Midpoint Type Rules from an Inequalities Point of View

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Midpoint Type Rules from an Inequalities Point of View

P.C. Cerone and S.S. Dragomir

Abstract. The article investigates interior point rules which contain the

mid-point as a special case, and obtains explicit bounds through the use of a Peano kernel approach and the modern theory of inequalities. Thus the simplest open Newton-Cotes rules are examined. Both Riemann-Stieltjes and Riemann inte-grals are evaluated with a variety of assumptions about the integrand enabling the characterisation of the bound in terms of a variety of norms. Perturbed quadrature rules are obtained through the use of Gr¨uss, Chebychev and Lupa¸s inequalities, producing a variety of tighter bounds. The implementation is demonstrated through the investigation of a variety of composite rules based on inequalities developed. The analysis allows the determination of the parti-tion required that would assure that the accuracy the result would be within a prescribed error tolerance. It is demonstrated that the bounds of the approx-imations are equivalent to those obtained from a Peano kernel that produces Trapezoidal type rules.

1. Introduction

The following inequality is well known in the literature as the midpoint inequal-ity:

Z b

a

f(x)dx₋(b₋a)f

a+b

2

≤

(b₋a)3 24 kf

00_k ∞, (1.1)

where the mapping f : [a, b] _⊂ _R_→_R is assumed to be twice differentiable on the interval (a, b) and having the second derivative bounded on (a, b). That is,

kf00_k_∞:= supx∈(a,b)|f00(x)|<∞.

Now, if we assume thatIn:a=x0< x1< ... < xn−1< xn=bis a partition of the interval [a, b] andf is as above, then we can approximate the integralR_abf(x)dx

by themidpoint quadratureformulaAM(f, In) having an error given byRM(f, In), where

AM(f, In) = n−1 X

i=0 f

xi+1+xi 2

hi (1.2)

1991Mathematics Subject Classification. Primary 41A55; Secondary 26D15.

Key words and phrases. Interior point type rules, Analytic Inequalities,A prioribounds.

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and the remainderRM(f, In) satisfies the estimation

|RM(f, In)| ≤ k

f00_k_∞

24 n−1 X

i=0 h3_i,

(1.3)

wherehi=xi+1−xi fori= 0,1,2, ..., n−1.

Equation (1.2) is known as the midpoint rule for n= 1 and as the composite midpoint rule forn >1.The midpoint rule is the most basic open Newton-Cotes quadrature in which function evaluations occur at the midpoints of equispaced intervals.

The current work investigates an interior point (which contains the midpoint as a special case) and obtains explicit bounds through the use of a Peano kernel approach and the modern theory of inequalities. This approach allows for the investigation of quadrature rules that place fewer restrictions on the behaviour of the integrand and thus allow us to cope with larger classes of functions. Expression (1.1) relies on the behaviour of the second derivative whereas bounds for the interior point are obtained in terms of Riemann-Stieltjes integrals in Sections 2, 3 and 4 for functions that are of bounded variation, Lipschitzian and monotonic respectively. In Section 5, interior point rules are obtained for f(n)_∈Lp[a, b], implying that

f(n)

p :=

Z b

a

f(n)(x)

p

dx !1

p

<_∞ forp_≥1

andf(n)

∞:= supx∈[a,b]

f(n)₍_x₎_.

Further, a generalised Taylor series representation is presented that enables an expansion about any point on an interval.

In Section 6, perturbed interior point rules are obtained using what are termed asprematurevariants of Gr¨uss, Chebychev and Lupa¸s inequalities. Atkinson [30] uses an asymptotic error estimate technique to obtain what he defines as a corrected rule. His approach, however, does not readily produce a bound on the error.

Further, in 6.2, alternate Gr¨uss type results are obtained to produce perturbed interior point rules with bounds given in terms of norms associated withf0₍_x₎₋_S, whereS= f(b_b)−f(a)

−a is the secant slope.

Finally, in Section 7, a perturbed interior point rule is obtained whose per-turbation involves S and not f0(b_b)−₋_af0(a). The bound relies on the behaviour of

f00₍_·_).

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2. The Ostrowski Inequality for Mappings of Bounded Variation

In this section we develop interior point type quadrature rules for functions that are of bounded variation. It includes the midpoint rule as a special case. Functions of bounded variation include a very large class in contrast to traditional interior or specifically midpoint rules which rely on the second derivative of the function for its error approximation.

2.1. Some Integral Inequalities. The following inequality for mappings of bounded variation holds [2]:

Theorem 1. Let u : [a, b] _→ _R be a mapping of bounded variation on [a, b].

Then for allx_∈[a, b],we have the inequality

Z b

a

u(t)dt₋(b₋a)u(x)

≤

1

2(b−a) +

x−

a+b

2

_{_}b

a (u),

(2.1)

whereWb_a(u)denotes the total variation ofu.

The constant 1₂ is the best possible one.

Proof. Using the integration by parts formula for Riemann-Stieltjes integrals

we have

Z x

a

(t₋a)du(t) =u(x) (x₋a)₋

Z x

a

u(t)dt

and

Z b

x

(t₋b)du(t) =u(x) (b₋x)₋

Z b

x

u(t)dt.

If we add the above two equalities, we obtain

(b₋a)u(x)₋

Z b

a

u(t)dt=

Z b

a

p(x, t)du(t),

(2.2)

where

p(x, t) :=

t₋aift_∈[a, x)

t₋b ifx_∈[x, b],

for allx, t_∈[a, b].

