(1)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

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. KLARI ˇCI ´C BAKULA, AND S. BANI ´C DEPARTMENT OFMATHEMATICS

UNIVERSITY OFHAIFA

HAIFA, 31905, ISRAEL

abramos@math.haifa.ac.il DEPARTMENT OFMATHEMATICS

FACULTY OFNATURALSCIENCES, MATHEMATICS ANDEDUCATION

UNIVERSITY OFSPLIT

TESLINA12, 21000 SPLIT

CROATIA

milica@pmfst.hr

FACULTY OFCIVILENGINEERING ANDARCHITECTURE

UNIVERSITY OFSPLIT

MATICE HRVATSKE15, 21000 SPLIT

CROATIA

Senka.Banic@gradst.hr

Received 03 January, 2006; accepted 20 January, 2006 Communicated by C.P. Niculescu

ABSTRACT. A variant of Jensen-Steffensen’s inequality is considered for convex and for su- perquadratic 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. INTRODUCTION.

LetI be an interval in Randf : I → Ra convex function on I. Ifξ = (ξ1, . . . , ξm)is any m-tuple inI^m andp= (p₁, . . . , p_m)any nonnegativem-tuple such thatPm

i=1p_i >0, then the well known Jensen’s inequality (see for example [7, p. 43])

(1.1) f 1

P_m

m

X

i=1

p_iξ_i

!

≤ 1 P_m

m

X

i=1

p_if(ξ_i) holds, whereP_m =Pm

i=1p_i.

ISSN (electronic): 1443-5756 c

2006 Victoria University. All rights reserved.

002-06

Iff is strictly convex, then(1.1)is strict unlessξ_i =cfor alli∈ {j :p_j >0}.

It is well known that the assumption “p is a nonnegative m-tuple” can be relaxed at the expense of more restrictions on them-tupleξ.

Ifpis a realm-tuple such that

(1.2) 0≤P_j ≤P_m, j = 1, . . . , m, P_m >0, wherePj = Pj

i=1pi , then for any monotonicm-tupleξ (increasing or decreasing) in I^m we get

ξ = 1 P_m

m

X

i=1

p_iξ_i ∈I,

and for any functionf convex onI (1.1)still holds. Inequality(1.1)considered under condi- tions(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 that0< a < banda≤ x₁ ≤x₂ ≤ · · · ≤x_n ≤bhold with at least one of thex_k satisfyinga < x_k < b. Ifw = (w₁, . . . , w_n)is a positiven-tuple withPn

i=1w_i = 1 and−∞< r < s <+∞, then

a < Q_r(a, b;x)< Q_s(a, b;x)< b , where

Qt(a, b;x)≡ a^t+b^t−

n

X

i=1

wix^t_i

!¹_t

for all realt6= 0, and

Q₀(a, b;x)≡ ab

G, G=

n

Y

i=1

x^w_iⁱ.

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 ann-tuplex = (x₁, . . . , x_n) such that0< x₁ ≤x₂ ≤ · · · ≤x_nandw = (w₁, . . . , w_n)is a positiven-tuple withPn

i=1w_i = 1, then

f x₁+x_n−

n

X

i=1

w_ix_i

!

≤f(x₁) +f(x_n)−

n

X

i=1

w_if(x_i).

This theorem is a special case of the following theorem proved in [4] by Abramovich, Klariˇci´c Bakula, Mati´c and Peˇcari´c:

Theorem C ([4, Th. 2]). Letf : I →R,whereI is an interval inRand let[a, b]⊆ I, a < b.

Let x = (x₁, . . . , x_n)be a monotonicn-tuple in [a, b]ⁿ and v = (v₁, . . . , v_n)a realn−tuple such that v_i 6= 0,i = 1, . . . , n,and0≤ V_j ≤ V_n, j = 1, . . . , n, V_n >0, whereV_j = Pj

i=1v_i. Iff is convex onI,then

(1.3) f a+b− 1

V_n

n

X

i=1

vixi

!

≤f(a) +f(b)− 1 V_n

n

X

i=1

vif(xi).

In casefis strictly convex, the equality holds in (1.3) iff one of the following two cases occurs:

(1) eitherx=aorx=b,

(2) there existsl ∈ {2, . . . , n−1} such that x=x₁+x_n−x_land (1.4)







x₁ =a, x_n=b or x₁ =b, x_n=a, V_j(xj−1−x_j) = 0, j= 2, . . . , l, V_j (x_j−x_j+1) = 0, j =l, . . . , n−1, whereV_j =Pn

i=jv_i, j= 1, . . . , nandx= (1/V_n)Pn i=1v_ix_i.

