Journal of Inequalities in Pure and
Applied Mathematics
REVERSE REARRANGEMENT INEQUALITIES VIA
MATRIX TECHNICS
volume 7, issue 2, article 43,
2006.
JEAN-CHRISTOPHE BOURIN
8 rue Henri Durel
78510 Triel, France
Received 01 August, 2005;
accepted 31 January, 2006.
Communicated by: F. Hansen
EMail: bourinjc@club-internet.fr
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
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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.
Contents
1
Reverse Rearrangement Inequalities . . . . . . . . . . . . . . . . . . . . . 3
2
Related Matrix Inequalities and Comments . . . . . . . . . . . . . . 10
References
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Jean-Christophe Bourin
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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 {ai }ni=1 and {bi }ni=1 be n-tuples of positive numbers with
p≥
ai
≥ q,
bi
i = 1, . . . , n,
for some p, q > 0. Then,
Reverse Rearrangement
Inequalities Via Matrix Technics
n
X
i=1
a↓i b↓i
n
p+q X
≤ √
ai b i .
2 pq i=1
The proof uses matrix arguments. Indeed, Theorem 1.1 is a byproduct of
some matrix inequalities which are given in Section 2.
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 ndimensional Hilbert space H. Let X = U |X| be the polar decomposition of X,
so U is unitary and |X| = (X ∗ X)1/2 . The trace norm of X is kXk1 = Tr |X|.
One may easily check that the trace norm is a norm: For any X, Y , consider the
polar decomposition X + Y = U |X + Y |. Then,
(1.1)
kX + Y k1 = Tr |X + Y | = Tr U ∗ (X + Y ) = Tr U ∗ X + Tr U ∗ Y.
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On the other hand, for all A,
J. Ineq. Pure and Appl. Math. 7(2) Art. 43, 2006
(1.2)
|Tr A| ≤ Tr |A|,
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as it is shown by computing |Tr A| in a basis of eigenvectors of |A|. From (1.1)
and (1.2) we infer that k · k1 is a norm.
We need a simple fact: Given two diagonal positive matrices X = diag(xi ),
Y = diag(yi ) and a permutation matrix V acting on the canonical basis {ei } by
V ei = eσ(i) , we have
kXV Y k1 =
(1.3)
X
xi yσ(i) .
Indeed, since
2
∗
2
|XV Y | ei = Y V X V Y ei =
(x2σ(i) yi2 )ei ,
we obtain |XV Y |ei = (xσ(i) yi )ei so that (1.3) holds.
Proof of Theorem 1.1. Introduce the diagonal matrices A = diag(ai ) and B =
diag(bi ). By the above discussion, we have
Reverse Rearrangement
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Jean-Christophe Bourin
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n
X
i=1
ai bi = kABk1
and
n
X
a↓i b↓i = kAV Bk1
i=1
for some permutation matrix V . Hence we have to show that
p+q
kAV Bk1 ≤ √ kABk1 .
2 pq
P
To this end consider the spectral representation V = i vi hi ⊗ hi where vi are
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the eigenvalues and hi the corresponding unit eigenvectors. We have
kAV Bk1 ≤
=
n
X
i=1
n
X
kA · vi hi ⊗ hi · Bk1
kAhi k kBhi k
i=1
n
p+q X
≤ √
hAhi , Bhi i
2 pq i=1
n
p+q X
= √
hhi , ABhi i
2 pq i=1
p+q
= √ kABk1 ,
2 pq
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 and b1 = 1/2, b2 = 1; then
with p = 4, q = 1,
5
a1 b 2 + a2 b 1
p+q
.
√ = =
2 pq
4
a1 b 1 + a2 b 2
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From the above, one easily derives:
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Corollary 1.2. Let {ai }ni=1 and {bi }ni=1 be n-tuples of positive numbers with
bi ≤ ai ≤ pbi ,
i = 1, . . . , n,
for some p > 0. Then,
n
X
i=1
a↓i b↓i
n
p+1 X
≤ √
ai b i .
2 p i=1
Moreover, for even n and each p, there are n-tuples for which equality occurs.
To obtain equality, consider an n-tuple {ai } for which the first half terms
√
equal p and the second half ones equal 1, and an n-tuple {bi } for which the
√
first half terms equal 1 and the second half ones equal 1/ 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 onto E, that is
the restriction of EZ to E where E is the orthoprojection onto E.
Lemma 1.3. Let Z > 0 with extremal eigenvalues a and b. Then, for every
norm one vector h,
a+b
kZhk ≤ √ hh, Zhi.
