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Ukrainian Mathematical Journal, Vol. 65, No. 7, December, 2013
(Ukrainian Original Vol. 65, No. 7, July, 2013)
NEW SHARP OSTROWSKI-TYPE INEQUALITIES AND
GENERALIZEDTRAPEZOID-TYPE INEQUALITIES FOR
RIEMANNSTIELTJESINTEGRALS AND THEIR APPLICATIONS
M. W. Alomari UDC 517.5
We prove new sharp weighted generalizations of Ostrowski-type
and generalized trapezoid-type inequal-ities for RiemannStieltjes
integrals. Several related inequalities are deduced and
investigated. NewSimpson-type inequalities are obtained for the RS
-integral. Finally, as an application, we estimate the er-ror of a
general quadrature rule for the RS -integral via the
Ostrowskigeneralized-trapezoid-quadratureformula.
1. Introduction
In order to approximate the RiemannStieltjes integral
baf(t)du(t), Dragomir [12] introduced the following
(general) quadrature rule:
D (f, u;x) := f(x) [u (b) u(a)]b
a
f(t)du(t).
Later, numerous authors studied this quadrature rule under
various assumptions imposed on the integrands andintegrators. In
what follows, we present a summary of these results. Let f, u : [a,
b] R be as follows:
(1) f is of the r -Hf -Holder type on [a, b], where Hf > 0
and r (0, 1] are given;(1 ) u is of the s-Hu -Holder type on [a,
b], where Hu > 0 and s (0, 1] are given;(2) f is of bounded
variation on [a, b];
(2 ) u is of bounded variation on [a, b];
(3) f is Lf -Lipschitz on [a, b];
(3 ) u is Lu -Lipschitz on [a, b];
(4) f is monotonically nondecreasing on [a, b];
(4 ) u is monotonically nondecreasing on [a, b];
(5) f is L1,f -Lipschitz on [a, x] and L2,f -Lipschitz on [x,
b];
(5 ) u is L1,u -Lipschitz on [a, x] and L2,u -Lipschitz on [x,
b];
(6) f is monotonically nondecreasing on [a, x] and [x, b];
(6 ) u is monotonically nondecreasing on [a, x] and [x, b];
(7) f is absolutely continuous on [a, b];
(8) |f | is convex on [a, b].Jerash University, Jordan.
Published in Ukrainskyi Matematychnyi Zhurnal, Vol. 65, No. 7,
pp. 894916, July, 2013. Original article submitted May 31,
2012.
0041-5995/13/65070995 c 2013 Springer Science+Business Media New
York 995
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996 M. W. ALOMARI
Then the following inequalities hold under the corresponding
assumptions:
|D (f, u;x)|
Hf
[b a2
+
x a+ b2]rba (u) , (1), (2) [13]
Hu
[(x a)s + (b x)s][1
2
ba(f) +
1
2
xa(f)bx(f)],[(x a)qs + (b x)qs]1/q
[(xa(f))
p +(b
x(f))p]1/p
, (1), (2) [14]
p > 1,1
p+
1
q= 1,[
b a2
+
x a+ b2]sba(f),
LuHfr + 1
[(x a)r+1 + (b x)r+1
], (1), (3) [6]
LfHus+ 1
[(x a)s+1 + (b x)s+1
], (1), (3) [6]
LuLf
14+
(x a+b2b a
)2 (b a)2 , (3), (3) [6]
max {L1,u, L2,u}
[b a2
+
x a+ b2] [f(b) f(a)] ,
(5), (6) [6][f(b) f(a)
2+
1
2
f(x) f(a) + f(b)2] (b a) ,
max {L1,f , L2,f}
[b a2
+
x a+ b2] [u(b) u(a)] ,
(5), (6) [6][u(b) u(a)
2+
1
2
u(x) u(a) + u(b)2] (b a) ,
Hf
[b a2
+
x a+ b2]r [u(b) u(a)] , (1), (4) [11]
Hu
[b a2
+
x a+ b2]s [f(b) f(a)] , (1), (4) [11]
supt[a,x]
{(x t) (f ;x, t)}xa(u) + supt[x,b]
{(t x) (f ;x, t)}bx(u), (2), (7) [7]1
2
[(x a)xa(u) f ,[a,x] + (b x)bx(u) f ,[x,b]]
+1
2|f (x)|
[(x a)xa(u) + (b x)bx(u)]. (2), (7), (8) [7]
(1.1)
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NEW SHARP OSTROWSKI-TYPE INEQUALITIES AND GENERALIZED
TRAPEZOID-TYPE INEQUALITIES 997
For the detailed desription of each inequality presented above,
the reader is referred to the correspondingreferences (and the
references therein).
