MATEMATIQKI VESNIK
originalni nauqni rad
research paper
68, 1 (2016), 45–57
March 2016
INTEGRAL INEQUALITIES OF JENSEN TYPE
FOR λ-CONVEX FUNCTIONS
S. S. Dragomir
Abstract. Some integral inequalities of Jensen type for λ-convex functions defined on real
intervals are given.
1. Introduction
Assume that I and J are intervals in R, (0, 1) ⊆ J and functions h and f are
real non-negative functions defined in J and I, respectively.
Definition 1. [20] Let h: J → [0, ∞) with h not identical to 0. We say that
f : I → [0, ∞) is an h-convex function if for all x, y ∈ I we have
f (tx + (1 − t)y) ≤ h(t)f (x) + h(1 − t)f (y)
(1.1)
for all t ∈ (0, 1).
For some results concerning this class of functions see [3, 13, 17–20].
This class of functions contains the class of Godunova-Levin type functions [9,
10, 14, 16]. It also contains the class of P functions and quasi-convex functions.
For some results on P -functions see [15] while for quasi convex functions, the reader
can consult [11].
Definition 2. [4] Let s be a real number, s ∈ (0, 1]. A function f : [0, ∞) →
[0, ∞) is said to be s-convex (in the second sense) or Breckner s-convex if
f (tx + (1 − t)y) ≤ ts f (x) + (1 − t)s f (y)
for all x, y ∈ [0, ∞) and t ∈ [0, 1].
For some properties of this class of functions see [1, 2, 4, 7, 8, 12].
2010 Mathematics Subject Classification: 26D15, 25D10
Keywords and phrases: Convex function; discrete inequality; λ-convex function; Jensen type
inequality.
45
46
S. S. Dragomir
We can introduce now another class of functions defined on a convex subset C
of a linear space X that contains as limiting cases the classes of Godunova-Levin
and P -functions.
Definition 3. We say that the function f : C ⊆ X → [0, ∞) is of s-GodunovaLevin type, with s ∈ [0, 1], if
1
1
f (tx + (1 − t)y) ≤ s f (x) +
f (y),
(1.2)
t
(1 − t)s
for all t ∈ (0, 1) and x, y ∈ C.
We observe that for s = 0 we obtain the class of P -functions while for s = 1
we obtain the class of Godunova-Levin. If we denote by Qs (C) the class of sGodunova-Levin functions defined on C, then we obviously have
P (C) = Q0 (C) ⊆ Qs1 (C) ⊆ Qs2 (C) ⊆ Q1 (C) = Q(C)
for 0 ≤ s1 ≤ s2 ≤ 1.
For different inequalities of Hermite-Hadamard or Jensen type related to these
classes of functions, see [1, 3, 13, 15–19].
A function h: J → R is said to be supermultiplicative if
h(ts) ≥ h(t)h(s) for any t, s ∈ J.
(1.3)
If the inequality (1.3) is reversed, then h is said to be submultiplicative. If the
equality holds in (1.3) then h is said to be a multiplicative function on J.
In [15], we introduced the following concept of functions:
Definition 4. Let λ: [0, ∞) → [0, ∞) be a function with the property that
λ(t) > 0 for all t > 0. A mapping f : C → R defined on convex subset C of a linear
space X is called λ-convex on C if
µ
¶
αx + βy
λ(α)f (x) + λ(β)f (y)
f
≤
(1.4)
α+β
λ(α + β)
for all α, β ≥ 0 with α + β > 0 and x, y ∈ C.
We observe that if f : C → R is λ-convex on C, then f is h-convex on C with
λ(t)
, t ∈ [0, 1]. If f : C → [0, ∞) is h-convex function with h supermultih(t) = λ(1)
plicative on [0, ∞), then f is λ-convex with λ = h.
We have the following result providing many examples of subadditive functions
λ: [0, ∞) → [0, ∞).
