Fejér-Type Inequalities (II)

(1)

K.-L. TSENG, SHIOW-RU HWANG, AND S.S. DRAGOMIR

Abstract. In this paper, we establish some Fej´er-type inequalities for convex functions. They complement the results from the previous recent paper [12].

1. Introduction

Throughout this paper, let f : [a, b] → R be convex, g : [a, b] → [0,∞) be

integrable and symmetric to a+₂b and define the following functions on [0,1] :

G(t) = 1 2

f

ta+ (1−t)a+b 2

+f

tb+ (1−t)a+b 2

;

H(t) = 1

b−a

Z b

a f

tx+ (1−t)a+b 2

dx;

Hg(t) =

Z b

a f

tx+ (1−t)a+b 2

g(x)dx;

L(t) = 1 2 (b−a)

Z b

a

[f(ta+ (1−t)x) +f(tb+ (1−t)x)]dx;

and

Lg(t) = 1 2

Z b

a

[f(ta+ (1−t)x) +f(tb+ (1−t)x)]g(x)dx.

Iff is defined as above, 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 [1].

For some results which generalize, improve, and extend this famous integral inequality see [2] – [17].

In [2], Dragomir established the following theorem which refines the first inequal-ity of (1.1).

Theorem A. Letf, H be defined as above. ThenH is convex, increasing on[0,1], and for allt∈[0,1], we have

(1.2) f

a+b

2

=H(0)≤H(t)≤H(1) = 1

b−a

Z b

a

f(x)dx.

In [7], Dragomir, Miloˇsevi´c and S´andor established the following inequalities related to (1.1):

1991Mathematics Subject Classification. 26D15.

Key words and phrases. Hermite-Hadamard inequality, Fej´er inequality, Convex function. This research was partially supported by grant NSC 97-2115-M-156-002.

(2)

Theorem B. Let f,H be defined as above. Then:

(1) The following inequality holds

f

_a₊_b

2

≤ 2

b−a

Z a+3₄b

3a+b

4

f(x)dx

(1.3)

≤ Z 1

0

H(t)dt

≤ 1 2

"

f

_a₊_b

2

+ 1

b−a

Z b

a

f(x)dx

#

.

(2) Iff is differentiable on[a, b],then, for allt∈[0,1], we have the inequalities

0≤ 1

b−a

Z b

a

f(x)dx−H(t) (1.4)

≤(1−t) "

f(a) +f(b)

2 −

1

b−a

Z b

a

f(x)dx

#

and

(1.5) 0≤f(a) +f(b)

2 −H(t)≤

(f0(b)−f0(a)) (b−a)

4 .

Theorem C. Letf, H, Gbe defined as above. Then:

(1) Gis convex and increasing on[0,1].

(2) We have

inf

t∈[0,1]

G(t) =G(0) =f

_a₊_b

2

and

sup

t∈[0,1]

G(t) =G(1) =f(a) +f(b)

2 .

(3) The following inequality holds for allt∈[0,1]:

(1.6) H(t)≤G(t).

(4) The following inequality holds:

2

b−a

Z a+34b

3a+b

4

f(x)dx≤ 1 2

f

₃_a₊_b

4

+f

_a_{+ 3}_b

4 (1.7)

≤ Z 1

0

G(t)dt

≤ 1 2

f

_a₊_b

2

+f(a) +f(b) 2

.

(5) Iff is differentiable on[a, b],then, for allt∈[0,1],we have the inequality

(1.8) 0≤H(t)−f

_a₊_b

2

≤G(t)−H(t).

Theorem D. Let f, H, G, Lbe defined as above. Then:

(3)

(2) We have the inequality:

(1.9) G(t)≤L(t)≤ 1−t

b−a

Z b

a

f(x)dx+t·f(a) +f(b)

2 ≤

f(a) +f(b) 2

for allt∈[0,1]and

sup

t∈[0,1]

L(t) =f(a) +f(b)

2 .

(3) For allt∈[0,1], we have the inequalities:

H(1−t)≤L(t) and H(t) +H(1−t)

2 ≤L(t).

In [8], Fej´er established the following weighted generalization of the Hermite-Hadamard inequality (1.1).

Theorem E. Letf, g be defined as above. Then

(1.10) f

_a₊_b

2 Z b

a

g(x)dx≤ Z b

a

f(x)g(x)dx≤ f(a) +f(b) 2

Z b

a

g(x)dx

is known as Fej´er inequality.

In [14], Yang and Tseng established the following theorem which refines the first inequality of (1.10) and generalizes Theorem A.