It is well known [30] that ifp: [a, b]_→_Ris continuous on [a, b] andv : [a, b] _→_R is of bounded variation on [a, b],then

Z b a

p(x)dv(x)

≤xsup∈[a,b]| p(x)_|

b

_

a (v).

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Applying the inequality (2.3) forp(x, t) as above andv(x) =u(x), x_∈[a, b],we get

Z b

a

p(x, t)du(t)

≤ t∈sup[a,b]

|p(x, t)_| b

_

a (u) (2.4)

= max_{x₋a, b₋x_}

b

_

a (u)

=

_b −a

2 +

x−

a+b

2

_{_}b

a (u)

and then by (2.4), via the identity (2.2), we deduce the desired inequality (2.1). Now to prove that 1₂ is the best possible constant assume that the inequality (2.1) holds with a constantC >0.That is,

Z b

a

u(t)dt₋u(x) (b₋a)

≤

C(b₋a) +

x−

a+b

2

_{_}b

a (u) (2.5)

for allx_∈[a, b].

Consider the mappingu: [a, b]_→_R,given by

u(x) =

0 ifx_∈[a, b]_\{a+b

2 }

1 ifx= a+₂b

in (2.5). Thenuis of bounded variation on [a, b],and

b

_

a

(u) = 2, Z b

a

u(t)dt= 0.

Forx=a+₂b,we get in (2.5)

1_≤2C ,

which implies thatC_≥1₂ and the theorem is completely proved.

The following corollary holds for monotonic mappings.

Corollary 1. Let u: [a, b]→Rbe a monotonic mapping on[a, b]. Then we

have the inequality

Z b

a

≤

1

2(b−a) +

x−

a+b

2

|u(b)₋u(a)_|.

The case of Lipschitzian mappings is embodied in the following corollary.

Corollary 2. Letu: [a, b]_→_Rbe anL-Lipschitzian mapping on[a, b].That is, we recall

|u(x)₋u(y)_{| ≤}L_|x₋y_| for all x, y_∈[a, b].

Then, for all x_∈[a, b]we have the inequality

Z b a

≤L

1

2(b−a) +

x−

a+b

2

(b₋a),

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Corollary 3. Let u: [a, b]→Rbe as above. Then we have the inequality:

Z b

a

u(t)dx₋(b₋a)u

a+b

2

≤

1 2(b−a)

b

_

a (u).

(2.6)

Similar inequalities can be found if we assume that u is monotonic or Lips-chitzian on [a, b] by takingx=a+₂b in Corollaries 1 and 2 respectively.

Remark 1. If we assume thatuis continuous differentiable on(a, b)andu0 is integrable on(a, b), then, by (2.1), we get

Z b

a

u(t)dx₋(b₋a)u(x)

≤

1

2(b−a) +

x−

a+b

2

ku0_k1 ,

which is the inequality obtained by Dragomir and Wang in the recent paper[7].

Remark 2. It is well known that iff : [a, b]→Ris a convex mapping on[a, b],

then theHermite-Hadamard inequality

f

a+b

2

≤ 1

b₋a Z b

a

f(x)dx_≤ f(a) +f(b)

2 (2.7)

holds[1].

Now, if we assume that f : I _⊂_R_→ _Ris convex on I and a, b_∈Int(I), a < b;

thenf0

+ is monotonic nondecreasing on [a, b]and, by Corollary 3, we obtain

0_≤ 1

b₋a Z b

a

f(x)dx₋f

a+b

2

≤ 1₂f+0

1 ,

(2.8)

which gives a counterpart for the first membership of Hadamard’s inequality.

Similar results can be obtained if we assume thatf is convex and monotonic or convex and Lipschitzian on [a, b].

2.2. A Quadrature Formula of Riemann Type. LetIn :a=x0< x1<

... < xn−1 < xn = b be a division of the interval [a, b] and ξi ∈ [xi, xi+1] (i =

0, ..., n₋1) a sequence of intermediate points forIn.Construct the Riemann sums

Rn(f, In,ξ) = n_X−1

i=0

f(ξ_i)hi,

wherehi:=xi+1−xi.

We have the following quadrature formula [2].

Theorem 2. Letf : [a, b]_→_Rbe a mapping of bounded variation on[a, b]and

In, ξi (i= 0, ..., n−1)be as above. Then we have the Riemann quadrature formula

Z b

a

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where the remainder satisfies the estimation

|Wn(f, In,ξ)| ≤ sup i=0,...,n

₁

2hi+

ξi−

xi+xi+1

2 _{_}b a (f) (2.10)

≤

1

2ν(h) +i=0sup,...,n

ξi−

xi+xi+1

2 _{_}b a (f)

≤ ν(h) b

_

a (f)

for allξi (i= 0, ..., n−1) as above, whereν(h) := max{hi|i= 0,1, ..., n}.

The constant 1₂ is sharp in (2.10).

Proof. Apply Theorem 1 on the interval [xi, xi+1] to get

Z xi+1

xi

f(x)dx₋f(ξi)hi

≤

₁

2hi+

ξi−

xi+xi+1

2

x_{_}i+1

xi

(f).

(2.11)

Summing over i from 0 ton₋1 and using the generalized triangle inequality we get

|Wn(f, In,ξ)| ≤ n_X−1

i=0

Z xi+1

xi

f(x)dx₋f(ξ_i)hi

≤

n_X−1

i=0

1 2hi+

ξi−

xi+xi+1

2

x_{_}i+1

xi

(f)

≤ sup i=0,...,n

1 2hi+

ξi−

xi+xi+1

2

n_X−1

i=0

x_{_}i+1 xi

(f)

= sup

i=0,...,n

1 2hi+

ξi−

xi+xi+1

2 _{_}b a (f).

