(1)http://jipam.vu.edu.au/ Volume 3, Issue 5, Article 68, 2002 OSTROWSKI TYPE INEQUALITIES FOR ISOTONIC LINEAR FUNCTIONALS S.S

http://jipam.vu.edu.au/

Volume 3, Issue 5, Article 68, 2002

OSTROWSKI TYPE INEQUALITIES FOR ISOTONIC LINEAR FUNCTIONALS

S.S. DRAGOMIR

SCHOOL OFCOMMUNICATIONS ANDINFORMATICS

VICTORIAUNIVERSITY OFTECHNOLOGY

PO BOX14428 MELBOURNECITYMC VICTORIA8001, AUSTRALIA. sever@matilda.vu.edu.au

URL:http://rgmia.vu.edu.au/SSDragomirWeb.html

Received 6 May, 2002; accepted 3 June, 2002 Communicated by P. Bullen

ABSTRACT. Some inequalities of Ostrowski type for isotonic linear functionals defined on a linear class of functionL := {f : [a, b]→R} are established. Applications for integral and discrete inequalities are also given.

Key words and phrases: Ostrowski Type Inequalities, Isotonic Linear Functionals.

2000 Mathematics Subject Classification. Primary 26D15, 26D10.

1. INTRODUCTION

The following result is known in the literature as Ostrowski’s inequality [13].

Theorem 1.1. Letf : [a, b] → Rbe a differentiable mapping on(a, b) with the property that

|f⁰(t)| ≤M for allt∈(a, b). Then (1.1)

f(x)− 1 b−a

Z b a

f(t)dt

≤

"

1

4+ x− ^a+b₂ 2

(b−a)²

#

(b−a)M for allx∈[a, b].

The constant ¹₄ is the best possible in the sense that it cannot be replaced by a smaller constant.

The following Ostrowski type result for absolutely continuous functions whose derivatives belong to the Lebesgue spacesL_p[a, b]also holds (see [9], [10] and [11]).

ISSN (electronic): 1443-5756

c 2002 Victoria University. All rights reserved.

047-02

Theorem 1.2. Letf : [a, b]→Rbe absolutely continuous on[a, b]. Then, for allx∈[a, b], we have:

(1.2)

f(x)− 1 b−a

Z b a

f(t)dt

≤























1

4 +_x−a+b 2

b−a

²

(b−a)kf⁰k_∞ if f⁰ ∈L∞[a, b] ;

1 (p+1)

1 p

h x−a b−a

p+1

+ ^b−x_b−ap+1i¹_p

(b−a)^1pkf⁰k_q if f⁰ ∈Lq[a, b],

1

p + ¹_q = 1, p >1;

h1 2 +

x−^a+b

2

b−a

ikf⁰k₁;

wherek·k_r (r ∈[1,∞]) are the usual Lebesgue norms onL_r[a, b], i.e., kgk_∞ :=ess sup

t∈[a,b]

|g(t)|

and

kgk_r :=

Z b a

|g(t)|^rdt

1 r

, r∈[1,∞).

The constants ¹₄, ¹

(p+1)^1p

and ¹₂ respectively are sharp in the sense presented in Theorem 1.1.

The above inequalities can also be obtained from Fink’s result in [12] on choosingn = 1and performing some appropriate computations.

If one drops the condition of absolute continuity and assumes that f is Hölder continuous, then one may state the result (see [7]):

Theorem 1.3. Letf : [a, b]→Rbe ofr−H−Hölder type, i.e.,

(1.3) |f(x)−f(y)| ≤H|x−y|^r, for all x, y ∈[a, b],

wherer∈(0,1]andH >0are fixed. Then for allx∈[a, b]we have the inequality:

(1.4)

f(x)− 1 b−a

Z b a

f(t)dt

≤ H r+ 1

"

b−x b−a

r+1

+

x−a b−a

r+1#

(b−a)^r. The constant _r+1¹ is also sharp in the above sense.

Note that if r = 1, i.e., f is Lipschitz continuous, then we get the following version of Ostrowski’s inequality for Lipschitzian functions (withLinstead ofH) (see [3])

(1.5)

f(x)− 1 b−a

Z b a

f(t)dt

≤



 1

4+ x− ^a+b₂ b−a

!2

(b−a)L.

Here the constant ¹₄ is also best.

Moreover, if one drops the continuity condition of the function, and assumes that it is of bounded variation, then the following result may be stated (see [4]).

