Some Numerical Radius Inequalities for Semi-Hilbert Space Operators 3

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From now on, A always stands for a positive linear operator in ( ). The cone of all positive operators of ( ) will be denoted by ( )+. InB thisH article, for a given A ( )+, we areB goingH to consider an additionalB H semi-inner product on ∈deﬁned B H by h· | ·iA H x y = Ax y , x, y . h | iA h | i ∀ ∈ H This makes into a semi-Hilbertian space. The seminorm induced by is H h· | ·iA given by x A = x x A for every x . It can be observed that ( , A) is a normedk spacek if andh | onlyi if A is one-to-one∈ H and that the semi-HilbertianH k·k space p ( , A) is complete if and only if the range of A is closed in . In what follows, byH ank·k operator we mean a bounded linear operator in ( ).H Also, the range of every operator T is denoted by (T ), its null space by B H(T ). Given a linear R N M subspace of , denotes the closure with respect to the norm topology of . If is a closedH subspaceM of , then P stands for the orthogonal projectionH onto M. H M M The semi-inner product A induces on the quotient / (A) an inner product which is not completeh· unless | ·i (A) is closed. However,H aN canonical construc- tion due to de Branges and RovnyakR [12] shows that the completion of / (A) is isometrically isomorphic to the Hilbert space (A1/2) equipped with theH followingN inner product R 1/2 1/2 1 2 A x, A y R(A / ) := P (A)x P (A)y , x, y . h i h R | R i ∀ ∈ H 1/2 1/2 For the sequel, the Hilbert space (A ), , R 1/2 will be denoted by R(A ). R h· ·i (A ) Deﬁnition 1.1. ([3]) Let T ( ). An operatorS ( ) is called an A- adjoint of T if for every x, y ∈ B, theH identity T x y ∈= BxHSy holds. That ∈ H h | iA h | iA is AS = T ∗A. Generally, the existence of an A-adjoint operator is not guaranteed. The set of all operators in ( ) admitting A-adjoints is denoted by A( ). By Douglas Theorem [15], we haveB H B H

( )= T ( ); (T ∗A) (A) . BA H { ∈ B H R ⊆ R } 1/2 Further, the set of all operators admitting A -adjoints is denoted by A1/2 ( ). Again, by applying Douglas Theorem, we obtain B H

1/2 ( )= T ( ); λ> 0 ; T x λ x , x . BA H { ∈ B H ∃ k kA ≤ k kA ∀ ∈ H} If T 1/2 ( ), we will say that T is A-bounded. Notice that ( ) and ∈ BA H BA H 1/2 ( ) are two subalgebras of ( ) which are, in general, neither closed nor BA H B H dense in ( ). Moreover, the following inclusions A( ) A1/2 ( ) ( ) hold withB equalityH if A is injective and has closedB range.H ⊆ For B anH account⊆ B H of results, we refer to [3, 4, 8, 16] and the references therein. Clearly, induces h· | ·iA a seminorm on 1/2 ( ). Indeed, if T 1/2 ( ), then BA H ∈ BA H T x A T A := sup k k = sup T x A ; x , x A =1 < . k k x (A), x A k k ∈ H k k ∞ ∈xR=0 k k 6 Some numerical radius inequalities for semi-Hilbert space operators 3

Notice that, if T 1/2 ( ), then T = 0 if and only if AT = 0. Further, it ∈ BA H k kA was proved in [20] that for T 1/2 ( ) we have ∈ BA H T = sup T x y ; x, y , x = y =1 . (1.2) k kA {|h | iA| ∈ H k kA k kA } It should be emphasized that it may happen that T = + for some T k kA ∞ ∈ ( ) 1/2 ( ) (see [16, Example 2]). B H \ BA H Before we move on, it should be mentioned that for T 1/2 ( ) we have ∈ BA H T x T x , x . (1.3) k kA ≤k kAk kA ∀ ∈ H Also, we would like to emphasize that (1.3) fails to hold in general for some 0 0 0 1 T ( ). In fact, one can take the operators A = and T = on ∈ B H 0 1 1 0 2 C . If x = (1, 0), then x A = 0 and T x A = 1. Thus, (1.3) fails to be true. Moreover, by applying (k1.3k) we show thatk k TS T S , (1.4) k kA ≤k kAk kA for every T,S 1/2 ( ). ∈ BA H If T A( ), then by Douglas theorem there exists a unique A-adjoint of T , ∈ B H♯ ♯ ♯ denoted by T A , which satisfies (T A ) (A). We observe that T A = A†T ∗A, R ⊆ R ♯ where A† is the Moore-Penrose inverse of A. For results concerning T A and A† see ♯A ♯A ♯A [3, 4]. Notice that if T A( ), then T A( ) and (T ) = P (A)TP (A). ∈ B H ∈ B H R R For more facts related to this class of operators, we invite the reader to [3, 4] and their references. Now, we recall that an operator T ( ) is said to be A-isometry if T x = x for all x . Further, an operator∈ B H U ( ) is k kA k kA ∈ H ∈ BA H called A-unitary if U and U ♯A are A-isometries. By an A-normal operator, we ♯A ♯A mean an operator T A( ) which satisfies T T = T T . For more details related to these classes∈ of B operators,H the reader can consult [3, 9]. Any operator T ( ) can be represented as T = (T )+ i (T ), where ∈ BA H ℜA ℑA T + T ♯A T T ♯A (T ) := and (T ) := − . ℜA 2 ℑA 2i Recently, the A-numerical range of an operator T ( ) is defined by Baklouti ∈ B H et al. in [8] as WA(T ) = T x x A ; x , x A = 1 . It was shown in [8] that W (T ) is a nonemptyh convex| i subset∈ of H Ck whichk is not necessarily closed A even if dim( ) < . Notice that supremum modulus of WA(T ) is called the A-numericalH radius∞ of T (see [8]). More precisely, we have ω (T ) = sup λ ; λ W (T ) = sup T x x ; x , x =1 . A | | ∈ A h | iA ∈ H k kA It should be mentioned that W (T )= C when T ( ) and satisfies T ( ( A)) A (A) ([8, Theorem 2.1.]). So, ω (T )=+ for∈ every B H operator T N( ) such6⊆ N A ∞ ∈ B H that T ( (A)) (A). Notice that ωA( ) is a seminorm on A1/2 ( ) with is equivalentN to the6⊆A N-operator seminorm. More· precisely, it was shownB H in [8] that for every T 1/2 ( ), we have ∈ BA H 1 T ω (T ) T . (1.5) 2 k kA ≤ A ≤k kA 4 Kais Feki

