Some A -spectral radius inequalities for A -bounded Hilbert space operators
aa r X i v : . [ m a t h . F A ] F e b Some A -spectral radius inequalities for A -boundedHilbert space operators Kais Feki Abstract.
Let r A ( T ) denote the A -spectral radius of an operator T whichis bounded with respect to the seminorm induced by a positive operator A on a complex Hilbert space H . In this paper, we aim to establish some A -spectral radius inequalities for products, sums and commutators of A -boundedoperators. Moreover, under suitable conditions on T and A we show that r A + ∞ X k =0 c k T k ! ≤ + ∞ X k =0 | c k | [ r A ( T )] k , where c k are complex numbers for all k ∈ N . Introduction and Preliminaries
Let B ( H , K ) denote the space of all bounded linear operators from a complexHilbert space ( H , h· | ·i ) into a Hilbert space K . We stand B ( H ) for B ( H , K ) with H = K as a C ∗ -algebra with the operator norm k · k and the unit I . If H = C d ,we identify B ( C d ) with the matrix algebra M d ( C ) of d × d complex matrices. Inall that follows, by an operator we mean a bounded linear operator. The rangeand the null space of an operator T are denoted by R ( T ) and N ( T ) , respectively.Also, T ∗ will be denoted to be the adjoint of T .An operator T is called positive if h T x | x i ≥ for all x ∈ H , and we thenwrite T ≥ . The cone of all positive operators will be denoted by B ( H ) + .Throughout this article, we shall assume that A ∈ B ( H ) is a positive operator.Such an A induces the following positive semi-definite sesquilinear form: h· | ·i A : H × H −→ C , ( x, y ) x | y i A := h Ax | y i . Notice that the induced seminorm is given by k x k A = h x | x i / A = k A / x k , forevery x ∈ H . Here, A / is denoted to be the square root of A . This makes H into a semi-Hilbertian space. One can check that k · k A is a norm on H if andonly if A is injective, and that ( H , k · k A ) is complete if and only if R ( A ) is closed.The following celebrated assertion is known as the Douglas theorem or Douglasmajorization theorem. Date : February 10, 2020.2010
Mathematics Subject Classification.
Primary 46C05, 47A12; Secondary 47B65, 47B15,47B20.
Key words and phrases.
Positive operator, semi-inner product, spectral radius, numericalradius.
Theorem 1.1. ( [13, Theorem 1] ) If T, S ∈ B ( H ) , then the following statementsare equivalent: (i) R ( S ) ⊆ R ( T ) ; (ii) T D = S for some D ∈ B ( H ) ; (iii) SS ∗ ≤ λ T T ∗ for some λ ≥ (or equivalently k S ∗ x k ≤ λ k T ∗ x k for all x ∈ H ).Moreover, if one of these conditions holds then, there exists a unique solution Q ∈ B ( H ) of the equation T X = S (known as the Douglas solution) so that (a) k Q k = inf { µ ; SS ∗ ≤ µT T ∗ } , (b) N ( Q ) = N ( S ) , (c) R ( Q ) ⊆ R ( T ∗ ) . Definition 1.1. ( [3] ) Let A ∈ B ( H ) + and T ∈ B ( H ) . An operator S ∈ B ( H ) iscalled an A -adjoint of T if for every x, y ∈ H , the identity h T x | y i A = h x | Sy i A holds. That is S is solution in B ( H ) of the equation AX = T ∗ A . The set of all operators which admit A / -adjoints is denoted by B A / ( H ) . Byapplying Theorem 1.1, it can observed that B A / ( H ) = { T ∈ B ( H ) ; ∃ λ > k T x k A ≤ λ k x k A , ∀ x ∈ H} . (1.1)Operators in B A / ( H ) are called A -bounded. Note that B A ( H ) is a subalgebra of B ( H ) which is neither closed nor dense in B ( H ) (see [3]). Further, clearly h· | ·i A induces the following seminorm on B A / ( H ) : k T k A := sup x ∈R ( A ) ,x =0 k T x k A k x k A = sup {k T x k A ; x ∈ H , k x k A = 1 } < ∞ . (1.2)In addition, it was shown in [15] that for T ∈ B A / ( H ) we have: k T k A = sup {|h T x | y i A | ; x, y ∈ H , k x k A = k y k A = 1 } . (1.3)We would like to mention that the inclusion B A / ( H ) ⊆ B ( H ) is in general strictas it is shown in the following example. Example 1.1.
