Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 2, Article 70, 2006
A VARIANT OF JENSEN-STEFFENSEN’S INEQUALITY FOR CONVEX AND
SUPERQUADRATIC FUNCTIONS
S. ABRAMOVICH, M. KLARIČIĆ BAKULA, AND S. BANIĆ
D EPARTMENT OF M ATHEMATICS
U NIVERSITY OF H AIFA
H AIFA , 31905, I SRAEL
abramos@math.haifa.ac.il
D EPARTMENT OF M ATHEMATICS
FACULTY OF NATURAL S CIENCES , M ATHEMATICS AND E DUCATION
U NIVERSITY OF S PLIT
T ESLINA 12, 21000 S PLIT
C ROATIA
milica@pmfst.hr
FACULTY OF C IVIL E NGINEERING AND A RCHITECTURE
U NIVERSITY OF S PLIT
M ATICE HRVATSKE 15, 21000 S PLIT
C ROATIA
Senka.Banic@gradst.hr
Received 03 January, 2006; accepted 20 January, 2006
Communicated by C.P. Niculescu
A BSTRACT. A variant of Jensen-Steffensen’s inequality is considered for convex and for superquadratic functions. Consequently, inequalities for power means involving not only positive
weights have been established.
Key words and phrases: Jensen-Steffensen’s inequality, Monotonicity, Superquadraticity, Power means.
2000 Mathematics Subject Classification. 26D15, 26A51.
1. I NTRODUCTION .
Let I be an interval in R and f : I → R a convex function on I. If ξ P
= (ξ1 , . . . , ξm ) is any
m-tuple in I m and p = (p1 , . . . , pm ) any nonnegative m-tuple such that m
i=1 pi > 0, then the
well known Jensen’s inequality (see for example [7, p. 43])
!
m
m
1 X
1 X
(1.1)
f
pi ξi ≤
pi f (ξi )
Pm i=1
Pm i=1
P
holds, where Pm = m
i=1 pi .
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
002-06
2
S. A BRAMOVICH , M. K LARI ČI Ć BAKULA ,
AND
S. BANI Ć
If f is strictly convex, then (1.1) is strict unless ξi = c for all i ∈ {j : pj > 0}.
It is well known that the assumption “p is a nonnegative m-tuple” can be relaxed at the
expense of more restrictions on the m-tuple ξ.
If p is a real m-tuple such that
0 ≤ Pj ≤ Pm , j = 1, . . . , m,
(1.2)
where Pj =
get
Pj
i=1
Pm > 0,
pi , then for any monotonic m-tuple ξ (increasing or decreasing) in I m we
m
1 X
ξ=
pi ξi ∈ I,
Pm i=1
and for any function f convex on I (1.1) still holds. Inequality (1.1) considered under conditions (1.2) is known as the Jensen-Steffensen’s inequality [7, p. 57] for convex functions.
In his paper [5] A. McD. Mercer considered some monotonicity properties of power means.
He proved the following theorem:
Theorem A. Suppose that 0 < a < b and a ≤ x1 ≤ x2 ≤ · · · ≤ xn ≤ b hold withPat least one
of the xk satisfying a < xk < b. If w = (w1 , . . . , wn ) is a positive n-tuple with ni=1 wi = 1
and −∞ < r < s < +∞, then
a < Qr (a, b; x) < Qs (a, b; x) < b ,
where
Qt (a, b; x) ≡
at + b t −
n
X
! 1t
wi xti
i=1
for all real t 6= 0, and
ab
Q0 (a, b; x) ≡ ,
G
G=
n
Y
i
xw
i .
i=1
In his next paper [6], Mercer gave a variant of Jensen’s inequality for which Witkowski
presented in [8] a shorter proof. This is stated in the following theorem:
Theorem B. If f is a convex function on an interval containing an n-tuple x = (xP
1 , . . . , xn )
such that 0 < x1 ≤ x2 ≤ · · · ≤ xn and w = (w1 , . . . , wn ) is a positive n-tuple with ni=1 wi =
1, then
!
n
n
X
X
f x1 + xn −
wi xi ≤ f (x1 ) + f (xn ) −
wi f (xi ) .
i=1
i=1
This theorem is a special case of the following theorem proved in [4] by Abramovich, Klaričić
Bakula, Matić and Pečarić:
Theorem C ([4, Th. 2]). Let f : I → R, where I is an interval in R and let [a, b] ⊆ I, a < b.
