FEJÉR INEQUALITIES FOR WRIGHT-CONVEX FUNCTIONS

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Fejér Inequalities for Wright-convex Functions

Ming-In Ho vol. 8, iss. 1, art. 9, 2007

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FEJÉR INEQUALITIES FOR WRIGHT-CONVEX FUNCTIONS

MING-IN HO

China Institute of Technology Nankang, Taipei

Taiwan 11522, China.

EMail:[email protected]

Received: 07 September, 2006

Accepted: 18 November, 2006

Communicated by: P.S. Bullen 2000 AMS Sub. Class.: 26D15.

Key words: Hermite-Hadamard inequality, Fejér inequality, Wright-convex functions.

Abstract: In this paper, we establish several inequalities of Fejér type for Wright- convex functions.

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1. Introduction

Iff : [a, b]→Ris a convex function, then

(1.1) f

a+b 2

≤ 1 b−a

Z b

a

f(x)dx≤ f(a) +f(b) 2 is known as the Hermite-Hadamard inequality ([5]).

In [4], Fejér established the following weighted generalization of the inequality (1.1):

Theorem A. Iff : [a, b]→Ris a convex function, then the inequality (1.2) f

a+b 2

Z b

a

p(x)dx≤ Z b

a

f(x)p(x)dx ≤ f(a) +f(b) 2

Z b

a

p(x)dx holds, wherep : [a, b] → R is nonnegative, integrable, and symmetric about x =

a+b 2 .

In recent years there have been many extensions, generalizations, applications and similar results of the inequalities(1.1)and(1.2)see [1] – [8], [10] – [16].

In [2], Dragomir established the following theorem which is a refinement of the first inequality of(1.1).

Theorem B. Iff : [a, b]→Ris a convex function, andH is defined on[0,1]by H(t) = 1

b−a Z b

a

f

tx+ (1−t)a+b 2

dx,

thenH is convex, increasing on[0,1],and for allt ∈[0,1], we have

(1.3) f

a+b 2

=H(0) ≤H(t)≤H(1) = 1 b−a

Z b

a

f(x)dx.

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In [11], Yang and Hong established the following theorem which is a refinement of the second inequality of(1.1):

Theorem C. Iff : [a, b]→Ris a convex function, andF is defined on[0,1]by

F (t) = 1 2 (b−a)

Z b

a

f

1 +t 2

a+

1−t 2

x

+f

1 +t 2

b+

1−t 2

x

dx,

thenF is convex, increasing on[0,1],and for allt ∈[0,1], we have

(1.4) 1

b−a Z b

a

f(x)dx=F (0)≤F (t)≤F (1) = f(a) +f(b)

2 .

We recall the definition of a Wright-convex function:

Definition 1.1 ([9, p. 223]). We say thatf : [a, b]→Ris a Wright-convex function, if, for allx,y+δ ∈[a, b]withx < yandδ ≥0, we have

(1.5) f(x+δ) +f(y)≤f(y+δ) +f(x).

LetC([a, b])be the set of all convex functions on[a, b]andW([a, b])be the set of all Wright-convex functions on [a, b]. Then C([a, b]) $ W([a, b]). That is, a convex function must be a Wright-convex function but the converse is not true. (see [9, p. 224]).

In [10], Tseng, Yang and Dragomir established the following theorems for Wright- convex functions related to the inequality(1.1), TheoremAand TheoremB:

Theorem D. Letf ∈W([a, b])∩L₁[a, b].Then the inequality(1.1)holds.

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Theorem E. Let f ∈ W([a, b])∩L₁[a, b] and let H be defined as in TheoremB.

Then H ∈ W([0,1]) is increasing on [0,1], and the inequality (1.3) holds for all t∈[0,1].

Theorem F. Letf ∈ W([a, b])∩L₁[a, b] and letF be defined as in TheoremC.

Then F ∈ W([0,1]) is increasing on [0,1], and the inequality (1.4) holds for all t∈[0,1].

