Fractional Integral Inequalities via s-Convex Functions

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by  M. Emin Ozdemir

FRACTIONAL INTEGRAL INEQUALITIES VIA s−CONVEX
FUNCTIONS
2
1 M.EMI˙NO¨ZDEMI˙RN,HAVVAKAVURMACIN,ANDC¸ETI˙NYILDIZN,⋆
0
2
Abstract. Inthispaper,weestablishseveralinequalitiesfors−convexmap-

n pingsthatareconnectedwiththeRiemann-Liouvillefractionalintegrals. Our
a resultshavesomerelationshipswithcertainintegral inequalitiesinthelitera-
J ture.

4
2

]
A 1. INTRODUCTION
C Letf :I ⊆R→RbeaconvexfunctiondefinedontheintervalI ofrealnumbers
h. and a<b. The following double inequality
t
a b
m a+b 1 f(a)+f(b)
f ≤ f(x)dx≤
2 b−a 2
[ (cid:18) (cid:19) Za

1 is well known in the literature as Hadamard’s inequality. Both inequalities hold in
v the reverseddirection if f is concave.
5 Let real function f be defined on some nonempty interval I of real line R. The
1
9 function f is said to be convex on I if inequality
4
f(tx+(1−t)y)≤tf(x)+(1−t)f(y)
.
1
0 holds for all x,y ∈I and t∈[0,1].
2 In [3], s−convex functions defined by Orlicz as following.
1
: Definition 1. A function f :R+ →R, where R+ =[0,∞), is said to be s−convex
v
in the first sense if
i
X f(αx+βy)≤αsf(x)+βsf(y)
ar for all x,y ∈ [0,∞), α,β ≥ 0 with αs+βs = 1 and for some fixed s ∈ (0,1]. We
denote by K1 the class of all s−convex functions.
s
Definition 2. A function f :R+ →R, where R+ =[0,∞), is said to be s−convex
in the second sense if
f(αx+βy)≤αsf(x)+βsf(y)
for all x,y ∈ [0,∞), α,β ≥ 0 with α+β = 1 and for some fixed s ∈ (0,1]. We
denote by K2 the class of all s−convex functions.
s
Date:January23,2012.
2000 Mathematics Subject Classification. 26A15,26A51,26D10.
Key words and phrases. Hadamard’s Inequality, Riemann-Liouville Fractional Integration,
H¨olderInequality, s−convexity.
⋆CorrespondingAuthor.
1
2 M.EMI˙NO¨ZDEMI˙RN,HAVVAKAVURMACIN,ANDC¸ETI˙NYILDIZN,⋆
Orlicz defined these class of functions in [3] and these definitions was used in
the theory of Orlicz spaces in [4] and [5]. Obviously, one can see that if we choose
s=1, both definitions reduced to ordinary concept of convexity.
Forseveralresultsrelatedto abovedefinitions wereferreadersto[2],[6],[7]and
[8].
In [6], Hadamard’s inequality which for s−convex functions in the second sence
is proved by Dragomir and Fitzpatrick.
Theorem 1. Suppose that f : [0,∞) → [0,∞) is an s−convex function in the
second sence, where s∈(0,1) and let a,b∈[0,∞), a<b. If f ∈L ([a,b]), then the
1
following inequalities hold:
b
a+b 1 f(a)+f(b)
(1.1) 2s−1f ≤ f(x)dx≤ .
2 b−a s+1
(cid:18) (cid:19) Z
a
The constant k = 1 is the best possible in the second inequality in (1.1).
s+1
In [7], Kırmacı et al. obtained Hadamard type inequalities which holds for
s−convex functions in the second sence. It is given in the next theorem.
Theorem 2. Let f : I →R, I ⊂[0,∞), be differentiable function on I◦ such that
f′ ∈ L ([a,b]), where a,b ∈ I, a < b. If |f′|q is s−convex on [a,b] for some fixed
1
s∈(0,1) and q ≥1, then:
(1.2)
f(a)+f(b) − 1 b f(x)dx ≤ b−a 1 q−q1 s+ 21 s q1 |f′(a)|q+|f′(b)|q q1 .
(cid:12) 2 b−a (cid:12) 2 2 (s+1)(s+2)
(cid:12)(cid:12) Za (cid:12)(cid:12) (cid:18) (cid:19) " (cid:0) (cid:1) # (cid:2) (cid:3)
(cid:12) (cid:12)
(cid:12) In [1], Dragomir and Agarw(cid:12)alproved the following inequality.
(cid:12) (cid:12)
Theorem 3. Let f : I◦ ⊆ R → R be a differentiable mapping on I◦,a,b ∈ I◦
with a < b, and let p > 1. If the new mapping |f′|p−p1 is convex on [a,b], then the
£ollowing inequality holds:
f(a)+f(b) 1 b b−a |f′(a)|p−p1 +|f′(b)|p−p1 p−p1
(1.3) − f(x)dx ≤ .
(cid:12)(cid:12) 2 b−aZ (cid:12)(cid:12) 2(p+1)p1 " 2 #
(cid:12) a (cid:12)
In [1(cid:12)(cid:12)2], Set et al. proved the follo(cid:12)(cid:12)wing Hadamard type inequality for s−convex
(cid:12) (cid:12)
functions in the second sense via Riemann-Liouville fractional integral.
Theorem 4. Let f :[a,b]⊂[0,∞)→R be a differentiable mapping on (a,b) with
a < b such that f′ ∈ L[a,b]. If |f′|q is s−convex in the second sense on [a,b] for