From a different view point, the authors of [14] considered the
problem of approximation of the Stieltjes
integral baf(t)du(t) by using the following generalized
trapezoid rule:
T (f, u;x) := [u(x) u(a)] f(a) + [(b) u(x)] f(b)b
a
f(t)du(t),
|T (f, u;x)|
Hu
[b a2
+
x a+ b2]rba (f) , (1), (2) [15]
Hf
[(x a)s + (b x)s][1
2
ba(u) +
1
2
xa(u)bx(u)] ,[(x a)qs + (b x)qs]1/q
[(xa(u))
p +(b
x(u))p]1/p
, (1), (2) [8]
p > 1,1
p+
1
q= 1,
[b a2
+
x a+ b2]sba (u) .
(1.2)
For new quadrature rules involving the RS -integral, see the
recent works [1, 2]. For the other results concern-ing various
approximations for RS -integrals under various assumptions on f and
u, see [3, 4, 8, 9, 1518] andthe references therein.
In the recent work [19], Z. Liu proved the following sharp
generalization of the weighted Ostrowski-typeinequality for
mappings of bounded variation (see also [20]):
Theorem 1.1. Let f : [a, b] R be a mapping of bounded variation
and let be g : [a, b] [0,) continu-ous and positive on (a, b).
Then, for any x [a, b] and [0, 1], we have
b
a
f(t)g(t)dt(1 ) f(x) b
a
g(t)dt +
f(a) xa
g (t) dt+ f(b)
bx
g(t)dt
[1
2+
12 ]12
ba
g(t)dt+
xa
g(t)dt 12
ba
g(t)dt
b
a
(f), (1.3)
whereba (f) denotes to the total variation of f over [a, b]. The
constant
[1
2+
12 ] is the best possible
constant.
For recent results concerning the Ostrowski inequality for
mappings of bounded variation, see [11, 1923].
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998 M. W. ALOMARI
The main aim of thie present paper is to introduce and discuss
new weighted generalizations of the Ostrowskiand the generalized
trapezoid inequalities for the RiemannStieltjes integrals.
2. Main Results
We begin with the following result:
Theorem 2.1. Let g, u : [a, b] [0,) be such that g is continuous
and positive on [a, b] and u is mono-tonically increasing on [a,
b]. If f : [a, b] R is a mapping of bounded variation on [a, b],
then, for any x [a, b]and [0, 1], the following inequality is
true:(1 )
f(x) (a+b)/2a
g(s)du(s) + f (a+ b x)b
(a+b)/2
g(s)du(s)
+
f(a) xa
g(s)du(s) + f(b)
bx
g(s)du(s)
ba
f(t)g(t)du(t)
[1
2+
12 ]12
ba
g(t)du(t) +
xa
g(t)du(t) 12
ba
g(t)du(t)
b
a
(f), (2.1)
whereba (f) denotes to the total variation of f over [a, b]
and
[1
2+
12 ] is the best possible constant.
Proof. We define a mapping
Kg,u (t;x) :=
(1 ) tag(s)du(s) +
tx
g(s)du(s), t [a, x] ,
(1 ) t(a+b)/2
g(s)du(s) +
tx
g(s)du(s), t (x, a+ b x] ,
(1 ) tbg(s)du(s) +
tx
g(s)du(s), t (a+ b x, b] .
Integrating by parts, we arrive at the following identity:
ba
Kg,u (t;x) df(t) =
xa
(1 ) ta
g(s)du(s) +
tx
g(s)du(s)
df(t)
+
a+bxx
(1 ) t(a+b)/2
g(s)du(s) +
tx
g(s)du(s)
df(t)
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NEW SHARP OSTROWSKI-TYPE INEQUALITIES AND GENERALIZED
TRAPEZOID-TYPE INEQUALITIES 999
+
ba+bx
(1 ) tb
g(s)du(s) +
tx
g(s)du(s)
df(t)
= (1 )
f(x) (a+b)/2a
g(s)du(s) + f (a+ b x)b
(a+b)/2
g(s)du(s)
+
f(a) xa
g(s)du(s) + f(b)
bx
g(s)du(s)
ba
f(t)g(t)du(t).
