A companion of Dragomir’s generalization of the Ostrowski inequality and applications to numerical integration
Some analogs of Dragomir’s generalization of the Ostrowski integral inequality are established. Some sharp inequalities are proved. An application to the composite quadrature rule is provided. Встановлено аналоги узагальнення Драгомiра iнтегральної нерiвностi Островського, отримано деякi точнi нерiв...
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Alomari, M.W. 2020-02-08T18:24:11Z 2020-02-08T18:24:11Z 2012 A companion of Dragomir’s generalization of the Ostrowski inequality and applications to numerical integration / M.W.Alomari // Український математичний журнал. — 2012. — Т. 64, № 4. — С. 435-450. — Бібліогр.: 20 назв. — англ. 1027-3190 https://nasplib.isofts.kiev.ua/handle/123456789/164167 517.5 Some analogs of Dragomir’s generalization of the Ostrowski integral inequality are established. Some sharp inequalities are proved. An application to the composite quadrature rule is provided. Встановлено аналоги узагальнення Драгомiра iнтегральної нерiвностi Островського, отримано деякi точнi нерiвностi. Наведено застосування до складеної квадратурної формули. en Інститут математики НАН України Український математичний журнал Статті A companion of Dragomir’s generalization of the Ostrowski inequality and applications to numerical integration Аналог узагальнення Драгомiра нерiвностi Островського та застосування до чисельного iнтегрування Article published earlier |
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A companion of Dragomir’s generalization of the Ostrowski inequality and applications to numerical integration |
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A companion of Dragomir’s generalization of the Ostrowski inequality and applications to numerical integration Alomari, M.W. Статті |
| title_short |
A companion of Dragomir’s generalization of the Ostrowski inequality and applications to numerical integration |
| title_full |
A companion of Dragomir’s generalization of the Ostrowski inequality and applications to numerical integration |
| title_fullStr |
A companion of Dragomir’s generalization of the Ostrowski inequality and applications to numerical integration |
| title_full_unstemmed |
A companion of Dragomir’s generalization of the Ostrowski inequality and applications to numerical integration |
| title_sort |
companion of dragomir’s generalization of the ostrowski inequality and applications to numerical integration |
| author |
Alomari, M.W. |
| author_facet |
Alomari, M.W. |
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Статті |
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Статті |
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2012 |
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English |
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Український математичний журнал |
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Інститут математики НАН України |
| format |
Article |
| title_alt |
Аналог узагальнення Драгомiра нерiвностi Островського та застосування до чисельного iнтегрування |
| description |
Some analogs of Dragomir’s generalization of the Ostrowski integral inequality are established. Some sharp inequalities are proved. An application to the composite quadrature rule is provided.
Встановлено аналоги узагальнення Драгомiра iнтегральної нерiвностi Островського, отримано деякi точнi нерiвностi. Наведено застосування до складеної квадратурної формули.
|
| issn |
1027-3190 |
| url |
https://nasplib.isofts.kiev.ua/handle/123456789/164167 |
| citation_txt |
A companion of Dragomir’s generalization of the Ostrowski inequality and applications to numerical integration / M.W.Alomari // Український математичний журнал. — 2012. — Т. 64, № 4. — С. 435-450. — Бібліогр.: 20 назв. — англ. |
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2025-11-25T16:15:08Z |
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2025-11-25T16:15:08Z |
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| fulltext |
UDC 517.5
M. W. Alomari (Jerash Univ., Jordan)
A COMPANION OF DRAGOMIR’S GENERALIZATION OF OSTROWSKI’S
INEQUALITY AND APPLICATIONS IN NUMERICAL INTEGRATION
АНАЛОГ УЗАГАЛЬНЕННЯ ДРАГОМIРА НЕРIВНОСТI ОСТРОВСЬКОГО
ТА ЗАСТОСУВАННЯ ДО ЧИСЕЛЬНОГО IНТЕГРУВАННЯ
Some analogs of Dragomir’s generalization of the Ostrowski integral inequality∣∣∣∣∣∣(b− a)
[
λ
f (a) + f (b)
2
+ (1− λ) f (x)
]
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤
≤
[
(b− a)2
4
(
λ2 + (1− λ)2
)
+
(
x− a+ b
2
)2
]∥∥f ′∥∥
∞
are established. Some sharp inequalities are proved. An application to a composite quadrature rule is provided.
Встановлено аналоги узагальнення Драгомiра iнтегральної нерiвностi Островського∣∣∣∣∣∣(b− a)
[
λ
f (a) + f (b)
2
+ (1− λ) f (x)
]
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤
≤
[
(b− a)2
4
(
λ2 + (1− λ)2
)
+
(
x− a+ b
2
)2
]∥∥f ′∥∥
∞ .
