Hermite–Hadamard-type inequalities for r-convex functions based on the use of Riemann–Liouville fractional integrals
By using two fundamental fractional integral identities, we deduce some new Hermite–Hadamard-type inequalities for differentiable r-convex functions and twice-differentiable r-convex functions involving Riemann–Liouville fractional integrals. Iз використанням двох фундаментальних дробових iнтегральн...
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2013
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| Цитувати: | Hermite–Hadamard-type inequalities for r-convex functions based on the use of Riemann–Liouville fractional integrals / J. Wang, J. Deng, M. Fečkan // Український математичний журнал. — 2013. — Т. 65, № 2. — С. 175-191. — Бібліогр.: 28 назв. — англ. |
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| citation_txt | Hermite–Hadamard-type inequalities for r-convex functions based on the use of Riemann–Liouville fractional integrals / J. Wang, J. Deng, M. Fečkan // Український математичний журнал. — 2013. — Т. 65, № 2. — С. 175-191. — Бібліогр.: 28 назв. — англ. |
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| description | By using two fundamental fractional integral identities, we deduce some new Hermite–Hadamard-type inequalities for differentiable r-convex functions and twice-differentiable r-convex functions involving Riemann–Liouville fractional integrals.
Iз використанням двох фундаментальних дробових iнтегральних тотожностей отримано новi нерiвностi типу Ермiта – Адамара для диференцiйовних r-опуклих функцiй та двiчi диференцiйовних r-опуклих функцiй, що мiстять дробовi iнтеграли Рiмана – Лiувiлля.
|
| first_indexed | 2025-12-07T16:17:22Z |
| format | Article |
| fulltext |
UDC 517.9
J. Wang, J. Deng (Guizhou Univ., China),
M. Fec̆kan (Comenius Univ. and Math. Inst. Slovak Acad.f Sci., Bratislava, Slovakia)
HERMITE – HADAMARD-TYPE INEQUALITIES FOR r-CONVEX FUNCTIONS
USING RIEMANN – LIOUVILLE FRACTIONAL INTEGRALS*
НЕРIВНОСТI ТИПУ ЕРМIТА – АДАМАРА ДЛЯ r-ОПУКЛИХ ФУНКЦIЙ
IЗ ВИКОРИСТАННЯМ ДРОБОВИХ IНТЕГРАЛIВ РIМАНА – ЛIУВIЛЛЯ
By using two fundamental fractional integral identities, we derive some new Hermite – Hadamard-type inequalities for
differentiable r-convex functions and twice-differentiable r-convex functions involving Riemann – Liouville fractional
integrals.
Iз використанням двох фундаментальних дробових iнтегральних тотожностей отримано новi нерiвностi типу Ер-
мiта – Адамара для диференцiйовних r-опуклих функцiй та двiчi диференцiйовних r-опуклих функцiй, що мiстять
дробовi iнтеграли Рiмана – Лiувiлля.
1. Introduction. It is well-known that one of the most fundamental and interesting inequalities for
classical convex functions is that associated with the name of Hermite – Hadamard inequality which
provides a lower and an upper estimations for the integral average of any convex functions defined
on a compact interval, involving the midpoint and the endpoints of the domain. More precisely, if
f : [a, b]→ R is a convex function, then it is integrable in sense of Riemann and
f
(
a+ b
2
)
≤ 1
b− a
b∫
a
f(t)dt ≤ f(a) + f(b)
2
. (1)
This above inequality (1) was firstly discovered by Hermite in 1881 in the journal Mathesis (see
Mitrinović and Lacković [1]). But, this beautiful result was nowhere mentioned in the mathematical
literature and was not widely known as Hermite’s result (see Pec̆arić et al. [2]). For more recent
results which generalize, improve, and extend this classical Hermite – Hadamard inequality, one can
see [3 – 13] and references therein.
Meanwhile, fractional integrals and derivatives provide an excellent tool for the description of
memory and hereditary properties of various materials and processes. It draws a great application
in nonlinear oscillations of earthquakes, many physical phenomena such as seepage flow in porous
media and in fluid dynamic traffic model. For more recent development on fractional calculus, one
can see the monographs [14 – 21].
Due to the widely application of Hermite – Hadamard-type inequalities and fractional integrals,
many researchers turn to study Hermite – Hadamard-type inequalities involving fractional integrals
not limited to integer integrals. Recently, more and more Hermite – Hadamard-type inequalities in-
volving fractional integrals have been obtained for different classes of functions; see for convex
functions [22] and nondecreasing functions [23], for m-convex functions and (s,m)-convex func-
tions [24, 25], for functions satisfying s-e-condition [26] and the references therein.
*The first and second authors acknowledge the support by National Natural Science Foundation of China (11201091)
and Key Projects of Science and Technology Research in the Chinese Ministry of Education (211169); the third author
acknowledges the support by Grants VEGA-MS 1/0507/11, VEGA-SAV 2/0124/12 and APVV-0134-10.
c© J. WANG, J. DENG, M. FEC̆KAN, 2013
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 2 175
176 J. WANG, J. DENG, M. FEC̆KAN
The notion of r-convexity undoubtedly plays a dominant role in almost all aspects of mathemati-
cal programming [27] and Hermite – Hadamard-type inequalities [28]. However, Hermite – Hadamard-
type inequalities for r-convex functions involving fractional integrals have not been studied. Thus,
the purpose of this paper is to establish Hermite – Hadamard-type inequalities for r-convex functions
via Riemann – Liouville fractional integral by using two fundamental fractional integrals identity in
Sarikaya et al. [22] and Wang et al. [25].
2. Preliminaries. In this section, we introduce notations, definitions, and preliminary facts.
In [28], Pearce et al. introduced the definition of r-convex function via power means.