In [12], Yang and Tseng established the following theorem which refines the inequality(1.2):

Theorem G ([12, Remark 6]). Letf andpbe defined as in TheoremA. IfP,Qare defined on[0,1]by

(1.6) P (t) = Z b

a

f

tx+ (1−t)a+b 2

p(x)dx (t∈(0,1))

and

(1.7) Q(t) = Z b

a

1 2

f

1 +t

2 a+1−t 2 x

p

x+a 2

+f

1 +t

2 b+1−t 2 x

p

x+b 2

dx (t∈(0,1)),

thenP,Qare convex and increasing on[0,1]and, for allt∈[0,1], (1.8) f

a+b 2

Z b

a

p(x)dx=P (0)≤P (t)≤P (1) = Z b

a

f(x)p(x)dx

and (1.9)

Z b

a

f(x)p(x)dx=Q(0)≤Q(t)≤Q(1) = f(a) +f(b) 2

Z b

a

p(x)dx.

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In this paper, we establish some results about Theorem A and Theorem G for Wright-convex functions which are weighted generalizations of TheoremD, Eand F.

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2. Main Results

In order to prove our main theorems, we need the following lemma [10]:

Lemma 2.1. Iff : [a, b]→R, then the following statements are equivalent:

1. f ∈W([a, b]) ;

2. for alls, t, u, v∈[a, b]withs ≤t≤u≤vandt+u=s+v, we have (2.1) f(t) +f(u)≤f(s) +f(v).

Theorem 2.2. Letf ∈ W([a, b])∩L₁[a, b]and letp : [a, b] → Rbe nonnegative, integrable, and symmetric aboutx= ^a+b₂ .Then the inequality(1.2)holds.

Proof. For the inequality (2.1) and the assumptions thatpis nonnegative, integrable, and symmetric aboutx= ^a+b₂ , we have

f

a+b 2

Z b

a

p(x)dx

= Z ^a+b₂

a

f

a+b 2

p(x)dx+ Z ^a+b₂

a

f

a+b 2

p(a+b−x)dx

= Z ^a+b₂

a

f

a+b 2

+f

a+b 2

p(x)dx

≤ Z ^a+b₂

a

[f(x) +f(a+b−x)]p(x)dx

x≤ a+b

2 ≤ a+b

2 ≤a+b−x

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= Z ^a+b₂

a

f(x)p(x)dx+ Z b

a+b 2

f(x)p(x)dx

= Z b

a

f(x)p(x)dx, and

f(a) +f(b) 2

Z b

a

p(x)dx

= Z ^a+b₂

a

f(a) +f(b) 2

p(x)dx+ Z ^a+b₂

a

f(a) +f(b) 2

p(a+b−x)dx

= Z ^a+b₂

a

[f(a) +f(b)]p(x)dx

≥ Z ^a+b₂

a

[f(x) +f(a+b−x)]p(x)dx (a≤x≤a+b−x≤b)

= Z ^a+b₂

a

f(x)p(x)dx+ Z b

a+b 2

f(x)p(x)dx= Z b

a

f(x)p(x)dx.

This completes the proof.

Remark 1. If we setp(x)≡1 (x∈[a, b])in Theorem2.2, then Theorem2.2gener- alizes TheoremD.

Remark 2. FromC([a, b])$W([a, b]), Theorem2.2generalizes TheoremA.

Theorem 2.3. Letf andpbe defined as in Theorem 2.2and letP be defined as in (1.6). ThenP ∈W([0,1])is increasing on[0,1], and the inequality(1.8)holds for allt ∈[0,1].

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Proof. Ifs, t, u, v ∈ [0,1]ands ≤t ≤ u≤ v,t+u =s+v, then forx ∈ a,^a+b₂ we have

b ≥sx+ (1−s)a+b

2 ≥tx+ (1−t)a+b 2

≥ux+ (1−u)a+b

2 ≥vx+ (1−v)a+b 2 ≥a and ifx∈_a+b

2 , b , then

a≤sx+ (1−s)a+b

2 ≤tx+ (1−t)a+b 2

≤ux+ (1−u)a+b

2 ≤vx+ (1−v)a+b 2 ≤b, where

tx+ (1−t)a+b 2

+

ux+ (1−u)a+b 2

=

sx+ (1−s)a+b 2

+

vx+ (1−v)a+b 2

.

