x) = Z
∞0
f (z)K(x, h, z)∆
(α,β)(z)dz, where K is an explicity known kernel function such that
ϕ
λ(x)ϕ
λ(y) = Z
∞0
ϕ
λ(z)K(x, y, z)∆
(α,β)(z)dz,
K(x, y, z) = 2
−2ρΓ(α + 1)(cosh x cosh y cosh z)
−α−β−1Γ(
12)Γ(α +
12)(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
2x + cosh
2y + cosh
2z − 1 2 cosh x cosh y cosh z and F is the Gauss hypergeometric function
F (a, b, c, z) = X
∞ k=0(a)
k(b)
k(c)
kk! z
k, |z| < 1, where (a)
0= 1 and (a)
k= a(a + 1)...(a + k − 1).
In [2], we have
(τ
[hf )(λ) = ϕ
λ(h) f b (λ) (2)
2. Generalization of Titchmarsh’s Theorem
Definition 2.1. A function f ∈ L
2(α,β)(
R+) is said to be in the ψ-Jacobi Lipschitz class, denote by Lip(ψ, 2), if
kτ
hf (x) − f (x)k
2,(α,β)= O(ψ(h)) as h −→ 0, where
1. ψ(t) is a continuous increasing function on [0, ∞) and ψ(0) = 0, 2. ψ(ts) = ψ(t)ψ(s), for all t, s ∈ [0, ∞),
3. h ≤ ψ(h) and R
1/h0
sψ(s
−2)ds = O(
h12ψ(h
2)) as h −→ 0.
Theorem 2.2. Let f ∈ L
2(α,β)(
R+). Then the following are equivalents 1. f ∈ Lip(ψ, 2),
2. R
∞r
| f(λ)| b
2dµ(λ) = O(ψ(r
−2)) as r −→ +∞.
Proof. 1 = ⇒ 2 Suppose that f ∈ Lip(ψ, 2). Then
kτ
hf (x) − f (x)k
2,(α,β)= O(ψ(h)) as h −→ 0 From formulas (1) and (2), we have
kτ
hf (x) − f (x)k
2,(α,β)= Z
∞0
|1 − ϕ
λ(h)|
2| f b (λ)|
2dµ(λ) If λ ∈ [
h1,
h2] then λh ≥ 1 and (3) of Lemma 1.1 implies that
1 ≤ 1
c
2|1 − ϕ
λ(h)|
2. Then
Z
2/h 1/h| f b (λ)|
2dµ(λ) ≤
Z
2/h 1/h|1 − ϕ
λ(h)|
2| f b (λ)|
2dµ(λ)
≤ Z
∞0
|1 − ϕ
λ(h)|
2| f b (λ)|
2dµ(λ)
= O(ψ
2(h)) = O(ψ(h
2)).
There exists then a positive constant C such that Z
2/h1/h
| f b (λ)|
2dµ(λ) ≤ Cψ(h
2).
Then Z
2rr
| f b (λ)|
2dµ(λ) ≤ Cψ(r
−2) as r −→ +∞
Furthermore, we have Z
∞r
| f(λ)| b
2dµ(λ) =
Z
2r r+ Z
4r2r
+ Z
8r4r
+...
| f b (λ)|
2dµ(λ)
≤ Cψ(r
−2) + Cψ((2r)
−2) + Cψ((4r)
−2) + ...
≤ Cψ(r
−2)(1 + ψ(2
−2) + ψ((2
−2)
2) + ψ((2
−2)
3) + ...)
≤ C(1 − ψ(2
−2))
−1ψ(r
−2), since ψ(2
−2) < 1.
This prove Z
∞r
| f(λ)| b
2dµ(λ) = O(ψ(r
−2)) as r −→ +∞.
2 = ⇒ 1 Suppose now that Z
∞r
| f(λ)| b
2dµ(λ) = O(ψ(r
−2)) as r −→ +∞.
We write
kτ
hf (x) − f (x)k
2,(α,β)= Z
∞0
|1 − ϕ
λ(h)|
2| f b (λ)|
2dµ(λ) = I
1+ I
2, where
I
1= Z
1/h0
|1 − ϕ
λ(h)|
2| f(λ)| b
2dµ(λ) and
I
2= Z
∞1/h
|1 − ϕ
λ(h)|
2| f b (λ)|
2dµ(λ)
Estimate the summands I
1and I
2from above. It follows from (1) of Lemma 1.1 that
I
2= Z
∞1/h
|1 − ϕ
λ(h)|
2| f b (λ)|
2dµ(λ) ≤ 4 Z
∞1/h
| f b (λ)|
2dµ(λ) = O(ψ(h
2)).
To estimate I
1, we use the inequalities (1) and (2) of Lemma 1.1 I
1=
Z
1/h0
|1 − ϕ
λ(h)|
2| f b (λ)|
2dµ(λ)
≤ 2 Z
1/h0
|1 − ϕ
λ(h)|| f(λ)| b
2dµ(λ)
≤ 2h
2Z
1/h0
(λ
2+ ρ
2)| f b (λ)|
2dµ(λ)
≤ 2ρ
2h
2Z
1/h0
| f b (λ)|
2dµ(λ) + 2h
2Z
1/h0
λ
2| f b (λ)|
2dµ(λ) Note that
2ρ
2h
2Z
1/h0
| f(λ)| b
2dµ(λ) ≤ 2ρ
2h
2Z
∞0
| f(λ)| b
2dµ(λ)
= 2ρ
2h
2kf k
22,(α,β)≤ 2ρ
2ψ(h
2)kf k
22,(α,β)= O(ψ(h
2)).
We put
φ(s) = Z
∞s
| f b (λ)|
2dµ(λ).
Integrating by parts, we obtain 2h
2Z
1/h 0λ
2| f b (λ)|
2dµ(λ) = 2h
2Z
1/h0
(−s
2φ
′(s))ds
= 2h
2− 1 h
2φ( 1
h ) + 2 Z
1/h0
sφ(s)ds
!
= −2φ( 1
h ) + 4h
2Z
1/h0
sφ(s)ds
≤ 4Kh
2Z
1/h0