Definition 2.1. The function f : [0, b∗]→ R is said to be r-convex, where r ≥ 0 and b∗ > 0, if
for every x, y ∈ [0, b∗] and t ∈ [0, 1], we have
f(tx+ (1− t)y) ≤
[
t(f(x))r + (1− t)(f(y))r
]1/r
, r 6= 0,
f(tx+ (1− t)y) ≤
(
f(x))t(f(y)
)(1−t)
, r = 0.
Remark 2.1. Clearly, a r-convex function must be a convex function, however, the inverse is
false. For example, f(x) = x1/2(x > 0) is a 4-convex function, but it is not a convex function in
anyway.
We also give some necessary definitions of fractional calculus which are used further in this
paper. For more details, one can see Kilbas et al. [16].
Definition 2.2. Let f ∈ L[a, b]. The symbols RLJ
α
a+f and RLJ
α
b−f denote the left-sided and
right-sided Riemann – Liouville fractional integrals of the order α ∈ R+ are defined by
(RLJ
α
a+f)(x) =
1
Γ(α)
x∫
a
(x− t)α−1f(t)dt, 0 ≤ a < x ≤ b,
and
(RLJ
α
b−f)(x) =
1
Γ(α)
b∫
x
(t− x)α−1f(t)dt, 0 ≤ a ≤ x < b,
respectively. Here Γ(·) is the Gamma function.
The following results will be used in the sequel.
Lemma 2.1 (see Lemma 4.1 [26]). For α > 0 and k > 0, we have
I(α, k) :=
1∫
0
tα−1ktdt = k
∞∑
i=1
(−1)i−1
(ln k)i−1
(α)i
< +∞,
where
(α)i = α(α+ 1)(α+ 2) . . . (α+ i− 1).
Moreover, it holds∣∣∣∣∣I(α, k)− k
m∑
i=1
(− ln k)i−1
(α)i
∣∣∣∣∣ ≤ | ln k|
α
√
2π(m− 1)
(
| ln k|e
m− 1
)m−1
.
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 2
FRACTIONAL HERMITE – HADAMARD-TYPE INEQUALITIES 177
Lemma 2.2. For α > 0 and k > 0, z > 0, we have
J(α, k) :=
1∫
0
(1− t)α−1ktdt =
∞∑
i=1
(ln k)i−1
(α)i
< +∞,
H(α, k, z) :=
z∫
0
tα−1ktdt = zαkz
∞∑
i=1
(−z ln k)i−1
(α)i
< +∞.
Proof. By using Lemma 2.1, we obtain
J(α, k) :=
1∫
0
(1− t)α−1ktdt =
1∫
0
tα−1k1−tdt = kI(α, k−1) =
∞∑
i=1
(ln k)i−1
(α)i
< +∞,
H(α, k, z) :=
z∫
0
tα−1ktdt = zα
1∫
0
Tα−1(kz)TdT = zαI(α, kz) = zαkz
∞∑
i=1
(−z ln k)i−1
(α)i
< +∞,
which implies the desired results.
Lemma 2.3. For α > 0 and k > 0, 1 ≥ z > 0, we have
R(α, k, z) :=
z∫
0
(1− t)α−1ktdt =
∞∑
i=1
(ln k)i−1
(α)i
(
1− kz(1− z)α+i−1
)
.
Proof. By using Lemma 2.1 and Lemma 2.2, we obtain
R(α, k, z) =
1∫
0
(1− t)α−1ktdt−
1∫
z
(1− t)α−1ktdt =
= J(α, k)−
1−z∫
0
tα−1k1−tdt = J(α, k)− kH(α, k−1, 1− z) =
=
∞∑
i=1
(ln k)i−1
(α)i
− k(1− z)αkz−1
∞∑
i=1
(1− z)i−1(ln k)i−1
(α)i
.
Lemma 2.3 is proved.
Finally, we recall the following elementary inequalities.
Lemma 2.4. For A ≥ 0, B ≥ 0, it holds
(A+B)θ ≤ 2θ−1(Aθ +Bθ) when θ ≥ 1,
(A+B)θ ≤ (Aθ +Bθ) when 0 < θ ≤ 1.
3. Hermite – Hadamard-type inequalities for differentiable r-convex functions. In this sec-
tion, we apply the most fundamental fractional integrals identity given by Sarikaya et al. [22] to
present some new Hermite – Hadamard-type inequalities for differentiable r-convex functions.
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 2
178 J. WANG, J. DENG, M. FEC̆KAN
Lemma 3.1 (see Lemma 2 [22]). 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)
2
− Γ(α+ 1)
2(b− a)α
[RLJ
α
a+f(b) +RL J
α
b−f(a)] =
=
b− a
2
1∫
0
[(1− t)α − tα]f ′(ta+ (1− t)b)dt. (2)
By using the above lemma, we can obtain the main results in this section.
Theorem 3.1. Let f : [0, b∗]→ R be a differentiable mapping with b∗ > 0. If |f ′| is measurable
and r-convex on [a, b] for some fixed 0 ≤ r < ∞, 0 ≤ a < b, then the following inequality for
fractional integrals holds:∣∣∣∣f(a) + f(b)
2
− Γ(α+ 1)
2(b− a)α
[RLJ
α
a+f(b) +RL J
α
b−f(a)]
∣∣∣∣ ≤ Kr,
where
Kr := 21/r−2
1− 2−α
1 + α
(b− a)
(
|f ′(a)|+ |f ′(b)|
)
for 0 < r ≤ 1,
Kr := 2−1/r
1− 2−α
1 + α
(b− a)
(
|f ′(a)|+ |f ′(b)|
)
for r > 1,
K0 :=
(b− a)|f ′(b)|
2
∞∑
i=1
[
(ln k)2i−1
(α+ 1)2i
(1− k) +
(ln k)2i−2
(α+ 1)2i−1
(
k + 1−
√
k
2α+2i−3
)]
,
and k =
|f ′(a)|
|f ′(b)|
.