Theorem 1.4. Assume that f : [a, b] → R is of bounded variation and denote by Wb

a(f)its total variation. Then

(1.6)

f(x)− 1 b−a

Z b a

f(t)dt

≤

"

1 2 +

x− ^a+b₂ b−a

# _b _

a

(f)

for allx∈[a, b].

The constant ¹₂ is the best possible.

If we assume more aboutf, i.e.,f is monotonically increasing, then the inequality (1.6) may be improved in the following manner [5] (see also [2]).

Theorem 1.5. Letf : [a, b]→Rbe monotonic nondecreasing. Then for allx∈[a, b], we have the inequality:

f(x)− 1 b−a

Z b a

f(t)dt (1.7)

≤ 1 b−a

[2x−(a+b)]f(x) + Z b

a

sgn(t−x)f(t)dt

≤ 1

b−a{(x−a) [f(x)−f(a)] + (b−x) [f(b)−f(x)]}

≤

"

1 2 +

x− ^a+b₂ b−a

#

[f(b)−f(a)].

All the inequalities in (1.7) are sharp and the constant ¹₂ is the best possible.

The version of Ostrowski’s inequality for convex functions was obtained in [6] and is incor- porated in the following theorem:

Theorem 1.6. Let f : [a, b] → Rbe a convex function on [a, b]. Then for any x ∈ (a, b)we have the inequality

1 2

(b−x)²f₊⁰ (x)−(x−a)²f_{_}⁰(x) (1.8)

≤ Z b

a

f(t)dt−(b−a)f(x)

≤ 1 2

(b−x)²f₋⁰ (b)−(x−a)²f₊⁰ (a) . In both parts of the inequality (1.8) the constant ¹₂ is sharp.

For other Ostrowski type inequalities, see [8].

In this paper we extend Ostrowski’s inequality for arbitrary isotonic linear functionals A : L → R, where L is a linear class of absolutely continuous functions defined on [a, b].Some applications for particular instances of linear functionalsAare also provided.

2. PRELIMINARIES

LetLbe a linear class of real-valued functions,g :E →Rhaving the properties (L1) f, g ∈Limply(αf +βg)∈Lfor allα, β ∈R;

(L2) 1∈L,i.e., iff(t) = 1,t∈E,thenf ∈L.

An isotonic linear functionalA:L→Ris a functional satisfying (A1) A(αf +βg) =αA(f) +βA(g)for allf, g ∈Landα, β ∈R; (A2) Iff ∈Landf ≥0, thenA(f)≥0.

The mappingAis said to be normalised if (A3) A(1) = 1.

Usual examples of isotonic linear functional that are normalised are the following ones A(f) := 1

µ(X) Z

X

f(x)dµ(x), if µ(X)<∞

or

A_w(f) := 1 R

Xw(x)dµ(x) Z

X

w(x)f(x)dµ(x), wherew(x)≥0,R

Xw(x)dµ(x)>0, X is a measurable space andµis a positive measure on X.

In particular, forx¯= (x₁, . . . , x_n),w¯ := (w₁, . . . , w_n)∈Rⁿwithw_i ≥0, W_n:=Pn

i=1w_i >

0we have

A(¯x) := 1 n

n

X

i=1

x_i and

Aw¯(¯x) := 1 W_n

n

X

i=1

wixi, are normalised isotonic linear functionals onRⁿ.

The following representation result for absolutely continuous functions holds.

Lemma 2.1. Let f : [a, b] → R be an absolutely continuous function on [a, b] and define e(t) =t,t∈ [a, b],g(t, x) =R1

0 f⁰[(1−λ)x+λt]dλ, t ∈[a, b]andx∈[a, b].IfA :L→R is a normalised linear functional on a linear classLof absolutely continuous functions defined on[a, b]and(x−e)·g(·, x)∈L,then we have the representation

(2.1) f(x) =A(f) +A[(x−e)·g(·, x)], forx∈[a, b].

Proof. For anyx, t∈[a, b]witht 6=x,one has f(x)−f(t)

x−t = Rx

t f⁰(u) x−t =

Z 1 0

f⁰[(1−λ)x+λt]dλ=g(t, x), giving the equality

(2.2) f(x) = f(t) + (x−t)g(t, x)

for anyt, x∈[a, b].

Applying the functionalA,we get

A(f(x)·1) =A(f + (x−e)g(·, x)), for anyx∈[a, b].

