Numerical radius inequalities for products and sums of semi-Hilbertian space operators
aa r X i v : . [ m a t h . F A ] D ec NUMERICAL RADIUS INEQUALITIES FOR PRODUCTS AND SUMS OFSEMI-HILBERTIAN SPACE OPERATORS
PINTU BHUNIA , KAIS FEKI A,B
AND KALLOL PAUL Abstract.
New inequalities for the A -numerical radius of the products andsums of operators acting on a semi-Hilbert space, i.e. a space generated by apositive semidefinite operator A , are established. In particular, it is proved foroperators T and S, having A -adjoint, that ω A ( T S ) ≤ ω A ( ST ) + 14 (cid:16) k T k A k S k A + k T S k A (cid:17) , where ω A ( T ) and k T k A denote the A -numerical radius and the A -operatorseminorm of an operator T . Introduction and Preliminaries
Let B ( H ) stand for the C ∗ -algebra of all bounded linear operators on a complexHilbert space H with inner product h· , ·i and the corresponding norm k · k . For T ∈ B ( H ), we denote by R ( T ), N ( T ) and T ∗ the range, the kernel and the adjointof T , respectively. For a given linear subspace M of H , its closure in the normtopology of H will be denoted by M . Further, let P S stand for the orthogonalprojection onto a closed subspace S of H . An operator T ∈ B ( H ) is called positiveif h T x, x i ≥ x ∈ H , and we then write T ≥
0. Furthermore, if T ≥ T is denoted by T / . For T ∈ B ( H ), the absolute valueof T , denoted by | T | , is defined as | T | = ( T ∗ T ) / . Throughout the article, A denotes a non-zero positive operator on H . The positive operator A induces thefollowing semi-inner product h· , ·i A : H × H −→ C , ( x, y ) x, y i A := h Ax, y i = h A / x, A / y i . The seminorm induced by h· , ·i A is given by k x k A = k A / x k for all x ∈ H . Itis easy to check that k · k A is a norm if and only if A is injective and that theseminormed space ( H , k · k A ) is complete if and only if R ( A ) = R ( A ). It iswell-known that the semi-inner product h· , ·i A induces an inner product on thequotient space H / N ( A ) which is not complete unless R ( A ) is closed. However, acanonical construction due to de Branges and Rovnyak [11] (see also [17]) showsthat the completion of H / N ( A ) is isometrically isomorphic to the Hilbert space R ( A / ) with the inner product h A / x, A / y i R ( A / ) := h P R ( A ) x, P R ( A ) y i , ∀ x, y ∈ H . Date : December 23, 2020.2010
Mathematics Subject Classification.
Key words and phrases.
Positive operator, A -numerical radius, Semi-inner product, sum,product. For the sequel, the Hilbert space (cid:0) R ( A / ) , h· , ·i R ( A / ) (cid:1) will be simply denotedby R ( A / ). For an account of results related to R ( A / ), we refer the readers to[6].Let T ∈ B ( H ). We recall that an operator S ∈ B ( H ) is called an A -adjoint of T if h T x, y i A = h x, Sy i A for all x, y ∈ H . One can observe that the existence of an A -adjoint of T is equivalent to the existence of a solution in B ( H ) of the equation AX = T ∗ A . Clearly, the existence of an A -adjoint operator is not guaranteed.If the set of all operators admitting A -adjoints is denoted by B A ( H ), then byDouglas theorem [13] we have, B A ( H ) = { T ∈ B ( H ) ; R ( T ∗ A ) ⊆ R ( A ) } . If T ∈ B A ( H ), the reduced solution of the equation AX = T ∗ A is a distinguished A -adjoint operator of T , which is denoted by T ♯ A and satisfies R ( T ♯ A ) ⊆ R ( A ).Note that T ♯ A = A † T ∗ A , where A † is the Moore-Penrose inverse of A (see [5]).If T ∈ B A ( H ), then T ♯ A ∈ B A ( H ), ( T ♯ A ) ♯ A = P R ( A ) T P R ( A ) and (( T ♯ A ) ♯ A ) ♯ A = T .Moreover, if S ∈ B A ( H ) then T S ∈ B A ( H ) and ( T S ) ♯ A = S ♯ A T ♯ A . For moreresults concerning T ♯ A , we invite the readers to see [4, 5]. An operator T is called A -bounded if there exists λ > k T x k A ≤ λ k x k A , for every x ∈ H . Invirtue of Douglas theorem, one can see that the set of all operators admitting A / -adjoints, denoted by B A / ( H ), is same as the collection of all A -boundedoperators, i.e., B A / ( H ) = { T ∈ B ( H ) ; ∃ λ > k T x k A ≤ λ k x k A , ∀ x ∈ H} . It is well-known that B A ( H ) and B A / ( H ) are two subalgebras of B ( H ) which are,in general, neither closed nor dense in B ( H ). Further, we have B A ( H ) ⊆ B A / ( H )(see [4, 15]). If T ∈ B A / ( H ), then the seminorm of T induces by h· , ·i A is givenby k T k A := sup x ∈R ( A ) ,x =0 k T x k A k x k A = sup (cid:8) k T x k A ; x ∈ H , k x k A = 1 (cid:9) < ∞ . (1.1)It was shown in [5, Proposition 2.3] that, for every T ∈ B A ( H ), we have k T ♯ A T k A = k T T ♯ A k A = k T k A = k T ♯ A k A . (1.2)Furthermore, Saddi [23] introduced the concept of the A -numerical radius of anoperator T ∈ B ( H ) as follows ω A ( T ) := sup {|h T x, x i A | ; x ∈ H , k x k A = 1 } . We mention here that it may happen that k T k A and ω A ( T ) are equal to + ∞ for some T ∈ B ( H ) \ B A / ( H ) (see [15]). However, it was shown in [7] that k · k A and ω A ( · ) are equivalent seminorms on B A / ( H ). More precisely, for every T ∈ B A / ( H ), the following inequalities hold k T k A ≤ ω A ( T ) ≤ k T k A . (1.3)Let T ∈ B A / ( H ). Then k T k A = 0 if and only if AT = 0. Furthermore, k T x k A ≤ k T k A k x k A , for every x ∈ H . This implies that for all
T, S ∈ B A / ( H ), k T S k A ≤ k T k A k S k A . An operator T ∈ B ( H ) is called A -selfadjoint if AT is UMERICAL RADIUS INEQUALITIES 3 selfadjoint and it is called A -positive if AT is a positive operator. An operator T ∈ B A ( H ) is said to be A -normal if T ♯ A T = T T ♯ A (see [6]). For the sequel, if A = I then k T k , r ( T ) and ω ( T ) denote respectively the classical operator norm,the spectral radius and the numerical radius of an operator T .For any operator T ∈ B A ( H ), we write ℜ A ( T ) := T + T ♯A . Following [25, Theo-rem 2.5], we have that if T ∈ B A ( H ) then ω A ( T ) = sup θ ∈ R (cid:13)(cid:13) ℜ A ( e iθ T ) (cid:13)(cid:13) A . (1.4)The A -spectral radius of an operator T ∈ B A / ( H ) was defined by the secondauthor in [15] as r A ( T ) := inf n ≥ k T n k n A = lim n →∞ k T n k n A . (1.5)The second equality in (1.5) is also proved in [15, Theorem 1]. In addition, it wasshown in [15] that r A ( · ) satisfies the commutativity property, i.e., r A ( T S ) = r A ( ST ) , ∀ T, S ∈ B A / ( H ) . (1.6)Also, the following relation between the A -spectral radius and the A -numericalradius of A -bounded operators is also proved in [15]. r A ( T ) ≤ ω A ( T ) , ∀ T ∈ B A / ( H ) . (1.7)Recently, many mathematicians have obtained different A -numerical radiusinequalities of semi-Hilbertian space operators, the interested readers are invitedto see [8, 9, 10, 16, 19, 21, 22] and references therein. Here, we obtain severalnew inequalities for the A -numerical radius of the products and the sums of semi-Hilbertian space operators. The bounds obtained here improve on the existingbounds. 2. On inequalities for product of operators
We begin this section with the following known lemma which can be found in[15].
