R E S E A R C H Open Access
Time scale Hardy-type inequalities with
?broken? exponent p
James A Oguntuase1, Olanrewaju O Fabelurin2, Abdulaziz G Adeagbo-Sheikh2and Lars-Erik Persson3,4*
*Correspondence: [email protected]
3Department of Mathematics, Luleå University of Technology, Luleå, 971 87, Sweden
4Narvik University College, P.O. Box 385, Narvik, 8505, Norway Full list of author information is available at the end of the article
Abstract
In this paper, some new Hardy-type inequalities involving ?broken? exponents are derived on arbitrary time scales. Our approach uses both convexity and
superquadracity arguments, and the results obtained generalize, complement and provide refinements of some known results in literature.
MSC: Primary 39B82; secondary 44B20; 46C05
Keywords: Hardy-type inequalities; ?broken? exponent; ?broken? time scale;
superquadracity
1 Introduction
In , Hardy [] stated (without proof ) the following inequality:
∞
x
x
f(t)dt p
dx≤ p
p– p ∞
fp(x)dx, p> , (.)
wherefis a non-negative measurable function. This result was finally proved by Hardy []
(see also Hardy []) in .
In , Hardy [] obtained and proved the following generalization of inequality (.):
∞
x
x
f(t)dt p
xαdx≤ p
p– –α p ∞
fp(x)xαdx, (.)
which holds for all measurable and non-negative functionsf on (,∞) wheneverα<p– , p≥.
In , Godunova [] discovered that inequality (.) can be proved via convexity ar- gument, but this result was not well known in western literature. The use of convexity argument to prove Hardy-type inequalities was independently rediscovered by Imoru []
and Kaijseret al.[] in and , respectively. After that a great number of papers based on this idea have been presented and applied (see [–]).
In a recent paper, Persson and Samko [] used the convexity argument to prove that inequality (.) is equivalent to the following inequality:
∞
x
x
f(t)dt p
dx x ≤
∞
fp(x)dx
x , (.)
©2015 Oguntuase et al.; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.
via the substitutionf(x) =g(x–/p)x–/p. In the same paper [] it was also shown that in- equality (.) is equivalent to inequality (.) via the substitutionf(t) =g(t(p––α)/p)t–(+/p). It thus follows that Hardy?s initial generalization (.) is not actually a generalization. Fur- thermore, in the same paper, sufficient conditions for a variant of inequality (.) to hold were given, namely the following inequality:
l
x
x
f(t)dt p
dx x ≤
l
fp(x)
–x l
dx
x (.)
for p < or p ≥ . The authors established the equivalence theorem for the one- dimensional Hardy-type inequalities. In particular, it was shown that inequality (.) is equivalent to the following variant of (.):
l
x
x
f(t)dt p
xαdx≤ p
p– –α p l
fp(x)xα
– x
l p–α–p
dx (.)
forp≤,α<p– orp< ,α>p– and ≤l≤ ∞.
A multidimensional version of this equivalence theorem concerning Hardy-type in- equalities was proved by Oguntuaseet al.[]. For the development of the use of convexity argument in obtaining Hardy-type inequalities, we refer interested readers to the review article by Oguntuase and Persson [] and the references cited therein. In a recent paper, Oguntuaseet al.[] stated and proved multidimensional Hardy-type inequalities with
?broken? exponent. In particular, the following result was established.
Theorem . Let b> , <l≤ ∞and p(x) =
p, ≤x≤b,
p, x>b, β(x) =
β, ≤x≤b, β, x>b,
where p,p,β,β∈ \ {}.If f is non-negative and measurable andβ(x) > ,then l
x
x
f(t)dt p(x)
x–β(x)dx≤ l
β(x)f(x)p(x)x–β(x)
– x
l β(x)
dx+I, (.) where I= ,if l≤b(so thatβ(x) =βand p(x) =p)and
I= β
b–β–l–β b
f(x)pdx– β
b–β–l–β b
f(x)pdx,
if l>b.If <p(x)≤,then(.)holds in the reversed direction(for the case l=∞, – (xl)β(x)≡and l–β=l–β≡).
