Volume 9 (2008), Issue 3, Article 62, 13 pp.
A VARIANT OF JESSEN’S INEQUALITY OF MERCER’S TYPE FOR
SUPERQUADRATIC FUNCTIONS
S. ABRAMOVICH, J. BARIĆ, AND J. PEČARIĆ
D EPARTMENT OF M ATHEMATICS
U NIVERSITY OF H AIFA
H AIFA , 31905, I SRAEL
abramos@math.haifa.ac.il
FAC .
OF
E LECTRICAL E NGINEERING , M ECHANICAL E NGINEERING AND NAVAL A RCHITECTURE
U NIVERSITY OF S PLIT, RUDJERA B OŠKOVI ĆA BB
21000 S PLIT, C ROATIA
jbaric@fesb.hr
FACULTY OF T EXTILE T ECHNOLOGY
U NIVERSITY OF Z AGREB
P IEROTTIJEVA 6, 10000 Z AGREB
C ROATIA
pecaric@hazu.hr
Received 14 December, 2007; accepted 03 April, 2008
Communicated by I. Gavrea
A BSTRACT. A variant of Jessen’s inequality for superquadratic functions is proved. This is a
refinement of a variant of Jessen’s inequality of Mercer’s type for convex functions. The result is
used to refine some comparison inequalities of Mercer’s type between functional power means
and between functional quasi-arithmetic means.
Key words and phrases: Isotonic linear functionals, Jessen’s inequality, Superquadratic functions, Functional quasi-arithmetic
and power means of Mercer’s type.
2000 Mathematics Subject Classification. 26D15, 39B62.
1. I NTRODUCTION
Let E be a nonempty set and L be a linear class of real valued functions f : E → R having
the properties:
L1: f, g ∈ L ⇒ (αf + βg) ∈ L for all α, β ∈ R;
L2: 1 ∈ L, i.e., if f (t) = 1 for t ∈ E, then f ∈ L.
An isotonic linear functional is a functional A : L → R having the properties:
A1: A (αf + βg) = αA(f ) + βA(g) for f, g ∈ L, α, β ∈ R (A is linear);
A2: f ∈ L, f (t) ≥ 0 on E ⇒ A(f ) ≥ 0 (A is isotonic).
The following result is Jessen’s generalization of the well known Jensen’s inequality for
convex functions [10] (see also [12, p. 47]):
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S. A BRAMOVICH , J. BARI Ć ,
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Theorem A. Let L satisfy properties L1, L2 on a nonempty set E, and let ϕ be a continuous
convex function on an interval I ⊂ R. If A is an isotonic linear functional on L with A(1) = 1,
then for all g ∈ L such that ϕ (g) ∈ L, we have A(g) ∈ I and
ϕ(A (g)) ≤ A(ϕ (g)).
Similar to Jensen’s inequality, Jessen’s inequality has a converse [7] (see also [12, p. 98]):
Theorem B. Let L satisfy properties L1, L2 on a nonempty set E, and let ϕ be a convex function
on an interval I = [m, M ] , −∞ < m < M < ∞. If A is an isotonic linear functional on L
with A(1) = 1, then for all g ∈ L such that ϕ (g) ∈ L so that m ≤ g(t) ≤ M for all t ∈ E, we
have
M − A (g)
A (g) − m
A(ϕ (g)) ≤
· ϕ(m) +
· ϕ(M ).
M −m
M −m
Inspired by I.Gavrea’s [9] result, which is a generalization of Mercer’s variant of Jensen’s
inequality [11], recently, W.S. Cheung, A. Matković and J. Pečarić, [8] gave the following
extension on a linear class L satisfying properties L1, L2.
Theorem C. Let L satisfy properties L1, L2 on a nonempty set E, and let ϕ be a continuous
convex function on an interval I = [m, M ] , −∞ < m < M < ∞. If A is an isotonic linear
functional on L with A(1) = 1, then for all g ∈ L such that ϕ (g) , ϕ (m + M − g) ∈ L so that
m ≤ g(t) ≤ M for all t ∈ E, we have the following variant of Jessen’s inequality
(1.1)
ϕ (m + M − A (g)) ≤ ϕ (m) + ϕ (M ) − A (ϕ (g)) .
