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volume 7, issue 2, article 43, 2006.

Received 01 August, 2005;

accepted 31 January, 2006.

Communicated by:F. Hansen

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Journal of Inequalities in Pure and Applied Mathematics

REVERSE REARRANGEMENT INEQUALITIES VIA MATRIX TECHNICS

JEAN-CHRISTOPHE BOURIN

8 rue Henri Durel 78510 Triel, France

EMail:bourinjc@club-internet.fr

c

2000Victoria University ISSN (electronic): 1443-5756 233-05

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Reverse Rearrangement Inequalities Via Matrix Technics

Jean-Christophe Bourin

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J. Ineq. Pure and Appl. Math. 7(2) Art. 43, 2006

http://jipam.vu.edu.au

Abstract

We give a reverse inequality to the most standard rearrangement inequality for sequences and we emphasize the usefulness of matrix methods to study classical inequalities.

2000 Mathematics Subject Classification:15A60.

Key words: Trace inequalities; Rearrangement inequalities.

1. Reverse Rearrangement Inequalities

We have the following reverse inequality to the most basic rearrangement inequality. Down arrows mean nonincreasing rearrangements.

Theorem 1.1. Let{a_i}ⁿ_i=1and{b_i}ⁿ_i=1 ben-tuples of positive numbers with p≥ a_i

b_i ≥q, i= 1, . . . , n, for somep, q >0. Then,

n

X

i=1

a^↓_ib^↓_i ≤ p+q 2√

pq

n

X

i=1

a_ib_i.

The proof uses matrix arguments. Indeed, Theorem 1.1 is a byproduct of some matrix inequalities which are given in Section2.

For the convenience of readers we recall some facts about the trace norm.

Capital letters A, B, . . . , Z, denote n-by-n matrices or operators on an n- dimensional Hilbert spaceH. LetX =U|X|be the polar decomposition ofX, soU is unitary and|X| = (X^∗X)^1/2. The trace norm ofX iskXk₁ = Tr|X|.

One may easily check that the trace norm is a norm: For anyX,Y, consider the polar decompositionX+Y =U|X+Y|. Then,

(1.1) kX+Yk₁ = Tr|X+Y|= TrU^∗(X+Y) = TrU^∗X+ TrU^∗Y.

On the other hand, for allA,

(1.2) |TrA| ≤Tr|A|,

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as it is shown by computing|TrA|in a basis of eigenvectors of|A|. From (1.1) and (1.2) we infer thatk · k1is a norm.

We need a simple fact: Given two diagonal positive matricesX = diag(x_i), Y = diag(y_i)and a permutation matrixV acting on the canonical basis{e_i}by V e_i =e_σ(i), we have

(1.3) kXV Yk₁ =X

x_iy_σ(i). Indeed, since

|XV Y|²e_i =Y V^∗X²V Y e_i = (x²_σ(i)y_i²)e_i, we obtain|XV Y|ei = (x_σ(i)yi)ei so that (1.3) holds.

Proof of Theorem1.1. Introduce the diagonal matricesA = diag(a_i)andB = diag(b_i). By the above discussion, we have

n

X

i=1

aibi =kABk1 and

n

X

i=1

a^↓_ib^↓_i =kAV Bk1

for some permutation matrixV. Hence we have to show that kAV Bk1 ≤ p+q

2√

pqkABk1. To this end consider the spectral representationV =P

iv_ih_i⊗h_i wherev_i are

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the eigenvalues andh_i the corresponding unit eigenvectors. We have kAV Bk1 ≤

n

X

i=1

kA·vihi⊗hi·Bk1

=

n

X

i=1

kAh_ik kBh_ik

≤ p+q 2√

pq

n

X

i=1

hAh_i, Bh_ii

= p+q 2√

pq

n

X

i=1

hh_i, ABh_ii

= p+q 2√

pqkABk₁,

where we have used the triangle inequality for the trace norm and Lemma 1.4 below.

The following example shows that equality can occur.