Отримано деякi точнi нерiвностi. Наведено застосування до складеної квадратурної формули.
1. Introduction. In 1938, Ostrowski established a very interesting inequality for differentiable
mappings with bounded derivatives, as follows:
Theorem 1. Let f : I ⊂ R→ R be a differentiable mapping on I◦, the interior of the interval
I, such that f ′ ∈ L[a, b], where a, b ∈ I with a < b. If |f ′ (x)| ≤M. Then the following inequality:
∣∣∣∣∣∣f (x)− 1
b− a
b∫
a
f (u) du
∣∣∣∣∣∣ ≤M (b− a)
14 +
(
x− a+ b
2
)2
(b− a)2
(1.1)
holds for all x ∈ [a, b]. The constant
1
4
is the best possible in the sense that it cannot be replaced by
a smaller constant.
In [16], Dragomir, Cerone and Roumeliotis proved the following generalization of Ostrowski’s
inequality.
Theorem 2. Let f : [a, b] → R be a continuous on [a, b], differentiable on (a, b) and whose
derivative f ′ is bounded on (a, b). Denote ‖f ′‖∞ := supt∈[a,b] |f ′(t)| <∞. Then
c© M. W. ALOMARI, 2012
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 4 435
436 M. W. ALOMARI∣∣∣∣∣∣(b− a)
[
λ
f (a) + f (b)
2
+ (1− λ) f (x)
]
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤
≤
[
(b− a)2
4
(
λ2 + (1− λ)2
)
+
(
x− a+ b
2
)2
]∥∥f ′∥∥∞ (1.2)
for all λ ∈ [0, 1] and a+ λ
b− a
2
≤ x ≤ b− λb− a
2
.
Using (1.2), the authors obtained estimates for the remainder term of the midpoint, trapezoid, and
Simpson formulae. They also gave applications of the mentioned results in numerical integration and
for special means. For recent results, generalizations and new inequalities of Hermite – Hadamard,
Ostrowski and Simpson’s type the reader may be refer to [1 – 20] and the references therein.
Motivated by [12], Dragomir in [14] has proved the following companion of the Ostrowski
inequality:
Theorem 3. Let f : [a, b] → R be an absolutely continuous function on [a, b]. Then we have
the inequalities ∣∣∣∣∣∣f (x) + f (a+ b− x)
2
− 1
b− a
b∫
a
f (t) dt
∣∣∣∣∣∣ ≤
≤
18 + 2
x− 3a+ b
4
b− a
2
(b− a) ‖f ′‖∞ , f ′ ∈ L∞ [a, b] ,
21/q
(q + 1)1/q
(x− ab− a
)q+1
+
a+ b
2
− x
b− a
q+1
1/q
(b− a)1/q ‖f ′‖[a,b],p ,
p > 1,
1
p
+
1
q
= 1, and f ′ ∈ Lp [a, b] ,1
4
+
∣∣∣∣∣∣∣
x− 3a+ b
4
b− a
∣∣∣∣∣∣∣
‖f ′‖[a,b],1
(1.3)
for all x ∈
[
a,
a+ b
2
]
.
In [15], Dragomir established some inequalities for this companion for mappings of bounded
variation.
Theorem 4. Let f : [a, b]→ R be a mapping of bounded variation on [a, b]. Then we have the
inequalities ∣∣∣∣∣∣f (x) + f (a+ b− x)
2
− 1
b− a
b∫
a
f (t) dt
∣∣∣∣∣∣ ≤
1
4
+
∣∣∣∣∣∣∣
x− 3a+ b
4
b− a
∣∣∣∣∣∣∣
· b∨
a
(f) (1.4)
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 4
A COMPANION OF DRAGOMIR’S GENERALIZATION OF OSTROWSKI’S INEQUALITY . . . 437
for any x ∈
[
a,
a+ b
2
]
, where
∨b
a(f) denotes the total variation of f on [a, b]. The constant 1/4 is
best possible.
In [19], Liu introduced some companions of an Ostrowski type inequality for functions whose
first derivative are absolutely continuous. In [9], Barnett, Dragomir and Gomma have proved some
companions for the Ostrowski inequality and the generalized trapezoid inequality. Recently, Alomari
[2] proved a companion inequality for differentiable mappings whose first derivatives are bounded.
In this paper, we prove a companion of Dragomir’s generalization of Ostrowski’s inequality (1.2).
Namely, inequalities for mappings of bounded variation and for absolutely continuous mappings
whose first derivatives are belong to L∞[a, b] and to Lp[a, b] are established.