Proof. Case 1: 0 < r ≤ 1. By Definition 2.1, Lemmas 2.4 and 3.1, we have∣∣∣∣f(a) + f(b)
2
− Γ(α+ 1)
2(b− a)α
[RLJ
α
a+f(b) +RL J
α
b−f(a)]
∣∣∣∣ ≤
≤ b− a
2
1∫
0
|(1− t)α − tα||f ′(ta+ (1− t)b)|dt ≤
≤ b− a
2
1∫
0
|(1− t)α − tα|
[
t|f ′(a)|r + (1− t)|f ′(b)|r
]1/r
dt ≤
≤ b− a
2
1∫
0
|(1− t)α − tα|21/r−1
[
t1/r|f ′(a)|+ (1− t)1/r|f ′(b)|
]
dt =
= 21/r−2(b− a)
|f ′(a)|
1∫
0
|(1− t)α − tα|t1/rdt +
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 2
FRACTIONAL HERMITE – HADAMARD-TYPE INEQUALITIES 179
+ |f ′(b)|
1∫
0
|(1− t)α − tα|(1− t)1/rdt
=
= 21/r−2(b− a)
(
|f ′(a)|+ |f ′(b)|
) 1∫
0
|(1− t)α − tα|t1/rdt =
= 21/r−2(b− a)
(
|f ′(a)|+ |f ′(b)|
) 1/2∫
0
((1− t)α − tα)(t1/r + (1− t)1/r)dt ≤
≤ 21/r−2(b− a)
(
|f ′(a)|+ |f ′(b)|
) 1/2∫
0
((1− t)α − tα)dt =
= 21/r−2
1− 2−α
1 + α
(b− a)
(
|f ′(a)|+ |f ′(b)|
)
.
Case 2: 1 < r. Like in Case 1, we obtain∣∣∣∣f(a) + f(b)
2
− Γ(α+ 1)
2(b− a)α
[RLJ
α
a+f(b) +RL J
α
b−f(a)]
∣∣∣∣ ≤
≤ b− a
2
1∫
0
|(1− t)α − tα|
[
t|f ′(a)|r + (1− t)|f ′(b)|r
]1/r
dt ≤
≤ b− a
2
1∫
0
|(1− t)α − tα|
[
t1/r|f ′(a)|+ (1− t)1/r|f ′(b)|
]
dt =
=
b− a
2
(
|f ′(a)|
1∫
0
|(1− t)α − tα|t1/rdt+
+|f ′(b)|
1∫
0
|(1− t)α − tα|(1− t)1/rdt
)
=
b− a
2
(
|f ′(a)|+ |f ′(b)|
) 1∫
0
|(1− t)α − tα|t1/rdt =
=
b− a
2
(
|f ′(a)|+ |f ′(b)|
) 1/2∫
0
((1− t)α − tα)(t1/r + (1− t)1/r)dt ≤
≤ 2−1/r(b− a)
(
|f ′(a)|+ |f ′(b)|
) 1/2∫
0
((1− t)α − tα)dt =
= 2−1/r
1− 2−α
1 + α
(b− a)
(
|f ′(a)|+ |f ′(b)|
)
.
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 2
180 J. WANG, J. DENG, M. FEC̆KAN
Case 3: r = 0. By using Definition 2.1 and Lemmas 2.1, 2.3, 3.1, we have∣∣∣∣f(a) + f(b)
2
− Γ(α+ 1)
2(b− a)α
[RLJ
α
a+f(b) +RL J
α
b−f(a)]
∣∣∣∣ ≤
≤ b− a
2
1∫
0
|(1− t)α − tα||f ′(ta+ (1− t)b)|dt ≤
≤ b− a
2
1∫
0
|(1− t)α − tα||f ′(a)|t|f ′(b)|1−tdt =
(b− a)|f ′(b)|
2
1∫
0
|(1− t)α − tα|ktdt =
=
(b− a)|f ′(b)|
2
1/2∫
0
(1− t)αktdt−
1/2∫
0
tαktdt+
1∫
1/2
tαktdt−
1∫
1/2
(1− t)αktdt
=
=
(b− a)|f ′(b)|
2
2
1/2∫
0
(1− t)αktdt− 2
1/2∫
0
tαktdt+
1∫
0
tαktdt−
1∫
0
(1− t)αktdt
=
=
(b− a)|f ′(b)|
2
∞∑
i=1
[
(ln k)2i−1
(α+ 1)2i
(1− k) +
(ln k)2i−2
(α+ 1)2i−1
(
k + 1−
√
k
2α+2i−3
)]
.
Theorem 3.1 is proved.
Theorem 3.2. Let f : [0, b∗] → R be a differentiable mapping with b∗ > 0. If |f ′|q, q > 1,
is measurable and r-convex on [a, b] for some fixed, 0 ≤ r < ∞, 0 ≤ a < b, then the following
inequality for fractional integrals holds:∣∣∣∣f(a) + f(b)
2
− Γ(α+ 1)
2(b− a)α
[RLJ
α
a+f(b) +RL J
α
b−f(a)]
∣∣∣∣ ≤ Kr,
where
Kr := (b− a)2
1−2r
qr
(
1− 2−pα
pα+ 1
)1/p [
r(|f ′(a)|q + |f ′(b)|q)
r + 1
]1/q
for 0 < r ≤ 1,
Kr :=
b− a
21−1/p
(
1− 2−pα
pα+ 1
)1/p [
r(|f ′(a)|q + |f ′(b)|q)
r + 1
]1/q
for r > 1,
K0 :=
(b− a)
21−1/p
(
1− 2−pα
pα+ 1
)1/p( |f ′(a)|q − |f ′(b)|q
q ln |f ′(a)| − q ln |f ′(b)|
)1/q
when |f ′(a)| 6= |f ′(b)|,
K0 :=
(b− a)
21−1/p
(
1− 2−pα
pα+ 1
)1/p
|f ′(a)| when |f ′(a)| = |f ′(b)|,
and
1
p
+
1
q
= 1.