By an A-selfadjoint, we mean an T ( ) which satisﬁes AT is selfadjoint, ∈ B H that is, AT = T ∗A. Further, for every A-selfadjoint operator T we have T = ω (T ) := sup T x x ; x , x =1 . (1.6) k kA A {|h | iA| ∈ H k kA } (see [16]). In addition, an operator T is said to be A-positive if AT 0 and we ♯A ♯A ≥ write T A 0. One can verify that T T A 0 and T T A 0. Moreover, in view of [4≥, Proposition 2.3.], we have ≥ ≥

2 T ♯A T = T T ♯A = T 2 = T ♯A . (1.7) k kA k kA k kA k kA For a given operator T , the A-Crawford number of T is deﬁned, as in [29], by c (T ) = inf T x x ; x , x =1 . A |h | iA| ∈ H k kA Recently, the present author proved in [16] some A-numerical radius inequalities for A-bounded operators. In particular, he showed that for every T A1/2 ( ) and all positive integer n, we have ∈ B H ω (T n) [ω (T )]n. (1.8) A ≤ A Also, it has been shown in [16] that for every T 1/2 ( ), we have ∈ BA H 1 ω (T ) T + T 2 1/2 . (1.9) A ≤ 2 k kA k kA Clearly, (1.9) is a reﬁnement of the second inequality in (1.5). For other facts and results related to the concept of A-numerical radius, the reader is referred to [8, 9, 16, 29] and the references therein. In recent years, several results covering some classes of operators on a complex Hilbert space , were extended H h· | ·i to , . The reader is invited to see [8, 9, 10, 18, 26, 28, 29, 30] and the A referencesH h· | therein.·i In this article, we will establish several results governing ω ( ) A · and A. Some of these results will be a natural extensions of the well-known case k·kA = I due to Kittaneh et al. [1, 2, 24, 6].

2. Results In this section, we present our results. In order to prove our ﬁrst result, we need the following lemmas. Lemma 2.1. ([29, Theorem 2.6.]) Let T ( ). Then ∈ BA H ω (T ) = sup α (T )+ β (T ) ; α, β R , α2 + β2 =1 . A k ℜA ℑA kA ∈ n o Lemma 2.2. ([29, Theorem 2.5.]) Let T ( ). Then ∈ BA H iθ iθ ωA(T ) = sup A(e T ) A = sup A(e T ) A θ R ℜ θ R ℑ ∈ ∈ Lemma 2.3. ([18, Lemma 1]) Let T ( ) be an A-selfadjoint operator. Then, ∈ B H T ♯A is A-selfadjoint and (T ♯A )♯A = T ♯A . Some numerical radius inequalities for semi-Hilbert space operators 5

A 0 T T Lemma 2.4. ([10, Lemma 6]) Let A = and T = 11 12 be such 0 A T21 T22 that T ( ) for all i, j 1, 2 . Then, T A( ) and ij ∈ BA H ∈{ } ∈ B H ⊕ H T ♯A T ♯A T♯A = 11 21 . T ♯A T ♯A 12 22 Now, we are in a position to prove our ﬁrst result in this paper. Theorem 2.1. Let T ( ). Then, ∈ BA H 1 T ♯A T + T T ♯A ω2 (T ) 1 T ♯A T + T T ♯A . (2.1) 4 k kA ≤ A ≤ 2 k kA Moreover, the inequalities in (2.1) are sharp.

♯ Proof. By using the property AT A = T ∗A we see that 1 1 ♯ 1 ♯ 1 A[ (T )] = AT + AT A = (T A )∗A + T ∗A = [ (T )]∗A. ℜA 2 2 2 2 ℜA Hence, A(T ) is an A-selfadjoint operator. Similarly, we prove that A(T ) is A-selfadjoint.ℜ So, by Lemma 2.3 we have ℑ ([ (T )]♯A )♯A = [ (T )]♯A and ([ (T )]♯A )♯A = [ (T )]♯A . (2.2) ℜA ℜA ℑA ℑA Now, it can be observed that