some fixed s∈(0,1] and q ≥1, then the following inequality for fractional integrals
holds
(1.4)
f(a)+f(b) Γ(α+1)
− [Jα f(b)+Jα f(a)]
2 2(b−a)α a+ b−
(cid:12) (cid:12)
(cid:12)(cid:12)b−a 2 1 1−q1 (cid:12)(cid:12)
≤ (cid:12) 1− (cid:12)
2 α+1 2α
(cid:20) (cid:18) (cid:19)(cid:21)
1 1 2α+s−1 q1 ′ q ′ q 1
× β ,s+1,α+1 −β ,α+1,s+1 + |f (a)| +|f (b)| q .
2 2 (α+s+1)2α+s
(cid:20) (cid:18) (cid:19) (cid:18) (cid:19) (cid:21)
(cid:0) (cid:1)
FRACTIONAL INEQUALITIES 3
Now,wegivesomenecessarydefinitions andmathematicalpreliminariesoffrac-
tional calculus theory which are used throughout this paper, see([9]).
Definition 3. Let f ∈L [a,b]. The Riemann-Liouville integrals Jα f and Jα f of
1 a+ b−
order α>0 with a≥0 are defined by
x
Jα f(x)= 1 (x−t)α−1f(t)dt, x>a
a+ Γ(α)
Z
a
and
b
Jα f(x)= 1 (t−x)α−1f(t)dt, x<b
b− Γ(α)
Z
x
∞
respectively where Γ(α)= e−uuα−1du. Here is J0 f(x)=J0 f(x)=f(x).
a+ b−
0
R
In the case of α=1, the fractionalintegralreduces to the classicalintegral. For
some recent results connected with fractional integral inequalities see ([10]-[17]).
In order to prove our main theorems, we need the following lemma:
Lemma 1. (see [18]) Let f : I ⊂ R → R be a differentiable mapping on I with
a < r, a,r ∈ I. If f′ ∈ L[a,r], then the following equality for fractional integral
holds:
f(a)+f(r) Γ(α+1)
− [Jα f(r)+Jα f(a)]
2 2(r−a)α a+ r−
r−a 1
= [(1−t)α−tα]f′(r+(a−r)t)dt.
2
Z0
2. MAIN RESULTS
Theorem 5. Let f : [a,b] ⊂ [0,∞) → R be a differentiable mapping on (a,b)
with a < r ≤ b such that f′ ∈ L[a,b]. If |f′| is s−convex on [a,b] for some fixed
s∈(0,1], then the following inequality for fractional integrals holds
f(a)+f(r) Γ(α+1)
− [Jα f(r)+Jα f(a)]
2 2(r−a)α a+ r−
(cid:12) (cid:12)
(cid:12)r−a 1 1 (cid:12) 2α+s−1
(cid:12) (cid:12) ′ ′
≤ (cid:12) β ,s+1,α+1 −β ,α+1,s+(cid:12)1 + [|f (a)|+|f (r)|].
2 2 2 (α+s+1)2α+s
(cid:20) (cid:18) (cid:19) (cid:18) (cid:19) (cid:21)
Proof. From Lemma 1 and using the properties of modulus, we get
f(a)+f(r) Γ(α+1)
− [Jα f(r)+Jα f(a)]
2 2(r−a)α a+ r−
(cid:12) (cid:12)
≤ (cid:12)(cid:12)(cid:12)r−a 1|(1−t)α−tα||f′(r+(a−r)t)|dt. (cid:12)(cid:12)(cid:12)
2
Z0
Since |f′| is s−convex on [a,b], we obtain inequality
|f′(r+(a−r)t)| =|f′(ta+(1−t)r)|≤ts|f′(a)|+(1−t)s|f′(r)|, t∈(0,1).
4 M.EMI˙NO¨ZDEMI˙RN,HAVVAKAVURMACIN,ANDC¸ETI˙NYILDIZN,⋆
Hence,
f(a)+f(r) Γ(α+1)
− [Jα f(r)+Jα f(a)]
2 2(r−a)α a+ r−
(cid:12) (cid:12)
(cid:12) 1 (cid:12)
≤ (cid:12)(cid:12)r−a 2 [(1−t)α−tα][ts|f′(a)|+(1−t)s(cid:12)(cid:12)|f′(r)|]dt
2
(Z0
1
+ [tα−(1−t)α][ts|f′(a)|+(1−t)s|f′(r)|]dt
1 )
Z2
and
21 1 1
ts(1−t)αdt = (1−t)stαdt=β ;s+1,α+1 ,
Z0 Z21 (cid:18)2 (cid:19)
12 1 1
(1−t)stαdt = ts(1−t)αdt=β ;α+1,s+1 ,
Z0 Z21 (cid:18)2 (cid:19)
12 1 1
ts+αdt = (1−t)s+αdt= ,
Z0 Z21 2s+α+1(s+α+1)
12 1 1 1
(1−t)s+αdt= ts+αdt= − .
Z0 Z12 s+α+1 2s+α+1(s+α+1)
We obtain
f(a)+f(r) Γ(α+1)
− [Jα f(r)+Jα f(a)]
2 2(r−a)α a+ r−
(cid:12) (cid:12)
(cid:12)r−a 1 1 (cid:12) 2α+s−1
(cid:12) (cid:12) ′ ′
≤ (cid:12) β ,s+1,α+1 −β ,α+1,s+(cid:12)1 + [|f (a)|+|f (r)|].
2 2 2 (α+s+1)2α+s
(cid:20) (cid:18) (cid:19) (cid:18) (cid:19) (cid:21)
(cid:3)
Theorem 6. Let f :[a,b]⊂[0,∞)→R be a differentiable mapping on (a,b) with
a < r ≤ b such that f′ ∈ L[a,b]. If |f′|q is s−convex in the second sense on [a,b]
for some fixed s∈(0,1] and q >1 with 1 + 1 =1, then the following inequality for
p q
fractional integrals holds
f(a)+f(r) Γ(α+1)
− [Jα f(r)+Jα f(a)]
2 2(r−a)α a+ r−
(cid:12) (cid:12)
(cid:12)(cid:12)r−a 1 p1 |f′(a)|q+|f′(r)|q q1 (cid:12)(cid:12)
≤ (cid:12) (cid:12)
2 αp+1 s+1
(cid:18) (cid:19) (cid:18) (cid:19)
where α∈[0,1].
FRACTIONAL INEQUALITIES 5
Proof. By Lemma 1 and using Ho¨lder inequality with the properties of modulus,
we have
f(a)+f(r) Γ(α+1)
− [Jα f(r)+Jα f(a)]
2 2(r−a)α a+ r−
(cid:12) (cid:12)
≤ (cid:12)(cid:12)(cid:12)r−a 1|(1−t)α−tα||f′(r+(a−r)t)|dt (cid:12)(cid:12)(cid:12)
2
Z0
1 1
1 p 1 q
r−a
≤ |(1−t)α−tα|pdt |f′(r+(a−r)t)|qdt .
2    
Z Z
0 0
   
