# Riemann-Liouville and Higher Dimensional Hardy Operators for NonNegative Decreasing Function inLp(·) Spaces

Authors: Muhammad Sarwar; Ghulam Murtaza; Irshaad Ahmed
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

One-weight inequalities with general weights for Riemann-Liouville transform andn-dimensional fractional integral operator in variable exponent Lebesgue spaces defined on Rn are investigated. In particular, we derive necessary and sufficient conditions governing one-weight inequalities for these operators on the cone of nonnegative decreasing functions in Lp(x) spaces.


---

## Body

## 1. Introduction

We derive necessary and sufficient conditions governing the one-weight inequality for the Riemann-Liouville operator(1)Rαf(x)=1xα∫0xf(t)(x-t)1-αdt0<α<1
and n-dimensional fractional integral operator
(2)Iαg(x)=1|x|α∫|y|<|x|g(t)|x-t|n-αdt0<α<n,
on the cone of nonnegative decreasing function in Lp(x) spaces.In the last two decades a considerable interest of researchers was attracted to the investigation of the mapping properties of integral operators in so-called Nakano spacesLp(·) (see, e.g., the monographs [1, 2] and references therein). Mathematical problems related to these spaces arise in applications to mechanics of the continuum medium. For example, Ružicka [3] studied the problems in the so-called rheological and electrorheological fluids, which lead to spaces with variable exponent.Weighted estimates for the Hardy transform(3)(H1f)(x)=∫0xf(t)dt,x>0,
in Lp(·) spaces were derived in the papers [4] for power-type weights and in [5–9] for general weights. The Hardy inequality for nonnegative decreasing functions was studied in [10, 11]. Furthermore Hardy type inequality was studied in [12, 13] by Rafeiro and Samko in Lebesgue spaces with variable exponent.Weighted problems for the Riemann-Liouville transform inLp(x) spaces were explored in the papers [5, 14–16] (see also the monograph [17]).Historically, one and two weight Hardy inequalities on the cone of nonnegative decreasing functions defined onR+ in the classical Lebesgue spaces were characterized by Arino and Muckenhoupt [18] and Sawyer [19], respectively.It should be emphasized that the operatorIαf(x) is the weighted truncated potential. The trace inequity for this operator in the classical Lebesgue spaces was established by Sawyer [20] (see also the monograph [21], Ch.6 for related topics).In general, the modular inequality(*)∫01|∫0xf(t)dt|q(x)v(x)dx≤c∫01|f(t)|p(t)w(t)dt
for the Hardy operator is not valid (see [22], Corollary 2.3, for details). Namely, the following fact holds: if there exists a positive constant c such that inequality (*) is true for all f≥0, where q; p; w; and v are nonnegative measurable functions, then there exists b∈[0,1] such that w(t)>0 for almost every t<b; v(x)=0 for almost every x>b, and p(t) and q(x) take the same constant values a.e. for t∈(0;b) and x∈(0;b)∩{v≠0}.To get the main result we use the following pointwise inequalities:(4)c1(Tf)(x)≤(Rαf)(x)≤c2(Tf)(x),c3(Hg)(x)≤(Iαg)(x)≤c4(Hg)(x),
for nonnegative decreasing functions, where c1, c2, c3, and c4 are constants and are independent of f, g, and x, and
(5)Tf(x)=1x∫0xf(t)dt,Hg(x)=1|x|n∫|y|<|x|g(y)dy.In the sequel by the symbolTf≈Tg we mean that there are positive constants c1 and c2 such that c1Tf(x)≤Tg(x)≤c2Tf(x). Constants in inequalities will be mainly denoted by c or C; the symbol R+ means the interval (0,+∞).