2 2 1 [ (T )]♯A + [ (T )]♯A = (T ♯A )♯A T ♯A + T ♯A (T ♯A )♯A ℜA ℑA 2 1 ♯ ♯ ♯A = T T A + T A T . (2.3) 2 A 0 Let A = and x be such that x = 1. Let also α, β R be such 0 A ∈ H k kA ∈ that α2 + β2 =1. It can seen that 2 α[ (T )]♯A + β[ (T )]♯A x ℜA ℑA A 2 ♯A ♯A [ A(T )] [ A(T )] αx = ℜ 0ℑ 0 βx A 2 ♯A ♯A [ A(T )] [ A(T )] ℜ ℑ (by (1.3)) ≤ 0 0 A ♯A [ (T )]♯A [ (T )]♯A [ (T )]♯A [ (T )]♯A = ℜA ℑA ℜA ℑA (by (1.7)) 0 0 0 0 A [ (T )]♯A [ (T )]♯A [ (T )]♯A 0 = ℜA ℑA ℜA 0 0 [ (T )]♯A 0 ℑA A ♯A 2 ♯A 2 = ([ A(T )] ) + ([ A(T )] ) ℜ ℑ A 1 ♯ = T T ♯A + T ♯A T A (by (2.3)) 2 A

1 ♯ ♯ = T T A + T A T (by (1.7)). 2 A

6 Kais Feki

Therefore, we obtain

2 2 ♯A ωA(T )= ωA(T ) 2 ♯A ♯A = sup α[ A(T )] + β[ A(T )] A (by Lemma 2.2) α2+β2=1 ℜ ℑ 1 T T ♯A + T ♯A T . (2.4) ≤ 2 A Let θ R. By making simple computations we see that ∈ 2 2 1 (eiθT ) + (eiθT ) = T T ♯A + T ♯A T . (2.5) ℜA ℑA 2 This implies, that

1 2 2 ♯A ♯A iθ iθ T T + T T = A(e T ) + A(e T ) 2 A ℜ ℑ A iθ 2 iθ 2 A(e T ) + A(e T ) ≤ ℜ A ℑ A 2 2 ω A(T ). (by (1.4) and Lemma 2 .2) ≤ Hence, we get 1 T T ♯A + T ♯A T ω2 (T ). (2.6) 4 A ≤ A Combining (2.4) together with (2.6) yields to (2.1) as desired. Now, to see that 1 the constant 2 is sharp we consider an arbitrary A-normal operator T . By [16, Theorem 4] we have T = ω (T ). So, we get k kA A 1 T ♯A T + T T ♯A = T T ♯A = T 2 = ω2 (T ). 2 k kA k kA k kA A The sharpness of the first inequality in (2.1) can be verified by considering the α 0 0 1 operators A = for some α =0 and T = 0 α 6 0 0 Remark 2.1. (1) If A = I, we obtain the well-known inequalities proved by F. Kittaneh in [24, Theorem 1]. Also, if A is an injective positive operator, we get the recent result proved by Bhunia et al. proved in [11, Corollary 2.7]. (2) The second inequality in Theorem 2.1 has recently been proved by Zamani in [29, Theorem 2.10] using a completely different argument. Moreover, it is not difficult to see that 1 1 √2 T T T ♯A + T ♯A T ω (T ) T T ♯A + T ♯A T T . 2k kA ≤ 2 A ≤ A ≤ 2 A ≤k kA q q So, the inequalities in Theorem 2.1 improve the inequalities in (1.5 ). In order to prove our next result in this paper, we need the following lemma. A 0 Lemma 2.5. ([19]) Let T,S ( ) and A = . Then, ∈ BA H 0 A 0 T 1 iθ iθ ♯ A A ω = sup e T + e− S A . S 0 2 θ R ∈

Some numerical radius inequalities for semi-Hilbert space operators 7

In particular, 0 T ωA = ω (T ). (2.7) T 0 A Now, we are in a position to prove our second result in this paper. A 0 Theorem 2.2. Let A = and T,S ( ). Then, 0 A ∈ BA H 0 T T + S A ωA min Ψ (T,S), Ψ (S, T ) k k , (2.8) S 0 ≤ A A ≤ 2 and 0 T ωA max Φ (T,S), Φ (S, T ) , (2.9) S 0 ≥ A A where 1 Ψ (T,S)= T T ♯A + S♯A S +2ω (TS), A 2 A A and q 1 Φ (T,S)= T T ♯A + S♯A S +2c (TS). A 2 A A q Proof. By Lemma 2.5, we see that 0 T 0 T ♯A ωA = ωA S 0 S 0 " # 1 iθ ♯A iθ ♯A ♯A = sup e S + e− (T ) A 2 θ R ∈ 1 ♯ 2 1 iθ ♯ iθ ♯ ♯ iθ ♯ iθ ♯ ♯ A = sup e S A + e− (T A ) A e S A + e− (T A ) A (by (1.7)) 2 θ R ∈ A 1 1 2 iθ ♯A iθ ♯A ♯A iθ ♯A ♯A iθ ♯A = sup e S + e− (T ) e− (S ) + e T 2 θ R A ∈ 1 1 ♯A ♯A ♯A ♯A ♯A ♯A 2iθ ♯A ♯A 2 = sup S (S ) +(T ) T +2 A e S T A 2 θ R ℜ ∈ 1 ♯A ♯A ♯A ♯A ♯A ♯A 2iθ ♯ A ♯A S (S ) +(T ) T A + 2sup A e S T A ≤ 2 k k θ R ℜ ∈ r 1 ♯ ♯ ♯A ♯ ♯ = T T A + S A S +2ωA S A T A (by Lemma 2.2) 2 A r 1 ♯ ♯ = T T A + S A S +2 ωA TS =ΨA(T,S), 2 k kA q where the last equality holds since X = X♯A and ω (X) = ω (X♯A ) for k kA k kA A A 0 T every X ( ). Hence, ωA Ψ (T,S). Moreover, by [19] we have ∈ BA H S 0 ≤ A 0 T 0 S 0 T ωA = ωA . This implies that ωA Ψ (S, T ). S 0 T 0 S 0 ≤ A 8 Kais Feki