2 ab
Proof. Let E be any subspace of H and let a0 and p
b0 be the extremal
p eigenvalues
0
0
0
of ZE . Then a ≥ a ≥ b ≥ b and, setting t = a/b, t = a0 /b0 , we have
t ≥ t0 ≥ 1. Since t −→ t + 1/t increases on [1, ∞) and
1
a0 + b 0
a+b
1
1 0 1
√ =
√
t+
,
t + 0 ,
=
2
t
2
t
2 ab
2 a0 b 0
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we infer
a+b
a0 + b 0
√ ≥ √
.
2 ab
2 a0 b 0
Therefore, it suffices to prove the lemma for ZE with E = span{h, Zh}.
√ Hence,
we may assume dim H = 2, Z = ae1 ⊗e1 +be2 ⊗e2 and h = xe1 +( 1 − x2 )e2 .
Setting x2 = y we have
p
a2 y + b2 (1 − y)
||Zh||
=
.
hh, Zhi
ay + b(1 − y)
The right hand side attains its maximum on [0, 1] at y = b/(a + b), and then
||Zh||
a+b
= √ ,
hh, Zhi
2 ab
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Jean-Christophe Bourin
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proving the lemma.
Lemma 1.4. Let A, B > 0 with AB = BA and pI ≥ AB −1 ≥ qI for some
p, q > 0. Then, for every vector h,
kAhk kBhk ≤
p+q
√ hAh, Bhi.
2 pq
Proof. Write h = B −1 f and apply Lemma 1.3.
Remark 1. Lemma 1.3 is nothing but a rephrasing of a Kantorovich inequality
and Lemma 1.4 a rephrasing of Cassel’s Inequality:
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Cassel’s inequality. For nonnegative n-tuples {ai }ni=1 , {bi }ni=1 and {wi }ni=1 with
p≥
ai
≥ q,
bi
i = 1, . . . , n,
for some p, q > 0; it holds that
! 12
! 12
n
n
n
X
X
p+q X
2
2
wi ai
wi bi
≤ √
wi ai bi .
2
pq
i=1
i=1
i=1
Of course it is a reverse inequality to the Cauchy-Schwarz inequality. To obtain
it from Lemma 1.4, one simply takes A = diag(a1 , . . . , an ), B = diag(b1 , . . . , bn )
√
√
and h = ( w1 , . . . , wn ). If one lets a = (a1 , . . . , an ) and b = (b1 , . . . , bn )
then Cassel’s inequality can be written as
(1.4)
Reverse Rearrangement
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Jean-Christophe Bourin
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p+q
kak kbk ≤ √ ha, bi
2 pq
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for a suitable inner product h·, ·i. It is then natural to search for conditions on
a, b ensuring that the above inequality remains valid with U a, U b for all orthogonal matrices U . This motivates a remarkable extension of Cassel’s inequality:
tDragomir’s inequality. For real vectors a, b such that ha − 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].
2
Taking squares in Cassel’s inequality and using the convexity of t we obtain:
√
√
n
n
n
X
X
( p + q)2 X
wi ai bi
wi ai
wi bi ≤
√
4 pq
i=1
i=1
i=1
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P
for all nonnegative n-tuples {ai }ni=1 , {bi }ni=1 and {wi }ni=1 with ni=1 wi = 1
and p ≥ ai /bi ≥ q for some p, q > 0. Though weaker than Cassel’s inequality,
this is also a sharp inequality: Taking
Pnbi = 1/ai we get the (sharp) Kantorovich
inequality: If p ≥ ai ≥ q > 0 and i=1 wi = 1, then
n
X
i=1
wi ai
n
X
i=1
wi a−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) ≥ q for some p, q > 0. Then,
Z
Z
Z
√
√
( p + q)2
f (ω) dP
g(ω) dP ≤
f g(ω) dP.
√
4 pq
Ω
Ω
Ω
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2.
Related Matrix Inequalities and Comments
We dicovered the statements of Theorem 1.1 and its corollaries while investigating some matrix inequalities. Among those are inequalities for symmetric
norms. Such a norm k · k is characterized by the property that kAk = kU AV k
for all A and all unitaries U , V . The most basic inequality for symmetric norms
is
kABk ≤ kBAk,
whenever the product AB is normal. In [1] (see also [2]) we established:
Theorem 2.1. Let A, B such that AB ≥ 0 and let Z > 0 with extremal eigenvalues a and b. Then, for every symmetric norm, the following sharp inequality
holds
a+b
kZABk ≤ √ kBZAk.