Let H = ℓ N ∗ ( C ) and A be the diagonal operator on ℓ N ∗ ( C ) definedas Ae n = e n n ! for all n ∈ N ∗ , where ( e n ) n ∈ N ∗ is denoted to be the canonical basisof ℓ N ∗ ( C ) . Let also T ℓ be the backward shift operator on ℓ N ∗ ( C ) (that is T ℓ e = 0 and T ℓ e n = e n − for all n ≥ ). It can observed that k e n k A = √ n ! for all n ∈ N ∗ and k T ℓ e n k A = √ ( n − = √ n k e n k A for n ≥ . Hence, we infer that k T ℓ k A = + ∞ and T ℓ ∈ B ( ℓ N ∗ ( C )) \ B A / ( ℓ N ∗ ( C )) . Before we move on, let us emphasize the following two facts. If T ∈ B A / ( H ) ,then k T x k A ≤ k T k A k x k A , ∀ x ∈ H . (1.4)Moreover for every T, S ∈ B A / ( H ) , we have k T S k A ≤ k T k A k S k A . (1.5) ome A -spectral radius inequalities for A -bounded Hilbert space operators 3 Recently, the present author introduced in [16] the concept of the A -spectral ra-dius of A -bounded operators. Henceforth, A is implicitly understood as a positiveoperator. His definition reads as follows: for T ∈ B A / ( H ) we have r A ( T ) := inf n ∈ N ∗ k T n k n A = lim n →∞ k T n k n A . (1.6)Notice that the second equality in (1.6) is also proved in [16]. If A = I , we getthe well-known spectral radius formula of an operator denoted simply by r ( T ) .The study of the spectral radius of Hilbert space operators received considerableattention in the last decades. The reader may consult [17, 1, 9, 14, 6] and thereferences therein.In the next proposition we collect some properties of the A -spectral radius. Proposition 1.1. ( [16] ) Let T, S ∈ B A / ( H ) . Then the following assertionshold: (1) If T S = ST , then r A ( T S ) ≤ r A ( T ) r A ( S ) . (2) If T S = ST , then r A ( T + S ) ≤ r A ( T ) + r A ( S ) . (3) r A ( T k ) = [ r A ( T )] k for all k ∈ N ∗ . It should be emphasized that r A ( · ) satisfies the commutativity property, whichasserts that r A ( T S ) = r A ( ST ) , (1.7)for every T, S ∈ B A / ( H ) (see [16]).Recently, the A -numerical range of T ∈ B ( H ) is introduced by H. Bakloutiet al. in [7] as W A ( T ) = (cid:8) h T x | x i A : x ∈ H , k x k A = 1 (cid:9) . This new conceptis a nonempty convex subset of C which is not necessarily closed. Moreover itssupremum modulus is called the A -numerical radius of T and it is given by ω A ( T ) = sup (cid:8) |h T x | x i A | : x ∈ H , k x k A = 1 (cid:9) . Notice that it may happen that ω A ( T ) = + ∞ for some T ∈ B ( H ) . Indeed, onecan consider the following operators A = (cid:18) (cid:19) ∈ M ( C ) + and T = (cid:18) (cid:19) ∈ M ( C ) . However, ω A ( T ) < + ∞ for every A -bounded operator T . More precisely,for all T ∈ B A / ( H ) we have k T k A ≤ ω A ( T ) ≤ k T k A . (1.8)On the other hand, the present author proved in [16] that for every T ∈ B A / ( H ) we have r A ( T ) ≤ ω A ( T ) . (1.9)Now, we mention that neither the existence nor the uniqueness of an A -adjointoperator is guaranteed. The set of all operators which admit A -adjoints is denotedby B A ( H ) . By applying Theorem 1.1 we see that B A ( H ) = { T ∈ B ( H ) ; R ( T ∗ A ) ⊂ R ( A ) } . Like B A / ( H ) , the subspace B A ( H ) is a subalgebra of B ( H ) which is neither closednor dense in B ( H ) . In addition, we have B A ( H ) ⊆ B A / ( H )) (see [5, Proposition1.2.]). Kais Feki
Let T ∈ B A ( H ) . The Douglas solution of the equation AX = T ∗ A is adistinguished A -adjoint operator of T , which is denoted by T ♯ A . Note that, T ♯ A = A † T ∗ A in which A † is denoted to be the Moore-Penrose inverse of A . Itis important to mention that if T ∈ B A ( H ) , then T ♯ A ∈ B A ( H ) , k T ♯ A k A = k T k A and ( T ♯ A ) ♯ A = P A T P A . Here, P A denotes the orthogonal projection onto R ( A ) .Furthermore, if T, S ∈ B A ( H ) , then ( T S ) ♯ A = S ♯ A T ♯ A . In addition, an operator U ∈ B A ( H ) is said to be A -unitary if U ♯ A U = ( U ♯ A ) ♯ A U ♯ A = P A . For proofs andmore facts about this class of operators, the reader is invited to consult [3, 4, 7, 8]and their references.Recently, many results covering some classes of operators on a complex Hilbertspace (cid:0) H , h· | ·i (cid:1) are extended to (cid:0) H , h· | ·i A (cid:1) (see, e.g., [19, 7, 8, 20, 18]).The remainder of this paper is organized as follows. Section 2 is meant toestablish several results governing r A ( · ) . Some of the obtained results will be anatural generalization of the well-known case A = I and extend the works of F.Kittaneh et al. [17, 1, 9, 14].In section 3 we consider the power series f ( z ) = P ∞ n =0 c n z n with complexcoefficients and f c ( z ) := P ∞ n =0 | c n | z n . Obviously, f and f c have the same radiusof convergence and if c n ≥ , for all n ∈ N ∗ , then f c = f . The main target of thissection in to establish, under some conditions on A and T , a relation between r A [ f ( T )] and f c [ r A ( T )] . The obtained results cover the work of S. S. Dragomir[14]. 2. A -spectral radius inequalities In this section, we will prove several inequalities related to r A ( T ) when T is an A -bounded operator. In all what follows, we consider the Hilbert space H = ⊕ di =1 H equipped with the following inner-product: h x, y i = d X k =1 h x k | y k i , for all x = ( x , · · · , x d ) ∈ H and y = ( y , · · · , y d ) ∈ H . Let A be a d × d operatordiagonal matrix with diagonal entries are the positive operator A , i.e. A = A · · · A · · · ... ... ... ... · · · A . Clearly, A ∈ B ( H ) + . So, the semi-inner product induced by A is given by h x, y i A = h A x, y i = d X k =1 h Ax k | y k i = d X k =1 h x k | y k i A , for all x = ( x , · · · , x d ) ∈ H and y = ( y , · · · , y d ) ∈ H .In order to prove our first main result in this section, we need the followinglemma. ome A -spectral radius inequalities for A -bounded Hilbert space operators 5 Lemma 2.1.