0 T Thus, ωA min Ψ (T,S), Ψ (S, T ) as desired. In addition, we S 0 ≤ A A have min Ψ (T,S), Ψ (S, T ) Ψ (T,S) A A ≤ A 1 = T T ♯A + S♯A S +2ω (TS) 2 A A q 1 ♯ ♯ T T A A + S A S A +2 TS A (by (1.5)) ≤ 2 k k k k k k 1p T 2 + S 2 +2 T S (by (1.7) and (1.4)) ≤ 2 k kA k kA k kAk kA 1q T + S = ( T + S )2 = k kA k kA . 2 k kA k kA 2 q Hence, (2.8) is proved. On the other hand, let x be such x A =1 and let β ♯A ♯A ∈ H 2iβ ♯A k♯Ak be a real number which satisﬁes S T x x A = e− S T x x A . So, we have h | i h | i

♯A 0 T 0 T ωA = ωA S 0 S 0 " # 1 iβ ♯ iβ ♯ ♯ e S A + e− (T A ) A ≥ 2 A 1 iβ ♯ iβ ♯ ♯ iβ ♯ ♯ iβ ♯ 2 = e S A + e− (T A ) A e− (S A ) A + e T A (by (1.7)) A 1 1 = S♯A (S♯A )♯A +(T ♯A )♯A T ♯A +2 e2iβS♯A T ♯A 2 . (2.10) 2 ℜA A

♯A ♯A ♯A ♯A ♯A ♯A 2iβ ♯A ♯A It can be veriﬁed that S (S ) +(T ) T +2 A e S T is A-selfadjoint, then by (1.6) we have ℜ S♯A (S♯A )♯A +(T ♯A )♯A T ♯A +2 e2iβS♯A T ♯A ℜA A S♯A (S♯A )♯A +(T ♯A )♯A T ♯A +2 e2iβS♯A T ♯A x x ≥ h ℜA | iA ♯A ♯A ♯A ♯A ♯A ♯A 2iβ ♯A ♯A = S (S ) +(T ) T x x A+2 A e S T x x A h | i hℜ | i = S♯A (S♯A )♯A +(T ♯A )♯A T ♯A x x +2 e2iβ S♯A T ♯A x x h | iA ℜ h | iA = S♯A (S♯A )♯A +(T ♯A )♯A T ♯A x x +2 S♯A T ♯A x x h | iA h | iA ♯A ♯A ♯A ♯A ♯A ♯A ♯A ♯A = S (S ) +(T ) T x x A +2 S T x x A , h | i h | i ♯A ♯A ♯A ♯A ♯A ♯A where the last equality follows since S (S ) +( T ) T A 0. So, by taking ♯A ≥ into account (2.10) and the fact that cA(X) = cA(X ) for all X A( ), we obtain ∈ B H 0 T 1 ωA S♯A (S♯A )♯A +(T ♯A )♯A T ♯A x x +2c (TS). S 0 ≥ 2 h | iA A q Some numerical radius inequalities for semi-Hilbert space operators 9

So, by taking the supremum over all x with x A =1 in the above inequality and then using (1.6) we get ∈ H k k 0 T 1 ωA S♯A (S♯A )♯A +(T ♯A )♯A T ♯A +2c (TS)=Φ(T,S), S 0 ≥ 2 A A q ♯A where the last equality follows since X A = X A for all X A( ). Now, by using an argument similar to thatk usedk ink the proofk of (2.8),∈ we B getH (2.9) as desired. This finishes the proof of the theorem. By letting T = S in the above theorem and then using (2.7), we reach the following corollary which generalizes [1, Theorem 2.4.] and considerably improves the inequalities in Theorem 2.1. Corollary 2.1. Let T ( ). Then ∈ BA H 1 1 T T ♯A + T ♯A T +2c (T 2) ω (T ) T T ♯A + T ♯A T +2ω (T 2). 2 A A ≤ A ≤ 2 A A q q Remark 2.2. Obviously, the first inequality in Corollary 2.1 is sharper than the first inequality in Theorem 2.1. Moreover, it can be observed that 1 1 ω2 (T ) T T ♯A + T ♯A T + ω (T 2) A ≤ 4 A 2 A 1 1 T T ♯A + T ♯A T + ω2 (T ) (by (1.8)) ≤ 4 A 2 A 1 1 T T ♯A + T ♯A T + T T ♯A + T ♯A T (by (2.1)) ≤ 4 A 4 A 1 = T T ♯A + T ♯A T . 2 A

This shows that the second inequality in Corollary 2.1 reﬁnes the second inequality in Theorem 2.1. As an application of Corollary 2.1 we state the following result. A 0 Theorem 2.3. Let A = . Let also T ( ) be such that (A)⊥ is 0 A ∈ BA H N an invariant subspace for T . Then, 1 I T 1 I T I T − ωA = + . (2.11) 0 I 2 0 I 0 I − − A − A !