Theorem F. Let f, g, Hg be defined as above. ThenHg is convex, increasing on

[0,1], and for allt∈[0,1], we have

(1.11) f

_a₊_b

2 Z b

a

g(x)dx=Hg(0)≤Hg(t)≤Hg(1) =

Z b

a

f(x)g(x)dx.

In this paper, we establish some Fej´er-type inequalities related to the functions

G, H, Hg, L, Lgand generalize Theorems B – D. They complement the results from the recent paper [12].

2. _{Main Results}

In order to prove our main results, we need the following lemma:

Lemma 1 (see [9]). Let f : [a, b]→_R be a convex function and let a≤A≤C≤

D≤B≤bwith A+B =C+D, then

f(C) +f(D)≤f(A) +f(B).

Now, we are ready to state and prove our results.

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f

_a₊_b

2 Z b

a

g(x)dx≤2 Z a+3₄b

3a+b

4

f(x)g

2x−a+b 2

dx

(2.1)

≤ Z 1

0

Hg(t)dt

≤ 1 2

"

f

a+b

2 Z b

a

g(x)dx+ Z b

a

f(x)g(x)dx

#

.

(2) If f is differentiable on [a, b] and g is bounded on [a, b], then, for all t ∈ [0,1], we have the inequality

0≤ Z b

a

f(x)g(x)dx−Hg(t)

≤(1−t) "

f(a) +f(b)

2 (b−a)− Z b

a

f(x)dx

# kgk_∞,

(2.2)

wherekgk_∞= sup

x∈[a,b] |g(x)|.

(3) If f is differentiable on[a, b], then, for allt∈[0,1],we have the inequality

0≤f(a) +f(b) 2

Z b

a

g(x)dx−Hg(t)

≤(f

0₍_b₎₋_f0₍_a_{)) (}_b₋_a₎

4

Z b

a

g(x)dx.

(2.3)

Proof. (1) Using simple techniques of integration and the hypothesis ofg, we have the following identities:

(2.4) f

_a₊_b

2 Z b

a

g(x)dx= 4 Z a+₂b

a

Z 1₂

0

f

_a₊_b

2

g(x)dtdx;

(2.5) 2 Z a+34b

3a+b

4

f(x)g

2x−a+b 2

dx

= 2 Z a+₂b

a

Z 1₂

0

f

_x

2 +

a+b

4

+f

_{3 (}_a₊_b₎

4 −

x

2

g(x)dtdx;

(2.6) Z 1

0

Hg(t)dt

= Z a+₂b

a

Z 1₂

0

f

ta+b

2 + (1−t)x

+f

tx+ (1−t)a+b 2

g(x)dtdx

+ Z a+₂b

a

Z 1₂

0

f

t(a+b−x) + (1−t)a+b 2

+f

ta+b

2 + (1−t) (a+b−x)

(5)

and

(2.7) 1 2

"

f

_a₊_b

2 Z b

a

g(x)dx+ Z b

a

f(x)g(x)dx

#

= Z a+2b

a

Z 12

0

f(x) +f

_a₊_b

2

g(x)dtdx

+ Z a+₂b

a

Z 1₂

0

f

_a₊_b

2

+f(a+b−x)

g(x)dtdx.

By Lemma 1, the following inequalities hold for allt∈ 0,1₂

andx∈

a,a+₂b .

(2.8) 4f

_a₊_b

2

≤2

f

_x

2 +

a+b

4

+f

_{3 (}_a₊_b₎

4 −

x

2

holds whenA= x₂ +a+₄b, C=D= a+₂b andB =3(a₄+b)−x

2 in Lemma 1.

(2.9) 2f

x

2 +

a+b

4

≤f

ta+b

2 + (1−t)x

+f

tx+ (1−t)a+b 2

holds whenA =ta+₂b + (1−t)x, C =D = x₂ +a+₄b and B =tx+ (1−t)a+₂b in Lemma 1.

(2.10) 2f

_{3 (}_a₊_b₎

4 −

x

2

≤f

t(a+b−x) + (1−t)a+b 2

+f

ta+b

2 + (1−t) (a+b−x)

holds whenA=t(a+b−x) + (1−t)a+₂b, C =D= 3(a₄+b)−x

2 andB =t

a+b

2 + (1−t) (a+b−x) in Lemma 1.

(2.11) f

ta+b

2 + (1−t)x

+f

tx+ (1−t)a+b 2

≤f(x) +f

_a₊_b

2

holds when A =x, C =ta+₂b + (1−t)x, D =tx+ (1−t)a+₂b and B = a+₂b in Lemma 1.