Example 1.1. Consider couples a1 = 2, a2 = 1 andb1 = 1/2, b2 = 1; then withp= 4,q= 1,

p+q 2√

pq = 5

4 = a1b2+a2b1

a₁b₁+a₂b₂. From the above, one easily derives:

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Corollary 1.2. Let{a_i}ⁿ_i=1and{b_i}ⁿ_i=1 ben-tuples of positive numbers with b_i ≤a_i ≤pb_i, i= 1, . . . , n,

for somep > 0. Then,

n

X

i=1

a^↓_ib^↓_i ≤ p+ 1 2√

p

n

X

i=1

a_ib_i.

Moreover, for evennand eachp, there aren-tuples for which equality occurs.

To obtain equality, consider an n-tuple {a_i} for which the first half terms equal √

pand the second half ones equal 1, and an n-tuple{b_i} for which the first half terms equal1and the second half ones equal1/√

p.

We turn to the lemmas necessary to complete the proof of Theorem 1.1.

Given a subspace E ⊂ H, denote by ZE the compression of Z ontoE, that is the restriction ofEZ toE whereEis the orthoprojection ontoE.

Lemma 1.3. Let Z > 0 with extremal eigenvalues a and b. Then, for every norm one vectorh,

kZhk ≤ a+b 2√

abhh, Zhi.

Proof. LetE be any subspace ofHand leta⁰andb⁰be the extremal eigenvalues of ZE. Then a ≥ a⁰ ≥ b⁰ ≥ b and, setting t = p

a/b, t⁰ = p

a⁰/b⁰, we have t ≥t⁰ ≥1. Sincet −→t+ 1/tincreases on[1,∞)and

a+b 2√

ab = 1 2

t+ 1

t

, a⁰+b⁰ 2√

a⁰b⁰ = 1 2

t⁰+ 1

t⁰

,

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we infer

a+b 2√

ab ≥ a⁰+b⁰ 2√

a⁰b⁰ .

Therefore, it suffices to prove the lemma forZEwithE = span{h, Zh}. Hence, we may assumedimH= 2,Z =ae₁⊗e₁+be₂⊗e₂andh=xe₁+(√

1−x²)e₂. Settingx² =ywe have

||Zh||

hh, Zhi =

pa²y+b²(1−y) ay+b(1−y) .

The right hand side attains its maximum on[0,1]aty=b/(a+b), and then

||Zh||

hh, Zhi = a+b 2√

ab, proving the lemma.

Lemma 1.4. Let A, B > 0 withAB = BAand pI ≥ AB⁻¹ ≥ qI for some p, q >0. Then, for every vectorh,

kAhk kBhk ≤ p+q 2√

pqhAh, Bhi.

Proof. Writeh =B⁻¹f and apply Lemma1.3.

Remark 1. Lemma1.3is nothing but a rephrasing of a Kantorovich inequality and Lemma1.4a rephrasing of Cassel’s Inequality:

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Cassel’s inequality. For nonnegativen-tuples{a_i}ⁿ_i=1,{b_i}ⁿ_i=1and{w_i}ⁿ_i=1with p≥ a_i

b_i ≥q, i= 1, . . . , n, for somep, q >0; it holds that

n

X

i=1

w_ia²_i

!¹₂ _n X

i=1

w_ib²_i

!¹₂

≤ p+q 2√

pq

n

X

i=1

w_ia_ib_i.

Of course it is a reverse inequality to the Cauchy-Schwarz inequality. To obtain it from Lemma1.4, one simply takesA= diag(a₁, . . . , a_n),B = diag(b₁, . . . , b_n) and h = (√

w₁, . . . ,√

w_n). If one lets a = (a₁, . . . , a_n)and b = (b₁, . . . , b_n) then Cassel’s inequality can be written as

(1.4) kak kbk ≤ p+q

2√

pqha, bi

for a suitable inner producth·,·i. It is then natural to search for conditions on a,bensuring that the above inequality remains valid withU a,U bfor all orthog- onal matricesU. This motivates a remarkable extension of Cassel’s inequality:

tDragomir’s inequality. For real vectorsa,b such thatha−qb, pb−ai ≥ 0 for some scalars p, q with pq > 0, inequality (1.4) holds. For this inequality and its complex version see [4], [5], [6].

