Hadamard Type Inequalities for m-Convex and (α,m)-Convex Functions via Fractional Integrals

0.14 MB·English

by  M. Emin Özdemir

HADAMARD TYPE INEQUALITIES FOR m−CONVEX AND
(α,m)−CONVEX FUNCTIONS VIA FRACTIONAL INTEGRALS
2
1 M.EMINO¨ZDEMIR(cid:7),MERVEAVCI♣,⋆,AHMETOCAKAKDEMIR♠,
0 AND♠ALPEREKINCI
2

n Abstract. In this paper, we established some new Hadamard-type integral
a inequalities for functions whose derivatives of absolute values are m−convex
J and(α,m)−convex functionsviaRiemann-Liouvillefractionalintegrals.

7
2

]
A
1. Introduction
F
. Letf :I ⊂R→RbeaconvexfunctiondefinedontheintervalI ofrealnumbers
h
t and a,b∈I with a<b. The following inequality is well known in the literature as
a
the Hermite–Hadamard inequality:
m
[ a+b 1 b f(a)+f(b)
f ≤ f(x)dx≤ .

1 (cid:18) 2 (cid:19) b−aZa 2
v
3 In [1], G.Toader defined the concept of m-convexity as the following:
8
7 Definition1. Thefunction f :[0,b]→Ris saidtobem−convex,wherem∈[0,1],
5 if for every x,y ∈[0,b] and t∈[0,1] we have:
.
1
f(tx+m(1−t)y)≤tf(x)+m(1−t)f(y).
0
2
1 Denote by Km(b) the set of the m−convex functions on [0,b] for which f(0)≤0.
:
v Severalpapershavebeenwrittenonm−convexfunctionsandwereferthepapers
i
X [2]-[11].
In [11], the following inequality of Hermite-Hadamard type for m−convex func-
r
a tions holds:
Theorem 1. Let f : [0,∞) → R be a m−convex function with m ∈ (0,1]. If
0≤a<b<∞ and f ∈L [a,b], then one has the inequality:
1
1 b f(a)+mf(b) f(b)+mf(a)
(1.1) f(x)dx≤min m , m .
b−a 2 2
Za ( )
In [19], S.S. Dragomir proved the following theorem.
Key words and phrases. m−convex functions, (α,m)−convex functions, Riemann-Liouville
fractionalintegral.
⋆CorrespondingAuthor.
1
2M.EMINO¨ZDEMIR(cid:7),MERVEAVCI♣,⋆,AHMETOCAKAKDEMIR♠,AND♠ALPEREKINCI
Theorem 2. Let f : [0,∞) → R be a m−convex function with m ∈ (0,1]. If
f ∈L [am,b] where 0≤a<b, then one has the inequality:
1
1 1 mb 1 b
(1.2) f(x)dx+ f(x)dx
m+1 mb−a b−ma
" Za Zma #
f(a)+f(b)
≤ .
2
In [18], Mihe¸san gave definition of (α,m)−convexity as following;
Definition 2. The function f : [0,b] → R, b > 0 is said to be (α,m)−convex,
where (α,m)∈[0,1]2, if we have
f(tx+m(1−t)y)≤tαf(x)+m(1−tα)f(y)
for all x,y ∈[0,b] and t∈[0,1].
Denote by Kα(b) the class of all (α,m)−convex functions on [0,b] for which
m
f(0)≤0. If we choose (α,m)=(1,m), it can be easily seen that (α,m)−convexity
reducestom−convexityandfor(α,m)=(1,1), wehaveordinaryconvexfunctions
on[0,b]. For the recentresults based onthe above definition see the papers [2], [9],
[6], [7], [4], and [5].
We give some necessarydefinitions andmathematicalpreliminaries of fractional
calculus theory which are used throughout this paper.
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
1 x
Jα f(x)= (x−t)α−1f(t)dt, x>a
a+ Γ(α)
Za
and
1 b
Jα f(x)= (t−x)α−1f(t)dt, x<b
b− Γ(α)
Zx
where Γ(α)= ∞e−tuα−1du, here is J0 f(x)=J0 f(x)=f(x).
