FEJÉR INEQUALITIES FOR WRIGHT-CONVEX FUNCTIONS
MING-IN HO
CHINAINSTITUTE OFTECHNOLOGY
NANKANG, TAIPEI
TAIWAN11522 mingin@cc.chit.edu.tw
Received 07 September, 2006; accepted 18 November, 2006 Communicated by P.S. Bullen
ABSTRACT. In this paper, we establish several inequalities of Fejér type for Wright-convex functions.
Key words and phrases: Hermite-Hadamard inequality, Fejér inequality, Wright-convex functions.
2000 Mathematics Subject Classification. 26D15.
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]→Ris nonnegative, integrable, and symmetric aboutx= a+b2 .
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 inequal- ity 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,
233-06
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.
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 all x,y+δ∈[a, b]withx < y andδ≥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]. ThenC([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), Theorem A and Theorem B:
Theorem D. Letf ∈W([a, b])∩L1[a, b].Then the inequality(1.1)holds.
Theorem E. Letf ∈ W([a, b])∩L1[a, b]and letH be defined as in Theorem B. ThenH ∈ W([0,1])is increasing on[0,1], and the inequality(1.3)holds for allt∈[0,1].
Theorem F. Let f ∈ W([a, b])∩L1[a, b] and letF be defined as in Theorem C. Then F ∈ W([0,1])is increasing on[0,1], and the inequality(1.4)holds for allt∈[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 Theorem A. 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.
In this paper, we establish some results about Theorem A and Theorem G for Wright-convex functions which are weighted generalizations of Theorem D, E and F.
2. MAINRESULTS
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 sym- metric 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
= 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 2.3. If we set p(x) ≡ 1 (x∈[a, b]) in Theorem 2.2, then Theorem 2.2 generalizes Theorem D.
Remark 2.4. FromC([a, b])$W([a, b]), Theorem 2.2 generalizes Theorem A.
Theorem 2.5. Letf andpbe defined as in Theorem 2.2 and letP be defined as in(1.6). Then P ∈W([0,1])is increasing on[0,1], and the inequality(1.8)holds for allt ∈[0,1].
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 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 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 2.6. If we set p(x) ≡ 1 (x∈[a, b]) in Theorem 2.5, then Theorem 2.2 generalizes Theorem E.
Theorem 2.7. Letf andpbe defined as in Theorem 2.2 and letQbe defined as in(1.7). Then Q∈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
. 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 nonnega- tive 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
≤ 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)
≤
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 thatpis 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
=
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
. 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 2.8. If we set p(x) ≡ 1 (x∈[a, b]) in Theorem 2.7, then Theorem 2.2 generalizes Theorem F.
Remark 2.9. FromC([a, b])$ W([a, b]), Theorem 2.5 and Theorem 2.7 generalize Theorem C.
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