In fact, to be more specific we have the following series of inequalities
ϕ (m + M − A (g))
≤ A (ϕ (m + M − g))
(1.2)
A (g) − m
M − A (g)
· ϕ(M ) +
· ϕ(m)
M −m
M −m
≤ ϕ (m) + ϕ (M ) − A (ϕ (g)) .
≤
If the function ϕ is concave, inequalities (1.1) and (1.2) are reversed.
In this paper we give an analogous result for superquadratic function (see also different analogous results in [6]). We start with the following definition.
Definition A ([1, Definition 2.1]). A function ϕ : [0, ∞) → R is superquadratic provided that
for all x ≥ 0 there exists a constant C(x) ∈ R such that
(1.3)
ϕ (y) − ϕ (x) − ϕ (|y − x|) ≥ C (x) (y − x)
for all y ≥ 0. We say that f is subquadratic if −f is a superquadratic function.
For example, the function ϕ(x) = xp is superquadratic for p ≥ 2 and subquadratic for
p ∈ (0, 2].
Theorem D ([1, Theorem 2.3]). The inequality
Z
Z Z
gdµ ≤
f
f (g (s)) − f g (s) − gdµ
dµ (s)
holds for all probability measures µ and all non-negative µ−integrable functions g, if and only
if f is superquadratic.
The following discrete version that follows from the above theorem is also used in the sequel.
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 62, 13 pp.
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J ESSEN ’ S I NEQUALITY OF M ERCER ’ S T YPE AND S UPERQUADRACITY
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Lemma A. Suppose
Pnthat f is superquadratic. Let xr ≥ 0, 1 ≤ r ≤ n and let x̄ =
where λr ≥ 0 and r=1 λr = 1. Then
n
X
λr f (xr ) ≥ f (x̄) +
r=1
n
X
Pn
r=1
λr x r
λr f (|xr − x̄|).
r=1
In [3] and [4] the following converse of Jensen’s inequality for superquadratic functions was
proved.
Theorem E. Let(Ω, A, µ) be a measurable space with 0 < µ(r) < ∞ and let f : [0, ∞) → R
be a superquadratic function. If g : Ω → [m, M ] ≤ [0, ∞) is such that g, f ◦ g ∈ L1 (µ), then
we have
Z
1
M − ḡ
ḡ − m
f (g)dµ ≤
f (m) +
f (M )
µ(Ω) Ω
M −m
M −m
Z
1
1
−
((M − g) f (g − m) + (g − m) f (M − g)) dµ,
µ(Ω) M − m Ω
R
1
for ḡ = µ(Ω)
gdµ.
Ω
The discrete version of this theorem is:
Theorem F. Let f : [0, ∞) → R be a superquadratic function. Let (x1 , . . . , xn ) be an n−tuple
n
in
Pn[m, M ] (0 ≤ m < M <1 ∞),
Pn and (p1 , . . . , pn ) be a non-negative n−tuple such that Pn =
p
>
0.
Denote
x̄
=
i=1 i
i=1 pi xi , then
Pn
n
1 X
M − x̄
x̄ − m
pi f (xi ) ≤
f (m) +
f (M )
Pn i=1
M −m
M −m
n
X
1
−
pi [(M − xi ) f (xi − m) + (xi − m) f (M − xi )] .
Pn (M − m) i=1
In Section 2 we give the main result of our paper which is an analogue of Theorem C for
superquadratic functions. In Section 3 we use that result to derive some refinements of the inequalities obtained in [8] which involve functional power means of Mercer’s type and functional
quasi-arithmetic means of Mercer’s type.
2. M AIN R ESULTS
Theorem 2.1. Let L satisfy properties L1, L2, on a nonempty set E, ϕ : [0, ∞) → R be a
continuous superquadratic function, and 0 ≤ m < M < ∞. Assume that A is an isotonic
linear functional on L with A(1) = 1. If g ∈ L is such that m ≤ g(t) ≤ M , for all t ∈ E, and
such that ϕ(g), ϕ(m + M − g), (M − g)ϕ(g − m), (g − m)ϕ(M − g) ∈ L, then we have
ϕ(m + M − A(g))
≤
A(g) − m
M − A(g)
ϕ(m) +
ϕ(M )
M −m
M −m
1
−
[(A(g) − m)ϕ(M − A(g)) + (M − A(g))ϕ(A(g) − m)]
M −m
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 62, 13 pp.
