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CHEBYSHEV TYPE INEQUALITIES FOR HILBERT SPACEOPERATORS
MOHAMMAD SAL MOSLEHIAN1 AND MOJTABA BAKHERAD2
Abstract. We establish several operator extensions of the Chebyshev inequality.
The main version deals with the Hadamard product of Hilbert space operators.
More precisely, we prove that if A is a C∗-algebra, T is a compact Hausdorff space
equipped with a Radon measure µ(t), α : T → [0,+∞) is a measurable function and
(At)t∈T , (Bt)t∈T are suitable continuous fields of operators in A having the synchro-
nous Hadamard property, then∫T
α(s)dµ(s)
∫T
α(t)(At ◦Bt)dµ(t) ≥(∫
T
α(t)Atdµ(t))◦(∫
T
α(s)Bsdµ(s)).
We apply states on C∗-algebras to obtain some versions related to synchronous func-
tions. We also present some Chebyshev type inequalities involving the singular values
of positive n× n matrices. Several applications are given as well.
1. Introduction and preliminaries
Let B(H ) denote the C∗-algebra of all bounded linear operators on a complex Hilbert
space H together with the operator norm ‖ · ‖. Let I stand for the identity operator.
In the case when dimH = n, we identify B(H ) with the matrix algebra Mn of all
n × n matrices with entries in the complex field C. An operator A ∈ B(H ) is called
positive (positive semidefinite for a matrix) if 〈Ax, x〉 ≥ 0 for all x ∈ H and then
we write A ≥ 0. By a strictly positive operator (positive definite for a matrix) A,
denoted by A > 0, we mean a positive invertible operator. For self-adjoint operators
A,B ∈ B(H ), we say B ≥ A (B > A, resp.) if B − A ≥ 0 (B − A > 0, resp.).
Let BJh(H ) be the set of all self-adjoint operators in B(H ), whose spectra are con-
tained in J . The Gelfand map f(t) 7→ f(A) is an isometrically ∗-isomorphism between
the C∗-algebra C(Sp(A)) of continuous functions on the spectrum Sp(A) of a self-
adjoint operator A and the C∗-algebra generated by I and A. If f, g ∈ C(Sp(A)), then
f(t) ≥ g(t) (t ∈ Sp(A)) implies that f(A) ≥ g(A).
If f be a continuous real valued function on an interval J . The function f is called
2010 Mathematics Subject Classification. Primary 47A63, Secondary 47A60.
Key words and phrases. Chebyshev inequality; Hadamard product; Bochner integral; super-
multiplicative function; singular value; operator mean.
1
2 M.S. MOSLEHIAN, M. BAKHERAD
operator monotone (operator decreasing, resp.) if A ≤ B implies f(A) ≤ g(B)
(f(B) ≤ g(A), resp.) for all A,B ∈ BJh(H ). For A ∈ Mn, the singular values
of A, denoted by s1(A), s2(A), · · · , sn(A), are the eigenvalues of the positive matrix
|A| = (A∗A)12 enumerated as s1(A) ≥ · · · ≥ sn(A) with their multiplicities counted.
Given an orthonormal basis {ej} of a Hilbert space H , the Hadamard product A◦Bof two operators A,B ∈ B(H ) is defined by 〈A ◦ Bei, ej〉 = 〈Aei, ej〉〈Bei, ej〉. It is
known that the Hadamard product can be presented by filtering the tensor product
A⊗B through a positive linear map. In fact,
A ◦B = U∗(A⊗B)U,
where U : H → H ⊗ H is the isometry defined by Uej = ej ⊗ ej; see [6]. For
matrices, one easily observe [14] that the Hadamard product of A = (aij) and B = (bij)
is A ◦B = (aijbij), a principal submatrix of the tensor product A⊗B = (aijB)1≤i,j≤n.
From now on when we deal with the Hadamard product of operators, we explicitly
assume that we fix an orthonormal basis.
The axiomatic theory of operator means has been developed by Kubo and Ando
[8]. An operator mean is a binary operation σ defined on the set of strictly positive
operators, if the following conditions hold:
(i) A ≤ C,B ≤ D imply AσB ≤ CσD;
(ii) An ↓ A,Bn ↓ B imply AnσBn ↓ AσB, where An ↓ A means that A1 ≥ A2 ≥ · · ·and An → A as n→∞ in the strong operator topology;
(iii) T ∗(AσB)T ≤ (T ∗AT )σ(T ∗BT ) (T ∈ B(H ));
(iv) IσI = I.
There exists an affine order isomorphism between the class of operator means and
the class of positive operator monotone functions f defined on (0,∞) with f(1) = 1
via f(t)I = Iσ(tI) (t > 0). In addition, AσB = A12f(A
−12 BA
−12 )A
12 for all strictly
positive operators A,B. The operator monotone function f is called the representing
function of σ. Using a limit argument by Aε = A+ εI, one can extend the definition of
AσB to positive operators. The operator means corresponding to the positive operator
monotone functions f]µ(t) = tµ and f!(t) = 2t1+t
on [0,∞) are the operator weighted
geometric mean A]µB = A12
(A
−12 BA
−12
)µA
12 and the operator harmonic mean A!B =
2(A−1 +B−1)−1, respectively.
Let us consider the real sequences a = (a1, · · · , an), b = (b1, · · · , bn) and the non-
negative sequence w = (w1, · · · , wn). Then the weighed Chebyshev function is defined
CHEBYSHEV TYPE INEQUALITIES FOR HILBERT SPACE OPERATORS 3
by
T (w; a, b) :=n∑j=1
wj
n∑j=1
wjajbj −n∑j=1
wjaj
n∑j=1
wjbj.