0 a+ b−
In the caseRof α = 1, the fractional integral reduces to the classical integral.
Properties of this operator can be found in the references [12]-[14].
In[15],Sarikayaetal. provedavariantoftheidentityisestablishedbyDragomir
and Agarwal in [16, Lemma 2.1] for fractional integrals as the following.
Lemma 1. Let f : [a,b] → R be a differentiable mapping on (a,b) with a < b. If
f′ ∈L[a,b], then the following equality for fractional integrals holds:
f(a)+f(b) Γ(α+1)
− [Jα f(b)+Jα f(a)]
2 2(b−a)α a+ b−
b−a 1
= [(1−t)α−tα]f′(ta+(1−t)b)dt.
2
Z0
The aim of this paper is to establish Hadamard type inequalities for m−convex
and (α,m)−convex functions via Riemann-Liouville fractional integrals.
INEQUALITIES VIA FRACTIONAL INTEGRALS 3
2. Inequalities for m−convex functions
Theorem 3. Let f : [a,b] → R be a positive function with 0 ≤ a < b and f ∈
L [a,b]. If f is a m−convex function on [a,b], then the following inequalities for
1
fractional integrals with α>0 and m∈(0,1] hold:
(2.1)
Γ(α) f(a) b Γ(α)Γ(2)
Jα f(b) ≤ +mf ,
(b−a)α a+ α+1 m Γ(α+2)
(cid:18) (cid:19)
Γ(α) f(b) a Γ(α)Γ(2)
Jα f(a) ≤ +mf .
(b−a)α b− α+1 m Γ(α+2)
(cid:16) (cid:17)
where Γ is Euler Gamma function.
Proof. Since f is a m−convex function on [a,b], we know that for any t∈[0,1]
b
(2.2) f(ta+(1−t)b)≤tf(a)+m(1−t)f
m
(cid:18) (cid:19)
and
a
(2.3) f(tb+(1−t)a)≤tf(b)+m(1−t)f .
m
By multiplying both sides of (2.2) by tα−1, then by int(cid:16)egra(cid:17)ting the resulting in-
equality with respect to t over [0,1], we obtain
1 1 b
tα−1f(ta+(1−t)b)dt ≤ tα−1 tf(a)+m(1−t)f
m
Z0 Z0 (cid:20) (cid:18) (cid:19)(cid:21)
f(a) b
= +mf β(α,2).
α+1 m
(cid:18) (cid:19)
It is easy to see that 1tα−1f(ta+(1−t)b)dt = Γ(α) Jα f(b) and we note that,
0 (b−a)α a+
the Beta and the Gamma function (see [17, pp 908-910])
R
1 ∞
β(x,y)= tx−1(1−t)y−1dt, x,y >0, Γ(x)= e−ttx−1dt, x>0
Z0 Z0
are used to evaluate the integral
1
tα−1(1−t)dt,
Z0
where
Γ(x)Γ(y)
β(x,y)= ,
Γ(x+y)
thus we can obtain that
Γ(α)Γ(2)
β(α,2)=
Γ(α+2)
which completes the proof.
For the proofofthe secondinequality in (2.1) wemultiply both sides of(2.3) by
tα−1, then integrate the resulting inequality with respect to t over [0,1]. (cid:3)
Remark 1. If we choose α = 1 in Theorem 3, then the inequalities (2.1) become
the inequality in (1.1).
4M.EMINO¨ZDEMIR(cid:7),MERVEAVCI♣,⋆,AHMETOCAKAKDEMIR♠,AND♠ALPEREKINCI
Theorem 4. Let f : I◦ ⊂ [0,∞) → R, be a differentiable function on I◦ such
that f′ ∈ L[a,b] where a,b ∈ I,a < b. If |f′|q is m−convex on [a,b] for some fixed
m∈(0,1] and q ≥1, then the following inequality for fractional integrals holds:
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)(cid:12)b−a21−q1 2α−1 |f′(a)|q +m f′ b (cid:12)(cid:12)(cid:12) q 1q .