Remark . Observe that under suitable substitutions, all the variants (.)-(.) can be recovered from (.). Thus (.) is more general than all the other inequalities above.
In , Řehák [, Lemma .] proved that ifTis any arbitrary time scale that is un- bounded above and containingaandα> , then the following estimates hold:
∞
a
s (σ(s))α ≤
∞
a
ds sα ≤
∞
a
s
sα. (.)
Řehák used inequality (.) to establish the time scale version of the Hardy inequality as follows.
Theorem . If a> ,p> ,and f is a non-negative function such that the delta integral ∞
a (f(s))ps exists as a finite number,then ∞
a
σ(x) –a
σ(x) a
f(t)t p
<
p p–
p ∞
a
fp(x)x. (.)
The above result by Řehák [] signaled the beginning of research on the time scale Hardy inequality. Since the publication of Řehák?s result on the time scale Hardy inequal- ity, other researchers (see, for instance, [–] and the references cited therein) have ob- tained its generalization both in the one-dimensional and multidimensional settings.
The aim of this paper is to obtain one-dimensional Hardy-type inequalities on a time scale with ?broken? exponent. It is a great interest of this subject (see,e.g., papers [–]
where a lot of interesting facts complementing this paper can be found).
Before we present our results, let us recall some essentials about time scales. In , Hilger introduced the calculus on time scales which unifies continuous and discrete anal- ysis. A time scaleTis an arbitrary nonempty closed subset of the real numbers. The two most popular examples areT=andT=Z. We define the forward jump operator σ byσ(t) :=inf{s∈T:s>t}and the graininessμof the time scaleTbyμ(t) :=σ(t) –t.
A pointt∈Tis said to be right-dense and right-scattered ifσ(t) =t,σ(t) >t, respectively.
We definefσ :=f◦σ. For a functionf:T→ , the delta derivative is defined by f(t) := lim
s→t,σ(s)=t
fσ(s) –f(t) σ(s) –t .
A functionf :T→Ris called rd-continuous provided it is continuous at all right-dense points inTand its left-sided limits exist (finite) at all left-dense points inT. Note that we have
σ(t) =t, μ(t) = , f=f,
b
a
f(t)t= b
a
f(t)dt, whenT=R,
σ(t) =t+ , μ(t) = , f=f,
b
a
f(t)t= b–
t=a
f(t), whenT=Z.
For more understanding of the theory of time scales, we refer the interested reader to [,
].
We recall the following definition of the well-known binomial theorem.
Definition .([, Definition .]) (Binomial theorem) Ifα,x∈ , the expansion of ( +x)αdefined by
( +x)α= +αx+α(α– )x
! +· · ·+α(α– )· · ·(α–β+ )xβ β! +· · ·
= ∞
β=
(α)(β)
(β+ )xβ (.)
is known as the binomial theorem.
Definition .([, Definition .]) A functionφ: [,∞)→ is called superquadratic provided that for allx≥ there exists a constantCx∈ such that
φ(y) –φ(x) –φ
|y–x| ≥Cx(y–x) for all y≥.
We say thatφis subquadratic if –φis superquadratic.
2 Time scale Hardy-type inequalities with ?broken? exponentpvia convexity Before we state our results in this section, we shall need the following lemmas.
Lemma . ([, Theorem .]) (Fubini?s theorem on time scales)Let(,M,μ)and ( ,L,λ)be two finite dimensional time scale measure spaces.If f :× → is a μ ×λ-integrable function and the function φ(y) :=
f(x,y)x for a.e. y∈ and ϕ(x) =
λf(x,y)y for a.e.x∈,thenφ isλ-integrable on ,ϕisμ-integrable on and
x
f(x,y)y=
y
f(x,y)x. (.)
Lemma .([, Theorem .]) Let a,b∈Tand c,d∈ .Suppose that f : [a,b]Tk → (c,d)is rd-continuous andφ: (c,d)→ is convex.Then
φ
b–a b
a
f(t)t
≤ b–a
b
a
φ
f(t)t. (.)
First, we give the following proposition which is an adaptation of Lemma . in [].