In 1882, Cebysev [3] proved that if a and b are monotone in the same sense, then
T (w; a, b) ≥ 0.
Some integral generalizations of this inequality was given by Barza, Persson and Soria
[1]. The Chebyshev inequality is a complement of the Gruss inequality; see [10] and
references therein.
A related notion is synchronicity. Recall that two continuous functions f, g : J → Rare synchronous on an interval J , if(
f(t)− f(s))(g(t)− g(s)
)≥ 0
for all s, t ∈ J . It is obvious that, if f, g are monotonic and have the same monotonicity,
then they are synchronic. Dragomir [4] generalized Chebyshev inequality for convex
functions on a real normed space and applied his results to show that if p1, · · · , pn is
a sequence of nonnegative numbers with∑n
j=1 pj = 1 and two sequences (v1, · · · , vn)
and (u1, · · · , un) in a real inner product space are synchronous, that is, 〈vk − vj, uk −uj〉 ≥ 0 for all j, k = 1, · · · , n, then
∑nk=1 pk〈vk, uk〉 ≥ 〈
∑nk=1 pkvk,
∑nk=1 pkuk〉. He
also presented some Chebyshev inequalities for self-adjoint operators acting on Hilbert
spaces in [5].
In this paper we provide several operator extensions of the Chebyshev inequality. In
the second section, we present our main results dealing with the Hadamard product
of Hilbert space operators and weighted operator geometric means. The key notion
is the so-called synchronous Hadamard property. More Chebyshev type inequalities
regarding operator means are presented in Section 3. In Section 4, we apply states on
C∗-algebras to obtained some versions related to synchronous functions. We present
some Chebyshev type inequalities involving the singular values of positive n×nmatrices
in the last section.
2. Chebyshev inequality dealing with Hadamard product
This section is devoted to presentation of some operator Chebyshev inequalities
dealing with the Hadamard product and weighted operator geometric means. The key
notion is the so-called synchronous Hadamard property.
4 M.S. MOSLEHIAN, M. BAKHERAD
Let A be a C∗-algebra of operators acting on a Hilbert space and let T be a compact
Hausdorff space. A field (At)t∈T of operators in A is called a continuous field of
operators if the function t 7→ At is norm continuous on T . If µ(t) is a Radon measure
on T and for a field (At) the function t 7→ ‖At‖ is integrable, one can form the Bochner
integral∫TAtdµ(t), which is the unique element in A such that
ϕ
(∫T
Atdµ(t)
)=
∫T
ϕ(At)dµ(t)
for every linear functional ϕ in the norm dual A ∗ of A .
By [13, Page 78], since t 7→ At is a continuous function from T to A , for every operator
At ∈ A we can consider an element of the form
Iλ(At) = Σnk=1At(sk)µ(Ek),
where the Ek’s form a partition of T into disjoint Borel subsets, and
sk ∈ Ek ⊆ {t ∈ T : ‖At − At(sk)‖ ≤ ε} (1 ≤ k ≤ n),
with λ = {E1, · · · , En, ε}. Then (Iλ(At))λ∈Λ is a uniformly convergent net to∫TAtdµ(t).
Let C(T,A ) denote the set of bounded continuous functions on T with values in A . It
is easy to see that the set C(T,A ) is a C∗-algebra under the pointwise operations and
the norm ‖(At)‖ = supt∈T ‖At‖; cf. [7]. Now since tensor product of two operators is
norm continuous, for any operator B ∈ A we have∫T
(At ⊗B)dµ(t) =( ∫
T
Atdµ(t))⊗B.
Also, for any operator C ∈ A∫T
(C∗AtC)dµ(t) = C∗( ∫
T
Atdµ(t))C.
Therefore ∫T
(At ◦B)dµ(t) =
∫T
V ∗(At ⊗B)V dµ(t) = V ∗∫T
(At ⊗B)dµ(t)V
= V ∗(
∫T
Atdµ(t)⊗B)V =
∫T
Atdµ(t) ◦B (At, B ∈ A ). (2.1)
Let us give our key definition.
Definition 2.1. Two fields (At)t∈T and (Bt)t∈T are said to have the synchronous
Hadamard property if (At − As
)◦(Bt −Bs
)≥ 0
for all s, t ∈ T .
CHEBYSHEV TYPE INEQUALITIES FOR HILBERT SPACE OPERATORS 5
The first result reads as follows.
Theorem 2.2. Let A be a C∗-algebra, T be a compact Hausdorff space equipped with
a Radon measure µ, let (At)t∈T and (Bt)t∈T be fields in C(T,A ) with the synchronous
Hadamard property and let α : T → [0,+∞) be a measurable function. Then∫T
α(s)dµ(s)
∫T
α(t)(At ◦Bt)dµ(t) ≥(∫
T
α(t)Atdµ(t))◦(∫
T
α(s)Bsdµ(s)). (2.2)
Proof. We have∫T
α(s)dµ(s)
∫T
α(t)(At ◦Bt)dµ(t)−(∫
T
α(t)Atdµ(t))◦(∫
T
α(s)Bsdµ(s))
=
∫T
∫T
α(s)α(t)(At ◦Bt)dµ(t)dµ(s)−∫T
(∫T
α(t)Atdµ(t))◦ α(s)Bsdµ(s)
(by 2.1)
=
∫T
∫T
α(s)α(t)(At ◦Bt)dµ(t)dµ(s)−∫T
∫T
α(t)α(s)(At ◦Bs)dµ(t)dµ(s)
(by 2.1)
=
∫T
∫T
(α(s)α(t)(At ◦Bt)− α(t)α(s)(At ◦Bs)
)dµ(t)dµ(s)
=1
2
∫T
∫T
[α(s)α(t)(At ◦Bt)− α(t)α(s)(At ◦Bs)
− α(s)α(t)(As ◦Bt) + α(t)α(s)(As ◦Bs)]dµ(t)dµ(s)
=1
2
∫T
∫T
[α(s)α(t)
(At − As
)◦(Bt −Bs
)]dµ(t)dµ(s)
≥ 0. (since the fields (At) and (Bt) have synchronous Hadamard property)
�
In the discrete case T = {1, · · · , n}, set α(i) = ωi ≥ 0 (1 ≤ i ≤ n). Then Theorem
2.2 forces the following result.