Let T = ( T ij ) d × d be such that T ij ∈ B A / ( H ) for all i, j . Then, T ∈ B A / ( H ) . Moreover, we have k T k A ≤ k b T A k , (2.1) where b T A = ( k T ij k A ) d × d .Proof. Let x = ( x , · · · , x d ) ∈ H . It can be seen that k T x k A = k A / T x k = d X k =1 (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) d X j =1 T kj x j (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) A ≤ d X k =1 d X j =1 k T kj x j k A ! . (2.2)On the other hand, since T ij ∈ B A / ( H ) for all i, j , then by (1.1) there exists λ ij > such that k T ij x k A ≤ λ ij k x k A , (2.3)for all x ∈ H and i, j ∈ { , · · · d } . So, by using (2.3) and the Cauchy-Shwarzinequality we get k T x k A ≤ d X k =1 d X j =1 λ kj k x j k A ! ≤ d (max k,j λ kj ) d X j =1 k x j k A ! ≤ d (max k,j λ kj ) d X j =1 k x j k A = λ k x k A , where λ := d p max k,j λ kj . Hence, by (1.1), we infer that T ∈ B A / ( H ) . So, byusing (1.2), we have k T k A = sup {k T x k A , x ∈ H , k x k A = 1 } . In order to prove (2.1), it suffices to show that k T x k A ≤ k b T A k k x k A , ∀ x ∈ H . (2.4) Kais Feki
Let x = ( x , · · · , x d ) ∈ H . Let b x A denote ( k x k A , · · · , k x d k A ) ∈ R d . Notice that k b x A k A = k x k A . By using (1.4) and (2.2), one can see that k T x k A ≤ d X k =1 d X j =1 k T kj x j k A ! ≤ d X k =1 d X j =1 k T kj k A k x j k A ! = (cid:13)(cid:13)(cid:13)b T A b x A (cid:13)(cid:13)(cid:13) ≤ k b T A k k b x A k = k b T A k k x k A . Hence, (2.4) is proved and thus the proof is complete. (cid:3)
Now, we are in a position to prove our first main result in this section.
Theorem 2.1.
Let T = ( T ij ) d × d be a d × d operator matrix be such that T ij ∈B A / ( H ) for all i, j and b T A = ( k T ij k A ) d × d . Then, r A ( T ) ≤ r ( b T A ) . That is r A T T · · · T d T T · · · T d ... ... ... ... T d T d · · · T dd ≤ r k T k A k T k A · · · k T d k A k T k A k T k A · · · k T d k A ... ... ... ... k T d k A k T d k A · · · k T dd k A . (2.5) Proof.
Notice, in general, that for operators T = ( T ij ) d × d and S = ( S ij ) d × d suchthat T ij ∈ B A / ( H ) for all i, j and S ij ∈ B A / ( H ) for all i, j respectively, we have k c TS A k ≤ k b T A kk b S A k . (2.6)Indeed, we have (cid:16) c TS A (cid:17) kj = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) d X ℓ =1 T kℓ S ℓj (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) A ≤ d X ℓ =1 k T kℓ k A k S ℓj k A = (cid:16)b T A b S A (cid:17) kj , for all k, j . Therefore, by the norm monotonicity of matrices with nonnegativeentries, we see that k c TS A k ≤ k b T A b S A k . This shows (2.6) since k b T A b S A k ≤ k b T A k k b S A k .Now, by using (2.1) together with (2.6) and an induction argument, we get k T n k A ≤ (cid:13)(cid:13)(cid:13)c T n A (cid:13)(cid:13)(cid:13) ≤ (cid:13)(cid:13)(cid:13)(cid:16)b T A (cid:17) n (cid:13)(cid:13)(cid:13) , for all n ∈ N ∗ . Thus, by using (1.6) we obtain r A ( T ) = lim n →∞ k T n k /n A ≤ lim n →∞ (cid:13)(cid:13)(cid:13)(cid:16)b T A (cid:17) n (cid:13)(cid:13)(cid:13) /n = r ( b T A ) . Therefore, we get (2.5) as desired. (cid:3)
Our second result in this section reads as follows. ome A -spectral radius inequalities for A -bounded Hilbert space operators 7 Theorem 2.2.
Let T , T , S , S ∈ B A / ( H ) . Then, r A ( T S + T S ) (2.7) ≤ (cid:20) k S T k A + k S T k A + q ( k S T k A − k S T k A ) + 4 k S T k A k S T k A (cid:21) . Proof.
Notice first that, in general, for T ∈ B A / ( H ) and A = (cid:18) A A (cid:19) , we have (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) T
00 0 (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) A = k T k A . (2.8)Indeed, it is not difficult to observe that (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) T
00 0 (cid:19) (cid:18) xy (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) A = k T x k A , for all ( x, y ) ∈ H ⊕ H . So, we get (2.8) by taking the supremum over all ( x, y ) ∈H ⊕ H with k x k A + k y k A = 1 and using (1.2).Now, let A = (cid:18) A A (cid:19) . By using (2.8) together with (1.6) we see that r A ( T S + T S ) = r A (cid:20)(cid:18) T S + T S
00 0 (cid:19)(cid:21) = r A (cid:20)(cid:18) T T (cid:19) (cid:18) S S (cid:19)(cid:21) = r A (cid:20)(cid:18) S S (cid:19) (cid:18) T T (cid:19)(cid:21) ( by (1.7) )= r A (cid:20)(cid:18) S T S T S T S T (cid:19)(cid:21) ≤ r (cid:20)(cid:18) k S T k A k S T k A k S T k A k S T k A (cid:19)(cid:21) ( by Theorem . (cid:16) k S T k A + k S T k A + q ( k S T k A − k S T k A ) + 4 k S T k A k S T k A (cid:17) (cid:3) Corollary 2.1.
Let
T, S ∈ B A / ( H ) . Then, we have r A ( T S ± ST ) ≤ (cid:18) k T S k A + k ST k A + q ( k T S k A − k ST k A ) + 4 k T k A k S k A (cid:19) Proof.
By letting T = S = T , S = S and T = ± S in Theorem 2.2 we get thedesired result. (cid:3) Notice that Corollary 2.1 provides an upper bound for the A -spectral radius ofthe commutator T S − ST . Kais Feki
Corollary 2.2.
Let U ∈ B A ( H ) be an A -unitary operator and T ∈ B A / ( H ) .Then, we have r A ( T U ± U T ) ≤ k T k A + (cid:13)(cid:13) T (cid:13)(cid:13) / A . Proof.