Proposition . Letα> andTbe any arbitrary time scale that is unbounded above.Let a,l∈Tbe such that <a<l≤ ∞.Then the following estimates hold:
l
a
σ(s) –αs≤ l
a
ds sα ≤
l
a
s
sα. (.)
Proof Supposel<∞and denote [a,l]T:={t∈T:a≤t≤l}. We prove only thatI≤I∗, where
I= b
a
σ(s)–αs
and I∗=
l
a
ds sα,
since the other inequality can be proven analogously. Suppose by contradiction that there exists a time scaleT∗ such thata,l∈T∗ andI>I∗, whereI∗ is taken over [a,l]T∗. This implies that there exists> such thatI–>I∗.
On the other hand, by virtue of the definition of the delta Riemann integrability, there exists a time scaleTDcontainingaand satisfying
TD={tk: ≤k≤n} with <a=t<t<t<· · ·<tn=l such that|I∗–ID|</, where
ID:=TD
l a
σ(s) –αs.
Here, the delta integral is taken over [a,l]TD. Thus we get I∗+≤ID<I∗+/,
a contradiction.
For the casel=∞, the proof is given in [].
Our first result in this section reads as follows.
Theorem . Letβ> andTbe any arbitrary time scale.If f :T→ is differentiable, then the following inequality
t
b
(s–a)β–
+
nβ+
k=
(β– )(k) (k+ )
μ(s) s–a
k s≤
β
(t–a)β– (b–a)β
(.)
holds for any a,b,t∈Tksuch that≤a<b≤t,where nβ:=inf
n∈
N∪ {}
:β–n≥
. (.)
Proof Letf :T→ be a function defined by f(t) :=
β
(t–a)β– (b–a)β
∀t∈T.
By Definition . and equation (.) we have that
f(t) = (t–a)β βμ(t)
+ βμ(t)
(t–a)+β(β– )
!
μ(t) t–a
+· · ·
–
= (t–a)β–
+(β– )
!
μ(t) t–a
+(β– )(β– )
!
μ(t) t–a
+· · ·
= (t–a)β–
+
nβ–
k=
(β– )(k) (k+ )
μ(t) t–a
k
+O
(β– )(β– )· · ·(β–nβ) (nβ+ )
μnβ(t)(t–a)β–(nβ+)
≥(t–a)β–
+
nβ–
k=
(β– )(k) (k+ )
μ(t) t–a
k .
Integrating, we get that t
b
(s–a)β–
+
nβ–
k=
(β– )(k) (k+ )
μ(s) s–a
k s≤
β
(t–a)β– (b–a)β
.
Remark . We observed that the chain rule can be applied to simplify the proof of Theo- rem .. The techniques for doing this can be found in the papers [–] and the details are left to interested readers. Also, a discrete version of Theorem . can easily be ob- tained, and interested readers can fill this gap since this is not the main focus of this paper.
Theorem . Let b> ,β(x) > , <l≤ ∞,and
p(x) =
p, ≤x≤b,
p, x>b, β(x) =
β, ≤x≤b,
β, x>b, (.)
where p,p,β,β∈ \{}.If f :T→ is non-negative-integrable and f∈Crd([a,b],) for which
l a
f(x)p(x) β(x)
– y–a
l–a β(x)
(y–a)–β(x)x<∞,
then l
a
(σ(x) –a)
σ(x) a
f(y)(y) p(x)
σ(x) –a –β(x)(x)
≤ l
a
f(x)p(x) β(x)
– x–a
l–a β(x)
(x–a)–β(x)(x) +I, (.)
where I= if l≤b(so thatβ(x) =βand p(x) =p)and
I= β
(b–a)–β– (l–a)–β
b a
f(x)px– β
(b–a)–β– (l–a)–β
b a
f(x)px.
Moreover,assume that p≥,p≥or p≥,p< or p< ,p≥or p< ,p< (for the case with negative parameters,we assume that the function f is strictly positive on the corresponding interval).
If <p(x)≤,then(.)holds in the reverse direction.