Note that, for a continuous function p : [a, b] R and a function
: [a, b] R of bounded variation, theRiemannStieltjes integral
bap(t)d(t) exists and the following inequality is true:
b
a
p(t)d(t)
supt[a,b] |p(t)|ba
() . (2.2)
Since f is of bounded variation on [a, b], by virtue of (2.2),
we find
(1 )f(x) (a+b)/2
a
g(s)du(s) + f (a+ b x)b
(a+b)/2
g(s)du(s)
+
f(a) xa
g(s)du(s) + f(b)
bx
g(s)du(s)
ba
f(t)g(t)du(t)
sup
t[a,b]|Kg,u (t;x)|
ba
(f). (2.3)
We now define mappings p, q : [a, b] R by the formulas
p1(t) := (1 )t
a
g(s)du(s) +
tx
g (s) du(s), t [a, x] ,
p2(t) := (1 )t
(a+b)/2
g(s)du(s) +
tx
g(s)du(s), t (x, a+ b x] ,
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1000 M. W. ALOMARI
p3(t) := (1 )t
b
g(s)du(s) +
tx
g (s) du(s), t (a+ b x, b] ,
for all [0, 1] and x [a, b]. Since g is positive and continuous
and u is monotonically increasing on [a, b], weconclude that the
RiemannStieltjes integral
bag(s)du(s) exists and is positive. In addition, since the
derivative
of the monotonically increasing function u is always positive
and, hence, (gu) (t) > 0 a.e., we conclude that,p1(t), p2(t),
p3(t) > 0 almost everywhere in their corresponding domains.
Thus, we get
supt[a,x]
|Kg,u (t;x)| = max(1 )
xa
g(s)du(s),
xa
g(s)du(s)
=[1
2+
12 ]
xa
g(s)du(s),
supt(x,a+bx]
|Kg,u (t;x)| = max
(1 )(a+b)/2x
g(s)du(s),
(a+b)/2x
g(s)du(s) +
a+bx(a+b)/2
g(s)du(s)
=1
2
a+bxx
g(s)du(s) + (1 )
a+bx
(a+b)/2
g(s)du(s)
,
and
supt(a+bx,b]
|Kg,u (t;x)| = max(1 )
bx
g(s)du(s),
bx
g(s)du(s)
=
[1
2+
12 ]
bx
g(s)du(s).
Hence,
supt[a,b]
|Kg,u (t;x)| =[1
2+
12 ]max
xa
g(s)du(s),
bx
g(s)du(s)
=
[1
2+
12 ]12
ba
g(s)du(s) +
xa
g(s)du(s) 12
ba
g(s)du(s)
. (2.4)
Therefore, by virtue of (2.3) and (2.4), we get (2.1). To prove
that1
2+
12 is the best possible constant
for all [0, 1], we take u(t) = t for all t [a, b] and consider
(1.3). Thus, the sharpness of the estimate followsfrom (1.3) (we
consider f and g defined as in [19]). Hence, the proof is
established and we omit the details.
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NEW SHARP OSTROWSKI-TYPE INEQUALITIES AND GENERALIZED
TRAPEZOID-TYPE INEQUALITIES 1001
Corollary 2.1. Let = 0 in relation (2.1). This yieldsf(x)b
a
g(t)du(t)b
a
f(t)g(t)du(t)
12
ba
g(t)du(t) +
xa
g(t)du(t) 12
ba
g(t)du(t)
b
a
(f). (2.5)
A general weighted version of the Ostrowski inequality for RS
-integrals presented above can be deduced asfollows:
f(x) baf(t)g(t)du(t) bag(t)du(t)
12 +
xag(t)du(t) b
ag(t)du(t)
12
b
a
(f) (2.6)
provided that g(t) 0 for almost all t [a, b] and
ba
g(t)du(t) 6= 0.
Remark 2.1. Choosing = 1 in (2.1), we findf(a)xa
g(s)du(s) + f(b)
bx
g(s)du(s)b
a
f(t)g(t)du(t)
12
ba
g(t)du(t) +
xa
g(t)du(t) 12
ba
g(t)du(t)
b
a
(f). (2.7)
This is the generalized trapezoid inequality for RS
-integrals.
Corollary 2.2. In relation (2.1), let g(t) = 1 for all t [a, b].
Then the following inequality is true: [(u(x) u(a)) f(a) + ((b)
u(x)) f(b)] + (1 ) [u(b) u(a)] f(x)
ba
f(t)du(t)
[1
2+
12 ] [u(b) u(a)2 +
u(x) u(a) + u(b)2 ] b
a
(f), (2.8)
where[1
2+
12 ] is the best possible constant.