The second inequality follows by the properties of sup (_·).

Now, as

ξi−

xi+xi+1

2 ≤ 1 2hi

for allξi∈[xi, xi+1](i= 0, ..., n−1) the last part of (2.10) is also proved.

Corollary 4. Let f : [a, b]→Rbe a monotonic mapping on [a, b]andIn, ξi (i= 0, ..., n₋1)be as above. Then we have the Riemann quadrature formula (2.9) where the remainder satisfies the estimation

|Wn(f, In,ξ)| ≤ sup i=0,...,n

1 2hi+

ξi−

xi+xi+1

2

|f(b)₋f(a)_|

≤

1

2ν(h) +i=0sup,...,n

ξi−

xi+xi+1

2

|f(b)₋f(a)_|

≤ ν(h)_|f(b)₋f(a)_|

for allξ_i (i= 0, ..., n₋1) as above.

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Corollary 5. Let f : [a, b]→R be anL-Lipschitzian mapping on[a, b] and In, ξi (i= 0, ..., n−1)be as above. Then we have the Riemann quadrature formula

(2.9) where the remainder satisfies the estimation

|Wn(f, In,ξ)| ≤ L n_X−1

i=0

1 2hi+

ξi−

xi+xi+1

2

hi

≤ L

n_X−1

i=0 h2i.

The proof is obvious by Corollary 2 applied on the intervals [xi, xi+1] and

summing the resulting inequalities. We shall omit the details.

Note that the best estimation we can get from (2.10) is that one for which

ξi =

xi+xi+1

2 , obtaining the following midpoint formula for functions of bounded

variation.

Corollary 6. Letf, In be as Theorem 2. Then we have the midpoint rule

Z b

a

f(x)dx=Mn(f, In) +Sn(f, In),

where

Mn(f, In) = n−1 X

i=0 f

xi+xi+1

2

hi

and the remainderSn(f, In)satisfies the estimation

|Sn(f, In)| ≤ 1 2ν(h)

b

_

a (f).

Similar results can be obtained from Corollaries 4 and 5.

Remark 3. If we assume that f : [a, b] _→ _R is differentiable on (a, b) and whose derivative f0 is integrable on (a, b) we can put instead of Wb_a(f) the L1₋

norm _kf0_k1 obtaining the estimation due to Dragomir and Wang from the paper

[7].

3. An Inequality for Monotonic Mappings

Bounds were obtained for monotonic mappings in Corollary 2 as a particular case in the development for functions of bounded variation. This section treats specifically monotonic functions and obtains tighter bounds.

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Theorem 3. Let u : [a, b] → R be a monotonic nondecreasing mapping on

[a, b].Then for allx_∈[a, b], we have the inequality

(b−a)u(x)− Z b

a

u(t)dt

(3.1)

≤ [2x₋(a+b)]u(x) +

Z b

a

sgn(t₋x)u(t)dt

≤ (x₋a)(u(x)₋u(a)) + (b₋x)(u(b)₋u(x))

≤

b₋a

2 +

x−

a+b

2

(u(b)₋u(a)).

All the inequalities in (3.1) are sharp and the constant 1₂ is the best possible one.

Proof. Using the integration by parts formula for Riemann-Stieltjes integrals,

we have the identity as given by (2.2). Now, assume that ∆n:a=x

(n) 0 < x

(n)

1 < ... < x (n)

n−1< x (n)

n =bis a sequence of divisions withν(∆n)→0 asn→ ∞,whereν(∆n) := maxi∈{0,...,n−1}(x

(n)

i+1−x (n)

i )

andξ(_in)_∈hx_i(n), x(_in₊₁)i.Ifp: [a, b]_→_Ris continuous on [a, b] andv: [a, b]_→_Ris monotonic nondecreasing on [a, b],then

Z b a

p(x)dv(x)

=

ν(∆limn)→0

n−1 X

i=0 p

ξ(_in)

h v

x(_in₊₁)

−v

x(_in)

i

≤ lim ν(∆n)→0

n−1 X

i=0 p

ξ(_in)

v

x(_in₊₁)

−v

x(_in)

≤ lim ν(∆n)→0

n−1 X

i=0 p

ξ(_in)

v

x(_in₊₁)

−v

x(_in)

=

Z b a |

p(x)_|dv(x).

Asuis monotonic nondecreasing on [a, b], andp(x,_·) is continuous on the intervals, then using the above inequality we can state that

Z b a

p(x, t)du(t)

≤

Z b a |

p(x, t)_|du(t).

(3.2)

Now, let us observe that

Z b

a

|p(x, t)_|du(t)

=

Z x

a

|t₋a_|du(t) +

Z b

x

|t₋b_|du(t)

=

Z x

a

(t₋a)du(t) +

Z b

x

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= (t₋a)u(t)x_a₋

Z x

a

u(t)dt₋(b₋t)u(t)b_x+

Z b

x

u(t)dt

= [2x₋(a+b)]u(x)₋

Z x

a

u(t)dt+

Z b

x

u(t)dt

= [2x₋(a+b)]u(x) +

Z b

a

sgn(t₋x)u(t)dt.

Using the inequality (3.2) and the identity (2.2) we get the first part of (3.1). Now let us observe that

Z b

a

sgn(t₋x)u(t)dt=₋

Z x

a

u(t)dt+

Z b

x

u(t)dt.