I T I 0 Proof. Let T = . It can be seen that T2 = . Hence, by Corollary 0 I 0 I 2.1 we have − 1 ωA(T)= TT♯A + T♯A T +2. (2.12) 2 A q P 0 T♯A (A) Now, by Lemma 2.4 we have = R♯A . Moreover, since (A)⊥ is T P (A) N − R an invariant subspace for T , then by [9, Lemma 1.1] we have TP (A) = P (A)T . R R 10 Kais Feki

Thus, a short calculation reveals that T T ♯A +2P 0 TT♯A T♯A T (A) + = R ♯A . 0 T T +2P (A) R This implies, by [19], that

TT♯A T♯A T ♯A ♯A 2 + A = max T T +2P (A) A, T T +2P (A) A = T A +2, k k k R k k R k k k

n o ♯A where the last equality follows by using (1.6) since the operators T T +2P (A) R ♯A and T T +2P (A) are A-positive. So, by taking into consideration (2.12) we get R 1 2 ωA(T)= T +4. (2.13) 2 k kA q Now, we will prove that

1 2 4 2 1 2 1 T A = 2+ T + T +4 T = T +4+ T A. (2.14) k k √2 k kA k kA k kA 2 k kA 2k k r q q By (1.2) there exists two sequences of A-unit vectors xn and yn in such that { } { } H

lim T xn yn A = T A. n + → ∞ |h | i | k k This implies, by applying the Cauchy-Schwarz inequality, that lim T xn A = n + k k T . Let (a, b) R2 be such that a2 + b2 =1 and → ∞ k kA ∈ 1 T A 1 T A a 2 2 k k = k k = (a + b T A) + b . (2.15) 0 1 0 1 b k k q iαn For n N, let T xn yn A = T xn yn A e for some αn R. Let Xn = iαn∈ Th | i |h | i | ∈ { } (ae yn, bxn) be a sequence in . It is not diﬃcult to see that Xn A =1. Moreover,{ a short} calculation revealsH⊕H that k k 2 2 iαn I T I T ae yn 0 I ≥ 0 I bxn − A − A iαn 2 2 = ae yn + bT xn A + bxn A k k k k = a2 + b2 T x 2 +2ab T x y + b 2 k nkA |h n | niA| | | n + → ∞ a2 + b2 T 2 +2ab T + b 2 −−−−→ k kA k kA | | 1 T 2 = A (by (2.15)). 0k 1k

Hence, I T 1 T k kA . 0 I ≥ 0 1 − A

Moreover, by [17] we have

I T 1 T k kA . 0 I ≤ 0 1 − A

Some numerical radius inequalities for semi-Hilbert space operators 11

1 T A Hence, T A = k k . On the other, by using (1.1), we see that k k 0 1

1 T 2 1 0 1 T A = r A 0k 1k T 1 0k 1k k kA 1 T = r A T Tk 2k+1 k kA k kA 1 = 2+ T 2 + T 4 +4 T 2 . 2 k kA k kA k kA q Thus, we prove the ﬁrst equality in (2.14). The second equality in (2.14) follows immediately. Therefore, (2.14) is proved. Now, by combining (2.13) together with (2.14), we get (2.11) as desired. A 0 I T Corollary 2.2. Let A = . Let also T = be such that T 0 A 0 I ∈ − A( ) and (A)⊥ is an invariant subspace for T . Then, the following assertions holdB H N

(1) A(T) A = ωA(T). kℜ k 1 1 (2) (T) A = T A T A− . kℑA k 2 k k −k k Proof. (1) It can be seen that

2 ♯A 2 (T) A = [ (T)] kℜA k ℜA A ♯ ♯ 2 P (T A ) A (A) 2 = TR♯A 2 P (A)! A − R ♯A ♯A ♯A ♯A P (T ) P (T ) (A) 2 (A) 2 = TR♯A TR♯A (by (1.7)) 2 P (A)! 2 P (A)! A − R − R ♯A ♯A ♯A P + (T ) T 0 (A) 4 = R T ♯A (T ♯A )♯A , 0 P (A) + 4 ! A R where the last equality follows by using the fact that (A) is an invariant ⊥ subspace for T . So, we obtain N

♯A ♯A ♯A ♯A ♯A ♯A 2 (T ) T T (T ) (T) A = max P + , P + kℜA k A 4 A 4 A A T ♯A 2 T 2 +4 = k kA +1= k kA . 4 4 Hence, we prove the desired property. 1 T A (2) By proceeding as in (1), one can prove that A( ) = 2 T A. So, the required result holds by using (2.14) and (2.11). kℑ k k k Now, we turn your attention to the study of some A-numerical radius for products and commutators of semi-Hilbert space operators. Our ﬁrst result in this 12 Kais Feki context generalizes [29, Corollary 4.3] since A1/2 ( ) A( ) and reads as follows. B H ⊆ B H

Theorem 2.4. Let T,S 1/2 ( ). Then, ∈ BA H ω (TS) 4 ω (T ) ω (S). (2.16) A ≤ A A If TS = ST , then ω (TS) 2ω (T ) ω (S). (2.17) A ≤ A A Proof. It follows from (1.5) and (1.4) that

ω (TS) TS T S 4 ω (T )ω (S). A ≤k kA ≤k kAk kA ≤ A A This proves (2.16). Now, in order to establish (2.17), we ﬁrst prove that

ω (TS + ST ) 4 ω (T )ω (S). (2.18) A ≤ A A

Assume that T,S A1/2 ( ) and satisfy ωA(T ) = ωA(S)=1. It is not diﬃcult to observe that ∈ B H 1 TS + ST = (T + S)2 (T S)2 . 2 − − So, by using the fact that ω ( ) is a seminorm and (1.8) we see that A · 1 ω (TS + ST ) ω2 (T + S)+ ω2 (T S) A ≤ 2 A A − [ω (T )+ ω (S)]2 =4. ≤ A A Hence,

ω (TS + ST ) 4, (2.19) A ≤ for all T,S 1/2 ( ) with ω (T ) = ω (S)=1. If ω (T ) = ω (S)=0, then ∈ BA H A A A A AT = AS = 0 and so (2.18) holds trivially. Now, assume that ωA(T ) = 0 and T S 6 ωA(S) = 0. By replacing T and S by and respectively in (2.19), we 6 ωA(T ) ωA(S) get (2.18) as required. Now, if TS = ST , then (2.17) follows immediately from (2.18).