Lemma 2.1.
Let T ∈ B ( H ) be an A -selfadjoint operator. Then, k T k A = ω A ( T ) = r A ( T ) . Our first result reads as:
Theorem 2.2.
Let
T, S ∈ B A ( H ) . Then, ω A ( T S ) ≤ k T k A ω A ( S ) + 12 ω A (cid:0) T S ± ST ♯ A (cid:1) . P. BHUNIA, K. FEKI AND K. PAUL
Proof.
Let θ ∈ R . Clearly, ℜ A ( e iθ T S ) is an A -selfadjoint operator. Therefore,from Lemma 2.1 we get, (cid:13)(cid:13) ℜ A ( e iθ T S ) (cid:13)(cid:13) A = ω A (cid:0) ℜ A ( e iθ T S ) (cid:1) = ω A (cid:18)
12 ( e iθ T S + e − iθ S ♯ A T ♯ A ) (cid:19) = ω A (cid:18)
12 ( e iθ T S + e − iθ T S ♯ A + e − iθ S ♯ A T ♯ A − e − iθ T S ♯ A ) (cid:19) = ω A (cid:18) T ℜ A ( e iθ S ) + 12 e − iθ ( S ♯ A T ♯ A − T S ♯ A ) (cid:19) ≤ ω A (cid:0) T ℜ A ( e iθ S ) (cid:1) + ω A (cid:18) e − iθ ( S ♯ A T ♯ A − T S ♯ A ) (cid:19) ≤ (cid:13)(cid:13) T ℜ A ( e iθ S ) (cid:13)(cid:13) A + 12 ω A (cid:0) S ♯ A T ♯ A − T S ♯ A (cid:1) ≤ k T k A (cid:13)(cid:13) ℜ A ( e iθ S ) (cid:13)(cid:13) A + 12 ω A (cid:0) S ♯ A T ♯ A − T S ♯ A (cid:1) ≤ k T k A ω A ( S ) + 12 ω A (cid:0) S ♯ A T ♯ A − T S ♯ A (cid:1) . Taking supremum over all θ ∈ R we get, ω A ( T S ) ≤ k T k A ω A ( S ) + 12 ω A (cid:0) S ♯ A T ♯ A − T S ♯ A (cid:1) . (2.1)Next, for x ∈ H we have, |h ( S ♯ A T ♯ A − T S ♯ A ) x, x i A | = |h ( T S − ST ♯ A ) x, x i A | . This implies that ω A (cid:0) S ♯ A T ♯ A − T S ♯ A (cid:1) = ω A (cid:0) T S − ST ♯ A (cid:1) . Thus, it follows from(2.1) that ω A ( T S ) ≤ k T k A ω A ( S ) + 12 ω A (cid:0) T S − ST ♯ A (cid:1) . (2.2)Also, replacing T by iT in (2.2) we get, ω A ( T S ) ≤ k T k A ω A ( S ) + 12 ω A (cid:0) T S + ST ♯ A (cid:1) . (2.3)The proof now follows from (2.2) and (2.3). (cid:3) Remark 2.3.
It is easy to verify that (also see in [9, Cor. 3.3]) ω A (cid:0) T S ± ST ♯ A (cid:1) ≤ k T k A ω A ( S ) . Therefore, k T k A ω A ( S ) + ω A (cid:0) T S ± ST ♯ A (cid:1) ≤ k T k A ω A ( S ) . Thus, the inequalityobtained in Theorem 2.2 is stronger than the well-know inequality ω A ( T S ) ≤ k T k A ω A ( S ) . In order to obtain our next inequality that gives an upper bound for the A -numerical radius of product of two operators, we need the following lemmas. Firstwe consider the 2 × A = (cid:18) A A (cid:19) . Clearly, A is a UMERICAL RADIUS INEQUALITIES 5 positive operator on
H ⊕ H . So, A induces the following semi-inner product on H ⊕ H defined as h x, y i A = h A x, y i = h x , y i A + h x , y i A , for all x = ( x , x ) , y = ( y , y ) ∈ H ⊕ H . Note that if T, S, X, Y ∈ B A ( H ) then,it was shown in [8, Lemma 3.1] that (cid:18) T SX Y (cid:19) ∈ B A ( H ⊕ H ) and (cid:18)
T SX Y (cid:19) ♯ A = (cid:18) T ♯ A X ♯ A S ♯ A Y ♯ A (cid:19) . (2.4) Lemma 2.4. ([19])
Let
T, S ∈ B ( H ) be A -positive operators. Then, ω A (cid:20)(cid:18) TS (cid:19)(cid:21) = 12 k T + S k A . (2.5) Lemma 2.5. ([18])
Let
T, S ∈ B A / ( H ) . Then, (a) ω A (cid:20)(cid:18) T S (cid:19)(cid:21) = max { ω A ( T ) , ω A ( S ) } . In particular, ω A (cid:20)(cid:18) T T (cid:19)(cid:21) = ω A (cid:20)(cid:18) T T ♯ A (cid:19)(cid:21) = ω A ( T ) . (2.6)(b) (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) TS (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) A = (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) T S (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) A = max {k T k A , k S k A } . Now we are in a position to obtain the following inequality.