(2.12) f

t(a+b−x) + (1−t)a+b 2

+f

ta+b

2 + (1−t) (a+b−x)

≤f

_a₊_b

2

+f(a+b−x)

holds for A= a+₂b, C =t(a+b−x) + (1−t)a+₂b, D=ta+₂b + (1−t) (a+b−x) and B = a+b−x in Lemma 1. Multiplying the inequalities (2.8)−(2.12) by

g(x) and integrating them overton 0,1

2

,overxon

a,a+b

2

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(2) By integration by parts, we have

Z a+₂b

a

a+b

2 −x

[f0(a+b−x)−f0(x)]dx

(2.13)

= Z b

a

x−a+b 2

f0(x)dx

= f(a) +f(b)

2 (b−a)− Z b

a

f(x)dx.

Using substitution rules for integration and the hypothesis of g, we have the fol-lowing identities

(2.14)

Z b

a

f(x)g(x)dx= Z a+₂b

a

[f(x) +f(a+b−x)]g(x)dx

and

(2.15) Hg(t) = Z a+₂b

a

f

tx+ (1−t)a+b 2

+f

t(a+b−x) + (1−t)a+b 2

g(x)dx.

Now, using the convexity off and the hypothesis of g, the inequality

f(x)−f

tx+ (1−t)a+b 2

g(x)

+

f(a+b−x)−f

t(a+b−x) + (1−t)a+b 2

g(x)

≤(1−t)

x−a+b 2

f0(x)g(x)

+ (1−t) _a₊_b

2 −x

f0(a+b−x)g(x)

= (1−t) _a₊_b

2 −x

[f0(a+b−x)−f0(x)]g(x)

≤(1−t)

a+b

2 −x

[f0(a+b−x)−f0(x)]kgk_∞

holds for all t ∈[0,1] and x∈

a,a+₂b

. Integrating the above inequalities overx

on

a,a+₂b

and using (2.13)−(2.15) and (1.11), we derive (2.2).

(3) Using the convexity off, we have

f(a)−f a+₂b

2 ≤

1 2

a−a+b 2

f0(a) = a−b 4 f

0₍_a₎

and

f(b)−f a+b

2

2 ≤

1 2

b−a+b 2

f0(b) = b−a 4 f

0₍_b₎

and taking their sum we obtain

f(a) +f(b)

2 −f

_a₊_b

2

≤(f

0₍_b₎₋_f0₍_a_{)) (}_b₋_a₎

(7)

Thus,

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

Z b

a

g(x)dx−f

a+b

2 Z b

a

g(x)dx

≤ (f

0₍_b₎₋_f0₍_a_{)) (}_b₋_a₎

4

Z b

a

g(x)dx.

Finally, (2.3) follows from (1.10),(1.11) and (2.16). This completes the proof.

Remark 3. Let g(x) = 1

b−a (x∈[a, b]) in Theorem 2. Then Hg(t) = H(t)

(t∈[0,1]) and Theorem 2 reduces to Theorem B.

In the following theorems, we point out some inequalities for the functions

H, Hg, G, Lg, Qconsidered above:

Theorem 4. Letf, g, G, Hgbe defined as above. Then we have the following Fej´ er-type inequalities:

(1) The following inequality holds for allt∈[0,1]:

(2.17) Hg(t)≤G(t)

Z b

a

g(x)dx.

2 Z a+3₄b

3a+b

4

f(x)g

2x−a+b 2

dx≤ 1 2

f

3a+b

4

+f

a+ 3b

4

Z b

a

g(x)dx

≤(b−a) Z 1

0

G(t)g((1−t)a+tb)dt

≤ 1 2

f

_a₊_b

2

+f(a) +f(b) 2

Z b

a

g(x)dx.

(2.18)

(3) If f is differentiable on [a, b] and g is bounded on [a, b], then, for all t ∈ [0,1], we have the inequality

(2.19) 0≤Hg(t)−f

_a₊_b

2 Z b

a

g(x)dx≤(b−a) [G(t)−H(t)]kgk_∞

wherekgk_∞= sup

x∈[a,b] |g(x)|.

Proof. (1) Using simple techniques of integration and the hypothesis ofg, we have that the following identity holds on [0,1]:

(2.20) G(t) Z b

a

g(x)dx= Z a+2b

a

f

ta+ (1−t)a+b 2

+f

tb+ (1−t)a+b 2

g(x)dx.