Taking squares in Cassel’s inequality and using the convexity oft²we obtain:

n

X

i=1

w_ia_i

n

X

i=1

w_ib_i ≤ (√ p+√

q)² 4√

pq

n

X

i=1

w_ia_ib_i

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for all nonnegative n-tuples {a_i}ⁿ_i=1, {b_i}ⁿ_i=1 and {w_i}ⁿ_i=1 with Pn

i=1w_i = 1 andp≥ai/bi ≥qfor somep, q > 0. Though weaker than Cassel’s inequality, this is also a sharp inequality: Takingb_i = 1/a_i we get the (sharp) Kantorovich inequality: Ifp≥a_i ≥q >0andPn

i=1w_i = 1, then

n

X

i=1

w_ia_i

n

X

i=1

w_ia⁻¹_i ≤ (p+q)² 4pq .

Let (Ω, P) be a probability space. The above discussions shows a sharp result:

Proposition 1.5. Let f(ω) and g(ω) be measurable functions on Ω such that p≥f(ω)/g(ω)≥qfor somep, q >0. Then,

Z

Ω

f(ω) dP Z

Ω

g(ω) dP ≤ (√ p+√

q)² 4√

pq Z

Ω

f g(ω) dP.

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2. Related Matrix Inequalities and Comments

We dicovered the statements of Theorem 1.1 and its corollaries while investi- gating some matrix inequalities. Among those are inequalities for symmetric norms. Such a normk · k is characterized by the property thatkAk=kU AVk for allAand all unitariesU,V. The most basic inequality for symmetric norms is

kABk ≤ kBAk,

whenever the productABis normal. In [1] (see also [2]) we established:

Theorem 2.1. LetA,B such thatAB ≥ 0and letZ >0with extremal eigen- valuesaandb. Then, for every symmetric norm, the following sharp inequality holds

kZABk ≤ a+b 2√

abkBZAk.

By sharpness, we mean that we can findAandB such that equality occurs.

Note that letting A = B be a rank one projectionh⊗hwe recapture Lemma 1.3 which is the starting point of Theorem 1.1. From this theorem we derived several known Kantorovich type inequalities and also a sharp operator inequality:

Corollary 2.2. Let0 ≤A ≤ I and letZ > 0with extremal eigenvaluesa and b. Then,

AZA≤ (a+b)² 4ab Z.

Next, let us note that an immediate consequence of Theorem1.1is:

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Corollary 2.3. LetZ ≥0and letA, B >0withAB =BAandpI ≥AB⁻¹ ≥ qI for somep, q >0. Then, for all symmetric norms,

kAZBk ≤ p+q 2√

pqkZABk.

Proof. From Theorem2.1we get

kAZBk=kAB⁻¹(BZ ·B)k ≤ p+q 2√

pqkABZk= p+q 2√

pqkZABk.

by the simple fact that kSTk = kT Sk for all Hermitians S, T, since kXk = kX^∗kfor allX.

The previous theorem cannot be extended to normal operatorsZ, except in the case of the trace norm:

Theorem 2.4. LetA, B > 0withAB = BAandpI ≥ AB⁻¹ ≥ qI for some p, q >0and letZ be normal. Then,

kAZBk₁ ≤ p+q 2√

pqkZABk₁.

The proof is quite similar to that of Theorem1.1. Clearly Theorem1.1is a corollary of Theorem2.4.

Some comments. One aim of the paper is to place stress on the power of matrix methods in dealing with classical inequalities. This is apparent in the quite natural statement and proof of Cassel’s inequality via Lemma 1.4. We also note that from the matrix inequality of Theorem 2.4 we infer our reverse

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rearrangement inequality stated in Theorem 1.1. Having now at our disposal the good statement, it remains to find a direct proof without matrix arguments (in particular without using complex numbers via the spectral decomposition).