Corollary 2.3. [9, Teorem 2.1] Let A1 ≥ · · · ≥ An, B1 ≥ · · · ≥ Bn be self-adjoint
operators and ω1, · · · , ωn be positive numbers. Then
n∑j=1
ωj
n∑j=1
ωj(Aj ◦Bj) ≥( n∑j=1
ωjAj
)◦( n∑j=1
ωjBj
).
Recall that a continuous function f : J → R is super-multiplicative if f(xy) ≥f(x)f(y), for all x, y ∈ J . In the next result we need the notion of increasing field. Let
T be a compact Hausdorff space as well as a totaly order set under an order �. We
6 M.S. MOSLEHIAN, M. BAKHERAD
say (At) is an increasing (decreasing, resp.) field, whenever t � s implies that At ≤ As
(At ≥ As, resp.).
Theorem 2.4. Let A be a C∗-algebra, T be a compact Hausdorff space equipped with a
Radon measure µ as a totaly order set, let (At)t∈T , (Bt)t∈T , (Ct)t∈T , (Dt)t∈T be positive
increasing fields in C(T,A ), let α : T → [0,+∞) be a measurable function and σ be
an operator mean with the super-multiplicative representing function. Then∫T
α(s)dµ(s)
∫T
α(t)(
(At ◦Bt)σ(Ct ◦Dt))dµ(t)
≥(∫
T
α(t)(AtσCt)dµ(t))◦(∫
T
α(s)(BsσDs)dµ(s)).
Proof. Let s, t ∈ T . Without loss of generality assume that s � t. Then by the property
(i) of the operator mean we have 0 ≤ (AtσBt)− (AsσBs). Then∫T
α(s)dµ(s)
∫T
(α(t)(At ◦Bt)σ(Ct ◦Dt)
)dµ(t)
−(∫
T
α(t)(AtσCt)dµ(t))◦(∫
T
α(s)(BsσDs)dµ(s))
=
∫T
∫T
α(s)α(t)(
(At ◦Bt)σ(Ct ◦Dt))dµ(t)dµ(s)
−∫T
∫T
α(t)α(s)(
(AtσCt) ◦ (BsσDs))dµ(t)dµ(s) (by 2.1)
≥∫T
∫T
α(s)α(t)(
(AtσCt) ◦ (BtσDt))dµ(t)dµ(s)
−∫T
∫T
α(t)α(s)(
(AtσCt) ◦ (BsσDs))dµ(t)dµ(s) (by [12, Theorem 6.7])
=1
2
∫T
∫T
α(s)α(t)[(
(AtσCt) ◦ (BtσDt))−(
(AtσCt) ◦ (BsσDs))
+(
(AsσCs) ◦ (BsσDs))−(
(AsσCs) ◦ (BtσDt))]dµ(t)dµ(s)
=1
2
∫T
∫T
α(s)α(t)[(AtσCt)− (AsσCs)
]◦[(BtσDt)− (BsσDs)
]dµ(t)dµ(s)
≥ 0. (by the property (i) of the operator mean)
�
A discrete version of the theorem above is the following result obtained by taking
T = {1, · · · , n}.
Corollary 2.5. Let Ai+1 ≥ Ai ≥ 0, Bi+1 ≥ Bi ≥ 0, Ci+1 ≥ Ci ≥ 0, Di+1 ≥ Di ≥0 (1 ≤ i ≤ n− 1), ω1, · · · , ωn be positive numbers and σ be an operator mean with the
CHEBYSHEV TYPE INEQUALITIES FOR HILBERT SPACE OPERATORS 7
super-multiplicative representing function. Then
n∑j=1
ωj
n∑j=1
ωj
[(Aj ◦Bj)σ(Cj ◦Dj)
]≥( n∑j=1
ωj(AjσCj))◦( n∑j=1
ωj(BjσDj)).
Theorem 2.6. Let A be a C∗-algebra, T be a compact Hausdorff space equipped with a
Radon measure µ as a totaly order set, let (At)t∈T , (Bt)t∈T be positive increasing fields
in C(T,A ) and let α : T → [0,+∞) be a measurable function. Then∫T
α(s)dµ(s)
∫T
α(t)(At ◦Bt)dµ(t) ≥(∫
T
α(t)(At]µBt)dµ(t))◦(∫
T
α(s)(As]1−µBs)dµ(s))
for all µ ∈ [0, 1].