Proof. Case 1: 0 < r ≤ 1. By Definition 2.1, Lemmas 2.4, 3.1 and using Hölder inequality, we
have
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 2
FRACTIONAL HERMITE – HADAMARD-TYPE INEQUALITIES 181∣∣∣∣f(a) + f(b)
2
− Γ(α+ 1)
2(b− a)α
[RLJ
α
a+f(b) +RL J
α
b−f(a)]
∣∣∣∣ ≤
≤ b− a
2
1∫
0
|(1− t)α − tα||f ′(ta+ (1− t)b)|dt ≤
≤ b− a
2
1∫
0
|(1− t)α − tα|pdt
1/p 1∫
0
|f ′(ta+ (1− t)b)|qdt
1/q
=
=
b− a
2
2
1/2∫
0
((1− t)α − tα)pdt
1/p 1∫
0
|f ′(ta+ (1− t)b)|qdt
1/q
≤
≤ b− a
21−1/p
1/2∫
0
((1− t)pα − tpα)pdt
1/p 1∫
0
|f ′(ta+ (1− t)b)|qdt
1/q
=
=
b− a
21−1/p
(
1− 2−pα
pα+ 1
)1/p
1∫
0
|f ′(ta+ (1− t)b)|qdt
1/q
≤
≤ b− a
21−1/p
(
1− 2−pα
pα+ 1
)1/p
1∫
0
[
t|f ′(a)|qr + (1− t)|f ′(b)|qr
]1/r
dt
1/q
≤
≤ b− a
21−1/p
(
1− 2−pα
pα+ 1
)1/p
21/r−1
1∫
0
[
t1/r|f ′(a)|q + (1− t)1/r|f ′(b)|q
]
dt
1/q
=
= (b− a)2
1−2r
qr
(
1− 2−pα
pα+ 1
)1/p [
r(|f ′(a)|q + |f ′(b)|q)
r + 1
]1/q
.
Case 2: 1 < r. By Definition 2.1, Lemmas 2.4, 3.1 and using Hölder inequality, like above we
obtain ∣∣∣∣f(a) + f(b)
2
− Γ(α+ 1)
2(b− a)α
[RLJ
α
a+f(b) +RL J
α
b−f(a)]
∣∣∣∣ ≤
≤ b− a
21−1/p
(
1− 2−pα
pα+ 1
)1/p
1∫
0
[
t|f ′(a)|qr + (1− t)|f ′(b)|qr
]1/r
dt
1/q
≤
≤ b− a
21−1/p
(
1− 2−pα
pα+ 1
)1/p
1∫
0
[
t1/r|f ′(a)|q + (1− t)1/r|f ′(b)|q
]
dt
1/q
=
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 2
182 J. WANG, J. DENG, M. FEC̆KAN
=
b− a
21−1/p
(
1− 2−pα
pα+ 1
)1/p [
r(|f ′(a)|q + |f ′(b)|q)
r + 1
]1/q
.
Case 3: r = 0. By Definition 2.1, Lemma 3.1 and using Hölder inequality, like above we have∣∣∣∣f(a) + f(b)
2
− Γ(α+ 1)
2(b− a)α
[RLJ
α
a+f(b) +RL J
α
b−f(a)]
∣∣∣∣ ≤
≤ (b− a)
21−1/p
(
1− 2−pα
pα+ 1
)1/p
1∫
0
|f ′(ta+ (1− t)b)|qdt
1/q
≤
≤ (b− a)
21−1/p
(
1− 2−pα
pα+ 1
)1/p
1∫
0
|f ′(a)|qt|f ′(b)|q(1−t)dt
1/q
.
Since
(
|f ′(a)|q − |f ′(b)|q
q ln |f ′(a)| − q ln |f ′(b)|
)1/q
=
(
|f ′(a)|q − |f ′(b)|q
q ln |f ′(a)| − q ln |f ′(b)|
)1/q
when |f ′(a)| 6= |f ′(b)|,
|f ′(a)|q when |f ′(a)| = |f ′(b)|.
Theorem 3.2 is proved.
4. Hermite – Hadamard-type inequalities for twice-differentiable r-convex functions. In this
section, we will first recall from our previous work an important fractional integrals identity including
the second order derivative of a function. Then, we apply this new fractional integrals identity to
present some new Hermite – Hadamard-type inequalities for twice-differentiable r-convex functions.
Lemma 4.1 (see Lemma 2.1, [25]). Let f : [a, b]→ R be twice-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)
2
− Γ(α+ 1)
2(b− a)α
[
RL
Jαa+f(b) +RL J
α
b−f(a)
]
=
=
(b− a)2
2
1∫
0
1− (1− t)α+1 − tα+1
α+ 1
f ′′(ta+ (1− t)b)dt. (3)
Now we are ready to present the main results in this section.
Theorem 4.1. Let f : [0, b∗]→ R be a twice-differentiable mapping with b∗ > 0. If |f ′′|q(q >
> 1) is measurable and r-convex on [a, b] for some fixed 0 ≤ r <∞, 0 ≤ a < b, then the following
inequality for fractional integrals holds:∣∣∣∣f(a) + f(b)
2
− Γ(α+ 1)
2(b− a)α
[RLJ
α
a+f(b) +RL J
α
b−f(a)]
∣∣∣∣ ≤ Ir,
where
Ir =
2
1−r−qr
qr (b− a)2
α+ 1
(
1− 2
pα+ p+ 1
)1/p
×
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FRACTIONAL HERMITE – HADAMARD-TYPE INEQUALITIES 183
×
(
|f ′′(a)|q + |f ′′(b)|q
)1/q ( r
r + 1
)1/q
for 0 < r ≤ 1,
Ir =
(b− a)2
2(α+ 1)
(
1− 2
pα+ p+ 1
)1/p (
|f ′′(a)|q + |f ′′(b)|q
)1/q ( r
r + 1
)1/q
for 1 < r,
I0 =
(b− a)2
2(α+ 1)
(
1− 2
pα+ p+ 1
)1/p
×
×
(
|f ′′(a)|q − |f ′′(b)|q
q ln |f ′′(a)| − q ln |f ′′(b)|
)1/q
when |f ′′(a)| 6= |f ′′(b)|,
I0 =
(b− a)2|f ′′(a)|
2(α+ 1)
(
1− 2
pα+ p+ 1
)1/p
when |f ′′(a)| = |f ′′(b)|,
and
1
p
+
1
q
= 1.