2 2α(α+1) m
(cid:20) (cid:21)(cid:20) (cid:12) (cid:18) (cid:19)(cid:12) (cid:21)
(cid:12) (cid:12)
Proof. Supposethatq =1.FromLemma1andbyusing(cid:12)theprope(cid:12)rtiesofmodulus,
(cid:12) (cid:12)
we have
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)(cid:12)b−a 1|(1−t)α−tα||f′(ta+(1−t)b)|dt. (cid:12)(cid:12)(cid:12)
2
Z0
Since |f′| is m−convex on [a,b], we have
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)(cid:12)b−a 1|(1−t)α−tα| t|f′(a)|+m(1−t) f(cid:12)(cid:12)(cid:12)′ b dt
2 m
Z0 (cid:20) (cid:12) (cid:18) (cid:19)(cid:12)(cid:21)
1 (cid:12) (cid:12)
= b−a 2 [(1−t)α−tα] t|f′(a)|+m(1−(cid:12)(cid:12)t) f′ b(cid:12)(cid:12) dt
2 m
(Z0 (cid:20) (cid:12) (cid:18) (cid:19)(cid:12)(cid:21)
(cid:12) (cid:12)
1 b(cid:12) (cid:12)
+ [tα−(1−t)α] t|f′(a)|+m(1−t) f′ (cid:12) dt (cid:12)
m
Z21 (cid:20) (cid:12) (cid:18) (cid:19)(cid:12)(cid:21) )
(cid:12) (cid:12)
b−a 2α−1 ′ ′ b (cid:12) (cid:12)
= |f (a)|+m f (cid:12) (cid:12)
2 2α(α+1) m
(cid:20) (cid:21)(cid:20) (cid:12) (cid:18) (cid:19)(cid:12)(cid:21)
(cid:12) (cid:12)
(cid:12) (cid:12)
where we use the facts that (cid:12) (cid:12)
12 (1−t)αtdt= 1tα(1−t)dt= 1 − α+3 ,
Z0 Z21 (α+1)(α+2) 2α+2(α+1)(α+2)
21 tα+1dt= 1(1−t)α+1dt= 1 ,
Z0 Z21 2α+2(α+2)
21 (1−t)α+1dt= 1tα+1dt= 1 − 1
Z0 Z12 (α+2) 2α+2(α+2)
and
12 tα(1−t)dt= 1(1−t)αtdt= α+3
Z0 Z21 2α+2(α+1)(α+2)
INEQUALITIES VIA FRACTIONAL INTEGRALS 5
which completes the proof for this case. Suppose now that q >1. From Lemma 1,
m−convexityof|f′|q andusing the well-knownHo¨lder’s inequality we havesucces-
sively
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)(cid:12)b−a 1|(1−t)α−tα||f′(ta+(1−t)b)|dt (cid:12)(cid:12)(cid:12)
2
Z0
= b−a 1|(1−t)α−tα|1−q1 |(1−t)α−tα|q1 |f′(ta+(1−t)b)|dt
2
Z0
q−1 1
b−a 1 q 1 q
≤ |(1−t)α−tα|dt |(1−t)α−tα||f′(ta+(1−t)b)|qdt
2
(cid:18)Z0 (cid:19) (cid:18)Z0 (cid:19)
≤ b−a21−q1 2α−1 |f′(a)|q+m f′ b q q1
2 2α(α+1) m
(cid:20) (cid:21)(cid:20) (cid:12) (cid:18) (cid:19)(cid:12) (cid:21)
(cid:12) (cid:12)
where we use the fact that (cid:12) (cid:12)
1 1 (cid:12) (cid:12) 1
|(1−t)α−tα|dt = 2 [(1−t)α−tα]dt+ [tα−(1−t)α]dt
Z0 Z0 Z12
2 1
= 1− .
α+1 2α
(cid:18) (cid:19)
(cid:3)
Corollary 1. Under the assumptions of Theorem 4 with α ∈ (0,1], the following
inequality holds:
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)(cid:12)b−a 1 p1 |f′(a)|q+m f′ mb q q1(cid:12)(cid:12)(cid:12).