Proof. Let s, t ∈ T . Without loss of generality assume that s � t. Then by the property
(i) of the operator mean, we have 0 ≤ (At]µBt) − (As]µBs) and 0 ≤ (At]1−µBt) −(As]1−µBs). Then∫T
α(s)dµ(s)
∫T
α(t)(At ◦Bt)dµ(t)−(∫
T
α(t)(At]µBt)dµ(t))◦(∫
T
α(s)(As]1−µBs)dµ(s))
=
∫T
∫T
α(s)α(t)(At ◦Bt)dµ(t)dµ(s)−∫T
∫T
α(t)α(s)(
(At]µBt) ◦ (As]1−µBs))dµ(t)dµ(s)
(by 2.1)
≥∫T
∫T
α(s)α(t)(
(At]µBt) ◦ (At]1−µBt))dµ(t)dµ(s)
−∫T
∫T
α(t)α(s)(
(At]µBt) ◦ (As]1−µBs))dµ(t)dµ(s) (by [12, Theorem 6.6])
=1
2
∫T
∫T
[α(s)α(t)
((At]µBt) ◦ (At]1−µBt)
)− α(t)α(s)
((At]µBt) ◦ (As]1−µBs)
)+ α(t)α(s)
((As]µBs) ◦ (As]1−µBs)
)− α(s)α(t)
((As]µBs) ◦ (At]1−µBt)
)]dµ(t)dµ(s)
=1
2
∫T
∫T
α(s)α(t)[(At]µBt)− (As]µBs)
]◦[(At]1−µBt)− (As]1−µBs)
]dµ(t)dµ(s)
≥ 0. (by the property (i) of the operator mean)
�
In the discrete case T = {1, · · · , n}, set α(i) = ωi ≥ 0 (1 ≤ i ≤ n) in Theorem 2.6.
Then
8 M.S. MOSLEHIAN, M. BAKHERAD
Corollary 2.7. Let An ≥ · · · ≥ A1 ≥ 0, Bn ≥ · · · ≥ B1 ≥ 0 and ω1, · · · , ωn be positive
numbers. Then
n∑j=1
ωj
n∑j=1
ωj(Aj ◦Bj
)≥( n∑j=1
ωj(Aj]µBj))◦( n∑j=1
ωj(Aj]1−µBj))
for all µ ∈ [0, 1].
Proposition 2.8. Let f : [0,∞)→ R be a super-multiplicative and operator monotone
function, A1 ≥ · · · ≥ An ≥ 0, B1 ≥ · · · ≥ Bn ≥ 0 and ω1, · · · , ωn be positive numbers.
Then
n∑j=1
ωj
n∑j=1
ωjf(Aj ◦Bj) ≥( n∑j=1
ωjf(Aj))◦( n∑j=1
ωjf(Bj)).
Proof.
n∑j=1
ωj
n∑j=1
ωjf(Aj ◦Bj)−( n∑j=1
ωjf(Aj))◦( n∑j=1
ωjf(Bj))
≥n∑j=1
ωj
n∑j=1
ωj
(f(Aj) ◦ f(Bj)
)−( n∑j=1
ωjf(Aj))◦( n∑j=1
ωjf(Bj))
(by [12, Theorem 6.3])
=n∑
i,j=1
[ωiωj
(f(Aj) ◦ f(Bj)
)− ωiωj
(f(Ai) ◦ f(Bj)
)]=
1
2
n∑i,j=1
ωiωj
[(f(Aj) ◦ f(Bj)
)−(f(Ai) ◦ f(Bj)
)+(f(Ai) ◦ f(Bi)
)−(f(Aj) ◦ f(Bi)
)]=
1
2
n∑i,j=1
ωiωj
[(f(Aj)− f(Ai)
)◦(f(Bj)− f(Bi)
)]≥ 0. (by the operator monotonicity of f)
�
Example 2.9. Let A1 ≥ · · · ≥ An ≥ 0, B1 ≥ · · · ≥ Bn ≥ 0 and ω1, · · · , ωn be positive
numbers. Then
n∑j=1
ωj
n∑j=1
ωj(Aj ◦Bj)p ≥
( n∑j=1
ωjApj
)◦( n∑j=1
ωjBpj
)for each p ∈ [0, 1].
In the finite dimensional case we get the following.
CHEBYSHEV TYPE INEQUALITIES FOR HILBERT SPACE OPERATORS 9
Corollary 2.10. Let A1 ≥ · · · ≥ Ak, B1 ≥ · · · ≥ Bk be positive n × n matrices and
ω1, · · · , ωk be positive numbers. Then
( k∑j=1
ωj)n
det( k∑j=1
ωj(Aj ◦Bj))≥( k∑j=1
ωnj det(Aj))( k∑
j=1
ωnj det(Bj)).
Proof.
( k∑j=1
ωj)n
det( k∑j=1
ωj(Aj ◦Bj))
= det( k∑j=1
ωj
k∑j=1
ωj(Aj ◦Bj))
≥ det(( k∑
j=1
ωjAj)◦( k∑j=1
ωjBj
))(by Corolary 2.3)
≥ det( k∑j=1
ωjAj
)det( k∑j=1
ωjBj
)(by the property of Hadamard product)
≥( k∑j=1
ωnj det(Aj))( k∑
j=1
ωnj det(Bj)). (by the property of det)
�
Proposition 2.11. Let A1 ≥ · · · ≥ Ak > 0, Bk ≥ · · · ≥ B1 ≥ 0 be n× n matrices and
ω1, · · · , ωk be positive numbers. Then
( k∑j=1
ωj)( k∑
j=1
ωjtr(A−1j Bj)
)≥( k∑j=1
ωjtr(Aj)−1)( k∑
j=1
ωjtr(Bj)).
Proof.