P∞
Theorem 1. [5] Let h(z) = n=0 an z n be a power series with nonnegative
coefficients an ≥ 0 for all n ∈ N and convergent on the open disk D(0, R) with
R > 0 or R = ∞. If r ∈ (0, R) then the function λr : [0, ∞) → [0, ∞) given by
·
¸
h(r)
λr (t) := ln
(1.5)
h(r exp(−t))
is nonnegative, increasing and subadditive on [0, ∞).
47
Integral inequalities of Jensen type
Now, if we take h(z) =
1
1−z ,
z ∈ D(0, 1), then
·
¸
1 − r exp(−t)
λr (t) = ln
1−r
(1.6)
is nonnegative, increasing and subadditive on [0, ∞) for any r ∈ (0, 1).
If we take h(z) = exp(z), z ∈ C then
λr (t) = r[1 − exp(−t)]
(1.7)
is nonnegative, increasing and subadditive on [0, ∞) for any r > 0.
P∞
Corollary 1. [5] Let h(z) = n=0 an z n be a power series with nonnegative
coefficients an ≥ 0 for all n ∈ N and convergent on the open disk D(0, R) with
R > 0 or R = ∞ and r ∈ (0, R). For a mapping f : C → R defined on convex subset
C of a linear space X, the following statements are equivalent:
(i) The function f is λr -convex with λr : [0, ∞) → [0, ∞),
·
¸
h(r)
λr (t) := ln
;
h(r exp(−t))
(ii) We have the inequality
¸f ( αx+βy
¸f (x) ·
¸f (y)
·
α+β )
h(r)
h(r)
h(r)
≤
h(r exp(−α − β))
h(r exp(−α))
h(r exp(−β))
for any α, β ≥ 0 with α + β > 0 and x, y ∈ C.
(iii) We have the inequality
·
[h(r exp(−α))]f (x) [h(r exp(−β))]f (y)
f ( αx+βy
α+β )
[h(r exp(−α − β))]
for any α, β ≥ 0 with α + β > 0 and x, y ∈ C.
≤ [h(r)]f (x)+f (y)−f (
αx+βy
α+β )
(1.8)
(1.9)
We observe that, in the case when
λr (t) = r[1 − exp(−t)], t ≥ 0
then the function f is λr -convex on convex subset C of a linear space X iff
µ
¶
αx + βy
[1 − exp(−α)]f (x) + [1 − exp(−β)]f (y)
f
≤
(1.10)
α+β
1 − exp(−α − β)
for any α, β ≥ 0 with α + β > 0 and x, y ∈ C. Notice that this definition is
independent of r > 0.
The inequality (1.10) is equivalent to
µ
¶
αx + βy
exp(β)[exp(α) − 1]f (x) + exp(α)[exp(β) − 1]f (y)
f
≤
(1.11)
α+β
exp(α + β) − 1
for any α, β ≥ 0 with α + β > 0 and x, y ∈ C.
Motivated by the large interest on Jensen and Hermite-Hadamard inequalities
that has been materialized in the last two decades by the publication of hundreds
of papers, we establish here some inequalities of these types for λ-convex functions
defined on real intervals.
48
S. S. Dragomir
2. Unweighted Jensen integral inequalities
The following discrete inequality of Jensen type has been obtained in [6]:
Theorem 2. Let λ: [0, ∞) → [0, ∞) be a function with the property that
λ(t) > 0 for all t > 0 and a mapping f : C → R defined on convex subset C of a
linear space X. The following statements are equivalent:
(i) f is λ-convex on C;
(ii) For all xi ∈ C and pi ≥ 0 with i ∈ {1, . . . , n}, n ≥ 2 so that Pn > 0, we
have the inequality
µ
¶
n
n
1 P
1 P
f
pi xi ≤
λ(pi )f (xi ).
(2.1)
Pn i=1
λ(Pn ) i=1
The proof can be done by induction over n ≥ 2.
Corollary 2.PLet f : C → R be a λ-convex function on C and αi ∈ [0, 1],
n
i ∈ {1, . . . , n} with i=1 αi = 1. Then for any xi ∈ C with i ∈ {1, . . . , n} we have
the inequality
µ n
¶
n
P
1 P
f
αi xi ≤
λ(αi )f (xi ).
