AN INTEGRAL INEQUALITY FOR 3-CONVEX
FUNCTIONS
Integral Inequality for
3-Convex Functions
VLAD CIOBOTARIU-BOER
“Avram Iancu” Secondary School
Cluj-Napoca, Romania
Vlad Ciobotariu-Boer
vol. 9, iss. 4, art. 98, 2008
EMail: vlad_ciobotariu@yahoo.com
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Received:
17 July, 2008
Accepted:
27 September, 2008
Communicated by:
JJ
II
J. Pečarić
J
I
2000 AMS Sub. Class.:
Primary 26D15, Secondary 26D10.
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Key words:
Chebyshev functional, Convex functions, Integral inequality.
Abstract:
In this paper, an integral inequality and an application of it, that imply the Chebyshev functional for two 3-convex (3-concave) functions, are given.
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1
Introduction
3
2
Results
4
3
An Application
15
Integral Inequality for
3-Convex Functions
Vlad Ciobotariu-Boer
vol. 9, iss. 4, art. 98, 2008
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1.
Introduction
For two Lebesgue functions f, g : [a, b] → R, consider the Chebyshev functional
Z b
Z b
Z b
1
1
C(f, g) :=
f (x)g(x)dx −
f (x)dx ·
g(x)dx.
b−a a
(b − a)2 a
a
In 1972, A. Lupaş [2] showed that if f, g are convex functions on the interval [a, b],
then
Z b
Z b
12
a+b
a+b
(1.1) C(f, g) ≥
x−
f (x)dx ·
x−
g(x)dx,
(b − a)4 a
2
2
a
Integral Inequality for
3-Convex Functions
Vlad Ciobotariu-Boer
vol. 9, iss. 4, art. 98, 2008
with equality when at least one of the functions f, g is a linear function on [a, b]. He
proved this result using the following lemma:
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Lemma 1.1. If f, g : [a, b] → R are convex functions on the interval [a, b], then
Contents
(1.2)
[F (e2 ) − F (e)2 ]F (f g) − F (e2 )F (f )F (g)
≥ F (ef )F (eg) − F (e)[F (f )F (eg) + F (ef )F (g)],
where F is an isotonic positive linear functional, defined by one of the following
relations:
Rb
Z b
n
X
p(x)f (x)dx
1
a
f (x)dx, F (f ) := R b
, F (f ) :=
pi f (xi )
(1.3) F (f ) :=
b−a a
p(x)dx
i=1
a
Pn
(xi ∈ [a, b]; pi ≥ 0, i = 1, 2, . . . , n,
i=1 pi = 1), p : [a, b] → R is a positive,
integrable function on [a, b] and e(x) = x, x ∈ [a, b]. If f or g is a linear function,
then the equality holds in (1.2).
In this note, we provide a lower bound for the Chebyshev functional in the case
of two 3-convex (3-concave) functions f and g.
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2.
Results
Note that the inequality (1.2) can be written in the form:
1
F
(e)
F
(g)
2
F
(e)
F
(e
)
F
(eg)
(2.1)
≥ 0.
F (f ) F (ef ) F (f g) The following lemma holds.
Lemma 2.1. If f, g : [a, b] →
[a, b], then
1
F (e)
(2.2)
F (e2 )
F (f )
Vlad Ciobotariu-Boer
R are 3-convex (3-concave) functions on the interval
F (e)
F (e2 )
F (e3 )
F (ef )
Integral Inequality for
3-Convex Functions
F (e2 )
F (e3 )
F (e4 )
F (e2 f )
F (g)
F (eg)
F (e2 g)
F (f g)
≥ 0,
where ei (x) = xi , x ∈ [a, b], i = 1, 4 and F is defined by (1.3).
If f is 3-convex (3-concave) and g is 3-concave (3-convex) then the reverse of the
inequality in (2.2) holds.