( k∑j=1
ωj)( k∑
j=1
ωjtr(A−1j Bj)
)≥( k∑j=1
ωj)( k∑
j=1
ωjtr(Aj)−1tr(Bj)
)(by [16, page 224])
≥( k∑j=1
ωjtr(Aj)−1)( k∑
j=1
ωjtr(Bj)). (by Chebyshev inequality)
�
3. More Chebyshev type inequalities regarding operator means
The first author and Najafi [11, Theorem 2.4] showed that for any non-negative
operator monotone function f on [0,+∞), positive operators Aj (j = 1, · · · , n) with
10 M.S. MOSLEHIAN, M. BAKHERAD
spectra in [λ, (1 + 2√
2)λ] for some λ ∈ R+ and operators Cj (j = 1, · · · , n) with∑nj=1C
∗jC = I, it holds that
f( n∑j=1
C∗jAjC)≤ 2
n∑j=1
C∗j f
(Aj2
)Cj. (3.1)
By [12, Theorem 5.7] for positive operators A,B,C,D ∈ B(H ) we have
AσC +BσD ≤ (A+B)σ(C +D). (3.2)
Now we show that the reverse of inequality (3.2) is true under some suitable conditions.
Theorem 3.1. Let A,B,C,D ∈ B(H ) be positive operators such that λA ≤ C ≤(1 + 2
√2)λA, λB ≤ D ≤ (1 + 2
√2)λB for some λ ∈ R+ and σ be an operator mean.
Then
(i) 2 (AσC +BσD) ≥ (A+B)σ(C +D)
(ii) 2α(AσC) + 2(1− α)(BσD) ≥ (αA+ (1− α)B)σ(αC + (1− α)D)
for all α ∈ [0, 1].
Proof. (i) Let A,B,C,D be strictly positive operators. Put X = A12 (A + B)−
12 , Y =
B12 (A+B)−
12 , V = A−
12CA−
12 and W = B−
12DB−
12 . It follows from X∗X + Y ∗Y = I
that S =
(X 0
Y 0
)is a contraction. Note that the spectrum of T =
(V 0
0 W
)is
contained in the interval [λ, (1 + 2√
2)λ]. Put S ′ = (I −S∗S)12 . Since S∗S +S ′∗S ′ = I,
by inequality (3.1) we have
f(S∗TS) ≤ f(S∗TS + S ′∗λS ′) ≤ 2S∗f
(T
2
)S + 2S ′∗f
(λ
2
)S ′. (3.3)
CHEBYSHEV TYPE INEQUALITIES FOR HILBERT SPACE OPERATORS 11
Hence(f(X∗V X + Y ∗WY ) 0
0 f(0)
)
= f
((X 0
Y 0
)∗(V 0
0 W
)(X 0
Y 0
))= f (S∗TS)
≤ 2S∗f
(T
2
)S + 2S ′∗f
(λ
2
)S ′ (by inequality (3.3))
= 2
(X 0
Y 0
)∗f
((V2
0
0 W2
))(X 0
Y 0
)+
(0 0
0 2f(λ2
) )
=
(2X∗f
(V2
)X + 2Y ∗f
(W2
)Y 0
0 2f(λ2
) ) ,whence
f(X∗V X + Y ∗WY ) ≤ 2X∗f
(V
2
)X + 2Y ∗f
(W
2
)Y. (3.4)
Moreover
2
((Aσ
C
2) + (Bσ
D
2)
)= 2
(A
12
(Iσ
(A−
12C
2A−
12
))A
12 +B
12
(Iσ
(B−
12D
2B−
12
))B
12
)(by the property (iii) of σ)
= 2
((A+B)
12 X∗
(IσV
2
)X(A+B)
12 + (A+B)
12Y ∗
(IσW
2
)Y (A+B)
12
)= 2
((A+B)
12
(X∗f
(V
2
)X
)(A+B)
12 + (A+B)
12
(Y ∗f
(W
2
)Y
)(A+B)
12
)= (A+B)
12
(2X∗f
(V
2
)X + 2Y ∗f
(W
2
)Y
)(A+B)
12
≥ (A+B)12f(X∗V X + Y ∗WY )(A+B)
12 (by 3.4)
= (A+B)12 (Iσ(X∗V X + Y ∗WY )) (A+B)
12
= (A+B)σ(A
12V A
12 +B
12WB
12
)(by the property (iii) of σ)
= (A+B)σ(C +D) . (3.5)
12 M.S. MOSLEHIAN, M. BAKHERAD
Since σ is upper continuous, we get the desired inequality.
(ii) For all α ∈ [0, 1] we have
2α(AσC) + 2(1− α)(BσD) = 2(αAσαC) + 2((1− α)Bσ(1− α)D)
≥ (αA+ (1− α)B)σ(αC + (1− α)D) .
�
Corollary 3.2. Let A,B,C,D ∈ B(H ) be positive operators such that A ≤ C ≤(1 + 2
√2)A and B ≤ D ≤ (1 + 2
√2)B. Then
2 (A]νC +B]νD) ≥ (A+B)]ν(C +D)
for all ν ∈ [0, 1].
Corollary 3.3. Let {Ai}ni=1 and {Bi}ni=1 be positive operators in B(H ) such that
λAj ≤ Bj ≤ (1 + 2√
2)λAj (j = 1, · · · , n) for some λ ∈ R+ and let σ be an operator
mean. Then
2n∑j=1
(AjσBj) ≥ (n∑j=1
Aj)σ(n∑j=1
Bj).
