(1)INEQUALITIES OF JENSEN-PE ˇCARI ´C-SVRTAN-FAN TYPE CHAOBANG GAO AND JIAJIN WEN COLLEGEOFMATHEMATICSANDINFORMATIONSCIENCE CHENGDUUNIVERSITY CHENGDU, 610106, P.R

INEQUALITIES OF JENSEN-PE ˇCARI ´C-SVRTAN-FAN TYPE

CHAOBANG GAO AND JIAJIN WEN

COLLEGEOFMATHEMATICSANDINFORMATIONSCIENCE

CHENGDUUNIVERSITY

CHENGDU, 610106, P.R. CHINA

kobren427@163.com wenjiajin623@163.com

Received 22 January, 2007; accepted 06 July, 2008 Communicated by S.S. Dragomir

ABSTRACT. By using the theory of majorization, the following inequalities of Jensen-Peˇcari´c- Svrtan-Fan type are established: LetIbe an interval,f :I→Randt∈I, x, a, b∈Iⁿ. Ifa1≤

· · · ≤an≤bn≤ · · · ≤b1, a1+b1≤ · · · ≤an+bn;f(t)>0, f⁰(t)>0, f⁰⁰(t)>0, f⁰⁰⁰(t)<0 for anyt∈I,then

f(A(a))

f(A(b)) = fn,n(a)

fn,n(b) ≤ · · · ≤ fk+1,n(a)

fk+1,n(b) ≤fk,n(a)

fk,n(b) ≤ · · · ≤ f1,n(a)

f1,n(b) =A(f(a)) A(f(b)), the inequalities are reversed forf⁰⁰(t) < 0, f⁰⁰⁰(t) > 0,∀t ∈ I, whereA(·)is the arithmetic mean and

fk,n(x) := 1

n k

X

1≤i1<···<ik≤n

f

xi₁+· · ·+xi_k

k

, k= 1, . . . , n.

Key words and phrases: Jensen’s inequality, Peˇcari´c-Svrtan’s inequality, Fan’s inequality, Theory of majorization, Hermite matrix.

2000 Mathematics Subject Classification. 26D15, 26E60.

1. INTRODUCTION

In what follows, we shall use the following symbols:

x:= (x₁, . . . , x_n); f(x) := (f(x₁), . . . , f(x_n)); G(x) := (x₁x₂· · ·x_n)^1/n;

A(x) := x1+x2+· · ·+xn

n ; Rⁿ+ := [0,+∞)ⁿ; Rⁿ++ := (0,+∞)ⁿ; Iⁿ :={x|x_i ∈I, i= 1, . . . , n, I is an interval};

This work is supported by the Natural Science Foundation of China (10671136) and the Natural Science Foundation of Sichuan Province Education Department (07ZA207).

The authors are deeply indebted to anonymous referees for many useful comments and keen observations which led to the present improved version of the paper as it stands, and also are grateful to their friend Professor Wan-lan Wang for numerous discussions and helpful suggestions in preparation of this paper.

033-07

f_k,n(x) := 1

n k

X

1≤i1<···<ik≤n

f

x_i1 +· · ·+x_ik k

, k = 1, . . . , n.

Jensen’s inequality states that: Letf :I →Rbe a convex function andx∈Iⁿ. Then

(1.1) f(A(x))≤A(f(x)).

This well-known inequality has a great number of generalizations in the literature (see [1] – [6]). An interesting generalization of(1.1)due to Peˇcari´c and Svrtan [5] is:

(1.2) f(A(x)) = f_n,n(x)≤ · · · ≤f_k+1,n(x)≤f_k,n(x)≤ · · · ≤f_1,n(x) = A(f(x)).

In 2003, Tang and Wen [6] obtained the following generalization of(1.2):

(1.3) f_r,s,n ≥ · · · ≥f_r,s,i ≥ · · · ≥f_r,s,s ≥ · · · ≥f_r,j,j ≥ · · · ≥f_r,r,r = 0, where

f_r,s,n :=

n r

n s

(f_r,n−f_s,n), f_k,n :=f_k,n(x), 1≤r≤s≤n.