(2.2)
λ(1) i=1
i=1
In particular, we have
¶
µ
f (x1 ) + · · · + f (xn )
x1 + · · · + xn
≤ c(n)
,
f
n
n
where
c(n) :=
(2.3)
nλ( n1 )
.
λ(1)
We have the following version of Jensen’s inequality as well:
Corollary 3. Let f : C → R be a λ-convex function on C and xi ∈ C and
pi ≥ 0 with i ∈ {1, . . . , n}, n ≥ 2 so that Pn > 0. Then we have the inequality
µ
¶
µ ¶
n
n
1 P
1 P
pi
f
pi xi ≤
λ
f (xi ).
(2.4)
Pn i=1
λ(1) i=1
Pn
The proof follows by (2.2) for αi =
pi
Pn ,
i ∈ {1, . . . , n}.
We are able now to state and prove the following unweighted Jensen inequality
for Riemann integral:
Theorem 3. Let u: [a, b] → [m, M ] be a Riemann integrable function on [a, b].
Let λ: [0, ∞) → [0, ∞) be a function with the property that λ(t) > 0 for all t > 0
49
Integral inequalities of Jensen type
and the function f : [m, M ] → [0, ∞) is λ -convex and Riemann integrable on the
interval [m, M ]. If the following limit exists
lim
t→0+
then
µ
1
f
b−a
Z
λ(t)
= k ∈ (0, ∞)
t
¶
b
u(t) dt
a
k
≤
λ(b − a)
Z
(2.5)
b
f (u(t)) dt.
(2.6)
a
Proof. Consider the sequence of divisions
(n)
dn : xi
=a+
i
(b − a), i ∈ {0, . . . , n}
n
and the intermediate points
(n)
ξi
=a+
i
(b − a), i ∈ {0, . . . , n}.
n
(n)
(n)
We observe that the norm of the division ∆n := maxi∈{0,...,n−1} (xi+1 − xi ) =
b−a
n → 0 as n → ∞ and since u is Riemann integrable on [a, b], then
µ
¶
Z b
n−1
P
P
i
b − a n−1
(n)
(n)
(n)
u a + (b − a) .
u(t) dt = lim
u(ξi )[xi+1 − xi ] = lim
n→∞ i=0
n→∞
n i=0
n
a
Also, since f : [m, M ] → [0, ∞) is Riemann integrable, then f ◦ u is Riemann integrable and
·
¸
Z b
P
i
b − a n−1
f (u(t)) dt = lim
f u(a + (b − a)) .
n→∞
n i=0
n
a
i
Utilising the inequality (2.1) for pi := b−a
n and xi := u(a + n (b − a)) we have
µ
µ
¶¶
P
1 b − a n−1
i
f
u a + (b − a)
b − a n i=0
n
µ
¶
µ µ
¶¶
P
n
b − a b − a n−1
i
≤
λ
f u a + (b − a)
λ(b − a)(b − a)
n
n i=0
n
(2.7)
for any n ≥ 1.
Observe that
lim
n→∞
λ( b−a
n )
b−a
n
= lim
t→0+
λ(t)
= k ∈ (0, ∞),
t
and by taking the limit over n → ∞ in the inequality (2.7), we deduce the desired
result (2.6).
CorollaryP
4. Let u: [a, b] → [m, M ] be a Riemann integrable function on
∞
[a, b] and h(z) = n=0 an z n be a power series with nonnegative coefficients an ≥ 0
50
S. S. Dragomir
for all n ∈ N and convergent on the open disk D(0, R) with R > 0 or R = ∞ and
r ∈ (0, R). Let λr : [0, ∞) → [0, ∞) be given by
·
¸
h(r)
λr (t) := ln
h(r exp(−t))
and the function f : [m, M ] → [0, ∞) be λr -convex and Riemann integrable on the
interval [m, M ]. Then
¶
µ
Z b
Z b
1
rh0 (r)
h
i
u(t) dt ≤
f
f (u(t)) dt.