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S. A BRAMOVICH , J. BARI Ć ,
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J. P E ČARI Ć
≤ ϕ(m) + ϕ(M ) − A (ϕ(g))
1
−
A ((g − m)ϕ(M − g) + (M − g)ϕ(g − m))
M −m
1
−
[(A(g) − m)ϕ(M − A(g)) + (M − A(g))ϕ(A(g) − m)] .
M −m
If the function ϕ is subquadratic, then all the inequalities above are reversed.
(2.1)
Proof. From Lemma A for n = 2, as well as from Theorem F, we get that for 0 ≤ m ≤ t ≤ M ,
M −t
t−m
M −t
t−m
ϕ(m) +
ϕ(M ) −
ϕ(t − m) −
ϕ(M − t).
M −m
M −m
M −m
M −m
Replacing t with M + m − t in (2.2) it follows that
(2.2)
ϕ(t) ≤
t−m
M −t
ϕ(m) +
ϕ(M )
M −m
M −m
t−m
M −t
−
ϕ(M − t) −
ϕ(t − m)
M − m
M −m
t−m
M −t
= ϕ(m) + ϕ(M ) −
ϕ(M ) +
ϕ(m)
M −m
M −m
t−m
M −t
−
ϕ(M − t) −
ϕ(t − m).
M −m
M −m
ϕ(M + m − t) ≤
Since m ≤ g(t) ≤ M for all t ∈ E, it follows that m ≤ A(g) ≤ M and we have
A(g) − m
M − A(g)
(2.3) ϕ(m + M − A(g)) ≤ ϕ(m) + ϕ(M ) −
ϕ(M ) +
ϕ(m)
M −m
M −m
A(g) − m
M − A(g)
−
ϕ(M − A(g)) −
ϕ(A(g) − m).
M −m
M −m
On the other hand, since m ≤ g(t) ≤ M for all t ∈ E it follows that
M − g(t)
g(t) − m
ϕ(m) +
ϕ(M )
M −m
M −m
M − g(t)
g(t) − m
−
ϕ(g(t) − m) −
ϕ(M − g(t)).
M −m
M −m
Using functional calculus we have
ϕ(g(t)) ≤
M − A(g)
A(g) − m
1
ϕ(m) +
ϕ(M ) −
A ((M − g(t)ϕ(g(t) − m))
M −m
M −m
M −m
1
−
A ((g(t) − m)ϕ(M − g(t))) .
M −m
Using inequalities (2.3) and (2.4), we obtain the desired inequality (2.1).
The last statement follows immediately from the fact that if ϕ is subquadratic then −ϕ is a
superquadratic function.
(2.4) A(ϕ(g)) ≤
Remark 1. If a function ϕ is superquadratic and nonnegative, then it is convex [1, Lema 2.2].
Hence, in this case inequality (2.1) is a refinement of inequality (1.1).
On the other hand, we can get one more inequality in (2.1) if we use a result of S. Banić and
S. Varosănec [5] on Jessen’s inequality for superquadratic functions:
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J ESSEN ’ S I NEQUALITY OF M ERCER ’ S T YPE AND S UPERQUADRACITY
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Theorem 2.2 ([5, Theorem 8, Remark 1]). Let L satisfy properties L1, L2, on a nonempty set
E, and let ϕ : [0, ∞) → R be a continuous superquadratic function. Assume that A is an
isotonic linear functional on L with A(1) = 1. If f ∈ L is nonnegative and such that ϕ(f ),
ϕ(|f − A(f )|) ∈ L, then we have
ϕ(A(f )) ≤ A(ϕ(f )) − A(ϕ(|f − A(f )|)).
(2.5)
If the function ϕ is subquadratic, then the inequality above is reversed.
Using Theorem 2.2 and some basic properties of superquadratic functions we prove the next
theorem.