Theorem 2.6.
Let
T, S ∈ B A ( H ) . Then, ω A ( T S ) ≤ ω A ( ST ) + 14 (cid:16) k T k A k S k A + k T S k A (cid:17) . (2.7) Proof.
Let θ ∈ R . Since ℜ A ( e iθ T S ) is an A -selfadjoint operator, so by Lemma2.1 we have, kℜ A ( e iθ T S ) k A = r A [ ℜ A ( e iθ T S )]= r A ( e iθ T S + e − iθ S ♯ A T ♯ A ) . (2.8)On the other hand, we have r A ( e iθ T S + e − iθ S ♯ A T ♯ A ) = r A (cid:20)(cid:18) e iθ T S + e − iθ S ♯ A T ♯ A
00 0 (cid:19)(cid:21) = r A (cid:20)(cid:18) e iθ T S ♯ A (cid:19) (cid:18) S e − iθ T ♯ A (cid:19)(cid:21) = r A (cid:20)(cid:18) S e − iθ T ♯ A (cid:19) (cid:18) e iθ T S ♯ A (cid:19)(cid:21) ( by (1.6))= r A (cid:20)(cid:18) e iθ ST SS ♯ A T ♯ A T e − iθ T ♯ A S ♯ A (cid:19)(cid:21) . P. BHUNIA, K. FEKI AND K. PAUL
By applying (1.7) we get r A ( e iθ T S + e − iθ S ♯ A T ♯ A ) ≤ ω A (cid:20)(cid:18) e iθ ST SS ♯ A T ♯ A T e − iθ T ♯ A S ♯ A (cid:19)(cid:21) ≤ ω A (cid:20)(cid:18) e iθ ST e − iθ T ♯ A S ♯ A (cid:19)(cid:21) + ω A (cid:20)(cid:18) SS ♯ A T ♯ A T (cid:19)(cid:21) = ω A ( ST ) + 12 k SS ♯ A + T ♯ A T k A , where the last equality follows from Lemma 2.4 together with (2.6). Therefore,from (2.8) we get, kℜ A ( e iθ T S ) k A ≤ ω A ( ST ) + k SS ♯ A + T ♯ A T k A . Hence, by taking supremum over all θ ∈ R and then using (1.4) we get, ω A ( T S ) ≤ ω A ( ST ) + k SS ♯ A + T ♯ A T k A , (2.9)If AT = 0 or AS = 0, then the inequality (2.7) holds trivially. Assume that AT =0 and AS = 0. By Replacing T and S by q k S k A k T k A T and q k T k A k S k A S , respectively, in(2.9) we obtain, ω A ( T S ) ≤ ω A ( ST ) + 14 (cid:13)(cid:13)(cid:13) k S k A k T k A T ♯ A T + k T k A k S k A SS ♯ A (cid:13)(cid:13)(cid:13) A . (2.10)It is easy to see that the operator k S k A k T k A T ♯ A T + k T k A k S k A SS ♯ A is A -positive. So, anapplication of Lemma 2.1 gives (cid:13)(cid:13)(cid:13) k S k A k T k A T ♯ A T + k T k A k S k A SS ♯ A (cid:13)(cid:13)(cid:13) A = r A (cid:16) k S k A k T k A T ♯ A T + k T k A k S k A SS ♯ A (cid:17) . (2.11)Next, r A (cid:16) k S k A k T k A T ♯ A T + k T k A k S k A SS ♯ A (cid:17) = r A (cid:20)(cid:18) k S k A k T k A T ♯ A T + k T k A k S k A SS ♯ A
00 0 (cid:19)(cid:21) = r A q k S k A k T k A T ♯ A q k T k A k S k A S ! q k S k A k T k A T q k T k A k S k A S ♯ A Further, by applying (1.6), we get r A (cid:16) k S k A k T k A T ♯ A T + k T k A k S k A SS ♯ A (cid:17) = r A q k S k A k T k A T q k T k A k S k A S ♯ A q k S k A k T k A T ♯ A q k T k A k S k A S ! = r A " k S k A k T k A T T ♯ A T SS ♯ A T ♯ A k T k A k S k A S ♯ A S ! . (2.12) UMERICAL RADIUS INEQUALITIES 7
In addition, we see that A k S k A k T k A T T ♯ A T SS ♯ A T ♯ A k T k A k S k A S ♯ A S ! = k S k A k T k A AT T ♯ A AT SAS ♯ A T ♯ A k T k A k S k A AS ♯ A S ! = k S k A k T k A ( T T ♯ A ) ∗ A ( S ♯ A T ♯ A ) ∗ A ( T S ) ∗ A k T k A k S k A ( S ♯ A S ) ∗ A ! = k S k A k T k A T T ♯ A T SS ♯ A T ♯ A k T k A k S k A S ♯ A S ! ∗ A . This implies that, k S k A k T k A T T ♯ A T SS ♯ A T ♯ A k T k A k S k A S ♯ A S ! is an A -selfadjoint operator. Hence,in view of Lemma 2.1 we have, (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k S k A k T k A T T ♯ A T SS ♯ A T ♯ A k T k A k S k A S ♯ A S !(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) A = r A " k S k A k T k A T T ♯ A T SS ♯ A T ♯ A k T k A k S k A S ♯ A S ! . (2.13)So, it follows from (2.11), (2.12) and (2.13) that (cid:13)(cid:13)(cid:13) k S k A k T k A T ♯ A T + k T k A k S k A SS ♯ A (cid:13)(cid:13)(cid:13) A = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k S k A k T k A T T ♯ A T SS ♯ A T ♯ A k T k A k S k A S ♯ A S !(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) A . Finally, by applying the triangle inequality and then using Lemma 2.5, we get (cid:13)(cid:13)(cid:13) k S k A k T k A T ♯ A T + k T k A k S k A SS ♯ A (cid:13)(cid:13)(cid:13) A ≤ (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k S k A k T k A T T ♯ A k T k A k S k A S ♯ A S !(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) A + (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) T SS ♯ A T ♯ A (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) A = max n k S k A k T k A k T T ♯ A k A , k T k A k S k A k S ♯ A S k A o + k T S k A = k S k A k T k A + k T S k A . Therefore, we get (2.7) as desired by taking (2.10) into account. (cid:3)
Remark 2.7.