Let x = (x1 , . . . , xn ) be a monotonic n-tuple in [a, b]n and v = (v1 , . . . , vn ) a real n−tuple
P
such that vi 6= 0, i = 1, . . . , n, and 0 ≤ Vj ≤ Vn , j = 1, . . . , n, Vn > 0, where Vj = ji=1 vi .
If f is convex on I, then
!
n
n
1 X
1 X
vi xi ≤ f (a) + f (b) −
vi f (xi ) .
(1.3)
f a+b−
Vn i=1
Vn i=1
In case f is strictly convex, the equality holds in (1.3) iff one of the following two cases occurs:
(1) either x = a or x = b,
J. Inequal. Pure and Appl. Math., 7(2) Art. 70, 2006
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J ENSEN -S TEFFENSEN ’ S INEQUALITY - S UPERQUADRATIC FUNCTIONS
3
(2) there exists l ∈ {2, . . . , n − 1} such that x = x1 + xn − xl and


 x1 = a, xn = b or x1 = b, xn = a,
(1.4)
V j (xj−1 − xj ) = 0, j = 2, . . . , l,


Vj (xj − xj+1 ) = 0, j = l, . . . , n − 1,
Pn
P
where V j = i=j vi , j = 1, . . . , n and x = (1/Vn ) ni=1 vi xi .
In the special case where v > 0 and f is strictly convex, the equality holds in (1.4) iff
xi = a, i = 1, . . . , n, or xi = b, i = 1, . . . , n.
Here, as in the rest of the paper, when we say that an n-tuple ξ is increasing (decreasing)
we mean that ξ1 ≤ ξ2 ≤ · · · ≤ ξn (ξ1 ≥ ξ2 ≥ · · · ≥ ξn ). Similarly, when we say that a
function f : I → R is increasing (decreasing) on I we mean that for all u, v ∈ I we have
u < v ⇒ f (u) ≤ f (v) (u < v ⇒ f (u) ≥ f (v)).
In Section 2 we refine Theorems A, B, and C. These refinements are achieved by superquadratic functions which were introduced in [1] and [2].
As Jensen’s inequality for convex functions is a generalization of Hölder’s inequality for
f (x) = xp , p ≥ 1, so the inequalities satisfied by superquadratic functions are generalizations
of the inequalities satisfied by the superquadratic functions f (x) = xp , p ≥ 2 (see [1], [2]).
First we quote some definitions and state a list of basic properties of superquadratic functions.
Definition 1.1. A function f : [0, ∞) → R is superquadratic provided that for all x ≥ 0 there
exists a constant C(x) ∈ R such that
(1.5)
f (y) − f (x) − f (|y − x|) ≥ C (x) (y − x)
for all y ≥ 0.
Definition 1.2. A function f : [0, ∞) → R is said to be strictly superquadratic if (1.5) is strict
for all x 6= y where xy 6= 0.
LemmaP
A ([2, Lemma 2.3]). Suppose that f is superquadratic.
Let ξi ≥ 0, i = 1, . . . , m, and
Pm
let ξ = m
p
ξ
,
where
p
≥
0,
i
=
1,
.
.
.
,
m,
and
p
=
1.
Then
i
i=1 i
i=1 i i
m
m
X
X
pi f (ξi ) − f ξ ≥
pi f ξi − ξ .
i=1
i=1
Lemma B ([1, Lemma 2.2]). Let f be superquadratic function with C(x) as in Definition 1.1.
Then:
(i) f (0) ≤ 0,
(ii) if f (0) = f 0 (0) = 0 then C(x) = f 0 (x) whenever f is differentiable at x > 0,
(iii) if f ≥ 0, then f is convex and f (0) = f 0 (0) = 0.
In [3] the following refinement of Jensen’s Steffensen’s type inequality for nonnegative superquadratic functions was proved:
Theorem D ([3, Th. 1]). Let f : [0, ∞) → [0, ∞) be a differentiable and superquadratic
function, let ξ be a nonnegative monotonic m-tuple in Rm and p a real m-tuple, m ≥ 3,
satisfying
0 ≤ Pj ≤ Pm , j = 1, . . . , m, Pm > 0.