If f or g is a polynomial function of degree at most two, then the equality holds
in (2.2).
vol. 9, iss. 4, art. 98, 2008
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Proof. Let [x, y, z, t; f ] be the divided difference of a certain function f . If f and g
are 3-convex (3-concave) on the interval [a, b], then we have
(2.3)
[x, y, z, t; f ] · [x, y, z, t; g] ≥ 0,
for all distinct points x, y, z, t from [a, b].
When f is 3-convex (3-concave) and g is 3-concave (3-convex) then the reverse
of the inequality in (2.3) holds.
In the following we prove (2.2) in the case when both functions f and g are 3convex (3-concave). The inequality (2.3) is equivalent to
1
1
1
1
1
1
1
1
x
x
y
z
t
y
z
t
(2.4)
·
2
≥ 0,
x
y2
z2
t2 x2
y2
z2
t2 f (x) f (y) f (z) f (t) g(x) g(y) g(z) g(t) with true equality holding when at least one of f and g is a polynomial function of
degree at most two.
Note that the function F defined by (1.3) has the property F (1) = 1. In order to
put in evidence the variable u, we write Fu instead of F .
Now, using the fact that F is a linear positive functional, by applying successively
on (2.4) the functionals Fx , Fy , Fz and then Ft , we obtain the inequality (2.2). For
instance, if
1 1 1 2
A = A(x, y, z, t, f, g) := y z t · f (x)g(x),
2 2 2 y z t then
Integral Inequality for
3-Convex Functions
Vlad Ciobotariu-Boer
vol. 9, iss. 4, art. 98, 2008
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1 1 1 2
Fx (A) = y z t · F (f g),
2 2 2 y z t 1
F (e) F (e2 )
Ft Fz Fy Fx (A) = 6 · F (e) F (e2 ) F (e3 )
F (e2 ) F (e3 ) F (e4 )
· F (f g)
and if
1 1 1 1 1 1 B = B(x, y, z, t, f, g) := y z t · x z t · f (x)g(y),
2 2 2 2 2 2 y z t x z t then
1 1 1 F (f )
1 1 Fx (B) = y z t · g(y) · F (ef ) z t ,
2 2 2 y z t F (e2 f ) z 2 t2 1
F
(e)
F
(g)
Ft Fz Fy Fx (B) = 2 · F (e) F (e2 ) F (eg) · F (e2 f )
F (e2 ) F (e3 ) F (e2 g) 2 1
F
(g)
F
(e
)
+ 2 · F (e) F (eg) F (e3 ) · F (ef )
F (e2 ) F (e2 g) F (e4 ) F (g)
F (e) F (e2 )
+ 2 · F (eg) F (e2 ) F (e3 )
F (e2 g) F (e3 ) F (e4 )
Integral Inequality for
3-Convex Functions
Vlad Ciobotariu-Boer
vol. 9, iss. 4, art. 98, 2008
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· F (f ).
Theorem 2.2. If f, g are 3-convex (3-concave) functions on the interval [a, b], then
Z b
Z b
180
(2.5) C(f, g) ≥
q(x)f (x)dx ·
q(x)g(x)dx
(b − a)6 a
a
Z b
Z b
a+b
12
a+b
+
x−
f (x)dx ·
x−
g(x)dx,
(b − a)4 a
2
2
a
where
a+b b−a
a+b b−a
x−
+ √
.
q(x) = x −
− √
2
2
2 3
2 3
If f is 3-convex (3-concave) and g is 3-concave (3-convex) then the reverse of the
inequality in (2.5) holds.
The equality in (2.5) holds when at least one of f or g is a polynomial function of
degree at most two on [a, b].
Proof. We choose
1
F (f ) =
b−a
(2.6)
Z
b
f (x)dx.
Integral Inequality for
3-Convex Functions
Vlad Ciobotariu-Boer
vol. 9, iss. 4, art. 98, 2008
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Then
a+b
,
2
F (e2 ) =
a2 + ab + b2
,
3
a3 + a2 b + ab2 + b3
,
4
F (e4 ) =
a4 + a3 b + a2 b2 + ab3 + b4
.