2 αp+1 2
(cid:18) (cid:19) (cid:12) (cid:0) (cid:1)(cid:12) !
(cid:12) (cid:12)
Proof. It is similar the proof of Theorem 4. In addition, we used the following
inequality
|tα−tα|≤|t −t |α,
1 2 1 2
where α∈(0,1] and t ,t ∈[0,1]. (cid:3)
1 2
Theorem 5. Let f : [0,∞) → R be a m−convex function with m ∈ (0,1]. If
f ∈L [am,b], where 0≤a<b, then the following inequality for fractional integrals
1
with α>0 holds:
(2.4)
Γ(α) 1 1
[Jα f(mb)+Jα f(mb)]+ [Jα f(mb)+Jα f(mb)]
m+1 (mb−a)α a+ mb− (b−ma)α ma+ b−
(cid:26) (cid:27)
f(a)+f(b)
≤ .
α
Proof. By the m−convexity of f we can write
f(ta+m(1−t)b)≤tf(a)+m(1−t)f(b),
f((1−t)a+mtb)≤(1−t)f(a)+mtf(b),
6M.EMINO¨ZDEMIR(cid:7),MERVEAVCI♣,⋆,AHMETOCAKAKDEMIR♠,AND♠ALPEREKINCI
f(tb+m(1−t)a)≤tf(b)+m(1−t)f(a)
and
f((1−t)b+mta)≤(1−t)f(b)+mtf(a)
for all t∈[0,1] and 0≤a<b.
If we add the above inequalities we get
(2.5)
f(ta+m(1−t)b)+f((1−t)a+mtb)+f(tb+m(1−t)a)+f((1−t)b+mta)
≤ (m+1)(f(a)+f(b)).
Bymultiplyingbothsidesof(2.5)withtα−1,thenintegratingtheresultinginequal-
ity with respect to t over [0,1], we obtain
1
tα−1[f(ta+m(1−t)b)+f((1−t)a+mtb)+f(tb+m(1−t)a)+f((1−t)b+mta)]dt
Z0
1
≤ (m+1)(f(a)+f(b)) tα−1dt.
Z0
It is easy to see that
1 Γ(α)
tα−1f(ta+m(1−t)b)dt= Jα f(mb),
(mb−a)α a+
Z0
1 Γ(α)
tα−1f((1−t)a+mtb)dt= Jα f(mb),
(mb−a)α mb−
Z0
1 Γ(α)
tα−1f(tb+m(1−t)a)dt= Jα f(mb)
(b−ma)α b−
Z0
and
1 Γ(α)
tα−1f((1−t)b+mta)dt= Jα f(mb).
(b−ma)α ma+
Z0
So the proof is completed. (cid:3)
Remark 2. If we choose α = 1 in Theorem 5, then the inequality (2.4) becomes
the inequality in (1.2).
3. Inequalities for (α,m)−convex functions
Theorem 6. Let f : [a,b] → R be a positive function with 0 ≤ a < b and f ∈
L [a,b]. If f is an (α ,m)−convex function on [a,b], then the following inequalities
1 1
hold for fractional integrals with α>0 and (α ,m)∈(0,1]2:
1
(3.1)
Γ(α) 1 mα b
Jα f(b) ≤ f(a)+ 1 f ,
(b−a)α a+ α+α α(α+α ) m
1 1 (cid:18) (cid:19)
Γ(α) 1 mα a
Jα f(a) ≤ f(b)+ 1 f .
(b−a)α b− α+α α(α+α ) m
1 1
(cid:16) (cid:17)
Proof. Since f is (α ,m)−convexfunction on [a,b], we know that for any t∈[0,1]
1
b
(3.2) f(ta+(1−t)b)≤tα1f(a)+m(1−tα1)f
m
(cid:18) (cid:19)
INEQUALITIES VIA FRACTIONAL INTEGRALS 7
and
a
(3.3) f(tb+(1−t)a)≤tα1f(b)+m(1−tα1)f .
m
(cid:16) (cid:17)
By multiplying both sides of (3.2) by tα−1, then by integrating the resulting in-
equality with respect to t over [0,1], we get
1 1 b
tα−1f(ta+(1−t)b)dt ≤ tα−1 tα1f(a)+m(1−tα1)f
m
Z0 Z0 (cid:20) (cid:18) (cid:19)(cid:21)
1 mα b
= f(a)+ 1 f .