By the inequality(2.1), we have (2.2) f

tx+ (1−t)a+b 2

+f

ux+ (1−u)a+b 2

≤f

sx+ (1−s)a+b 2

+f

vx+ (1−v)a+b 2

for allx ∈[a, b]. Now, using the inequality(2.2)andpis nonnegative on [a, b], we

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have (2.3)

f

tx+ (1−t)a+b 2

+f

ux+ (1−u)a+b 2

p(x)

≤

f

sx+ (1−s)a+b 2

+f

vx+ (1−v)a+b 2

p(x)

for allx∈[a, b]. Integrating the inequality(2.3)overxon[a, b], we have P (t) +P(u)≤P (s) +P (v).

HenceP ∈W([0,1]).

Next, if0≤s≤t≤1andx∈ a,^a+b₂

, then tx+ (1−t)a+b

2 ≤sx+ (1−s)a+b 2

≤s(a+b−x) + (1−s)a+b 2

≤t(a+b−x) + (1−t)a+b 2 , where

sx+ (1−s)a+b 2

+

s(a+b−x) + (1−s)a+b 2

=

tx+ (1−t)a+b 2

+

t(a+b−x) + (1−t)a+b 2

. By the inequality(2.1) and the assumptions that p is nonnegative, integrable, and

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symmetric aboutx= ^a+b₂ , we have P (s) =

Z b

a

f

sx+ (1−s)a+b 2

p(x)dx

= Z ^a+b₂

a

f

sx+ (1−s)a+b 2

p(x)dx

+ Z ^a+b₂

a

f

s(a+b−x) + (1−s)a+b 2

p(a+b−x)dx

= Z ^a+b₂

a

f

sx+ (1−s)a+b 2

+f

s(a+b−x) + (1−s)a+b 2

p(x)dx

≤ Z ^a+b₂

a

f

tx+ (1−t)a+b 2

+f

t(a+b−x) + (1−t)a+b 2

p(x)dx

= Z ^a+b₂

a

f

tx+ (1−t)a+b 2

p(x)dx

+ Z ^a+b₂

a

f

t(a+b−x) + (1−t)a+b 2

p(a+b−x)dx

= Z b

a

f

tx+ (1−t)a+b 2

p(x)dx=P (t).

Thus,P is increasing on[0,1], and the inequality(1.8)holds for allt∈[0,1]. This completes the proof.

Remark 3. If we setp(x)≡1 (x∈[a, b])in Theorem2.3, then Theorem2.2gener- alizes TheoremE.

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Theorem 2.4. Letf andpbe defined as in Theorem2.2 and letQbe defined as in (1.7). ThenQ∈W([0,1])is increasing on[0,1], and the inequality(1.9)holds for allt ∈[0,1].

Proof. Ifs, t, u, v ∈ [0,1]ands ≤ t≤ u≤ v,t+u=s+v, then for allx ∈[a, b]

we have a≤

1 +v 2

a+

1−v 2

x≤

1 +u 2

a+

1−u 2

x

≤

1 +t 2

a+

1−t 2

x≤

1 +s 2

a+

1−s 2

x≤b and

a≤

1 +s 2

b+

1−s 2

x≤

1 +t 2

b+

1−t 2

x

≤

1 +u 2

b+

1−u 2

x≤

1 +v 2

b+

1−v 2

x≤b, where

1 +u 2

a+

1−u 2

x

+

1 +t 2

a+

1−t 2

x

=

1 +v 2

a+

1−v 2

x

+

1 +s 2

a+

1−s 2

x

and

1 +t 2

b+

1−t 2

x

+

1 +u 2

b+

1−u 2

x

=

1 +s 2

b+

1−s 2

x

+

1 +v 2

b+

1−v 2

x

.