5
F (e) =
(2.7)
F (e3 ) =
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Note that the inequality (2.2) can be written as
2 1
F
(e)
F
(e
)
1
F (e) 2
2
2
3 (2.8)
F (e) F (e ) F (e ) · F (f g) − 2 · F (e f )F (e g)
F
(e)
F
(e
)
F (e2 ) F (e3 ) F (e4 ) 1
F (e2 ) F (e3 ) F (e2 ) · F (f )F (g)
−
· F (ef )F (eg) − F (e2 ) F (e4 ) F (e3 ) F (e4 ) 1
F (e) +
· [F (e2 f )F (eg) + F (ef )F (e2 g)]
F (e2 ) F (e3 ) F (e) F (e2 ) 2
2
− 2
3 · [F (e f )F (g) + F (f )F (e g)]
F (e ) F (e )
F (e) F (e2 ) · [F (ef )F (g) + F (f )F (eg)] ≥ 0.
+ F (e3 ) F (e4 ) By calculation, we find
1
F (e) F (e2 )
(2.9)
F (e) F (e2 ) F (e3 )
F (e2 ) F (e3 ) F (e4 )
(2.10)
(2.11)
1
F (e2 )
F (e2 )
F (e3 )
(b − a)6
,
=
2160
F (e2 ) (b − a)2 (4a2 + 7ab + 4b2 )
=
,
45
F (e4 ) F (e3 ) (b − a)2 (a4 + 4a3 b + 10a2 b2 + 4ab3 + b4 )
=
,
240
F (e4 ) Integral Inequality for
3-Convex Functions
Vlad Ciobotariu-Boer
vol. 9, iss. 4, art. 98, 2008
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(2.12)
(2.13)
(2.14)
1
F (e2 )
F (e)
F (e2 )
F (e)
F (e3 )
F (e) (b − a)2 (a + b)
=
,
12
F (e3 ) F (e2 ) (b − a)2 (a2 + 4ab + b2 )
=
,
72
F (e3 ) F (e2 ) (b − a)2 (a3 + 4a2 b + 4ab2 + b3 )
=
.
60
F (e4 ) The relations (2.7) – (2.14) give us
Z
(b − a)5 b
(2.15)
f (x)g(x)dx
2160 a
Z
Z b
a4 + 4a3 b + 10a2 b2 + 4ab3 + b4 b
−
f (x)dx
g(x)dx
240
a
a
Z b
Z b
1
2
x f (x)dx
x2 g(x)dx
≥
12 a
a
Z
Z b
4a2 + 7ab + 4b2 b
+
xf (x)dx
xg(x)dx
45
a
a
Z b
Z b
Z b
Z b
a+b
2
2
−
x f (x)dx
xg(x)dx +
xf (x)dx
x g(x)dx
12
a
a
a
a
Z b
Z b
a2 + 4ab + b2
2
+
x f (x)dx
g(x)dx
72
a
a
Z b
Z b
2
+
f (x)dx
x g(x)dx
a
a
Integral Inequality for
3-Convex Functions
Vlad Ciobotariu-Boer
vol. 9, iss. 4, art. 98, 2008
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Z b
Z b
a3 + 4a2 b + 4ab2 + b3
−
xf (x)dx
g(x)dx
60
a
a
Z b
Z b
+
f (x)dx
xg(x)dx ,
a
a
or
(2.16)
Integral Inequality for
3-Convex Functions
C(f, g)
Z b
Z b
180
2
≥
x f (x)dx
x2 g(x)dx
(b − a)6
a
a
Z
Z b
4(4a2 + 7ab + 4b2 ) b
+
xf (x)dx
xg(x)dx
15
a
a
Z b
Z b
2a4 + 10a3 b + 21a2 b2 + 10ab3 + 2b4
·
f (x)dx
g(x)dx
+
540
a
a
Z b
Z b
Z b
Z b
a+b
2
2
x f (x)dx
xg(x)dx +
xf (x)dx
x g(x)dx
−
12
a
a
a
a
Z b
Z b
a2 + 4ab + b2
2
+
x f (x)dx
g(x)dx
72
a
a
Z b
Z b
2
+
f (x)dx
x g(x)dx
a
a
Z b
Z b
a3 + 4a2 b + 4ab2 + b3
xf (x)dx
−
g(x)dx
60
a
a
Z b
Z b
+
f (x)dx
xg(x)dx .