## 2. Preliminaries

We say that a radial functionf:Rn→R+ is decreasing if there is a decreasing function g:R+→R+ such that g(|x|)=f(x), x∈Rn. We will denote g again by f. Let p:Rn→Rn be a measurable function, satisfying the conditions p-=essinf⁡x∈Rn⁡p(x)>0, p+=esssup⁡x∈Rn⁡p(x)<∞.Givenp:Rn→R+ such that 0<p-≤p+<∞ and a nonnegative measurable function (weight) u in Rn, let us define the following local oscillation of p:
(6)φp(·),u(δ)=esssup⁡x∈B(0,δ)∩supp⁡u⁡p(x)-essinf⁡x∈B(0,δ)∩supp⁡u⁡p(x),
where B(0,δ) is the ball with center 0 and radius δ.We observe thatφp(·),u(δ) is nondecreasing and positive function such that
(7)lim⁡δ→∞φp(·),u(δ)=pu+-pu-,
where pu+ and pu- denote the essential infimum and supremum of p on the support of u, respectively.By the similar manner (see [10]) the function ψp(·),u(η) is defined for an exponent p:R+↦R+ and weight v on R+:
(8)ψp(·),v(η)=esssup⁡x∈(0,η)∩supp⁡v⁡p(x)-essinf⁡x∈(0,η)∩supp⁡v⁡p(x).LetD(R+) be the class of nonnegative decreasing functions on R+ and let DR(Rn) be the class of all nonnegative radially decreasing functions on Rn. Suppose that u is measurable a.e. positive function (weight) on Rn. We denote by Lp(x)(u,Rn) the class of all nonnegative functions on Rn for which
(9)Sp(f)=∫Rn|f(x)|p(x)u(x)dμ(x)<∞.For essential properties ofLp(x) spaces we refer to the papers [23, 24] and the monographs [1, 2].Under the symbolLdecp(x)(u,R+) we mean the class of nonnegative decreasing functions on R+ from Lp(x)(u,Rn)∩DR(Rn).Now we list the well-known results regarding one-weight inequality for the operatorT. For the following statement we refer to [18].Theorem A.
Letr be constant such that 0<r<∞. Then the inequity
(10)∫0∞v(x)(Tf(x))rdx≤c∫0∞v(x)(f(x))rdx,f∈Lr(v,R+),f↓
for a weight v holds, if and only if there exists a positive constant C such that for all s>0(11)∫s∞(sx)rv(x)dx≤C∫0sv(x)dx.Condition (11) is called Br condition and was introduced in [18].Theorem B (see [10]).
Letv be a weight on (0,∞) and p:R+→R+ such that 0<p-≤p+<∞, and assume that ψp(·),v(0+)=0. The following facts are equivalent:(a)
there exists a positive constantc such that, for any f∈D(R+),
(12)∫0∞(Tf(x))p(x)v(x)dx≤C∫0∞(f(x))p(x)v(x)dx;(b)
for anyr,s>0,
(13)∫r∞(rsx)p(x)v(x)dx≤C∫0rv(x)sp(x)dx;(c)
p|supp⁡v≡p0 a.e. and v∈Bp0.Proposition 1.
For the operatorsT,H,Rα, and Iα, the following relations hold:(a)