Asuis monotonic nondecreasing on [a, b],we can state that

Z x

a

u(t)dt_≥(x₋a)u(a) and

Z b

x

u(t)dt_≤(b₋x)u(b)

so that

Z b

a

sgn(t₋x)u(t)dt_≤(b₋x)u(b)₋(x₋a)u(a).

Consequently

[2x₋(a+b)]u(x) +

Z b

a

≤ [2x₋(a+b)]u(x) + (b₋x)u(b)₋(x₋a)u(a) = (b₋x)(u(b)₋u(x)) + (x₋a)(u(x)₋u(a)) and the second part of (3.1) is proved.

Finally, let us observe that

(b₋x)(u(b)₋u(x)) + (x₋a)(u(x)₋u(a))

≤ max_{b₋x, x₋a_}[u(b)₋u(x) +u(x)₋u(a)]

=

b₋a

2 +

x−

a+b

2

(u(b)₋u(a))

and the inequality (3.1) is thus proved.

Now for the sharpness of the inequalities, assume that (3.1) holds with a constant

C >0 instead of 1₂.That is,

(b−a)u(x)− Z b

a

u(t)dt

(3.3)

≤ [2x₋(a+b)]u(x) +

Z b

a

≤ (x₋a)(u(x)₋u(a)) + (b₋x)(u(b)₋u(x))

≤

C(b₋a) +

x−

a+b

2

(u(b)₋u(a)).

Consider the mappingu0: [a, b]_→_Rgiven by

u0(x) :=

−1 if x=a

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Putting in (3.3)u=u0 andx=a,we have

u(x)−

Z b

a

u(t)dt

= [2x₋(a+b)]u(x) +

Z b

a

= (x₋a)(u(x)₋u(a)) + (b₋x)(u(b)₋u(x)) = 1

≤

C(b₋a) +

x−

a+b

2

(u(b)₋u(a))

=

C+1

2

(b₋a),

which proves the sharpness of the first two inequalities and the fact thatC should not be less than 1₂.

The following corollaries are interesting.

Corollary 7. Letube as above. Then we have the midpoint inequality:

(b−a)u

a+b

2

−

Z b

a

u(t)dt

(3.4)

≤ Z b

a

sgn

t₋a+b

2

u(t)dt

≤ b−₂a[u(b)₋u(a)].

Also, we have the following“trapezoid inequality”for monotonic nondecreasing mappings.

Corollary 8. Under the above assumptions, we have

b₋a

2 [u(a) +u(b)]−

Z b

a

u(t)dt ≤

b₋a

2 [u(b)−u(a)]. (3.5)

Proof. Takingx=a and x=b in Theorem 3, summing, using the triangle

inequality and dividing by 2,we get the desired inequality (3.5).

3.2. A Quadrature Formula. LetIn :a=x0< x1 < ... < xn−1< xn=b be a division of the interval [a, b] and ξi∈[xi, xi+1] (i= 0, ..., n−1) a sequence of

intermediate points forIn.Construct the Riemann sums

Rn(f, In,ξ) = n_X−1

i=0

f(ξ_i)hi,

wherehi:=xi+1−xi.

We have the following quadrature formula.

Theorem 4. Let f : [a, b] → R be a monotonic nondecreasing mapping on

[a, b]andIn, ξi(i= 0, ..., n−1)be as above. Then we have the Riemann quadrature

formula

Z b

a

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|Wn(f, In,ξ)| (3.7)

≤ 2 n_X−1

i=0

ξ_i₋xi+xi+1

2

f(ξ_i) +

Z b

a

S(t, In,ξ)f(t)dt

≤

n_X−1

i=0

[(ξ_i₋xi) (f(ξi)−f(xi)) + (xi+1−ξi)(f(xi+1)−f(ξi))]

≤ sup i=0,...,n

1 2hi+

ξi−

xi+xi+1

2

(f(b)₋f(a))

≤

1

2ν(h) +i=0sup,...,n

ξi−

xi+xi+1

2

(f(b)₋f(a))

≤ ν(h)(f(b)₋f(a))

for allξ_i (i= 0, ..., n₋1) as above, whereν(h) := maxi=0,...,n{hi} and

S(t, In,ξ) =sgn(t−ζi) if t∈[xi, xi+1)(i= 0, ..., n−1).

Z xi+1

xi

≤ 2(ξ_i₋xi+xi+1

2 )f(ξi) +

Z xi+1

xi

S(t, In,ξ)f(t)dt

≤ (ξi−xi) (f(ξi)−f(xi)) + (xi+1−ξi)(f(xi+1)−f(ξi))

≤

1 2hi+

ξi−

xi+xi+1

2

(f(xi+1)−f(xi)).

|Wn(f, In,ξ)|

≤

n_X−1

i=0

Z xi+1 xi

f(x)dx₋f(ξi)hi

≤ 2 n_X−1

i=0

ξi−

xi+xi+1

2

f(ξi) +

Z xi+1 xi

S(t, In,ξ)f(t)dt

≤

n_X−1

i=0

[(ξ_i₋xi) (f(ξi)−f(xi)) + (xi+1−ξi)(f(xi+1)−f(ξi))]

≤

n_X−1

i=0

1 2hi+

ξi−

xi+xi+1

2

(f(xi+1)−f(xi))

≤ sup i=0,...,n

1 2hi+

ξi−

xi+xi+1

2

n_X−1

i=0

(f(xi+1)−f(xi))

= sup

i=0,...,n

₁

2hi+

ξi−

xi+xi+1

2

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The fourth inequality follows by the properties of sup(_·).