Theorem 2.3. Let L satisfy properties L1, L2, on a nonempty set E, and let ϕ : [0, ∞) → R be
a continuous superquadratic function, and let 0 ≤ m < M < ∞. Assume that A is an isotonic
linear functional on L with A(1) = 1. If g ∈ L is such that m ≤ g(t) ≤ M , for all t ∈ E, and
such that ϕ(g), ϕ(m + M − g), (M − g)ϕ(g − m), (g − m)ϕ(M − g), ϕ (|g − A (g)|) ∈ L,
then we have
ϕ(m + M − A(g))
(2.6)
(2.7)
(2.8)
≤ A(ϕ(m + M − g)) − A(ϕ(|g − A(g)|))
M − A(g)
A(g) − m
ϕ(m) +
ϕ(M )
M −m
M −m
1
−
A ((g − m)ϕ(M − g) + (M − g)ϕ(g − m)) − A(ϕ(|g − A(g)|))
M −m
≤ ϕ(m) + ϕ(M ) − A (ϕ(g))
2
−
A ((g − m)ϕ(M − g) + (M − g)ϕ(g − m)) − A(ϕ(|g − A(g)|)).
M −m
≤
If the function ϕ is subquadratic, then all the inequalities above are reversed.
Proof. Notice that (m + M − g) ∈ L. Since m ≤ g(t) ≤ M for all t ∈ E, it follows that
m ≤ m + M − g(t) ≤ M for all t ∈ E. Applying (2.5) to the function f = m + M − g we get
ϕ(A(m + M − g))
= ϕ(m + M − A(g))
≤ A(ϕ(m + M − g)) − A(ϕ(|m + M − g − A(m + M − g)|))
= A(ϕ(m + M − g)) − A(ϕ(|m + M − g − m − M + A(g)|))
= A(ϕ(m + M − g)) − A(ϕ(|g − A(g)|)),
which is the inequality (2.6).
From the discrete Jensen’s inequality for superquadratic functions we get for all m ≤ x ≤ M ,
(2.9)
ϕ(x) ≤
M −x
x−m
M −x
x−m
ϕ(m) +
ϕ(M ) −
ϕ(x − m) −
ϕ(M − x).
M −m
M −m
M −m
M −m
Replacing x in (2.9) with m + M − g(t) ∈ [m, M ] for all t ∈ E, we have
ϕ(m + M − g(t)) ≤
g(t) − m
M − g(t)
ϕ(m) +
ϕ(M )
M −m
M −m
g(t) − m
M − g(t)
−
ϕ(M − g(t)) −
ϕ(g(t) − m).
M −m
M −m
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 62, 13 pp.
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S. A BRAMOVICH , J. BARI Ć ,
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Since A is linear, isotonic and satisfies A (1) = 1, from the above inequality it follows that
(2.10) A(ϕ(m + M − g)) ≤
A(g) − m
M − A(g)
ϕ(m) +
ϕ(M )
M −m
M −m
1
A ((g − m)ϕ(M − g) + (M − g)ϕ(g − m)) .
−
M −m
Adding −A(ϕ(|g − A(g)|)) on both sides of (2.10) we get
(2.11) A(ϕ(m + M − g)) − A(ϕ(|g − A(g)|)) ≤
−
M − A(g)
A(g) − m
ϕ(m) +
ϕ(M )
M −m
M −m
1
A ((g − m)ϕ(M − g) + (M − g)ϕ(g − m)) − A(ϕ(|g − A(g)|)),
M −m
which is the inequality (2.7).
The right hand side of (2.11) can be written as follows
(2.12) ϕ(m) + ϕ(M ) −
−
M − A(g)
A(g) − m
ϕ(m) −
ϕ(M )
M −m
M −m
1
A ((g − m)ϕ(M − g) + (M − g)ϕ(g − m)) − A(ϕ(|g − A(g)|)).
M −m
On the other hand, replacing x, in (2.9), with g(t) ∈ [m, M ], for all t ∈ E, we get
ϕ(g(t)) ≤
(2.13)
g(t) − m
M − g(t)
ϕ(m) +
ϕ(M )
M −m
M −m
g(t) − m
M − g(t)
−
ϕ(g(t) − m) −
ϕ(M − g(t)).