In the special case where v > 0 and f is strictly convex, the equality holds in (1.4) iff xi =a, i= 1, . . . , n,orxi =b,i= 1, . . . , n.

Here, as in the rest of the paper, when we say that an n-tupleξ is increasing (decreasing) we mean that ξ₁ ≤ ξ₂ ≤ · · · ≤ ξ_n (ξ₁ ≥ ξ₂ ≥ · · · ≥ ξ_n). 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 su- perquadratic 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) = x^p, p≥ 1,so the inequalities satisfied by superquadratic functions are generalizations of the inequalities satisfied by the superquadratic functionsf(x) =x^p, p≥2(see [1], [2]).

First we quote some definitions and state a list of basic properties of superquadratic functions.

Definition 1.1. A functionf : [0,∞)→ R is superquadratic provided that for allx ≥0there exists a constantC(x)∈Rsuch that

(1.5) f(y)−f(x)−f(|y−x|)≥C(x) (y−x) for ally≥0.

Definition 1.2. A functionf : [0,∞) →Ris said to be strictly superquadratic if (1.5) is strict for allx6=ywherexy 6= 0.

Lemma A ([2, Lemma 2.3]). Suppose thatf is superquadratic. Letξ_i ≥ 0, i = 1, . . . , m,and letξ =Pm

i=1p_iξ_i,wherep_i ≥0, i= 1, . . . , m,andPm

i=1p_i = 1.Then

m

X

i=1

p_if(ξ_i)−f ξ

≥

m

X

i=1

p_if ξ_i−ξ

.

Lemma B ([1, Lemma 2.2]). Letf be superquadratic function withC(x)as in Definition 1.1.

Then:

(i) f(0)≤0,

(ii) iff(0) =f⁰(0) = 0thenC(x) = f⁰(x)wheneverf is differentiable atx >0, (iii) iff ≥0,thenf is convex andf(0) =f⁰(0) = 0.

In [3] the following refinement of Jensen’s Steffensen’s type inequality for nonnegative su- perquadratic 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 R^m and p a real m-tuple, m ≥ 3, satisfying

0≤P_j ≤P_m, j = 1, . . . , m, P_m >0.

Letξbe defined as

ξ= 1 Pm

m

X

i=1

p_iξ_i.

Then

m

X

i=1

p_if(ξ_i)−P_mf ξ

≥

k−1

X

i=1

P_if(ξ_i+1−ξ_i) +P_kf ξ −ξ_k (1.6)

+Pk+1f ξk+1−ξ +

m

X

i=k+2

Pif(ξi−ξi−1)

≥

" _k X

i=1

P_i+

m

X

i=k+1

P_i

# f

Pm i=1p_i

ξ−ξ_i Pk

i=1P_i+Pm i=k+1P_i

!

≥(m−1)P_mf Pm

i=1p_i ξ−ξ_i

(m−1)Pm

! , whereP_i =Pm

j=ip_j andk ∈ {1, . . . , m−1}satisfies ξ_k≤ξ ≤ξ_k+1. In casef is also strictly superquadratic, inequality

m

X

i=1

p_if(ξ_i)−P_mf ξ

>(m−1)P_mf Pm

i=1p_i ξ_i−ξ

(m−1)Pm

!

holds forξ>0unless one of the following two cases occurs:

(1) eitherξ=ξ₁ orξ =ξ_m,

(2) there existsk ∈ {3, . . . , m−2}such thatξ=ξ_k and (1.7)

( P_j(ξ_j −ξ_j+1) = 0, j = 1, . . . , k−1 P_j(ξ_j−ξj−1) = 0, j =k+ 1, . . . , m.

In these two cases

m

X

i=1

p_if(ξ_i)−P_mf ξ

= 0.

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. LetI be an interval inR, andξ,ηtwo decreasingm-tuples such thatξ,η ∈I^m. Letpbe a realm-tuple such that

(1.8)

k

X

i=1

p_iξ_i ≤

k

X

i=1

p_iη_i fork = 1,2, . . . , m−1, and

(1.9)

m

X

i=1

p_iξ_i =

m

X

i=1

p_iη_i . Then for every continuous convex functionf :I →Rwe have (1.10)

m

X

i=1

p_if(ξ_i)≤

m

X

i=1

p_if(η_i).