Now, as

ξi−

xi+xi+1

2

≤

1 2hi

for allξi∈[xi, xi+1](i= 0, ..., n−1) the last part of (3.7) is also proved.

Corollary 9. Letf, In be as in Theorem 4. Then we have the midpoint rule

Z b

a

f(x)dx=Mn(f, In) +Sn(f, In),

where

Mn(f, In) = n−1 X

i=0 f

xi+xi+1

2

hi

|Sn(f, In)| ≤

Z b

a

µ(In)f(t)dt≤ 1

2ν(h)(f(b)−f(a)),

where

µ(In) =sgn

t₋xi+xi+1

2

if t_∈[xi, xi+1) (i= 0, ..., n−1).

4. Ostrowski Inequality for Lipschitzian Mappings

In Corollary 2, bounds were obtained for an interior point rule for Lipschitzian mappings as a special instance of functions of bounded variation. Treating specifi-cally Lipschitzian mappings, tighter bounds are now obtained.

4.1. Integral Inequalities. The following inequality for Lipschitzian map-pings holds [38].

Theorem 5. Let u: [a, b]→R be anL−Lipschitzian mapping on [a, b]. That

is,

|u(x)₋u(y)_{| |≤}L_|x₋y_| for all x, y_∈[a, b].

Then we have the inequality

Z b

a

≤L

₍_b −a)2

4 + (x−

a+b

2 )

2

,

(4.1)

for allx_∈[a, b].

The constant 1₄ is the best possible one.

Proof. Using the integration by parts formula for Riemann-Stieltjes integrals

we have the identity as given in (2.2). Now, assume that ∆n : a=x

(n) 0 < x

(n)

1 < ... < x (n)

n−1 < x (n)

n =b is a sequence of

divisions withν(∆n)→0 asn→ ∞,whereν(∆n) := maxi∈{0,...,n−1}

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andξ(_in)_∈hx(_in), x_i(n₊₁)i.Ifp: [a, b]_→_Ris Riemann integrable on [a, b] andv: [a, b]

→RisL-Lipschitzian on [a, b],then

Z b

a

p(x)dv(x)

=

ν(∆n)lim→0

n_X−1

i=0

pξ(_in)[vx(_i₊₁n)₋vx(_in)]

≤ lim ν(∆n)→0

n_X−1

i=0

pξ(_in)x_i(₊₁n) ₋x(_in)

vx(_i₊₁n)₋vx(_in) x(_i₊₁n) ₋x(_in)

≤ L lim ν(∆n)→0

n−1 X

i=0

pξ(_in)x_i(n₊₁) ₋x(_in)

and so

Z b

a

p(x)dv(x)

≤L

Z b

a |

p(x)_|dx.

(4.2)

Applying the inequality (4.2) forp(x, t) as given in (2.2) andv(x) =u(x), x_∈[a, b],

we get

Z b

a

p(x, t)du(t)

≤ L "Z x

a

|t₋a_|dt+

Z b

x

|t₋b_|dt #

= L

2

(x₋a)2+ (b₋x)2

= L

"

(b₋a)2

4 +

x₋a+b

2

2#

(4.3)

and so by (4.3), via the identity (2.2), we deduce the desired inequality (4.1). Now to determine the best constant, assume that the inequality (4.1) holds with a constantC >0. That is

Z b

a

≤L

"

C(b₋a)2+

x₋a+b

2

2#

(4.4)

Consider the mappingf : [a, b]_→_R, f(x) =xin (4.4). Then

x−

a+b

2

≤C(b−a)2+

x₋a+b

2

for allx_∈[a, b]; and then forx=a,we get

b₋a

2 ≤

C+1

4

(b₋a)

which implies thatC_≥1₄ and the theorem is completely proved.

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Corollary 10. Let u: [a, b]→R be as above. Then we have the inequality:

Z b

a

u(t)dx₋(b₋a)u

a+b

2

≤

1

4L(b−a)

2_.

(4.5)

Remark 4. It is well known that iff : [a, b]→Ris a convex mapping on[a, b],

then theHermite-Hadamard inequality holds

f

a+b

2

≤_b 1

−a Z b

a

f(x)dx_≤ f(a) +f(b)

2 .

(4.6)

Now, if we assume thatf :I_⊂_R_→_Ris convex onI anda, b_∈Int(I), a < b;then

f₊0 is monotonic nondecreasing on[a, b] and by Theorem 5 we obtain

0_≤ 1

b₋a Z b

a

f(x)dx₋f _a₊_b

2

≤1

4f 0

+(b)(b−a),

(4.7)

which gives a counterpart for the first membership of Hadamard’s inequality.

4.2. A Quadrature Formula of Riemann Type. LetIn :a=x0< x1<

... < xn−1 < xn = b be a division of the interval [a, b] and ξi ∈ [xi, xi+1] (i =

0, ..., n₋1) a sequence of intermediate points forIn.Construct the Riemann sums

Rn(f, In, ξ) = n−1 X

i=0

f(ξ_i)hi,

wherehi:=xi+1−xi.

We have the following quadrature formula [38].

Theorem 6. Let f : [a, b] → R be an L−Lipschitzian mapping on [a, b] and In, ξi (i= 0, ..., n−1)be as above. Then we have the Riemann quadrature formula

Z b

a

f(x)dx=Rn(f, In, ξ) +Wn(f, In, ξ), (4.8)

|Wn(f, In, ξ)| ≤ 1 4L

n_X−1

i=0 h2_i +L

n−1 X

i=0

ξ_i₋xi+xi+1

2

(4.9)

≤ 1₂L

n_X−1

i=0 h2_i

for allξ_i (i= 0, ..., n₋1) as above. The constant 1

4 is sharp in (4.9).