a
a
Vlad Ciobotariu-Boer
vol. 9, iss. 4, art. 98, 2008
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The last inequality can be written as
(2.17)
C(f, g)
2
2
Z b
a+b
a+b
x−
f (x)dx ·
x−
g(x)dx
2
2
a
a
"Z 2
Z b
b
15
a+b
·
x−
f (x)dx ·
g(x)dx+
(b − a)4
2
a
a
#
2
Z b
Z b
a+b
g(x)dx
f (x)dx ·
x−
2
a
a
Z b
Z b
5
f (x)dx ·
g(x)dx
4(b − a)2 a
a
Z b
Z b
a+b
12
a+b
x−
f (x)dx ·
x−
g(x)dx,
(b − a)4 a
2
2
a
180
≥
(b − a)6
−
+
+
+
Z b
Integral Inequality for
3-Convex Functions
Vlad Ciobotariu-Boer
vol. 9, iss. 4, art. 98, 2008
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which is equivalent to (2.5).
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Corollary 2.3. Let f and g be as in Theorem 2.2 and assume that
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(2.18)
f (x) = −f (a + b − x)
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or
(2.19)
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g(x) = −g(a + b − x)
for all x from [a, b]. Then Lupaş’ inequality holds.
Proof. Note that the function denoted by q in Theorem 2.2 is symmetric about x =
a+b
, namely
2
q(x) = q(a + b − x),
Close
for all x from [a, b].
Assume that (2.18) is satisfied. Then we have
Z b
Z
1 b
(2.20)
q(x)f (x)dx =
q(x)[f (x) − f (a + b − x)]dx
2 a
a
Z
Z
1 b
1 b
q(x)f (x)dx −
q(a + b − x)f (a + b − x)dx
=
2 a
2 a
Z
Z
1 b
1 b
=
q(x)f (x)dx −
q(t)f (t)dt = 0.
2 a
2 a
Integral Inequality for
3-Convex Functions
Vlad Ciobotariu-Boer
vol. 9, iss. 4, art. 98, 2008
From (2.5) and (2.20), we deduce (1.1).
Note that the condition (2.3) is important. The same results are valid if we suppose that this (or its reverse) is satisfied. Thus, we obtain a more general result:
Theorem 2.4. If the functions f and g are integrable on the interval [a, b] and satisfy
(2.3) (or its reverse), then we have (2.5) (or its reverse).
The equality in (2.5) holds when at least one of f or g is a polynomial function of
degree at most two on [a, b].
Corollary 2.5. If the function f is integrable on [a, b], then we have
b
Z
2
(2.21) (b − a)
f (x)dx −
a
12
≥
(b − a)2
Z
f (x)dx
a
a+b
x−
2
where q(x) is defined in Theorem 2.2.
f (x)dx
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a
Z b Contents
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2
b
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180
+
(b − a)4
Z
2
b
q(x)f (x)dx
a
,
Proof. Considering g(x) = f (x) in (2.5), we find the inequality (2.21).
Remark 1. The inequality (2.21) is better than the well-known inequality
Z b
2
Z b
2
(2.22)
(b − a)
f (x)dx ≥
f (x)dx ,
a
a
valid for all integrable functions f on [a, b].