Notice first that since U is an A -unitary operator, then k U x k A = k U ♯ A x k A = k x k A . (2.9)This implies, by using (1.2), that k U T k A = k U ♯ A T k A = k T k A , ∀ T ∈ B A / ( H ) . (2.10)Now, we will prove that k T U k A = k T k A . Clearly, we have {k T U x k A ; x ∈ H , k x k A } ⊆ {k T y k A ; y ∈ H , k y k A } . So, by (1.2) we get k T U k A ≤ k T k A . On the other hand, let λ ∈ {|h T ♯ A x | y i A | ; x, y ∈ H , k x k A = k y k A = 1 } , then there exist x, y ∈ H such that k x k A = k y k A = 1 and λ = |h T ♯ A x | y i A | . Let x = P A x + z and y = P A y + z with z , z ∈ N ( A ) . Since T ∈ B A / ( H ) , then N ( A ) is an invariant subspace for each T . Hence, we obtain λ = |h x | T y i A | = |h P A x + z | AT ( P A y + z ) i| = |h P A x | AT ( P A y + z ) i| = |h P A x | AT ( P A y ) i| = |h P A x | T P A y i A | = |h T ♯ A P A x | P A y i A | . Moreover, since P A = ( U ♯ A ) ♯ A U ♯ A , it follows that λ = |h T ♯ A ( U ♯ A ) ♯ A U ♯ A x | ( U ♯ A ) ♯ A U ♯ A y i A | = |h U ♯ A T ♯ A ( U ♯ A ) ♯ A U ♯ A x | U ♯ A y i A | = |h [ U ♯ A T U ] ♯ A U ♯ A x | U ♯ A y i A | . Hence, λ ∈ (cid:8) |h [ U ♯ A T U ] ♯ A z | t i A | ; z, t ∈ H , k z k A = k t k A = 1 (cid:9) . This yields, by (1.3), that k T ♯ A k A ≤ k [ U ♯ A T U ] ♯ A k A which in turn implies that k T k A ≤ k U ♯ A T U k A . So, by (2.10) we get k T k A ≤ k T U k A . Consequently, we have k T k A = k T U k A . (2.11)Therefore, by letting S = U in Corollary 2.1 and then using (2.10) together with(2.11) we see that r A ( T U ± U T ) ≤ k T k A + (cid:13)(cid:13) T (cid:13)(cid:13) / A (cid:13)(cid:13) U (cid:13)(cid:13) / A . This leads to the desired inequality since k U k A = 1 . (cid:3) ome A -spectral radius inequalities for A -bounded Hilbert space operators 9 In order to prove our next result, we need the following lemma.
Lemma 2.2.
Let
T, S ∈ B A / ( H ) and A = (cid:18) A A (cid:19) . Then, the following asser-tions hold: (1) r A (cid:20)(cid:18) T S (cid:19)(cid:21) = max { r A ( T ) , r A ( S ) } . (2) r A (cid:20)(cid:18) TS (cid:19)(cid:21) = p r A ( T S ) .Proof. (1) It can observed that (cid:20)(cid:18) T S (cid:19)(cid:21) n = (cid:18) T n S n (cid:19) , ∀ n ∈ N ∗ . Moreover, for every ( x, y ) ∈ H ⊕ H we have (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) T n S n (cid:19) (cid:18) xy (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) A = (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) T n xT n y (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) A = k T n x k A + k S n y k A ≤ max (cid:8) k T n k A , k S n k A (cid:9) ( k x k A + k y k A ) . This implies, by using (1.2), that (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) T n S n (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) A ≤ max {k T n k A , k S n k A } . Let ( x, ∈ H ⊕ H be such that k x k A = 1 . Then (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) T n S n (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) A ≥ (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) T n S n (cid:19) (cid:18) x (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) A = k T n x k A . So, by taking the supremum over all x ∈ H with k x k A = 1 , we get (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) T n S n (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) A ≥ k T n k A . Similarly, if we take (0 , y ) ∈ H ⊕ H with k y k A = 1 , we get (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) T n S n (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) A ≥ k S n k A . Hence, (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) T n S n (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) A = max {k T n k A , k S n k A } , for all n ∈ N ∗ . Hence, the proof of the first assertion is finished by using (1.6).(2) By using the first assertion and Proposition 1.1 we see that r A (cid:20)(cid:18) TS (cid:19)(cid:21) = r A "(cid:18) TS (cid:19) = r A (cid:20)(cid:18) T S ST (cid:19)(cid:21) = max { r A ( T S ) , r A ( ST ) } . However, by (1.7) we have r A ( T S ) = r A ( ST ) . Therefore, the proof is complete. (cid:3) Now, we state the following theorem.
Theorem 2.3.
Let T = (cid:18) P QR S (cid:19) be such that
P, Q, R, S ∈ B A / ( H ) and A = (cid:18) A A (cid:19) . Then, max np r A ( QR ) , max { r A ( P ) , r A ( S ) } o ≤ (cid:16) k T k A + k T k / A (cid:17) . (2.12) Proof.
Let U = (cid:18) I OO − I (cid:19) . In view of [11, Lemma 3.1.], we have U ∈ B A ( H⊕H ) and U ♯ A = (cid:18) I ♯ A OO ( − I ) ♯ A (cid:19) . So, ones get U ♯ A U = (cid:18) P A OO − P A (cid:19) (cid:18) I OO − I (cid:19) = (cid:18) P A OO P A (cid:19) = P A , Similarly, we show that ( U ♯ A ) ♯ A U ♯ A = P A . Hence, U is an A -unitary operator.Moreover, it can verified that TU + UT = 2 (cid:18) P − S (cid:19) and TU − UT = 2 (cid:18) − QR (cid:19) . (2.13)So, by using Lemma 2.27 together with (2.13) we get { r A ( P ) , r A ( S ) } = r A (cid:20)(cid:18) P − S (cid:19)(cid:21) = r A ( TU + UT ) ≤ k T k A + k T k / A , ( by Corollary . . Thus, we get max { r A ( P ) , r A ( S ) } ≤ (cid:16) k T k A + k T k / A (cid:17) . (2.14)On the other hand, again by using Lemma 2.27 together with (2.13) we get p r A ( QR ) = r A (cid:20)(cid:18) − Q R (cid:19)(cid:21) = r A ( TU − UT ) ≤ k T k A + k T k / A , ( by Corollary . . Thus, we get p r A ( QR ) ≤ (cid:16) k T k A + k T k / A (cid:17) . (2.15)Combining (2.14) together with (2.15) yields to the desired result. (cid:3) In order to establish a new A -spectral radius inequality, we need to recall from[10] the following lemma. ome A -spectral radius inequalities for A -bounded Hilbert space operators 11 Lemma 2.3.
