Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 5, Article 192, 2006
A SHARP INEQUALITY OF OSTROWSKI-GRÜSS TYPE
ZHENG LIU
I NSTITUTE OF A PPLIED M ATHEMATICS
FACULTY OF S CIENCE
A NSHAN U NIVERSITY OF S CIENCE AND T ECHNOLOGY
A NSHAN 114044, L IAONING
P EOPLE ’ S R EPUBLIC OF C HINA .
lewzheng@163.net
Received 9 February, 2006; accepted 23 May, 2006
Communicated by J. Sándor
A BSTRACT. The main purpose of this paper is to use a Grüss type inequality for RiemannStieltjes integrals to obtain a sharp integral inequality of Ostrowski-Grüss type for functions
whose first derivative are functions of Lipschitizian type and precisely characterize the functions
for which equality holds.
Key words and phrases: Ostrowski-Grüss type inequality, Grüss type inequality for Riemann-Stieltjes integrals, Lipschitzian
type function, Sharp bound.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
In 1935, G. Grüss (see [4, p. 296]) proved the following integral inequality which gives an
approximation for the integral of a product of two functions in terms of the product of integrals
of the two functions.
Theorem A. Let h, g : [a, b] → R be two integrable functions such that φ ≤ h(x) ≤ Φ and
γ ≤ g(x) ≤ Γ for all x ∈ [a, b], where φ, Φ, γ, Γ are real numbers. Then we have
Z b
Z b
Z b
1
1
1
(1.1)
|T (h, g)| := h(x)g(x)dx −
h(x)dx ·
g(x)dx
b−a a
b−a a
b−a a
1
≤ (Φ − φ)(Γ − γ),
4
and the inequality is sharp, in the sense that the constant
one.
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
163-06
1
4
cannot be replaced by a smaller
2
Z HENG L IU
It is clear that the constant
1
4
is achieved for
a+b
h(x) = g(x) = sgn x −
.
2
From then on, (1.1) has been known in the literature as the Grüss inequality.
In 1998, S.S. Dragomir and I. Fedotov [2] established the following Grüss type inequality for
Riemann-Stieltjes integrals:
Theorem B. Let h, u : [a, b] → R be so that u is L-Lipschitzian on [a, b], i.e.,
|u(x) − u(y)| ≤ L |x − y|
for all x, y ∈ [a, b], h is Riemann integrable on [a, b] and there exists the real numbers m, M so
that m ≤ h(x) ≤ M for all x ∈ [a, b]. Then we have the inequality
Z b
Z
1
u(b) − u(a) b
≤ L(M − m)(b − a)
(1.2)
h(x)du(x)
−
h(t)dt
2
b−a
a
and the constant
1
2
a
is sharp.
In a recent paper [3], the inequality (1.2) has been improved and refined as follows:
Theorem C. Let h, u : [a, b] → R be so that u is L-Lipschitzian on [a, b], h is Riemann
integrable on [a, b] and there exist the real numbers m, M so that m ≤ h(x) ≤ M for all
x ∈ [a, b]. Then we have
Z b
Z b
u(b)
−
u(a)
(1.3)
h(x)du(x)
−
h(t)dt
b−a
a
a
Z b
Z b
1
≤L
h(t)dtdx
h(x) − b − a
a
a
p
≤ L(b − a) T (h, h)
1
≤ L(M − m)(b − a).
2
All the inequalities in (1.3) are sharp and the constant 12 is the best possible one.
Theorem D. Let h, u : [a, b] → R be so that u is (l, L)-Lipschitzian on [a, b], i.e., it satisfies the
condition
l(x2 − x1 ) ≤ u(x2 ) − u(x1 ) ≤ L(x2 − x1 )
for a ≤ x1 ≤ x2 ≤ b with l < L, h is Riemann integral on [a, b] and there exist the real numbers
m, M so that m ≤ h(x) ≤ M for all x ∈ [a, b]. Then we have the inequality
Z b
Z b
u(b)
−
u(a)
(1.4)
h(x)du(x)
−
h(t)dt
b−a
a
a
Z b
Z b
L−l
1
≤
h(x) −
h(t)dtdx
2
b−a a
a
p
L−l
≤
(b − a) T (h, h)
2
1
≤ (L − l)(M − m)(b − a).
4
All the inequalities in (1.4) are sharp and the constant 41 is the best possible one.
In [1], L.J. Dedić et al. have proved the following Ostrowski type inequality as
J. Inequal. Pure and Appl. Math., 7(5) Art. 192, 2006
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A S HARP I NEQUALITY OF O STROWSKI -G RÜSS T YPE
3
Theorem E. If u0 is L-Lipschitzian on [a, b], then for every x ∈ [a, b] we have
Z b
b−a
u(a) + u(b)
a+b
0
u(t)dt −
u(x) +
+ x−
u (x) (1.5) 2
2
2
a
!
3
a + b (b − a)3
1 .
≤L
x−
+
3
2 48
In this paper, we will use Theorem C and Theorem D to obtain some sharp integral inequalities of Ostrowski-Grüss type for functions whose first derivative are functions of Lipschitzian
type. Thus a further generalization of the Ostrowski type inequality and a perturbed version of
the inequality (1.5) is obtained.
2. T HE R ESULTS
Theorem 2.1. Let u : [a, b] → R be a differentiable function so that u0 is (l, L)-Lipschitzian on
[a, b], i.e., satisfies the condition
l(x2 − x1 ) ≤ u0 (x2 ) − u0 (x1 ) ≤ L(x2 − x1 )
(2.1)
for a ≤ x1 ≤ x2 ≤ b with l < L.Then for all x ∈ [a, b] we have
Z b
b−a
u(a) + u(b)
a+b
0
(2.2) u(t)dt −
(u(x) +
+ x−
u (x)
2
2
2
a
a + b 2 (b − a)2 L − l
u0 (b) − u0 (a)
x−
−
−
≤ 4 I(a, b, x),
4
2
12
where
(2.3)
I(a, b, x) =





















