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1002 M. W. ALOMARI
In particular,
if = 0, then we get
[u(b) u(a)] f(x)b
a
f(t)du(t)
[u(b) u(a)
2+
u(x) u(a) + (b)2] b
a
(f); (2.9)
if =1
3, then
13 {[u(x) u(a)] f(a) + 2 [u(b) u(a)] f(x) + [u(b) u(x)] f(b)}
b
a
f(t)du(t)
2
3
[u(b) u(a)
2+
u(x) u(a) + u(b)2] b
a
(f); (2.10)
if =1
2, then
12 {[u(x) u(a)] f(a) + [u(b) u(a)] f(x) + [u(b) u(x)] f(b)}
b
a
f(t)du(t)
1
2
[u(b) u(a)
2+
u(x) u(a) + u(b)2] b
a
(f); (2.11)
if = 1, then
[u(x) u(a)] f(a) + [u(b) u(x)] f(b)b
a
f(t)du(t)
[u(b) u(a)
2+
u(x) u(a) + (b)2] b
a
(f). (2.12)
Proof. The results follow from Theorem 2.1. It remains to prove
the sharpness of (2.8). Suppose that
0 12
and that (2.8) holds with a constant C1 > 0, i.e.,
[(u(x) u(a)) f(a) + ((b) u(x)) f(b)]+ (1 ) [u(b) u(a)] f(x)b
a
f(t)du(t)
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NEW SHARP OSTROWSKI-TYPE INEQUALITIES AND GENERALIZED
TRAPEZOID-TYPE INEQUALITIES 1003
C1[u(b) u(a)
2+
u(x) u(a) + u(b)2] b
a
(f). (2.13)
Let f, u : [a, b] R be defined as follows: u(t) = t and
f(t) =
0, t [a, b] \
{a+ b
2
},
1
2, t =
a+ b
2.
This yieldsba(f) = 1 and
baf(t)du(t) = 0. We now set x =
a+ b
2. It follows from (2.13) that
(1 ) b a2 C1 b a
2,
which proves that C1 1 . Hence, 1 is the best possible constant
for all 0 12.
We now assume that1
2 1 and (2.8) holds with a constant C2 > 0, i.e.,
[(u(x) u(a)) f(a) + ((b) u(x)) f(b)]+ (1 ) [u(b) u(a)] f(x)b
a
f(t)du(t)
C2
[u(b) u(a)
2+
u(x) u(a) + u(b)2] b
a
(f). (2.14)
Let f, u : [a, b] R be defined as follows: u(t) = t and
f(t) =
0, t (a, b],
1, t = a.
This yieldsba(f) = 1 and
baf(t)du(t) = 0. Setting x =
a+ b
2, by (2.14), we find
b a2 C2 b a
2,
which proves that C2 and, therefore, is the best possible
constant for all 12 1. This enables us to
conclude conclude that[1
2+
12 ] is the best possible constant for all [0, 1].
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1004 M. W. ALOMARI
Corollary 2.3. Let x =a+ b
2in (2.10). The following Simpson-type inequality for the
RiemannStieltjes
integral is true: 13{[u
(a+ b
2
) u(a)
]f(a) + 2 [u(b) u(a)] f
(a+ b
2
)
+
[u(b) u
(a+ b
2
)]f(b)
}
ba
f(t)du(t)
2
3
[u(b) u(a)
2+
u(a+ b2) u(a) + u(b)
2
] ba
(f). (2.15)
Here,2
3is the best possible constant.
Remark 2.2. For recent three-point quadrature rules and the
corresponding inequalities for RiemannStieltjesintegrals, we refer
the reader to [2].
Corollary 2.4. In (2.8), let u(t) = t for all t [a, b]. Then ((x
a) f(a) + (b x) f(b)) + (1 ) (b a) f(x)
ba
f(t)dt
[1
2+
12 ] [b a2 +
x a+ b2] b
a
(f). (2.16)
For x =a+ b
2, the following inequality is true:
(b a)[f(a) + f(b)
2+ (1 ) f
(a+ b
2
)]
ba
f (t) dt
[1
2+
12 ] b a2
ba
(f). (2.17)
Remark 2.3. Under the assumptions of Theorem 2.1, a weighted
generalization of the Montgomery-typeidentity for RiemannStieltjes
integrals can be deduced as follows:
f(x) =1 b
ag(s)du(s)
ba
Kg,u (t;x) df(t) +1 b
ag(s)du(s)
ba
f(t)g(t)du(t)
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NEW SHARP OSTROWSKI-TYPE INEQUALITIES AND GENERALIZED
TRAPEZOID-TYPE INEQUALITIES 1005
for all x [a, b], where
Kg,u (t;x) :=
tag(s)du(s), t [a, x],
tbg(s)du(s), t (x, b] ,
provided that bag(s)du(s) 6= 0.
3. On L-Lipschitz Integrators
Theorem 3.1. Let g be as in Theorem (2.1). Let u : [a, b] [0,)
be of bounded variation on [a, b]. Iff : [a, b] R is L-Lipschitzian
on [a, b], then, for any x [a, b] and [0, 1],
f(a) x
a
g(s)du(s) + f(b)
bx
g(s)du(s)
+(1 ) f(x)b
a
g(s)du(s)b
a
f(t)g(t)du(t)
Lmax
{(x a) sup
t[a,x]{M(t)}, (b x) sup
t[x,b]{N(t)}
}ba
(u), (3.1)
where
M(t) := max
{(1 ) sup
s[a,t]|g(s)|, sup
s[t,x]|g(s)|
}
and
N(t) := max
{(1 ) sup
s[t,b]|g(s)|, sup
s[t,x]|g(s)|
}.