(2.8)
h(r)
b−a a
a
h(r) ln
h(r exp(−(b−a)))
Proof. We observe that λr is differentiable on (0, ∞) and
λ0r (t) :=
for t ∈ (0, ∞), where h0 (z) =
r exp(−t)h0 (r exp(−t))
h(r exp(−t))
P∞
n=1
nan z n−1 . Since λr (0) = 0, therefore
λ(s)
rh0 (r)
= λ0+ (0) =
> 0 for r ∈ (0, R).
s→0+ s
h(r)
k = lim
Utilising (2.6) we get the desired result (2.8).
The following Hermite-Hadamard type inequality holds:
Corollary 5. With the assumptions of Theorem 3 for f and λ and if [a, b] =
[m, M ], we have the Hermite-Hadamard type inequality
µ
¶
Z b
a+b
k
f
≤
f (t) dt.
(2.9)
2
λ(b − a) a
Remark 1. Assume that the function f : [m, M ] → [0, ∞) is λ-convex and
Riemann integrable on the interval [m, M ] with
λ(t) = 1 − exp(−t), t ≥ 0.
If u: [a, b] → [m, M ] is a Riemann integrable function on [a, b], then
¶
µ
Z b
Z b
1
1
f
u(t) dt ≤
f (u(t)) dt.
b−a a
1 − exp(−(b − a)) a
In particular, for [a, b] = [m, M ] and u(t) = t we have the Hermite-Hadamard type
inequality
µ
¶
Z b
a+b
1
f
≤
f (t) dt.
2
1 − exp(−(b − a)) a
The proof follows from (2.6) observing that
k = lim
t→0+
λ(t)
= λ0+ (0) = 1.
t
Integral inequalities of Jensen type
51
Utilising a similar argument and the inequality (2.4) we can state the following
result as well:
Theorem 4. Let u: [a, b] → [m, M ] be a Riemann integrable function on [a, b].
Let λ: [0, ∞) → [0, ∞) be a function with the property that λ(t) > 0 for all t > 0
and the function f : [m, M ] → [0, ∞) is λ -convex and Riemann integrable on the
interval [m, M ]. If the limit (2.5) exists, then
¶
µ
Z b
Z b
k
1
u(t) dt ≤
f (u(t)) dt.
(2.10)
f
b−a a
λ(1)(b − a) a
Examples of such inequalities are incorporated below:
CorollaryP
6. Let u: [a, b] → [m, M ] be a Riemann integrable function on
∞
[a, b] and h(z) = n=0 an z n be a power series with nonnegative coefficients an ≥ 0
for all n ∈ N and convergent on the open disk D(0, R) with R > 0 or R = ∞ and
r ∈ (0, R). Let λr : [0, ∞) → [0, ∞) be given by
·
¸
h(r)
λr (t) := ln
h(r exp(−t))
and the function f : [m, M ] → [0, ∞) be λr -convex and Riemann integrable on the
interval [m, M ]. Then
µ
¶
Z b
Z b
1
rh0 (r)
h
i
f
u(t) dt ≤
f (u(t)) dt.
(2.11)
b−a a
a
(b − a)h(r) ln h(r)
−1
h(re
)
We also have the Hermite-Hadamard type inequality:
Corollary 7. With the assumptions of Theorem 4 for f and λ and if [a, b] =
[m, M ], we have the Hermite-Hadamard type inequality
µ
¶
Z b
a+b
k
f
≤
f (t) dt.
(2.12)
2
λ(1)(b − a) a
Remark 2. Assume that the function f : [m, M ] → [0, ∞) is λ-convex and
Riemann integrable on the interval [m, M ] with λ(t) = 1 − exp(−t), t ≥ 0. If
u: [a, b] → [m, M ] is a Riemann integrable function on [a, b], then
µ
¶
Z b
Z b
1
e
1
f
u(t) dt ≤
·
f (u(t)) dt.
b−a a
e−1 b−a a
In particular, for [a, b] = [m, M ] and u(t) = t we have the Hermite-Hadamard type
inequality
¶
µ
Z b
e
1
a+b
≤
·
f (t) dt.
f
2
e−1 b−a a
52
S. S. Dragomir
3. Weighted Jensen integral inequalities
We can prove now a weighted version of Jensen inequality.