Proof. Case 1: 0 < r ≤ 1. By Definition 2.1, Lemmas 2.4, 4.1 and using Hölder inequality, we
have ∣∣∣∣f(a) + f(b)
2
− Γ(α+ 1)
2(b− a)α
[RLJ
α
a+f(b) +RL J
α
b−f(a)]
∣∣∣∣ ≤
≤ (b− a)2
2
1∫
0
1− (1− t)α+1 − tα+1
α+ 1
|f ′′(ta+ (1− t)b)|dt ≤
≤ (b− a)2
2(α+ 1)
1∫
0
(1− (1− t)α+1 − tα+1)pdt
1/p 1∫
0
|f ′′(ta+ (1− t)b)|qdt
1/q
≤
≤ (b− a)2
2(α+ 1)
1∫
0
(1− (1− t)p(α+1) − tp(α+1))dt
1/p
×
×
1∫
0
[t|f ′′(a)|qr + (1− t)|f ′′(b)|qr]1/rdt
1/q
≤
≤ (b− a)2
2(α+ 1)
(
1− 2
pα+ p+ 1
)1/p21/r−1
1∫
0
[t1/r|f ′′(a)|q + (1− t)1/r|f ′′(b)|q]dt
1/q
=
=
2
1−r−qr
qr (b− a)2
α+ 1
(
1− 2
pα+ p+ 1
)1/p (
|f ′′(a)|q + |f ′′(b)|q
)1/q ( r
r + 1
)1/q
.
Case 2: 1 < r. By Definition 2.1, Lemmas 2.4, 4.1 and using Hölder inequality, like above we
obtain
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 2
184 J. WANG, J. DENG, M. FEC̆KAN∣∣∣∣f(a) + f(b)
2
− Γ(α+ 1)
2(b− a)α
[
RL
Jαa+f(b) +RL J
α
b−f(a)
]∣∣∣∣ ≤
≤ (b− a)2
2(α+ 1)
(
1− 2
pα+ p+ 1
)1/p
1∫
0
[
t|f ′′(a)|qr + (1− t)|f ′′(b)|qr
]1/r
dt
1/q
≤
≤ (b− a)2
2(α+ 1)
(
1− 2
pα+ p+ 1
)1/p
1∫
0
[
t1/r|f ′′(a)|q + (1− t)1/r|f ′′(b)|q
]
dt
1/q
=
=
(b− a)2
2(α+ 1)
(
1− 2
pα+ p+ 1
)1/p (
|f ′′(a)|q + |f ′′(b)|q
)1/q ( r
r + 1
)1/q
.
Case 3: r = 0. By Definition 2.1, Lemma 4.1 and using Hölder inequality, like above we have∣∣∣∣f(a) + f(b)
2
− Γ(α+ 1)
2(b− a)α
[
RL
Jαa+f(b) +RL J
α
b−f(a)
]∣∣∣∣ ≤
≤ (b− a)2
2(α+ 1)
(
1− 2
pα+ p+ 1
)1/p
1∫
0
|f ′′(ta+ (1− t)b)|qdt
1/q
≤
≤ (b− a)2
2(α+ 1)
(
1− 2
pα+ p+ 1
)1/p
1∫
0
|f ′′(a)|qt|f ′′(b)|q(1−t)dt
1/q
.
Since
1∫
0
|f ′′(a)|qt|f ′′(b)|q(1−t)dt =
|f ′′(a)|q − |f ′′(b)|q
q ln |f ′′(a)| − q ln |f ′′(b)|
when |f ′′(a)| 6= |f ′′(b)|,
|f ′′(a)|q when |f ′′(a)| = |f ′′(b)|.
Theorem 4.1 is proved.
Corollary 4.1. Suppose that the assumptions of Theorem 4.1 hold. Moreover, |f ′′(x)| ≤ M on
[a, b]. Then ∣∣∣∣f(a) + f(b)
2
− Γ(α+ 1)
2(b− a)α
[
RL
Jαa+f(b) +RL J
α
b−f(a)
]∣∣∣∣ ≤ I ′r,
where
I ′r =
2
1−qr
qr M(b− a)2
(α+ 1)
(
1− 2
pα+ p+ 1
)1/p( r
r + 1
)1/q
for 0 < r ≤ 1,
I ′2 =
2
1−q
q M(b− a)2
(α+ 1)
(
1− 2
pα+ p+ 1
)1/p( r
r + 1
)1/q
for 1 < r,
I ′0 =
M(b− a)2
2(α+ 1)
(
1− 2
pα+ p+ 1
)1/p
,
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FRACTIONAL HERMITE – HADAMARD-TYPE INEQUALITIES 185
and
1
p
+
1
q
= 1.
Next, we give another Hermit – Hadamard-type inequality for powers in terms of the second
derivatives.