Proof. By Theorem 2.1, we arrive at the identity
ba
Kg,u (t;x) df(t) =
f(a) xa
g(s)du(s) + f(b)
bx
g(s)du(s)
+ (1 ) f(x)b
a
g(s)du(s)b
a
f(t)g(t)du(t).
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1006 M. W. ALOMARI
Note that, for a Riemann integrable function p : [c, d] R and a
L-Lipschitzian function : [c, d] R, thefollowing inequality is
true:
dc
p(t)d(t)
Ldc
|p(t)| dt. (3.2)
Since f is a L-Lipschitz mapping on [a, b], by (3.2), we
obtainb
a
Kg,u (t;x) df(t)
Lb
a
|Kg,u (t;x)| dt = L xa
|p(t)| dt+b
x
|q(t)| dt. (3.3)
However, since u is of bounded variation on [a, b] and g is
continuous, by (2.2), we find
|p(t)| (1 )t
a
g(s)du(s)
+ t
x
g(s)du(s)
(1 ) sup
s[a,t]|g(s)|
ta
(u) + sups[t,x]
|g(s)|xt
(u)
max{(1 ) sup
s[a,t]|g(s)| , sup
s[t,x]|g(s)|
}xa
(u)
:=M(t)
xa
(u). (3.4)
Similarly, we get
|q(t)| max{(1 ) sup
s[t,b]|g(s)|, sup
s[t,x]|g(s)|
}bx
(u) := N(t)bx
(u). (3.5)
Thus, in view of (3.3)(3.5), we obtainb
a
Kg,u (t;x) df(t)
L xa
|p(t)| dt+b
x
|q(t)| dt
L x
a
M(t)dt
xa
(u) +
bx
N(t)dt
bx
(u)
L[(x a) sup
t[a,x]{M(t)}
xa
(u) + (b x) supt[x,b]
{N(t)}bx
(u)
]
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NEW SHARP OSTROWSKI-TYPE INEQUALITIES AND GENERALIZED
TRAPEZOID-TYPE INEQUALITIES 1007
Lmax{(x a) sup
t[a,x]{M(t)}, (b x) sup
t[x,b]{N(t)}
}ba
(u),
which gives the required result.
Remark 3.1. In Theorem 3.1, if g(t) = 1 for all t [a, b],
then
M(t) = N(t) =
[1
2+
12 ] for all t [a, b].
Corollary 3.1. In estimate (3.1), let g(t) = 1 for all t [a, b].
Then the following inequality is true: [(u(x) u(a)) f(a) + ((b)
u(x)) f(b)] + (1 ) [u(b) u(a)] f(x)
ba
f(t)du(t)
L
[1
2+
12 ] [b a2 +
x a+ b2] b
a
(u), (3.6)
where[1
2+
12 ] is the best possible constant.
Thus, in particular,
if = 0, then we get
[u(b) u(a)] f(x)b
a
f(t)du(t)
L[b a2
+
x a+ b2] b
a
(u); (3.7)
if =1
3, then
13 {[u(x) u(a)] f(a) + 2 [u (b) u(a)] f(x) + [u(b) u(x)] f(b)}
b
a
f(t)du(t)
2
3L
[b a2
+
x a+ b2] b
a
(u); (3.8)
if =1
2, then
12 {[u(x) u(a)] f(a) + [u(b) u(a)] f(x) + [u(b) u(x)] f(b)}
b
a
f(t)du(t)
-
1008 M. W. ALOMARI
12L
[b a2
+
x a+ b2] b
a
(u); (3.9)
if = 1, then
[u(x) u(a)] f(a) + [u(b) u(x)] f(b)b
a
f(t)du(t)
L[b a2
+
x a+ b2] b
a
(u). (3.10)
Proof. The results follow from Theorem 3.1. It remains to prove
the sharpness of (3.6). Suppose that
0 12
and that (3.6) holds with a constant C1 > 0, i.e.,
[(u(x) u(a)) f(a) + ((b) u(x)) f(b)]+ (1 ) [u(b) u(a)] f(x)b
a
f(t)du(t)
LC1
[b a2
+
x a+ b2] b
a
(u). (3.11)
Let f, u : [a, b] R be defined as follows f(t) = t b and let
u(t) =
0, t [a, b),1, t = b.
Hence, f is L-Lipschitz with L = 1,ba(u) = 1, and
ba
f(t)du(t) = 0.
Thus, in view of (3.11), by setting x =a+ b
2, we find
(1 ) b a2 C1 b a
2.