We know that for α∈[0,1] and ∀t ,t ∈[0,1],
1 2
|tα−tα|≤|t −t |α,
1 2 1 2
therefore
1 1
|(1−t)α−tα|pdt ≤ |1−2t|αpdt
Z Z
0 0
12 1
= [1−2t]αpdt+ [2t−1]αpdt
Z Z
0 1
2
1
= .
αp+1
Since |f′|q is s−convex on [a,b],we get
f(a)+f(r) Γ(α+1)
− [Jα f(r)+Jα f(a)]
2 2(r−a)α a+ r−
(cid:12) (cid:12)
(cid:12) (cid:12) 1
(cid:12) 1 1 (cid:12) q
≤ (cid:12)r−a 1 p ts|f′(a)|q+(1−t)s|f(cid:12)′(r)|q dt
2 αp+1  
(cid:18) (cid:19) Z
0 (cid:2) (cid:3)
r−a 1 p1 |f′(a)|q+|f′(r)|q q1 
=
2 αp+1 s+1
(cid:18) (cid:19) (cid:18) (cid:19)
which completes the proof. (cid:3)
Corollary 1. If in Theorem 6, we choose r =b then, we have
f(a)+f(b) Γ(α+1)
(2.1) − [Jα f(b)+Jα f(a)]
2 2(b−a)α a+ b−
(cid:12) (cid:12)
(cid:12)(cid:12)b−a 1 p1 |f′(a)|q+|f′(b)|q q1 (cid:12)(cid:12)
≤ (cid:12) . (cid:12)
2 αp+1 s+1
(cid:18) (cid:19) (cid:18) (cid:19)
Remark 1. If we choose α=1 ve s=1 in Corollary 6 then, we have
f(a)+f(b) 1 b b−a |f′(a)|q+|f′(b)|q 1q
− f(x)dx ≤
(cid:12)(cid:12)(cid:12) 2 b−aZa (cid:12)(cid:12)(cid:12) 2(p+1)p1 (cid:20) 2 (cid:21)
(cid:12) (cid:12)
which is(cid:12)the inequality in (1.3). (cid:12)
(cid:12) (cid:12)
6 M.EMI˙NO¨ZDEMI˙RN,HAVVAKAVURMACIN,ANDC¸ETI˙NYILDIZN,⋆
Theorem 7. Let f :[a,b]⊂[0,∞)→R be a differentiable mapping on (a,b) with
a<r ≤b such that f′ ∈L[a,b]. If |f′|q is s−convex in the second sense on [a,b] for
some fixed s∈(0,1] and q ≥1, then the following inequality for fractional integrals
holds
f(a)+f(r) Γ(α+1)
− [Jα f(r)+Jα f(a)]
2 2(r−a)α a+ r−
(cid:12) (cid:12)
(cid:12)(cid:12)r−a 2 1 1−1q (cid:12)(cid:12)
≤ (cid:12) 1− (cid:12)
2 α+1 2α
(cid:20) (cid:18) (cid:19)(cid:21)
1 1 2α+s−1 q1 ′ q ′ q 1
× β ,s+1,α+1 −β ,α+1,s+1 + |f (a)| +|f (r)| q .
2 2 (α+s+1)2α+s
(cid:20) (cid:18) (cid:19) (cid:18) (cid:19) (cid:21)
(cid:0) (cid:1)
Proof. From Lemma 1 and using the well-known power mean inequality with the
properties of modulus, we have
f(a)+f(r) Γ(α+1)
− [Jα f(r)+Jα f(a)]
2 2(r−a)α a+ r−
(cid:12) (cid:12)
≤ (cid:12)(cid:12)(cid:12)r−a 1|(1−t)α−tα||f′(r+(a−r)t)|dt (cid:12)(cid:12)(cid:12)
2
Z0
1 1−1q 1 1q
r−a
≤ |(1−t)α−tα|dt |(1−t)α−tα||f′(r+(a−r)t)|qdt
2    
Z Z
0 0
   