Proof Letl≤b. Applying Jensen?s inequality (.), Fubini?s Theorem. and Proposi- tion ., we obtain that
l a
(σ(x) –a)
σ(x) a
f(y)y p(x)
σ(x) –a–β(x)x
≤ l
a
(σ(x) –a)
σ(x)
a
f(y)py
σ(x) –a –βx
≤ l
a
f(y)p(y)
β(y) (y–a)–β(y)
–
(y–a) (l–a)
β(y) y.
Next, for the caseb<l, by applying Jensen?s inequality (.), Fubini?s Theorem., Propo- sitions . and ., we find that
l
a
(σ(x) –a)
σ(x) a
f(y)y p(x)
σ(x) –a–β(x)x
= b
a
(σ(x) –a)
σ(x) a
f(y)y p
σ(x) –a –βx +
l
b
(σ(x) –a)
b
a
f(y)y p
σ(x) –a–βx +
l b
(σ(x) –a)
σ(x) b
f(y)y p
σ(x) –a –βx
≤ b
a
f(y)p β
(y–a)–β
– y–a
l–a β
y
+ l
b
f(y)p
β (y–a)–β
– y–a
y–a β
+ b
a
(b–a)–β– (l–a)–β f(y)p β –
(b–a)–β– (l–a)–β f(y)p β
y
= l
a
f(y)p(x)
β(x) (y–a)–β(x)
– y–a
l–a β(x)
y+I.
For the proof of the case <p(x)≤, we first note that the functions involving exponents pandpare concave. Therefore the two inequalities above hold in the reverse direction
so also this case is proved.
Remark . By takingT=anda= in Theorem ., inequality (.) coincides with inequality (.) obtained in [].
Next, we state a dual version of Theorem ., when the Hardy operator
H:f(x)→ σ(x) –a
σ(x) a
f(y)y is replaced by the dual Hardy operator
H∗:f(x)→
σ(x) –a ∞
σ(x)
f(t)t (σ(t) –a)(t–a). Hence, our result in this direction reads as follows.
Theorem . Let b> ,β> , ≤l<∞and
p(x) =
p, ≤x≤b,
p, x>b, β(x) =
β, ≤x≤b,
β, x>b, (.)
where p,p,β,β∈ \{}.Moreover,assume that p≥,p≥or p≥,p< or p< , p≥or p< ,p< (for the case with negative parameters,we assume that the function
f is strictly positive on the corresponding interval).If f is a non-negative delta integrable function and f ∈Crd([a,b],)for which
∞
l
β(t)
f(t) p(t)(t–a)β(t)
– l–a
t–a β(t)
t
(σ(t) –a)(t–a)<∞, then
∞
l
σ(x) –a ∞
σ(x)
f(t)t (σ(t) –a)(t–a)
p(x)
(x–a)β(x)
×
+
nβ(x)–
k=
(β(x) – )(k) (k+ )
μ(x) x–a
k
x (σ(x) –a)(x–a)
≤ ∞
l
β(x)
f(x)p(x)(x–a)β(x)
– l–a
x–a β(x)
x
(σ(x) –a)(x–a)+I, (.) where I= if l≤b(so thatβ(x) =βand p(x) =p)and
I= β
(b–a)β– (l–a)β ∞
b
(f(x))p (σ(x) –a)(x–a)x –
β
(b–a)β– (l–a)β ∞
b
(f(x))p
(σ(x) –a)(x–a)x.
If <p(x)≤,then(.)holds in the reverse direction.
Proof Letl>b. By utilizing Jensen?s inequality (.), Fubini?s Theorem. and Lemma .
and taking into account (.), we find that ∞
l
σ(x) –a ∞
σ(x)
f(t)t (σ(t) –a)(t–a)
p(x)
(x–a)β(x)
×
+
nβ(x)–
k=
(β(x) – )(k) (k+ )
μ(x) x–a
k
x (σ(x) –a)(x–a)
≤ ∞
l
∞
σ(x)
(f(t))pt (σ(t) –a)(t–a)
+
nβ–
k=
(β– )(k) (k+ )
μ(x) x–a
k
(x–a)β–x
= ∞
l
(f(t))p (σ(t) –a)(t–a)
t l
(x–a)β–
+
nβ–
k=
(β– )(k) (k+ )
μ(x) x–a
k x
t
≤ ∞
l
β(t)
f(t) p(t)(t–a)β(t)
– l–a
t–a β(t)
t (σ(t) –a)(t–a).