By taking A = I in Theorem 2.6 we get a recent result proved byKittaneh et al. in [2].The following corollary is an immediate consequence of Theorem 2.6. Corollary 2.8.
Let
T, S ∈ B A ( H ) . Then ω A ( T S ) ≤
12 ( ω A ( ST ) + k T k A k S k A ) . (2.14)Next, we obtain the following inequalities assuming T to be A -positive. Theorem 2.9.
Let
T, S ∈ B A / ( H ) . If T is A -positive, then ω A ( T S ) ≤ k T k A ω A ( S ) and ω A ( ST ) ≤ k T k A ω A ( S ) . P. BHUNIA, K. FEKI AND K. PAUL
Proof.
For all α ∈ [0 ,
1] we have, ω A ( T S ) = ω A (( T − α k T k A I ) S + α k T k A S ) ≤ ω A (( T − α k T k A I ) S ) + α k T k A ω A ( S ) ≤ k ( T − α k T k A I ) S k A + α k T k A ω A ( S ) ≤ k T − α k T k A I k A k S k A + α k T k A ω A ( S ) . Since T is A -positive, so we observe that k T − α k T k A I k A = (1 − α ) k T k A for all α ∈ [0 , ω A ( T S ) ≤ k T k A (cid:18) (1 − α ) k S k A + αω A ( S ) (cid:19) . (2.15)This holds for all α ∈ [0 , α = 1 in (2.15) we get, ω A ( T S ) ≤ k T k A ω A ( S ) . Similarly, we can prove that ω A ( ST ) ≤ k T k A ω A ( S ) . Thus, we complete the proof. (cid:3)
Considering A = I in Theorem 2.9 we get the following numerical radius in-equalities for the product of Hilbert space operators. Corollary 2.10.
Let
T, S ∈ B ( H ) with T positive. Then, ω ( T S ) ≤ k T k ω ( S ) and ω ( ST ) ≤ k T k ω ( S ) . Remark 2.11.
1. We would like to note that the numerical radius ω ( . ) satisfies ω ( T S ) ≤ ω ( T ) ω ( S ) if either T or S is positive.2. Abu-Omar and Kittaneh in [3, Cor. 2.6] obtained that if T, S ∈ B ( H ) with T positive, then ω ( T S ) ≤ k T k ω ( S ). Thus, Corollary 2.10 is stronger than [3, Cor.2.6]. 3. On inequalities for sum of operators
We begin this section with the following lemma.
Lemma 3.1.
For any x, y, z ∈ H , we have |h x, y i A | + |h x, z i A | ≤ k x k A (cid:16) max {k y k A , k z k A } + |h y, z i A | (cid:17) . (3.1) Proof.
First note that, by the proof of [14, Th. 3] we have, |h x, y i| + |h x, z i| ≤ k x k (cid:16) max {k y k , k z k } + |h y, z i| (cid:17) , (3.2)for every x, y, z ∈ H . Now, |h x, y i A | + |h x, z i A | = |h A / x, A / y i| + |h A / x, A / z i| . So, by applying (3.2), we obtain |h x, y i A | + |h x, z i A | ≤ k A / x k (cid:16) max {k A / y k , k A / z k } + |h A / y, A / z i| (cid:17) . Hence, we get (3.1) as required. (cid:3)
UMERICAL RADIUS INEQUALITIES 9
Now, we are in a position to prove the following theorem.
Theorem 3.2.
Let
T, S ∈ B A ( H ) . Then ω A ( T + S ) ≤ r (cid:16) k T T ♯ A + SS ♯ A k A + k T T ♯ A − SS ♯ A k A (cid:17) + ω A ( ST ♯ A ) + 2 ω A ( T ) ω A ( S ) . Proof.
Recall first that for every t, s ∈ R it holdsmax { t, s } = 12 (cid:16) t + s + | t − s | (cid:17) . (3.3)Now, let x ∈ H with k x k A = 1. Using Lemma 3.1 we get, |h ( T + S ) x, x i A | ≤ |h x, T ♯ A x i A | + |h x, S ♯ A x i A | + 2 |h T x, x i A | |h Sx, x i A |≤ max (cid:8) k T ♯ A x k A , k S ♯ A x k A (cid:9) + |h ST ♯ A x, x i A | + 2 |h T x, x i A | |h Sx, x i A | = 12 (cid:16) k T ♯ A x k A + k S ♯ A x k A + (cid:12)(cid:12) k T ♯ A x k A − k S ♯ A x k A (cid:12)(cid:12) (cid:17) + |h ST ♯ A x, x i A | + 2 |h T x, x i A | |h Sx, x i A | (by (3.3))= 12 (cid:16) h ( T T ♯ A + SS ♯ A ) x, x i A + (cid:12)(cid:12) h ( T T ♯ A − SS ♯ A ) x, x i A (cid:12)(cid:12) (cid:17) + |h ST ♯ A x, x i A | + 2 |h T x, x i A | |h Sx, x i A |≤ (cid:16) ω A ( T T ♯ A + SS ♯ A ) + ω A ( T T ♯ A − SS ♯ A ) (cid:17) + ω A ( ST ♯ A ) + 2 ω A ( T ) ω A ( S )= 12 (cid:16) k T T ♯ A + SS ♯ A k A + k T T ♯ A − SS ♯ A k A (cid:17) + ω A ( ST ♯ A ) + 2 ω A ( T ) ω A ( S ) , where the last equality follows from Lemma 2.1, since the operators T T ♯ A ± SS ♯ A are A -selfadjoint. So, we infer that |h ( T + S ) x, x i A | ≤ (cid:16) k T T ♯ A + SS ♯ A k A + k T T ♯ A − SS ♯ A k A (cid:17) + ω A ( ST ♯ A ) + 2 ω A ( T ) ω A ( S ) . Therefore, the desired result follows by taking supremum over all x ∈ H with k x k A = 1 in the last inequality. (cid:3) Our next objective is to refine the triangle inequality related to ω A ( · ). To dothis, we need to recall the following lemma from [20]. Lemma 3.3.
Let T , T , S , S ∈ B A / ( H ) . Then, r A ( T S + T S ) ≤ (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) k S T k A p k S T k A k S T k A p k S T k A k S T k A k S T k A (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) . (3.4)Now, we are in a position to prove the following theorem which covers andgeneralizes a recent result proved by Abu-Omar and Kittaneh in [1]. Theorem 3.4.
Let
T, S ∈ B A ( H ) . Then, ω A ( T + S ) ≤ " ω A ( T ) + ω A ( S ) + r ( ω A ( T ) − ω A ( S )) + 4 sup θ ∈ R kℜ A ( e iθ T ) ℜ A ( e iθ S ) k A ≤ ω A ( T ) + ω A ( S ) . (3.5) Proof.
