Some generalized numerical radius inequalities involving Kwong ...

SOME GENERALIZED NUMERICAL RADIUS INEQUALITIES INVOLVING KWONG FUNCTIONS

arXiv:1801.07619v1 [math.FA] 23 Jan 2018

MOJTABA BAKHERAD Abstract. We prove several numerical radius inequalities involving positive semidefinite matrices via the Hadamard product and Kwong functions. Among other inequalities, it is shown that if X is a arbitrary n × n matrix and A, B are positive

semidefinite, then

ω(Hf,g (A)) ≤ k ω(AX + XA), which is equivalent to ω Hf,g (A, B) ± Hf,g (B, A)

≤ k ′ {ω((A + B)X + X(A + B)) + ω((A − B)X − X(A − B))} ,

(t) where f and g are two continuous functions on (0, ∞) such that h(t) = fg(t) is Kwong, n o n o f (λ)g(λ) f (λ)g(λ) ′ k = max : λ ∈ σ(A) and k = max : λ ∈ σ(A) ∪ σ(B) . λ λ

1. Introduction Let Mn be the C ∗ -algebra of all n × n complex matrices and h · , · i be the standard scalar product in Cn . A capital letter means an n × n matrix in Mn . For Hermitian

matrices A and B, we write A ≥ 0 if A is positive semidefinite, A > 0 if A is positive definite, and A ≥ B if A − B ≥ 0. The numerical radius of A ∈ Mn is defined by ω(A) := sup{| hAx, xi |: x ∈ Cn , k x k= 1}.

It is well known that ω( · ) defines a norm on Mn , which is equivalent to the usual

operator norm k · k. In fact, for any A ∈ Mn , 21 kAk ≤ ω(A) ≤ kAk; see [12]. For

further information about numerical radius inequalities we refer the reader to [12, 17]

and references therein. We use the notation J for the matrix whose entries are equal to one. The Hadamard product (Schur product) of two matrices A, B ∈ Mn is the matrix

A ◦ B whose (i, j) entry is aij bij (1 ≤ i, j ≤ n). The Schur multiplier operator SA on

2010 Mathematics Subject Classification. Primary 47A12, Secondary 47A30, 47A63. Key words and phrases. Numerical radius, Hadamard product, Operator monotone, Kwong function. 1

2

M. BAKHERAD

Mn is defined by SA (X) = A ◦ X (X ∈ Mn ). The induced norm of SA with respect

to the numerical radius norm will be denoted by

ω(SA (X)) ω(A ◦ X) = sup . ω(X) ω(X) X6=0 X6=0

kSA kω = sup

A continuous real valued function f on an interval (a, b) ⊆ R is called operator monotone if A ≤ B implies f (A) ≤ f (B) for all Hermitian matrices A, B ∈ Mn with spectra

(a, b) in (a, b). Following [3], a continuous real-valued function f defined on an interval f (λi )+f (λj ) with a > 0 is called a Kwong function if the matrix Kf = is λi +λj i,j=1,2,··· ,n

positive semidefinite for any (distinct) λ1 , · · · , λn in (a, b). It is easy to see that if f is a nonzero Kwong function, then f is positive and

1 f

is Kwong. Kwong [14] showed that

the set of all Kwong functions on (0, ∞) is a closed cone and includes all non-negative operator monotone functions on (0, ∞). Also, Audenaert [3] gave a characterization

of Kwong functions by showing that, for given 0 ≤ a < b, a function f on an interval √ √ (a, b) is Kwong if and only if the function g(x) = xf ( x) is operator monotone on (a2 , b2 ). The Heinz means are defined as Hν (a, b) =

a1−ν bν +aν b1−ν 2

for a, b > 0 and 0 ≤ ν ≤ 1.

These interesting means interpolate between the geometric and arithmetic means. In √ , where a, b > 0 and fact, the Heinz inequalities assert that ab ≤ Hν (a, b) ≤ a+b 2 0 ≤ ν ≤ 1. There have been obtained several Heinz type inequalities for Hilbert space

operators and matrices; see [?] and references therein.

For two continuous functions f and g on (0, ∞) we denote Hf,g (A, B) = f (A)Xg(B) + g(A)Xf (B) and Hf,g (A) = f (A)Xg(A) + g(A)Xf (A), where A, B, X ∈ Mn such that A, B are positive semidefinite. In particular, f (t) = tα

and g(t) = t1−α (α ∈ [0, 1])

Hα (A, B) = Aα XB 1−α + A1−α XB α

and Hα (A) = Aα XA1−α + A1−α XAα .