Next, letl≤b. Then, by applying Jensen?s inequality (.), Fubini?s Theorem. and Lemma ., and taking into account (.), we find that
∞
l
σ(x) –a ∞
σ(x)
f(t)t (σ(t) –a)(t–a)
p(x)
(x–a)β(x)
×
+
nβ(x)–
k=
(β(x) – )(k) (k+ )
μ(x) x–a
k
x (σ(x) –a)(x–a)
≤ b
l
∞
σ(x)
(f(t))pt (σ(t) –a)(t–a)
+
nβ–
k=
(β– )(k) (k+ )
μ(x) x–a
k
(x–a)β–x
+ ∞
b
∞
σ(x)
(f(t))pt (σ(t) –a)(t–a)
+
nβ–
k=
(β– )(k) (k+ )
μ(x) x–a
k
(x–a)β–x
= b
l
(f(t))p (σ(t) –a)(t–a)
t
l
(x–a)β–
+
nβ–
k=
(β– )(k) (k+ )
μ(x) x–a
k x
t
+ ∞
b
(f(t))p (σ(t) –a)(t–a)
b l
(x–a)β–
+
nβ–
k=
(β– )(k) (k+ )
μ(x) x–a
k x
t
+ ∞
b
(f(t))p (σ(t) –a)(t–a)
t b
(x–a)β–
+
nβ–
k=
(β– )(k) (k+ )
μ(x) x–a
k x
t
≤ b
l
(f(t))p (σ(t) –a)(t–a)
β
(t–a)β– (l–a)β t +
∞
b
(f(t))p (σ(t) –a)(t–a)
β
(b–a)β– (l–a)β t +
∞
b
(f(t))p (σ(t) –a)(t–a)
β
(t–a)β– (b–a)β t
= ∞
l
β(t)
f(t)p(t)(t–a)β(t)
– b–a
t–a β(t)
t
(σ(t) –a)(t–a)+I. Remark . By takingT=anda= in Theorem ., inequality (.) coincides with Theorem . in [].
3 Time scale Hardy-type inequalities with ?broken? exponentpvia superquadracity
Refined Jensen?s inequality on time scales for superquadratic functions has been recently obtained by Barić et al.This inequality is very useful in the proof of our results in this section.
Lemma . ([, Theorem .]) Let a,b∈T. Suppose that f : [a,b]Tk →[,∞]is rd- continuous andφ: [,∞]→ is continuous and superquadratic.Then
φ
b–a b
a
f(t)t
≤ b–a
b
a
φ
f(s) –φ
f(s) – b–a
b
a
f(t)t
s. (.)
Proof For the proof, see [].
Our first result in this section reads as follows.
Theorem . Let the assumptions of Theorem.be satisfied.Moreover,let u∈Crd([a,b], )be a non-negative function such that the-integrall
t u(x)
(σ(x)–a)β(x)+x<∞and define the weight functionυby
v(t) := (t–a) l
t
u(x)
(σ(x) –a)β(x)+x, t∈(a,b). (.)
()Ifis a non-negative superquadratic function on(a,c), <a<c≤ ∞,then l
a
u(x)
(σ(x) –a) σ(x)
a
f(t)t
σ(x) –a –β(x)x +
l
a
l
t
f(t) – (σ(x) –a)
σ(x) a
f(t)t
u(x)
(σ(x) –a)β(x)+xt
≤ l
a
υ(x)
f(x) x
x–a (.)
holds for all-integrable functions and f ∈Crd([a,b],R)such that f(x)∈(a,c).
()If the real-valued functionis subquadratic on(a,c), <a<c≤ ∞,then(.)holds in the reversed direction.
Proof () Letl≤b. Applying refined Jensen?s inequality (.), after taking into account Definition ., we get that
l a
u(x)
(σ(x) –a) σ(x)
a
f(t)t
σ(x) –a –β(x)x
≤ l
a
u(x) (σ(x) –a)β+
σ(x) a
f(t)tx –
l
a
u(x) (σ(x) –a)β+
σ(x) a
f(t) – (σ(x) –a)
σ(x) a
f(t)t
tx. (.)
