# Bessel Transform of(k,γ)-Bessel Lipschitz Functions

Authors: Radouan Daher; Mohamed El Hamma
Journal: Journal of Mathematics
 (2013)

---

## Abstract

Using a generalized translation operator, we obtain an analog of Theorem 5.2 in Younis (1986) for the Bessel transform for functions satisfying the(k,γ)-Bessel Lipschitz condition in L2,α(ℝ+).


---

## Body

## 1. Introduction and Preliminaries

Younis Theorem 5.2 [1] 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 transforms; namely, we have the following.Theorem 1 (see [1]).
Letf∈L2(ℝ). Then the followings are equivalent:(1)
∥f(x+h)-f(x)∥2=O(hα/(log(1/h))β)ash→0, 0<α<1, β>0,(2)
∫|x|≥r‍|ℱ(f)(x)|2dx=O(r-2α(logr)-2β)asr→+∞, where ℱ stands for the Fourier transform of f.In this paper, we obtain a generalization of Theorem1 for the Bessel transform. For this purpose, we use a generalized translation operator.Assume thatL2,α(ℝ+); α>-1/2 is the Hilbert space of measurable functions f(t) on ℝ+ with finite norm
(1)∥f∥2,α=(∫0∞‍|f(x)|2x2α+1dx)1/2.Let(2)B=d2dt2+(2α+1)tddt
be the Bessel differential operator.Forα≥-1/2, we introduce the Bessel normalized function of the first kind jα defined by
(3)jα(z)=Γ(α+1)∑n=0∞‍(-1)nn!Γ(n+α+1)(z2)2n,
where Γ is the gamma function (see [2]).The functiony=jα(x) satisfies the differential equation
(4)By+y=0,
with the initial conditions y(0)=1 and y′(0)=0. jα(z) is function infinitely differentiable, even, and, moreover, entirely analytic.Lemma 2.
Forx∈ℝ+ the following inequality is fulfilled:
(5)|1-jα(x)|≥c,
with x≥1, where c>0 is a certain constant which depends only on α.Proof.
Analog of Lemma 2.9 is in [3].Lemma 3.
The following inequalities are valid for Bessel functionjα:(1)
|jα(x)|≤1, for all x∈ℝ+,(2)
1-jα(x)=O(x2),0≤x≤1.Proof.
See [4].The Bessel transform we call the integral transform from [2, 5, 6]
(6)f^(λ)=∫0∞‍f(t)jα(λt)t2α+1dt,λ∈ℝ+.The inverse Bessel transform is given by the formula(7)f(t)=(2αΓ(α+1))-2∫0∞‍f^(λ)jα(λt)λ2α+1dλ.We have the Parseval's identity(8)∥f^∥2,α=2αΓ(α+1)∥f∥2,α.InL2,α(ℝ+), consider the generalized translation operator Th defined by
(9)Thf(t)=cα∫0π‍f(t2+h2-2thcosφ)sin2αφdφ,
where
(10)cα=(∫0π‍sin2αφdφ)-1=Γ(α+1)Γ(1/2)Γ(α+(1/2)).The following relations connect the generalized translation operator and the Bessel transform; in [7] we have
(11)(Thf^)(λ)=jα(λh)f^(λ).


## 2. Main Result

In this section we give the main result of this paper. We need first to define(k,γ)-Bessel Lipschitz class.Definition 4.
Let0<k<1 and γ≥0. A function f∈L2,α(ℝ+) is said to be in the (k,γ)-Bessel Lipschitz class, denoted by Lip(k, γ, 2), if
(12)∥Thf(t)-f(t)∥2,α=O(hk(log(1/h))γ),ash⟶0.
Our main result is as follows.Theorem 5.
Letf∈L2,α(ℝ+). Then the followings are equivalents(1)
f∈  Lip(k,γ,2).(2)
∫r∞‍|f^(λ)|2λ2α+1dλ=O(r-2k/(logr)2γ),asr→+∞.Proof.
(
1
)
⇒
(
2
) Assume that f∈ Lip(k, γ, 2). Then we have
(13)∥Thf(t)-f(t)∥2,α2=1(2αΓ(α+1))2∫0∞‍|1-jα(λh)|2|f^(λ)|2λ2α+1dλ.
Ifλ∈[1/h,2/h] then λh≥1 and Lemma 2 implies that
(14)1≤1c2|1-jα(λh)|.
Then(15)∫1/h2/h‍|f^(λ)|2λ2α+1dλ≤1c2∫1/h2/h‍|1-jα(λh)|2|f^(λ)|2λ2α+1dλ≤1c2∫0∞‍|1-jα(λh)|2|f^(λ)|2λ2α+1dλ=O(h2k(log(1/h))2γ).
We obtain(16)∫r2r‍|f^(λ)|2λ2α+1dλ≤Cr-2k(logr)2γ,
where C is a positive constant.
So that(17)∫r∞‍|f^(λ)|2λ2α+1dλ=[∫r2r‍+∫2r4r‍+∫4r8r‍+⋯]|f^(λ)|2λ2α+1dλ≤Cr-2k(logr)2γ+C(2r)-2k(log2r)2γ+C(4r)-2k(log4r)2γ+⋯≤Cr-2k(logr)2γ(1+2-2k+(2-2k)2+(2-2k)3+⋯)≤CKr-2k(logr)2γ,
where K=(1-2-2k)-1 since 2-2k<1.
This proves that(18)∫r∞‍|f^(λ)|2λ2α+1dλ=O(r-2k(logr)2γ)asr⟶+∞.
(
2
)
⇒
(
1
) Suppose now that
(19)∫r∞‍|f^(λ)|2λ2α+1dλ=O(r-2k(logr)2γ)asr⟶+∞.
We write(20)∫0∞‍|1-jα(λh)|2|f^(λ)|2λ2α+1dλ=I1+I2,
where
(21)I1=∫01/h‍|1-jα(λh)|2|f^(λ)|2λ2α+1dλ,I2=∫1/h∞‍|1-jα(λh)|2|f^(λ)|2λ2α+1dλ.
Estimate the summandsI1 and I2 from above. It follows from the inequality |jα(λh)|≤1 that
(22)I2=∫1/h∞‍|1-jα(λh)|2|f^(λ)|2λ2α+1dλ≤4∫1/h∞‍|f^(λ)|2λ2α+1dλ=O(h2k(log(1/h))2γ).
To estimateI1, we use the inequality (2) of Lemma 3.
Set(23)ϕ(x)=∫x∞‍|f^(λ)|2λ2α+1dλ.
Using integration by parts, we obtain(24)I1≤-C1h2∫01/h‍s2ϕ′(s)ds≤-C1ϕ(1h)+2C1h2∫01/h‍sϕ(s)ds≤C2h2∫01/h‍sϕ(s)ds≤C2h2∫01/h‍ss-2k(logs)-2γds≤C3h2k(log(1/h))-2γ,
where C1,C2,andC2 are positive constants and this ends the proof.



---

*Source: 418546-2013-04-03.xml*