A norm ||| · ||| on Mn is called unitarily invariant if |||UAV ||| = |||A||| for all A ∈ Mn

and all unitary matrices U, V ∈ Mn . Let A, B, X ∈ Mn such that A and B are positive

semidefinite. In [15] it was conjectured a general norm inequality of the Heinz inequality

|||Hf,g (A, B)||| ≤ |||AX + XB|||, where f and g are two continuous functions on (0, ∞) such that f (t)g(t) ≤ t and the function h(t) =

f (t) g(t)

is Kwong. In particular, if f (t) = tα

and g(t) = t1−α (α ∈ [0, 1]), then we state a Heinz type inequality |||Hα(A, B)||| ≤

|||AX + XB|||, where A, B, X ∈ Mn such that A, B are positive semidefinite. For further information, we refer the reader to [5, 7] and references therein.

SOME GENERALIZED NUMERICAL RADIUS INEQUALITIES

3

The numerical radius ω( · ) is a weakly unitarily invariant norm on Mn , that is

ω(U ∗ AU) = ω(A) for every A ∈ Mn and every unitary U ∈ Mn . In [1], the authors proved a Heinz type inequality for the numerical radius as follows ω(Hα(A)) ≤ ω(AX + XA),

(1.1)

in which A, X ∈ Mn such that A is positive semidefinite. They also showed that the inequality ω(Hα(A, B)) ≤ ω(AX + XB) is not true in general.

Our research aim is to show some numerical radius inequalities via the Hadamard

product and Kwong functions. By using some ideas of [10, 11] and [15], we obtain some extensions and generalizations of inequality (1.1), which are generalizations of a Hienz type inequality for the numerical radius. For instance, we prove if A, X ∈ Mn such that A is positive semidefinite, then

ω(Hf,g (A)) ≤ k ω(AX + XA), where f and g are two continuous functions on (0, ∞) such that o n : λ ∈ σ(A) . k = max f (λ)g(λ) λ

f (t) g(t)

is Kwong and

2. main results For our purpose we need the following lemmas. Lemma 2.1. [19, Theorem 3.4] (Spectral Decomposition) Let A ∈ Mn with eigenvalues

λ1 , · · · , λn . Then A is normal if and only if there exists a unitary matrix U such that U ∗ AU = diag(λ1 , · · · , λn ). In particular, A is positive definite if and only if the λj (1 ≤ j ≤ n) are positive. Lemma 2.2. [2, Corollary 4] Let A = [aij ] ∈ Mn be positive semidefinite. Then kSA kω = max aii . i

Lemma " 2.3. [13] #!Let X, Y ∈ Mn . Then X 0 (i) ω = max{ω(X), ω(Y )}; 0 Y " #! 0 X max(ω(X+Y ),ω(X−Y )) ≤ω ≤ (ii) 2 Y 0

ω(X+Y )+ω(X−Y ) . 2

Now, we are in position to demonstrate the first result of this section by using some ideas of [10, 11, 15].

4

M. BAKHERAD

Theorem 2.4. Let A, B ∈ Mn be positive semidefinite, X ∈ Mn , and let f , g be two continuous functions on (0, ∞) such that h(t) =

f (t) g(t)

is Kwong. Then

ω(Hf,g (A)) ≤ k ω(AX + XA), n o where k = maxλ∈σ(A) f (λ)g(λ) . λ

(2.1)

Moreover, inequality (2.1) is equivalent to the inequality ω Hf,g (A, B) ± Hf,g (B, A)

≤ k ′ {ω((A + B)X + X(A + B)) + ω((A − B)X − X(A − B))} , (2.2) n o f (λ)g(λ) ′ where k = maxλ∈σ(A)∪σ(B) . λ Proof. Assume that A is positive definite. Since the numerical radius is weakly uni-

tarily invariant, we may assume that A is diagonal matrix with positive eigenvalues λ1 , · · · , λn . It follows from

f g

is a Kwong function that f (λi )g −1 (λj ) + f (λj )g −1 (λi ) Λ Z = [zij ] = Λ λi + λj (i,j=1,··· ,n)

is positive semidefinite, where Λ = diag (g(λ1), · · · , g(λn )). It follows from Lemma 2.2

that

kSZ kω = max zii = max i

or equivalently,

ω(Z◦X) ω(X)

i

f (λi )g(λi ) ≤k λi

≤ k (0 6= X ∈ Mn ).