α+α α(α+α ) m
1 1 (cid:18) (cid:19)
It is easy to see that 1tα−1f(ta+(1−t)b)dt= Γ(α) Jα f(b), by using this fact
0 (b−a)α a+
the proof of the first inequality is completed.
R
Similarly, by multiplying both sides of (3.3) by tα−1, then by integration, the
proof of the second inequality is completed. (cid:3)
Corollary 2. If we choose α = α with α,α ∈ (0,1] in Theorem 6, we have the
1 1
following inequalities;
Γ(α) 1 b
Jα f(b) ≤ f(a)+mf ,
(b−a)α a+ 2α m
(cid:20) (cid:18) (cid:19)(cid:21)
Γ(α) 1 a
Jα f(a) ≤ f(b)+mf .
(b−a)α b− 2α m
h (cid:16) (cid:17)i
Remark 3. If we choose α = α = 1 in Theorem 6, then the inequalities reduces
1
to the inequality (1.1).
Theorem 7. Let f :I◦ ⊂[0,∞)→R, be a differentiable function on I◦ such that
f′ ∈L[a,b] where a,b∈I, a<b. If |f′|q is an (α ,m)−convex function and |f′| is
1
decreasing on [a,b], then the following inequalities hold for fractional integrals with
α>0, (α ,m)∈(0,1]2 ;
1
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)(cid:12)b−a 2α−1 q−q1 2α+α1 −1 (cid:12)(cid:12)(cid:12) |f′(a)|q−m f′ b q
2 (cid:20)2α−1(α+1)(cid:21) (cid:20)(cid:18)2α+α1(α+α1+1)(cid:19)(cid:18) (cid:12) (cid:18)m(cid:19)(cid:12) (cid:19)
m b q 1 q1 (cid:12)(cid:12) (cid:12)(cid:12)
+ f′ 1− . (cid:12) (cid:12)
α+1 m 2α
(cid:12) (cid:18) (cid:19)(cid:12) (cid:18) (cid:19)(cid:21)
(cid:12) (cid:12)
Proof. Suppose(cid:12)thatq =(cid:12)1.FromLemma1andbyusingthepropertiesofmodulus,
(cid:12) (cid:12)
we have
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)(cid:12)b−a 1|(1−t)α−tα||f′(ta+(1−t)b)|dt. (cid:12)(cid:12)(cid:12)
2
Z0
8M.EMINO¨ZDEMIR(cid:7),MERVEAVCI♣,⋆,AHMETOCAKAKDEMIR♠,AND♠ALPEREKINCI
Since |f′| is (α ,m)−convex on [a,b], we have
1
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)(cid:12)b−a 1|(1−t)α−tα| tα|f′(a)|+m(1−tα(cid:12)(cid:12)(cid:12)) f′ b dt
2 m
Z0 (cid:20) (cid:12) (cid:18) (cid:19)(cid:12)(cid:21)
1 (cid:12) (cid:12)
= b−a 2 [(1−t)α−tα] tα|f′(a)|+m(1−tα(cid:12)(cid:12)) f′ b(cid:12)(cid:12) dt
2 m
(Z0 (cid:20) (cid:12) (cid:18) (cid:19)(cid:12)(cid:21)
(cid:12) (cid:12)
1 b(cid:12) (cid:12)
+ [tα−(1−t)α] tα|f′(a)|+m(1−tα) f′ (cid:12) dt (cid:12).