Let ξ be defined as
m
1 X
ξ=
pi ξi .
Pm i=1
J. Inequal. Pure and Appl. Math., 7(2) Art. 70, 2006
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4
S. A BRAMOVICH , M. K LARI ČI Ć BAKULA ,
AND
S. BANI Ć
Then
(1.6)
m
X
pi f (ξi ) − Pm f ξ ≥
i=1
k−1
X
Pi f (ξi+1 − ξi ) + Pk f ξ − ξk
i=1
+ P k+1 f ξk+1 − ξ +
m
X
P i f (ξi − ξi−1 )
i=k+2
!
p
ξ
−
ξ
i
i
≥
Pi +
P i f Pk i=1 Pm
i=1 Pi +
i=k+1 P i
i=1
i=k+1
!
Pm
i=1 pi ξ − ξi
≥ (m − 1) Pm f
,
(m − 1) Pm
"
where P i =
Pm
j=i
k
X
m
X
#
Pm
pj and k ∈ {1, . . . , m − 1} satisfies
ξk ≤ ξ ≤ ξk+1 .
In case f is also strictly superquadratic, inequality
m
X
!
ξi − ξ p
i
i=1
(m − 1) Pm
Pm
pi f (ξi ) − Pm f ξ > (m − 1) Pm f
i=1
holds for ξ > 0 unless one of the following two cases occurs:
(1) either ξ = ξ1 or ξ = ξm ,
(2) there exists k ∈ {3, . . . , m − 2} such that ξ = ξk and
(
Pj (ξj − ξj+1 ) = 0, j = 1, . . . , k − 1
(1.7)
P j (ξj − ξj−1 ) = 0, j = k + 1, . . . , m.
In these two cases
m
X
pi f (ξi ) − Pm f ξ = 0.
i=1
In Section 2 we refine Theorem B and Theorem C for functions which are superquadratic and
positive. One of the refinements is derived easily from Theorem D.
We use in Section 3 the following theorem [7, p. 323] to give an alternative proof of Theorem
B.
Theorem E. Let I be an interval in R, and ξ, η two decreasing m-tuples such that ξ, η ∈ I m .
Let p be a real m-tuple such that
k
X
(1.8)
pi ξi ≤
i=1
k
X
p i ηi
i=1
for k = 1, 2, . . . , m − 1, and
m
X
(1.9)
i=1
pi ξi =
m
X
p i ηi .
i=1
Then for every continuous convex function f : I → R we have
m
m
X
X
(1.10)
pi f (ξi ) ≤
pi f (ηi ) .
i=1
J. Inequal. Pure and Appl. Math., 7(2) Art. 70, 2006
i=1
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J ENSEN -S TEFFENSEN ’ S INEQUALITY - S UPERQUADRATIC FUNCTIONS
5
2. VARIANTS OF J ENSEN -S TEFFENSEN ’ S I NEQUALITY FOR P OSITIVE
S UPERQUADRATIC F UNCTIONS
In this section we refine in two ways Theorem C for functions which are superquadratic and
positive. The refinement in Theorem 2.1 follows by showing that it is a special case of Theorem
D for specific p. The refinement in Theorem 2.2 follows the steps in the proof of Theorem B
given by Witkowski in [8]. Therefore the second refinement is confined only to the specific p
given in Theorem B, which means that what we get is a variant of Jensen’s inequality and not
of the more general Jensen-Steffensen’s inequality.
Theorem 2.1. Let f : [0, ∞) → [0, ∞) and let [a, b] ⊆ [0, ∞) . Let x = (x1 , . . . , xn ) be a
monotonic n−tuple in [a, b]n and v = (v1 , . . . , vn ) a real n-tuple such that vi 6= 0, i = 1, . . . , n,
Pj
0 ≤ Vj ≤ Vn , j = 1, . . . , n, and Vn > 0, where Vj =
i=1 vi . If f is differentiable and
superquadratic, then
!
n
n
1 X
1 X
(2.1) f (a) + f (b) −
vi f (xi ) − f a + b −
vi xi
Vn i=1
Vn i=1


P
P
b − a − V1n ni=1 vi a + b − xi − V1n nj=1 vj xj .