Let T = (cid:18) P QR S (cid:19) be such that
P, Q, R, S ∈ B A / ( H ) and A = (cid:18) A A (cid:19) . Then ω A ( T ) ≤ (cid:20) ω A ( P ) + ω A ( S ) + q(cid:0) ω A ( P ) − ω A ( S ) (cid:1) + (cid:0) k Q k A + k R k A (cid:1) (cid:21) . Theorem 2.4.
Let T , T , S , S ∈ B A / ( H ) . Then, r A ( T S + T S ) ≤ (cid:20) ω A ( S T ) + ω A ( S T ) (cid:1) + q ( ω A ( S T ) − ω A ( S T )) + 4 k S T k A k S T k A (cid:21) . Moreover, this inequality refines (2.7) .Proof.
Let A = (cid:18) A A (cid:19) . By proceeding as in the proof of Theorem 2.2 andusing (1.9) we get r A ( T S + T S ) = r A (cid:20)(cid:18) S T S T S T S T (cid:19)(cid:21) ≤ ω A (cid:20)(cid:18) S T S T S T S T (cid:19)(cid:21) . So, by applying Lemma 2.3 we obtain r A ( T S + T S ) ≤ (cid:0) ω A ( S T ) + ω A ( S T ) (cid:1) + 12 (cid:16)q ( ω A ( S T ) − ω A ( S T )) + (cid:0) k S T k A + k S T k A (cid:1) (cid:17) . On the other hand, it can be observed that for every positive real numbers α and β we have inf (cid:8) αt + βt : t ∈ R , t > (cid:9) = 2 p αβ. (2.16)If k S T k A = 0 or k S T k A = 0 , then the required inequality holds trivially.Assume that k S T k A = 0 and k S T k A = 0 . By replacing T and S by εT and ε S in the last inequality respectively, and then taking the infimum over ε > and using (2.16) we get the desired inequality.Now, in order to see that the obtained inequality refines (2.7), we let a = 12 (cid:20) k S T k + k S T k + q ( k S T k − k S T k ) + 4 k S T k k S T k (cid:21) and a = 12 (cid:20) ω A ( S T ) + ω A ( S T ) (cid:1) + q ( ω A ( S T ) − ω A ( S T )) + 4 k S T k A k S T k A (cid:21) . It is not difficult to observe that a = r (cid:20)(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:21) = (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) , and a = r (cid:20)(cid:18) ω A ( S T ) p k S T k A k S T k A p k S T k A k S T k A ω A ( S T ) (cid:19)(cid:21) = (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) ω A ( S T ) p k S T k A k S T k A p k S T k A k S T k A ω A ( S T ) (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) Since ω A ( X ) ≤ k X k A for all X ∈ B A / ( H ) , then it follows from the norm mono-tonicity of matrices with nonnegative entries that a ≤ a . (cid:3) Corollary 2.3.
Let
T, S ∈ B A / ( H ) . Then, we have r A ( T + S ) ≤ (cid:18) ω A ( T ) + ω A ( S ) + q [ ω A ( T ) − ω A ( S )] + 4 µ ( T, S ) (cid:19) , where µ ( T, S ) = min {k T S k A , k ST k A } .Proof. By letting T = T , S = S and T = S = I in Theorem 2.4 we get r A ( T + S ) ≤ (cid:18) ω A ( T ) + ω A ( S ) + q [ ω A ( T ) − ω A ( S )] + 4 k ST k A (cid:19) . (2.17)This implies, by symmetry, that r A ( T + S ) ≤ (cid:18) ω A ( T ) + ω A ( S ) + q [ ω A ( T ) − ω A ( S )] + 4 k T S k A (cid:19) . (2.18)So, we get the desired result by combining (2.17) together with (2.18). (cid:3) Corollary 2.4.
Let
T, S ∈ B A / ( H ) . Then, we have r A ( T S ) ≤ (cid:18) ω A ( T S ) + ω A ( ST ) + q [ ω A ( T S ) − ω A ( ST )] + 4 ν ( T, S ) (cid:19) , where ν ( T, S ) = min {k T k A k ST S k A , k S k A k T ST k A } .Proof. By letting T = T , S = S , T = T S and S = I in Theorem 2.4 we obtain r A ( T S ) ≤ (cid:18) ω A ( T S ) + ω A ( ST ) + q [ ω A ( T S ) − ω A ( ST )] + 4 k T k A k ST S k A (cid:19) . This in turn implies, by symmetry, that r A ( T S ) ≤ (cid:18) ω A ( T S ) + ω A ( ST ) + q [ ω A ( T S ) − ω A ( ST )] + 4 k S k A k T ST k A (cid:19) . Therefore, the desired inequality follows immediately from the above two inequal-ities. (cid:3) ome A -spectral radius inequalities for A -bounded Hilbert space operators 13 Corollary 2.5.
Let
T, S ∈ B A / ( H ) . Then, we have r A ( T S ± ST ) (2.19) ≤ (cid:18) ω A ( T S ) + ω A ( ST ) + q [ ω A ( T S ) − ω A ( ST )] + 4 k T k A k S k A (cid:19) Moreover, if
T S = ST , then r A ( T S ) ≤ (cid:16) ω A ( T S ) + (cid:13)(cid:13) T (cid:13)(cid:13) / A (cid:13)(cid:13) S (cid:13)(cid:13) / A (cid:17) . (2.20) Proof.
The inequality (2.19) follows from Theorem 2.4 by letting T = S = T , S = S and T = ± S . Moreover, if T S = ST , then (2.20) holds immediately byusing (2.19). (cid:3) The following proposition is also an immediate consequence of Theorem 2.4.
Proposition 2.1.
Let
T, S ∈ B A / ( H ) . Then, we have r A ( T S ± ST ) ≤ ω A ( T S ) + min n k T k / A (cid:13)(cid:13) T S (cid:13)(cid:13) / A , (cid:13)(cid:13) T S (cid:13)(cid:13) / A k S k / A o . (2.21) and r A ( T S ± ST ) ≤ ω A ( ST ) + min n k S k / A (cid:13)(cid:13) ST (cid:13)(cid:13) / A , (cid:13)(cid:13) S T (cid:13)(cid:13) / A k T k / A o . (2.22) Proof.