2. The case when f is of bounded variation.
Theorem 5. Let f : [a, b] → R be a mapping of bounded variation on [a, b]. Then for all
λ ∈ [0, 1] and a+ λ
b− a
2
≤ x ≤ a+ b
2
, we have the inequality∣∣∣∣∣∣(b− a)
[
λ
f (a) + f (b)
2
+ (1− λ) f (x) + f (a+ b− x)
2
]
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤
≤
max
{
λ
b− a
2
,
(
x− (2− λ) a+ λb
2
)
,
(
a+ b
2
− x
)}
·
∨b
a (f) ,
b− a
2
max
{∨x
a (f) ,
∨a+b−x
x (f) ,
∨b
a+b−x (f)
}
,
(2.1)
where
∨b
a (f) denotes to the total variation of f over [a, b]. The constant
1
2
in the second inequality
is the best possible in the sense that it cannot be replaced by a smaller one.
Proof. Using the integration by parts formula for Riemann – Stieltjes integral, we have
x∫
a
(
t−
(
a+ λ
b− a
2
))
df (t) =
(
x− a− λb− a
2
)
f (x) + λ
b− a
2
f (a)−
x∫
a
f(t)dt,
a+b−x∫
x
(
t− a+ b
2
)
df (t) =
(
a+ b
2
− x
)
(f (x) + f (a+ b− x))−
a+b−x∫
x
f(t)dt,
and
b∫
a+b−x
(
t−
(
b− λb− a
2
))
df (t) =
= λ
b− a
2
f (b) +
(
x− a− λb− a
2
)
f (a+ b− x)−
b∫
a+b−x
f(t)dt.
Adding the above equalities, we get
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 4
438 M. W. ALOMARI
b∫
a
K (x, t) f ′(t)dt = (b− a)
[
λ
f (a) + f (b)
2
+ (1− λ) f (x) + f (a+ b− x)
2
]
−
b∫
a
f(t)dt,
where
K (x, t) =
t−
(
a+ λ
b− a
2
)
, t ∈ [a, x] ,
t− a+ b
2
, t ∈ (x, a+ b− x] ,
t−
(
b− λb− a
2
)
, t ∈ (a+ b− x, b] ,
for all λ ∈ [0, 1] and a+ λ
b− a
2
≤ x ≤ a+ b
2
.
Now, we use the fact that for a continuous function p : [c, d] → R and a function ν : [c, d] → R
of bounded variation, one has the inequality∣∣∣∣∣∣
d∫
c
p(t)dν (t)
∣∣∣∣∣∣ ≤ sup
t∈[c,d]
|p(t)|
b∨
a
(ν) . (2.2)
Applying the inequality (2.2) for p(t) = K (x, t) , as above and ν(t) = f(t), t ∈ [a, b], we get∣∣∣∣∣∣
b∫
a
K (x, t) df(t)
∣∣∣∣∣∣ ≤
∣∣∣∣∣∣
x∫
a
K (x, t) df (t)
∣∣∣∣∣∣+
∣∣∣∣∣∣
a+b−x∫
x
K (x, t) df(t)
∣∣∣∣∣∣+
∣∣∣∣∣∣
b∫
a+b−x
K (x, t) df(t)
∣∣∣∣∣∣ ≤
≤ sup
t∈[a,x]
|K (x, t)| ·
x∨
a
(f) + sup
t∈[x,a+b−x]
|K (x, t)| ·
a+b−x∨
x
(f)+
+ sup
t∈[a+b−x,b]
|K (x, t)| ·
b∨
a+b−x
(f) =
= max
{
λ
b− a
2
,
(
x− (2− λ) a+ λb
2
)}
·
x∨
a
(f) +
(
a+ b
2
− x
)
·
a+b−x∨
x
(f)+
+max
{
λ
b− a
2
,
(
x− (2− λ) a+ λb
2
)}
·
b∨
a+b−x
(f) :=M(x).
Now, observe that
M(x) ≤ max
{
λ
b− a
2
,
(
x− (2− λ) a+ λb
2
)
,
(
a+ b
2
− x
)}
×
×
[
x∨
a
(f) +
a+b−x∨
x
(f) +
b∨
a+b−x
(f)
]
=
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 4
A COMPANION OF DRAGOMIR’S GENERALIZATION OF OSTROWSKI’S INEQUALITY . . . 439
= max
{
λ
b− a
2
,
(
x− (2− λ) a+ λb
2
)
,
(
a+ b
2
− x
)}
·
b∨
a
(f) ,
which proves the first inequality in (2.1). Also,
M(x) ≤ max
{
x∨
a
(f) ,
a+b−x∨
x
(f) ,
b∨
a+b−x
(f)
}
×
×
[
λ
b− a
2
+
(
x− a− λb− a
2
)
+
(
a+ b
2
− x
)]
=
=
b− a
2
max
{
x∨
a
(f) ,
a+b−x∨
x
(f) ,
b∨
a+b−x
(f)
}
for all λ ∈ [0, 1] and a+ λ
b− a
2
≤ x ≤ a+ b
2
, thus the second inequality in (2.1) is proved. To
prove that the constant
1
2
in the second inequality is sharp, assume that the second inequality holds
with constant C > 0, i.e.,∣∣∣∣∣∣(b− a)
[
λ
f (a) + f (b)
2
+ (1− λ) f (x) + f (a+ b− x)
2
]
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤
≤ C (b− a) ·max
{
x∨
a
(f) ,
a+b−x∨
x
(f) ,
b∨
a+b−x
(f)
}
(2.3)
for all λ ∈ [0, 1] and a+ λ
b− a
2
≤ x ≤ a+ b
2
. Consider the mapping
f(t) =
0, t ∈ (a, b),
1, t = a, b,
then for x = a and λ = 0, we have
∫ b
a
f(t)dt = 0,
∨b
a (f) = 2, making of use (2.3), we get
(b− a) ≤ 2C (b− a) ,
which gives
1
2
≤ C and thus
1
2
is the best possible, which completes the proof.