Our next result reads as follows.

Theorem 2.5. Let T , T ,S ,S ( ). Then 1 2 1 2 ∈ BA H

ω (T S S T ) T T ♯A + T ♯A T S♯A S + S S♯A . A 1 1 ± 2 2 ≤ 1 1 2 2 A 1 1 2 2 A q q

Some numerical radius inequalities for semi-Hilbert space operators 13

Proof. Let x be such that x A =1. An application of the Cauchy-Schwarz inequality gives∈ H k k 2 2 (T S S T )x x T S x x + S T x x 1 1 ± 2 2 | A ≤ h 1 1 | iA h 2 2 | iA 2 ♯A ♯A = S1x T x + T2x S x h | 1 iA h | 2 iA 2 ♯A ♯A S1x T x + T2x S x ≤ k kAk 1 kA k kAk 2 kA 2 2 T x 2 + T ♯A x S x 2 + S♯A x ≤ k 2 kA k 1 kA k 1 kA k 2 kA = T ♯A T + T T ♯A x x S♯A S + S S♯A x x h 2 2 1 1 | iAh 1 1 2 2 | iA ♯A ♯A ♯A ♯A T1T1 + T2 T2 S1 S1+ S2S2 . ≤ A A Thus 1/2 1/2 (T S S T )x x T T ♯A + T ♯A T S♯A S + S S♯A , 1 1 ± 2 2 | A ≤ 1 1 2 2 A 1 1 2 2 A for all x with x A = 1 . By taking the supremum over all x with x =1 ∈in H the abovek k inequality we obtain ∈ H k kA ω (T S S T ) T T ♯A + T ♯A T S♯A S + S S♯A , A 1 1 ± 2 2 ≤ 1 1 2 2 A 1 1 2 2 A as desired. q q

The following corollary is an immediate consequence of Theorem 2.5 and generalizes the well-known result proved by Fong and Holbrook in [21]. Corollary 2.3. Let T,S ( ). Then ∈ BA H ω (TS ST ) 2√2 min T ω (S), S ω (T ) , (2.20) A ± ≤ k kA A k kA A n o Proof. By letting T1 = T2 = T and S1 = S2 = S in Theorem 2.5 and then using the second inequality in (2.1) we get

ω (TS ST ) T T ♯A + T ♯A T SS♯A + S♯A S A ± ≤ A A q q ♯ 2 ♯ 2 2 T T A + T A T ωA(S) ≤ A A q 2 2 =2 T A + T A ωA(S ) (by (1.7)). Hence, q

ω (TS ST ) 2√2 T ω (S). (2.21) A ± ≤ A A By replacing T and S by S and T respectively in (2.21), we get the desired result. Clearly (2.20) provides an upper bound for the A-numerical radius of the commutator TS ST . Next, we state− the following result which is a natural generalization of another well-known theorem proved by Fong and Holbrook in [21]. 14 Kais Feki

Theorem 2.6. Let T,S ( ). Then, ∈ BA H ω (TS ST ♯A ) 2 T ω (S). A ± ≤ k kA A Proof. We ﬁrst prove that ω (TS + ST ♯A ) 2 T ω (S). (2.22) A ≤ k kA A In view of Lemma 2.2, we have

♯A ♯A ♯A ωA(TS + ST )= ωA([TS + ST ] )

iθ ♯A ♯A = sup A e [TS + ST ] A. (2.23) θ R ℜ ∈ Moreover, a short calculation reveals that eiθ[TS + ST ♯A ]♯A =(T ♯A )♯A eiθS♯A + eiθS♯A T ♯A , ℜA ℜA ℜA for every θ R. So, by using (2.23) together with Lemma (2.2 ) we get ∈ ♯A ♯A ♯A iθ ♯A iθ ♯A ♯A ωA(TS + ST ) = sup (T ) A e S + A e S T A θ R ℜ ℜ ∈ ♯ A ♯A iθ ♯A ♯A iθ ♯A (T ) A sup A e S A + T A sup A e S A ≤k k θ R ℜ k k θ R ℜ ∈ ∈ =2 T ω (S♯A ) ( by (1.7 )) k kA A =2 T ω (S). k kA A This proves (2.22). By replacing T by iT in (2.22) we get ω (TS ST ♯A ) 2 T ω (S). (2.24) A − ≤ k kA A Therefore, we prove the desired result. Our next aim is to give an improvement of (1.9). In order to achieve this goal and other goals in the rest of this paper, we need the following lemmas.