≥ (n + 1) f 
n+1
In case f is also strictly superquadratic and a > 0, inequality (2.1) is strict unless one of the
following two cases occurs:
(1) either x = a or x = b,
(2) there exists l ∈ {2, . . . , n − 1} such that x = x1 + xn − xl and

x1 = a, xn = b or x1 = b, xn = a



(2.2)
V j (xj−1 − xj ) = 0 , j = 2, . . . , l,



Vj (xj − xj+1 ) = 0 , j = l, . . . , n − 1,
P
P
where V j = ni=j vi , j = 1, . . . , n, and x = V1n ni=1 vi xi .
In these two cases we have
n
1 X
f (a) + f (b) −
vi f (xi ) − f
Vn i=1
n
1 X
a+b−
vi xi
Vn i=1
!
= 0.
In the special case where v > 0 and f is also strictly superquadratic, the equality holds in (2.1)
iff xi = a, i = 1, . . . , n, or xi = b, i = 1, . . . , n.
Proof. Suppose that x is an increasing n-tuple in [a, b]n . The proof of the theorem is an immediate result of Theorem D, by defining the (n + 2)-tuples ξ and p as
ξ1 = a,
ξi+1 = xi , i = 1, . . . , n,
ξn+2 = b
p1 = 1,
pi+1 = −vi /Vn , i = 1, . . . , n,
pn+2 = 1.
Then we get (2.1) from the last inequality in (1.6) and from the fact that in our special case
we have
k
n+2
X
X
Pi +
P i ≤ n + 1,
i=1
J. Inequal. Pure and Appl. Math., 7(2) Art. 70, 2006
i=k+1
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6
S. A BRAMOVICH , M. K LARI ČI Ć BAKULA ,
for Pi =
Pi
j=1
Pn+2
pj and P i =
j=i
AND
S. BANI Ć
pj , and
m
n
1 X
1 X
pi ξi = a + b −
vi xi = a + b − x.
ξ=
Pm i=1
Vn i=1
The proof of the equality case and the special case where v > 0 follows also from Theorem
D.
We have
Vj
, j = 1, . . . , n, Pn+1 = 0, Pn+2 = 1,
Vn
Vj−2
P 1 = 1, P 2 = 0, P j =
, j = 3, . . . , n + 2.
Vn
Obviously, ξ = ξ1 is equivalent to x = b and ξ = ξn+2 is equivalent to x = a. Also, the
existence of some k ∈ {3, . . . , m − 2} such that ξ = ξk and that (1.7) holds is equivalent
to the existence of some l ∈ {2, . . . , n − 1} such that x = x1 + xn − xl = a + b − xl and
that (2.2) holds. Therefore, applying Theorem D we get the desired conclusions. In the case
e = (xn , . . . , x1 ) and v
e = (vn , . . . , v1 ),
when x is decreasing we simply replace x and v with x
respectively, and then argue in the same manner.
In the special case that v > 0 also, Vi > 0 and V j > 0, i = 1, . . . , n, and therefore according
to (2.2) equality holds in (2.1) only when either x1 = · · · = xn = a or x1 = · · · = xn = b. Pj =
In the following
P theorem we will prove a refinement of Theorem B. Without loss of generality
we assume that ni=1 vi = 1.
Theorem 2.2. Let f : [0, ∞) → [0, ∞) and let [a, b] ⊆ [0, ∞) , a < b. Let x =
P(x1 , . . . , xn ) be
an n-tuple in [a, b]n and v = (v1 , . . . , vn ) a real n-tuple such that v ≥ 0 and ni=1 vi = 1. If f
is superquadratic we have
!
n
n
X
X
f (a) + f (b) −
vi f (xi ) − f a + b −
vi xi
i=1
≥
n
X
vi f
i=1
(2.3)
≥
n
X
i=1
vi f
n
! i=1 n
X
X xi − a
b − xi
vj xj − xi + 2
vi
f (b − xi ) +
f (xi − a)
b
−
a
b
−
a
j=1
i=1
n
!
n
X
X
2 (xi − a) (b − xi )
vj xj − xi + 2
vi f
.
b−a
j=1
i=1
If f is strictly superquadratic and v > 0 equality holds in (2.3) iff xi = a, i = 1, . . . , n, or
xi = b, i = 1, . . . , n.