Remark 1. In Theorem 5, choose λ = 0, then we get∣∣∣∣∣∣(b− a) f (x) + f (a+ b− x)
2
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤
≤ (x− a) ·
x∨
a
(f) +
(
a+ b
2
− x
)
·
a+b−x∨
x
(f) + (x− a) ·
b∨
a+b−x
(f) ≤
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 4
440 M. W. ALOMARI
≤ max
{
(x− a) ,
(
a+ b
2
− x
)}
·
b∨
a
(f) =
[
1
4
(b− a) +
∣∣∣∣x− 3a+ b
4
∣∣∣∣] · b∨
a
(f) ,
which gives (1.4).
Corollary 1. Let f as in Theorem 5, then we have∣∣∣∣∣∣(b− a)
[
λ
f (a) + f (b)
2
+ (1− λ) f
(
a+ b
2
)]
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤
≤ b− a
2
[
1
2
+
∣∣∣∣λ− 1
2
∣∣∣∣] · b∨
a
(f) (2.4)
for all λ ∈ [0, 1]. The ‘first’ constant
1
2
is the best possible in the sense that it cannot be replaced by
a smaller one.
Proof. In Theorem 5, choose x =
a+ b
2
, we get∣∣∣∣∣∣(b− a)
[
λ
f (a) + f (b)
2
+ (1− λ) f
(
a+ b
2
)]
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤
≤ max
{
λ
b− a
2
, (1− λ) b− a
2
}
·
b∨
a
(f) =
=
[
b− a
2
·max {λ, (1− λ)}
]
·
b∨
a
(f) =
1
2
[
1
2
+
∣∣∣∣λ− 1
2
∣∣∣∣] (b− a) · b∨
a
(f)
which proves the inequality (2.4). To prove that the constant
1
2
is sharp, assume that the inequal-
ity (2.4) holds with constant C > 0, i.e.,∣∣∣∣∣∣(b− a)
[
λ
f (a) + f (b)
2
+ (1− λ) f
(
a+ b
2
)]
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤
≤ C
[
1
2
+
∣∣∣∣λ− 1
2
∣∣∣∣] (b− a) · b∨
a
(f) (2.5)
for all λ ∈ [0, 1]. Consider the mapping
f(t) =
0, t ∈ [a, b]\
{
a+ b
2
}
,
1, t =
a+ b
2
,
then we have
∫ b
a
f(t)dt = 0,
∨b
a (f) = 2, and choose λ = 0, making of use (2.5), we get
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 4
A COMPANION OF DRAGOMIR’S GENERALIZATION OF OSTROWSKI’S INEQUALITY . . . 441
b− a ≤ 2C (b− a) ,
which gives
1
2
≤ C and thus
1
2
is the best possible, which completes the proof.
Corollary 2. In Corollary 1, if we choose
(1) λ = 0, then we get
∣∣∣∣∣∣(b− a) f
(
a+ b
2
)
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤ 1
2
(b− a) ·
b∨
a
(f) ,
(2) λ =
1
3
, then we get
∣∣∣∣∣∣b− a6
[
f (a) + 4f
(
a+ b
2
)
+ f (b)
]
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤ 1
3
(b− a) ·
b∨
a
(f) ,
(3) λ =
1
2
, then we get
∣∣∣∣∣∣b− a2
[
f (a) + f (b)
2
+ f
(
a+ b
2
)]
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤ 1
4
(b− a) ·
b∨
a
(f) ,
(4) λ = 1, then we get
∣∣∣∣∣∣(b− a) f (a) + f (b)
2
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤ 1
2
(b− a) ·
b∨
a
(f) .
The constants
1
2
,
1
3
,
1
4
and
1
2
are the best possible.