(14)Rαf≈Tf,0<α<1,f∈D(R+);(b)
(15)Iαg≈Hg,0<α<n,g∈DR(Rn).Proof.
(a) Upper estimate: representRαf as follows:
(16)Rαf(x)=1xα∫0x/2f(t)(x-t)1-αdt+1xα∫x/2xf(t)(x-t)1-αdt=S1(x)+S2(x).
Observe that if t<x/2, then x/2<x-t. Hence
(17)S1(x)≤c1x∫0x/2f(t)dt≤cTf(x),
where the positive constant c does not depend on f and x. Using the fact that f is decreasing we find that
(18)S2(x)≤cf(x2)≤cTf(x).
Lower estimate follows immediately by using the fact thatf is nonnegative and the obvious estimate x-t≤x and 0<t<x.
(b) Upper estimate: let us represent the operatorIα as follows:
(19)Iαg(x)=1|x|α∫|y|<|x|/2g(y)|x-y|n-αdy+1|x|α∫|x|/2<|y|<|x|g(y)|x-y|n-αdy≕S1′(x)+S2′(x).
Since |x|/2≤|x-y| for |y|<|x|/2 we have that
(20)S1′(x)≤c|x|n∫|y|<|x|/2g(y)dy≤cHg(x).
Taking into account the fact that f is radially decreasing on Rn we find that there is a decreasing function f:R+→R+ such that
(21)S2′(x)≤f(|x|2)·1|x|α∫|x|/2<|y|<|x||x-y|α-ndy.
Let Fx={y:|x|/2<|y|<|x|}. Then we have
(22)∫Fx|x-y|α-ndy=∫0∞|{y∈Fx:|x-y|α-n>t}|dt≤∫0|x|α-n|{y∈Fx:|x-y|α-n>t}|dt+∫|x|α-n∞|{y∈Fx:|x-y|α-n>t}|dt≕I1+I2.
It is easy to see that
(23)I1≤∫0|x|α-n|B(0,|x|)|dt=c|x|α;
while using the fact that n/(n-α)>1 we find that
(24)I2≤∫|x|α-n∞|{y∈Fx:|x-y|≤t1/(α-n)}|dt≤c∫|x|α-n∞tn/(α-n)dt=cα,n|x|α.
Finally we conclude that
(25)S2′(x)≤cf(|x|2)≤cHf(x).
Lower estimate follows immediately by using the fact that f is nonnegative and the obvious estimate |x-y|≤|x|, where 0<|y|<|x|.We will also need the following statement.Lemma 2.
Letr be a constant such that 0<r<∞. Then the inequality
(26)∫Rn(Hf(x))ru(x)dx≤C∫Rn(f(x))ru(x)dx,f∈Ldecr(u,Rn),
holds, if and only if there exists a positive constant C such that, for all s>0,
(27)∫|x|>s(s|x|)r|x|r(1-n)u(x)dx≤C∫|x|<s|x|r(1-n)u(x)dx.Proof.
We will see that inequality (26) is equivalent to the inequality
(28)∫0∞u~(t)(Tf¯(t))rdt≤C∫0∞u~(t)(f¯(t))rdt,
where u~(t)=t(n-1)(1-r)u¯(t), f¯(t)=tn-1f(t), and u¯(t)=∫Sn-1u(tx¯)dσ(x¯).
Indeed, using polar coordinates inRn we have
(29)∫Rn(Hf(x))ru(x)dx=∫Rnu(x)(1tn∫|y|<tf(y)dy)rdx=∫0∞tn-1(1tn∫|y|<tf(y)dy)r(∫Sn-1u(tx¯)dσx¯)dt=C∫0∞tn-1t-nrtr(1t∫0tτn-1f(τ)dτ)ru¯(t)dt=C∫0∞tn-1tr(1-n)u¯(t)(1t∫0tf¯(τ)dτ)rdt≤C∫0∞u~(t)(f¯(t))rdt=C∫0∞t(n-1)(1-r)t(n-1)r(f(t))rdt=C∫Rn(f(x))ru(x)dx.
Conversely taking the test functionfr(x)=χB(0,r)(x)|x|1-n, r>0, in modular inequality (26), one can easily obtain inequality (27).