Theorem 4.2. Let f : [0, b∗]→ R be a twice-differentiable mapping with b∗ > 0. If |f ′′|q(q >
> 1) is measurable and r-convex on [a, b] for some fixed 0 ≤ r <∞, 0 ≤ a < b, then the following
inequality for fractional integrals holds:∣∣∣∣f(a) + f(b)
2
− Γ(α+ 1)
2(b− a)α
[
RL
Jαa+f(b) +RL J
α
b−f(a)
]∣∣∣∣ ≤ Ir,
where
Ir =
2
1−r−qr
qr (b− a)2
(α+ 1)
(
|f ′′(a)|q + |f ′′(b)|q
)1/q
×
×
(
r
r + 1
− β
(
r
r + 1
, αq + q + 1
)
− r
αqr + qr + r + 1
)1/q
for 0 < r ≤ 1,
Ir =
(b− a)2
2(α+ 1)
(
|f ′′(a)|q + |f ′′(b)|q
)1/q
×
×
(
r
r + 1
− β
(
r
r + 1
, αq + q + 1
)
− r
αqr + qr + r + 1
)1/q
for 1 < r,
I0 =
(b− a)2
2(α+ 1)
(
|f ′′(a)|q − |f ′′(b)|q
q (ln |f ′′(a)| − ln |f ′′(b)|)
−
−
∞∑
i=1
(ln |f ′′(a)| − ln |f ′′(b)|)i−1
(αq + q + 1)i
[
qi−1|f ′′(b)|q + (−q)i−1|f ′′(a)|q
])1/q
when |f ′′(a)| 6= |f ′′(b)|,
I0 =
(b− a)2|f ′′(a)|
2(α+ 1)
(
1− 2
αq + q + 1
)1/q
when |f ′′(a)| = |f ′′(b)|,
and
1
p
+
1
q
= 1.
Proof. Case 1: 0 < r ≤ 1. By Definition 2.1, Lemmas 2.4, 4.1 and using Hölder inequality, we
have ∣∣∣∣f(a) + f(b)
2
− Γ(α+ 1)
2(b− a)α
[RLJ
α
a+f(b) +RL J
α
b−f(a)]
∣∣∣∣ ≤
≤ (b− a)2
2
1∫
0
1− (1− t)α+1 − tα+1
α+ 1
|f ′′(ta+ (1− t)b)|dt ≤
≤ (b− a)2
2(α+ 1)
1∫
0
1dt
1/p 1∫
0
(1− (1− t)α+1 − tα+1)q|f ′′(ta+ (1− t)b)|qdt
1/q
≤
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 2
186 J. WANG, J. DENG, M. FEC̆KAN
≤ (b− a)2
2(α+ 1)
( 1∫
0
(1− (1− t)(α+1)q − t(α+1)q)[t|f ′′(a)|qr + (1− t)|f ′′(b)|qr]1/rdt
)1/q
≤
≤ 2
1−r−qr
qr (b− a)2
α+ 1
1∫
0
(1− (1− t)(α+1)q − t(α+1)q)
[
t1/r|f ′′(a)|q + (1− t)1/r|f ′′(b)|q
]
dt
1/q
=
=
2
1−r−qr
qr (b− a)2
(α+ 1)
(
|f ′′(a)|q + |f ′′(b)|q
)1/q
×
×
(
r
r + 1
− β
(
r
r + 1
, αq + q + 1
)
− r
αqr + qr + r + 1
)1/q
.
Case 2: 1 < r. By Definition 2.1, Lemmas 2.4, 4.1 and using Hölder inequality, like above we
obtain ∣∣∣∣f(a) + f(b)
2
− Γ(α+ 1)
2(b− a)α
[RLJ
α
a+f(b) +RL J
α
b−f(a)]
∣∣∣∣ ≤
≤ (b− a)2
2(α+ 1)
( 1∫
0
(1− (1− t)(α+1)q − t(α+1)q)[t|f ′′(a)|qr + (1− t)|f ′′(b)|qr]1/rdt
)1/q
≤
≤ (b− a)2
2(α+ 1)
( 1∫
0
(1− (1− t)(α+1)q − t(α+1)q)[t1/r|f ′′(a)|q + (1− t)1/r|f ′′(b)|q]dt
)1/q
=
=
(b− a)2
2(α+ 1)
(
|f ′′(a)|q + |f ′′(b)|q
)1/q( r
r + 1
− β
(
r
r + 1
, αq + q + 1
)
− r
αqr + qr + r + 1
)1/q
.
Case 3: r = 0. According to Lemmas 2.1, 2.2, 4.1, using Hölder inequality, like above we have∣∣∣∣f(a) + f(b)
2
− Γ(α+ 1)
2(b− a)α
[RLJ
α
a+f(b) +RL J
α
b−f(a)]
∣∣∣∣ ≤
≤ (b− a)2
2(α+ 1)
1∫
0
(1− (1− t)α+1 − tα+1)q|f ′′(ta+ (1− t)b)|qdt
1/q
≤
≤ (b− a)2
2(α+ 1)
( 1∫
0
(1− (1− t)(α+1)q − t(α+1)q)|f ′′(a)|qt|f ′′(b)|q(1−t)dt
)1/q
.
Since
1∫
0
(1− (1− t)(α+1)q − t(α+1)q)|f ′′(a)|qt|f ′′(b)|q(1−t)dt =
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 2
FRACTIONAL HERMITE – HADAMARD-TYPE INEQUALITIES 187
=
|f ′′(a)|q − |f ′′(b)|q
q (ln |f ′′(a)| − ln |f ′′(b)|)
−
∞∑
i=1
(ln |f ′′(a)| − ln |f ′′(b)|)i−1
(αq + q + 1)i
×
×
[
qi−1|f ′′(b)|q + (−q)i−1|f ′′(a)|q
]
when |f ′′(a)| 6= |f ′′(b)|,
|f ′′(a)|q when |f ′′(a)| = |f ′′(b)|.
Theorem 4.2 is proved.