Theorem 5. Let u, w: [a, b] → [m, M ] be Riemann integrable functions on
Rb
[a, b] and w(t) ≥ 0 for any t ∈ [a, b] with a w(t) dt > 0. Let λ: [0, ∞) → [0, ∞) be a
function with the property that λ(t) > 0 for all t > 0 and the function f : [m, M ] →
[0, ∞) is λ-convex and Riemann integrable on the interval [m, M ]. If the following
limit exists, is finite and
t
= ` > 0,
(3.1)
lim
t→∞ λ(t)
then
µ
¶
Z b
Z b
1
1
f Rb
w(t)u(t) dt ≤ ` R b
λ(w(t))f (u(t)) dt.
(3.2)
a
a
w(t)
dt
w(t)
dt
a
a
Proof. Consider the sequence of divisions
i
(n)
dn : xi = a + (b − a), i ∈ {0, . . . , n}
n
and the intermediate points
i
(n)
ξi = a + (b − a), i ∈ {0, . . . , n}.
n
(n)
(n)
We observe that the norm of the division ∆n := maxi∈{0,...,n−1} (xi+1 − xi ) =
b−a
n → 0 as n → ∞.
If we write the inequality (2.1) for the sequences
¶
µ
¶
µ
i
i
pi = w a + (b − a) and xi = u a + (b − a) , i ∈ {0, . . . , n}
n
n
we get
Ã
µ
¶ µ
¶!
n−1
P
1
i
i
f Pn−1
w a + (b − a) u a + (b − a)
i
n
n
i=0 w(a + n (b − a)) i=0
1
≤ Pn−1
λ( i=0 w(a + ni (b − a)))
¶¶ µ µ
¶¶
µ µ
n−1
P
i
i
×
f u a + (b − a)
,
(3.3)
λ w a + (b − a)
n
n
i=0
for n ≥ 1.
Observe that
Ã
¶ µ
¶!
µ
n−1
P
i
1
i
f Pn−1
w a + (b − a) u a + (b − a)
i
n
n
i=0 w(a + n (b − a)) i=0
Ã
µ
¶
µ
¶!
b−a
n−1
P
i
i
n
= f b−a Pn−1
w a + (b − a) u a + (b − a)
i
n
n
i=0 w(a + (b − a)) i=0
n
n
Integral inequalities of Jensen type
53
and
1
Pn−1
i
i=0 w(a + n (b − a)))
µ
µ
¶¶ µ µ
¶¶
n−1
P
i
i
×
λ w a + (b − a)
f u a + (b − a)
n
n
i=0
Pn−1
i
w(a + n (b − a))
1
= Pi=0
× b−a Pn−1
n−1
i
i
λ( i=0 w(a + n (b − a)))
i=0 w(a + n (b − a))
n
¶¶
µ
µ
¶¶
µ
µ
P
i
b − a n−1
i
f u a + (b − a)
.
×
λ w a + (b − a)
n i=0
n
n
λ(
Then from (3.3) we get
Ã
¶ µ
¶!
µ
i
i
f b−a Pn−1
w a + (b − a) u a + (b − a)
i
n
n
i=0 w(a + n (b − a)) i=0
n
Pn−1
w(a + ni (b − a))
1
≤ Pi=0
× b−a Pn−1
n−1
i
i
λ( i=0 w(a + n (b − a)))
i=0 w(a + n (b − a))
n
¶¶
µ
µ
¶¶
µ
µ
P
i
b − a n−1
i
×
λ w a + (b − a)
f u a + (b − a)
(3.4)
n i=0
n
n
b−a
n
n−1
P
for all n ≥ 1. Since
µ
¶
µ
¶
n−1
P
P
i
b − a n−1
i
n
lim
w a + (b − a) = lim
w a + (b − a) × lim
n→∞ i=0
n→∞
n→∞ b − a
n
n i=0
n
Z b
=
w(t) dt × ∞ = ∞,
a
then
Pn−1
i
t
i=0 w(a + n (b − a))
= lim
=`
Pn−1
i
n→∞ λ(
n→∞
λ(t)
i=0 w(a + n (b − a)))
lim
and by letting n → ∞ in (3.4) we get the desired result (3.2).
