Linear Algebra and its Applications 434 (2011) 636–640
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Linear Algebra and its Applications
journal homepage: www.elsevier.com/locate/laa
Generalized matrix version of reverse Hölder inequality
Rupinderjit Kaur a ,∗ , Mandeep Singh a , Jaspal Singh Aujla b
a
b
Department of Mathematics, Sant Longowal Institute of Engineering and Technology, Longowal 148106, Punjab, India
Department of Mathematics, National Institute of Technology, Jalandhar 144011, Punjab, India
A R T I C L E
I N F O
Article history:
Received 13 July 2010
Accepted 9 September 2010
Available online 13 October 2010
A B S T R A C T
A generalized matrix version of reverse Cauchy–Schwarz/Hölder inequality is proved. This includes the recent results proved by Bourin,
Fujii, Lee, Niezgoda and Seo.
© 2010 Elsevier Inc. All rights reserved.
Submitted by C.K. Li
AMS classification:
15A18
15A42
15A60
47A30
47A63
47A64
Keywords:
Positive definite matrices
Matrix means
Positive linear maps
Hölder/Cauchy–Schwarz reverse inequality
1. Introduction
In what follows, the capital letters A, B, C, . . . denote the n × n (n arbitrary but fixed) matrices over
the algebra of complex numbers, i.e. elements of Mn (C). By A B (A > B) we mean that A − B is
positive semidefinite (positive definite). A linear map φ : Mn (C) → Mm (C) is called positive if it maps
positive definite matrices into positive definite matrices. We use the notation ·, · to denote the usual
inner product in Cn and || · || denotes the vector norm of a vector in Cn with respect to this inner
∗ Corresponding author.
E-mail addresess: rupinder− grewal− [email protected] (R. Kaur), [email protected] (M. Singh), [email protected] (J. Singh
Aujla).
0024-3795/$ - see front matter © 2010 Elsevier Inc. All rights reserved.
doi:10.1016/j.laa.2010.09.027
R. Kaur et al. / Linear Algebra and its Applications 434 (2011) 636–640
637
product. It is well known that Mn (C) is an inner product space where the canonical inner product is
given by (A, B) = Tr(AB∗ ) (Here Tr(X ) denotes trace of X).
The axiomatic theory for connections and means of pairs of positive matrices have been developed
by Nishio and Ando [8] and Kubo and Ando [5]. A binary operation σ defined on the set of positive
definite matrices is called a connection if
(i) A C, B D implies Aσ B C σ D,
(ii) C (Aσ B)C (CAC )σ (CBC ),
(iii) Ak ↓ A and Bk ↓ B imply Ak σ Bk ↓ Aσ B.
A mean is a connection with normalization condition,
(iv) I σ I = I .
Kubo and Ando [5] proved the existence of an affine order isomorphism between the class of connections and the class of positive operator monotone functions on R+ . This isomorphism σ ↔ f is
characterized by the relation
Aσ B = A1/2 f A−1/2 BA−1/2 A1/2 .
The operator means corresponding to operator monotone functions t α , 0 < α < 1 are called weighted
geometric means and are denoted by #α . In particular #1/2 is called geometric mean.
In Section 2 we shall prove general results which unify and include the recent results proved by
Bourin, Fujii, Lee, Niezgoda and Seo in [3,4,6,7].
2. Main results
We begin this section by stating our main results.
Theorem 1. Let A, B > 0 be such that aA B bA for some scalars 0
map. Then for any connection σ ,
φ(A)σ φ(B) where ω
=
1
ω
< b a and let φ be a positive linear
φ(Aσ B),
f (a)−f (b)
(a−b)f (c ) for some fixed c
∈ (b, a) and f is the representing function of σ.
Theorem 2. Let A, B > 0 be such that aA B bA for some scalars 0
map. Then for any connection σ ,
< b a and let φ be a positive linear
φ(A)σ φ(B) − φ(Aσ B) −g (t0 )φ(A),
where g (t )
= μt + ν − f (t ), t0 a fixed point in (b, a) and f is the representing function of σ , μ =
af (b)−bf (a)
and ν =
.
a−b
f (a)−f (b)
a −b
To prove Theorems 1 and 2, we need the following lemmas.
