Fejér Inequalities for Wright-convex Functions
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FEJÉR INEQUALITIES FOR WRIGHT-CONVEX FUNCTIONS
MING-IN HO
China Institute of Technology Nankang, Taipei
Taiwan 11522, China.
EMail:mingin@cc.chit.edu.tw
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 func- tions.
Abstract: In this paper, we establish several inequalities of Fejér type for Wright- convex functions.
Fejér Inequalities for Wright-convex Functions
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Contents
1 Introduction 3
2 Main Results 7
Fejér Inequalities 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.
Fejér Inequalities for Wright-convex Functions
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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.
Fejér Inequalities for Wright-convex Functions
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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.
Fejér Inequalities for Wright-convex Functions
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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
Fejér Inequalities for Wright-convex Functions
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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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