This proves that C1 1 and, therefore 1 is the best possible
constant for all 0 12.
We now suppose that1
2 1 and inequality (3.6) holds with a constant C2 > 0,
i.e.,
[(u(x) u(a)) f(a) + ((b) u(x)) f(b)] + (1 ) [u(b) u(a)]
f(x)b
a
f(t)du(t)
-
NEW SHARP OSTROWSKI-TYPE INEQUALITIES AND GENERALIZED
TRAPEZOID-TYPE INEQUALITIES 1009
LC2[b a2
+
x a+ b2] b
a
(u). (3.12)
Let f, u : [a, b] R be defined as follows: f(t) = t a and
u(t) =
0, t [a, b] \
{a+ b
2
},
1
2, t =
a+ b
2.
Hence, f is L-Lipschitz with L = 1 andba(u) = 1 and
ba
f(t)du(t) = 0.
Setting x =a+ b
2, in view of (3.12), we find
b a2 C2 b a
2,
which proves that C2 and, therefore, is the best possible
constant for all 12 1. Thus, we can
conclude that[1
2+
12 ] is the best possible constant for all [0, 1].
Corollary 3.2. In (3.8), let x =a+ b
2. Then the following Simpson-type inequality for RS -integrals
is true:
13{[u
(a+ b
2
) u(a)
]f(a) + 2 [u(b) u(a)] f
(a+ b
2
)
+
[u(b) u
(a+ b
2
)]f(b)
}
ba
f(t)du(t)
1
3L (b a)
ba
(u), (3.13)
where1
3is the best possible constant.
Corollary 3.3. In (3.6), let u(t) = t for all t [a, b]. Then ((x
a) f(a) + (b x) f(b)) + (1 ) (b a) f(x)b
a
f(t)dt
-
1010 M. W. ALOMARI
L (b a)[1
2+
12 ] [b a2 +
x a+ b2]. (3.14)
For x =a+ b
2, the following inequality is true:
(b a)[f(a) + f(b)
2+ (1 ) f
(a+ b
2
)]
ba
f (t) dt
L[1
2+
12 ] (b a)22 . (3.15)
4. On Monotonic Nondecreasing Integrators
Theorem 4.1. Let g and u be as in Theorem 3.1. If f : [a, b] R
is monotonically nondecreasing on [a, b],then, for any x [a, b] and
[0, 1],
f(a) x
a
g(s)du(s) + f(b)
bx
g(s)du(s)
+(1 ) f(x)b
a
g(s)du(s)b
a
f(t)g(t)du(t)
sup
t[a,x]{M(t)} [f(x) f(a)]
xa
(u) + supt[x,b]
{N(t)} [f(b) f(x)]bx
(u), (4.1)
where M(t) and N(t) are defined in Theorem (3.1).
Proof. We now use the identity
f(a) xa
g(s)du(s) + f(b)
bx
g(s)du(s)
+ (1 ) f(x)b
a
g(s)du(s)b
a
f(t)g(t)du(t) =
ba
Kg,u (t;x) df(t).
It is well known that, for a monotonic nondecreasing function :
[a, b] R and a continuous functionp : [a, b] R, the following
inequality is true:
b
a
p(t)d(t)
b
a
|p(t)| d(t). (4.2)
-
NEW SHARP OSTROWSKI-TYPE INEQUALITIES AND GENERALIZED
TRAPEZOID-TYPE INEQUALITIES 1011
As f is monotonically nondecreasing on [a, b], by (4.2), we
get
b
a
Kg,u (t;x) df(t)
b
a
|Kg,u (t;x)| df(t) =xa
|p(t)| df(t) +b
x
|q(t)| df(t). (4.3)
Further, since u is of bounded variation on [a, b] and g is
continuous, by (3.4) and (3.5), we obtain
|p(t)| M(t)xa
(u),
|q(t)| N(t)bx
(u).
(4.4)
Thus, by (4.3) and (4.4), we find
b
a
Kg,u (t;x) df(t)
xa
|p(t)| df(t) +b
x
|q(t)| df(t)
x
a
M(t)df(t)
xa
(u) +
bx
N(t)df(t)
bx
(u)
supt[a,x]
{M(t)} [f(x) f(a)]xa
(u) + supt[x,b]
{N(t)} [f(b) f(x)]bx
(u),
which gives the required result.
Corollary 4.1. In Theorem (4.1), let g(t) = 1 for all t [a, b].
Then the following inequality is true: [(u(x) u(a)) f(a) + ((b)
u(x)) f(b)]+ (1 ) [u(b) u(a)] f(x)
ba
f(t)du(t)
[1
2+
12 ]{[f(x) f(a)]
xa
(u) + [f(b) f(x)]bx
(u)
}
[1
2+
12 ][f(b) f(a)2 +
f(x) f(a) + f(b)2] b
a
(u). (4.5)
For the last inequality,[1
2+
12 ] is the best possible constant.