Corollary 2.6. If the functions f, g satisfy the following conditions:
(i) f, g are 3-convex (3-concave) functions on [a, b];
Integral Inequality for
3-Convex Functions
Vlad Ciobotariu-Boer
vol. 9, iss. 4, art. 98, 2008
(ii) f, g are differentiable functions on [a, b],
then we have
Z b
[f (b) − f (a)][g(b) − g(a)]
(2.23)
f 0 (x)g 0 (x)dx ≥
b−a
a
Z b
1
f (a) + f (b)
12
+
f (x)dx −
b−a b−a a
2
Z b
1
g(a) + g(b)
×
g(x)dx −
.
b−a a
2
Proof. In (1.1), we use the fact that if f, g are 3-convex functions on [a, b], then f 0 , g 0
are convex functions on [a, b]. We have
Z b
Z b
Z b
1
1
0
0
0
f (x)g (x)dx −
f (x)dx ·
g 0 (x)dx
b−a a
(b − a)2 a
a
Z b
Z b
12
a+b
a+b
0
≥
x−
f (x)dx ·
x−
g 0 (x)dx,
4
(b − a) a
2
2
a
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which is equivalent to (2.23).
Remark 2. If, in addition, f and g are convex (concave) on [a, b], then the inequality
(2.23) is better than the inequality
Z b
[f (b) − f (a)][g(b) − g(a)]
(2.24)
f 0 (x) · g 0 (x)dx ≥
,
b−a
a
which is valid for all convex (concave) functions f, g on [a, b].
Remark 3. Lemma 2.1 can be generalized for n-convex functions, obtaining a result
similar to (2.2), from where the inequality
Rb
b−a
b2 −a2
bn −an
.
.
.
g(x)dx
2
n
a
Rb
b2 −a2
3 −a3
n+1 −an+1
b
b
...
xg(x)dx
3
n+1
a
2
≥ 0,
...
... ...
...
(2.25) . . .
n n
R
n+1
n+1
2n−1
2n−1
b
b −a
b
−a
b
−a
n−1
.
.
.
x
g(x)dx
n
n+1
2n−1
a
Rb
R b n−1
Rb
Rb
f (x)dx
xf (x)dx . . .
x f (x)dx
f (x)g(x)dx a
a
a
a
holds for all integer numbers n ≥ 3.
Integral Inequality for
3-Convex Functions
Vlad Ciobotariu-Boer
vol. 9, iss. 4, art. 98, 2008
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Some similar results related to the Chebyshev functional are given in [1] – [6].
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3.
An Application
Let f, g be two 3-time differentiable functions defined on a nonempty interval [a, b].
Denote
m1 = inf f (3) (x),
x∈[a,b]
M1 = sup f (3) (x),
x∈[a,b]
(3)
m2 = inf g (x),
x∈[a,b]
M2 = sup g (3) (x).
Integral Inequality for
3-Convex Functions
x∈[a,b]
Vlad Ciobotariu-Boer
Considering the functions F1 , G1 , F2 , G2 : [a, b] → R, defined by
m1 x3
F1 (x) =
− f (x),
6
M1 x 3
F2 (x) =
− f (x),
6
vol. 9, iss. 4, art. 98, 2008
m2 x3
G1 (x) =
− g(x),
6
M2 x 3
G2 (x) =
− g(x),
6
(3)
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(3)
we note that these are 3-differentiable on [a, b] and F1 (x) ≤ 0, G1 (x) ≤ 0,
(3)
(3)
F2 (x) ≥ 0, G2 (x) ≥ 0 for all x ∈ [a, b]. Therefore F1 , G1 are 3-concave on [a, b]
and F2 , G2 are 3-convex on [a, b].