Proof. The proof follows the technique in [8] and refines the result to positive superquadratic
functions. From Lemma A we know that for any λ ∈ [0, 1] the following holds:
λf (a) + (1 − λ) f (b) − f (λa + (1 − λ) b)
≥ λf (|a − λa − (1 − λ) b|) + (1 − λ) f (|b − λa − (1 − λ) b|)
= λf (|(1 − λ) (a − b)|) + (1 − λ) f (|λ (b − a)|)
(2.4)
= λf ((1 − λ) (b − a)) + (1 − λ) f (λ (b − a)) .
Also, for any xi ∈ [a, b] there exists a unique λi ∈ [0, 1] such that xi = λi a + (1 − λi ) b. We
have
n
n
X
X
(2.5)
f (a) + f (b) −
vi f (xi ) = f (a) + f (b) −
vi f (λi a + (1 − λi ) b) .
i=1
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J ENSEN -S TEFFENSEN ’ S INEQUALITY - S UPERQUADRATIC FUNCTIONS
7
Applying (2.4) on every xi = λi a + (1 − λi ) b in (2.5) we obtain
f (a) +f (b) −
n
X
vi f (xi )
i=1
n
X
vi − λi f (a) − (1 − λi ) f (b)
≥ f (a) + f (b) +
i=1
(2.6)
+ λi f ((1 − λi ) (b − a)) + (1 − λi ) f (λi (b − a))
n
X
=
vi [(1 − λi ) f (a) + λi f (b)]
i=1
+
n
X
vi [λi f ((1 − λi ) (b − a)) + (1 − λi ) f (λi (b − a))] .
i=1
Applying again (2.4) on (2.6) we get
(2.7) f (a) + f (b) −
n
X
vi f (xi ) ≥
n
X
i=1
vi f ((1 − λi ) a + λi b)
i=1
+2
n
X
vi [λi f ((1 − λi ) (b − a)) + (1 − λi ) f (λi (b − a))] .
i=1
Applying again Lemma A on (2.7) we obtain
f (a) +f (b) −
≥f
n
X
i=1
n
X
+
vi f (xi )
!
vi [(1 − λi ) a + λi b]
i=1
n
X
!
n
X
vj [(1 − λj ) a + λj b]
(1 − λi ) a + λi b −
vi f
i=1
n
X
+2
j=1
vi [(1 − λi ) f (λi (b − a)) + λi f ((1 − λi ) (b − a))]
i=1
!
n
X
=f a+b−
vi xi +
vi f vj xj − xi i=1
i=1
j=1
n
X
xi − a
b − xi
+2
vi
f (b − xi ) +
f (xi − a) ,
b
−
a
b
−
a
i=1
n
X
(2.8)
!
n
X
and this is the first inequality in (2.3).
Since f is a nonnegative superquadratic function, from Lemma B we know that it is also
convex, so from (2.8) we have
X
n
n
X
xi − a
b − xi
2 (b − xi ) (xi − a)
vi
f (b − xi ) +
f (xi − a) ≥
vi f
,
b−a
b−a
b−a
i=1
i=1
hence, the second inequality in (2.3) is proved.
J. Inequal. Pure and Appl. Math., 7(2) Art. 70, 2006
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8
S. A BRAMOVICH , M. K LARI ČI Ć BAKULA ,
AND
S. BANI Ć
For the case when f is strictly superquadratic and v > 0 we may deduce that inequalities
(2.6) and (2.7) become equalities iff each of the λi , i = 1, . . . , n, is either equal to 1 or equal to
0, which means that xi ∈ {a, b} , i = 1, . . . , n. However, since we also have
n
X
vj xj − xi = 0,
i = 1, . . . , n,
j=1
we deduce that xi = a, i = 1, . . . , n, or xi = b, i = 1, . . . , n.
This completes the proof of the theorem.