On the other hand, we have
1 1 1
2
|(1−t)α−tα|dt = [(1−t)α−tα]dt+ [tα−(1−t)α]dt
Z0 Z0 Z12
2 1
= 1− .
α+1 2α
(cid:18) (cid:19)
Since |f′|q is s−convex on [a,b], we obtain
|f′(r+(a−r)t)|q =|f′(ta+(1−t)r)|q ≤ts|f′(a)|q +(1−t)s|f′(r)|q, t∈(0,1)
and
1 1
|(1−t)α−tα||f′(r+(a−r)t)|qdt ≤ |(1−t)α−tα| ts|f′(a)|q+(1−t)s|f′(r)|q dt
Z0 Z0
1 (cid:2) (cid:3)
= 2 [(1−t)α−tα] ts|f′(a)|q+(1−t)s|f′(r)|q dt
Z0
1 (cid:2) (cid:3)
+ [tα−(1−t)α] ts|f′(a)|q +(1−t)s|f′(r)|q dt
1
Z2
(cid:2) (cid:3)
FRACTIONAL INEQUALITIES 7
Since
12 1 1
ts(1−t)αdt = (1−t)stαdt=β ;s+1,α+1 ,
Z0 Z21 (cid:18)2 (cid:19)
12 1 1
(1−t)stαdt = ts(1−t)αdt=β ;α+1,s+1 ,
Z0 Z21 (cid:18)2 (cid:19)
12 1 1
ts+αdt = (1−t)s+αdt=
Z0 Z21 2s+α+1(s+α+1)
and
21 1 1 1
(1−t)s+αdt= ts+αdt= − .
Z0 Z12 s+α+1 2s+α+1(s+α+1)
Therefore, we have
f(a)+f(r) Γ(α+1)
− [Jα f(r)+Jα f(a)]
2 2(r−a)α a+ r−
(cid:12) (cid:12)
(cid:12)(cid:12)r−a 2 1 1−1q (cid:12)(cid:12)
≤ (cid:12) 1− (cid:12)
2 α+1 2α
(cid:20) (cid:18) (cid:19)(cid:21)
1 1 2α+s−1 ′ q ′ q q1
× β ,s+1,α+1 −β ,α+1,s+1 − |f (a)| +|f (r)| .
2 2 (α+s+1)2α+s
(cid:26)(cid:20) (cid:18) (cid:19) (cid:18) (cid:19) (cid:21) (cid:27)
(cid:0) (cid:1)
(cid:3)
Remark 2. If we choose r =b in Theorem 7, we obtain the inequality in (1.4) of
Theorem 4.
Remark 3. If we choose r =b and α= 1 in Theorem 7, we obtain the inequality
in (1.2) of Theorem 2.
Theorem 8. Let f :[a,b]⊂[0,∞)→R be a differentiable mapping on (a,b) with
a < r ≤ b such that f′ ∈ L[a,b]. If |f′|q is s−concave on [a,b] and q > 1 with
1 + 1 =1, then the following inequality for fractional integrals holds
p q
f(a)+f(r) Γ(α+1)
− [Jα f(r)+Jα f(a)]
2 2(r−a)α a+ r−
(cid:12) (cid:12)
(cid:12) 1 (cid:12)
≤ (cid:12)(cid:12)r−a 1 p f′ a+r . (cid:12)(cid:12)
22−qs (cid:18)αp+1(cid:19) (cid:12) (cid:18) 2 (cid:19)(cid:12)
(cid:12) (cid:12)
Proof. From Lemma 1 and using Ho¨lde(cid:12)r inequality,(cid:12)we have
(cid:12) (cid:12)
f(a)+f(r) Γ(α+1)
− [Jα f(r)+Jα f(a)]
2 2(r−a)α a+ r−
(cid:12) (cid:12)
≤ (cid:12)(cid:12)(cid:12)r−a 1|(1−t)α−tα||f′(r+(a−r)t)|dt (cid:12)(cid:12)(cid:12)
2
Z0
1 1
1 p 1 q
≤ r−a |(1−t)α−tα|pdt |f′(r+(a−r)t)|qdt .
2    
Z Z
0 0
   