1
6
a+b
2
−x
1
6
a+b
2
−x
a+3b
4
− x [3(x − a) + (b − x)]
h
i3
(b−a)2 2
4 1
a+b 2
+ 3 2 x− 2
+ 48
,
x−
3a+b
4
+4
16
3
h
1
2
x− a+b
2
1
6
x− a+b
2
1
6
x− a+b
2
with
2
+
h
1
2
(b−a)2
48
i3
2
,
− x [3(x − a) + (b − x)]
h
2 (b−a)2 i 32
+4 12 x− a+b
+ 48
,
2
3a+b
4
+ 43
h
1
2
ξ < x < ζ,
ζ ≤ x ≤ θ,
θ < x < η,
[(x − a) + 3(b − x)]
2 (b−a)2 i 32
,
x− a+b
+ 48
2
√
a+b
3(b − a)
ξ=
−
,
2√
6
6(b − a)
,
ζ =a+
6
< ζ < a+b
< θ < a+3b
< η < b.
and a < ξ < 3a+b
4
2
4
J. Inequal. Pure and Appl. Math., 7(5) Art. 192, 2006
[(x − a) + 3(b − x)]
i3
(b−a)2 2
a+b 2
x− 2
+ 48
,
a+3b
4
x−
a ≤ x ≤ ξ,
η≤x≤b
√
a+b
3(b − a)
η=
+
,
2√
6
6(b − a)
θ =b−
6
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4
Z HENG L IU
Proof. Integrating by parts produces the identity
Z b
(2.4)
K(x, t)du0 (t)
a
Z b
1
u(a) + u(b)
a+b
0
=
u(t)dt − (b − a) u(x) +
+ x−
u (x) ,
2
2
2
a
where
( 1
, t ∈ [a, x],
(t − a) t − a+b
2
2
(2.5)
K(x, t) =
1
, t ∈ (x, b].
(t − b) t − a+b
2
2
Moreover,
1
b−a
(2.6)
Z
a
b
1
K(x, t)dt =
4
"
a+b
x−
2
2
#
(b − a)2
−
.
12
Applying the Grüss type inequality (1.4) gives
Z b
Z b
0
0
u
(b)
−
u
(a)
0
K(x, t)du (t) −
K(x, t)dt
b−a
a
a
Z Z b
L − l b 1
dt.
≤
K(x,
t)
−
K(x,
s)ds
2
b
−
a
a
a
Then for any fixed x ∈ [a, b] we can derive from (2.4), (2.5) and (2.6) that
Z b
b
−
a
u(a)
+
u(b)
a
+
b
0
(2.7) u(t)dt −
u(x) +
+ x−
u (x)
2
2
2
a
"
#
2
u0 (b) − u0 (a)
a+b
(b − a)2 L − l
x−
−
I(a, b, x),
−
≤
4
2
12
4
where
x
a+b
1
a + b 2 (b − a)2 I(a, b, x) =
−
x−
−
(t − a) t− 2
dt
2
2
12
a
Z b
a+b
1
a + b 2 (b − a)2 +
−
x−
−
(t − b) t− 2
dt.
2
2
12
x
The last two integrals can be calculated as follows:
For brevity, we put
"
#
2
a+b
1
a+b
(b − a)2
p1 (t) := (t − a) t −
−
x−
−
, t ∈ [a, x],
2
2
2
12
"
#
2
a+b
1
a+b
(b − a)2
p2 (t) := (t − b) t −
−
x−
−
, t ∈ [x, b].
2
2
2
12
Z
Then we have
"
2
2 #
1 (b − a)
a+b
− x−
;
2
12
2
1
b−a
a+b
(b − a)2
p1 (x) =
x+
x−
+
,
2
2
2
24
p1 (a) = p2 (b) =
J. Inequal. Pure and Appl. Math., 7(5) Art. 192, 2006
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A S HARP I NEQUALITY OF O STROWSKI -G RÜSS T YPE
1
p2 (x) =
2
Set
b−a
x−
2
5
a+b
(b − a)2
x−
+
.
2
24
√
a+b
3(b − a)
ξ=
−
,
6
2
√
6(b − a)
ζ =a+
,
6
√
a+b
3(b − a)
η=
+
,
2
6
√
6(b − a)
θ =b−
.
6