-
1012 M. W. ALOMARI
Thus, in particular,
if = 0, then we get
[u(b) u(a)] f(x)b
a
f(t)du(t)
[f(x) f(a)]xa
(u) + [f(b) f (x)]bx
(u)
[f(b) f(a)
2+
f(x) f(a) + f (b)2] b
a
(u); (4.6)
if =1
3, then
13 {[u(x) u(a)] f(a) + 2 [u (b) u(a)] f(x) + [u(b) u(x)] f(b)}
b
a
f(t)du(t)
2
3
{[f(x) f(a)]
xa
(u) + [f(b) f(x)]bx
(u)
}
23
[f(b) f(a)
2+
f(x) f(a) + f(b)2] b
a
(u); (4.7)
if =1
2, then
12 {[u(x) u(a)] f(a) + [u(b) u(a)] f(x) + [u(b) u(x)] f(b)}
b
a
f(t)du(t)
1
2
{[f(x) f(a)]
xa
(u) + [f(b) f(x)]bx
(u)
}
12
[f(b) f(a)
2+
f(x) f(a) + f(b)2] b
a
(u). (4.8)
if = 1, then
[u(x) u(a)] f(a) + [u(b) u(x)] f(b)b
a
f(t)du(t)
-
NEW SHARP OSTROWSKI-TYPE INEQUALITIES AND GENERALIZED
TRAPEZOID-TYPE INEQUALITIES 1013
[f(x) f(a)]xa
(u) + [f(b) f (x)]bx
(u)
[f(b) f(a)
2+
f(x) f(a) + f (b)2] b
a
(u). (4.9)
Proof. The required results follow from Theorem 4.1. It remains
to prove the sharpness of (4.5). Suppose
that 0 12
and that (4.5) holds with a constant C1 > 0, i.e.,
[(u(x) u(a)) f(a) + ((b) u(x)) f(b)]
+ (1 ) [u(b) u(a)] f(x)b
a
f(t)du(t)
C1
[f(b) f(a)
2+
f(x) f(a) + f (b)2] b
a
(u). (4.10)
Let f, u : [a, b] R be defined as follows:
f(t) =
1, t = a,0, t = (a, b],
and
u(t) =
0, t [a, b) ,1, t = b.
Thus, f is monotonically nondecreasing on [a, b],ba(u) = 1,
and
ba
f(t)du(t) = 0.
Setting x = a, in view of (4.10), we conclude that 1 C1. This
proves that 1 is the best possible constantfor all 0 1
2.
We now assume that1
2 1 and (4.5) holds with a constant C2 > 0, i.e.,
[(u(x) u(a)) f(a) + ((b) u(x)) f(b)]
-
1014 M. W. ALOMARI
+(1 ) [u(b) u(a)] f(x)b
a
f(t)du(t)
C2
[f(b) f(a)
2+
f(x) f(a) + f (b)2] b
a
(u). (4.11)
Let f, u : [a, b] R be defined as f(t) above and let u(t) = t.
This yields ba(u) = b a andb
a
f(t)du(t) = 0.
Thus, setting x = b, by (4.11), we see that C2 and, therefore,
is the best possible constant for all1
2 1. Hence, we conclude that
[1
2+
12 ] is the best possible constant for all [0, 1].
Corollary 4.2. In (4.7), let x =a+ b
2. Then the following Simpson-type inequality for RS -integrals
is true:
13{[u
(a+ b
2
) u(a)
]f(a) + 2 [u(b) u(a)] f
(a+ b
2
)
+
[u(b) u
(a+ b
2
)]f(b)
}
ba
f(t)du(t)
2
3
[f
(a+ b
2
) f(a)
] (a+b)/2a
(u) +
[f(b) f
(a+ b
2
)] b(a+b)/2
(u)
2
3
[f(b) f(a)
2+
f (a+ b2) f(a) + f(b)
2
] ba
(u). (4.12)
For the last inequality,2
3is the best possible constant.