Since

A(f(x)·1) = f(x)A(1) =f(x) and

A(f + (x−e)·g(·, x)) =A(f) +A((x−e)·g(·, x)),

the equality (2.1) is obtained.

The following particular cases are of interest:

Corollary 2.2. Letf : [a, b]→Rbe an absolutely continuous function on[a, b].Then we have the representation:

(2.3) f(x) = 1 Rb

a w(t)dt Z b

a

w(t)f(t)dt

+ 1

Rb

a w(t)dt Z b

a

w(t) (x−t) Z 1

0

f⁰[(1−λ)x+λt]dλ

dt

for anyx∈[a, b],wherep: [a, b]→Ris a Lebesgue integrable function withRb

aw(t)dt 6= 0.

In particular, we have (2.4) f(x) = 1

b−a Z b

a

f(t)dt+ 1 b−a

Z b a

(x−t) Z 1

0

f⁰[(1−λ)x+λt]dλ

dt for eachx∈[a, b].

The proof is obvious by Lemma 2.1 applied for the normalised linear functionals A_w(f) := 1

Rb

a w(t)dt Z b

a

w(t)f(t)dt, A(f) := 1 b−a

Z b a

f(t)dt defined on

L:={f : [a, b]→R, f is absolutely continuous on [a, b]}. The following discrete case also holds.

Corollary 2.3. Letf : [a, b]→Rbe an absolutely continuous function on[a, b].Then we have the representation:

(2.5) f(x) = 1 W_n

n

X

i=1

w_if(x_i) + 1 W_n

n

X

i=1

w_i(x−x_i) Z 1

0

f⁰[(1−λ)x+λx_i]dλ

for anyx∈[a, b],wherex_i ∈[a, b], w_i ∈R (i={1, . . . , n})withW_n :=Pn

i=1w_i 6= 0.

In particular, we have (2.6) f(x) = 1

n

n

X

i=1

f(xi) + 1 n

n

X

i=1

(x−xi) Z 1

0

f⁰[(1−λ)x+λxi]dλ

for anyx∈[a, b].

3. OSTROWSKI TYPE INEQUALITIES

The following theorem holds.

Theorem 3.1. With the assumptions of Lemma 2.1, and assuming thatA : L →Ris isotonic, then we have the inequalities

(3.1) |f(x)−A(f)|

≤

















 A

|x−e| kf⁰k_[x,·],∞

if |x−e| kf⁰k_[x,·],∞∈L, f⁰ ∈L∞[a, b] ; A

|x−e|^1q kf⁰k_[x,·],p

if |x−e|^1q kf⁰k_[x,·],p ∈L, f⁰ ∈L_p[a, b], p > 1, ¹_p + ¹_q = 1;

A

kf⁰k_[x,·],1

if kf⁰k_[x,·],1 ∈L, where

khk_[m,n],∞:=ess sup

t∈[m,n]

(t∈[n,m])

|h(t)| and

khk_[m,n],p :=

Z n m

|h(t)|^pdt

1 p

, p≥1.

If we denote

M∞(x) := A

|x−e| kf⁰k_[x,·],∞

, M_p(x) := A

|x−e|^1q kf⁰k_[x,·],p , M₁(x) := A

kf⁰k_[x,·],1 , then we have the inequalities:

(3.2) M∞(x)

≤



















kf⁰k_[a,b],∞A(|x−e|) if |x−e| ∈L, f⁰ ∈L∞[a, b] ; h

A

kf⁰k^β_[x,·],∞i_β¹

[A(|x−e|^α)]^α1 if kf⁰k^β_[x,·],∞,|x−e|^α ∈L,

f⁰ ∈L_∞[a, b], α >1, _α¹ + ¹_β = 1;

₁

2(b−a) +

x− ^a+b₂

A

kf⁰k_[x,·],∞

if kf⁰k_[x,·],∞∈L, f⁰ ∈L∞[a, b].

(3.3) M_p(x)

≤























maxn

kf⁰k_[a,x],p,kf⁰k_[x,b],po A

|x−e|^1q

if |x−e|^1q ∈L, f⁰ ∈L_p[a, b] ; h

A

kf⁰k^β_[x,·],pi¹_β h A

|x−e|^αqi_α¹

if kf⁰k^β_[x,·],p,|x−e|^αq ∈L, f⁰ ∈L_p[a, b], α >1, _α¹ + ¹_β = 1;

₁

2(b−a) +

x− ^a+b₂

¹_q A

kf⁰k_[x,·],p

if kf⁰k_[x,·],p ∈L, f⁰ ∈L_p[a, b]

and

(3.4) M₁(x)≤











1

2kf⁰k_[a,b],1+¹₂

kf⁰k_[a,x],1− kf⁰k_[x,b],1 , h

A

kf⁰k^β_[x,·],1i¹_β

, β >1.