Z xi+1

xi

≤L

"

1 4h

2

i +

ξ_i₋xi+xi+1

2

2# .

(4.10)

|Wn(f, In, ξ)| ≤ n−1 X

i=0

Z xi+1

xi

≤ L

n_X−1

i=0

1 4h

2

i + (ξi−

xi+xi+1

2 )

2

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Now, as

ξ_i₋xi+xi+1

2

2 ≤1₄h2_i

for allξ_i_∈[xi, xi+1](i= 0, ..., n−1) the second part of (4.9) is also proved.

Note that the best estimation we can get from (4.9) is that one for which

ξ_i= xi+xi+1

2 obtaining the following midpoint formula.

Corollary 11. Let f, In be as above. Then we have the midpoint rule

Z b

a

f(x)dx=Mn(f, In) +Sn(f, In)

where

Mn(f, In) = n_X−1

i=0 f

xi+xi+1

2

hi

|Sn(f, In)| ≤ 1 4L

n−1 X

i=0 h2_i.

Remark 5. If we assume that f : [a, b] → R is differentiable on (a, b) and

whose derivative f0 is bounded on (a, b) we can put instead ofL the infinity norm

kf0_k_∞, obtaining the estimation due to Dragomir and Wang from the paper[4].

5. A Generalisation for Derivatives that are Absolutely Continuous

In 1938, Ostrowski (see for example [1, p.468]) proved the following integral inequality.

Letf :I _⊆_R_→_Rbe a differentiable mapping on I◦₍ _I◦ _{is the interior of} _I_), and let a, b _∈I◦ with a < b.If f0 : (a, b)_→_R is bounded on (a, b), i.e., _kf0_k_∞ :=

sup t∈(a,b)|

f0(t)_|<_∞, then we have the inequality:

f(x)−

1

b₋a Z b

a

f(t)dt ≤

"

1 4 +

x₋a+₂b2

(b₋a)2

#

(b₋a)_kf0_k_∞

The constant 1₄ is sharp in the sense that it can not be replaced by a smaller one.

For applications of Ostrowski’s inequality to some special means and some numerical quadrature rules, we refer the reader to the recent paper [4] by S.S. Dragomir and S. Wang.

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Let f : [a, b] _→ _R be an n-times differentiable function, n _≥ 1, such that

f(n)

∞:= sup_t ∈(a,b)

f(n)₍_t₎_<

∞.Then

1

n f(x) +

n_X−1

k=1 n₋k

k! ·

f(k−1)(a)(x₋a)k₋f(k−1)(b)(x₋b)k

b₋a

!

−_b 1 −a

Z b

a

f(t)dt ≤

f(n)

∞

n(n+ 1)!·

(x₋a)n+1_{+ (}_b

−x)n+1 b₋a

In [8], P. Cerone, S.S. Dragomir and J. Roumeliotis proved the following Os-trowski type inequality for twice differentiable mappings:

Letf : [a, b]_→_Rbe a twice differentiable mapping on (a, b) and f00_{: (}_{a, b}₎_→_R is bounded, i.e.,_kf00_k

∞= sup t∈(a,b)|

f00₍_t₎_|_<_∞_. _{Then we have the inequality:}

f(x)−

1

b₋a Z b

a

f(t)dt₋

x₋a+b

2

f0(x)

≤ "

1

24(b−a)

2

+1 2

x₋a+b

2

2#

kf00_k_∞_≤ (b−a) 2

6 kf

00_k ∞

In this section we establish another generalization of the Ostrowski inequality forn-time differentiable mappings which naturally generalizes the result from [8]. Further, work on representation in terms of power series expansion of functions on an interval is presented, giving a generalisation of Taylor series which produces an expansion about a point.

5.1. Integral identities. The following theorem holds [15] (see also [39]).

Theorem 7. Let f : [a, b] _→ _R be a mapping such that f(n−1)is absolutely continuous on [a, b]. Then for all x_∈[a, b] we have the identity:

Z b

a

f(t)dt = n_X−1

k=0 "

(b₋x)k+1+ (₋1)k(x₋a)k+1 (k+ 1)!

# f(k)(x) (5.1)

+ (₋1)n

Z b

a

Kn(x, t)f(n)(t)dt

where the kernel Kn : [a, b]

2

→Ris given by

Kn(x, t) :=

        

(t₋a)n

n! ift∈[a, x]

(t₋b)n

n! if t∈(x, b]

, x_∈[a, b] (5.2)

andnis a natural number, n_≥1.

Proof. The proof is by mathematical induction [15]. For a different argument,

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Forn= 1,we have to prove the equality

Z b

a

f(t)dt= (b₋a)f(x)₋

Z b

a

K1(x, t)f(1)(t)dt

(5.3)

where

K1(x, t) :=

  

t₋a ift_∈[a, x]

t₋b ift_∈(x, b]

.

Integrating by parts, we have:

Z b

a

K1(x, t)f(1)(t)dt

=

Z x

a

(t₋a)f0(t)dt+

Z b

x

(t₋b)f0(t)dt

= (t₋a)f(t)_|x_a₋

Z x

a

f(t)dt+ (t₋b)f(t)_|b_x₋

Z b

x

f(t)dt

= (x₋a)f(x) + (b₋x)f(x)₋

Z b

a

f(t)dt

= (b₋a)f(x)₋

Z b

a

f(t)dt

and the identity (5.3) is proved.