Ky Fan’s arithmetic-geometric mean inequality is (see [7]): Letx∈(0,1/2]ⁿ. Then

(1.4) A(x)

A(1−x) ≥ G(x) G(1−x).

In this paper, we shall establish further extensions of(1.2)and(1.4)as follows:

Theorem 1.1. LetIbe an interval. Iff :I →R, a, b∈Iⁿ(n≥2)and (i) a1 ≤ · · · ≤an≤bn ≤ · · · ≤b1, a1+b1 ≤ · · · ≤an+bn; (ii) f(t)>0, f⁰(t)>0, f⁰⁰(t)>0, f⁰⁰⁰(t)<0for anyt ∈I,

then (1.5) f(A(a))

f(A(b)) = fn,n(a)

f_n,n(b) ≤ · · · ≤ fk+1,n(a)

f_k+1,n(b) ≤ fk,n(a)

f_k,n(b) ≤ · · · ≤ f1,n(a)

f_1,n(b) = A(f(a)) A(f(b)). The inequalities are reversed forf⁰⁰(t)<0, f⁰⁰⁰(t)>0,∀t∈I. The equality signs hold if and only ifa₁ =· · ·=a_nandb₁ =· · ·=b_n.

In Section 3, several interesting results of Ky Fan shall be deduced. In Section 4, the matrix variant of(1.5)will be established.

2. PROOF OFTHEOREM1.1

Lemma 2.1. Letf :I →Rbe a function whose second derivative exists andx∈Iⁿ, α∈Ω_n={α∈Rⁿ+:α₁ +· · ·+α_n= 1}.

Writing

S(α, x) := 1 n!

X

i1···in

f(α₁x_i1 +· · ·+α_nx_in),

whereP

i1···in denotes summation over all permutations of{1,2, . . . , n}, F(α) := log

S(α, a) S(α, b)

, a, b∈Iⁿ,

u_i(x) := α₁x_i1 +α₂x_i2 +

n

X

j=3

α_jx_ij,

vi(x) := α1xi2 +α2xi1 +

n

X

j=3

αjxij, i= (i1, i2, . . . , in).

Then there existξ_i(a)betweenu_i(a)andv_i(a),andξ_i(b)betweenu_i(b)andv_i(b)such that

(2.1) (α₁−α₂) ∂F

∂α1

− ∂F

∂α2

= 1 n!

X

i3···in

X

1≤i1<i2≤in

f⁰⁰(ξ_i(a))(u_i(a)−v_i(a))²

S(α, a) − f⁰⁰(ξ_i(b))(u_i(b)−v_i(b))² S(α, b)

,

whereP

i3···i_n denotes the summation over all permutations of{1,2, . . . , n} \ {i₁, i₂}.

Proof. Note the following identities:

S(α, x) = 1 n!

X

i3···i_n

X

1≤i16=i2≤n

f(α₁x_i1 +· · ·+α_nx_in)

= 1 n!

X

i3···in

X

1≤i₁<i2≤n

[f(u_i(x)) +f(v_i(x))];

∂

∂α₁[f(u_i) +f(v_i)]− ∂

∂α₂[f(u_i) +f(v_i)] = [f⁰(u_i)−f⁰(v_i)](x_i1 −x_i2);

(α₁ −α₂) ∂S

∂α1

− ∂S

∂α2

= 1 n!

X

i3···i_n

X

1≤i1<i2≤n

[f⁰(u_i)−f⁰(v_i)](α₁−α₂)(x_i1 −x_i2)

= 1 n!

X

i3···in

X

1≤i1<i2≤n

[f⁰(u_i)−f⁰(v_i)](u_i −v_i).

(2.2)

ByF(α) = logS(α, a)−logS(α, b)and(2.2), we have (α₁−α₂)

∂F

∂α₁ − ∂F

∂α₂

= (α₁−α₂)

[S(α, a)]⁻¹

∂S(α, a)

∂α₁ − ∂S(α, a)

∂α₂

−[S(α, b)]⁻¹

∂S(α, b)

∂α₁ − ∂S(α, b)

∂α₂

= 1 n!