Theorem 4.3. Let f : [0, b∗] → R be a twice-differentiable mapping with b∗ > 0. If |f ′′|q,
q > 1, is measurable and r-convex on [a, b] for some fixed 0 ≤ r <∞, 0 ≤ a < b, then the following
inequality for fractional integrals holds:∣∣∣∣f(a) + f(b)
2
− Γ(α+ 1)
2(b− a)α
[RLJ
α
a+f(b) +RL J
α
b−f(a)]
∣∣∣∣ ≤ Ir,
where
Ir = 2
n−r−nqr
qnr
α
1
p+
1
qm (b− a)2
(α+ 1)(α+ 2)
1
p+
1
qm
(
|f ′′(a)|qn + |f ′′(b)|qn
) 1
qn
×
×
(
r
n+ r
− β
(
n+ r
r
, α+ 2
)
− r
αr + 2r + n
) 1
qn
for 0 < r ≤ n,
Ir =
α
1
p+
1
qm (b− a)2
2(α+ 1)(α+ 2)
1
p+
1
qm
(
|f ′′(a)|qn + |f ′′(b)|qn
) 1
qn
×
×
(
r
n+ r
− β
(
n+ r
r
, α+ 2
)
− r
αr + 2r + n
) 1
qn
for n < r,
I0 =
α
1
p+
1
qm (b− a)2
2(α+ 1)(α+ 2)
1
p+
1
qm
(
|f ′′(a)|nq − |f ′′(b)|nq
nq[ln |f ′′(a)| − ln |f ′′(b)|]
−
−
∞∑
i=1
(nq)i−1
(
ln |f ′′(a)| − ln |f ′′(b)|
)i−1 [
|f ′′(b)|nq + |f ′′(a)|nq(−1)i−1
]
(α+ 2)i
) 1
qn
when |f ′′(a)| 6= |f ′′(b)|,
I0 =
α
1
p+
1
qm+
1
qn (b− a)2|f ′′(a)|
2(α+ 1)(α+ 2)
1
p+
1
qm+
1
qn
when |f ′′(a)| = |f ′′(b)|,
and
1
p
+
1
q
= 1,
1
m
+
1
n
= 1, n > 1.
Proof. Case 1: 0 < r ≤ n. By Lemmas 2.4, 4.1 and using Hölder inequality, we have∣∣∣∣f(a) + f(b)
2
− Γ(α+ 1)
2(b− a)α
[RLJ
α
a+f(b) +RL J
α
b−f(a)]
∣∣∣∣ ≤
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 2
188 J. WANG, J. DENG, M. FEC̆KAN
≤ (b− a)2
2
1∫
0
1− (1− t)α+1 − tα+1
α+ 1
|f ′′(ta+ (1− t)b)|dt ≤
≤ (b− a)2
2(α+ 1)
1∫
0
(1− (1− t)α+1 − tα+1)dt
1/p×
×
1∫
0
(1− (1− t)α+1 − tα+1)|f ′′(ta+ (1− t)b)|qdt
1/q ≤
≤ α1/p(b− a)2
2(α+ 1)(α+ 2)1/p
1∫
0
(1− (1− t)(α+1) − t(α+1))[t|f ′′(a)|qr + (1− t)|f ′′(b)|qr]1/rdt
1/q
≤
≤ α1/p(b− a)2
2(α+ 1)(α+ 2)1/p
1∫
0
(1− (1− t)(α+1) − t(α+1))dt
1
qm
×
×
( 1∫
0
(1− (1− t)(α+1) − t(α+1))
[
t|f ′′(a)|qr + (1− t)|f ′′(b)|qr
]n
r dt
) 1
qn
≤
≤ α
1
p+
1
qm (b− a)2
2(α+ 1)(α+ 2)
1
p+
1
qm
×
×
2
n
r−1
1∫
0
(1− (1− t)(α+1) − t(α+1))
[
tn/r|f ′′(a)|qn + (1− t)n/r|f ′′(b)|qn
]
dt
1
qn
=
= 2
n−r−nqr
qnr
α
1
p+
1
qm (b− a)2
(α+ 1)(α+ 2)
1
p+
1
qm
(
|f ′′(a)|qn + |f ′′(b)|qn
) 1
qn
×
×
(
r
n+ r
− β
(
n+ r
r
, α+ 2
)
− r
αr + 2r + n
) 1
qn
.
Case 2: For 1 < n < r, like in Case 1, we have∣∣∣∣f(a) + f(b)
2
− Γ(α+ 1)
2(b− a)α
[RLJ
α
a+f(b) +RL J
α
b−f(a)]
∣∣∣∣ ≤ α
1
p+
1
qm (b− a)2
2(α+ 1)(α+ 2)
1
p+
1
qm
×
×
1∫
0
(1− (1− t)(α+1) − t(α+1))
[
t|f ′′(a)|qr + (1− t)|f ′′(b)|qr
]n
r dt
1
qn
≤
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 2
FRACTIONAL HERMITE – HADAMARD-TYPE INEQUALITIES 189
≤ α
1
p+
1
qm (b− a)2
2(α+ 1)(α+ 2)
1
p+
1
qm
×
×
1∫
0
(1− (1− t)(α+1) − t(α+1))
[
tn/r|f ′′(a)|qn + (1− t)n/r|f ′′(b)|qn
]
dt
1
qn
≤
≤ α
1
p+
1
qm (b− a)2
2(α+ 1)(α+ 2)
1
p+
1
qm
×
×
1∫
0
(1− (1− t)(α+1) − t(α+1))[t
n
r |f ′′(a)|qn + (1− t)
n
r |f ′′(b)|qn]dt
1
qn
≤
≤ α
1
p+
1
qm (b− a)2
2(α+ 1)(α+ 2)
1
p+
1
qm
(
|f ′′(a)|qn + |f ′′(b)|qn
) 1
qn ×
×
(
r
n+ r
− β
(
n+ r
r
, α+ 2
)
− r
αr + 2r + n
) 1
qn
.
Case 3: r = 0. Denote
k =
(
|f ′′(a)|
|f ′′(a)|
)nq
.