2. VARIANTS OFJENSEN-STEFFENSEN’S INEQUALITY FORPOSITIVE

SUPERQUADRATICFUNCTIONS

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 monotonicn−tuple in[a, b]ⁿandv = (v₁, . . . , v_n)a realn-tuple such thatv_i 6= 0, i= 1, . . . , n, 0 ≤ V_j ≤ V_n, j = 1, . . . , n, and V_n > 0, where V_j = Pj

i=1v_i. If f is differentiable and superquadratic, then

(2.1) f(a) +f(b)− 1 V_n

n

X

i=1

v_if(x_i)−f a+b− 1 V_n

n

X

i=1

v_ix_i

!

≥(n+ 1)f





b−a−_V¹

n

Pn i=1v_i

a+b−x_i− _V¹

n

Pn j=1v_jx_j

n+ 1



. In casef is also strictly superquadratic and a > 0, inequality (2.1) is strict unless one of the following two cases occurs:

(1) eitherx=aorx=b,

(2) there existsl ∈ {2, . . . , n−1}such thatx=x₁+x_n−x_land

(2.2)











x1 =a, xn=b or x1 =b, xn=a Vj(xj−1−xj) = 0, j= 2, . . . , l, V_j(x_j −x_j+1) = 0, j =l, . . . , n−1, whereV_j =Pn

i=jv_i,j = 1, . . . , n, andx= _V¹

n

Pn i=1v_ix_i. In these two cases we have

f(a) +f(b)− 1 V_n

n

X

i=1

v_if(x_i)−f a+b− 1 V_n

n

X

i=1

v_ix_i

!

= 0.

In the special case wherev>0andf is also strictly superquadratic, the equality holds in (2.1) iffx_i =a, i= 1, . . . , n,orx_i =b, i= 1, . . . , n.

Proof. Suppose thatxis an increasingn-tuple in[a, b]ⁿ.The proof of the theorem is an imme- diate result of Theorem D, by defining the(n+ 2)-tuplesξandpas

ξ₁ =a, ξ_i+1 =x_i, i= 1, . . . , n, ξ_n+2 =b p₁ = 1, p_i+1 =−v_i/V_n, i= 1, . . . , n, p_n+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

X

i=1

Pi +

n+2

X

i=k+1

Pi ≤n+ 1,

forP_i =Pi

j=1p_j andP_i =Pn+2

j=i p_j,and ξ= 1

P_m

m

X

i=1

piξi =a+b− 1 V_n

n

X

i=1

vixi =a+b−x.

The proof of the equality case and the special case wherev > 0follows also from Theorem D.

We have

P_j = Vj

V_n, j = 1, . . . , n, P_n+1 = 0, P_n+2 = 1, P₁ = 1, P₂ = 0, P_j = Vj−2

Vn

, j = 3, . . . , n+ 2.

Obviously, ξ = ξ₁ is equivalent to x = b and ξ = ξ_n+2 is equivalent to x = a. Also, the existence of some k ∈ {3, . . . , m−2} such that ξ = ξ_k and that (1.7) holds is equivalent to the existence of some l ∈ {2, . . . , n−1}such that x = x₁ +x_n−x_l = a +b −x_l and that(2.2)holds. Therefore, applying Theorem D we get the desired conclusions. In the case whenxis decreasing we simply replacexandvwithxe = (xn, . . . , x1)andev = (vn, . . . , v1), respectively, and then argue in the same manner.

In the special case thatv >0also,V_i >0andV_j >0, i= 1, . . . , n, and therefore according to(2.2)equality holds in(2.1)only when eitherx₁ =· · ·=x_n =a or x₁ =· · ·=x_n=b.

In the following theorem we will prove a refinement of Theorem B. Without loss of generality we assume thatPn

i=1v_i = 1.

Theorem 2.2. Letf : [0,∞)→[0,∞)and let[a, b]⊆[0,∞), a < b.Letx= (x₁, . . . , x_n)be ann-tuple in[a, b]ⁿandv = (v₁, . . . , v_n)a realn-tuple such thatv ≥0andPn

i=1v_i = 1.Iff is superquadratic we have

f(a) +f(b)−

n

X

i=1

v_if(x_i)−f a+b−

n

X

i=1

v_ix_i

!