The following unweighted version of Jensen inequality holds:
Corollary 8. Let u: [a, b] → [m, M ] be a Riemann integrable function on
[a, b]. Let λ: [0, ∞) → [0, ∞) be a function with the property that λ(t) > 0 for all
t > 0 and the function f : [m, M ] → [0, ∞) be λ -convex and Riemann integrable on
the interval [m, M ]. If the limit (3.1) exists, then
µ
¶
Z b
Z b
1
1
f
u(t) dt ≤ `λ(1)
f (u(t)) dt.
(3.5)
b−a a
b−a a
Moreover, if [a, b] = [m, M ], then by taking u(t) = t, t ∈ [a, b], we have the HermiteHadamard inequality
µ
¶
Z b
a+b
1
f
≤ `λ(1)
f (t) dt.
(3.6)
2
b−a a
54
S. S. Dragomir
Remark 3. In order to give examples of subadditive functions λ: [0, ∞) →
[0, ∞) with the property that λ(t) > 0 for all t > 0 and for which the following
limit exists, is finite and
t
= ` > 0,
(3.7)
lim
t→∞ λ(t)
P∞
we consider the power series h(z) = n=1 an z n with nonnegative coefficients an ≥ 0
for all n ≥ 1, a1 > 0 and convergent on the open disk D(0, R) with R > 0 or R = ∞.
Let λr : [0, ∞) → [0, ∞) be given by
·
¸
h(r)
λr (t) := ln
.
h(r exp(−t))
We know that λr is differentiable on (0, ∞) and
λ0r (t) =
for t ∈ (0, ∞), where h0 (z) =
r exp(−t)h0 (r exp(−t))
h(r exp(−t))
P∞
n=1
lim
t→∞
nan z n−1 . By l’Hospital’s rule we have
t
1
= lim
.
λr (t) t→∞ λ0r (t)
Since for the power series h(z) = a1 z + a2 z 2 + a3 z 3 + · · · we have h0 (z) = a1 +
2a2 z + 3a3 z 2 + · · · , then
r exp(−t)(a1 + 2a2 r exp(−t) + 3a3 (r exp(−t))2 + · · · )
r exp(−t)(a1 + a2 r exp(−t) + a3 (r exp(−t))2 + · · · )
a1 + 2a2 r exp(−t) + 3a3 (r exp(−t))2 + · · ·
=
, t ∈ (0, ∞).
a1 + a2 r exp(−t) + a3 (r exp(−t))2 + · · ·
λ0r (t) =
Therefore limt→∞ λ0r (t) = 1 and limt→∞
t
λr (t)
= 1.
Corollary 9. Let u, w: [a, b] → [m, M ] be Riemann integrable functions on
Rb
[a, b] and w(t) ≥ 0 for any t ∈ [a, b] with a w(t) dt > 0. Consider the power series
P∞
h(z) = n=1 an z n with nonnegative coefficients an ≥ 0 for all n ≥ 1, a1 > 0 and
convergent on the open disk D(0, R) with R > 0 or R = ∞. Let r ∈ (0, R) and
assume that the function f : [m, M ] → [0, ∞) is λr -convex and Riemann integrable
on the interval [m, M ] with
·
¸
h(r)
λr (t) := ln
.
h(r exp(−t))
Then we have the inequality
µ
¶
¸
Z b
Z b ·
1
1
h(r)
f Rb
w(t)u(t) dt ≤ R b
ln
f (u(t)) dt.
h(r exp(−w(t)))
w(t) dt a
w(t) dt a
a
a
(3.8)
Integral inequalities of Jensen type
55
The proof follows by Theorem 5 observing that ` = 1.