A first immediate simplification consists in noting that we can assume that a₁, . . . , a_n=a^↓₁, . . . , a^↓_n and b₁, . . . , b_n=b^↓_σ(1), . . . , b^↓_σ(n)

for a permutation σ. By decomposingσ in cycles we may assume that σ is a cycle. Equivalently we may assume that

a₁, . . . , a_n=a^↓_σ(1), . . . , a^↓_σ(n) and b₁, . . . , b_n=b^↓_σ(2), . . . , b^↓_σ(n), b^↓_σ(1) for a permutationσ. However, does it really simplify the problem ?

It is tempting to try to reduce the problem to the casen= 2. We have no idea of how to proceed. The casen = 2can be easily solved by elementary methods as it is shown in the next proposition. The proof shows that the inequality of Theorem1.1is sharp (and equality can occur whennis even).

Proposition 2.5. Leta^∗ ≥a∗ >0andb^∗ ≥b∗ >0with

p≥ a^∗

b_∗ and a∗

b^∗ ≥q for somep, q >0. Then,

a^∗b^∗+a∗b∗

a^∗b∗+a∗b^∗ ≤ p+q 2√

pq.

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Proof. First, fix a^∗, a∗ and renamed b^∗, b∗ by x, y respectively. We want to maximize

f(x, y) = a^∗x+a∗y a^∗y+a∗x on the domain

∆ =

(x, y) : x≥y, q ≤ a_∗

x ≤p, q≤ a^∗ y ≤p

that is,

∆ =

(x, y) : x≥y, a∗

p ≤x≤ a∗

q , a^∗

p ≤y≤ a^∗ q

. Thus∆is a triangle (more precisely a half-square) with vertices

(a^∗/p, a^∗/p) (a∗/q, a∗/q) (a∗/q, a^∗/p).

On∆we have∂f /∂x > 0and∂f /∂y <0. This shows thatftakes its maximun in∆at(a∗/q, a^∗/p). The value is then

a^∗a∗

q +^a^∗_p^a^∗

a^∗a^∗

p +^a^∗_q^a^∗.

Next, observe that in our inequality we can take a^∗ = 1. Hence, lettinga∗ = t, we have to check that

max

t∈[0,1]

(¹_p +¹_q)t

1

p +^t_q² = p+q 2√

pq.

Considering the derivative, we see that the maximum is attained att = p q/p and we obtain the expected value.

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We close with two open problems:

Problem 1. Find a direct proof of Theorem1.1.

Problem 2. Let {a_i}ⁿ_i=1 and {b_i}ⁿ_i=1 be n-tuples of positive numbers. Find a suitable bound for the difference

n

X

i=1

a^↓_ib^↓_i −

n

X

i=1

a_ib_i.

In the research/survey paper [3] we consider matrix proofs and several ex- tensions of some classical inequalities of Chebyshev, Grüss and Kantorovich type.

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References

[1] J.-C. BOURIN, Symmetric norms and reverse inequalities to Davis and Hansen-Pedersen characterizations of operator convexity, Math. Ineq.

Appl., 9(1) (2006), xxx-xxx.

[2] J.-C. BOURIN, Compressions, Dilations and Matrix Inequalities, RGMIA Monographs, Victoria University, Melbourne 2004. [ONLINE: http://

rgmia.vu.edu.au/monographs].

[3] J.-C. BOURIN, Matrix versions of some classical inequalities, Linear Alge- bra Appl., (2006), in press.

[4] S.S. DRAGOMIR, Reverse of Schwarz, triangle and Bessel Inequalities, RGMIA Res. Rep. Coll., 6(Supp.) (2003), Art. 19. [ONLINE: http://

rgmia.vu.edu.au/v6(E).html]

[5] S.S. DRAGOMIR, A survey on Cauchy-Bunyakowsky-Schwarz type dis- crete inequalities, J. Inequal. Pure and Appl. Math., 4(3) (2003), Art.

63. [ONLINE: http://jipam.vu.edu.au/article.php?sid=

301]

[6] N. ELEZOVI Ć, L. MARANGUNI Ć AND J.E. PE ˇCARI Ć, Unified treate- ment of complemented Schwarz and Grüss inequalities in inner product spaces, Math. Ineq. Appl., 8(2) (2005), 223–231.