Proof. Using (2.1) and taking the modulus, we have

|f(x)−A(f)|=|A((x−e)·g(·, x))|

(3.5)

≤A(|(x−e)·g(·, x)|)

=A(|x−e| |g(·, x)|). Fort6=x(t, x ∈[a, b])we may state

|g(t, x)| ≤ Z 1

0

|f⁰((1−λ)x+λt)|dλ

≤ess sup

λ∈[0,1]

|f⁰((1−λ)x+λt)|

=kf⁰k_[x,t],∞.

Hölder’s inequality will produce

|g(t, x)| ≤ Z 1

0

|f⁰((1−λ)x+λt)|dλ

≤ Z 1

0

|f⁰((1−λ)x+λt)|^pdλ _p¹

= 1

x−t Z x

t

|f⁰(u)|^pdu ¹_p

=|x−t|^−1pkf⁰k_[x,t],p, p > 1, 1 p +1

q = 1;

and finally

|g(t, x)| ≤ Z 1

0

|f⁰((1−λ)x+λt)|dλ = 1

t−xkf⁰k_[x,t],1. Consequently

(3.6) |(x−e)| |g(·, x)| ≤















|x−e| kf⁰k_[x,·],∞ if f⁰ ∈L∞[a, b] ;

|x−e|^1q kf⁰k_[x,·],p if f⁰ ∈L_p[a, b], kf⁰k_[x,·],1

for anyx∈[a, b].

Applying the functionalAto (3.6) and using (3.5) we deduce the inequality (3.1).

We have

M_∞(x)≤ sup

t∈[a,b]

nkf⁰k_[x,t],∞o

A(|x−e|)

= maxn

kf⁰k_[a,x],∞,kf⁰k_[x,b],∞o

A(|x−e|)

=kf⁰k_[a,b],∞A(|x−e|) and the first inequality in (3.2) is proved.

Using Hölder’s inequality for the functionalA,i.e., (3.7) |A(hg)| ≤[A(|h|^α)]^α1

h A

|g|^βi¹_β

, α >1, 1 α + 1

β = 1, wherehg,|h|^α,|g|^β ∈L, we have

M∞(x)≤[A(|x−e|^α)]^α1 h A

kf⁰k^β_[x,·],∞i_β¹ and the second part of (3.2) is proved.

In addition,

M∞(x)≤ sup

t∈[a,b]

|x−t|A

kf⁰k_[x,·],∞

= max{x−a, b−x}A

kf⁰k_[x,·],∞

= 1

2(b−a) +

x− a+b 2

A

kf⁰k_[x,·],∞

and the inequality (3.2) is completely proved.

We also have

M_p(x)≤ sup

t∈[a,b]

nkf⁰k_[x,t],po A

|x−e|^1q

= maxn

kf⁰k_[a,x],p,kf⁰k_[x,b],po A

|x−e|^1q . Using Hölder’s inequality (3.7) one has

Mp(x)≤h A

|x−e|^αqi_α¹ h A

kf⁰k^β_[x,·],pi¹_β

, α >1, 1 α + 1

β = 1 and

M_p(x)≤ sup

t∈[a,b]

n|x−t|^1qo A

kf⁰k_[x,·],p

= maxn

(x−a)^1q ,(b−x)^1qo A

kf⁰k_[x,·],p

= 1

2(b−a) +

x−a+b 2

¹_q A

kf⁰k_[x,·],p , proving the inequality (3.3).

Finally,

A

kf⁰k_[x,·],1

≤ sup

t∈[a,b]

n

kf⁰k_[x,t],1o A(1)

= maxn

kf⁰k_[a,x],1,kf⁰k_[x,b],1o

= 1

2kf⁰k_[a,b],1+ 1 2

kf⁰k_[a,x],1− kf⁰k_[x,b],1 . By Hölder’s inequality, we have

A

kf⁰k_[x,·],1

≤h A

kf⁰k^β_[x,·],1i_β¹

, β >1,

and the last part of (3.4) is also proved.