By utilizing Fubini?s Theorem. and taking into account Definition . of the weight functionv, we obtain that the right-hand side of (.) is not greater than
l
a
υ(t)
f(t) t t–a –
l
a
l
t
f(t) – (σ(x) –a)
σ(x) a
f(t)t
u(x)
(σ(x) –a)β+xt.
Letl>b. Applying again refined Jensen?s inequality (.), after taking into account Defini- tion ., we find that
l
a
u(x)
(σ(x) –a) σ(x)
a
f(t)t
σ(x) –a –β(x)x
≤ l
a
u(x) (σ(x) –a)β+
σ(x) a
f(t)tx –
l
a
u(x) (σ(x) –a)β+
σ(x) a
f(t) – (σ(x) –a)
σ(x) a
f(t)t
tx
+ l
a
u(x) (σ(x) –a)β+
σ(x) a
f(t) tx –
l a
u(x) (σ(x) –a)β+
σ(x) a
f(t) – (σ(x) –a)
σ(x) a
f(t)t
tx. (.)
Finally, utilizing Fubini?s Theorem. and taking into account Definition . of the weight functionv, we obtain that the right-hand side of (.) equals
l
a
υ(t)
f(t) t t–a –
l
a
l
t
f(t) – (σ(x) –a)
σ(x) a
f(t)t
u(x)
(σ(x) –a)β+xt +
l
a
υ(t)
f(t) t t–a –
l
a
l
t
f(t) – (σ(x) –a)
σ(x) a
f(t)t
u(x)
(σ(x) –a)β+xt. (.) The proof for the case whenφ is subquadratic is similar except that the only inequality above holds in the reverse direction. The proof is now complete.
We now give some applications of Theorem ..
Example . If we letβ(x) = and apply Theorem . to (σ(x)–ax–a )u(x) instead ofu(x), then we get
l
a
u(x)
(σ(x) –a) σ(x)
a
f(t)t x
x–a +
l
a
l
t
f(t) – (σ(x) –a)
σ(x) a
f(t)t
u(x)
(x–a)(σ(x) –a)xt
≤ l
a
υ(x)
f(x) x
x–a. (.)
The sign of inequality (.) is reversed for the case <p(x)≤.
Remark . Inequality (.) coincides with Theorem . in [].
Now we will use the well-known fact thatφ(u) =up(x)is superquadratic forp(x)≥ and (subquadratic if <p(x)≤) in the next example.
Example . Let u(x) = andp(x)≥. By Proposition ., we get that
v(x)≤(x–a)–β(x) β(x)
– x–a
l–a β(x)
ifl<∞.
Under these conditions, inequality (.) yields l
a
(σ(x) –a)
σ(x) a
f(t)t p(x)
σ(x) –a –β(x)x +
l a
l t
f(t) – (σ(x) –a)
σ(x) a
f(t)t
p(x) u(x)
(σ(x) –a)β(x)+xt
≤ l
a
fp(x)(x) β(x)
– x–a
l–a β(x)
(x–a)–β(x)x. (.)
The sign of inequality (.) is reversed for the case <p(x)≤.
Remark . Sinceφis a non-negative function, the second term on the left-hand side of inequality (.) is non-negative.
Theorem . Let the assumptions of Theorem.be satisfied.Moreover,let u∈Crd([a,b], R)be a non-negative function such that
t l
+
nβ–
k=
(β– )(k) (k+ )
μ(x) x–a
k
u(x)(x–a)β(x)–x<∞,
and define the weight functionυby
v(t) :=
t l
+
nβ–
k=
(β– )(k) (k+ )
μ(x) x–a
k
u(x)(x–a)β(x)–x, t∈(a,b). (.)