M −m
M −m
Applying the functional A on (2.13) we have
A(ϕ(g)) ≤
(2.14)
M − A(g)
A(g) − m
ϕ(m) +
ϕ(M )
M −m
M −m
1
−
A ((M − g)ϕ(g − m) + (g − m)ϕ(M − g)) ,
M −m
The inequality (2.14) can be written as follows
−
M − A(g)
A(g) − m
ϕ(m) −
ϕ(M )
M −m
M −m
1
≤ −A(ϕ(g)) −
A ((g − m)ϕ(M − g) + (M − g)ϕ(g − m)) .
M −m
Using (2.12) we get
A(g) − m
M − A(g)
ϕ(m) +
ϕ(M )
M −m
M −m
1
−
A ((g − m)ϕ(M − g) + (M − g)ϕ(g − m)) − A(ϕ(|g − A(g)|))
M −m
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J ESSEN ’ S I NEQUALITY OF M ERCER ’ S T YPE AND S UPERQUADRACITY
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≤ ϕ(m) + ϕ(M ) − A(ϕ(g))
1
−
A ((g − m)ϕ(M − g) + (M − g)ϕ(g − m))
M −m
1
−
A ((g − m)ϕ(M − g) + (M − g)ϕ(g − m)) − A(ϕ(|g − A(g)|))
M −m
= ϕ(m) + ϕ(M ) − A(ϕ(g))
2
−
A ((g − m)ϕ(M − g) + (M − g)ϕ(g − m)) − A(ϕ(|g − A(g)|)).
M −m
Now, it follows that
M − A(g)
A(g) − m
ϕ(m) +
ϕ(M )
M −m
M −m
1
−
A ((g − m)ϕ(M − g) + (M − g)ϕ(g − m)) − A(ϕ(|g − A(g)|))
M −m
≤ ϕ(m) + ϕ(M ) − A(ϕ(g))
2
A ((g − m)ϕ(M − g) + (M − g)ϕ(g − m)) − A(ϕ(|g − A(g)|)),
−
M −m
which is the inequality (2.8).
3. A PPLICATIONS
Throughout this section we suppose that:
(i) L is a linear class having properties L1, L2 on a nonempty set E.
(ii) A is an isotonic linear functional on L such that A(1) = 1.
(iii) g ∈ L is a function of E to [m, M ] (0 < m < M < ∞) such that all of the following
expressions are well defined.
Let ψ be a continuous and strictly monotonic function on an interval I = [m, M ], (0 < m <
M < ∞).
For any r ∈ R, a power mean of Mercer’s type functional

1
r
r
r r

 [m + M − A(g )] , r 6= 0
Q(r, g) :=
mM


,
r = 0,
exp (A(log g))
and a quasi-arithmetic mean functional of Mercer’s type
fψ (g, A) = ψ −1 (ψ(m) + ψ(M ) − A(ψ(g)))
M
are defined in [8] and the following theorems are proved.
Theorem G. If r, s ∈ R and r ≤ s, then
Q(r, g) ≤ Q(s, g).
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 62, 13 pp.
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S. A BRAMOVICH , J. BARI Ć ,
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Theorem H.
(i) If either χ ◦ ψ −1 is convex and χ is strictly increasing, or χ ◦ ψ −1 is concave and χ is
strictly decreasing, then
fψ (g, A) ≤ M
fχ (g, A) .
M
(3.1)
(ii) If either χ ◦ ψ −1 is concave and χ is strictly increasing, or χ ◦ ψ −1 is convex and χ is
strictly decreasing, then the inequality (3.1) is reversed.
Applying the inequality (2.1) to the adequate superquadratic functions we shall give some
refinements of the inequalities in Theorems G and H. To do this, we will define following
functions.
1
r
r
r
r rs
♦(m, M, r, s, g, A) = r
A
(M
−
g
)(g
−
m
)
M − mr
1
r
r
r
r rs
A
(g
−
m
)(M
−
g
)
+ r
M − mr
s
1
+ r
(A(g r ) − mr ) (M r − A(g r )) r
r
M −m
s
1
r
r
r
r r
+ r
(M
−
A(g
))
(A(g
)
−
m
)
.