4. THECASE WHERE|f⁰|IS CONVEX

The following theorem also holds.

Theorem 4.1. Letf : [a, b]→Rbe an absolutely continuous function such thatf⁰ : (a, b)→R is convex in absolute value, i.e.,|f⁰|is convex on(a, b).IfA :L→Ris a normalised isotonic linear functional and|x−e|,|x−e| |f⁰| ∈L,then

(4.1) |f(x)−A(f)| ≤ 1

2[|f⁰(x)|A(|x−e|) +A(|x−e| |f⁰|)]

≤



















1 2

hkf⁰k_[a,b],∞+|f⁰(x)|i

A(|x−e|), if f⁰ ∈L∞[a, b] ;

1 2

|f⁰(x)|A(|x−e|) + [A(|x−e|^α)]^α1 h A

|f⁰|^βi_β¹

if |x−e|^α,|f⁰|^β ∈L, α >1, _α¹ + ¹_β = 1;

1 2

|f⁰(x)|A(|x−e|) +₁

2(b−a) +

x− ^a+b₂

A(|f⁰|)

if |f⁰| ∈L.

Proof. Since|f⁰|is convex, we have

|g(t, x)| ≤ Z 1

0

|f⁰((1−λ)x+λt)|dλ

=|f⁰(x)|

Z 1 0

(1−λ)dλ+|f⁰(t)|

Z 1 0

λdλ

= |f⁰(x)|+|f⁰(t)|

2 .

Thus,

|f(x)−A(f)| ≤A

|x−e| · |f⁰(x)|+|f⁰(t)|

2

= 1

2[|f⁰(x)|A(|x−e|) +A(|x−e| |f⁰|)]

and the first part of (4.1) is proved.

We have

A(|x−e| |f⁰|)≤ess sup

t∈[a,b]

{|f⁰(t)|} ·A(|x−e|)

=kf⁰k_[a,b],∞A(|x−e|). By Hölder’s inequality for isotonic linear functionals, we have

A(|x−e| |f⁰|)≤[A(|x−e|^α)]^α1 h A

|f⁰|^βi¹_β

, α >1, 1 α + 1

β = 1 and finally,

A(|x−e| |f⁰|)≤ sup

t∈[a,b]

|x−t| ·A(|f⁰|)

= max (x−a, b−x)·A(|f⁰|)

= 1

2(b−a) +

x−a+b 2

A(|f⁰|).

The theorem is thus proved.

5. SOMEINTEGRAL INEQUALITIES

If we consider the normalised isotonic linear functionalA(f) = _b−a¹ Rb

a f, then by Theorem 3.1 forf : [a, b] → Ran absolutely continuous function, we may state the following integral inequalities

f(x)− 1 b−a

Z b a

f(t)dt (5.1)

≤ 1 b−a

Z b a

|x−t| kf⁰k_[x,t],∞dt

≤











































kf⁰k_[a,b],∞



 1

4+ x− ^a+b₂ b−a

!2

(b−a) (Ostrowski’s inequality) providedf⁰ ∈L∞[a, b] ; 1

b−a Z b

a

kf⁰k^β_[x,t],∞dt _β¹ "

(b−x)^α+1+ (x−a)^α+1 (α+ 1) (b−a)

#_α¹

iff⁰ ∈L∞[a, b], kf⁰k_[x,·],∞∈L_β[a, b], α >1, _α¹ +_β¹ = 1;

"

1 2+

x−^a+b₂ b−a

#Z b a

kf⁰k_[x,t],∞dt

iff⁰ ∈L∞[a, b], and if kf⁰k_[x,·],∞ ∈L₁[a, b], for eachx∈[a, b] ;

f(x)− 1 b−a

Z b a

f(t)dt (5.2)

≤ 1 b−a

Z b a

|x−t|^1q kf⁰k_[x,t],pdt

≤







































qmaxn

kf⁰k_[a,x],p,kf⁰k_[x,b],po

(b−x)^1q+1+(x−a)^1q+1 (b−a)(q+1)

,

p > 1, ¹_p + ¹_q = 1 andf⁰ ∈L_p[a, b] ; q^α1

1 b−a

Rb

a kf⁰k^β_[x,t],pdt_β¹

(b−x)^αq+1+(x−a)^αq+1 (b−a)(q+α)