If |f ′′(a)| 6= |f ′′(b)|, we obtain∣∣∣∣f(a) + f(b)
2
− Γ(α+ 1)
2(b− a)α
[RLJ
α
a+f(b) +RL J
α
b−f(a)]
∣∣∣∣ ≤
≤ (b− a)2
2
1∫
0
1− (1− t)α+1 − tα+1
α+ 1
|f ′′(ta+ (1− t)b)|dt ≤
≤ (b− a)2
2(α+ 1)
1∫
0
(1− (1− t)α+1 − tα+1)dt
1/p
×
×
1∫
0
(1− (1− t)α+1 − tα+1)|f ′′(ta+ (1− t)b)|qdt
1/q
≤
≤ α1/p(b− a)2
2(α+ 1)(α+ 2)1/p
1∫
0
(1− (1− t)(α+1) − t(α+1))[|f ′′(a)|qt|f ′′(b)|q(1−t)]dt
1/q
≤
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 2
190 J. WANG, J. DENG, M. FEC̆KAN
≤ α1/p(b− a)2
2(α+ 1)(α+ 2)1/p
1∫
0
(1− (1− t)(α+1) − t(α+1))dt
1
qm
×
×
1∫
0
(1− (1− t)(α+1) − t(α+1))[|f ′′(a)|qt|f ′′(b)|q(1−t)]ndt
1
qn
=
=
α
1
p+
1
qm (b− a)2
2(α+ 1)(α+ 2)
1
p+
1
qm
|f ′′(b)|
1∫
0
(1− (1− t)(α+1) − t(α+1))ktdt
1
qn
=
=
α
1
p+
1
qm (b− a)2
2(α+ 1)(α+ 2)
1
p+
1
qm
|f ′′(b)|
(
k − 1
ln k
−
∞∑
i=1
(ln k)i−1
(α+ 2)i
− k
∞∑
i=1
(−1)i−1
(ln k)i−1
(α+ 2)i
) 1
qn
=
=
α
1
p+
1
qm (b− a)2
2(α+ 1)(α+ 2)
1
p+
1
qm
[
|f ′′(a)|nq − |f ′′(b)|nq
nq
[
ln |f ′′(a)| − ln |f ′′(b)|
]−
−
∞∑
i=1
(nq)i−1
(
ln |f ′′(a)| − ln |f ′′(b)|
)i−1 (
|f ′′(b)|nq + |f ′′(a)|nq(−1)i−1
)
(α+ 2)i
] 1
qn
.
If |f ′′(a)| = |f ′′(b)|, we have∣∣∣∣f(a) + f(b)
2
− Γ(α+ 1)
2(b− a)α
[RLJ
α
a+f(b) +RL J
α
b−f(a)]
∣∣∣∣ ≤
≤ α
1
p+
1
qm (b− a)2
2(α+ 1)(α+ 2)
1
p+
1
qm
1∫
0
(1− (1− t)(α+1) − t(α+1))|f ′′(a)|qndt
1
qn
=
=
α
1
p+
1
qm+
1
qn (b− a)2|f ′′(a)|
2(α+ 1)(α+ 2)
1
p+
1
qm+
1
qn
.
Theorem 4.3 is proved.
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Received 14.12.12
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 2
|
| id | nasplib_isofts_kiev_ua-123456789-164980 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1027-3190 |
| language | English |
| last_indexed | 2025-12-07T16:17:22Z |
| publishDate | 2013 |
| publisher | Інститут математики НАН України |
| record_format | dspace |
| spelling | Wang, J. Deng, J. Fečkan, M. 2020-02-11T11:56:03Z 2020-02-11T11:56:03Z 2013 Hermite–Hadamard-type inequalities for r-convex functions based on the use of Riemann–Liouville fractional integrals / J. Wang, J. Deng, M. Fečkan // Український математичний журнал. — 2013. — Т. 65, № 2. — С. 175-191. — Бібліогр.: 28 назв. — англ. 1027-3190 https://nasplib.isofts.kiev.ua/handle/123456789/164980 517.9 By using two fundamental fractional integral identities, we deduce some new Hermite–Hadamard-type inequalities for differentiable r-convex functions and twice-differentiable r-convex functions involving Riemann–Liouville fractional integrals. Iз використанням двох фундаментальних дробових iнтегральних тотожностей отримано новi нерiвностi типу Ермiта – Адамара для диференцiйовних r-опуклих функцiй та двiчi диференцiйовних r-опуклих функцiй, що мiстять дробовi iнтеграли Рiмана – Лiувiлля. en Інститут математики НАН України Український математичний журнал Статті Hermite–Hadamard-type inequalities for r-convex functions based on the use of Riemann–Liouville fractional integrals Нерiвностi типу Ермiта-Адамара для r-опуклих функцiй iз використанням дробових iнтегралiв Рiмана-Лiувiлля Article published earlier |
| spellingShingle | Hermite–Hadamard-type inequalities for r-convex functions based on the use of Riemann–Liouville fractional integrals Wang, J. Deng, J. Fečkan, M. Статті |
| title | Hermite–Hadamard-type inequalities for r-convex functions based on the use of Riemann–Liouville fractional integrals |
| title_alt | Нерiвностi типу Ермiта-Адамара для r-опуклих функцiй iз використанням дробових iнтегралiв Рiмана-Лiувiлля |
| title_full | Hermite–Hadamard-type inequalities for r-convex functions based on the use of Riemann–Liouville fractional integrals |
| title_fullStr | Hermite–Hadamard-type inequalities for r-convex functions based on the use of Riemann–Liouville fractional integrals |
| title_full_unstemmed | Hermite–Hadamard-type inequalities for r-convex functions based on the use of Riemann–Liouville fractional integrals |
| title_short | Hermite–Hadamard-type inequalities for r-convex functions based on the use of Riemann–Liouville fractional integrals |
| title_sort | hermite–hadamard-type inequalities for r-convex functions based on the use of riemann–liouville fractional integrals |
| topic | Статті |
| topic_facet | Статті |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/164980 |
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