X

i3···in

X

1≤i1<i2≤n

[f⁰(ui(a))−f⁰(vi(a))][ui(a)−vi(a)]

S(α, a)

− [f⁰(u_i(b))−f⁰(v_i(b))][u_i(b)−v_i(b)]

S(α, b)

= 1 n!

X

i3···i_n

X

1≤i1<i2≤n

f⁰⁰(ξ_i(a))[u_i(a)−v_i(a)]²

S(α, a) −f⁰⁰(ξ_i(b))[u_i(b)−v_i(b)]² S(α, b)

.

Here we used the Mean Value Theorem forf⁰(t). This completes the proof.

Lemma 2.2. Under the hypotheses of Theorem 1.1, F is a Schur-convex function or a Schur- concave function onΩ_n, whereF is defined by Lemma 2.1.

Proof. It is easy to see thatΩ_n is a symmetric convex set andF is a differentiable symmetric function onΩ_n. To prove thatF is a Schur-convex function onΩ_n, it is enough from [8, p .57]

to prove that

(2.3) (α₁−α₂)

∂F

∂α₁ − ∂F

∂α₂

≥0, ∀α∈Ω_n.

To prove(2.3), it is enough from Lemma 2.1 to prove (2.4) f⁰⁰(ξ_i(a))[u_i(a)−v_i(a)]²

S(α, a) ≥ f⁰⁰(ξ_i(b))[u_i(b)−v_i(b)]²

S(α, b) .

Using the given conditionsa₁ ≤ · · · ≤a_n ≤b_n≤ · · · ≤ b₁, f(t)>0andf⁰(t)>0, we obtain thata_j ≤b_j (j = 1,2, . . . , n)and the inequalities:

(2.5) 1

S(α, a) ≥ 1

S(α, b) >0.

By the given condition (i) of Theorem 1.1 and1≤i₁ < i₂ ≤n, we have a_i2 −a_i1 ≥b_i1 −b_i2 ≥0

and

(2.6) [u_i(a)−v_i(a)]² ≥[u_i(b)−v_i(b)]² ≥0.

From(2.5)and(2.6), we get

(2.7) [u_i(a)−v_i(a)]²

S(α, a) ≥ [u_i(b)−v_i(b)]² S(α, b) ≥0.

Note thata, b∈Iⁿ, u_i(a), v_i(a), u_i(b), v_i(b)∈I, and min{u_i(a), v_i(a)} ≤ξ_i(a)

≤max{ui(a), vi(a)}

≤min{u_i(b), v_i(b)}

≤ξ_i(b)

≤max{u_i(b), v_i(b)}.

It follows that

(2.8) ξ_i(a)≤ξ_i(b) (ξ_i(a), ξ_i(b)∈I).

Iff⁰⁰(t)>0, f⁰⁰⁰(t)<0for anyt∈I, from these and(2.8)we get (2.9) f⁰⁰(ξ_i(a))≥f⁰⁰(ξ_i(b))>0.

Combining with(2.7)and (2.9), we have proven that (2.4) holds, hence, F is a Schur-convex function onΩ_n.

Similarly, iff⁰⁰(t)<0, f⁰⁰⁰(t)>0for anyt ∈I, we obtain (2.10) −f⁰⁰(ξ_i(a))≥ −f⁰⁰(ξ_i(b))>0.

Combining with(2.7)and(2.10), we know that the inequalities are reversed in(2.4)and(2.3).

Therefore,F is a Schur-concave function onΩ_n. This ends the proof of Lemma 2.2.

Remark 1. When α₁ 6= α₂, there is equality in (2.3)if a₁ = · · · = a_n and b₁ = · · · = b_n. In fact, there is equality in (2.3) if and only if there is equality in (2.5), (2.8), (2.9) and the first inequality in(2.6)or all the equality signs hold in(2.6). For the first case, bya₁ ≤ · · · ≤ a_n ≤ b_n ≤ · · · ≤ b₁, we get a₁ = · · · = a_n, b₁ = · · · = b_n. For the second case, we have a_i1 −a_i2 = 0 = b_i1 −b_i2. Since1≤i₁ < i₂ ≤nandi₁, i₂ are arbitrary, we geta₁ =· · ·=a_n, b₁ =· · ·=b_n. Clearly, ifa₁ =· · ·=a_n, b₁ =· · ·=b_n, then(2.3)reduces to an equality.