≥

n

X

i=1

vif

n

X

j=1

vjxj−xi

! + 2

n

X

i=1

vi

x_i−a

b−af(b−xi) + b−x_i

b−af(xi−a)

≥

n

X

i=1

vif

n

X

j=1

vjxj−xi

! + 2

n

X

i=1

vif

2 (x_i−a) (b−x_i) b−a

. (2.3)

If f is strictly superquadratic and v > 0 equality holds in (2.3) iffx_i = 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 anyλ∈[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)|)

=λf((1−λ) (b−a)) + (1−λ)f(λ(b−a)). (2.4)

Also, for anyx_i ∈ [a, b]there exists a unique λ_i ∈ [0,1]such thatx_i = λ_ia+ (1−λ_i)b. We have

(2.5) f(a) +f(b)−

n

X

i=1

v_if(x_i) =f(a) +f(b)−

n

X

i=1

v_if(λ_ia+ (1−λ_i)b).

Applying(2.4)on everyx_i =λ_ia+ (1−λ_i)bin(2.5)we obtain f(a) +f(b)−

n

X

i=1

v_if(x_i)

≥f(a) +f(b) +

n

X

i=1

v_i

−λ_if(a)−(1−λ_i)f(b) +λ_if((1−λ_i) (b−a)) + (1−λ_i)f(λ_i(b−a))

=

n

X

i=1

v_i[(1−λ_i)f(a) +λ_if(b)]

(2.6)

+

n

X

i=1

v_i[λ_if((1−λ_i) (b−a)) + (1−λ_i)f(λ_i(b−a))]. Applying again(2.4)on(2.6)we get

(2.7) f(a) +f(b)−

n

X

i=1

v_if(x_i)≥

n

X

i=1

v_if((1−λ_i)a+λ_ib)

+ 2

n

X

i=1

v_i[λ_if((1−λ_i) (b−a)) + (1−λ_i)f(λ_i(b−a))]. Applying again Lemma A on(2.7)we obtain

f(a) +f(b)−

n

X

i=1

v_if(x_i)

≥f

n

X

i=1

v_i[(1−λ_i)a+λ_ib]

!

+

n

X

i=1

v_if

(1−λ_i)a+λ_ib−

n

X

j=1

v_j[(1−λ_j)a+λ_jb]

!

+ 2

n

X

i=1

v_i[(1−λ_i)f(λ_i(b−a)) +λ_if((1−λ_i) (b−a))]

=f a+b−

n

X

i=1

v_ix_i

! +

n

X

i=1

v_if

n

X

j=1

v_jx_j−x_i

!

+ 2

n

X

i=1

v_i

x_i−a

b−a f(b−x_i) + b−x_i

b−af(x_i−a)

, (2.8)

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

n

X

i=1

v_i

x_i−a

b−a f(b−x_i) + b−x_i

b−af(x_i−a)

≥

n

X

i=1

v_if

2 (b−x_i) (x_i−a) b−a

, hence, the second inequality in (2.3) is proved.

For the case whenf is strictly superquadratic andv > 0 we may deduce that inequalities (2.6)and(2.7)become equalities iff each of theλ_i, i= 1, . . . , n,is either equal to 1 or equal to 0, which means thatx_i ∈ {a, b}, i = 1, . . . , n. However, since we also have

n

X

j=1

v_jx_j−x_i = 0, i= 1, . . . , n, we deduce thatx_i =a, i= 1, . . . , n,orx_i =b, i= 1, . . . , n.

This completes the proof of the theorem.

Corollary 2.3. Let v = (v₁, . . . , v_n)be a realn-tuple such that v ≥ 0, Pn

i=1v_i = 1 and let x= (x1, . . . , xn)be ann-tuple in[a, b]ⁿ, 0< a < b. Then for any real numbersrandssuch that ^s_r ≥2we have

Q_s(a, b;x) Qr(a, b;x)

s

−1

≥ 1

Q_r(a, b;x)^s





n

X

i=1

v_i

n

X

j=1

v_jx^r_j −x^r_i

s r

+2

n

X

i=1

vi

x^r_i −a^r

b^r−a^r (b^r−x^r_i)^sr +b^r−x^r_i

b^r−a^r (x^r_i −a^r)^sr # (2.9)

≥ 1

Q_r(a, b;x)^s





n

X

i=1

v_i

n

X

j=1

v_jx^r_j −x^r_i

s r

+ 2

n

X

i=1

v_i

2 (x^r_i −a^r) (b^r−x^r_i) b^r−a^r

^s_r



, whereQ_p(a, b;x) = (a^p+b^p −Pn

i=1v_ix_i)^1p,p∈R\ {0}.

If ^s_r >2andv >0,the equalities hold in(2.9)iffx_i =a, i= 1, . . . , norx_i =b, i= 1, . . . , n.