By Lemma 1, the following inequality holds for allx∈

a,a+₂b :

(2.21) f

tx+ (1−t)a+b 2

+f

t(a+b−x) + (1−t)a+b 2

≤f

ta+ (1−t)a+b 2

+f

tb+ (1−t)a+b 2

(8)

It holds when

A=ta+ (1−t)a+b

2 , C=tx+ (1−t)

a+b

2 ,

D=t(a+b−x) + (1−t)a+b

2 and B=tb+ (1−t)

a+b

2

in Lemma 1. Multiplying the inequality (2.21) byg(x),integrating both sides over

xon

a,a+₂b

and using identities (2.15) and (2.20), we derive (2.17).

(2) As for (1), we have the following identities:

(2.22) 2 Z a+3₄b

3a+b

4

f(x)g

2x−a+b 2

dx

= Z a+₂b

a

f

_x

2 +

a+b

4

+f

_{3 (}_a₊_b₎

4 −

x

2

g(x)dx;

(2.23) 1 2

f

₃_a₊_b

4

+f

_a_{+ 3}_b

4

Z b

a

g(x)dx

= Z a+₂b

a

f

₃_a₊_b

4

+f

_a_{+ 3}_b

4

g(x)dx;

(b−a) Z 1

0

G(t)g((1−t)a+tb)dt

=b−a 2 " Z 1 1 2 f

ta+ (1−t)a+b 2

g(ta+ (1−t)b)dt

+ Z 12

0

f

ta+ (1−t)a+b 2

g((1−t)a+tb)dt

+ Z 1₂

0

f

tb+ (1−t)a+b 2

g((1−t)a+tb)dt

+ Z 1 1 2 f

tb+ (1−t)a+b 2

g(ta+ (1−t)b)dt

#

= Z a+₂b

a

1 2

f

_x₊_a

2

+f

₂_a₊_b₋_x

2

+f

_b₊_x

2

+f

_a_{+ 2}_b₋_x

2

g(x)dx; (2.24) and (2.25) 1 2 f

a+b

2

+f(a) +f(b) 2

Z b

a

g(x)dx

= Z a+₂b

a

f

_a₊_b

2

+f(a) +f(b) 2

g(x)dx.

By Lemma 1, the following inequalities hold for allx∈

a,a+₂b.

(2.26) f

_x

2 +

a+b

4

+f

_{3 (}_a₊_b₎

4 −

x

2

≤f

₃_a₊_b

4

+f

_a_{+ 3}_b

(9)

holds whenA= 3a₄+b, C=x₂+a+₄b, D=3(a₄+b)−x

2 andB=

a+3b

4 in Lemma 1.

(2.27) f

3a+b

4

≤ 1 2

f

x+a

2

+f

2a+b−x

2

holds whenA= x+₂a, C=D= 3a₄+b andB =2a+₂b−x in Lemma 1.

(2.28) f

_a_{+ 3}_b

4

≤1 2

f

_b₊_x

2

+f

_a_{+ 2}_b₋_x

2

holds whenA= b+x

2 , C=D=

a+3b

4 andB=

a+2b−x

2 in Lemma 1.

(2.29) f

_x₊_a

2

+f

₂_a₊_b₋_x

2

≤f(a) +f

_a₊_b

2

holds whenA=a, C= x+₂a, D= 2a+₂b−x andB =a+₂b in Lemma 1.

(2.30) f

_b₊_x

2

+f

_a_{+ 2}_b₋_x

2

≤f

_a₊_b

2

+f(b)

holds when A= a+₂b, C = b+₂x, D= a+2₂b−x andB =b in Lemma 1. Multiplying the inequalities (2.26)−(2.30) byg(x), integrating both sides overxona,a+₂b

and using identities (2.22)−(2.25), we derive (2.18).

(3) By integration by parts, we have

t

Z a+2b

a

x−a+b 2

f0

tx+ (1−t)a+b 2

+ _a₊_b

2 −x

f0

t(a+b−x) + (1−t)a+b 2

dx

=t

Z b

a

x−a+b 2

f0

tx+ (1−t)a+b 2

dx

= (b−a) [G(t)−H(t)].

(2.31)

Now, using the convexity off and the hypothesis of g, the inequality

f

tx+ (1−t)a+b 2

−f

a+b

2

g(x)

+

f

t(a+b−x) + (1−t)a+b 2

−f

_a₊_b

2

g(x)

≤t

x−a+b 2

f0

tx+ (1−t)a+b 2

g(x)

+t

_a₊_b

2 −x

f0

t(a+b−x) + (1−t)a+b 2

(10)

=t

_a₊_b

2 −x f

0

t(a+b−x) + (1−t)a+b 2

−f0

tx+ (1−t)a+b 2

g(x)

≤t

_a₊_b

2 −x f

0_t₍_a₊_b₋_x_{) + (1}₋_t₎a+b 2

−f0

tx+ (1−t)a+b 2

kgk_∞

holds for allt∈[0,1] andx∈

a,a+b

2

. Integrating the above inequality over xon

a,a+₂b

and using (2.31) and (1.11), we derive (2.17).This completes the proof.