8 M.EMI˙NO¨ZDEMI˙RN,HAVVAKAVURMACIN,ANDC¸ETI˙NYILDIZN,⋆
Since |f′|q is s−concave on [a,b],we get
1
a+r q
|f′(r+(a−r)t)|qdt≤2s−1 f′ ,
2
Z (cid:12) (cid:18) (cid:19)(cid:12)
0 (cid:12) (cid:12)
so (cid:12) (cid:12)
(cid:12) (cid:12)
f(a)+f(r) Γ(α+1)
− [Jα f(r)+Jα f(a)]
2 2(r−a)α a+ r−
(cid:12) (cid:12)
(cid:12) 1 (cid:12)
≤ (cid:12)(cid:12)r−a 1 p f′ a+r . (cid:12)(cid:12)
22−qs (cid:18)αp+1(cid:19) (cid:12) (cid:18) 2 (cid:19)(cid:12)
which completes the proof. (cid:12)(cid:12) (cid:12)(cid:12) (cid:3)
(cid:12) (cid:12)
Corollary 2. If we choose r =b in Theorem 8, we obtain
f(a)+f(b) Γ(α+1)
(2.2) − [Jα f(b)+Jα f(a)]
2 2(b−a)α a+ b−
(cid:12) (cid:12)
(cid:12) 1 (cid:12)
≤ (cid:12)(cid:12)b−a 1 p f′ a+b . (cid:12)(cid:12)
22−qs (cid:18)αp+1(cid:19) (cid:12) (cid:18) 2 (cid:19)(cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
Refe(cid:12)rences (cid:12)
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NATATU¨RKUNIVERSITY,K.K.EDUCATIONFACULTY, DEPARTMENTOFMATH-
EMATICS,25240,CAMPUS,ERZURUM, TURKEY
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