Proof of Theorem 1.1. First we note that if

α=α_k :=



k⁻¹, k⁻¹, . . . , k⁻¹

| {z }

k

,0, . . . ,0



,

we obtain that

S(α_k, x) =f_k,n(x) and

(2.11) F(α_k) = logfk,n(a)

f_k,n(b).

By Lemma 2.2, we observe that F(α) is a Schur-convex(concave) function on Ωn. Using α_k+1 ≺α_kforα_k, α_k+1 ∈ Ω_nand the definition of Schur-convex(concave) functions, we have [8]

(2.12) F(α_k+1)≤(≥)F(α_k), k= 1, . . . , n−1.

It follows from(2.11)and(2.12)that(1.5)holds. Sinceα_k+1 6= α_k, combining this fact with Remark 1, we observe that the equality signs hold in (1.5) if and only if a₁ = · · · = a_n, b₁ =· · ·=b_n. This completes the proof of Theorem 1.1.

3. COROLLARY OF THEOREM1.1

Corollary 3.1. Let0< r <1, s≥1,0< a_i ≤2^−1/s, b_i = (1−a^s_i)^1/s, i= 1, . . . , n, f(t) =t^r, t∈(0,1). Then the inequalities in(1.5)are reversed.

Proof. Without loss of generality, we can assume that0 < a₁ ≤ · · · ≤ a_n. Byb_i = (1−a^s_i)^1/s and0< ai ≤2^−1/s(i= 1, . . . , n), we have

0< a₁ ≤ · · · ≤a_n ≤2^−1/s ≤b_n≤ · · · ≤b₁ <1.

Now we takeg(t) :=t+ (1−t^s)^1/s(0 < t ≤ 2^−1/s), sog⁰(t) = 1−(1−t^s)^(1/s)−1t^s−1 ≥ 0, i.e.,gis an increasing function. Thus

a₁+b₁ ≤ · · · ≤a_n+b_n.

It is easy to see thatf(t) = t^r > 0, f⁰(t) = rt^r−1 > 0, f⁰⁰(t) = r(r−1)t^r−2 < 0, f⁰⁰⁰(t) = r(r−1)(r−2)t^r−3 >0for anyt∈(0,1). By Theorem 1.1, Corollary 3.1 can be deduced. This

completes the proof.

Corollary 3.2. Leta∈(0,1/2]ⁿ. Writing

[AG;x]_k,n := Y

1≤i₁<···<i_k≤n

x_i1 +· · ·+x_ik k

! ¹ (ⁿk)

,

we have

A(a)

A(1−a) = [AG;a]_n,n [AG; 1−a]_n,n

≥ · · · ≥ [AG;a]_k+1,n

[AG; 1−a]_k+1,n ≥ [AG;a]_k,n [AG; 1−a]_k,n

≥ · · · ≥ [AG;a]_1,n

[AG; 1−a]_1,n = G(a) G(1−a). (3.1)

Equalities hold throughout if and only ifa₁ =· · ·=a_n. (Compare(3.1)with [7, 10, 11]) Proof. We chooses = 1in Corollary 3.1. Raising each term to the power of1/r and letting r→0in(1.5),(3.1)can be deduced. This ends the proof.

Corollary 3.3. Letf :I →Rbe such thatf(t) >0, f⁰(t) >0, f⁰⁰(t)>0, f⁰⁰⁰(t)<0for any t ∈I. LetΦ :I₀ → Ibe increasing andΨ :I₀ → Ibe decreasing, and suppose that Φ + Ψis increasing andsup Φ≤inf Ψ. Then

(3.2)

f

|I₀|⁻¹R

I0Φdt f

|I₀|⁻¹R

I0Ψdt ≤

R

I0f(Φ)dt R

I0f(Ψ)dt,

where|I0|is the length of the intervalI0. The inequality is reversed forf⁰⁰(t) <0, f⁰⁰⁰(t)> 0,

∀t∈I.

In fact, since (3.2)is an integral version of the inequality ^f(A(a))_f(A(b)) ≤ ^A(f(a))_A(f(b)), therefore(3.2) holds by Theorem 1.1.