2 ab
By sharpness, we mean that we can find A and B such that equality occurs.
Note that letting A = B be a rank one projection h ⊗ h we 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:
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Corollary 2.2. Let 0 ≤ A ≤ I and let Z > 0 with extremal eigenvalues a and
b. Then,
(a + b)2
AZA ≤
Z.
4ab
Next, let us note that an immediate consequence of Theorem 1.1 is:
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Corollary 2.3. Let Z ≥ 0 and let A, B > 0 with AB = BA and pI ≥ AB −1 ≥
qI for some p, q > 0. Then, for all symmetric norms,
kAZBk ≤
p+q
√ kZABk.
2 pq
Proof. From Theorem 2.1 we get
p+q
p+q
kAZBk = kAB −1 (BZ · B)k ≤ √ kABZk = √ kZABk.
2 pq
2 pq
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by the simple fact that kST k = kT Sk for all Hermitians S, T , since kXk =
kX ∗ k for all X.
Jean-Christophe Bourin
The previous theorem cannot be extended to normal operators Z, except in
the case of the trace norm:
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Theorem 2.4. Let A, B > 0 with AB = BA and pI ≥ AB
p, q > 0 and let Z be normal. Then,
kAZBk1 ≤
−1
≥ qI for some
p+q
√ kZABk1 .
2 pq
The proof is quite similar to that of Theorem 1.1. Clearly Theorem 1.1 is a
corollary of Theorem 2.4.
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 , . . . , an = a↓1 , . . . , a↓n
and
b1 , . . . , bn = b↓σ(1) , . . . , b↓σ(n)
for a permutation σ. By decomposing σ in cycles we may assume that σ is a
cycle. Equivalently we may assume that
a1 , . . . , an = a↓σ(1) , . . . , a↓σ(n)
and
b1 , . . . , bn = b↓σ(2) , . . . , b↓σ(n) , b↓σ(1)
for a permutation σ. However, does it really simplify the problem ?
It is tempting to try to reduce the problem to the case n = 2. We have no idea
of how to proceed. The case n = 2 can be easily solved by elementary methods
as it is shown in the next proposition. The proof shows that the inequality of
Theorem 1.1 is sharp (and equality can occur when n is even).
Proposition 2.5. Let a∗ ≥ a∗ > 0 and b∗ ≥ b∗ > 0 with
p≥
a∗
b∗
and
a∗
≥q
b∗
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for some p, q > 0. Then,
a∗ b ∗ + a∗ b ∗
p+q
≤ √ .
∗
∗
a b ∗ + a∗ b
2 pq
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Proof. First, fix a∗ , a∗ and renamed b∗ , b∗ by x, y respectively. We want to
maximize
a∗ x + a∗ y
f (x, y) = ∗
a y + a∗ x
on the domain
a∗
a∗
∆ = (x, y) : x ≥ y, q ≤
≤ p, q ≤
≤p
x
y
that is,
∆=
a∗
a∗ a∗
a∗
(x, y) : x ≥ y,
≤x≤ ,
≤y≤
p
q p
q
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.
Jean-Christophe Bourin
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 > 0 and ∂f /∂y < 0. This shows that f takes its maximun
in ∆ at (a∗ /q, a∗ /p). The value is then
a∗ a∗
q
a∗ a∗
p
+
+
a∗ a∗
p
a∗ a∗
q
.
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∗
Next, observe that in our inequality we can take a = 1. Hence, letting a∗ = t,
we have to check that
( p1 + 1q )t
p+q
max 1 t2 = √ .
t∈[0,1]
2 pq
+ q
p
p
Considering the derivative, we see that the maximum is attained at t = 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 Theorem 1.1.
Problem 2. Let {ai }ni=1 and {bi }ni=1 be n-tuples of positive numbers. Find a
suitable bound for the difference
n
X
i=1
a↓i b↓i
−
n
X
ai b i .
i=1
In the research/survey paper [3] we consider matrix proofs and several extensions 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 Algebra 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 discrete inequalities, J. Inequal. Pure and Appl. Math., 4(3) (2003), Art.
63. [ONLINE: http://jipam.vu.edu.au/article.php?sid=
301]
[6] N. ELEZOVIĆ, L. MARANGUNIĆ AND J.E. PEČARIĆ, Unified treatement of complemented Schwarz and Grüss inequalities in inner product
spaces, Math. Ineq. Appl., 8(2) (2005), 223–231.
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