Corollary 3. In (2.1), choose λ =
1
4
and x =
2a+ b
3
, then we get the following 3/8-Simpson’s
inequality:
∣∣∣∣∣∣b− a8
[
f (a) + 3f
(
2a+ b
3
)
+ 3f
(
a+ 2b
3
)
+ f (b)
]
−
b∫
a
f (t) dt
∣∣∣∣∣∣ ≤
≤
5b− a
24
·
∨b
a (f) ,
b− a
2
·max
{∨ 2a+b
3
a (f) ,
∨a+2b
3
2a+b
3
(f) ,
∨b
a+2b
3
(f)
}
.
(2.6)
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442 M. W. ALOMARI
3. The case when f ′ ∈ L∞[a, b].
Theorem 6. Let f : I ⊂ R→ R be an absolutely continuous mapping on I◦, the interior of the
interval I, where a, b ∈ I with a < b. If f ′ is bounded on [a, b], i.e., ‖f ′‖∞ := supt∈[a,b] |f ′(t)| <∞.
Then the inequality∣∣∣∣∣∣(b− a)
[
λ
f (a) + f (b)
2
+ (1− λ) f (x) + f (a+ b− x)
2
]
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤
≤
[
(b− a)2
8
(
2λ2 + (1− λ)2
)
+ 2
(
x− (3− λ) a+ (1 + λ) b
4
)2
]∥∥f ′∥∥∞ (3.1)
holds, for all λ ∈ [0, 1] and a+ λ
b− a
2
≤ x ≤ a+ b
2
.
Proof. Defining the mapping
K (x, t) =
t−
(
a+ λ
b− a
2
)
, t ∈ [a, x] ,
t− a+ b
2
, t ∈ (x, a+ b− x] ,
t−
(
b− λb− a
2
)
, t ∈ (a+ b− x, b] ,
(3.2)
for all λ ∈ [0, 1] and a+ λ
b− a
2
≤ x ≤ a+ b
2
.
Integrating by parts, we obtain
b∫
a
K (x, t) f ′(t)dt = (b− a)
[
λ
f (a) + f (b)
2
+ (1− λ) f (x) + f (a+ b− x)
2
]
−
b∫
a
f(t)dt.
Since, f ′ is bounded, we can state that∣∣∣∣∣∣(b− a)
[
λ
f (a) + f (b)
2
+ (1− λ) f (x) + f (a+ b− x)
2
]
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤
≤
b∫
a
|K (x, t)|
∣∣f ′(t)∣∣ dt ≤ ∥∥f ′∥∥∞
b∫
a
|K (x, t)| dt.
Now, since
r∫
p
|t− q| dt =
q∫
p
(q − t) dt+
r∫
q
(t− q) dt = (q − p)2 + (r − q)2
2
=
=
1
4
(p− r)2 +
(
q − r + p
2
)2
(3.3)
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for all r, p, q such that p ≤ q ≤ r. Then, we observe that
x∫
a
∣∣∣∣t− (a+ λ
b− a
2
)∣∣∣∣ dt = 1
4
(x− a)2 +
(
λ
b− a
2
− x− a
2
)2
,
a+b−x∫
x
∣∣∣∣t− a+ b
2
∣∣∣∣ dt = (x− a+ b
2
)2
,
and
b∫
a+b−x
∣∣∣∣t− (b− λb− a2
)∣∣∣∣ dt = 1
4
(x− a)2 +
(
x− a
2
− λb− a
2
)2
.
Then, we have
b∫
a
|K (x, t)| dt = (x− a)2 + ((x− a)− λ (b− a))2
2
+
(
x− a+ b
2
)2
=
=
1
4
λ2 (b− a)2 +
(
x− (2− λ) a+ λb
2
)2
︸ ︷︷ ︸
by (3.3)
+
(
x− a+ b
2
)2
=
=
λ2
4
(b− a)2 + (1− λ)2
8
(b− a)2 + 2
(
x− (3− λ) a+ (1 + λ) b
4
)2
︸ ︷︷ ︸
by (3.3)
=
=
(b− a)2
8
(
2λ2 + (1− λ)2
)
+ 2
(
x− (3− λ) a+ (1 + λ) b
4
)2
,
which gives that∣∣∣∣∣∣(b− a)
[
λ
f (a) + f (b)
2
+ (1− λ) f (x) + f (a+ b− x)
2
]
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤
≤
[
(b− a)2
8
(
2λ2 + (1− λ)2
)
+ 2
(
x− (3− λ) a+ (1 + λ) b
4
)2
]∥∥f ′∥∥∞
for all λ ∈ [0, 1] and a+ λ
b− a
2
≤ x ≤ a+ b
2
, which gives the required result.