Assume that (5.1) holds for “n” and let us prove it for “n+ 1”. That is, we have to prove the equality

Z b

a

f(t)dt = n

X

k=0 "

(b₋x)k+1+ (₋1)k(x₋a)k+1 (k+ 1)!

#

f(k)(x) (5.4)

+ (₋1)n+1

Z b

a

Kn+1(x, t)f(n+1)(t)dt.

We have, using (5.2),

Z b

a

Kn+1(x, t)f(n+1)(t)dt

=

Z x

a

(t₋a)n+1 (n+ 1)! f

(n+1)

(t)dt+

Z b

x

(t₋b)n+1 (n+ 1)! f

(n+1)

(t)dt

and integrating by parts gives

Z b

a

Kn+1(x, t)f(n+1)(t)dt

= (t−a) n+1

(n+ 1)! f

(n)

(t)

x

a

−_n1_! Z x

a

(t₋a)nf(n)(t)dt

+ (t−b) n+1

(n+ 1)! f

(n)₍_t₎

b

x

−_n1_! Z b

x

(t₋b)nf(n)(t)dt

= (x−a) n+1

+ (₋1)n+2(b₋x)n+1

(n+ 1)! f

(n)₍_x₎ −

Z b

a

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That is

Z b

a

Kn(x, t)f(n)(t)dt =

(x₋a)n+1+ (₋1)n+2(b₋x)n+1

(n+ 1)! f

(n)₍_x₎

− Z b

a

Kn+1(x, t)f(n+1)(t)dt.

Now, using the mathematical induction hypothesis, we get

Z b

a

f(t)dt = n_X−1

k=0 "

(b₋x)k+1+ (₋1)k(x₋a)k+1 (k+ 1)!

# f(k)(x)

+(b−x) n+1

+ (₋1)n(x₋a)n+1

(n+ 1)! f

(n)₍_x₎

−(₋1)n

Z b

a

Kn+1(x, t)f(n+1)(t)dt

= n

X

k=0 "

(b₋x)k+1+ (₋1)k(x₋a)k+1 (k+ 1)!

# f(k)(x)

+ (₋1)n+1

Z b

a

Kn+1(x, t)f(n+1)(t)dt.

That is, identity (5.4) and the theorem is thus proved.

Corollary 12. With the above assumptions, we have the representation Z b

a

f(t)dt = n−1 X

k=0 "

1 + (₋1)k (k+ 1)!

#

(b₋a)k+1 2k+1 f

(k)

a+b

2

(5.5)

+ (₋1)n

Z b

a

Mn(t)f(n)(t)dt

where

Mn(t) :=

        

(t₋a)n

n! if t∈

a,a+₂b

(t₋b)n

n! ift∈ a+b

2 , b

.

The proof follows by Theorem 7 by choosing x = a+b

2 so that Mn(t) = Kn a+2b, t

.

Corollary 13. With the above assumptions, we have the representation: Z b

a

f(t)dt = n−1 X

k=0

(b₋a)k+1 (k+ 1)!

"

f(k)₍_a_{) + (}

−1)kf(k)₍_b₎

2

#

+

Z b

a

Tn(t)f(n)(t)dt (5.6)

where

Tn(t) := 1

n!

(b₋t)n+ (₋1)n(t₋a)n 2

, t_∈[a, b].

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Proof. Choosex=aandx=bin (5.1),then summing the resulting identities

and dividing by 2, gives

Z b

a

f(t)dt = n

X

k=0

(b₋a)k+1 (k+ 1)!

"

f(k)(a) + (₋1)kf(k)(b) 2

#

+

Z b

a

Tn(t)f(n)(t)dt

and the corollary is proved.

The following Taylor-like formula with integral remainder also holds.

Corollary 14. Let g : [a, y] → R be a mapping such that g(n) is absolutely

continuous on [a, y].Then for all x_∈[a, y],we have the identity

g(y) = g(a) + n−1 X

k=0 h

(y₋x)k+1+ (₋1)k(x₋a)k+1

i

(k+ 1)! g

(k+1)₍_x₎

+ (₋1)n

Z y

a

Kn(y, t)g(n+1)(t)dt. (5.8)

The proof is obvious by Theorem 7 choosingf =g0_,_and_b₌_y.

Remark 6. If we choose n = 1 in (5.1), we get the identity (5.3) which is

the identity employed by S.S. Dragomir and S. Wang to prove an Ostrowski type inequality in paper [4].

If in (5.5) we choosen= 1, then we get

Z b a

f(t)dt= (b₋a)f

a+b

2

−

Z b a

M1(t)f0(t)dt

(5.9)

where

M1(t) =

  

t₋a ift_∈a,a+₂b t₋b ift_∈ a+₂b, b

which gives the midpoint type identity useful in Numerical Analysis, although here only the first derivative is involved.

Also, if we putn= 1 in (5.6), we get the trapezoid identity

Z b

a

f(t)dt=b−a

2 (f(a) +f(b)) +

Z b

a

T1(t)f0(t)dt

(5.10)

where

T1(t) = a+b

2 −t, t∈[a, b].

Finally, if in the Taylor-like formula (5.8) we putn= 1,we get

g(y) =g(a) + (y₋a)g0(x)₋

Z y

a

K1(y, t)g(2)(t)dt

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Remark 7. If we choose n= 2in (5.1), we get the identity:

Z b

a

f(t)dt = (b₋a)f(x)₋

x₋a+b

2

f0(x) (5.11)

+

Z b

a

K2(x, t)f00(t)dt

where K2(x, t) is as given in (5.2), which is the identity employed by P. Cerone, S.S. Dragomir and J. Roumeliotis to prove some Ostrowski type inequalities for twice differentiable mappings in the paper [8].