1 If we put E = [ λi +λ ] and F = j

[f (λi )g(λj ) + f (λj )g(λi )] ∈ Mn , then

ω(E ◦ F ◦ X) = ω(Z ◦ X) ≤ k ω(X)

(X ∈ Mn ).

Let the matrix C be the entrywise inverse of E, i.e., C ◦ E = J. Thus ω(F ◦ X) ≤ k ω(C ◦ X)

(X ∈ Mn )

or equivalently ω(Hf,g (A)) = ω(f (A)Xg(A) + g(A)Xf (A)) ≤ k ω(AX + XA). (2.3) " # A1 0 Now, if A is positive semidefinite, we may assume that A = , where A1 ∈ 0 0 " # X1 X2 Mk (k < n) is a positive definite matrix. Let X = , where X1 ∈ Mk and X3 X4


5

X4 ∈ Mn−k . Then we have "

ω(Hf,g (A)) = ω

f (A1 )X1 g(A1) + g(A1 )X1 f (A1 ) 0 0

"

≤ kω

0

#!

(by Lemma 2.3(i)) #! A1 X1 + X1 A1 0 (by (2.3)) 0 0

= k ω(A1 X1 + X1 A1 ) ≤ k ω(AX + XA)

(by Lemma 2.3(i)) (by [8, Lemma 2.1]).

Hence, we reach inequality (2.1). Moreover, if we replace A and X by ! 0 X in inequality (2.1), respectively, then X 0 " #! 0 Hf,g (A, B) ω Hf,g (B, A) 0 " #! 0 AX + XB ′ ≤k ω , XA + BX 0

A 0 0 B

(2.4) ! and

whence n o max ω Hf,g (A, B) ± Hf,g (B, A) " #! 0 f (A)Xg(B) + g(A)Xf (B) ≤ 2ω g(B)Xf (A) + f (B)Xg(A) 0

≤ 2k ′ ω

"

0

(by Lemma 2.3(ii)) #! AX + XB

XA + BX

0

(by (2.4))

≤ k ′ (ω(AX + XB + XA + BX) + ω(AX + XB − XA − BX)) (by Lemma 2.3(ii)). Thus, we have inequality (2.2). Also, if we put B = A in inequality (2.2), then we reach inequality (2.1).

If we take f (t) = tα and g(t) = t1−α in Theorem 2.4 for each 0 ≤ α ≤ 1, then we get

the next result.

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M. BAKHERAD

Corollary 2.5. [1, Theorem 2.4] Let A, B ∈ Mn be positive semidefinite, X ∈ Mn ,

and let 0 ≤ α ≤ 1. Then

ω(Hα(A)) ≤ ω(AX + XA).

(2.5)

Moreover, inequality (2.5) is equivalent to the inequality ω Hα (A, B) ± Hα (B, A)

≤ ω((A + B)X + X(A + B)) + ω((A − B)X − X(A − B)).

Corollary 2.6. Let A, B ∈ Mn be positive semidefinite, X ∈ Mn , and let f be a non-

negative operator monotone function on [0, ∞) such that f ′ (0) = limx→0+ f ′ (x) < ∞ and f (0) = 0. Then

ω(f (A)X + Xf (A)) ≤ f ′ (0) ω(AX + XA).

(2.6)

Moreover, inequality (2.6) is equivalent to the inequality ω(X(f (A) + f (B)) + (f (A) + f (B))X) ′ ≤ f (0) ω((A + B)X + X(A + B)) + ω((A − B)X − X(A − B)) .

Proof. A function g is non-negative operator increasing on [0, ∞) if and only if

non-negative operator increasing on [0, ∞); see [9]. Hence

is Then f (t) t f (t) have t ≤ ′ ′

decreasing. If 0 ≤ x ≤ t, then

f (t) t

≤

f (x) . x

t f (t)

t g(t)

is

is operator increasing.

Now, by taking x → 0+ we

f ′ (0). If we put g(t) = 1 (t ∈ [0, ∞)) in Theorem 2.4, it follows from

k = k ≤ f (0) that we get the required result.