According to Theorem 1.1, (1.5)implies inequalities (1.1), (1.2)and (3.1), and the impli- cation (3.1) to (1.4) is obvious. Consequently, Theorem 1.1 is a generalization of Jensen’s inequality(1.1), Peˇcari´c-Svrtan’s inequalities(1.2)and Fan’s inequality(1.4). Note that The- orem 1.1 contains a great number of inequalities as special cases. To save space we omit the details.

4. A MATRIX VARIANT

LetA= (a_ij)n×n(n ≥2)be a Hermite matrix of ordern. ThentrA= Pn

i=1a_iiis the trace ofA. As is well-known, there exists a unitary matrixU such thatA = Udiag(λ₁, . . . , λ_n)U^∗, whereU^∗ is the transpose conjugate matrix ofU and the components ofλ = (λ₁, . . . , λ_n)are the eigenvalues of A. ThustrA = λ₁ +· · ·+λ_n. Let λ ∈ Iⁿ. Then, for f : I → R, we definef(A) := Udiag(f(λ₁), . . . , f(λ_n))U^∗ (see [9]). Note thatdiag(λ₁, . . . , λ_n) = U^∗AU. Based on the above, we may use the following symbols: If, forA, we keep the elements on the cross points of thei₁, . . . , i_kth rows and thei₁, . . . , i_kth columns; replacing the other elements by nulls, then we denote this new matrix by A_i1···i_k. Clearly, we have tr[U^∗AU]_i1···i_k = λ_i1 +

· · ·+λ_ik. Thus we also define that f_k,n(A) := 1

n k

X

1≤i1<···<ik≤n

f

λ_i1 +· · ·+λ_ik k

= 1

n k

X

1≤i1<···<i_k≤n

f 1

ktr[U^∗AU]_i1···i_k

.

In particular, we have

f_1,n(A) = 1 n

n

X

i=1

f(λ_i) = 1

ntr(f(A));

f_n,n(A) = f

λ₁+· · ·+λ_n n

=f 1

ntrA

;

fn−1,n(A) = 1 n

n

X

i=1

f

trA−λ_i n−1

= 1 ntrf

E·trA−A n−1

,

whereE is a unit matrix. In fact, from U^∗

E·trA−A n−1

U = diag

trA−λ₁

n−1 , . . . ,trA−λ_n n−1

, we get

trf

E·trA−A n−1

=

n

X

i=1

f

trA−λ_i n−1

.

Based on the above facts and Theorem 1.1, we observe the following.

Theorem 4.1. Let I be an interval and let λ, µ ∈ Iⁿ. Suppose the components ofλ,µare the eigenvalues of Hermitian matricesAandB. If

(i) λ₁ ≤ · · · ≤λ_n≤µ_n≤ · · · ≤µ₁, λ₁+µ₁ ≤ · · · ≤λ_n+µ_n;

(ii) the functionf : I → R satisfiesf(t) > 0, f⁰(t) > 0, f⁰⁰(t) > 0, f⁰⁰⁰(t) < 0 for any t ∈I, and we have

f _n¹trA

f ¹_ntrB ≤ trf ^E·trA−A_n−1

trf ^E·trB−B_n−1 ≤ · · · ≤ f_k+1,n(A)

f_k+1,n(B) ≤ f_k,n(A)

f_k,n(B) ≤ · · · ≤ trf(A) trf(B).

The inequalities are reversed for f⁰⁰(t) < 0, f⁰⁰⁰(t) > 0, ∀t ∈ I. Equalities hold throughout if and only ifλ₁ =· · ·=λ_nandµ₁ =· · ·=µ_n.

Remark 2. IfI = (0,1/2],0 < λ₁ ≤ · · · ≤ λ_n ≤ 1/2, B =E −A, then the precondition (i) of Theorem 4.1 can be satisfied.

Remark 3. Lemma 2.2 possesses a general and meaningful result that should be an important theorem. Theorem 1.1 is only an application of Lemma 2.2.

Remark 4. Iff(t)<0, f⁰(t)<0for anyt∈I, then we can apply Theorem 1.1 to−f. Remark 5. In [12, 13], several applications on Jensen’s inequalities are displayed.

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