If in (5.5) we choosen= 2, then we get

Z b

a

f(t)dt= (b₋a)f

a+b

2

+

Z b

a

M2(t)f00(t)dt

(5.12)

where

M2(t) =

        

(t₋a)2

2 ift∈

a,a+₂b

(t₋b)2

2 ift∈

a+b

2 , b

which is the classical midpoint identity.

Also, if we putn= 2 in (5.6),we get the identity

Z b

a

f(t)dt = b−a

2 (f(a) +f(b)) +

(b₋a)2

2 ·

f0₍_a₎₋_f0₍_b₎ 2 (5.13)

+

Z b

a

T2(t)f00(t)dt

whereT2(t) is as given in (5.7).

Finally, if we putn= 2 in (5.8), we get

g(y) = g(a) + (y₋a)g0(x)₋(y₋a)

x₋a+y

2

g00(x) (5.14)

+

Z y

a

K2(y, t)g(3)(t)dt,

whereK2 is given in (5.2) anda≤x≤y.

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Theorem 8. Let f : [a, b] → R be a mapping such that f(n−1) is absolutely

continuous on [a, b]. Then for allx_∈[a, b],we have the inequalities

Z b a

f(t)dt₋

n_X−1

k=0 "

(b₋x)k+1+ (₋1)k(x₋a)k+1 (k+ 1)!

# f(k)(x)

≤                   

kf(n)_k

∞

(n+1)! h

(x₋a)n+1+ (b₋x)n+1i if f(n)

∈L_∞[a, b],

kf(n)

kp

n! h₍_x

−a)nq+1+(b−x)nq+1

nq+1

i1

q

if p >1, 1_p+1_q = 1

and f(n)

∈Lp[a, b],

kf(n)_k 1 n!

1

2(b−a) + x₋a+b

2 n

if f(n)_∈L1[a, b],

(5.15)

where

f(n)

∞

:= sup t∈[a,b]

f(n)(t)

<_∞and

f(n)

p := Z b a f(n)(t)

p dt !1 p .

Proof. Using the identity (5.1),we have: Z b a

f(t)dt₋

n_X−1

k=0 "

(b₋x)k+1+ (₋1)k(x₋a)k+1 (k+ 1)!

#

f(k)(x)

= Z b a

Kn(x, t)f(n)(t)dt

:=Q(x).

Now, observe that

Q(x)_≤

f(n)

∞k

Kn(x,·)k1=

f(n)

∞ Z b a

|Kn(x, t)|dt

and so using (5.2),

Q(x) _≤

f(n)

∞ "Z x a

(t₋a)n

n! dt+

Z b x

(b₋t)n

n! dt

#

=

f(n)

∞ (n+ 1)!

h

(x₋a)n+1+ (b₋x)n+1i,

and the first part of inequality (5.15) is proved.

Further, using H¨older’s integral inequality, we have

Q(x) _≤

f(n)

_p

Z b

a |

Kn(x, t)| q dt !1 q = f(n)

p

n!

"Z x

a

(t₋a)nqdt+

Z b

x

(b₋t)nqdt #1

q ,

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Finally, let us observe that,

Q(x) _{≤ k}Kn(x,·)k_∞

f(n)

₁

=

f(n)

1

sup t∈[a,b]

|Kn(x, t)|

=

f(n)

1

n! [max{x−a, b−x}] n

=

f(n)

1 n!

b₋a

2 +

x−

a+b

2

n

and the theorem is completely proved.

Remark 8. It may be noticed that the expressions for the bounds of a

gener-alised interior point rule such as that given by (5.15) are upper bounded by taking

x= a or x= b in the bound while keeping a general x_∈ [a, b] for the rule. The sharpest bound is obtained by taking x = a+₂b, giving the result expressed in the following corollary.

Corollary 15. Let the conditions of Theorem 8 hold. Then

Z b

a

f(t)dt₋

n_X−1

k=0

b₋a

2

k+1_h

1 + (₋1)kif

(k) a+b

2

(k+ 1)!

≤                 

(b−a)n+1

2n₍_n_+1)! f(n)

∞ if f

(n)

∈L_∞[a, b],

(b−a)n+ 1q

2n_n_!(_nq₊₁₎1q f(n)

p if p >1,

1

p +

1

q = 1

and f(n)

∈Lp[a, b],

(b−a)n 2n_n_!

f(n)

1 if f

(n)

∈L1[a, b].

(5.16)

Remark 9. Taking n = 1 in Theorem 8 and Corollary 14 reproduces some

of the results obtained by Dragomir and Wang ( [4]-[7]) while n = 2 reproduces the results of Cerone, Dragomir and Roumeliotis ([8]-[11]). It is important to note that assuming that the behaviour of the first derivatives determines the bound on the rule allows greater flexibility than assumptions about the second derivative. Taking

n = 2 allows a comparison with traditional midpoint x=a+₂b or interior point rules. It should further be noted that only even derivatives occur in the rule given in (5.16).

Remark 10. It is most interesting to observe that the bounds given in (5.15)

for a generalised interior point method obtained from investigating various norms of

Kn(x, t)as given by (5.2) are the same as the bounds obtained from the generalised

trapezoidal type rule resulting from various norms of the Peano kernel given by

(x−t)n

n! .