M − mr
and
♦(m, M, ψ, χ, g, A)
1
A (ψ(M ) − ψ(g))χ ψ −1 (ψ(g) − ψ(m)))
=
ψ(M ) − ψ(m)
1
+
A (ψ(g) − ψ(m))χ ψ −1 (ψ(M ) − ψ(g))
ψ(M ) − ψ(m)
1
+
(A(ψ(g)) − ψ(m)) χ ψ −1 (ψ(M ) − A(ψ(g)))
ψ(M ) − ψ(m)
1
+
(ψ(M ) − A(ψ(g))) χ ψ −1 (A(ψ(g)) − ψ(m)) .
ψ(M ) − ψ(m)
Now, the following theorems are valid.
Theorem 3.1. Let r, s ∈ R.
(i) If 0 < 2r ≤ s, then
1
(3.2)
Q(r, g) ≤ [(Q(s, g))s − ♦(m, M, r, s, g, A)] s .
(ii) If 2r ≤ s < 0, then for (Q (s, g))s − ♦ (M, m, r, s, g, A) > 0
1
(3.3)
Q(r, g) ≤ [(Q(s, g))s − ♦(M, m, r, s, g, A)] s ,
where we used ♦(M, m, r, s, g, A) to denote the new function derived from the function
♦(m, M, r, s, g, A) by changing the places of m and M .
(iii) If 0 < s ≤ 2r, then for (Q (s, g))s − ♦ (M, m, r, s, g, A) > 0 the reverse inequality (3.2)
holds.
(iv) If s ≤ 2r < 0, then the reversed inequality (3.3) holds.
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 62, 13 pp.
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J ESSEN ’ S I NEQUALITY OF M ERCER ’ S T YPE AND S UPERQUADRACITY
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Proof.
(i) It is given that
0 < m ≤ g ≤ M < ∞.
Since 0 < 2r ≤ s, it follows that
0 < mr ≤ g r ≤ M r < ∞.
Applying Theorem 2.1, or more precisely inequality (2.1) to the superquadratic function
s
ϕ(t) = t r (note that rs ≥ 2 here) and replacing g, m and M with g r , mr and M r ,
respectively, we have
s
[mr + M r − A(g r )] r
s
1
(A(g r ) − mr ) (M r − A(g r )) r
+ r
r
M −m
s
1
+ r
(M r − A(g r )) (A(g r ) − mr ) r
r
M −m
s
≤ m + M s − A(g s )
s
1
− r
A (M r − g r )(g r − mr ) r
r
M −m
1
r
r
r
r rs
A
(g
−
m
)(M
−
g
)
− r
.
M − mr
i.e.
[Q(r, g)]s ≤ [Q(s, g)]s − ♦(m, M, r, s, g, A).
(3.4)
Raising both sides of (3.4) to the power 1s > 0, we get desired inequality (3.2).
(ii) In this case we have
0 < M r ≤ g r ≤ mr < ∞.
Applying Theorem 2.1 or, more precisely, the reversed inequality (2.1) to the subs
quadratic function ϕ(t) = t r (note that now we have 0 < rs ≤ 2) and replacing g,
m and M with g r , mr and M r , respectively, we get
s
[M r + mr − A(g r )] r
s
1
r
r
r
r r
+ r
(A(g
)
−
M
)
(m
−
A(g
))
m − Mr
s
1
+ r
(mr − A(g r )) (A(g r ) − M r ) r
r
m −M
s
≥ M + ms − A(g s )
1
r
r
r
r rs
− r
A
(m
−
g
)(g
−
M
)
m − Mr
s
1
− r
A (g r − M r )(mr − g r ) r .
r
m −M
Since 2r ≤ s < 0, raising both sides to the power 1s , it follows that
1
1
[M r + mr − A(g r )] r ≤ [M s + ms − A(g s ) − ♦(M, m, r, s, g, A)] s ,
or
1
Q(r, g) ≤ [(Q(s, g))s − ♦(M, m, r, s, g, A)] s .