Applying Theorem 2.2 we shall prove the following result:
Theorem 3.1. Let f, g be two 3-differentiable functions on the nonempty interval
[a, b]. Then, we have
Z b
1
a+b
(3.1) L(f, g) −
x−
r(x)h(x)dx
6(b − a) a
2
(m1 + M1 )(m2 + M2 )
6
· (b − a) +
403200
(M1 − m1 )(M2 − m2 )
≤
· (b − a)6 ,
403200
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where
(3.2) L(f, g) = C(f, g)
Z b
Z b
a+b
12
a+b
−
x−
f (x)dx ·
x−
g(x)dx
(b − a)4 a
2
2
a
Z b
Z b
180
−
q(x)f (x)dx ·
q(x)g(x)dx,
(b − a)6 a
a
Integral Inequality for
3-Convex Functions
Vlad Ciobotariu-Boer
(3.3)
(3.4)
m 1 + M1
m 2 + M2
h(x) =
· g(x) +
· f (x),
2
2
√ !
√ !
a + b (b − a) 15
a + b (b − a) 15
x−
.
r(x) = x −
−
+
2
10
2
10
Proof. Applying Theorem 2.2, we have
Z b
Z b
180
q(x)F1 (x)dx ·
q(x)G1 (x)dx
(3.5) C(F1 , G1 ) ≥
(b − a)6 a
a
Z b
Z b
12
a+b
a+b
+
x−
F1 (x)dx ·
G1 (x)dx,
x−
(b − a)4 a
2
2
a
Z b
Z b
180
(3.6) C(F2 , G2 ) ≥
q(x)F2 (x)dx ·
q(x)G2 (x)dx
(b − a)6 a
a
Z b
Z b
12
a+b
a+b
+
x−
F2 (x)dx ·
x−
G2 (x)dx,
(b − a)4 a
2
2
a
vol. 9, iss. 4, art. 98, 2008
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Z b
Z b
180
(3.7) C(F1 , G2 ) ≤
q(x)F1 (x)dx ·
q(x)G2 (x)dx
(b − a)6 a
a
Z b
Z b
12
a+b
a+b
+
x−
F1 (x)dx ·
x−
G2 (x)dx,
(b − a)4 a
2
2
a
Z b
Z b
180
(3.8) C(F2 , G1 ) ≤
q(x)F2 (x)dx ·
q(x)G1 (x)dx
(b − a)6 a
a
Z b
Z b
12
a+b
a+b
+
x−
F2 (x)dx ·
x−
G1 (x)dx,
(b − a)4 a
2
2
a
where q(x) is defined in Theorem 2.2.
By calculation, we find
m1 m2
(3.9) C(F1 , G1 ) = C(f, g)+
(b − a)2 (9a4 +20a3 b+26a2 b2 +20ab3 +9b4 )
4032
Z b
1
−
x3 [m1 g(x) + m2 f (x)]dx
6(b − a) a
Z
a3 + a2 b + ab2 + b3 b
[m1 g(x) + m2 f (x)]dx,
+
24(b − a)
a
Integral Inequality for
3-Convex Functions
Vlad Ciobotariu-Boer
vol. 9, iss. 4, art. 98, 2008
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Z b
(3.10)
Z b
12
a+b
a+b
x−
F1 (x)dx ·
x−
G1 (x)dx
4
(b − a) a
2
2
a
Z b
Z b
12
a+b
a+b
=
x−
f (x)dx ·
x−
g(x)dx
(b − a)4 a
2
2
a
m1 m2
+
(b − a)2 (3a2 + 4ab + 3b2 )2
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3a2 + 4ab + 3b2
−
20(b − a)
Z b
a
a+b
x−
2
[m1 g(x) + m2 f (x)]dx,
Z b
Z b
180
q(x)F1 (x)dx ·
q(x)G1 (x)dx
(3.11)
(b − a)6 a
a
Z b
Z b
180
m1 m2
=
q(x)f (x)dx ·
q(x)g(x)dx +
(b − a)4 (a + b)2
6
(b − a) a
2880
a
Z b
a+b
−
·
q(x)[m1 g(x) + m2 f (x)]dx.