Let θ ∈ R . It can be seen that ℜ A [ e iθ ( T + S )] is an A -selfadjoint operator.So, by Lemma 2.1 we get, (cid:13)(cid:13) ℜ A [ e iθ ( T + S )] (cid:13)(cid:13) A = r A (cid:16) ℜ A [ e iθ ( T + S )] (cid:17) . By letting T = I , S = ℜ A ( e iθ T ), T = ℜ A ( e iθ S ) and S = I in Lemma 3.3 andthen using the norm monotonicity of matrices with nonnegative entries we get, (cid:13)(cid:13) ℜ A [ e iθ ( T + S )] (cid:13)(cid:13) A = r A (cid:16) ℜ A ( e iθ T ) + ℜ A ( e iθ S ) (cid:17) ≤ (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) kℜ A ( e iθ T ) k A (cid:13)(cid:13) ℜ A ( e iθ T ) ℜ A ( e iθ S ) (cid:13)(cid:13) / A (cid:13)(cid:13) ℜ A ( e iθ T ) ℜ A ( e iθ S ) (cid:13)(cid:13) / A kℜ A ( e iθ S ) k A !(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≤ (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ω A ( T ) r sup θ ∈ R (cid:13)(cid:13) ℜ A ( e iθ T ) ℜ A ( e iθ S ) (cid:13)(cid:13) A r sup θ ∈ R (cid:13)(cid:13) ℜ A ( e iθ T ) ℜ A ( e iθ S ) (cid:13)(cid:13) A ω A ( S ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) = 12 " ω A ( T ) + ω A ( S ) + r ( ω A ( T ) − ω A ( S )) + 4 sup θ ∈ R kℜ A ( e iθ T ) ℜ A ( e iθ S ) k A . By taking supremum over all θ ∈ R we get, ω A ( T + S ) ≤ " ω A ( T ) + ω A ( S ) + r ( ω A ( T ) − ω A ( S )) + 4 sup θ ∈ R kℜ A ( e iθ T ) ℜ A ( e iθ S ) k A . (3.6)This proves the first inequality. Moreover, r ( ω A ( T ) − ω A ( S )) + 4 sup θ ∈ R kℜ A ( e iθ T ) ℜ A ( e iθ S ) k A ≤ q ( ω A ( T ) − ω A ( S )) + 4 ω A ( T ) ω A ( S )= q ( ω A ( T ) + ω A ( S )) = ω A ( T ) + ω A ( S ) . So, by using (3.6) we easily get the second inequality. (cid:3)
The following lemma (see in [23]) plays a crucial role in proving our next result.
UMERICAL RADIUS INEQUALITIES 11
Lemma 3.5.
Let x, y, e ∈ H with k e k A = 1 . Then, |h x, e i A h e, y i A | ≤ (cid:0) |h x, y i A | + k x k A k y k A (cid:1) . Now, we prove the following theorem.
Theorem 3.6.
Let
T, S ∈ B A ( H ) . Then ω A ( T + S ) ≤ r ω A ( T ) + ω A ( S ) + 12 k T ♯ A T + SS ♯ A k A + ω A ( ST ) . Proof.
Let x ∈ H be such that k x k A = 1. One can verify that |h ( T + S ) x, x i A | ≤ |h T x, x i A | + |h Sx, x i A | + 2 |h T x, x i A | |h Sx, x i A | = |h T x, x i A | + |h Sx, x i A | + 2 |h T x, x i A | |h x, S ♯ A x i A | . Using Lemma 3.5 we get, |h ( T + S ) x, x i A | ≤ |h T x, x i A | + |h Sx, x i A | + k T x k A k S ♯ A x k A + |h T x, S ♯ A x i A | = |h T x, x i A | + |h Sx, x i A | + p h T ♯ A T x, x i A h SS ♯ A x, x i A + |h ST x, x i A | . By using the arithmetic-geometric mean inequality we get, |h ( T + S ) x, x i A | ≤ ω A ( T ) + ω A ( S ) + 12 (cid:0) h T ♯ A T x, x i A + h SS ♯ A x, x i A (cid:1) + ω A ( ST )= ω A ( T ) + ω A ( S ) + 12 h ( T ♯ A T + SS ♯ A ) x, x i A + ω A ( ST ) ≤ ω A ( T ) + ω A ( S ) + 12 ω A (cid:0) T ♯ A T + SS ♯ A (cid:1) + ω A ( ST )= ω A ( T ) + ω A ( S ) + 12 (cid:13)(cid:13) T ♯ A T + SS ♯ A (cid:13)(cid:13) A + ω A ( ST ) , where the last equality follows from Lemma 2.1. So, we infer that |h ( T + S ) x, x i A | ≤ ω A ( T ) + ω A ( S ) + 12 (cid:13)(cid:13) T ♯ A T + SS ♯ A (cid:13)(cid:13) A + ω A ( ST ) , for all x ∈ H with k x k A = 1. Thus, by taking the supremum over all x ∈ H with k x k A = 1, we get ω A ( T + S ) ≤ ω A ( T ) + ω A ( S ) + 12 (cid:13)(cid:13) T ♯ A T + SS ♯ A (cid:13)(cid:13) A + ω A ( ST ) . This proves the desired result. (cid:3)
As an application of the above theorem, we get the following corollary.
Corollary 3.7.
Let T ∈ B A ( H ) . Then ω A ( T ) ≤ q(cid:13)(cid:13) T T ♯ A + T ♯ A T (cid:13)(cid:13) A + 2 ω A ( T ) ≤ √ p k T ♯ A T + T T ♯ A k A . Proof.
Clearly, the first inequality follows by taking S = T in Theorem 3.6. More-over, it is well-known that ω A ( T ) ≤ ω A ( T ) (see [15]) and ω A ( T ) ≤ (cid:13)(cid:13) T T ♯ A + T ♯ A T (cid:13)(cid:13) A .So, we get that14 (cid:13)(cid:13) T T ♯ A + T ♯ A T (cid:13)(cid:13) A + 12 ω A ( T ) ≤ (cid:13)(cid:13) T T ♯ A + T ♯ A T (cid:13)(cid:13) A + 12 ω A ( T ) ≤ (cid:13)(cid:13) T T ♯ A + T ♯ A T (cid:13)(cid:13) A + 14 (cid:13)(cid:13) T T ♯ A + T ♯ A T (cid:13)(cid:13) A = 12 (cid:13)(cid:13) T T ♯ A + T ♯ A T (cid:13)(cid:13) A . This proves that the second inequality in Corollary 3.7. (cid:3)
Remark 3.8.