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 62, 13 pp.
http://jipam.vu.edu.au/
10
S. A BRAMOVICH , J. BARI Ć ,
AND
J. P E ČARI Ć
(iii) In this case we have 0 < rs ≤ 2. Since 0 < mr ≤ g r ≤ M r < ∞, we can apply Theorem
2.1, or more precisely, the reversed inequality (2.1) to the subquadratic function ϕ(t) =
s
t r . Replacing g, m and M with g r , mr and M r , respectively, it follows that
s
[mr + M r − A(g r )] r
s
1
r
r
r
r r
+ r
(A(g
)
−
m
)
(M
−
A(g
))
M − mr
s
1
(M r − A(g r )) (A(g r ) − mr ) r
+ r
r
M −m
s
≥ m + M s − A(g s )
1
r
r
r
r rs
− r
A
(M
−
g
)(g
−
m
)
M − mr
s
1
− r
A (g r − mr )(M r − g r ) r ,
r
M −m
i.e.
(3.5)
[Q(r, g)]s ≥ [Q(s, g)]s − ♦(m, M, r, s, g, A).
Raising both sides of (3.5) to the power
1
s
> 0 we get
1
Q(r, g) ≥ [(Q(s, g))s − ♦(m, M, r, s, g, A)] s .
(iv) Since r < 0, from 0 < m ≤ g ≤ M < ∞ it follows that 0 < M r ≤ g r ≤ mr < ∞.
s
Now, we are applying Theorem 2.1 to the superquadratic function ϕ(t) = t r , because
s
≥ 2 here, and analogous to the previous theorem we get
r
[Q(r, g)]s ≤ [Q(s, g)]s − ♦(M, m, r, s, g, A).
Raising both sides to the power
1
s
< 0 it follows that
1
Q(r, g) ≥ [(Q(s, g))s − ♦(M, m, r, s, g, A)] s .
Theorem 3.2. Let r, s ∈ R.
(i) If 0 < 2s ≤ r, then
(3.6)
1
Q(r, g) ≥ [(Q(s, g))r + ♦(m, M, s, r, g, A)] r ,
where we used ♦(m, M, s, r, g, A) to denote the new function derived from the function
♦(m, M, r, s, g, A) by changing the places of r and s.
(ii) If 2s ≤ r < 0, then
(3.7)
1
Q(r, g) ≤ [(Q(s, g))r + ♦(M, m, s, r, g, A)] r .
(iii) If 0 < r ≤ 2s, then the reversed inequality (3.6) holds.
(iv) If r ≤ 2s < 0, then the reversed inequality (3.7) holds.
Proof.
r
(i) Applying inequality (2.1) to the superquadratic function ϕ(t) = t s (note that rs ≥ 2
here) and replacing g, m and M with g s , ms and M s , (0 < ms ≤ g s ≤ M s < ∞)
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 62, 13 pp.
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J ESSEN ’ S I NEQUALITY
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M ERCER ’ S T YPE AND S UPERQUADRACITY
11
respectively, we have
r
[ms + M s − A(g s )] s
r
1
s
s
s
s s
+ s
(A(g
)
−
m
)
(M
−
A(g
))
M − ms
r
1
(M s − A(g s )) (A(g s ) − ms ) s
+ s
s
M −m
r
≥ m + M r − A(g r )
1
s
s
s
s rs
− s
A
(M
−
g
)(g
−
m
)
M − ms
r
1
− s
A (g s − ms )(M s − g s ) s ,
s
M −m
i.e.
[Q(s, g)]r ≤ [Q(r, g)]r − ♦(m, M, s, r, g, A).
Raising both sides to the power 1r > 0, the inequality (3.6) follows.
(ii) Since s < 0, we have 0 < M s ≤ g s ≤ ms < ∞ so the function ♦ will be of the form
♦(M, m, s, r, g, A). Since 0 < rs ≤ 2, we will apply Theorem 2.1 to the subquadratic
r
function ϕ(t) = t s and, as in previous case, it follows that
[Q(s, g)]r + ♦(M, m, s, r, g, A) ≥ [Q(r, g)]r .
Raising both sides to the power 1r < 0, the inequality (3.7) follows.
r
(iii) Since 0 < rs ≤ 2, we will apply Theorem 2.1 to the subquadratic function ϕ(t) = t s
and then it follows that
[Q(s, g)]r + ♦(m, M, s, r, g, A) ≥ [Q(r, g)]r .