Remark 2. In (3.1), choose λ = 0, then we have∣∣∣∣∣∣(b− a) f (x) + f (a+ b− x)
2
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤
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444 M. W. ALOMARI
≤
[
(b− a)2
8
+ 2
(
x− 3a+ b
4
)2]∥∥f ′∥∥∞ ,
which is equivalent to the first inequality in (1.3), and if we choose x =
3a+ b
4
, then we have∣∣∣∣∣∣b− a2
[
f
(
3a+ b
4
)
+ f
(
a+ 3b
4
)]
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤ (b− a)2
8
∥∥f ′∥∥∞ .
Corollary 4. Let f as in Theorem 6, then we get∣∣∣∣∣∣(b− a)
[
λ
f (a) + f (b)
2
+ (1− λ) f
(
a+ b
2
)]
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤
≤
(
λ2 + (1− λ)2
) (b− a)2
4
∥∥f ′∥∥∞ . (3.4)
The constant
1
4
is the best possible in the sense that it cannot be replaced by a smaller one.
Proof. In the proof of Theorem 6, choose x =
a+ b
2
we get the required result. To show that
1/4 is the best possible (3.4). Assume (3.4) holds with constant C > 0, i.e.,∣∣∣∣∣∣(b− a)
[
λ
f (a) + f (b)
2
+ (1− λ) f
(
a+ b
2
)]
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤
≤ C
(
λ2 + (1− λ)2
)
(b− a)2 ·
∥∥f ′∥∥∞ (3.5)
for all λ ∈ [0, 1]. Consider the function f (t) =
∣∣∣∣t− a+ b
2
∣∣∣∣, t ∈ [a, b], then
b∫
a
f(t)dt =
(b− a)2
4
and ‖f ′‖∞ = 1. Using (3.5) with λ = 1, we get
1
4
≤ C, which shows that 1/4 is the best possible,
which completes the proof.
Corollary 5. In Corollary 4, if we choose
(1) λ = 0, then we get∣∣∣∣∣∣(b− a) f
(
a+ b
2
)
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤ 1
4
(b− a)2
∥∥f ′∥∥∞ ,
(2) λ =
1
3
, then we get∣∣∣∣∣∣b− a6
[
f (a) + 4f
(
a+ b
2
)
+ f (b)
]
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤ 5
36
(b− a)2
∥∥f ′∥∥∞ ,
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(3) λ =
1
2
, then we get∣∣∣∣∣∣b− a2
[
f (a) + f (b)
2
+ f
(
a+ b
2
)]
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤ 1
8
(b− a)2
∥∥f ′∥∥∞ ,
(4) λ = 1, then we get∣∣∣∣∣∣(b− a) f (a) + f (b)
2
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤ 1
4
(b− a)2
∥∥f ′∥∥∞ .
The constants
1
4
,
5
36
,
1
8
and
1
4
are the best possible.
Corollary 6. In (3.1), choose λ =
1
4
and x =
2a+ b
3
, then we get the following 3/8-Simpson’s
inequality ∣∣∣∣∣∣b− a8
[
f (a) + 3f
(
2a+ b
3
)
+ 3f
(
a+ 2b
3
)
+ f (b)
]
−
b∫
a
f (t) dt
∣∣∣∣∣∣ ≤
≤ 25
288
(b− a)2 ·
∥∥f ′∥∥∞ . (3.6)
4. The case when f ′ ∈ Lp[a, b].
Theorem 7. Let f : I ⊂ R→ R be an absolutely continuous mapping on I◦, the interior of the
interval I, where a, b ∈ I with a < b. If f ′ is belong to Lp[a, b], p > 1. Then we have the following
inequality: ∣∣∣∣∣∣(b− a)
[
λ
f (a) + f (b)
2
+ (1− λ) f (x) + f (a+ b− x)
2
]
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤
≤
(
2
q + 1
)1/q ∥∥f ′∥∥
p
[(
λ
b− a
2
)q+1
+
(
a+ b
2
− x
)q+1
+
(
x− (2− λ) a+ λb
2
)q+1
]1/q
(4.1)
for all λ ∈ [0, 1], a+ λ
b− a
2
≤ x ≤ a+ b
2
, and
1
p
+
1
q
= 1, p > 1.
Proof. Using Hölder inequality, we have∣∣∣∣∣∣b− a2
[λ (f (a) + f (b)) + (1− λ) (f (x) + f (a+ b− x))]−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤
≤
b∫
a
|K (x, t)|q dt
1/q b∫
a
∣∣f ′ (t)∣∣p dt
1/p =
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446 M. W. ALOMARI
=
∥∥f ′∥∥
p
x∫
a
∣∣∣∣t− (a+ λ
b− a
2
)∣∣∣∣q dt+
a+b−x∫
x
∣∣∣∣t− a+ b
2
∣∣∣∣q dt+
b∫
a+b−x
∣∣∣∣t− (b− λb− a2
)∣∣∣∣q dt
=
=
(
2
q + 1
)1/q ∥∥f ′∥∥
p
[(
λ
b− a
2
)q+1
+
(
a+ b
2
− x
)q+1
+
(
x− (2− λ) a+ λb
2
)q+1
]1/q
for all λ ∈ [0, 1], a+ λ
b− a
2
≤ x ≤ a+ b
2
, and
1
p
+ 1
q = 1, p > 1.