Example 3.4. (i) A reverse of Holder’s inequality: If 0 ≤ Bqj ≤ Apj ≤ (1+2
√2)Bq
j (j =
1, · · · , n) and p, q > 0 with p−1 + q−1 = 1, then
2n∑j=1
(Apj] 1pBqj ) ≥
( n∑j=1
Apj)] 1p
( n∑j=1
Bqj
).
(ii) A reverse of Minkowski’s inequality: If 0 < Bj ≤ Aj ≤ (1 + 2√
2)Bj (j = 1, · · · , n),
then by utilizing A!B = 2(A−1 +B−1)−1 we get
2n∑j=1
(A−1j +B−1
j )−1 ≥[( n∑
j=1
A−1j
)−1+( n∑j=1
B−1j
)−1]−1
.
4. Chebyshev inequality for synchronous functions involving states
In this section, we apply the continuous functional calculus to synchronous functions
and present some Chebyshev type inequalities involving states on C∗-algebras. Our
main result of this section reads as follows.
Theorem 4.1. Let A be a unital C∗-algebra, τ1, τ2 be states on A and f, g : J → Rbe synchronous functions. Then
τ1
(f(A)g(A)
)+ τ2
(f(B)g(B)
)≥ τ1
(f(A)
)τ2
(g(B)
)+ τ2
(f(B)
)τ1
(g(A)
)(4.1)
for all A,B ∈ BJh(H ).
CHEBYSHEV TYPE INEQUALITIES FOR HILBERT SPACE OPERATORS 13
Proof. For the synchronous functions f, g and for each s, t ∈ J
f(t)g(t) + f(s)g(s)− f(t)g(s)− f(s)g(t) ≥ 0.
Fix s ∈ J . By the functional calculus for the operator A we have
f(A)g(A) + f(s)g(s)− f(A)g(s)− f(s)g(A) ≥ 0,
whence
τ1
(f(A)g(A)
)+ f(s)g(s)− τ1
(f(A)
)g(s)− f(s)τ1
(g(A)
)≥ 0.
Now for the operator B
τ1
(f(A)g(A)
)+ f(B)g(B)− τ1
(f(A)
)g(B)− f(B)τ1
(g(A)
)≥ 0.
For the state τ2 we have
τ1
(f(A)g(A)
)+ τ2
(f(B)g(B)
)≥ τ1
(f(A)
)τ2
(g(B)
)+ τ2
(f(B)
)τ1
(g(A)
).
�
Using Theorem 4.1 we obtain two next corollaries.
Corollary 4.2. Let A be a unital C∗-algebra, τ be a state on A and f, g : J → R be
synchronous functions. Then
τ(f(A)g(A)
)≥ τ
(f(A)
)τ(g(A)
)for all operator A ∈ BJh(H ). In particular
τ(f(A)2
)≥ τ
(f(A)
)2.
Proof. Put B = A in inequality (4.1) to get the result. �
Corollary 4.3. [5, Theorem 1] Let f, g : J → R be synchronous functions. Then
〈f(A)g(A)x, x〉+ 〈f(B)f(B)y, y〉 ≥ 〈f(A)x, x〉〈g(B)y, y〉+ 〈f(B)y, y〉〈g(A)x, x〉
for all operators A,B ∈ BJh(H ) and all unit vectors x, y ∈H .
Proof. Apply Theorem 4.1 to the states τ1, τ2 defined by τ1(A) = 〈Ax, x〉, τ2(A) =
〈Ay, y〉 (A ∈ B(H )) for fixed unit vectors x, y ∈H . �
14 M.S. MOSLEHIAN, M. BAKHERAD
Example 4.4. (i) Let τ be a state on B(H ) and p, q > 0. Since f(t) = tp and g(t) = tq
are synchronous
τ(Ap+q) + τ(Bp+q) ≥ τ(Ap)τ(Bq) + τ(Bp)τ(Aq) (A,B ≥ 0).
In a similar fashion, for self-adjoint operators A,B ∈ B(H )
τ(eαA+βA) + τ(eαB+βB) ≥ τ(eαA)τ(eβB) + τ(eβB)τ(eαA) (α, β ≥ 0).
(ii) Let f, g : J → R be synchronous functions. Then for n × n matrices A,B with
spectra in J
tr(f(A)g(A) + f(B)g(B)
)≥ 1
n
(tr(f(A)
)tr(g(B)
)+ tr
(g(A)
)tr(f(B)
)).
(iii) Let A,B be positive matrices, C be a positive definite matrix with tr(C) = α and
p, q ≥ 0. Utilizing τ(A) = tr(A ◦ C) we have
tr(Ap+q ◦ C +Bp+q ◦ C) ≥ 1
α
(tr(Ap ◦ C)tr(Bq ◦ C) + tr(Aq ◦ C)tr(Bp ◦ C)
).
Using the same strategy as in the proof of [15, Lemma 2.1] we get the next theorem.
Theorem 4.5. Let A be a unital C∗-algebra, τ be a state on A and f : J → [0,+∞),
g : J → R be continuous functions such that f is decreasing and g is operator decreasing
on a compact interval J . Then
τ(f(A)g(A)
)≥ τ
(f(B)
)τ(g(A)
)for all A,B ∈ BJh(H ) such that A ≤ B.
Proof. Put α = infx∈J g(x) and β = supx∈J g(x). Then α ≤ g(x) ≤ β (x ∈ J). So
αI ≥ g(B) ≥ βI, whence α ≥ τ(g(B)
)≥ β. Therefore, there exists a number t0 ∈ J
satisfying either
g(x) ≤ τ(g(B)
)if x ∈ J, x ≥ t0
and
g(x) ≥ τ(g(B)
)if x ∈ J, x < t0
or satisfying
g(x) ≤ τ(g(B)
)if x ∈ J, x > t0
and
g(x) ≥ τ(g(B)
)if x ∈ J, x ≤ t0.