m
Z21 (cid:20) (cid:12) (cid:18) (cid:19)(cid:12)(cid:21) )
(cid:12) (cid:12)
(cid:12) (cid:12)
By computing the above integrals, we obtain (cid:12) (cid:12)
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 2α+α1 −1 ′ (cid:12)(cid:12)′ b
≤ (cid:12) |f (a)|−m f(cid:12)
2 (cid:20)(cid:18)2α+α1(α+α1+1)(cid:19)(cid:18) (cid:12) (cid:18)m(cid:19)(cid:12)(cid:19)
m b 1 (cid:12) (cid:12)
+ f′ 1− (cid:12)(cid:12) (cid:12)(cid:12)
α+1 m 2α
(cid:12) (cid:18) (cid:19)(cid:12)(cid:18) (cid:19)(cid:21)
(cid:12) (cid:12)
(cid:12) (cid:12)
where we used the fact th(cid:12)at (cid:12)
21 1 1
tα1(1−t)αdt= tα1(1−t)αdt=β( ;α +1,α+1).
1
Z0 Z12 2
Thiscompletestheproofofthiscase. Supposenowthatq >1.Again,fromLemma
1 and by applying well-known Ho¨lder’s inequality, we have
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) q−1 (cid:12) 1
(cid:12)b−a 1 q 1 (cid:12) q
≤ (cid:12) |(1−t)α−tα|dt |(1−t)(cid:12)α−tα||f′(ta+(1−t)b)|qdt
2
(cid:18)Z0 (cid:19) (cid:18)Z0 (cid:19)
By computing the above integrals and by using (α ,m)−convexity of |f′|q, we
1
deduce
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)(cid:12)b−a 2α−1 q−q1 2α+α1 −1 (cid:12)(cid:12)(cid:12) |f′(a)|q−m f′ b q
2 (cid:20)2α−1(α+1)(cid:21) (cid:20)(cid:18)2α+α1(α+α1+1)(cid:19)(cid:18) (cid:12) (cid:18)m(cid:19)(cid:12) (cid:19)
m b q 1 q1 (cid:12)(cid:12) (cid:12)(cid:12)
+ f′ 1− . (cid:12) (cid:12)
α+1 m 2α
(cid:12) (cid:18) (cid:19)(cid:12) (cid:18) (cid:19)(cid:21)
(cid:12) (cid:12)
Which complet(cid:12)(cid:12)es the pr(cid:12)(cid:12)oof. (cid:3)
INEQUALITIES VIA FRACTIONAL INTEGRALS 9
Corollary 3. If we choose α = α with α,α ∈ (0,1] in Theorem 7, we have the
1 1
inequality;
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)(cid:12)b−a 2α−1 q−q1 22α−1 |f(cid:12)(cid:12)(cid:12)′(a)|q−m f′ b q
2 2α−1(α+1) 22α(2α+1) m
(cid:20) (cid:21) (cid:20)(cid:18) (cid:19)(cid:18) (cid:12) (cid:18) (cid:19)(cid:12) (cid:19)
m b q 1 q1 (cid:12)(cid:12) (cid:12)(cid:12)
+ f′ 1− . (cid:12) (cid:12)
α+1 m 2α
(cid:12) (cid:18) (cid:19)(cid:12) (cid:18) (cid:19)(cid:21)
(cid:12) (cid:12)
Remark 4. Ident(cid:12)ical resul(cid:12)t of Theorem 5, can be stated for (α,m)−convex func-
(cid:12) (cid:12)
tions, but we omit the details.
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10M.EMINO¨ZDEMIR(cid:7),MERVEAVCI♣,⋆,AHMETOCAKAKDEMIR♠,AND♠ALPEREKINCI
(cid:7)Ataturk University, K.K. Education Faculty, Department of Mathematics, 25240,
Erzurum,Turkey
E-mail address: [e-mail]
♣Adıyaman University, Faculty of Science and Arts, Department of Mathematics,
02040,Adıyaman,Turkey
E-mail address: [e-mail]
♠Ag˘rıI˙brahimC¸ec¸enUniversity,FacultyofScienceandArts,DepartmentofMath-
ematics,04100,Ag˘rı,Turkey
E-mail address: [e-mail]
E-mail address: [e-mail]