Remark 3. In Theorem 7, choose λ = 0, then we have∣∣∣∣∣∣(b− a) f (x) + f (a+ b− x)
2
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤
≤
(
2
q + 1
)1/q ∥∥f ′∥∥
p
[(
a+ b
2
− x
)q+1
+ (x− a)q+1
]1/q
,
which is equivalent to the second inequality in (1.3), and if x =
3a+ b
4
, then we have∣∣∣∣∣∣b− a2
[
f
(
3a+ b
4
)
+ f
(
a+ 3b
4
)]
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤ (b− a)(q+1)/q
4 (q + 1)1/q
∥∥f ′∥∥
p
.
Corollary 7. In Theorem (7), choose x =
a+ b
2
, we get∣∣∣∣∣∣(b− a)
[
λ
f (a) + f (b)
2
+ (1− λ) f
(
a+ b
2
)]
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤
≤ 1
2
(
λq+1 + (1− λ)q+1
q + 1
)1/q
(b− a)(q+1)/q
∥∥f ′∥∥
p
. (4.2)
The constant
1
2
is the best possible in the sense that it cannot be replaced by a smaller one.
Proof. In the proof of Theorem 7, choose x =
a+ b
2
we get the required result. To show that
1/2 is the best possible (4.2). Assume (4.2) holds with constant C > 0, i.e.,∣∣∣∣∣∣(b− a)
[
λ
f (a) + f (b)
2
+ (1− λ) f
(
a+ b
2
)]
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤
≤ C
(
λq+1 + (1− λ)q+1
q + 1
)1/q
(b− a)
q+1
q
∥∥f ′∥∥
p
(4.3)
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for all λ ∈ [0, 1]. Consider the function f(t) =
∣∣∣∣t− a+ b
2
∣∣∣∣, t ∈ [a, b], then
∫ b
a
f(t)dt =
(b− a)2
4
and ‖f ′‖p = (b− a)1/p . Using (4.3) with λ = 0, we get
(b− a)2
4
≤ C 1
(q + 1)1/q
(b− a)(q+1)/q (b− a)1/p ,
which gives
1
4
≤ C
(q + 1)1/q
for any q > 1. Letting q → 1+, we deduce that C ≥ 1
2
, and the sharpness of the constant in (4.2) is
proved, which completes the proof.
Corollary 8. In Corollary 7, if we choose
(1) λ = 0, then we get∣∣∣∣∣∣(b− a) f
(
a+ b
2
)
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤ (b− a)(q+1)/q
2 (q + 1)1/q
∥∥f ′∥∥
p
,
(2) λ =
1
3
, then we get
∣∣∣∣∣∣b− a6
[
f (a) + 4f
(
a+ b
2
)
+ f (b)
]
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤
≤ 1
6
(
1 + 2q+1
3 (q + 1)
)1/q
(b− a)(q+1)/q
∥∥f ′∥∥
p
,
(3) λ =
1
2
, then we get
∣∣∣∣∣∣b− a2
[
f (a) + f (b)
2
+ f
(
a+ b
2
)]
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤ (b− a)(q+1)/q
4 (q + 1)1/q
∥∥f ′∥∥
p
,
(4) λ = 1, then we get∣∣∣∣∣∣b− af (a) + f (b)
2
−
b∫
a
f(t)dt
∣∣∣∣∣∣ ≤ (b− a)(q+1)/q
2 (q + 1)1/q
∥∥f ′∥∥
p
.
The constants
1
2 (q + 1)1/q
,
1
6
(
1 + 2q+1
3 (q + 1)
)1/q
,
1
4 (q + 1)1/q
and
1
2 (q + 1)1/q
are the best pos-
sible.
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448 M. W. ALOMARI
Corollary 9. In (4.1), choose λ =
1
4
and x =
2a+ b
3
, then we get the following 3/8-Simpson’s
inequality: ∣∣∣∣∣∣b− a8
[
f (a) + 3f
(
2a+ b
3
)
+ 3f
(
a+ 2b
3
)
+ f (b)
]
−
b∫
a
f (t) dt
∣∣∣∣∣∣ ≤
≤
(
2
q + 1
)1/q
[(
1
8
)q+1
+
(
1
6
)q+1
+
(
5
24
)q+1
]1/q
(b− a)(q+1)/q
∥∥f ′∥∥
p
. (4.4)
5. A composite quadrature formula. Let In : a = x0 < x1 < . . . < xn = b be a division of the
interval [a, b] and hi = xi+1 − xi, i = 0, 1, 2, . . . , n− 1.