CHEBYSHEV TYPE INEQUALITIES FOR HILBERT SPACE OPERATORS 15
Hence (f(x)− f(t0)
)(g(x)− τ
(g(B)
))≥ 0
for all x ∈ J . Thus
f(x)(g(x)− τ
(g(B)
))≥ f(t0)
(g(x)− τ
(g(B)
))for all x ∈ J . Hence
f(A)(g(A)− τ
(g(B)
))≥ f(t0)
(g(A)− τ
(g(B)
)).
Now
τ(f(A)g(A)
)− τ(g(B)
)τ(f(A)
)= τ(f(A)
(g(A)− τ(g(B))
))≥ τ
(f(t0)
(g(A)− τ(g(B))
))= f(t0)
(τ(g(A)
)− τ(g(B)
))≥ 0. (since g is operator decreasing)
�
Remark 4.6. Assumption A ≤ B is necessary in Theorem 4.5, since if τ(A) = 12tr(A),
f(t) = g(t) = 1t, A =
(2 0
0 3
)and B =
(1 0
0 1
), then we observe that A � B and
τ(A−2) = 1372< 5
12= τ(A−1)τ(B−1).
Corollary 4.7. Suppose that f : J → [0,+∞) and g : J → R are continuous functions
such that f is decreasing and g is operator decreasing. Then
〈f(A)g(A)x, x〉 − 〈f(B)x, x〉〈g(A)x, x〉 ≥ 0
for all operators A,B ∈ BJh(H ) such that A ≤ B and all unit vector x ∈H .
Proof. Apply Theorem 4.5 to the state τ defined by τ(A) = 〈Ax, x〉 (A ∈ B(H )) for a
fixed unit vector x ∈ H. �
Using the same strategy as in the proof of Theorem 4.1 we get the next result.
Theorem 4.8. Let A be a unital C∗-algebra, τ1, τ2 be states on A and f, g : R → Rbe synchronous functions. Then
τ2
(f(A)g(A)
)+ f(τ1(B)
)g(τ1(B)
)≥ f
(τ1(A)
)τ2
(g(B)
)+ τ1
(f(B)
)g(τ2(A)
)(4.2)
for all self-adjoint operators A,B.
16 M.S. MOSLEHIAN, M. BAKHERAD
Corollary 4.9. Let f, g : J → R be synchronous functions. Then
〈f(A)g(A)x, x〉+ f(〈By, y〉)g(〈By, y〉) ≥ f(〈Ax, x〉)〈g(B)y, y〉+ 〈f(B)y, y〉g(〈Ax, x〉)
for all operators A,B ∈ BJh(H ) and all unit vectors x, y ∈H .
Proof. Apply Theorem 4.8 to the states τ1, τ2 defined by τ1(A) = 〈Ax, x〉, τ2(A) =
〈Ay, y〉 (A ∈ B(H )) for fixed unit vectors x, y ∈H . �
Corollary 4.10. [5, Theorem 2] Let f, g : J → R are synchronous functions. Then
〈f(A)g(A)x, x〉 − f(〈Ax, x〉)g(〈Ax, x〉) ≥ [〈f(A)x, x〉 − f(〈Ax, x〉)][g(〈Ax, x〉)− 〈g(A)x, x〉]
for all operator A ∈ BJh(H ) and any unit vector x ∈H .
Corollary 4.11. Let A be a unital C∗-algebra, τ be a state on A and f, g : R → Rbe synchronous functions. Then
τ(f(B)g(B)
)− τ(f(B)
)τ(g(B)
)≥(τ(f(B)
)− f
(τ(A)
))(g(τ(A)
)− τ(g(B)
))for all self-adjoint operators A,B.
Proof. By using inequality (4.2) we have
τ(f(B)g(B)
)− τ(f(B)
)τ(g(B)
)≥ f
(τ(A)
)τ(g(B)
)+ τ(f(B)
)g(τ(A)
)− f
(τ(A)
)g(τ(A)
)− τ(f(B)
)τ(g(B)
)=(τ(f(B)
)− f
(τ(A)
))(g(τ(A)
)− τ(g(B)
)).
�
By using Corollary 4.11 and the Davis–Choi–Jensen inequality [12] to get the next
result.
Corollary 4.12. Let A be a unital C∗-algebra, τ be a state on A and f, g : R → Rbe synchronous and one is convex while the other is concave on R. Then
τ(f(A)g(A)
)− τ(f(A)
)τ(g(A)
)≥(τ(f(A)
)− f
(τ(A)
))(g(τ(A)
)− τ(g(A)
))≥ 0
for all self-adjoint operator A.
In the next proposition we establish a version of Aczel–Chebyshev type inequality.
CHEBYSHEV TYPE INEQUALITIES FOR HILBERT SPACE OPERATORS 17
Proposition 4.13. Let A be a unital C∗-algebra, τ be a state on A and f, g be
continuous functions such that 0 ≤ f(x) ≤ α and 0 ≤ g(x) ≤ β for some non-negative
real numbers α, β. Then(αβ − τ
(f(B)g(B)
))≥(α− τ
(f(B)
))(β − τ
(g(A)
))(4.3)
for all positive operators A,B ∈ A .
