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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

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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

R

with equals 1 at 0 and satisfies the differential equation

(D

α,β

+ λ

2

+ ρ

2

λ

(t) = 0, where ρ = α + β + 1 and

D

α,β

= d

2

dt

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

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

ρ

Γ(iλ)Γ(

12

(1 + iλ)) Γ(

12

(ρ + iλ))Γ(

12

(ρ + iλ) − β)) , where α ≥ −

12

and α > β ≥ −

12

.

The Jacobi transform of a function f ∈ L

2(α,β)

(

R+

) is defined by f b (λ) =

Z

0

f(t)ϕ

λ

(t)∆

(α,β)

(t)dt

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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)|

2

dλ.

The Jacobi transform a unitary isomorphism from L

2(α,β)

(

R+

) onto L

2

(

R+

,

1

dµ(λ)), i.e.

kf k

2,(α,β)

= k f b k

L2(R+,1

dµ(λ))

(1)

Recall from [4] the generalized translation operator τ

h

of a suitable function f on

R+

, is defined by

τ

h

f (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

Γ(α + 1)(cosh x cosh y cosh z)

α−β−1

Γ(

12

)Γ(α +

12

)(sinh x sinh y sinh z)

(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 and F is the Gauss hypergeometric function

F (a, b, c, z) = X

k=0

(a)

k

(b)

k

(c)

k

k! z

k

, |z| < 1, where (a)

0

= 1 and (a)

k

= a(a + 1)...(a + k − 1).

In [2], we have

[h

f )(λ) = ϕ

λ

(h) f b (λ) (2)

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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

h

f (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/h

0

sψ(s

2

)ds = O(

h12

ψ(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

2

dµ(λ) = O(ψ(r

2

)) as r −→ +∞.

Proof. 1 = ⇒ 2 Suppose that f ∈ Lip(ψ, 2). Then

h

f (x) − f (x)k

2,(α,β)

= O(ψ(h)) as h −→ 0 From formulas (1) and (2), we have

h

f (x) − f (x)k

2,(α,β)

= Z

0

|1 − ϕ

λ

(h)|

2

| f b (λ)|

2

dµ(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)|

2

dµ(λ) ≤

Z

2/h 1/h

|1 − ϕ

λ

(h)|

2

| f b (λ)|

2

dµ(λ)

≤ Z

0

|1 − ϕ

λ

(h)|

2

| f b (λ)|

2

dµ(λ)

= O(ψ

2

(h)) = O(ψ(h

2

)).

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There exists then a positive constant C such that Z

2/h

1/h

| f b (λ)|

2

dµ(λ) ≤ Cψ(h

2

).

Then Z

2r

r

| f b (λ)|

2

dµ(λ) ≤ Cψ(r

2

) as r −→ +∞

Furthermore, we have Z

r

| f(λ)| b

2

dµ(λ) =

Z

2r r

+ Z

4r

2r

+ Z

8r

4r

+...

| f b (λ)|

2

dµ(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

))

1

ψ(r

2

), since ψ(2

2

) < 1.

This prove Z

r

| f(λ)| b

2

dµ(λ) = O(ψ(r

2

)) as r −→ +∞.

2 = ⇒ 1 Suppose now that Z

r

| f(λ)| b

2

dµ(λ) = O(ψ(r

2

)) as r −→ +∞.

We write

h

f (x) − f (x)k

2,(α,β)

= Z

0

|1 − ϕ

λ

(h)|

2

| f b (λ)|

2

dµ(λ) = I

1

+ I

2

, where

I

1

= Z

1/h

0

|1 − ϕ

λ

(h)|

2

| f(λ)| b

2

dµ(λ) and

I

2

= Z

1/h

|1 − ϕ

λ

(h)|

2

| f b (λ)|

2

dµ(λ)

Estimate the summands I

1

and I

2

from above. It follows from (1) of Lemma 1.1 that

I

2

= Z

1/h

|1 − ϕ

λ

(h)|

2

| f b (λ)|

2

dµ(λ) ≤ 4 Z

1/h

| f b (λ)|

2

dµ(λ) = O(ψ(h

2

)).

(6)

To estimate I

1

, we use the inequalities (1) and (2) of Lemma 1.1 I

1

=

Z

1/h

0

|1 − ϕ

λ

(h)|

2

| f b (λ)|

2

dµ(λ)

≤ 2 Z

1/h

0

|1 − ϕ

λ

(h)|| f(λ)| b

2

dµ(λ)

≤ 2h

2

Z

1/h

0

2

+ ρ

2

)| f b (λ)|

2

dµ(λ)

≤ 2ρ

2

h

2

Z

1/h

0

| f b (λ)|

2

dµ(λ) + 2h

2

Z

1/h

0

λ

2

| f b (λ)|

2

dµ(λ) Note that

2

h

2

Z

1/h

0

| f(λ)| b

2

dµ(λ) ≤ 2ρ

2

h

2

Z

0

| f(λ)| b

2

dµ(λ)

= 2ρ

2

h

2

kf k

22,(α,β)

≤ 2ρ

2

ψ(h

2

)kf k

22,(α,β)

= O(ψ(h

2

)).

We put

φ(s) = Z

s

| f b (λ)|

2

dµ(λ).

Integrating by parts, we obtain 2h

2

Z

1/h 0

λ

2

| f b (λ)|

2

dµ(λ) = 2h

2

Z

1/h

0

(−s

2

φ

(s))ds

= 2h

2

− 1 h

2

φ( 1

h ) + 2 Z

1/h

0

sφ(s)ds

!

= −2φ( 1

h ) + 4h

2

Z

1/h

0

sφ(s)ds

≤ 4Kh

2

Z

1/h

0

sψ(s

2

)ds

≤ 4Kh

2

1 h

2

ψ(h

2

)

= O(ψ(h

2

)).

Finally, then

h

f (x) − f (x)k

2,(α,β)

= O(ψ(h)) as h −→ 0,

which complets the proof of theorem.

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References

[1] J.P. Anker, E. Damek, C. Yacoub, Spherical analysis on harmonoc AN groups, Ann. Scuola. Norm. Sup. Pisa, 23 (1996), 643-679.

[2] W.O. Bray, M.A. Pinsky, Growth properties of Fourier transforms via mod- uli of continuity, J. Funct. Analysis, 255 (2008), 2265-2285.

[3] R. Daher, M.El Hamma, An analog of Titchmarsh’s theorem of Jacobi transform, Int. Journal of Math. Analysis, 6, No. 20 (2012), 975-981.

[4] M. Flensted-Jensen, T.H. Koornwinder, The convolution structure for Ja- cobi expansions, Ark. Math., 11 (1973), 245-262.

[5] T.H. Koornwinder, Jacobi functions and analysis on noncompact semi sim- ple Lie groups, In: Special Functions: Group Theoretical Aspect and Ap- plications (Ed-s: R.A. Askey et al), Dordrect-Boston, Reidel (1984), 1-85.

[6] S.S. Platonov, Approximation of functions in L

2

-metric on noncompact

rank 1 symmetric spaces, Algebra Analiz, 11, No. 1 (1999), 244-270.

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