We first cite the following lemma due to Fujii et al. [10], which will be needed in the next theorem. Lemma 2.7. [10, Lemma 3.1] Let λ1 , · · · , λn be any positive real numbers and −2 < t ≤ 2. If f and g are two continuous functions on (0, ∞) such that

the n × n matrix

Y =

f (λi )g −1 (λj ) + f (λj )g −1 (λi ) λ2i + tλi λj + λ2j

f (t) g(t)

is Kwong, then

i,j=1,··· ,n

is positive semidefinite. Theorem 2.8. Let A, B ∈ Mn be positive semidefinite, X ∈ Mn , f , g be two continuous functions on (0, ∞) such that

f (t) g(t)

is Kwong, and let −2 < t ≤ 2. Then

1 1 2k ω A2 X + tAXA + XA2 , ω A 2 Hf,g (A) A 2 ≤ t+2

(2.7)


where k = maxλ∈σ(A)

n

f (λ)g(λ) λ

o

7

.

Moreover, inequality (2.7) is equivalent to the inequality 1 1 4k ′ ω A 2 Hf,g (A, B) B 2 ≤ ω A2 X + tAXB + XB 2 , t+2 o n . where k ′ = maxλ∈σ(A)∪σ(B) f (λ)g(λ) λ

(2.8)

Proof. First, we show inequality (2.7). It is enough to show the inequality in the case A is positive definite. Since the numerical radius is weakly unitarily invariant, we may assume that A is diagonal 1 matrix with positive eigenvalues λ1 , · · · , λn . Let 1 Σ = diag λ12 g(λ1 ), · · · , λn2 g(λn ) . It follows from Lemma 2.7 that (t + 2) (f (λ )g −1(λ ) + f (λ )g −1(λ )) i j j j Z = [zij ] = Σ Σ 2 2 2(λi + tλi λj + λj ) i,j=1,··· ,n

is positive semidefinite for −2 < t ≤ 2. In addition, all diagonal entries of Z are no

more than k. Therefore,

f (λi )g(λi ) ≤ k, i i λi i h ω(Z◦X) 1 and P = whence ω(X) ≤ k (0 6= X ∈ Mn ). Now, let M = λ2 +tλi λj +λ2 i j i,j=1,··· ,n h 1 1i t+2 2 λi f (λi )g(λj ) + f (λj )g(λi )λj2 . Then 2 kSZ kω = max zii = max

i,j=1,··· ,n

ω(M ◦ P ◦ X) = ω(Z ◦ X) ≤ k ω(X)

(0 6= X ∈ Mn ).

Let the matrix N be the entrywise inverse of M, i.e., M ◦ N = J. Hence ω(P ◦ X) ≤ k ω(N ◦ X)

(0 6= X ∈ Mn )

or equivalently 2k ω(A2 X + tAXA + XA2 ), t+2 o n : λ ∈ σ(A) . Hence we have where X ∈ Mn , −2 < t ≤ 2 and k = max f (λ)g(λ) λ 1

1

ω(A 2 (Hf,g (A)) A 2 ) ≤

inequality (2.7).

A 0

Now, if we replace A and X by

0 B

!

and

0 X 0

0

!

inequality (2.7), respectively,

then ω

"

1

1

0 A 2 (Hf,g (A, B)) B 2 0

0

#!

2k ′ ≤ ω t+2

"

0 A2 X + tAXB + XB 2 0

0

#!

.

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M. BAKHERAD

Hence "

1 1 1 ω A 2 Hf,g (A, B) B 2 ≤ ω 2

1 #! 1 0 A 2 Hf,g (A, B) B 2

0

0

(by Lemma 2.3)

2k ′ ≤ ω t+2

"

0 A2 X + tAXB + XB 2 0

0

#!

2k ′ ω A2 X + tAXB + XB 2 (by Lemma 2.3). t+2 Thus, we reach inequality (2.8). Also, if we put B = A in inequality (2.7), then we get ≤

inequality (2.8).

Corollary 2.9. Let A ∈ Mn be positive semidefinite. If f is a positive operator

monotone function on (0, ∞), then

3 3 1 1 ω A 2 f (A)Xf (A)−1A 2 + A 2 f (A)−1 Xf (A)A 2 ≤

where X ∈ Mn and −2 < t ≤ 2.