_α¹

iff⁰ ∈L_p[a, b], and kf⁰k_[x,·],p ∈L_β[a, b], whereα >1, _α¹ + ¹_β = 1

1

2 +|^x−a+b₂ |

b−a

¹_q

1 b−a

Rb

a kf⁰k_[x,t],pdt

iff⁰ ∈L_p[a, b], and kf⁰k_[x,·],p∈L₁[a, b], for eachx∈[a, b]and

f(x)− 1 b−a

Z b a

f(t)dt (5.3)

≤ 1 b−a

Z b a

kf⁰k_[x,t],1dt

≤













 1

2kf⁰k_[a,b],1+1 2

kf⁰k_[a,x],1− kf⁰k_[x,b],1

iff⁰ ∈L₁[a, b] ; 1

b−a Z b

a

kf⁰k^β_[x,t],1dt ¹β

iff⁰ ∈L₁[a, b], kf⁰k_[x,.],1 ∈L_β[a, b], whereβ >1, for eachx∈[a, b].

If we assume now thatf : [a, b]→Ris absolutely continuous and such that|f⁰|is convex on (a, b),then by Theorem 4.1 we obtain the following integral inequalities established in [1]

f(x)− 1 b−a

Z b a

f(t)dt (5.4)

≤ 1 2



|f⁰(x)|



 1

4 + x− ^a+b₂ b−a

!2

(b−a) + 1 b−a

Z b a

|x−t| |f⁰(t)|dt





≤







































1 2

hkf⁰k_[a,b],∞+|f⁰(x)|i

1

4 +_x−a+b 2

b−a

²

(b−a) iff⁰ ∈L∞[a, b] ;

1 2

|f⁰(x)|

1

4 +_x−a+b 2

b−a

2

(b−a) +h_(b−x)α+1+(x−a)^α+1

(α+1)(b−a)

i_α¹ h

1 b−a

Rb

a |f⁰(t)|^βdti¹_β if f⁰ ∈Lβ[a, b], α >1, _α¹ +_β¹ = 1;

1 2

|f⁰(x)|

1

4 +_x−a+b 2

b−a

²

(b−a) +

1

2 +|^x−a+b2 |

b−a

Rb

a |f⁰(t)|dt

iff⁰ ∈L₁[a, b], for eachx∈[a, b].

6. SOMEDISCRETE INEQUALITIES

For a given interval[a, b],consider the division

I_n :a =x₀ < x₁ <· · ·< xn−1 < x_n =b

and the intermediate pointsξ_i ∈[x_i, x_i+1], i= 0, n−1.Ifh_i :=x_i+1−x_i >0 i= 0, n−1 we may define the following functionals

A(f;I_n, ξ) := 1 b−a

n−1

X

i=0

f(ξ_i)h_i (Riemann Rule)

A_T (f;I_n) := 1 b−a

n−1

X

i=0

f(x_i) +f(x_i+1)

2 ·h_i (Trapezoid Rule) A_M(f;I_n) := 1

b−a

n−1

X

i=0

f

x_i+x_i+1 2

·h_i (Mid-point Rule)

A_S(f;I_n) := 1

3A_T (f;I_n) + 2

3A_M(f;I_n). (Simpson Rule)

We observe that, all the above functionals are obviously linear, isotonic and normalised.

Consequently, all the inequalities obtained in Sections 2 and 3 may be applied for these functionals.

If, for example, we use the following inequality (see Theorem 3.1) (6.1) |f(x)−A(f)| ≤ kf⁰k_[a,b]A(|x−e|), x∈[a, b],

providedf : [a, b]→Ris absolutely continuous andf⁰ ∈L∞[a, b],then we get the inequalities

(6.2)

f(x)− 1 b−a

n−1

X

i=0

f(ξ_i)h_i

≤ kf⁰k_[a,b],∞ 1 b−a

n−1

X

i=0

|x−ξ_i|h_i,

(6.3)

f(x)− 1 b−a

n−1

X

i=0

f(x_i) +f(x_i+1)

2 ·h_i

≤ kf⁰k_[a,b],∞· 1 b−a

n−1

X

i=0

|x−x_i|+|x−x_i+1|

2 h_i,

(6.4)

f(x)− 1 b−a

n−1

X

i=0

f

x_i+x_i+1 2

·h_i

≤ kf⁰k_[a,b],∞ 1 b−a

n−1

X

i=0

x− x_i+x_i+1 2

h_i,

for eachx∈[a, b].

Similar results may be stated if one uses for example Theorem 4.1. We omit the details.

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