Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 2, Article 43, 2006
REVERSE REARRANGEMENT INEQUALITIES VIA MATRIX TECHNICS
JEAN-CHRISTOPHE BOURIN
8 RUE H ENRI D UREL
78510 T RIEL , F RANCE
bourinjc@club-internet.fr
Received 01 August, 2005; accepted 31 January, 2006
Communicated by F. Hansen
A BSTRACT. 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.
Key words and phrases: Trace inequalities; Rearrangement inequalities.
2000 Mathematics Subject Classification. 15A60.
1. R EVERSE R EARRANGEMENT I NEQUALITIES
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
ai
p≥
≥ q, i = 1, . . . , n,
bi
for some p, q > 0. Then,
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 n-dimensional 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.
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
233-05
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J EAN -C HRISTOPHE B OURIN
On the other hand, for all A,
|Tr A| ≤ Tr |A|,
(1.2)
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
X
(1.3)
kXV Y k1 =
xi yσ(i) .
Indeed, since |XV Y |2 ei = Y V ∗ X 2 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
n
X
ai bi = kABk1
and
i=1
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 the eigenvalues
and hi the corresponding unit eigenvectors. We have
kAV Bk1 ≤
n
X
kA · vi hi ⊗ hi · Bk1
i=1
=
n
X
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,
p+q
5
a1 b 2 + a2 b 1
.
√ = =
2 pq
4
a1 b 1 + a2 b 2
From the above, one easily derives:
Corollary 1.2. Let {ai }ni=1 and {bi }ni=1 be n-tuples of positive numbers with
bi ≤ ai ≤ pbi ,
J. Inequal. Pure and Appl. Math., 7(2) Art. 43, 2006
i = 1, . . . , n,
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R EVERSE R EARRANGEMENT I NEQUALITIES
for some p > 0. Then,
n
X
i=1
a↓i b↓i
3
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 p
H and let a0p
and b0 be the extremal eigenvalues of ZE . Then
0
0
0
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
1 0 1
a+b
1
a0 + b 0
√ =
√
t+
,
=
t + 0 ,
2
t
2
t
2 ab
2 a0 b 0
we infer
a+b
a0 + b 0
√ ≥ √
.
2 ab
2 a0 b 0
Therefore, it suffices to prove the lemma for ZE with E =
we may assume
√ span{h, Zh}. Hence,
2
2
dim H = 2, Z = ae1 ⊗ e1 + be2 ⊗ e2 and h = xe1 + ( 1 − x )e2 . Setting x = 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
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,
p+q
kAhk kBhk ≤ √ hAh, Bhi.
2 pq
Proof. Write h = B −1 f and apply Lemma 1.3.
Remark 1.5. Lemma 1.3 is nothing but a rephrasing of a Kantorovich inequality and Lemma
1.4 a rephrasing of Cassel’s Inequality:
Cassel’s inequality. For nonnegative n-tuples {ai }ni=1 , {bi }ni=1 and {wi }ni=1 with
ai
p≥
≥ q, i = 1, . . . , n,
bi
for some p, q > 0; it holds that
! 21
! 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
J. Inequal. Pure and Appl. Math., 7(2) Art. 43, 2006
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J EAN -C HRISTOPHE B OURIN
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
p+q
(1.4)
kak kbk ≤ √ ha, bi
2 pq
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].
Taking squares in Cassel’s inequality and using the convexity of t2 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
P
n
n
n
for all nonnegative n-tuples {ai }i=1 , {bi }i=1 and {wi }i=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 P
inequality:
Taking bi = 1/ai we get the (sharp) Kantorovich inequality: If p ≥ ai ≥ q > 0 and ni=1 wi =
1, then
n
n
X
X
(p + q)2
.
wi ai
wi a−1
≤
i
4pq
i=1
i=1
Let (Ω, P ) be a probability space. The above discussions shows a sharp result:
Proposition 1.6. 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 g(ω) dP.
f (ω) dP
g(ω) dP ≤
√
4 pq
Ω
Ω
Ω
2. R ELATED M ATRIX I NEQUALITIES AND C OMMENTS
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
kZABk ≤
a+b
√ 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:
J. Inequal. Pure and Appl. Math., 7(2) Art. 43, 2006
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R EVERSE R EARRANGEMENT I NEQUALITIES
5
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:
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,
p+q
kAZBk ≤ √ 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
by the simple fact that kST k = kT Sk for all Hermitians S, T , since kXk = kX ∗ k for all
X.
The previous theorem cannot be extended to normal operators Z, except in the case of the
trace norm:
Theorem 2.4. Let A, B > 0 with AB = BA and pI ≥ AB −1 ≥ qI for some p, q > 0 and let
Z be normal. Then,
p+q
kAZBk1 ≤ √ 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 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
a∗
a∗
and
≥q
p≥
b∗
b∗
for some p, q > 0. Then,
p+q
a∗ b ∗ + a∗ b ∗
≤ √ .
∗
∗
a b ∗ + a∗ b
2 pq
J. Inequal. Pure and Appl. Math., 7(2) Art. 43, 2006
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6
J EAN -C HRISTOPHE B OURIN
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
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∗
+ a pa∗
q
.
a∗ a∗
+ a∗qa∗
p
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.
We close with two open problems:
Problem 2.1. Find a direct proof of Theorem 1.1.
Problem 2.2. Let {ai }ni=1 and {bi }ni=1 be n-tuples of positive numbers. Find a suitable bound
for the difference
n
n
X
X
a↓i b↓i −
ai b i .
i=1
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.
R EFERENCES
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[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/
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J. Inequal. Pure and Appl. Math., 7(2) Art. 43, 2006
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