By letting T = I , T = S , S = T S and S = ± T in Theorem 2.4 we get r A ( T S ± ST ) ≤ ω A ( T S ) + k T k / A (cid:13)(cid:13) T S (cid:13)(cid:13) / A . (2.23)On the other hand, similarly by letting T = T S , T = S , S = I and S = ± T in Theorem 2.4 we obtain r A ( T S ± ST ) ≤ ω A ( T S ) + (cid:13)(cid:13) T S (cid:13)(cid:13) / A k S k / A . (2.24)So, the inequality (2.21) follows immediately by using (2.23) and (2.24). Inaddition, the inequality (2.22) follows from (2.21) by symmetry. (cid:3) Corollary 2.6.
Let
T, S ∈ B A / ( H ) be such that T S = ST . Then, r A ( T S ) ≤ h ω A ( T S ) + k T k / A k S k / A k T S k / A i . (2.25) and r A ( T S ) ≤ h ω A ( T S ) + min n k T k A (cid:13)(cid:13) S (cid:13)(cid:13) / A , (cid:13)(cid:13) T (cid:13)(cid:13) / A k S k A oi . (2.26) Proof.
Since
T S = ST , then it follows from (2.21) that r A ( T S ) ≤ h k T S k A + min n k T k / A (cid:13)(cid:13) T S (cid:13)(cid:13) / A , (cid:13)(cid:13) T S (cid:13)(cid:13) / A k S k / A oi . (2.27)On the other hand, by using (1.5) we see that k T k / A (cid:13)(cid:13) T S (cid:13)(cid:13) / A ≤ k T k / A k S k / A k T S k / A , and (cid:13)(cid:13) T S (cid:13)(cid:13) / A k S k / A ≤ k T k / A k S k / A k T S k / A . So, we infer that min n k T k / A (cid:13)(cid:13) T S (cid:13)(cid:13) / A , (cid:13)(cid:13) T S (cid:13)(cid:13) / A k S k / A o ≤ k T k / A k S k / A k T S k / A . Hence, by taking into account (2.27), we get (2.25) as required. Moreover, (2.26)follows immediately by using (2.27) together with (1.5). (cid:3)
Now, in order to prove our next result which also a consequence of Theorem2.4 we need to recall from [20] the following lemma.
Lemma 2.4.
Let 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 , where ℜ A ( e iθ T ) = e iθ T + e − iθ T ♯ A . Our next result is stated as follows.
Theorem 2.5.
Let
T, S ∈ B A ( H ) . Then, ω A ( T S ) ≤ (cid:16) ω A ( ST ) + k T k A k S k A (cid:17) . (2.28) Proof.
Let θ ∈ R . It can be seen that ℜ A ( e iθ T S ) is an A -self-adjoint operator(that is A ℜ A ( e iθ T S ) is a self-adjoint operator). So, by [16] we deduce that kℜ A ( e iθ T S ) k A = r A (cid:16) ℜ A ( e iθ T S ) (cid:17) . On the other hand, we have r A (cid:16) ℜ A ( e iθ T S ) (cid:17) = 12 r A (cid:16) e iθ T S + e − iθ ( T S ) ♯ A (cid:17) = 12 r A (cid:16) e iθ T S + e − iθ S ♯ A T ♯ A (cid:17) . By letting T = e iθ T , S = S , T = e − iθ S ♯ A and S = T ♯ A in Theorem 2.4, we get r A (cid:16) e iθ T S + e − iθ S ♯ A T ♯ A (cid:17) ≤ ω A ( ST ) + k T k A k S k A . Hence, ω A ( T S ) = sup θ ∈ R (cid:13)(cid:13) ℜ A ( e iθ T S ) (cid:13)(cid:13) A ≤ (cid:16) ω A ( ST ) + k T k A k S k A (cid:17) . (cid:3) Now, we turn your attention to establish an estimate for the A -spectral radiusof the sum of product of a d -pairs of operators. In order to achieve our goal, weneed some prerequisites.The semi-inner product h· | ·i A induces on the quotient H / N ( A ) an inner prod-uct which is not complete unless R ( A ) is closed. However, it was shown in [12]that the completion of H / N ( A ) is isometrically isomorphic to the Hilbert space R ( A / ) := (cid:0) R ( A / ) , h· , ·i R ( A / ) (cid:1) such that h A / x, A / y i R ( A / ) := h P A x | P A y i , ∀ x, y ∈ H . For more information related to the Hilbert space R ( A / ) , we refer the readerto [5, 18] and the references therein. The following proposition is taken from [5]. ome A -spectral radius inequalities for A -bounded Hilbert space operators 15 Proposition 2.2.
Let T ∈ B ( H ) . Then T ∈ B A / ( H ) if and only if there existsa unique e T ∈ B ( R ( A / )) such that Z A T = e T Z A , where Z A : H → R ( A / ) , x Z A x := Ax.
In addition, for T ∈ B A / ( H ) , we have k T k A = k e T k B ( R ( A / )) , (2.29)(see [5, Proposition 3.9]). Also, we need the following Lemma. Lemma 2.5. ( [16] ) If T ∈ B A / ( H ) , then ω A ( T ) = ω ( e T ) . Now, we are in a position to prove the following upper bound for the A -numerical radius of d × d operator matrices which allows us to establish an esti-mate for the A -spectral radius of the sum of products of a d -pairs of operators. Theorem 2.6.
Let T = ( T ij ) d × d be such that T ij ∈ B A / ( H ) for all i, j . Then, ω A ( T ) ≤ max ≤ i ≤ d ( ω A ( T ii ) + 12 d X j =1 ,j = i ( k T ij k A + k T ji k A ) ) . Proof.