Remark 4. With the assumptions of Corollary 9 for u, h and f we have
µ
¶
·
¸
Z b
Z b
1
h(r)
1
f
u(t) dt ≤ ln
f (u(t)) dt.
(3.9)
b−a a
h(re−1 ) b − a a
In particular, for [a, b] = [m, M ] we have the Hermite-Hadamard inequality
¶
·
¸
µ
Z b
h(r)
1
a+b
f
≤ ln
f (t) dt.
(3.10)
2
h(re−1 ) b − a a
4. Interval dependency
Let u: [a, b] → [m, M ] be a Riemann integrable function on [a, b]. Let
λ: [0, ∞) → [0, ∞) be a function with the property that λ(t) > 0 for all t > 0
and the function f : [m, M ] → [0, ∞) be λ-convex and Riemann integrable on the
interval [m, M ]. Assume also that the following limit exists
lim
t→0+
λ(t)
= k ∈ (0, ∞).
t
By Theorem 3 we have that
µ
¶
Z b
Z b
1
1
∆(f, u, λ; [a, b]) :=
f (u(t)) dt − λ(b − a)f
u(t) dt ≥ 0.
k
b−a a
a
(4.1)
Theorem 6. With the above assumptions for u, λ, f and k we have:
(i) For any c ∈ (a, b) we have
∆(f, u, λ; [a, b]) ≥ ∆(f, u, λ; [a, c]) + ∆(f, u, λ; [c, b]) ≥ 0,
(4.2)
i.e., ∆(f, u, λ; ·) is a superadditive function of intervals.
(ii) For any c, d ∈ (a, b) with c < d we have
∆(f, u, λ; [a, b]) ≥ ∆(f, u, λ; [c, d]) ≥ 0,
i.e., ∆(f, u, λ; ·) is a monotonic nondecreasing function of intervals.
Proof. (i) By the λ-convexity of f we have for c ∈ (a, b) that
µ
¶
Z b
1
f
u(t) dt
b−a a
µ
µ
¶
µ
¶¶
Z c
Z b
1
b−c
1
c−a
u(t) dt +
u(t) dt
=f
b−a c−a a
b−a b−c c
Rc
Rb
1
1
λ(c − a)f ( c−a a u(t) dt) + λ(b − c)f ( b−c
u(t) dt)
c
≤
.
λ(b − a)
(4.3)
56
S. S. Dragomir
Therefore
∆(f, u, λ; [a, b])
Z c
Z
=
f (u(t)) dt +
a
Z
c
Z
c
≥
f (u(t)) dt +
"
λ(c −
µ
¶
Z b
1
1
f (u(t)) dt − λ(b − a)f
u(t) dt
k
b−a a
b
f (u(t)) dt −
c
a
×
b
1
a)f ( c−a
Rc
a
1
λ(b − a)
k
1
u(t) dt) + λ(b − c)f ( b−c
Rb
c
u(t) dt)
#
λ(b − a)
= ∆(f, u, λ; [a, c]) + ∆(f, u, λ; [c, b])
and the inequality (4.2) is proved.
(ii) Obvious by the property (4.2).
Remark 5. If [a, b] = [m, M ] and u(t) = t, t ∈ [a, b] then the functional
µ
¶
Z b
1
a+b
δ(f, λ; [a, b]) :=
f (t) dt − λ(b − a)f
≥0
k
2
a
is a superadditive and monotonic nondecreasing function of intervals.
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(received 01.05.2015; in revised form 19.10.2015; available online 15.11.2015)
Mathematics, College of Engineering & Science, Victoria University, PO Box 14428, Melbourne
City, MC 8001, Australia
School of Computational & Applied Mathematics, University of the Witwatersrand, Private Bag
3, Johannesburg 2050, South Africa
E-mail: sever.dragomir@vu.edu.au