P
Corollary 2.3. Let v = (v1 , . . . , vn )be a real n-tuple such that v ≥ 0, ni=1 vi = 1 and let
x = (x1 , . . . , xn ) be an n-tuple in [a, b]n , 0 < a < b. Then for any real numbers r and s such
that rs ≥ 2 we have
s
Qs (a, b; x)
−1
Qr (a, b; x)

n
s
n
r
X
X
1
r
r

vi ≥
vj xj − xi Qr (a, b; x)s i=1 j=1
r
#
n
r
r
X
s
s
x i − ar r
b
−
x
i
(2.9)
+2
vi
(b − xri ) r + r
(xri − ar ) r
r
r
r
b −a
b −a
i=1


s
rs
n
n
n
X
r
r
r
r
r
X
X
2
(x
−
a
)
(b
−
x
)
1
i
i

,
≥
vi vj xrj − xri + 2
vi
s
r
r
Qr (a, b; x) i=1
b
−
a
j=1
i=1
1
P
where Qp (a, b; x) = (ap + bp − ni=1 vi xi ) p , p ∈ R \ {0} .
If rs > 2 and v > 0, the equalities hold in (2.9) iff xi = a, i = 1, . . . , n or xi = b, i = 1, . . . , n.
s
Proof. We define a function f : (0, ∞) → (0, ∞) as f (x) = x r . It can be easily checked that
for any real numbers r and s such that rs ≥ 2 the function f is superquadratic. We define a new
positive n-tuple ξ in [ar , br ] as ξi = xri , i = 1, . . . , n. From Theorem 2.2 we have
! rs
n
n
X
X
as + b s −
vi xsi − ar + br −
vi xri
i=1
(2.10)
i=1
s
r
n
n
n
X
r
X
X
s
s
br − xri r
x i − ar r
r
r
r r
r r
≥
vi vj xj − xi + 2
vi r
(b − xi ) + r
(xi − a )
r
r
b
−
a
b
−
a
i=1
j=1
i=1
s
s
n
n
n
X
r
X
X
2 (xri − ar ) (br − xri ) r
r
r
vi
≥
vi vj xj − xi + 2
≥ 0.
b r − ar
i=1
i=1
j=1
We have
as + b s −
n
X
vi xsi −
ar + b r −
i=1
n
X
! rs
vi xri
= Qs (a, b; x)s − Qr (a, b; x)s
i=1
so from (2.10) the inequalities in (2.9) follow.
s
The equality case follows from the equality case in Theorem 2.2, as the function f (x) = x r
is strictly superquadratic for rs > 2.
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J ENSEN -S TEFFENSEN ’ S INEQUALITY - S UPERQUADRATIC FUNCTIONS
9
Remark 2.4. It is an immediate result of Corollary 2.3 that if rs > 2 and there is at least one
j ∈ {1, . . . , n} such that
vj xrj − ar br − xrj > 0,
then for this j we have
! rs
s
2 xrj − ar br − xrj
Qs (a, b; x)
2vj
> 0.
−1>
Qr (a, b; x)
Qr (a, b; x)s
b r − ar
3. A N A LTERNATIVE P ROOF OF T HEOREM B
In this section we give an interesting alternative proof of Theorem B based on Theorem E.
To carry out that proof we need the following technical lemma.
Lemma 3.1. Let y = (y1 , . . P
. , ym ) be a decreasing real m-tuple and p = (p1 , . . . , pm ) a
nonnegative real m-tuple with m
i=1 pi = 1 . We define
m
X
y=
pi yi
i=1
and the m-tuple
y = (y, y, . . . , y) .
Then the m-tuples η = y, ξ = y and p satisfy conditions (1.8) and (1.9).
Proof. Note that y is a convex combination of y1 , y2 , . . . , ym , so we know that
ym ≤ y ≤ y1 .
