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

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Jensen-Pe ˇcari ´c-Svrtan-Fan Type Inequalities

Chaobang Gao and Jiajin Wen vol. 9, iss. 3, art. 74, 2008

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INEQUALITIES OF JENSEN-PE ˇ CARI ´ C-SVRTAN-FAN TYPE

CHAOBANG GAO AND JIAJIN WEN

College Of Mathematics And Information Science Chengdu University

Chengdu, 610106, P.R. China

EMail:kobren427@163.com wenjiajin623@163.com

Received: 22 January, 2007

Accepted: 06 July, 2008

Communicated by: S.S. Dragomir 2000 AMS Sub. Class.: 26D15, 26E60.

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

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Close Abstract: By using the theory of majorization, the following inequalities of Jensen-

Peˇcari´c-Svrtan-Fan type are established: LetI be an interval, f : I → R andt∈I, x, a, b∈Iⁿ. Ifa1 ≤ · · · ≤an ≤bn≤ · · · ≤b1, a1+b1 ≤ · · · ≤ an+bn;f(t)>0, f⁰(t)>0, f⁰⁰(t)>0, f⁰⁰⁰(t)<0for anyt∈I,then

f(A(a))

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

fn,n(b) ≤ · · · ≤ f_k+1,n(a)

fk+1,n(b) ≤ f_k,n(a) fk,n(b)

≤ · · · ≤ f_1,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.

Acknowledgements: This work is supported by the Natural Science Foundation of China (10671136) and the Natural Science Foundation of Sichuan Province Edu- cation Department (07ZA207).

The authors are deeply indebted to anonymous referees for many useful com- ments 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.

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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+· · ·+x_n

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

f_k,n(x) := 1

n k

X

1≤i₁<···<i_k≤n

f

x_i₁ +· · ·+x_i_k k

, k= 1, . . . , n.

Jensen’s inequality states that: Let f : I → R be 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:

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

≤ · · · ≤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.

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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. LetI be an interval. Iff :I →R, a, b∈Iⁿ(n ≥2)and (i) a₁ ≤ · · · ≤a_n ≤b_n ≤ · · · ≤b₁, a₁+b₁ ≤ · · · ≤a_n+b_n;

(ii) f(t)>0, f⁰(t)>0, f⁰⁰(t)>0, f⁰⁰⁰(t)<0for anyt ∈I, then

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) (1.5)

≤ · · · ≤ f1,n(a)

f_1,n(b) = A(f(a)) A(f(b)).

The inequalities are reversed for f⁰⁰(t) < 0, f⁰⁰⁰(t) > 0,∀t ∈ I. The equality signs hold if and only ifa₁ =· · ·=a_nandb₁ =· · ·=b_n.

In Section3, several interesting results of Ky Fan shall be deduced. In Section4, the matrix variant of(1.5)will be established.

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2. Proof of Theorem 1.1

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(α1xi1 +· · ·+αnxin),

whereP

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

S(α, a) S(α, b)

, a, b∈Iⁿ,

ui(x) := α1xi1 +α2xi2+

n

X

j=3

αjxij,

v_i(x) := α₁x_i₂ +α₂x_i₁+

n

X

j=3

α_jx_i_j, i= (i₁, i₂, . . . , i_n).

Then there existξi(a) between ui(a) and vi(a), and ξi(b) between ui(b) and vi(b) such that

(2.1) (α₁−α₂) ∂F

∂α₁ − ∂F

∂α₂

= 1 n!

X

i3···i_n

X

1≤i₁<i2≤i_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)

,

whereP

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

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Proof. Note the following identities:

S(α, x) = 1 n!

X

i3···in

X

1≤i16=i2≤n

f(α₁x_i₁ +· · ·+α_nx_i_n)

= 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

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= 1 n!

X

i3···in

X

1≤i1<i2≤n

[f⁰(u_i(a))−f⁰(v_i(a))][u_i(a)−v_i(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···in

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 Theorem1.1,F is a Schur-convex function or a Schur-concave function onΩ_n, whereF is defined by Lemma2.1.

Proof. It is easy to see thatΩn is a symmetric convex set andF is a differentiable symmetric function onΩn. To prove thatF is a Schur-convex function onΩn, 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 Lemma2.1to 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 thataj ≤bj (j = 1,2, . . . , n)and the inequalities:

(2.5) 1

S(α, a) ≥ 1

S(α, b) >0.

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By the given condition (i) of Theorem1.1and1≤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{ui(a), vi(a)} ≤ξi(a)

≤max{u_i(a), v_i(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 onΩn.

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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 Lemma2.2.