Corollary 4.3. In (4.5), let u(t) = t for all t [a, b]. Then ((x
a) f(a) + (b x) f(b)) + (1 ) (b a) f(x)
ba
f(t)dt
[1
2+
12 ] {(x a) [f (x) f(a)] + (b x) [f(b) f(x)]}
-
NEW SHARP OSTROWSKI-TYPE INEQUALITIES AND GENERALIZED
TRAPEZOID-TYPE INEQUALITIES 1015
[1
2+
12 ] [f(b) f(a)2 +
f(x) f(a) + f(b)2] (b a) . (4.13)
For x =a+ b
2,
(b a)[f(a) + f(b)
2+ (1 ) f
(a+ b
2
)]
ba
f (t) dt
1
2(b a)
[1
2+
12 ] [f (b) f(a)]
[1
2+
12 ] [f(b) f(a)2 +
f (a+ b2) f(a) + f(b)
2
](b a) . (4.14)Remark 4.1. Note that, in Theorems 2.1, 3.1, 4.1,
one may get various new inequalities by replacing the
assumptions imposed on u, e.g., by the assumptions of bounded
variation, Lu -Lipschitz, or monotonically nonde-creasing behavior
on [a, b]. In some cases, this gives inequalities dual to the
inequalities established above.
It remains to note that, in Theorem 3.1 and according to the
assumptions imposed on u, one may obtain severalestimations for the
functions p(t) and q(t) which, therefore, gives different functions
M(t) and N(t).
Remark 4.2. In Theorems 2.1, 3.1, and 4.1, different result(s)
in terms of the Lp norms can be formulated ifwe apply the
well-known Holder integral inequality and note that
dc
g(s)du(s)
qu (d) u (c) p d
c
|g(s)|p du (s),
where p > 1 and1
p+
1
q= 1.
Remark 4.3. We can now obtain some results for the Riemann
integral of a product in terms of L1 -, Lp -,and L -norms by using
the reasoning similar to that considered in [12] (see also [1,
2]).
5. Applications to the Ostrowski Generalized Trapezoid
Quadrature Formula for RS -Integrals
Let In : a = x0 < x1 < s < xn = b be a partition of the
interval [a, b]. We define the general RiemannStieltjes sum as
follows:
S (f, u, In, ) =n1i=0
{[u(i) u(xi)] f(xi) + [u(xi+1) u(i)] f(xi+1)}
+ (1 ) [u(xi+1) u(xi)] f (i) . (5.1)
In what follows, we establish the upper bound for the error of
approximation of the RiemannStieltjes integral baf(t)du(t) by its
RiemannStieltjes sums S (f, u, In, ). As an example, we apply
inequality (2.8).
-
1016 M. W. ALOMARI
Theorem 5.1. Under the assumptions of Corollary (2.2), the
following equality is true:
ba
f(t)du(t) = S (f, u, In, ) +R (f, u, In, ) ,
where S (f, u, In, ) is given in (5.1) and the remainder R (f,
u, In, ) has the bound
|R (f, u, In, )| [1
2+
12 ] [u(b) u(a)] b
a
(f). (5.2)
Proof. Applying Corollary 2.2 on the intervals [xi, xi+1], we
conclude that{[u(i) u(xi)] f(xi) + [u(xi+1) u(i)] f(xi+1)}
+(1 ) [u (xi+1) u(xi)] f (i)xi+1xi
f(t)du(t)
[1
2+
12 ] [u(xi+1) u(xi)2 +
u(i) u(xi) + u(xi+1)2] xi+1
xi
(f)
for all i {0, 1, 2, s, n 1}.We now sum this inequality over i
from 0 to n 1 and apply the generalized triangle inequality. This
gives
|R (f, u, In, )| =n1i=0
{[u(i) u(xi)] f(xi) + [u (xi+1) u(i)] f (xi+1)}
+(1 ) [u (xi+1) u(xi)] f (i)xi+1xi
f(t)du(t)
[1
2+
12 ] n1
i=0
[u (xi+1) u(xi)
2+
u (i) u(xi) + u (xi+1)2] xi+1
xi
(f)
[1
2+
12 ][n1i=0
u (xi+1) u(xi)2
+n1i=0
u(i) u(xi) + u(xi+1)2]n1i=0
xi+1xi
(f)
[1
2+
12 ][u(b) u(a)
2+ sup
i=0,1,...,n1
u(i) u(xi) + u (xi+1)2]
ba
(f)
-
NEW SHARP OSTROWSKI-TYPE INEQUALITIES AND GENERALIZED
TRAPEZOID-TYPE INEQUALITIES 1017
[1
2+
12 ][u(b) u(a)] b
a
(f).
Since
supi=0,1,...,n1
u(i) u(xi) + u (xi+1)2 sup
i=0,1,...,n1u(xi+1) u(xi)
2=u(b) u(a)
2,
we get
n1i=0
xi+1xi
(f) =
ba
(f).
This completes the proof.
Remark 5.1. Note that one can use the remaining inequalities
from Section 2 to establish other bounds forR (f, u, In, ). We omit
the details.
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Abstract1. Introduction2. Main Results3. On L-Lipschitz
Integrators4. On Monotonic Nondecreasing Integrators5. Applications
to the Ostrowski Generalized Trapezoid Quadrature Formula for
RS-IntegralsREFERENCES