Notice first that since T ij ∈ B A / ( H ) for all i, j , then by Lemma 2.1 wehave T ∈ B A / ( H ) . So, by Proposition 2.2 there exists a unique e T ∈ B ( R ( A / )) such that Z A T = e T Z A . On the other hand, since T ij ∈ B A / ( H ) for all i, j , then byProposition 2.2 there exists a unique f T ij ∈ B ( R ( A / )) such that Z A T ij = f T ij Z A .Let S = ( f T ij ) d × d . It is not difficult to verify that Z A T = S Z A . So, we infer that e T = S = ( f T ij ) d × d . (2.30)Now, since S = ( f T ij ) d × d is a d × d operator matrix with f T ij ∈ B ( R ( A / )) for all i, j , then by using [2, Theorem 2] together with [2, Remark 1] we observe that ω ( S ) ≤ max ≤ i ≤ d ( ω ( f T ii ) + 12 d X j =1 ,j = i ( k f T ij k B ( R ( A / )) + k f T ji k B ( R ( A / )) ) ) . This implies, by applying Lemma 2.5 in combination with (2.30) and (2.29), that ω A ( T ) = ω ( S ) ≤ max ≤ i ≤ d ( ω A ( T ii ) + 12 d X j =1 ,j = i ( k T ij k A + k T ji k A ) ) , as required. Hence, the proof is complete. (cid:3) Now we can state and prove the following result which extends [9, Theorem2.10.].
Theorem 2.7.
Let ( T , · · · , T d ) ∈ B A / ( H ) d and ( S , · · · , S d ) ∈ B A / ( H ) d .Then, r A d X k =1 T k S k ! ≤ max ≤ i ≤ d ( ω A ( S i T i ) + 12 d X j =1 ,j = i ( k S i T j k A + k S j T i k A ) ) . Proof.
Notice first that, it can be verified that (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X . . .
00 0 . . . ... . . . (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) A = k X k A , for all X ∈ B A / ( H ) . So, we see that r A d X i =1 T i S i ! = r A P di =1 T i S i . . .
00 0 . . . ... . . . = r A T T . . . T d . . . ... . . . S . . . S . . . ...S d . . . = r A S . . . S . . . ...S d . . . T T . . . T d . . . ... . . . ( by (1.7) ) . Hence, by using (1.9) and Theorem 2.6 we get r A d X k =1 T k S k ! = r A S T S T . . . S T d S T S T . . . S T d ...S d T S d T . . . S d T d ≤ ω A S T S T . . . S T d S T S T . . . S T d ...S d T S d T . . . S d T d ≤ max ≤ i ≤ d { ω A ( S i T i ) + 12 d X j =1 ,j = i ( k S i T j k A + k S j T i k A ) } . ome A -spectral radius inequalities for A -bounded Hilbert space operators 17 Hence, the proof is complete. (cid:3) A -spectral radius for functions of operators In this section, we discuss the A -spectral radius of an operator which is definedby the help of power series f ( z ) = P ∞ n =0 c n z n such that c k are complex coefficientsfor all k . In order to achieve our goal in this section, we need some lemmas. Lemma 3.1.
Let
T, S ∈ B A / ( H ) be such that T S = ST . Then, | r A ( T ) − r A ( S ) | ≤ r A ( T − S ) . (3.1) Proof.
Since
T S = ST , then clearly T − S and S commute. So by Proposition1.1 we have r A ( T ) = r A ( T − S + S ) ≤ r A ( T − S ) + r A ( S ) . This immediately gives r A ( T ) − r A ( S ) ≤ r A ( T − S ) . (3.2)On the other hand, since S − T and T commute, then again by applying Propo-sition 1.1, we get r A ( S ) ≤ r A ( S − T ) + r A ( T ) , which in turn implies that r A ( S ) − r A ( T ) ≤ r A ( S − T ) = r A ( T − S ) . So, we get − r A ( T − S ) ≤ r A ( T ) − r A ( S ) . (3.3)Therefore, we obtain the desired inequality (3.1) by combining (3.2) together with(3.3). (cid:3) Lemma 3.2.
Let ( T n ) n ∈ N ⊂ B A / ( H ) be a sequence of A -bounded operators suchthat T i T j = T j T i for any i, j ∈ N . Then, for every p ∈ N ∗ we have r A p X k =0 T k ! ≤ p X k =0 r A ( T k ) . (3.4) Proof.
We proceed by induction over p . If p = 1 , then (3.4) is true by usingProposition 1.1. Now, assume that (3.4) holds for some p > . Since the operators T j are commuting, then the operators P pk =1 T k and T p +1 commute. So, by usingthe induction hypothesis and applying Proposition 1.1 we see that r A p +1 X k =0 T k ! = r A p X k =0 T k + T p +1 ! ≤ r A p X k =0 T k ! + r A ( T p +1 ) ≤ p X k =0 r A ( T k ) + r A ( T p +1 ) = p +1 X k =0 r A ( T k ) . Hence, (3.4) is proved for any p ∈ N ∗ . (cid:3) Lemma 3.3.
Let ( T n ) n ∈ N ⊂ B A / ( H ) be a sequence of A -bounded operators suchthat T i T j = T j T i for any i, j ∈ N . Let also T ∈ B A / ( H ) . Then, (cid:16) lim n →∞ k T n − T k A = 0 (cid:17) ⇒ (cid:16) lim n →∞ r A ( T n ) = r A ( T ) (cid:17) . (3.5) Proof.
Notice first that, in general, by using (1.6) together with (1.5) it can bechecked that r A ( X ) ≤ k X k A , ∀ X ∈ B A / ( H ) . (3.6)Since T n − T ∈ B A / ( H ) for all n ∈ N , then an application of (3.1) together with(3.6) gives | r A ( T n ) − r A ( T ) | ≤ r A ( T n − T ) ≤ k T n − T k A , for any n ∈ N . Hence, the property (3.5) follows immediately. (cid:3) Now, we are in a position to state and prove the following result.
Theorem 3.1.