Remark 1. Whenα₁ 6=α₂, there is equality in(2.3)ifa₁ =· · ·=a_nandb₁ =· · ·= 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 geta₁ =· · ·=a_n, b₁ =· · ·=b_n. For the second case, we havea_i₁ −a_i₂ = 0 =b_i₁ −b_i₂. Since1 ≤i₁ < i₂ ≤ nand i₁, 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 Theorem1.1. First we note that if

α=α_k :=



k⁻¹, k⁻¹, . . . , k⁻¹

| {z }

k

,0, . . . ,0



,

we obtain that

S(αk, x) = fk,n(x) and

(2.11) F(α_k) = logf_k,n(a)

f_k,n(b).

By Lemma2.2, we observe thatF(α)is a Schur-convex(concave) function on Ω_n. Usingαk+1 ≺ αk for αk, αk+1 ∈ Ωn and the definition of Schur-convex(concave)

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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 Remark1, 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 Theorem1.1.

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3. Corollary of Theorem 1.1

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 that 0 < a₁ ≤ · · · ≤ a_n. By b_i = (1−a^s_i)^1/sand0< a_i ≤2^−1/s(i= 1, . . . , n), we have

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

Now we take g(t) := t + (1 − t^s)^1/s(0 < t ≤ 2^−1/s), so g⁰(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 Theorem1.1, Corollary3.1 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_i_k 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)

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Equalities hold throughout if and only if a₁ = · · · = a_n. (Compare (3.1) with [7,10,11])

Proof. We chooses= 1in Corollary3.1. Raising each term to the power of1/rand lettingr →0in(1.5),(3.1)can be deduced. This ends the proof.

Corollary 3.3. Let f : I → R be such that f(t) > 0, f⁰(t) > 0, f⁰⁰(t) > 0, f⁰⁰⁰(t) < 0 for any t ∈ I. Let Φ : I₀ → I be increasing and Ψ : I₀ → I be decreasing, and suppose thatΦ + 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(f(b))^(a)), therefore(3.2)holds by Theorem1.1.

According to Theorem1.1, (1.5)implies inequalities(1.1), (1.2)and(3.1), and the implication (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 Theorem 1.1 contains a great number of inequalities as special cases. To save space we omit the details.

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4. A Matrix Variant

LetA = (a_ij)n×n(n ≥ 2)be a Hermite matrix of ordern. Then trA = Pn i=1a_ii is the trace ofA. As is well-known, there exists a unitary matrix U such thatA = Udiag(λ₁, . . . , λ_n)U^∗, where U^∗ is the transpose conjugate matrix of U and the components ofλ= (λ1, . . . , λn)are the eigenvalues ofA. ThustrA=λ1+· · ·+λn. Letλ ∈Iⁿ. Then, forf :I → R, we definef(A) := Udiag(f(λ1), . . . , f(λn))U^∗ (see [9]). Note that diag(λ₁, . . . , λ_n) = U^∗AU. Based on the above, we may use the following symbols: If, for A, we keep the elements on the cross points of the i₁, . . . , i_kth rows and thei₁, . . . , i_kth columns; replacing the other elements by nulls, then we denote this new matrix byA_i₁···i_k. Clearly, we havetr[U^∗AU]_i₁···i_k =λ_i₁ +

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

n k

X

1≤i1<···<i_k≤n

f

λi1 +· · ·+λi_k

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

,

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

E·trA−A n−1

U = diag

trA−λ1

n−1 , . . . ,trA−λn

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 Theorem1.1, we observe the following.

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

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

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

f _n¹trA

f _n¹trB ≤ trf ^E·trA−A_n−1

trf ^E·trB−B_n−1 ≤ · · · ≤ fk+1,n(A)

f_k+1,n(B) ≤ fk,n(A)

f_k,n(B) ≤ · · · ≤ trf(A) trf(B). The inequalities are reversed forf⁰⁰(t)<0, f⁰⁰⁰(t)>0,∀t∈I. Equalities hold throughout if and only ifλ₁ =· · ·=λ_nandµ₁ =· · ·=µ_n.

Remark 2. If I = (0,1/2], 0 < λ1 ≤ · · · ≤ λn ≤ 1/2, B = E −A, then the precondition (i) of Theorem4.1can be satisfied.

Remark 3. Lemma2.2 possesses a general and meaningful result that should be an important theorem. Theorem1.1is only an application of Lemma2.2.

Remark 4. Iff(t)< 0, f⁰(t)< 0for anyt ∈ I, then we can apply Theorem1.1 to

−f.

Remark 5. In [12,13], several applications on Jensen’s inequalities are displayed.

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