From the definitions of the m-tuples ξ and η we have
m
m
m
m
X
X
X
X
pi = y =
pi yi =
p i ηi .
pi ξi = y
i=1
i=1
i=1
i=1
Hence, condition (1.9) is satisfied. Furthermore, for k = 1, 2, . . . , m − 1 we have
k
X
p i ηi −
k
X
i=1
pi ξi =
k
X
i=1
pi yi − y
i=1
=
Since
k
X
i=1
i=1
pi
i=1
k
X
pi yi −
i=1
Pm
k
X
m
X
p j yj
j=1
k
X
pi .
i=1
pi = 1 , we can write
p i ηi −
k
X
pi ξi =
i=1
k
X
pj +
j=1
=
=
m
X
pj
k
X
j=k+1
k
X
m
X
i=1
pj
pi yi −
i=1
k
X
!
pi
m
X
k
X
pi yi −
pj yi −
k
X
j=1
m
X
pi
i=1
j=k+1
pi
k
X
i=1
j=k+1
i=1
=
m
X
pj yj +
m
X
j=k+1
!
pj yj
k
X
pi
i=1
pj yj
j=k+1
m
X
!
pj yj
j=k+1
pj (yi − yj ) .
j=k+1
J. Inequal. Pure and Appl. Math., 7(2) Art. 70, 2006
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10
S. A BRAMOVICH , M. K LARI ČI Ć BAKULA ,
AND
S. BANI Ć
Since p is nonnegative and y is decreasing, we obtain
k
X
p i ηi −
i=1
k
X
pi ξi ≥ 0,
k = 1, 2, . . . , m − 1,
i=1
which means that condition (1.8) is satisfied as well.
Now we can give an alternative proof of Theorem B which is mainly based on Theorem E.
P
Proof of Theorem B. Since x = ni=1 wi xi is a convex combination of x1 , x2 , . . . , xn it is clear
that there is an s ∈ {1, 2, . . . , n − 1} such that
x1 ≤ · · · ≤ xs ≤ x ≤ xs+1 ≤ · · · ≤ xn ,
that is,
−x1 ≥ · · · ≥ −xs ≥ −x ≥ −xs+1 ≥ · · · ≥ −xn .
(3.1)
Adding x1 + xn to all the inequalities in (3.1) we obtain
xn ≥ · · · ≥ x1 + xn − xs ≥ x1 + xn − x ≥ x1 + xn − xs+1 ≥ · · · ≥ x1 ,
which gives us
(3.2)
x1 + xn − x = x1 + xn −
n
X
wi xi ∈ [x1 , xn ] .
i=1
We use (1.10) to prove the theorem. For this, we define the (n + 2)-tuples ξ, η and p as
follows:
η1 = xn , η2 = xn , η3 = xn−1 , . . . , ηn = x2 , ηn+1 = x1 , ηn+2 = x1 ,
p1 = 1, p2 = −wn , p3 = −wn−1 , . . . , pn = −w2 , pn+1 = −w1 , pn+2 = 1,
ξ1 = ξ2 = · · · = ξn+2 = η,
η=
n+2
X
p i ηi = x 1 + x n −
i=1
wj xj .
j=1
It is easily verified that ξ and η are decreasing and that
η and p satisfy conditions (1.8) and (1.9).
Condition (1.9) is trivially fulfilled since
n+2
X
n
X
pi ξi = η
n+2
X
pi = η =
i=1
i=1
Pn+2
i=1
n+2
X
pi = 1. It remains to see that ξ,
p i ηi .
i=1
Further, we have ξi = η, i = 1, 2, . . . , n + 2 . To prove (1.8) , we need to demonstrate that
(3.3)
η
k
X
pi ≤
i=1
k
X
pi ηi , k = 1, 2, . . . , n + 1.
i=1
For k = 1, (3.3) becomes η ≤ xn , and this holds because of (3.2) . On the other hand, for
k = n + 1, (3.3) becomes
!
n
n
X
X
η 1−
wi ≤ xn −
wi xi ,
i=1
i=1
that is,
0 ≤ xn − x,
and this holds because of (3.2) .
J. Inequal. Pure and Appl. Math., 7(2) Art. 70, 2006
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J ENSEN -S TEFFENSEN ’ S INEQUALITY - S UPERQUADRATIC FUNCTIONS
11
If k ∈ {2, . . . , n} , (3.3) can be rewritten and in its stead we have to prove that
!
n
n
X
X
wi xi .
(3.4)
η 1−
wi ≤ xn −
i=n+2−k
i=n+2−k
Let us consider the decreasing n-tuple y, where
yi = x1 + xn − xi , i = 1, 2, . . . , n.