Let T ∈ B A / ( H ) be such that A is an invertible operator. Let f ( z ) = P ∞ n =0 c n z n be a power series with complex coefficients and convergent onthe open disk D (0 , R ) ⊂ C with R > . If k T k A < R, then r A [ f ( T )] ≤ f c ( r A ( T )) , (3.7) where f c ( z ) := P ∞ k =0 | c n | z n .Proof. Let T ∈ B A / ( H ) and consider the sequence { S n } in B A / ( H ) such that S n := P nk =0 c k T k for all n ∈ N ∗ . For any p, q ∈ N ∗ with p > q we have k S p − S q k A ≤ p X k = q +1 | c k | k T k kA . (3.8)Since k T k A < R , then P k ∈ N | c k | k T k kA is convergent. So, it follows from (3.8) that { S n } is a Cauchy sequence in B A / ( H ) . On the other hand, since A is invertible,then it can checked that ( B A / ( H ) , k · k A ) is complete. This implies that lim n →∞ k S n − f ( T ) k A = 0 . (3.9)Let n ∈ N ∗ and consider the operators M k := c k T k for k ∈ { , · · · , n } . Since f ( T ) commutes with M k for all k ∈ { , · · · , n } , then we deduce that f ( T ) S n = S n f ( T ) . So, since (3.9) holds, it follows from Lemma 3.3 that lim n →∞ r A ( S n ) = r A [ f ( T )] . (3.10)On the other hand, it can observed that M i M j = M j M i for any i, j ∈ { , · · · , n } . So, by applying Lemma 3.2 we get r A n X k =0 c k T k ! ≤ n X k =0 r A (cid:0) c k T k (cid:1) = n X k =0 | c k | r A (cid:0) T k (cid:1) . This implies, by using Proposition 1.1, that r A n X k =0 c k T k ! ≤ n X k =0 | c k | [ r A ( T )] k . (3.11) ome A -spectral radius inequalities for A -bounded Hilbert space operators 19 Since r A ( T ) ≤ k T k A < R, then f c ( r A ( T )) = P ∞ k =0 | c k | [ r A ( T )] k is convergent.Hence, by letting n → ∞ in (3.11) and then using (3.10), we obtain (3.7) asrequired. (cid:3) Example 3.1.
Let A be the diagonal positive operator on ℓ N ∗ ( C ) given by Ae n − = e n − and Ae n = 2 e n for all n ≥ , where ( e n ) n ∈ N ∗ denotes the canonical basis of ℓ N ∗ ( C ) . Clearly, A is an invertible operator. On the other hand, it is well-knownthat
11 + z = ∞ X n =0 ( − n z n , ∀ z ∈ D (0 , and exp ( z ) = ∞ X n =0 z n n ! ∀ z ∈ C . Now, let T ∈ B A / ( H ) be such that k T k A < . In view of Theorem 3.1 we have r A (cid:2) ( I ± T ) − (cid:3) ≤ [1 − r A ( T )] − . Moreover, if T ∈ B A / ( H ) then again by using Theorem 3.1 we get r A [exp ( T )] ≤ exp [ r A ( T )] . 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 n × n operator matrices,Linear Algebra and its Application, 468 (2015), 18-26.[3] M.L. Arias, G. Corach, M.C. Gonzalez, Partial isometries in semi-Hilbertian spaces, LinearAlgebra Appl. 428 (7) (2008) 1460-1475.[4] M.L. Arias, G. Corach, M.C. Gonzalez, Metric properties of projections in semi-Hilbertianspaces, Integral Equations and Operator Theory, 62 (2008), pp.11-28.[5] M.L. Arias, G. Corach, M.C. Gonzalez, Lifting properties in operator ranges, Acta Sci.Math. (Szeged) 75:3-4(2009), 635-653.[6] H. Baklouti, K.Feki, On joint spectral radius of commuting operators in Hilbert spaces,Linear Algebra Appl. 557 (2018), 455-463.[7] H. Baklouti, K.Feki, O.A.M. Sid Ahmed, Joint numerical ranges of operators in semi-Hilbertian spaces, Linear Algebra Appl. 555 (2018) 266-284.[8] H. Baklouti, K.Feki, O.A.M. Sid Ahmed, Joint normality of oper-ators in semi-Hilbertian spaces, Linear Multilinear Algebra (2019), https://doi.org/10.1080/03081087.2019.1593925 , in press.[9] P. Bhunia, S. Bag and K. Paul, Numerical radius inequalities and its applications in esti-mation of zeros of polynomials, Linear Algebra Appl. 573 (2019) 166-177.[10] P. Bhunia and K. Paul, Some improvements of numerical radius inequalities of operatorsand operator matrices, arXiv:1910.06775v1 [math.FA] (to appear in Linear and MultilinearAlgebra).[11] P. Bhunia, K.Feki, K. Paul, A -Numerical radius orthogonality and parallelism of semi-Hilbertian space operators and their applications, arXiv:2001.04522v1 [math.FA] 13 Jan2020.[12] L. de Branges, J. Rovnyak, Square Summable Power Series, Holt, Rinehert and Winston,New York, 1966.[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, Spectral Radius Inequalities for Functions of Operators Defined by PowerSeries, Filomat 30:10 (2016), 2847-2856.[15] M. Faghih-Ahmadi, F. Gorjizadeh, A-numerical radius of A-normal operators in semi-Hilbertian spaces, Italian journal of pure and applied mathematics n. 36-2016 (73-78).[16] K. Feki, Spectral radius of semi-Hilbertian space operators and its applications, Annals ofFunctionnal Analysis (2020), https://doi.org/10.1007/s43034-020-00064-y .[17] F. Kittaneh, Spectral radius inequalities for Hilbert space operators, Proc. Amer. Math.Soc. 134 (2006), 385-390.[18] W. Majdak, N.A. Secelean, L. Suciu, Ergodic properties of operators in some semi-Hilbertian spaces, Linear and Multilinear Algebra, 61:2, (2013) 139-159.[19] M.S. Moslehian, Q. Xu, A. Zamani, Seminorm and numerical radius inequalities of opera-tors in semi-Hilbertian spaces, Linear Algebra Appl. 591 (2020) 299-321.[20] A. Zamani, A -numerical radius inequalities for semi-Hilbertian space operators, LinearAlgebra Appl. 578(2019) 159-183. [1] University of Sfax, Tunisia.
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