It is easy to find that p1 (a) = p2 (b) ≤ 0 for x ∈ [a, ξ] ∪ [η, b], p1 (a) = p2 (b) > 0 for x ∈ (ξ, η)
and p1 (x) ≤ 0 for x ∈ [a, ζ], p1 (x) > 0 for x ∈ (ζ, b], p2 (x) > 0 for x ∈ [a, θ), p2 (x) ≤ 0 for
x ∈ [θ, b]. Notice that
a<ξ<
a+b
a + 3b
3a + b
<ζ<
<θ<
< η < b,
4
2
4
we see that there are five possible cases to be determined.
(i) In case x ∈ [ζ, θ]. p1 (a) = p2 (b) > 0, p1 (x) ≥ 0, p2 (x) ≥ 0 and it is easy to find by elementary calculus
that the function p1 (t) is strictly decreasing in a, 3a+b
and strictly increasing in
4 3a+b
a+3b
, x , also, as the function
strictly
4
p2 (t) isa+3b
decreasing in x, 4 and strictly increasing in
a+3b
3a+b
, b . Moreover, p1 4 = p2 4 < 0. So, p1 (t) has two zeros in (a, x) at the points
4
" # 12
2
3a + b
1
a+b
(b − a)2
−
x−
+
t1 =
4
2
2
48
and
" # 12
2
1
a+b
(b − a)2
3a + b
t2 =
+
x−
+
.
4
2
2
48
Also p2 (t) has two zeros in (x, b) at the points
" # 12
2
a + 3b
1
a+b
(b − a)2
t3 =
−
x−
+
4
2
2
48
and
" # 12
2
a + 3b
1
a+b
(b − a)2
t4 =
+
x−
+
.
4
2
2
48
Thus we have
(2.8)
"
#
2
a+b
1
a+b
(b − a)2
I(a, b, x) =
(t − a) t−
−
x−
+
dt
2
2
2
24
a
2
#
Z t2 " 1
a+b
(b − a)2
a+b
+
x−
−
− (t − a) t−
dt
2
2
24
2
t1
Z x
a+b
1
a + b 2 (b − a)2
+
(t − a) t−
−
x−
+
dt
2
2
2
24
t2
Z
t1
J. Inequal. Pure and Appl. Math., 7(5) Art. 192, 2006
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6
Z HENG L IU
"
#
2
a+b
(b − a)2
1
a+b
+
(t − b) t−
−
x−
+
dt
2
2
2
24
x
2
#
Z t4 " 1
a+b
(b − a)2
a+b
+
x−
−
− (t − b) t−
dt
2
2
24
2
t3
#
2
Z b"
a+b
1
a+b
(b − a)2
+
−
x−
+
dt
(t − b) t−
2
2
2
24
t4
" # 32
2
a+b
(b − a)2
16 1
+
.
=
x−
3 2
2
48
Z
t3
(ii) In case x ∈ [a, ξ], p1 (a) = p2 (b) ≤ 0, p1 (x) < 0,p2 (x) > 0 and p1 (t) is strictly decreasing
in (a, x) as well as p2 (t) is strictly decreasing in (x, a+3b
) and strictly increasing in ( a+3b
, b) with
4
4
a+3b
t3 ∈ (x, 4 ) such that p2 (t3 ) = 0. Thus we have
" 2
#
a+b
1
(b − a)2
a+b
x−
−
− (t − a) t−
dt
I(a, b, x) =
2
2
24
2
a
#
2
Z t3 "
a+b
1
a+b
(b − a)2
+
(t − b) t−
−
x−
+
dt
2
2
2
24
x
2
#
Z b" a+b
(b − a)2
a+b
1
x−
−
− (t − b) t−
dt
+
2
2
24
2
t3
1 a+b
a + 3b
=
−x
− x [3(x − a) + (b − x)]
6
2
4
" #3
2
2 2
4 1
a+b
(b − a)
+
x−
+
.
3 2
2
48
x
Z
(2.9)
(iii) In case (ξ, ζ), p1 (a) = p2 (b) > 0, p1 (x) < 0,p2 (x) > 0 and p1 (t) has a unique zero t1 ∈