Proof. We define a functionf : (0,∞)→ (0,∞)asf(x) = x^sr.It can be easily checked that for any real numbersrandssuch that ^s_r ≥2the functionf is superquadratic. We define a new positiven-tupleξ in[a^r, b^r]asξ_i =x^r_i, i= 1, . . . , n.From Theorem 2.2 we have

a^s+b^s−

n

X

i=1

v_ix^s_i − a^r+b^r−

n

X

i=1

v_ix^r_i

!^s_r

≥

n

X

i=1

v_i

n

X

j=1

v_jx^r_j −x^r_i

s r

+ 2

n

X

i=1

v_i

x^r_i −a^r

b^r−a^r (b^r−x^r_i)^sr +b^r−x^r_i

b^r−a^r (x^r_i −a^r)^sr

≥

n

X

i=1

v_i

n

X

j=1

v_jx^r_j −x^r_i

s r

+ 2

n

X

i=1

v_i

2 (x^r_i −a^r) (b^r−x^r_i) b^r−a^r

^s_r

≥0.

(2.10) We have

a^s+b^s−

n

X

i=1

v_ix^s_i − a^r+b^r−

n

X

i=1

v_ix^r_i

!^s_r

=Q_s(a, b;x)^s−Q_r(a, b;x)^s so from(2.10)the inequalities in(2.9)follow.

The equality case follows from the equality case in Theorem 2.2, as the functionf(x) =x^sr

is strictly superquadratic for ^s_r >2.

Remark 2.4. It is an immediate result of Corollary 2.3 that if ^s_r > 2and there is at least one j ∈ {1, . . . , n}such that

v_j x^r_j −a^r

b^r−x^r_j

>0, then for thisj we have

Q_s(a, b;x) Q_r(a, b;x)

s

−1> 2v_j Q_r(a, b;x)^s

2 x^r_j−a^r

b^r−x^r_j b^r−a^r

!^s_r

>0.

3. AN ALTERNATIVEPROOF OFTHEOREMB

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, . . . , y_m) be a decreasing real m-tuple and p = (p₁, . . . , p_m) a nonnegative realm-tuple withPm

i=1p_i = 1. We define y=

m

X

i=1

p_iy_i and them-tuple

y= (y, y, . . . , y).

Then them-tuplesη =y,ξ=yandpsatisfy conditions(1.8)and(1.9).

Proof. Note thatyis a convex combination ofy₁, y₂, . . . , y_m, so we know that ym ≤y≤y1.

From the definitions of them-tuplesξ andηwe have

m

X

i=1

p_iξ_i =y

m

X

i=1

p_i =y=

m

X

i=1

p_iy_i =

m

X

i=1

p_iη_i.

Hence, condition(1.9)is satisfied. Furthermore, fork = 1,2, . . . , m−1we have

k

X

i=1

p_iη_i−

k

X

i=1

p_iξ_i =

k

X

i=1

p_iy_i−y

k

X

i=1

p_i

=

k

X

i=1

p_iy_i−

m

X

j=1

p_jy_j

k

X

i=1

p_i. SincePm

i=1p_i = 1,we can write

k

X

i=1

p_iη_i−

k

X

i=1

p_iξ_i =

k

X

j=1

p_j+

m

X

j=k+1

p_j

! _k X

i=1

p_iy_i−

k

X

j=1

p_jy_j+

m

X

j=k+1

p_jy_j

! _k X

i=1

p_i

=

m

X

j=k+1

p_j

k

X

i=1

p_iy_i−

k

X

i=1

p_i

m

X

j=k+1

p_jy_j

=

k

X

i=1

p_i

m

X

j=k+1

p_jy_i −

m

X

j=k+1

p_jy_j

!

=

k

X

i=1

pi m

X

j=k+1

pj(yi−yj).

Sincepis nonnegative andyis decreasing, we obtain

k

X

i=1

p_iη_i−

k

X

i=1

p_iξ_i ≥0, k = 1,2, . . . , m−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.

Proof of Theorem B. Sincex=Pn

i=1w_ix_iis a convex combination ofx₁, x₂, . . . , x_nit is clear that there is ans∈ {1,2, . . . , n−1}such that

x1 ≤ · · · ≤xs ≤x≤xs+1 ≤ · · · ≤xn, that is,

(3.1) −x₁ ≥ · · · ≥ −x_s ≥ −x≥ −x_s+1 ≥ · · · ≥ −x_n. Addingx₁+x_nto all the inequalities in(3.1)we obtain

x_n≥ · · · ≥x₁+x_n−x_s ≥x₁+x_n−x≥x₁+x_n−x_s+1 ≥ · · · ≥x₁, which gives us

(3.2) x₁+x_n−x=x₁+x_n−

n

X

i=1

w_ix_i ∈[x₁, x_n].