Lemma 2.6. ([5]) Let T ( ). Then T 1/2 ( ) if and only if there exists ∈ B H ∈ BA H a unique T (R(A1/2)) such that Z T = TZ . Here, Z : R(A1/2) is ∈ B A A A H → deﬁned by ZAx = Ax. e e Lemma 2.7. ([16]) Let T 1/2 ( ). Then ∈ BA H (a) T A = T (R(A1/2)). k k k kB (b) ωA(T )= ω(T ). e Lemma 2.8. ([27, Proposition 2.9]) Let T ( ). Then e ∈ BA H ^ T ♯A = T ∗ and (T ♯A )♯A = T. Now, we are ready to prove the following proposition. g e e Proposition 2.1. Let T ( ). Then, ∈ BA H 2 T 2 T ♯A T + T T ♯A T 2 + T 2 . (2.25) k kA ≤k kA ≤k kA k kA Some numerical radius inequalities for semi-Hilbert space operators 15

Proof. Firstly, we shall prove that for every X,Y 1/2 ( ) we have ∈ BA H XY = XY and X^+ Y = X + Y. (2.26)

Since, X,Y A1/2 ( ), then by Lemma 2.6 there exists a unique X, Y 1/2 ∈ B Hg e e e e ∈ (R(A )) such that ZAX = XZA and ZAY = YZA. So, B e e Z (XY )= XZ Y = XYZ . A e A e A On the other hand, since XY 1/2 ( ), then again in view of Lemma 2.6 we ∈ BA eH e e have ZA(XY )=(XY )ZA. Thus, since XY is unique, then we conclude that XY = XY . Similarly, we can prove that X^+ Y = X + Y . Now, by using Lemmag 2.6 (a) togetherg with (2.26) and Lemma 2.8 we get g e e e e ♯A ♯A ♯A ^ ♯A T T + T T A = T T + T T (R(A1/2)) k k k kB = (T )∗T + T (T )∗ (R(A1/2)). (2.27) k kB Moreover, by using basic properties of the spectral radius of Hilbert space operators, we see that e e e e

(T )∗T + T (T )∗ (R(A1/2)) = r (T )∗T + T (T )∗ k kB (T ) T + T (T ) 0 e e e e = r e e∗ e e ∗ 0 0 e e e e T (T )∗ T 0 = r | | | | | | 0 0 (T ) 0 ∗ e e | e | T 0 T (T )∗ = r | | | |e | | (T )∗ 0 0 0 | | Hence, we get e e e e (T )∗T T (T)∗ (T )∗T + T (T )∗ (R(A1/2)) = r | || | k kB (T ) T T (T ) ∗ ∗ | e ||e | e e Thus, by usinge [22e, Theoreme e 1.1.] we obtain e e e e (T )∗T (R(A1/2)) T (T)∗ (R(A1/2)) (T )∗T + T (T )∗ (R(A1/2)) r k kB k| || |kB k kB ≤ " (T )∗ T (R(A1/2)) T (T )∗ (R(A1/2)) !# k| e ||e |kB ke e kB 2 2 e e e e 1 2 T (R(A1/2)) (T ) (R(A / )) = r k ekB e k kB e e 2 1 2 2 " (T ) (R(A / )) T (R(A1/2)) !# k e kB ke kB 2 2 1 2 = T (R(A1/2)) + (T ) (R(A / )). k kB e k kB e So, by taking into account (2.27) we get e e ♯ ♯ 2 2 A A 1 2 T T + T T A T (R(A1/2)) + (T ) (R(A / )) k k ≤k kB k kB 2 2 1 2 = T (R(A1/2)) + T (R(A / )) (by (2.26)) k ekB k ekB 2 2 = T + T A (by Lemma 2.7(b)). k ekA k k f 16 Kais Feki

Now, by [23, Lemma 7] we have 2 2 (T ) (R(A1/2)) (T )∗T + T (T )∗ (R(A1/2)). k kB ≤k kB This implies, by using (2.26) together with Lemmas 2.8 and 2.7 (b), that e e e e e 2 T 2 T ♯A T + T T ♯A . k kA ≤k kA Hence, the proof is finished. Now, we state the following two corollaries. The first one give a considerable improvement of (1.9). Corollary 2.4. Let T ( ). Then ∈ BA H 1 1 ω (T ) T 2 + T 2 +2ω (T 2) T + T 2 1/2 . A ≤ 2 k kA k kA A ≤ 2 k kA k kA q Proof. Observe first that, in view of (1.4), we have T 2 = T 2 1/2 T 2 1/2 T T 2 1/2. (2.28) k kA k kA k kA ≤k kAk kA Moreover, by Corollary 2.1, one has 1 ω (T ) T T ♯A + T ♯A T +2ω (T 2) A ≤ 2 A A q 1 2 2 2 T A + T +2ωA(T ) (by (2.25)) ≤ 2 k k k kA 1q T 2 + T 2 +2 T 2 (by (1.5)) ≤ 2 k kA k kA k kA 1q T 2 +2 T T 2 1/2 + T 2 (by (1.4) and (2.28)) ≤ 2 k kA k kAk kA k kA q 2 1 2 1/2 = T A + T A 2r k k k k 1 = T + T 2 1/2 . 2 k kA k kA This completes the proof. Corollary 2.5. Let T ( ). Then the following assertions hold: ∈ BA H √2 (1) T = T T ♯A + T ♯A T if and only if T 2 = T 2 . k kA 2 A k kA k kA (2) If AT 2 =0,q then T = T T ♯A + T ♯A T . k kA A q Proof. (1) By Proposition 2.1 we have 2 T 2 T ♯A T + T T ♯A T 2 + T 2 2 T 2 . k kA ≤k kA ≤k kA k kA ≤ k kA So, the desired property follows.