Raising both sides to the power
1
r
> 0, we get
1
Q(r, g) ≤ [(Q(s, g))r + ♦(m, M, s, r, g, A)] r .
r
(iv) Since rs ≥ 2, we will apply Theorem 2.1 to the superquadratic function ϕ(t) = t s and
use the function ♦(M, m, s, r, g, A) instead of ♦(m, M, s, r, g, A). Then we get
[Q(s, g)]r + ♦(M, m, s, r, g, A) ≤ [Q(r, g)]r .
Raising both sides to the power
1
r
< 0, it follows that
1
Q(r, g) ≥ [(Q(s, g))r + ♦(M, m, s, r, g, A)] r .
Remark 2. Notice that some cases in the last theorems have common parts. In some of them
we can establish double inequalities. For example, if 0 < r ≤ 2s and 0 < s ≤ 2r, then for
(Q (s, g))s − ♦ (M, m, r, s, g, A) > 0
1
1
[(Q (s, g))r + ♦ (m, M, s, r, g, A)] r ≥ Q (r, g) ≥ [(Q (s, g))s − ♦ (m, M, r, s, g, A)] s .
Theorem 3.3. Let ψ ∈ C([m, M ]) be strictly increasing and let χ ∈ C([m, M ]) be strictly
monotonic functions.
(i) If either χ◦ψ −1 is superquadratic and χ is strictly increasing, or χ◦ψ −1 is subquadratic
and χ is strictly decreasing, then
−1
f
f
(3.8)
Mψ (g, A) ≤ χ
χ Mχ (g, A) − ♦(m, M, ψ, χ, g, A) ,
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 62, 13 pp.
http://jipam.vu.edu.au/
12
S. A BRAMOVICH , J. BARI Ć ,
AND
J. P E ČARI Ć
(ii) If either χ◦ψ −1 is subquadratic and χ is strictly increasing or χ◦ψ −1 is superquadratic
and χ is strictly decreasing, then the inequality (3.8) is reversed.
Proof. Suppose that χ ◦ ψ −1 is superquadratic. Letting ϕ = χ ◦ ψ −1 in Theorem 2.1 and
replacing g, m and M with ψ(g), ψ(m) and ψ(M ) respectively, we have
χ ψ −1 (ψ(m) + ψ(M ) − A(ψ(g)))
1
+
(A(ψ(g)) − ψ(m)) χ ψ −1 (ψ(M ) − A(ψ(g)))
ψ(M ) − ψ(m)
1
+
(ψ(M ) − A(ψ(g))) χ ψ −1 (A(ψ(g)) − ψ(m))
ψ(M ) − ψ(m)
≤ χ(m) + χ(M ) − A(χ(g))
1
A (ψ(M ) − ψ(m)) χ ψ −1 (ψ(g) − ψ(m))
−
ψ(M ) − ψ(m)
1
−
A (ψ(g) − ψ(m)) χ ψ −1 (ψ(M ) − ψ(g)) ,
ψ(M ) − ψ(m)
i.e.,
fψ (g, A) ≤ χ(m) + χ(M ) − A(χ(g)) − ♦(m, M, ψ, χ, g, A)
χ M
(3.9)
≤ χ ◦ χ−1 (χ(m) + χ(M ) − A(χ(g))) − ♦(m, M, ψ, χ, g, A)
fχ (g, A) − ♦(m, M, ψ, χ, g, A).
≤χ M
If χ is strictly increasing, then the inverse function χ−1 is also strictly increasing and inequality
(3.9) implies the inequality (3.8). If χ is strictly decreasing, then the inverse function χ−1 is
also strictly decreasing and in that case the reverse of (3.9) implies (3.8). Analogously, we get
the reverse of (3.8) in the cases when χ ◦ ψ −1 is superquadratic and χ is strictly decreasing, or
χ ◦ ψ −1 is subquadratic and χ is strictly increasing.
Remark 3. If the function ψ in Theorem 3.3 is strictly decreasing, then the inequality (3.8) and
its reversal also hold under the same assumptions, but with m and M interchanged.
Remark 4. Obviously, Theorem 3.1 and Theorem 3.2 follow from Theorem 3.3 and Remark 3
by choosing ψ(t) = tr and χ(t) = ts , or vice versa.
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