4(b − a) a
From (3.5) and (3.9) – (3.11), we obtain
(3.12) L(f, g) +
m1 m2
(b − a)6
100800
Z b
1
a+b
≥
x−
r(x)[m1 g(x) + m2 f (x)]dx.
6(b − a) a
2
In a similar way we can prove that the inequality (3.6) is equivalent to
Integral Inequality for
3-Convex Functions
Vlad Ciobotariu-Boer
vol. 9, iss. 4, art. 98, 2008
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M1 M2
(3.13) L(f, g) +
(b − a)6
100800
Z b
1
a+b
≥
x−
r(x)[M1 g(x) + M2 f (x)]dx,
6(b − a) a
2
the inequality (3.7) is equivalent to
(3.14) L(f, g) +
m 1 M2
(b − a)6
100800
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1
≤
6(b − a)
Z b
a
a+b
x−
2
r(x)[m1 g(x) + M2 f (x)]dx,
and the inequality (3.8) is equivalent to
(3.15) L(f, g) +
M1 m 2
(b − a)6
100800
Z b
1
a+b
≤
x−
r(x)[M1 g(x) + m2 f (x)]dx.
6(b − a) a
2
Integral Inequality for
3-Convex Functions
Vlad Ciobotariu-Boer
vol. 9, iss. 4, art. 98, 2008
From (3.12) and (3.13) we deduce
Z b
1
a+b
(3.16) L(f, g) −
x−
r(x)h(x)dx
6(b − a) a
2
m 1 m 2 + M1 M2
≥−
(b − a)6 .
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From (3.14) and (3.15) we find
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I
1
(3.17) L(f, g) −
6(b − a)
Z b
a
a+b
x−
2
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Page 19 of 21
r(x)h(x)dx
≤−
m 1 M2 + M1 m 2
(b − a)6 .
201600
The inequalities (3.16) and (3.17) prove (3.1).
Corollary 3.2. If f, g are 3-time differentiable on [a, b] and symmetric about x =
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a+b
,
2
then we have
(3.18)
(m
+
M
)(m
+
M
)
1
1
2
2
6
L(f, g) +
(b
−
a)
403200
≤
(M1 − m1 )(M2 − m2 )
(b − a)6 .
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Proof. Note that the functions h and r defined on [a, b] by (3.3) and (3.4) are symmetric about x = a+b
. Hence, their product h · r is symmetric about x = a+b
and
2
2
Integral Inequality for
3-Convex Functions
Vlad Ciobotariu-Boer
vol. 9, iss. 4, art. 98, 2008
Z b
(3.19)
a
a+b
x−
2
r(x)h(x)dx = 0.
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From (3.1) and (3.19), we obtain (3.18).
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References
[1] E.V. ATKINSON, An inequality, Univ. Beograd. Publ. Elektrotehn. Fak. Ser.
Mat. Fiz., (357-380) (1971), 5–6.
[2] A. LUPAŞ, An integral inequality for convex functions, Univ. Beograd. Publ.
Elektrotehn. Fak. Ser. Mat. Fiz., (381-409) (1972), 17–19.
[3] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Classical and New Inequalitites in Analysis, Kluwer Academic Publishers, Dordrecht, 1992.
Integral Inequality for
3-Convex Functions
Vlad Ciobotariu-Boer
vol. 9, iss. 4, art. 98, 2008
[4] D.S. MITRINOVIĆ AND P.M. VASIĆ, Analytic Inequalities, Springer-Verlag,
Berlin and New-York, 1970.
Title Page
[5] J.E. PEČARIĆ, F. PROSCHAN AND Y.I. TONG, Convex Functions, Partial Orderings, and Statistical Applications, Academic Press, San Diego, 1992.
[6] P.M. VASIĆ AND I.B. LACKOVIĆ, Notes on convex functions VI: On an inequality for convex functions proved by A. Lupaş, Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. Fiz., (634-677) (1979), 36–41.
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