We have
y=
n
X
wi yi
i=1
=
n
X
wi (x1 + xn − xi )
i=1
= x1 + xn −
n
X
wi xi = x1 + xn − x = η.
i=1
If we apply Lemma 3.1 to the n-tuple y and to the weights w, then m = n and for all l ∈
{1, 2, . . . , n − 1} the inequality
l
X
y
wi ≤
i=1
l
X
wi (x1 + xn − xi )
i=1
P
P
holds. Taking into consideration that y = η, li=1 wi = 1 − ni=l+1 wi and changing indices as
l = n + 1 − k, we deduce that
! n+1−k
n
X
X
(3.5)
η 1−
wi ≤
wi (x1 + xn − xi ) ,
i=1
i=n+2−k
for all k ∈ {2, . . . , n}. The difference between the right side of (3.4) and the right side of (3.5)
is
n
n+1−k
X
X
xn −
wi xi −
wi (x1 + xn − xi )
i=1
i=n+2−k
n
X
= xn −
wi xi − xn
= xn 1 −
!
wi
= xn
i=n+2−k
n
X
=
i=n+2−k
wi −
wi xi −
wi xi −
wi (x1 − xi )
n+1−k
X
wi (x1 − xi )
i=1
i=n+2−k
wi (xn − xi ) +
n+1−k
X
i=1
i=n+2−k
n
X
wi (x1 − xi )
i=1
n
X
−
i=1
n
X
wi −
n+1−k
X
i=1
i=n+2−k
n+1−k
X
n+1−k
X
n+1−k
X
wi (xi − x1 ) ≥ 0,
i=1
since w is nonnegative and x is increasing. Therefore, the inequality
(3.6)
n+1−k
X
wi (x1 + xn − xi ) ≤ xn −
i=1
J. Inequal. Pure and Appl. Math., 7(2) Art. 70, 2006
n
X
wi xi
i=n+2−k
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12
S. A BRAMOVICH , M. K LARI ČI Ć BAKULA ,
AND
S. BANI Ć
holds for all k ∈ {2, . . . , n}. From (3.5) and (3.6) we obtain (3.4) . This completes the proof
that the m-tuples ξ, η and p satisfy conditions (1.8) and (1.9) and we can apply Theorem E to
obtain
n
n+2
X
X
wi f (xi ) + f (x1 ) .
pi f (η) ≤ f (xn ) −
i=1
i=1
Taking into consideration that
f
Pn+2
x1 + xn −
i=1
n
X
pi = 1 and η = x1 + xn −
!
wi xi
≤ f (x1 ) + f (xn ) −
i=1
Pn
j=1
n
X
wj xj we finally get
wi f (xi ) .
i=1
R EFERENCES
[1] S. ABRAMOVICH, G. JAMESON AND G. SINNAMON, Refining Jensen’s inequality, Bull. Math.
Soc. Math. Roum., 47 (2004), 3–14.
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superquadratic functions, J. Inequal. in Pure and Appl. Math., 5(4) (2004), Art. 91. [ONLINE:
http://jipam.vu.edu.au/article.php?sid=444].
[3] S. ABRAMOVICH, S. BANIĆ, M. MATIĆ AND J. PEČARIĆ, Jensen-Steffensen’s and related inequalities for superquadratic functions, submitted for publication.
[4] S. ABRAMOVICH, M. KLARIČIĆ BAKULA, M. MATIĆ AND J. PEČARIĆ, A variant of JensenSteffensen’s inequality and quasi-arithmetic means, J. Math. Anal. Applics., 307 (2005), 370–385.
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3(3) (2002), Art. 40. [ONLINE: http://jipam.vu.edu.au/article.php?sid=192].
[6] A. McD. MERCER, A variant of Jensen’s inequality, J. Inequal. in Pure and Appl. Math., 4(4)
(2003), Art. 73. [ONLINE: http://jipam.vu.edu.au/article.php?sid=314].
[7] J.E. PEČARIĆ, F. PROSCHAN AND Y.L. TONG, Convex Functions, Partial Orderings, and Statistical Applications, Academic Press, Inc. (1992).
[8] A. WITKOWSKI, A new proof of the monotonicity property of power means, J. Inequal. in Pure
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php?sid=425].
J. Inequal. Pure and Appl. Math., 7(2) Art. 70, 2006
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