Lemma 3. Let Z
Then
> 0 be such that aI Z bI for some scalars 0 < b a and let h ∈ Cn be a unit vector.
ωf (Zh, h) f (Z )h, h,
where ω
=
f (a)−f (b)
(a − b )f (c ) , c
∈ (b, a) and f is a positive operator monotone function on R+ .
Proof. First notice that the result is trivially true if b = a. Indeed a → b implies c → b and hence
ω → 1. Moreover, Z reduces to a scalar matrix in this case and so f (Zh, h) = f (Z )h, h = f (b). We
may therefore assume without loss of generality that 0 < b < a.
f (a)−f (b)
af (b)−bf (a)
Consider the line μt + ν where μ =
and ν =
. Note that f (t ) and the line μt +
a −b
a −b
ν intersect at the points (a, f (a)) and (b, f (b)). f being operator monotone, is concave and strictly
increasing on R+ . Thus, using the concavity of f (t ) we see that
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R. Kaur et al. / Linear Algebra and its Applications 434 (2011) 636–640
μt + ν f (t )
(1)
∈ [b, a]. By Lagrange’s Mean Value Theorem there exists a point d ∈
We can write d = β a + (1 − β)b for some β ∈ (0, 1). Let
for all t
f (a)−f (b)
.
a− b
Fα (t )
(b, a) with f (d)
=
= μt + ν − ωα f (t ),
f (a)−f (b)
(a−b)f (α a+(1−α)b) for 0 α 1. Fα (t ) is a convex function of t with minima at t = α a +
(1 − α)b, since Fα (α a + (1 − α)b) = 0. Thus F0 (t ) attains its minimum at t = b in [b, a]. Note that
F0 (b) = f (b)(1 − ω0 ) is non-negative. This follows from the facts that f (t ) is decreasing in [b, a] and
so f (b) f (d) which imply
where ωα
=
f (a) − f (b)
f (a) − f (b)
= 1.
(a − b)f (b) (a − b)f (d)
ω0 =
Thus F0 (t ) 0 for all t ∈ [b, a]. Further note that ωβ = 1, and therefore it follows from (1) that Fβ (t ) 0
for all t ∈ [b, a]. Since the function α → Fα is continuous, it follows that there exists at least one
α ∈ (0, 1) such that Fα (t ) 0 for all t ∈ [b, a]. Let α0 = Sup{α ∈ (0, 1) : Fα (t ) 0 for all t ∈ [b, a]}.
f (a)−f (b)
Choose c = α0 a + (1 − α0 )b and ω = ωα0 = (a−b)f (c) . Then Fα0 (t ) 0 for all t ∈ [b, a]. This implies
ωf (t ) μt + ν
(2)
Since aI Z bI and h is a unit vector, we have a h∗ Zh b, i.e,
for all t ∈ [b, a]. Let t = Zh, h.
a t b. Therefore putting this t in (2) we have
ωf (Zh, h) μZh, h + ν f (Z )h, h.
The second inequality in the above inequality follows from inequality (1). This completes the proof.
Lemma 4. Let A, B
vector. Then
> 0 be such that aA B bA for some scalars 0 < b a and let h ∈ Cn be a unit
(Aσ B)h, h Ah, hσ Bh, h ω−1 (Aσ B)h, h
for all connections σ with representing function f and ω is as in Lemma 3.
∈ Cn , x =
/ 0 and h = ||xx|| . Let Z = A−1/2 BA−1/2 . Since aA B bA which implies
−
1/2
−
1/2
bI, so aI Z bI. By concavity of f (t ), we have
BA
aI A
Proof. Let x
f (Z )
x
,
x
||x|| ||x||
f
Z
x
,
x
(3)
||x|| ||x||
(see [2, p. 281]). On using Lemma 3, we have
1
ω −1
f
Zx, x f (Z )x, x.
2
||x||
||x||2
Combining (3) and (4) we obtain,
1
1
f (Z )x, x f
Zx, x ||x||2
||x||2
or equivalently
f (Z )x, x ||x|| f
2
1
||x||2
(4)
ω −1
f (Z )x, x,
||x||2
Zx, x ω−1 f (Z )x, x.