4 ω A2 X + tAXA + XA2 , t+2

Proof. Since f positive operator monotone on (0, ∞), then g(t) = monotone on (0, ∞) and also

f (t) g(t)

t f (t)

is operator

= tf 2 (t) is Kwong function [15]. So f and g satisfy

the conditions of Theorem 2.8. Hence we have the desired inequality.

Example 2.10. The function f (t) = log(1 + t) is operator monotone on (0, ∞); see [9]. If we put g(t) = 1, then

f (t) g(t)

= log(1 + t) is Kwong [14]. Using Theorem 2.4 we

have 2 ω A2 X + tAXA + XA2 , t+2 where A, X ∈ Mn such that A is positive semidefinite and −2 < t ≤ 2. 1 1 ω A 2 log(I + A)X + X log(I + A) A 2 ≤

Now, we infer the following lemma due to Zhan [18], which will be needed in the next theorem. Lemma 2.11. [18, Lemma 5] Let λ1 , · · · , λn be any positive real numbers, r ∈ [−1, 1]

and −2 < t ≤ 2. Then the n × n matrix λri + λrj L= λ2i + tλi λj + λ2j i,j=1,··· ,n is positive semidefinite. Now, we shall show the following result related to [10].


9

Proposition 2.12. Let A, X ∈ Mn such that A is positive semidefinite, β > 0 and

1 ≤ 2r ≤ 3. Then

ω Ar XA2−r +A2−r XAr 4β(1 − r0 ) 2 2 ≤ ω 2(1 − 2β + 2βr0 )AXA + (A X + tAXA + XA ) , t+2 where −2 < t ≤ 2β − 2 and r0 = min{ 21 + |1 − r|, 1 − |1 − r|}. Proof. Since the numerical radius is weakly unitarily invariant, we may assume that A is diagonal matrix with positive eigenvalues λ1 , · · · , λn . Since 1 ≤ 2r ≤ 3, then 1 2 1 4

1−2β+2βr0 (t + 2) + t. It follows from −2 < t ≤ 2β − 2 2β(1−r0 ) t+2 t 1 > 0 and −2 < t0 ≤ 2, where t0 = 2β(1−r + β(1−r 4β(1−r0 ) 0) 0)

≤ r0 ≤ 43 . Let t0 =

≤ 1 − r0 ≤

1 , 4

that

Hence, by using Lemma 2.11, the n × n matrix W = [wij ] = is positive semidefinite for

1 2

and

− 2.

λ2−2r + λ2−2r t+2 i j Λr 2 Λr 4β(1 − r0 ) λi + t0 λi λj + λ2j i,j=1,··· ,n ≤ r ≤ 23 , where Λ = diag (λ1 , · · · , λn ). Therefore,

(t + 2)λri (2λ2−2r )λri i =1 i i 4β(1 − r0 )(t0 + 2)λ2i ≤ 1 (0 6= X ∈ Mn ). Now, let O = λ2i + t0 λi λj + λ2j i,j=1,··· ,n and 1 . Then 4β(1−r0 ) 2 2 kSW kω = max wii = max

whence M=

ω(W ◦X) ω(X)

2(1−2β+2βr0 )λi λj +

t+2

(λi X+tλi λj +λj )

i,j=1,··· ,n

ω(O ◦ M ◦ X) = ω(W ◦ X) ≤ ω(X)

(0 6= X ∈ Mn ).

Let the matrix N be the entrywise inverse of M, i.e., M ◦ N = J. Hence ω(O ◦ X) ≤ ω(N ◦ X)

(0 6= X ∈ Mn )

or equivalently ω Ar XA2−r +A2−r XAr 4β(1 − r0 ) 2 2 ≤ ω 2(1 − 2β + 2βr0 )AXA + (A X + tAXA + XA ) , t+2 where −2 < t ≤ 2β − 2 and r0 = min{ 21 + |1 − r|, 1 − |1 − r|}.

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M. BAKHERAD

3. Acknowledgement. The author would like to sincerely thank the referee for some useful comments and suggestions.

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18. Zhan, X. Inequalities for unitarily invariant norms, SIAM J. Matrix Anal. Appl. 20 (2), 466-470, 1999. 19. Zhang, F. Matrix Theory, Second edition, Springer, New York, 2011. Department of Mathematics, Faculty of Mathematics, University of Sistan and Baluchestan, Zahedan, I.R.Iran. E-mail address: [email protected]; [email protected]