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By the inequality(2.1), we have (2.4) f

1 +u 2

a+

1−u 2

x

+f

1 +t 2

a+

1−t 2

x

≤f

1 +v 2

a+

1−v 2

x

+f

1 +s 2

a+

1−s 2

x

and (2.5) f

1 +t 2

b+

1−t 2

x

+f

1 +u 2

b+

1−u 2

x

≤f

1 +s 2

b+

1−s 2

x

+f

1 +v 2

b+

1−v 2

x

for allx∈[a, b]. Now, using the inequality(2.4),(2.5)and the assumptions thatpis nonnegative on[a, b], we have

1 2f

1 +u 2

a+

1−u 2

x

p

x+a 2

(2.6)

+ 1 2f

1 +t 2

a+

1−t 2

x

p

x+a 2

+ 1 2f

1 +t 2

b+

1−t 2

x

p

x+b 2

+ 1 2f

1 +u 2

b+

1−u 2

x

p

x+b 2

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≤ 1 2f

1 +v 2

a+

1−v 2

x

p

x+a 2

+1 2f

1 +s 2

a+

1−s 2

x

p

x+a 2

+1 2f

1 +s 2

b+

1−s 2

x

p

x+b 2

+1 2f

1 +v 2

b+

1−v 2

x

p

x+b 2

Integrating the inequality(2.6)overxon[a, b], we have Q(t) +Q(u)≤Q(s) +Q(v). HenceQ∈W([0,1]).

Next, if0≤s≤t≤1andx∈[a, b], then 1 +t

2

a+

1−t 2

x≤

1 +s 2

a+

1−s 2

x

≤

1 +s 2

b+

1−s 2

(a+b−x)

≤

1 +t 2

b+

1−t 2

(a+b−x) and

1 +t 2

a+

1−t 2

(a+b−x)≤

1 +s 2

a+

1−s 2

(a+b−x)

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≤

1 +s 2

b+

1−s 2

x≤

1 +t 2

b+

1−t 2

x, where

1 +s 2 a

+

1−s 2

x

+

1 +s 2

b+

1−s 2

(a+b−x)

=

1 +t 2

a+

1−t 2

x

+

1 +t 2

b+

1−t 2

(a+b−x)

, and

1 +s 2

a+

1−s 2

(a+b−x)

+

1 +s 2

b+

1−s 2

x

=

1 +t 2

a+

1−t 2

(a+b−x)

+

1 +t 2

b+

1−t 2

x

. By the inequality (2.1) and the assumptions that p is nonnegative and symmetric aboutx= ^a+b₂ , we have

f

1 +s 2

a+

1−s 2

x

p

x+a 2

(2.7)

+f

1 +s 2

b+

1−s 2

(a+b−x)

p

2a+b−x 2

+f

1 +s 2

a+

1−s 2

(a+b−x)

p

a+ 2b−x 2

+f

1 +s 2

b+

1−s 2

x

p

x+b 2

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=

f

1 +s 2

a+

1−s 2

x

+ f

1 +s 2

a+

1−s 2

(a+b−x)

p

x+a 2

+

f

1 +s 2

b+

1−s 2

(a+b−x)

+ f

1 +s 2

b+

1−s 2

x

p

x+b 2

≤

f

1 +t 2

a+

1−t 2

x

+ f

1 +t 2

a+

1−t 2

(a+b−x)

p

x+a 2

+

f

1 +t 2

b+

1−t 2

(a+b−x)

+ f

1 +t 2

b+

1−t 2

x

p

x+b 2

=f

1 +t 2

a+

1−t 2

x

p

x+a 2

+f

1 +t 2

b+

1−t 2

(a+b−x)

p

2a+b−x 2

+f

1 +t 2

a+

1−t 2

(a+b−x)

p

a+ 2b−x 2

+f

1 +t 2

b+

1−t 2

x

p

x+b 2

.

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Integrating the inequality(2.7)overxon[a, b], we have 4Q(s)≤4Q(t)

HenceQis increasing on[0,1], and the inequality(1.9)holds for allt ∈[0,1].

This completes the proof.

Remark 4. If we setp(x)≡1 (x∈[a, b])in Theorem2.4, then Theorem2.2gener- alizes TheoremF.

Remark 5. FromC([a, b]) $ W([a, b]), Theorem 2.3 and Theorem2.4 generalize TheoremC.

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