Ukrainian Mathematical Journal Volume 65 issue 7 2013 [doi 10.1007_s11253-013-0837-z] Alomari, M. W....

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

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)

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].

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)


+

ba+bx

(1 ) tb

g(s)du(s) +

tx

g(s)du(s)

df(t)

= (1 )

f(x) (a+b)/2a


(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


(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] ,

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.


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+


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)


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.,


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].

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)


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).

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)

]


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+


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


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



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)


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+


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


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)


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+


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


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)


[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


[(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+


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


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)]}


[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)


[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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1018 M. W. ALOMARI

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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

Ukrainian Mathematical Journal Volume 65 issue 7 2013 [doi 10.1007_s11253-013-0837-z] Alomari, M. W....

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