(5)
R. Kaur et al. / Linear Algebra and its Applications 434 (2011) 636–640
639
Replacing Z by A−1/2 BA−1/2 and x by A1/2 h, in (5) we get,
A1/2 f (A−1/2 BA−1/2 )A1/2 h, h Ah, hf (Ah, h−1 Bh, h)
ω−1 A1/2 f (A−1/2 BA−1/2 )A1/2 h, h,
i.e.
(Aσ B)h, h Ah, hσ Bh, h ω−1 (Aσ B)h, h.
This gives the desired inequality.
We need the next lemma to prove Theorem 2.
Lemma 5. Let A, B > 0 be such that aA B bA for some scalars 0
σ be a connection. Then
< b a, h ∈ Cn be a unit vector and
Ah, hσ Bh, h − (Aσ B)h, h −g (t0 )Ah, h,
where f is the representing function of
and t0 ∈ (b, a).
(6)
σ , g (t ) = μt + ν − f (t ), where μ =
f (a)−f (b)
,ν
a −b
=
af (b)−bf (a)
a−b
Proof. Again, as in Lemma 3, we may take without loss of generality that 0 < b < a, since a → b
implies t0 → b and hence g (t0 ) → 0, while on using the definition of σ a connection, the left side of
the inequality (6) approaches to 0.
f (a)−f (b)
af (b)−bf (a)
Consider the line μt + ν where μ =
and ν =
. As in the proof of Lemma 3 we have
a− b
a−b
μt + ν f (t )
for all t ∈ [b, a]. By Lagrange’s Mean Value Theorem there exists t0
= μ. Let
F (t )
∈ (b, a) such that f (t0 ) =
f (a)−f (b)
a −b
= μt + ν − f (t ) − g (t0 ).
This is a convex function of t. Thus we see that F (t0 ) = 0. This along with the convexity of F (t ) imply
that F (t ) has minimum value at t = t0 . Further note that F (t0 ) = 0. Thus F (t ) 0 for all t ∈ [b, a].
Hence
f (t ) − (μt
Let Z
=
+ ν) −g (t0 ), t ∈ [b, a].
A−1/2 BA−1/2 and x
||x||2 f
and hence,
∈ C ,x =
/ 0. Taking t =
n
1
A−1/2 BA−1/2 ||xx|| , ||xx||
(7)
∈ [b, a], in (7) we obtain
A−1/2 BA−1/2 x, x − f A−1/2 BA−1/2 x, x −g (t0 )||x||2 ,
||x||2
x, xσ A−1/2 BA−1/2 x, x − f A−1/2 BA−1/2 x, x −g (t0 )x, x.
Now if we replace x by A1/2 h, in (8) we get the desired inequality.
Proof of Theorem 1. First suppose that φ is given by φ(A)
Lemma 4,
φ(A)σ φ(B) = Ah, hσ Bh, h
ω−1 (Aσ B)h, h
ω−1 φ(Aσ B),
which proves the theorem in this case.
(8)
= Ah, h for any vector h ∈ Cn . Then by
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R. Kaur et al. / Linear Algebra and its Applications 434 (2011) 636–640
Now consider the case of general positive linear map
inequality of Lemma 4,
φ . Let h ∈ Cn be any vector. Then, by first
(φ(A)σ φ(B))h, h φ(A)h, hσ φ(B)h, h = ψ(A)σ ψ(B)
(9)
where ψ(A) = φ(A)h, h.
Note that ψ is a positive linear functional on Mn (C). Therefore there exists X 0 such that ψ(A)
TrAX . Hence, if π(A) : Mn (C) → Mn (C) is left multiplication by A, then
=
ψ(A) = (π(A)h, h)
where h
yields
= X 1/2 . Since aA B bA implies aπ(A) π(B) bπ(A), the second inequality of Lemma 4
ψ(A)σ ψ(B) ω−1 ψ(Aσ B)
(10)
Combining (9) and (10) we have,
φ(A)σ φ(B) 1
ω
φ(Aσ B).
Proof of Theorem 2. The proof of this theorem is similar to the proof of Theorem 1 on using Lemma
5 and is therefore not included. Remark 6. The main results proved in [3,6,7] follow by taking σ = #α in Theorem 1 and main results
proved in [4] follow on taking σ = #α in Theorem 2. It is further remarked that the inequality
φ(Aσ B) φ(A)σ φ(B)
is proved in [1].
Acknowledgement
The authors like to thank a referee for useful comments.
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