We use (1.10) to prove the theorem. For this, we define the (n+ 2)-tuplesξ, η and pas follows:

η1 =xn, η2 =xn, η3 =xn−1, . . . , ηn =x2, ηn+1 =x1, ηn+2 =x1, p₁ = 1, p₂ =−w_n, p₃ =−wn−1, . . . , p_n =−w₂, p_n+1 =−w₁, p_n+2 = 1, ξ₁ =ξ₂ =· · ·=ξ_n+2 =η, η=

n+2

X

i=1

p_iη_i =x₁+x_n−

n

X

j=1

w_jx_j. It is easily verified thatξ andη are decreasing and thatPn+2

i=1 p_i = 1. It remains to see thatξ, ηandpsatisfy conditions(1.8)and(1.9).

Condition(1.9)is trivially fulfilled since

n+2

X

i=1

p_iξ_i =η

n+2

X

i=1

p_i =η=

n+2

X

i=1

p_iη_i.

Further, we haveξ_i =η, i= 1,2, . . . , n+ 2. To prove(1.8),we need to demonstrate that

(3.3) η

k

X

i=1

pi ≤

k

X

i=1

piηi, k= 1,2, . . . , n+ 1.

For k = 1, (3.3) becomes η ≤ x_n, and this holds because of (3.2). On the other hand, for k =n+ 1,(3.3)becomes

η 1−

n

X

i=1

w_i

!

≤x_n−

n

X

i=1

w_ix_i, that is,

0≤x_n−x, and this holds because of(3.2).

Ifk ∈ {2, . . . , n},(3.3)can be rewritten and in its stead we have to prove that

(3.4) η 1−

n

X

i=n+2−k

wi

!

≤xn−

n

X

i=n+2−k

wixi. Let us consider the decreasingn-tupley, where

y_i =x₁+x_n−x_i, i= 1,2, . . . , n.

We have

y=

n

X

i=1

w_iy_i

=

n

X

i=1

w_i(x₁+x_n−x_i)

=x1+xn−

n

X

i=1

wixi =x1+xn−x=η.

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

y

l

X

i=1

w_i ≤

l

X

i=1

w_i(x₁+x_n−x_i) holds. Taking into consideration thaty=η,Pl

i=1w_i = 1−Pn

i=l+1w_iand changing indices as l =n+ 1−k, we deduce that

(3.5) η 1−

n

X

i=n+2−k

w_i

!

≤

n+1−k

X

i=1

w_i(x₁+x_n−x_i),

for allk∈ {2, . . . , n}. The difference between the right side of(3.4)and the right side of(3.5) is

x_n−

n

X

i=n+2−k

w_ix_i−

n+1−k

X

i=1

w_i(x₁+x_n−x_i)

=xn−

n

X

i=n+2−k

wixi−xn n+1−k

X

i=1

wi−

n+1−k

X

i=1

wi(x1−xi)

=x_n 1−

n+1−k

X

i=1

w_i

!

−

n

X

i=n+2−k

w_ix_i−

n+1−k

X

i=1

w_i(x₁−x_i)

=x_n

n

X

i=n+2−k

w_i−

n

X

i=n+2−k

w_ix_i−

n+1−k

X

i=1

w_i(x₁−x_i)

=

n

X

i=n+2−k

w_i(x_n−x_i) +

n+1−k

X

i=1

w_i(x_i−x₁)≥0, sincewis nonnegative andxis increasing. Therefore, the inequality (3.6)

n+1−k

X

i=1

wi(x1+xn−xi)≤xn−

n

X

i=n+2−k

wixi

holds for allk ∈ {2, . . . , n}. From(3.5)and(3.6)we obtain(3.4).This completes the proof that them-tuplesξ,ηandpsatisfy conditions(1.8)and(1.9)and we can apply Theorem E to obtain

n+2

X

i=1

p_if(η)≤f(x_n)−

n

X

i=1

w_if(x_i) +f(x₁). Taking into consideration thatPn+2

i=1 p_i = 1andη =x₁+x_n−Pn

j=1w_jx_j we finally get f x1+xn−

n

X

i=1

wixi

!

≤f(x1) +f(xn)−

n

X

i=1

wif(xi).

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