Note that Corollary 3.7 has been recently proved in [25].Our next improvement reads as:
Theorem 3.9.
Let
T, S ∈ B A ( H ) be A -selfadjoint. Then, ω A ( T + S ) ≤ q ω A ( T + i S ) + ω A ( ST ) + k T k A k S k A ≤ ω A ( T ) + ω A ( S ) . Proof.
Let x ∈ H be such that k x k A = 1 . Then we have, |h ( T + S ) x, x i A | ≤ ( |h T x, x i A | + |h T x, x i A | ) = |h T x, x i A | + |h Sx, x i A | + 2 |h T x, x i A ||h Sx, x i A | = |h T x, x i A + i h Sx, x i A | + 2 |h T x, x i A h Sx, x i A | = |h ( T + i S ) x, x i A | + 2 |h T x, x i A h x, S ♯ A x i A |≤ |h ( T + i S ) x, x i A | + k T x k A k S ♯ A x k A + |h T x, S ♯ A x i A | (by Lemma 3.5)= |h ( T + i S ) x, x i A | + k T x k A k S ♯ A x k A + |h ST x, x i A |≤ ω A ( T + i S ) + k T k A k S k A + ω A ( ST ) . Taking supremum over all x ∈ H with k x k A = 1 we get, ω A ( T + S ) ≤ ω A ( T + i S ) + k T k A k S k A + ω A ( ST ) . Thus, we have the first inequality of the theorem. Now we prove the secondinequality. It is not dificult to show that ω A ( T +i S ) ≤ k T k A + k S k A . Also we have ω A ( ST ) ≤ k T k A k S k A . So, ω A ( T + i S ) + k T k A k S k A + ω A ( ST ) ≤ ( k T k A + k S k A ) . Since ω A ( T ) = k T k A and ω A ( S ) = k S k A , so we get the required second inequalityof the theorem. (cid:3) Next we obtain the inequalities for the sum of k operators. First we recallthe following results: For T ∈ B ( H ), it was shown in [6, Proposition 3.6.] that T ∈ B A / ( H ) if and only if there exists a unique e T ∈ B ( R ( A / )) such that Z A T = e T Z A . Here, Z A : H → R ( A / ) is defined by Z A x = Ax . Also, it hasbeen proved in [15] that for every T ∈ B A / ( H ) we have, k T k A = k e T k B ( R ( A / )) and ω A ( T ) = ω ( e T ) . (3.7)On the basis of the above results we obtain the following theorems. UMERICAL RADIUS INEQUALITIES 13
Theorem 3.10.
For i = 1 , , . . . , k, let S i ∈ B A ( H ) . Then, ω nA k X i =1 S i ! ≤ k n − "(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k X i =1 (cid:18)(cid:16) S ♯ A i S i (cid:17) n + (cid:16) S i S ♯ A i (cid:17) n (cid:19)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) A + 2 k X i =1 ω A (cid:16)(cid:16) S ♯ A i S i (cid:17) n (cid:16) S i S ♯ A i (cid:17) n (cid:17) , for all n = 1 , , , . . . . Proof.
Let x ∈ H be such that k x k = 1. Since S i ∈ B A ( H ), so S i ∈ B ( H ). Thenwe have, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)* k X i =1 S i ! x, x +(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n ≤ k X i =1 | h S i x, x i | ! n ≤ k n − k X i =1 | h S i x, x i | n ≤ k n − k X i =1 h| S i | x, x i n h| S ∗ i | x, x i n , (cid:16) | h S i x, x i | ≤ h| S i | x, x ih| S ∗ i | x, x i (cid:17) ≤ k n − k X i =1 h| S i | n x, x ih| S ∗ i | n x, x i , (cid:16) h Sx, x i r ≤ h S r x, x i , S ≥ , r ≥ (cid:17) = k n − k X i =1 h| S i | n x, x ih x, | S ∗ i | n x i≤ k n − k X i =1 (cid:0)(cid:13)(cid:13) | S i | n x (cid:13)(cid:13) (cid:13)(cid:13) | S ∗ i | n x (cid:13)(cid:13) + |h| S i | n x, | S ∗ i | n x i| (cid:1) ≤ k n − k X i =1 (cid:18)
12 ( (cid:13)(cid:13) | S i | n x (cid:13)(cid:13) + (cid:13)(cid:13) | S ∗ i | n x (cid:13)(cid:13) ) + |h| S i | n | S ∗ i | n x, x i| (cid:19) = k n − k X i =1 (cid:10) ( | S i | n + | S ∗ i | n ) x, x (cid:11) + k n − k X i =1 |h| S i | n | S ∗ i | n x, x i| = k n − * k X i =1 ( | S i | n + | S ∗ i | n ) ! x, x + + k n − k X i =1 |h| S i | n | S ∗ i | n x, x i|≤ k n − (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k X i =1 ( | S i | n + | S ∗ i | n ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) + k n − k X i =1 ω (cid:0) | S i | n | S ∗ i | n (cid:1) , where the second inequality follows from Bohr’s inequality ([24]), i.e., if for i =1 , , . . . , n , a i be a positive real number then k X i =1 a i ! r ≤ k r − k X i =1 a ri , r ≥ x, y, e ∈ H with k e k = 1 then |h x, e ih e, y i| ≤
12 ( k x kk y k + |h x, y i| ) . Taking supremum over all x ∈ H with k x k = 1 we get, ω n k X i =1 S i ! ≤ k n − (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k X i =1 ( | S i | n + | S ∗ i | n ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) + k n − k X i =1 ω (cid:0) | S i | n | S ∗ i | n (cid:1) . (3.8)Now since B A ( H ) ⊆ B A / ( H ), so for each i = 1 , , . . . , k , S i ∈ B A / ( H ) . Therefore,there exists unique e S i in B ( R ( A / )) such that Z A S i = e S i Z A . Now, R ( A / ) beinga complex Hilbert space, so (3.8) implies that ω n k X i =1 e S i ! ≤ k n − (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k X i =1 ( | e S i | n + | e S i ∗ | n ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) B ( R ( A / )) + k n − k X i =1 ω (cid:16) | e S i | n | e S i ∗ | n (cid:17) . It is well-known that for
S, T ∈ B A / ( H ), we have ^ S + λT = e S + λ e T and f ST = e S e T for all λ ∈ C (see [17]). So, from the above inequality we have, ω n ^ k X i =1 S i ≤ k n − (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k X i =1 (cid:18)(cid:16) ( e S i ) ∗ e S i (cid:17) n + (cid:16) e S i ( e S i ) ∗ (cid:17) n (cid:19)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) B ( R ( A / )) + k n − k X i =1 ω (cid:16)(cid:16) ( e S i ) ∗ e S i (cid:17) n (cid:16) e S i ( e S i ) ∗ (cid:17) n (cid:17) . UMERICAL RADIUS INEQUALITIES 15
Also since ( e S i ) ∗ = g S ♯ A i , so ω n ^ k X i =1 S i ≤ k n − (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k X i =1 (cid:18)g S ♯ A i e S i (cid:19) n + (cid:18) e S i g S ♯ A i (cid:19) n !(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) B ( R ( A / )) + k n − ω (cid:18)(cid:18)g S ♯ A i e S i (cid:19) n (cid:18) e S i g S ♯ A i (cid:19) n (cid:19) = k n − (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ^ k X i =1 (cid:18)(cid:16) S ♯ A i S i (cid:17) n + (cid:16) S i S ♯ A i (cid:17) n (cid:19)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) B ( R ( A / )) + k n − k X i =1 ω ^ (cid:16) S ♯ A i S i (cid:17) n (cid:16) S i S ♯ A i (cid:17) n ! . Hence, ω nA k X i =1 S i ! ≤ k n − (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k X i =1 (cid:18)(cid:16) S ♯ A i S i (cid:17) n + (cid:16) S i S ♯ A i (cid:17) n (cid:19)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) A + k n − k X i =1 ω A (cid:16)(cid:16) S ♯ A i S i (cid:17) n (cid:16) S i S ♯ A i (cid:17) n (cid:17) . Thus, we complete the proof. (cid:3)
In particular considering k = 1 and n = 1 in Theorem 3.10 we get the followingresult. Corollary 3.11.