♯A ♯A ♯A ♯A (2) Clearly, we have (T T + T T ) T T = T T A 0. Hence, by using (1.6) together with (1.7) and Proposition −2.1 we obtain ≥ T 2 T ♯A T + T T ♯A T 2 + T 2 . k kA ≤k kA ≤k kA k kA 2 2 So, if AT = then T A =0. Therefore, the required assertion follows immediately. k k Some numerical radius inequalities for semi-Hilbert space operators 17

By using the connection between A-bounded operators and operators in (R(A1/2)) we will generalize some well-known numerical radius inequalities. We beginB with the following result which generalizes [26, Theorem 3.3.] since ( ) 1/2 ( ). BA H ⊆ BA H Theorem 2.7. Let T 1/2 ( ). Then, ∈ BA H ω2 (T ) 1 T 2 + ω T 2 . (2.29) A ≤ 2 k kA A Proof. By applying Lemma 2.7 and [13, Theorem1.] we see that 2 2 ωA(T )= ω (T )

1 2 2 eT (R(A1/2)) + ω (T ) ≤ 2 k kB h i 1 2 2 = Te (R(A1/2)) + ω Te . 2 k kB h i So, we get (2.29) by applying Lemmae 2.7. f

Remark 2.3. For a given T A1/2 ( ), it follows from Theorem 2.7, (1.5) and (1.4) that ∈ B H ω (T ) 1 T 2 + ω T 2 1/2 1 T 2 + T 2 1/2 T . A ≤ 2 k kA A ≤ 2 k kA k kA ≤k kA Hence, (2.29) reﬁnes the second inequality in (1.5). Next, we state the following theorem.

Theorem 2.8. Let U A( ) be an A-unitary operator and T A1/2 ( ) be such that UT = T U. Then,∈ B H ∈ B H ω (UT ) ω (T ). (2.30) A ≤ A If U is an A-isometry operator which commutes with an operator T A1/2 ( ), then (2.30) also holds true. ∈ B H

Proof. Notice ﬁrst that, it was proved in [3], that an operator U A( ) is A-unitary if and only if ∈ B H

♯A ♯A ♯A ♯A U U =(U ) U = P (A). R This implies that

^♯A ♯^A ♯A ♯A ^ U U = (U ) U = P (A). R

^ 1 2 On the other hand, it can be seen that P (A) = IR(A / . So, by using Lemma 2.8 we get R

U ∗U = UU ∗ = IR(A1/2 . So, U is an unitary operator on the Hilbert space R(A1/2). Moreover, since e e e e UT = T U, then UT = T U. Thus, by applying [14, Theorem 1.9.], we obtain e ω(UT ) ω(T ). e e e e ≤ This proves (2.30) by observing that UT = UT and using Lemma 2.7 (b). Now, e e e let U A( ) is an A-isometry operator. Then, by [3, Corollary 3.7.], we ∈♯A B H have U U = P (A). By using similare e argumentsg as above, one can see that R 18 Kais Feki

U (R(A1/2)) is an isometry operator. So, the proof of the theorem is complete by∈ proceeding B as above and using [14, Theorem 1.9.]. e Our next result reads as follows.

Theorem 2.9. Let T ( ) and S 1/2 ( ) be such that TS = ST and ∈ BA H ∈ BA H T ♯A S = ST ♯A . Then, ω (TS) ω (S) T , (2.31) A ≤ A k kA and ω (ST ) ω (T ) S . A ≤ A k kA Proof. Since TS = ST and T ♯A S = ST ♯A , then by the same arguments used in the proof of Theorem 2.8, we infer that

T S = ST and T ∗S = ST ∗. So, an application of [14, Theorem 1.10] shows that e e e e e e e e ω(T S) ω(S) T (R(A1/2)) and ω(ST ) ω(T ) S (R(A1/2)). ≤ k kB ≤ k kB Therefore, the desired results by applying (2.26) together with Lemma 2.7. ee e e e e e e Corollary 2.6. Let T,S ( ) be such that T is an A-isometry operator. ∈ BA H Assume that TS = ST and T ♯A S = ST ♯A . Then, ω (TS) ω (S). A ≤ A Proof. Since T is an A-isometry operator, then clearly we have T A =1. There- fore, we get the desired result by applying Theorem 2.9 k k We end this paper by the following theorem.

Theorem 2.10. Let T A( ) be an A-normal operator such that TS = ST . Then, ∈ B H ω (TS) ω (T )ω (S). (2.32) A ≤ A A Proof. Let T A( ) is an A-normal. It is not difficult to see that T (R(A1/2)) is a∈ normal B H operator. So, since T is an A-normal operator and satis-∈ B fies TS = ST , then T is a normal operator on R(A1/2) and satisfies T S = eST . Therefore, by Feglede’s theorem [25] we deduce that T ∗S = ST ∗. This implies, by taking adjoints, thate T S∗ = S∗T . Hence, by applying [14, Theoreme e 1.10]e wee get e e e e

eωe(T S)e eω(S) T (R(A1/2)). ≤ k kB On the other hand, since T is a normal operator in (R(A1/2)), then ω(T ) = ee e e B T (R(A1/2)) (see [7]). So, we deduce that k kB e e ω(T S) ω(S)ω(T ). e ≤ Hence, by using (2.26) and Lemma 2.7 (b), we obtain ee e e ω (TS) ω (S)ω (S), A ≤ A A as required. Some numerical radius inequalities for semi-Hilbert space operators 19 References

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