Consider the general quadrature formula
Qn (In, f) :=
n−1∑
i=0
hi
2
[
λ (f (xi) + f (xi+1)) + (1− λ) (f (αi) + f (xi + xi+1 − αi))
]
(5.1)
for all λ ∈ [0, 1] and xi + λ
xi+1 − xi
2
≤ αi ≤
xi + xi+1
2
.
The following result holds.
Theorem 8. Let f as in Theorem 5, then we have
b∫
a
f(t)dt = Qn (In, f) +Rn (In, f),
where Qn (In, f) is defined by formula (5.1), and the remainder satisfies the estimates
|Rn (In, f)| ≤
n−1∑
i=0
max
{
λ
hi
2
,
(
αi −
(2− λ)xi + λxi+1
2
)
,
(
xi + xi+1
2
− αi
)}
·
xi+1∨
xi
(f),
n−1∑
i=0
hi
2
·max
αi∨
xi
(f) ,
xi+xi+1−αi∨
αi
(f) ,
xi+1∨
xi+xi+1−αi
(f)
for all λ ∈ [0, 1] and xi + λ
xi+1 − xi
2
≤ αi ≤
xi + xi+1
2
.
Proof. Applying inequality (2.1) on the intervals [xi, xi+1], we may state that
Ri (Ii, f) =
xi+1∫
xi
f (t) dt− hi
2
[
λ (f (xi) + f (xi+1)) + (1− λ) (f (αi) + f (xi + xi+1 − αi))
]
.
Summing the above inequality over i from 0 to n− 1, we get
Rn (In, f) =
=
n−1∑
i=0
xi+1∫
xi
f(t)dt−
n−1∑
i=0
hi
2
[λ (f (xi) + f (xi+1)) + (1− λ) (f (αi) + f (xi + xi+1 − αi))] =
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A COMPANION OF DRAGOMIR’S GENERALIZATION OF OSTROWSKI’S INEQUALITY . . . 449
=
b∫
a
f(t)dt−
n−1∑
i=0
hi
2
[
λ (f (xi) + f (xi+1)) + (1− λ) (f (αi) + f (xi + xi+1 − αi))
]
,
which follows form (2.1), that
|Rn (In, f)| =
=
∣∣∣∣∣∣
b∫
a
f(t)dt−
n−1∑
i=0
hi
2
[
λ (f (xi) + f (xi+1)) + (1− λ) (f (αi) + f (xi + xi+1 − αi))
]∣∣∣∣∣∣ ≤
≤
n−1∑
i=0
max
{
λ
hi
2
,
(
αi −
(2− λ)xi + λxi+1
2
)
,
(
xi + xi+1
2
− αi
)}
·
xi+1∨
xi
(f),
n−1∑
i=0
hi
2
·max
αi∨
xi
(f) ,
xi+xi+1−αi∨
αi
(f),
xi+1∨
xi+xi+1−αi
(f)
,
which completes the proof.
Theorem 9. Let f as in Theorem 6, then we have
b∫
a
f(t)dt = Qn (In, f) +Rn (In, f) ,
where Qn (In, f) is defined by formula (5.1), and the remainder satisfies the estimates
|Rn (In, f)| ≤
≤
∥∥f ′∥∥∞ n−1∑
i=0
[
h2i
8
(
2λ2 + (1− λ)2
)
+ 2
(
αi −
(3− λ)xi + (1 + λ)xi+1
4
)2
]
for all λ ∈ [0, 1] and xi + λ
xi+1 − xi
2
≤ αi ≤
xi + xi+1
2
.
Proof. The proof is similar to that of Theorem 7, using Theorem 6. We shall omit the details.
Theorem 10. Let f as in Theorem 7, then we have
b∫
a
f(t)dt = Qn (In, f) +Rn (In, f) ,
where Qn (In, f) is defined by formula (5.1), and the remainder satisfies the estimates
|Rn (In, f)| ≤
(
2
q + 1
)1/q ∥∥f ′∥∥
p
×
×
n−1∑
i=0
[(
λ
hi
2
)q+1
+
(
xi + xi+1
2
− αi
)q+1
+
(
αi −
(2− λ)xi + λxi+1
2
)q+1
]1/q
for all λ ∈ [0, 1] and xi + λ
xi+1 − xi
2
≤ αi ≤
xi + xi+1
2
.
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 4
450 M. W. ALOMARI
Proof. The proof is similar to that of Theorem 8, using Theorem 8. We shall omit the details.
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Received 23.11.11
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 4
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