## 3. The Main Results

To formulate the main results we need to prove the following proposition.Proposition 3.
Letu be a weight on Rn and p:Rn→R+ such that 0<p-≤p+<∞, and assume that φp(·),u(0+)=0. The following statements are equivalent:(a)
there exists a positive constantC such that, for any f∈DR(Rn),
(30)∫Rn(Hf(x))p(x)u(x)dx≤C∫Rn(f(x))p(x)u(x)dx;(b)
for anyr,s>0,
(31)∫|x|>r(rs|x|n)p(x)u(x)dx≤C∫B(0,r)|x|(1-n)p(x)u(x)sp(x)dx;(c)
p|supp⁡u≡p0 a.e. and u∈Bp0.Proof.
We use the arguments of [10]. To show that (a) implies (b) it is enough to test the modular inequality (30) for the function fr,s(x)=(1/s)χB(0,r)(x)|x|1-n, s,r>0. Indeed, it can be checked that
(32)Hfr,s(x)={1|x|ns∫|y|≤|x||y|1-ndy,if|x|≤r;1|x|ns∫|y|≤r|y|1-ndy,if|x|>r.
Further, we find that(33)∫|x|>ru(x)(Hfr,s)p(x)dx≤∫Rnu(x)(Hfr,s)p(x)dx≤C∫Rnu(x)(1sχB(0,r)(x)|x|1-n)p(x)dx.
Therefore
(34)∫|x|>ru(x)(rs|x|n)p(x)dx≤C∫B(0,r)|x|(1-n)p(x)u(x)sp(x)dx.
To obtain (c) from (b) we are going to prove that condition (b) implies that φp(·),u(δ) is a constant function; namely, φp(·),u(δ)=pu+-pu- for all δ>0. This fact and the hypothesis on φp(·),u(δ) imply that φp(·),u(δ)≡0, and hence, due to (7),
(35)p|supp⁡u≡pu+-pu-≡p0a.e.
Finally (31) means that u∈Bp0. Let us suppose that φp(·),u is not constant. Then one of the following conditions holds:(i)
there existsδ>0 such that
(36)α=esssup⁡x∈B(0,δ)∩supp⁡u⁡p(x)<pu+<∞,
and, hence, there exists ϵ>0 such that
(37)|{|x|>δ:p(x)≥α+ϵ}∩supp⁡u|>0,
or(ii)
there existsδ>0 such that
(38)β=essinf⁡x∈B(0,δ)∩suppu⁡p(x)>pu->0,
and then, for some ϵ>0,
(39)|{|x|>δ:p(x)≤β-ϵ}∩supp⁡u|>0.In case (i) we observe that condition (b), for r=δ, implies that
(40)∫|x|>δ(δs)p(x)u(x)|x|np(x)dx≤C∫B(0,δ)|x|(1-n)p(x)u(x)sp(x)dx.
Then using (36) we obtain, for s<min⁡(1,δ),
(41)(δs)α+ϵ∫{|x|≥δ:p(x)≥α+ϵ}u(x)|x|np(x)dx≤Csα∫B(0,δ)u(x)|x|(1-n)p(x)dx,
which is clearly a contradiction if we let s↓0. Similarly in case (ii) let us consider the same condition (b), for r=δ, and fix now s>1. Taking into account (38) we find that
(42)1sβ-ϵ∫{|x|≥δ:p(x)≤β-ϵ}(δ|x|n)p(x)u(x)dx≤Csβ∫B(0,δ)|x|(1-n)p(x)u(x)dx,
which is a contradiction if we let s↑∞.
Finally, the fact that condition (c) implies (a) follows from [18,Theorem 1.7].Theorem 4.
Letu be a weight on (0,∞) and p:R+→R+ such that 0<p-≤p+<∞. Assume that ψp(·),v(0+)=0. The following facts are equivalent:(i)
there exists a positive constantC such that, for any f∈D(R+),
(43)∫R+(Rαf(x))p(x)u(x)dx≤C∫R+(f(x))p(x)u(x)dx;(ii)
condition (13) holds;(iii)
condition(c) of Theorem B is satisfied.Proof.
Proof follows by using Theorem B and Proposition1(a).Theorem 5.
Letu be a weight on Rn and p:Rn→R+ such that 0<p-≤p+<∞, and assume that φp(·),u(0+)=0. The following facts are equivalent:(i)
there exists a positive constantC such that, for any f∈DR(Rn),
(44)∫Rn(Iαf(x))p(x)u(x)dx≤C∫Rn(f(x))p(x)u(x)dx;(ii)
condition (31) holds;(iii)
condition (c) of Proposition3 holds.Proof.
Proof follows by using Propositions3 and 1(b).



---

*Source: 621857-2014-08-26.xml*