()If the real-valued functionis a superquadratic on(a,c), <a<c≤ ∞,then ∞
l
u(x)
σ(x) –a ∞
σ(x)
f(t)t (σ(t) –a)(t–a)
(x–a)β(x)
×
+
nβ(x)–
k=
(β(x) – )(k) (k+ )
μ(x) x–a
k
x (σ(x) –a)(x–a)
+ ∞
l
t
l
+
nβ–
k=
(β– )(k) (k+ )
μ(x) x–a
k
u(x)(x–a)β–
× f(t) –
σ(x) –a ∞
σ(x)
f(t)t (σ(t) –a)(t–a)
xt
≤ ∞
l
υ(x)
f(x) x
(σ(x) –a)(x–a) (.)
holds for all-integrable functions f ∈Crd([a,b],R)such that f(x)∈(a,c).
()If the real-valued functionis subquadratic on(a,c), <a<c≤ ∞,then(.)holds in the reversed direction.
Proof Letl>b. Applying refined Jensen?s inequality (.), we find that the first term on the left-hand side of inequality (.) is not greater than
∞
l
u(x)(x–a)β–
∞
σ(x)
(f(t))t (σ(t) –a)(t–a)
×
+
nβ–
k=
(β– )(k) (k+ )
μ(x) x–a
k x
– ∞
l
u(x)(x–a)β–
+
nβ–
k=
(β– )(k) (k+ )
μ(x) x–a
k
× ∞
σ(x)
f(t) –
σ(x) –a ∞
σ(x)
f(t)t (σ(t) –a)(t–a)
tx. (.)
Finally, employing Fubini?s Theorem. and Proposition ., we obtain that the right-hand side of (.) is not greater than
∞
l
υ(t)
f(t) t
(σ(t) –a)(t–a) –
∞
l
t
l
+
nβ–
k=
(β– )(k) (k+ )
μ(x) x–a
k
u(x)(x–a)β–
× f(t) –
σ(x) –a ∞
σ(x)
f(t)t (σ(t) –a)(t–a)
xt.
Now we consider Theorem . in some special cases. First we note that if we setu(x) = , then we find that
v(x)≤(x–a)β(x) β(x)
– l–a
x–a β(x)
ifl<∞ by Proposition ..
Corollary . Let the assumptions of Theorem .be satisfied.Ifφ is non-negative su- perquadratic,then
∞
l
σ(x) –a ∞
σ(x)
f(t)t (σ(t) –a)(t–a)
(x–a)β(x)
×
+
nβ(x)–
k=
(β(x) – )(k) (k+ )
μ(x) x–a
k
x (σ(x) –a)(x–a)
+ ∞
l
t
l
+
nβ–
k=
(β– )(k) (k+ )
μ(x) x–a
k
(x–a)β–
× f(t) –
σ(x) –a ∞
σ(x)
f(t)t (σ(t) –a)(t–a)
xt
≤ ∞
l
(x–a)β(x) β(x)
– l–a
x–a β(x)
f(x) x
(σ(x) –a)(x–a). (.) Example . Assume thatT=,a= and(x) =xp. Then inequality (.) yields the inequality
∞
l
x
∞
x
f(t)dt t
p(x)
xβ(x)dx x +
∞
l
t
l
(x–a)β(x)–
f(t) –x ∞
x
f(t)dt t
p(x)
dx dt
≤ ∞
l
β(x)
f(x)p(x)xβ(x)
– l
x β
dx
x. (.)
Remark . Sinceφ is a non-negative function, the second term on the left-hand side of inequality (.) is non-negative. Hence inequality (.) provides a refinement of in- equality (.) in [] if written forl≤b.
Competing interests
The authors declare that they have no competing interests.
Authors? contributions
All the authors have contributed in all parts to equal extent. Also, all the authors read and approved the final manuscript.
Author details
1Department of Mathematics, Federal University of Agriculture, P.M.B. 2240, Abeokuta, Ogun, Nigeria.2Department of Mathematics, Obafemi Awolowo University, Ile-Ife, Osun, Nigeria.3Department of Mathematics, Luleå University of Technology, Luleå, 971 87, Sweden.4Narvik University College, P.O. Box 385, Narvik, 8505, Norway.
Acknowledgements
The authors are grateful to the careful referees for useful comments and advice that have led to the improvement of this work. The first author who is a senior associate of the Abdus Salam International Centre for Theoretical Physics (ICTP), Trieste, Italy would like to thank the Abdus Salam ICTP for support.
Received: 23 August 2014 Accepted: 17 December 2014
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