(a, x), p2 (t) has two zeros t3 , t4 ∈ (x, b). Thus we have
(2.10)
I(a, b, x) =
t1
"
2
2
#
a+b
1
a+b
(b − a)
−
x−
+
dt
2
2
2
24
a
Z x 1
a + b 2 (b − a)2
a+b
+
x−
−
− (t − a) t−
dt
2
2
24
2
t1
#
2
Z t3 "
a+b
1
a+b
(b − a)2
+
(t − b) t−
−
x−
+
dt
2
2
2
24
x
2
#
Z t4 " 1
a+b
(b − a)2
a+b
+
x−
−
− (t − b) t−
dt
2
2
24
2
t3
Z b
a+b
1
a + b 2 (b − a)2
+
(t − b) t−
−
x−
+
dt
2
2
2
24
t4
Z
(t − a) t−
J. Inequal. Pure and Appl. Math., 7(5) Art. 192, 2006
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A S HARP I NEQUALITY OF O STROWSKI -G RÜSS T YPE
1
=
6
7
a+b
3a + b
−x
x−
[(x − a) + 3(b − x)]
2
4
# 32
" 2
a+b
(b − a)2
1
+
+4
x−
.
2
2
48
(iv) In case x ∈ (θ, η), p1 (a) = p2 (b) > 0, p1 (x) > 0, p2 (x) < 0 and p1 (t) has two zeros
t1 , t2 ∈ (a, x), p2 (t) has a unique zero t4 ∈ (x, b). Thus we have
Z t1 1
a + b 2 (b − a)2
a+b
(t − a) t−
(2.11)
I(a, b, x) =
−
x−
+
dt
2
2
2
24
a
2
#
Z t2 " a+b
(b − a)2
a+b
1
−
+
x−
− (t − a) t−
dt
2
2
24
2
t1
#
2
Z x"
a+b
1
a+b
(b − a)2
+
(t − a) t−
−
x−
+
dt
2
2
2
24
t2
2
#
Z t4 " 1
a+b
(b − a)2
a+b
+
x−
−
− (t − b) t−
dt
2
2
24
2
x
#
2
Z b"
a+b
1
a+b
(b − a)2
+
(t − b) t−
−
x−
+
dt
2
2
2
24
t4
1
a+b
a + 3b
=
x−
− x [3(x − a) + (b − x)]
6
2
4
# 32
" 2
a+b
(b − a)2
1
x−
+
+4
.
2
2
48
(v) In case x ∈ [η, b], p1 (a) = p2 (b) ≤ 0, p1 (x) > 0, p2 (x) < 0 and p1 (t) has a unique zero
t2 ∈ (a, x), p2 (t) ≤ 0 for t ∈ [x, b]. Thus we have
2
#
Z t2 " 1
a+b
(b − a)2
a+b
(2.12)
I(a, b, x) =
x−
−
− (t − a) t−
dt
2
2
24
2
a
#
2
Z x"
a+b
1
a+b
(b − a)2
+
(t − a) t−
−
x−
+
dt
2
2
2
24
t2
2
#
Z b" 1
a+b
(b − a)2
a+b
+
x−
−
− (t − b) t −
dt
2
2
24
2
x
1
a+b
3a + b
=
x−
x−
[(x − a) + 3(b − x)]
6
2
4
" # 32
2
4 1
a+b
(b − a)2
+
x−
+
.
3 2
2
48
Consequently, the inequality (2.2) with (2.3) follows from (2.7), (2.8), (2.9), (2.10), (2.11) and
(2.12).
The proof is completed.
J. Inequal. Pure and Appl. Math., 7(5) Art. 192, 2006
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8
Z HENG L IU
Remark 2.2. It is not difficult to prove that the inequality (2.2) with (2.3) is sharp in the sense
that we can construct the function u to attain the equality in (2.2) with (2.3). Indeed, we may
choose u such that
 1
(t − a)2 ,
a ≤ t < x,