Let S ∈ B A ( H ) . Then ω A ( S ) ≤ (cid:13)(cid:13)(cid:13)(cid:0) S ♯ A S (cid:1) + (cid:0) SS ♯ A (cid:1) (cid:13)(cid:13)(cid:13) A + 12 ω A (cid:0) S ♯ A S S ♯ A (cid:1) . Next result reads as:
Theorem 3.12.
For i = 1 , , . . . , n, let S i ∈ B A ( H ) . Then ω nA k X i =1 S i ! ≤ k n − (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k X i =1 (cid:16)(cid:16) S ♯ A i S i (cid:17) n + (cid:16) S i S ♯ A i (cid:17) n (cid:17)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) A . Proof.
Let x ∈ H with k x k = 1. S i ∈ B A ( H ) implies S i ∈ B ( H ). So we have, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)* k X i =1 S i ! x, x +(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n ≤ k X i =1 | h S i x, x i | ! n ≤ k n − k X i =1 | h S i x, x i | n ≤ k n − k X i =1 h| S i | x, x i n h| S ∗ i | x, x i n , (cid:16) | h S i x, x i | ≤ h| S i | x, x ih| S ∗ i | x, x i (cid:17) ≤ k n − k X i =1 h| S i | n x, x ih| S ∗ i | n x, x i , (cid:16) h Sx, x i r ≤ h S r x, x i , S ≥ , r ≥ (cid:17) ≤ k n − k X i =1 (cid:0) h| S i | n x, x i + h| S ∗ i | n x, x i (cid:1) ≤ k n − k X i =1 (cid:0) h| S i | n x, x i + h| S ∗ i | n x, x i (cid:1) = k n − k X i =1 h (cid:0) | S i | n + | S ∗ i | n (cid:1) x, x i = k n − * k X i =1 (cid:0) | S i | n + | S ∗ i | n (cid:1) x, x + ≤ k n − (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k X i =1 (cid:0) | S i | n + | S ∗ i | n (cid:1)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) . Taking supremum over all x ∈ H with k x k = 1 we get, ω n k X i =1 S i ! ≤ k n − (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k X i =1 (cid:0) | S i | n + | S ∗ i | n (cid:1)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) . (3.9)Now for each i = 1 , , . . . , k , S i ∈ B A / ( H ) . So, there exists unique e S i in B ( R ( A / )) such that Z A S i = e S i Z A . Now R ( A / ) being a complex Hilbertspace, we have from (3.9) that ω n k X i =1 e S i ! ≤ k n − (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k X i =1 ( | e S i | n + | e S i ∗ | n ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) B ( R ( A / )) . UMERICAL RADIUS INEQUALITIES 17
Using the property that if
S, T ∈ B A / ( H ) then ^ S + λT = e S + λ e T and f ST = e S e T for all λ ∈ C , we have ω n ^ k X i =1 S i ≤ k n − (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k X i =1 (cid:16)(cid:16) ( e S i ) ∗ e S i (cid:17) n + (cid:16) e S i ( e S i ) ∗ (cid:17) n (cid:17)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) B ( R ( A / )) . Also ( e S i ) ∗ = g S ♯ A i , so ω n ^ k X i =1 S i ≤ k n − (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k X i =1 (cid:18)(cid:18)g S ♯ A i e S i (cid:19) n + (cid:18) e S i g S ♯ A i (cid:19) n (cid:19)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) B ( R ( A / )) = k n − (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ^ k X i =1 (cid:16)(cid:16) S ♯ A i S i (cid:17) n + (cid:16) S i S ♯ A i (cid:17) n (cid:17)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) B ( R ( A / )) . Hence, ω nA k X i =1 S i ! ≤ k n − (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k X i =1 (cid:16)(cid:16) S ♯ A i S i (cid:17) n + (cid:16) S i S ♯ A i (cid:17) n (cid:17)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) A . Hence we complete the proof. (cid:3)
Finally we obtain the following result.
Theorem 3.13.
For i = 1 , , . . . , n, let S i ∈ B A ( H ) . Then, ω nA k X i =1 S i ! ≤ k n − √ k X i =1 ω A (cid:16)(cid:16) S ♯ A i S i (cid:17) n + i (cid:16) S i S ♯ A i (cid:17) n (cid:17) . Proof.