Remark 5. Let g(x) = 1

b−a (x∈[a, b]) in Theorem 4. Then Hg(t) = H(t)

(t∈[0,1]) and Theorem 4 reduces to Theorem C.

Theorem 6. Let f, g, G, Hg, Lg be defined as above. Then we have the following results:

(1) Lg is convex on [0,1].

(2) The following inequalities hold for allt∈[0,1]:

G(t) Z b

a

g(x)dx≤Lg(t) (2.32)

≤(1−t) Z b

a

f(x)g(x)dx+t· f(a) +f(b) 2

Z b

a

g(x)dx

≤f(a) +f(b) 2

Z b

a

g(x)dx;

and

(2.33) Hg(1−t)≤Lg(t) ;

(2.34) Hg(t) +Hg(1−t)

2 ≤Lg(t).

(3) The following bound is true:

(2.35) sup

t∈[0,1]

Lg(t) = f(a) +f(b) 2

Z b

a

g(x)dx.

Proof. (1) It is easily observed from the convexity off thatLg is convex on [0,1].

(2) As for (1) in Theorem 4, we have that the following identity holds on [0,1]:

(2.36) Lg(t) =1 2

Z a+₂b

a

[f(ta+ (1−t)x) +f(ta+ (1−t) (a+b−x))

+f(tb+ (1−t)x) +f(tb+ (1−t) (a+b−x))]g(x)dx.

By Lemma 1, the following inequalities hold for allx∈

a,a+₂b .

(2.37) 2f

ta+ (1−t)a+b 2

(11)

holds whenA=ta+(1−t)x, C=D=ta+(1−t)a+₂bandB=ta+(1−t) (a+b−x) in Lemma 1.

(2.38) 2f

tb+ (1−t)a+b 2

≤f(tb+ (1−t)x) +f(tb+ (1−t) (a+b−x))

holds whenA=tb+(1−t)x, C=D=tb+(1−t)a+₂b andB=tb+(1−t) (a+b−x) in Lemma 1. Multiplying the inequalities (2.37)−(2.38) byg(x),integrating them overxona,a+₂band using identities (2.20) and (2.36), we derive the first inequal-ity of (2.32). Using the convexity off and the inequality (1.10), the last part of (2.32) holds. Again from the convexity off, we get

Hg(1−t) =

Z b

a f

(1−t)x+ta+b

2

g(x)dx

(2.39)

= Z b

a f

_ta_{+ (1}₋_t₎_x

2 +

tb+ (1−t)x

2

g(x)dx

≤Lg(t)

and (2.33) is proved. From (2.17),(2.32) and (2.33),we get (2.34).

(3) Using (2.32),the inequality (2.35) holds. This completes the proof.

Remark 7. Let g(x) = _b₋1_a (x∈[a, b]) in Theorem 6. Then Hg(t) = H(t) (t∈[0,1]), Lg(t) =L(t) (t∈[0,1]) and Theorem 6 reduces to Theorem D.

References

[1] J. Hadamard, Étude sur les propriétés des fonctions entières en particulier d’une fonction considérée par Riemann,J. Math. Pures Appl.,58(1893), 171–215.

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[9] D.Y. Hwang, K.L. Tseng and G.S. Yang, Some Hadamard’s inequalities for co-ordinated convex functions in a rectangle from the plane,Taiwanese J. Math.,11(1) (2007), 63–73. [10] K.C. Lee and K.L. Tseng, On a weighted generalization of Hadamard’s inequality for G

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[16] G.S. Yang and K.L. Tseng, On certain multiple integral inequalities related to Hermite-Hadamard inequalities,Utilitas Math.,62(2002), 131–142.

[17] G.S. Yang and K.L. Tseng, Inequalities of Hermite-Hadamard-Fej´er type for convex functions and Lipschitzian functions,Taiwanese J. Math.,7(3) (2003), 433–440.

Department of Mathematics, Aletheia University, Tamsui, Taiwan 25103.

E-mail address:[email protected]

China University of Science and Technology, Nankang, Taipei, Taiwan 11522

School of Engineering and Science, Victoria University, PO Box 14428, Melbourne City MC, Victoria 8001, Australia.