2




 L (t − x)2 + l [2(x − a)t − (x2 − a2 )], x ≤ t < t3 ,
2
2
u(t) =

l

[(t − t3 )2 + 2(x − a)t − (x2 − a2 )]


2


+ L2 [2(t3 − x)t − (t23 − x2 )],
t3 ≤ t ≤ b,
which follows

l(t − a),
a ≤ t < x,



L(t − x) + (x − a)l,
x ≤ t < t3,
u0 (t) =



l(t − t3 + x − a) + (t3 − x)L, t3 ≤ t ≤ b,
for any x ∈ [a, ξ], and
u(t) =
 L
(t − a)2 ,

2




l

(t − t1 )2 + L2 [2(t1 − a)t − (t21 − a2 )],


2




L

[(t − x)2 + 2(t1 − a)t − (t21 − a2 )]

2




+ l [2(x − t1 )t − (x2 − t2 )],
2
1
a ≤ t < t1 ,
t1 ≤ t < x,
x ≤ t < t3 ,

l

[(t − t23 ) + 2(x − t1 )t − (x2 − t21 )]


2




t3 ≤ t < t4 ,
+ L2 [2(t3 − x + t1 − a)t − (t23 − x2 + t21 − a2 )],





L


[(t − t4 )2 + 2(t3 − x + t1 − a)t − (t23 − x2 + t21 − a2 )]

2


+ 2l [2(t4 − t3 + x − t1 )t − (t24 − t23 + x2 − t21 )], t4 ≤ t ≤ b,
which follows

L(t − a),






l(t − t1 ) + (t1 − a)L,



L(t − x + t1 − a) + (x − t1 )l,
u0 (t) =




l(t − t3 + x − t1 ) + (t3 − x + t1 − a)L,





L(t − t4 + t3 − x + t1 − a) + (t4 − t3 + x − t1 )l,
J. Inequal. Pure and Appl. Math., 7(5) Art. 192, 2006
a ≤ t < t1 ,
t1 ≤ t < x,
x ≤ t < t3 ,
t3 ≤ t < t4 ,
t4 ≤ t ≤ b,
http://jipam.vu.edu.au/
A S HARP I NEQUALITY OF O STROWSKI -G RÜSS T YPE
9
for any x ∈ (ξ, ζ), and
 L
(t − a)2 ,

2





 2l (t − t1 )2 + L2 [2(t1 − a)t − (t21 − a2 )],





L

[(t − t2 )2 + 2(t1 − a)t − (t21 − a2 )]

2




+ 2l [2(t2 − t1 )t − (t22 − t21 )],
u(t) =

l

[(t − t23 ) + 2(t2 − t1 )t − (t22 − t21 )]


2




+ L2 [2(t3 − t2 + t1 − a)t − (t23 − t22 + t21 − a2 )],





L


[(t − t4 )2 + 2(t3 − t2 + t1 − a)t − (t23 − t22 + t21 − a2 )]

2


+ 2l [2(t4 − t3 + t2 − t1 )t− (t24 − t23 + t22 − t21 ) ],
a ≤ t < t1 ,
t1 ≤ t < t2 ,
t2 ≤ t < t3 ,
t3 ≤ t < t4 ,
t4 ≤ t ≤ b,
which follows

L(t − a),







l(t − t1 ) + (t1 − a)L,



L(t − t2 + t1 − a) + (t2 − t1 )l,
u0 (t) =





l(t − t3 + t2 − t1 ) + (t3 − t2 + t1 − a)L,





L(t − t4 + t3 − t2 + t1 − a) + (t4 − t3 + t2 − t1 )l,
a ≤ t < t1 ,
t1 ≤ t < t2 ,
t2 ≤ t < t3 ,
t3 ≤ t < t4 ,
t4 ≤ t ≤ b,
for any x ∈ (ξ, ζ), and
 L
(t − a)2 ,