Let x ∈ H with k x k = 1. Then we have, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)* k X i =1 S i ! x, x +(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n ≤ k n − k X i =1 (cid:0) h| S i | n x, x i + h| S ∗ i | n x, x i (cid:1) . Now we observe that | a + b | ≤ √ | a + i b | for all a, b ∈ R . Using this inequalitywe have, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)* k X i =1 S i ! x, x +(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n ≤ k n − √ k X i =1 (cid:12)(cid:12) h| S i | n x, x i + i h| S ∗ i | n x, x i (cid:12)(cid:12) ≤ k n − √ k X i =1 ω (cid:0) | S i | n + i | S ∗ i | n (cid:1) . Taking supremum over all x ∈ H with k x k = 1 we get, ω n k X i =1 S i ! ≤ k n − √ k X i =1 ω (cid:0) | S i | n + i | S ∗ i | n (cid:1) . (3.10) Now for each i = 1 , , . . . , k , S i ∈ B A / ( H ) . So, there exists unique e S i in B ( R ( A / )) such that Z A S i = e S i Z A . Now R ( A / ) being a complex Hilbertspace, from (3.10) we get, ω n k X i =1 e S i ! ≤ k n − √ k X i =1 ω (cid:16) | e S i | n + i | e S i ∗ | n (cid:17) . (3.11)Using the property that if S, T ∈ B A / ( H ) then ^ S + λT = e S + λ e T and f ST = e S e T for all λ ∈ C and ( e S i ) ∗ = g S ♯ A i , we have form (3.11) that ω n ^ k X i =1 e S i ≤ k n − √ k X i =1 ω (cid:18) ^ ( S ♯ A i S i ) n + i( S i S ♯ A i ) n (cid:19) . Hence, ω nA k X i =1 S i ! ≤ k n − √ k X i =1 ω A (cid:16)(cid:16) S ♯ A i S i (cid:17) n + i (cid:16) S i S ♯ A i (cid:17) n (cid:17) , as required. (cid:3) The following corollary is an easy consequence of Theorem 3.13.
Corollary 3.14.
Let S ∈ B A ( H ) . Then ω A ( S ) ≤ √ ω A (cid:0) S ♯ A S + i SS ♯ A (cid:1) . References [1] A. Abu-Omar, F. Kittaneh, Notes on some spectral radius and numerical radius inequali-ties, Studia Math. 227 (2015), no. 2, 97–109.[2] A. Abu-Omar and F. Kittaneh, Numerical radius inequalities for products and commuta-tors of operators, Houston J. Math. 41(4) (2015) 1163-1173.[3] A. Abu-Omar and F. Kittaneh, Numerical radius inequalities for products of Hilbert spaceoperators, J. Operator Theory 72:2 (2014) 521-527.[4] M.L. Arias, G. Corach and M.C. Gonzalez, Partial isometries in semi-Hilbertian spaces,Linear Algebra Appl. 428 (7) (2008) 1460-1475.[5] M.L. Arias, G. Corach and M.C. Gonzalez, Metric properties of projections in semi-Hilbertian spaces, Integral Equations Operator Theory, 62 (2008) 11-28.[6] M.L. Arias, G. Corach and M.C. Gonzalez, Lifting properties in operator ranges, Acta Sci.Math. (Szeged) 75:3-4(2009) 635-653.[7] H. Baklouti, K. Feki and O.A.M. Sid Ahmed, Joint numerical ranges of operators in semi-Hilbertian spaces, Linear Algebra Appl. 555 (2018) 266-284.[8] P. Bhunia, K. Feki and K. Paul, A -Numerical radius orthogonality and parallelism ofsemi-Hilbertian space operators and their applications, Bull. Iran. Math. Soc. (2020). https://doi.org/10.1007/s41980-020-00392-8 [9] P. Bhunia, K. Paul and R.K. Nayak, On inequalities for A -numerical radius of operators,Electron. J. Linear Algebra, 36 (2020) 143-157.[10] P. Bhunia, R.K. Nayak and K. Paul, Refinements of A-numerical radius inequalities andtheir applications, Adv. Oper. Theory 5 (2020) 1498-1511.[11] L. de Branges and J. Rovnyak, Square Summable Power Series, Holt, Rinehert and Win-ston, New York, 1966. UMERICAL RADIUS INEQUALITIES 19 [12] M.L. Buzano, Generalizzatione della diseguaglianza di Cauchy-Schwarz, Rend. Sem. Mat.Univ. e Politech. Torino 31(1971/73) (1974) 405-409.[13] R.G. Douglas, On majorization, factorization and range inclusion of operators in Hilbertspace, Proc. Amer. Math. Soc. 17 (1966) 413-416.[14] S.S. Dragomir,
Some inequalities for the Euclidean operator radius of two operators inHilbert spaces , Linear Algebra Appl. 419 (2006) 256-264.[15] K. Feki, Spectral radius of semi-Hilbertian space operators and its applications, Ann.Funct. Anal. 11 (2020) 929-946. https://doi.org/10.1007/s43034-020-00064-y [16] K. Feki, A note on the A -numerical radius of operators in semi-Hilbert spaces, Arch. Math.115 (2020) 535-544. https://doi.org/10.1007/s00013-020-01482-z [17] K. Feki, On tuples of commuting operators in positive semidefinite inner product spaces,Linear Algebra Appl. 603 (2020) 313-328.[18] K. Feki, Generalized numerical radius inequalities of operators in Hilbert spaces, Adv.Oper. Theory, (2021). https://doi.org/10.1007/s43036-020-00099-x [19] K. Feki, Some bounds for the A -numerical radius of certain 2 × A -spectral radius inequalities for A -bounded Hilbert space operators,arXiv:2002.02905v1 [math.FA].[21] N.C. Rout, S. Sahoo and D. Mishra, Some A -numerical radius inequali-ties for semi-Hilbertian space operators , Linear Multilinear Algebra (2020) https://doi.org/10.1080/03081087.2020.1774487 [22] N.C. Rout, S. Sahoo and D. Mishra, On A -numerical radius inequal-ities for 2 × https://doi.org/10.1080/03081087.2020.1810201. [23] A. Saddi, A -Normal operators in Semi-Hilbertian spaces, The Australian Journal of Math-ematical Analysis and Applications, 9 (2012) 1-12.[24] M.P. Vasi´c and D.J. Keˆcki´c, Some inequalities for complex numbers, Math. Balkanica 1(1971) 282-286.[25] A. Zamani, A -numerical radius inequalities for semi-Hilbertian space operators, LinearAlgebra Appl. 578 (2019) 159-183. [1 , Department of Mathematics, Jadavpur University, Kolkata 700032, WestBengal, India.
Email address : [email protected] ; [email protected] [2 a ] University of Monastir, Faculty of Economic Sciences and Managementof Mahdia, Mahdia, Tunisia [2 b ] Laboratory Physics-Mathematics and Applications (LR/13/ES-22), Facultyof Sciences of Sfax, University of Sfax, Sfax, Tunisia
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