Proof. If α = 0 or β = 0, inequality (4.3) is travail. Now assume that α > 0 and β > 0.
Then (4.3) is equivalent to the inequality(1− τ
(f(B)g(B)
))≥(1− τ
(f(B)
))(1− τ
(g(A)
)),
with 0 ≤ f(x) ≤ 1 and 0 ≤ g(x) ≤ 1. Then we have(1− τ
(f(B)g(B)
))≥(1− τ
(f(B)
))≥(1− τ
(f(B)
))(1− τ
(g(A)
))≥ 0.
�
5. Chebyshev type inequalities involving singular values
In this section we deal with some singular value versions of the Chebyshev inequality
for positive n× n matrices. We need the following known result.
Lemma 5.1. [2, Corollary III.2.2] Let A,B be n× n Hermitian matrices. Then
λ↓j(A+B) ≥ λ↓n(A) + λ↓j(B) (1 ≤ j ≤ n).
Theorem 5.2. Let f, g : [0,+∞)→ [0,+∞) be synchronous functions. Then
sj(f(A)g(A)
)+ sj
(f(B)g(B)
)≥ sn
(f(A)
)sn(g(B)
)+
1
2
(sj(g(A)
)sj(f(B)
)+ sj
(g(B)
)sj(f(A)
))for all positive matrices A,B ∈Mn and all j = 1, 2, · · · , n.
Proof. For synchronous functions f, g we have
f(t)g(t) + f(s)g(s) ≥ f(t)g(s) + f(s)g(t) (s, t ≥ 0).
If we fix s ∈ [0,+∞), then
f(A)g(A) + f(s)g(s)I ≥ f(A)g(s) + f(s)g(A).
18 M.S. MOSLEHIAN, M. BAKHERAD
Hence
sj(f(A)g(A)
)+ f(s)g(s) = sj
(f(A)g(A) + f(s)g(s)
)≥ sj
(f(A)g(s) + f(s)g(A)
)≥ sn
(f(A)g(s)
)+ sj
(f(s)g(A)
)(by inequality (5.1) )
= sn(f(A)
)g(s) + f(s)sj
(g(A)
)(1 ≤ j ≤ n).
Using functional calculus for B we get
sj(f(A)g(A)
)+ f(B)g(B) ≥ sn
(f(A)
)g(B) + sj
(g(A)
)f(B) (1 ≤ j ≤ n).
Thus
sj(f(A)g(A)
)+ sj
(f(B)g(B)
)≥ sj
(sn(f(A)
)g(B) + sj
(g(A)
)f(B)
)≥ sn
(sn(f(A)g(B)
))+ sj
(sj(g(A)
)f(B)
)(by inequality (5.1) )
= sn(f(A)
)sn(g(B)
)+ sj
(g(A)
)sj(f(B)
)(1 ≤ j ≤ n).
(5.1)
In inequality (5.1), if we interchange the roles of A and B, then we get
sj(f(B)g(B)
)+ sj
(f(A)g(A)
)≥ sn
(f(B)
)sn(g(A)
)+ sj
(g(B)
)sj(f(A)
)(1 ≤ j ≤ n).
(5.2)
By (5.1) and (5.2)
sj(f(A)g(A)
)+ sj
(f(B)g(B)
)≥ sn
(f(A)
)sn(g(B)
)+
1
2
(sj(g(A)
)sj(f(B)
)+ sj
(g(B)
)sj(f(A)
))(1 ≤ j ≤ n).
�
In the following example we show that the constant 12
is the best possible one.
Example 5.3. For arbitrary synchronous functions f, g : [0,+∞) → [0,+∞), let us
putA = B = In×n. Then sj(f(A)g(B)
)= sj
(f(B)g(B)
)= f(1)g(1) and sj
(f(B)g(A)
)=
sj(g(B)f((A)
)= f(1)g(1), (1 ≤ j ≤ n). Thus
sj(f(A)g(A)
)+ sj
(f(B)g(B)
)= sn
(f(A)
)sn(g(B)
)+
1
2
(sj(g(A)
)sj(f(B)
)+ sj
(g(B)
)sj(f(A)
))for all j = 1, 2, · · · , n.
Using the same strategy as in the proof of Theorem 5.2 we get the next result.
CHEBYSHEV TYPE INEQUALITIES FOR HILBERT SPACE OPERATORS 19
Theorem 5.4. Let f, g : [0,+∞)→ [0,+∞) be synchronous functions. Then
f(sj(A)
)g(sj(A)
)+ sj
(f(B)g(B)
)≥ f
(sj(A)
)sn(g(B)
)+ sj
(f(B)
)g(sj(A)
)for all positive matrices A,B ∈Mn and for all j = 1, 2, · · · , n.
Example 5.5. Let A,B be positive n× n matrices and p, q > 0. Then
sj(Ap+q) + sj(B)psj(B)q ≥ sn(Bq)sj(A)p + sj(A)qsj(B
p) (1 ≤ j ≤ n).
If A be a positive definite n× n matrix, then
sj((logA)2
)+(
log sj(A))2 ≥ sn(logA) log sj(A) + log sj(A)sj(logA) (1 ≤ j ≤ n).
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1 Department of Pure Mathematics, Center of Excellence in Analysis on Alge-
braic Structures (CEAAS), Ferdowsi University of Mashhad, P. O. Box 1159, Mashhad
91775, Iran
E-mail address: [email protected], [email protected]
2 Department of Pure Mathematics, Ferdowsi University of Mashhad, P. O. Box
1159, Mashhad 91775, Iran.
E-mail address: [email protected]; [email protected]