2




 l (t − t1 )2 + L [2(t1 − a)t − (t2 − a2 )],

1

2
2




L

[(t − t2 )2 + 2(t1 − a)t − (t21 − a2 )]

2




+ 2l [2(t2 − t1 )t − (t22 − t21 )],
u(t) =

l

[(t − x2 ) + 2(t2 − t1 )t − (t22 − t21 )]


2




+ L2 [2(x − t2 + t1 − a)t − (x2 − t22 + t21 − a2 )],





L


[(t − t4 )2 + 2(x − t2 + t1 − a)t − (x2 − t22 + t21 − a2 )]

2


+ 2l [2(t4 − x + t2 − t1 )t − (t24 − x2 + t22 − t21 )],
a ≤ t < t1 ,
t1 ≤ t < t2 ,
t2 ≤ t < x,
x ≤ t < t4 ,
t4 ≤ t ≤ b,
which follows

L(t − a),







l(t − t1 ) + (t1 − a)L,



L(t − t2 + t1 − a) + (t2 − t1 )l,
u0 (t) =





l(t − x + t2 − t1 ) + (x − t2 + t1 − a)L,





L(t − t4 + x − t2 + t1 − a) + (t4 − x + t2 − t1 )l,
J. Inequal. Pure and Appl. Math., 7(5) Art. 192, 2006
a ≤ t < t1 ,
t1 ≤ t < t2 ,
t2 ≤ t < x,
x ≤ t < t4 ,
t4 ≤ t ≤ b,
http://jipam.vu.edu.au/
10
Z HENG L IU
for any x ∈ (θ, η), and
 l
(t − a)2 ,
a ≤ t < t2 ,

2




 L (t − t2 )2 + l [2(t2 − a)t − (t2 − a2 )], t2 ≤ t < x,
2
2
2
u(t) =

l

[(t − x)2 + 2(t2 − a)t − (t22 − a2 )]


2


+ L2 [2(x − t2 )t − (x2 − t22 )],
x ≤ t ≤ b,
which follows

l(t − a),
a ≤ t < t2 ,



L(t − t2 ) + (t2 − a)l,
t2 ≤ t < x,
u0 (t) =



l(t − x + t2 − a) + (x − t2 )L, x ≤ t ≤ b.
for any x ∈ [η, b].
It is clear that all the above u0 (t) satisfy the condition (2.1) on [a, b].
Remark 2.3. For x = a+b
, we have
2
Z b
2
a
+
b
u(a)
+
u(b)
(b
−
a)
b
−
a
0
0
u
+
+
[u
(b)
−
u
a)]
u(t)dt
−
2
2
2
48
a
(L − l)(b − a)3
√
≤
.
144 3
Corollary 2.4. If u0 is L-Lipschitzian on [a, b], then for all x ∈ [a, b] we have
Z b
b−a
u(a) + u(b)
a+b
0
(2.13) u(x) +
+ x−
u (x)
u(t)dt −
2
2
2
a
a + b 2 (b − a)2 L
u0 (b) − u0 (a)
−
x−
−
≤ 2 I(a, b, x),
4
2
12
where I(a, b, x) is as defined in (2.3).
Proof. It is immediate by taking l = −L in the theorem.
R EFERENCES
[1] L.J. DEDIĆ, M. MATIĆ, J. PEČARIĆ AND A. VUKELIĆ, On generalizations of Ostrowski inequality via Euler harmonic identities, J. of Inequal. & Appl., 7(6) (2002), 787–805.
[2] S.S. DRAGOMIR AND I. FEDOTOV, An inequality of Grüss type for Riemann-Stieltjes integral and
applications for special means, Tamkang J. of Math., 29(4) (1998), 286–292.
[3] Z. LIU, Refinement of an inequality of Grüss type for Riemann-Stieltjes integral, Soochow J. of
Math., 30(4) (2004), 483–489.
[4] D.S. MITRINOVIĆ, J. PEČARIĆ AND A.M. FINK, Classical and New Inequalities in Analysis,
Kluwer Academic Publishers, Dordrecht, 1993.
J. Inequal. Pure and Appl. Math., 7(5) Art. 192, 2006
http://jipam.vu.edu.au/