EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 15, No. 3, 2022, 1067-1089 ISSN 1307-5543 – ejpam.com Published by New York Business Global The Fuglede-Putnam theorem and quasinormality for class p-wA(s, t) operators M.H.M.Rashid1,∗, N. H. Altaweel2 1 Department of Mathematics, Faculty of Science P.O. Box(7), Mu’tah university, Al-Karak- Jordan 2 Department of Mathematics-Faculty of Science, University of Tabuk, P.O.Box 741- Tabuk 71491, Saudi Arabia Abstract. In this work, we demonstrate that (i) if T is a class p-wA(s, t) operator and T (s, t) is quasinormal (resp., normal), then T is also quasinormal (resp., normal) (ii) If T and T∗ are class p-wA(s, t) operators, then T is normal; (iii) the normal portions of quasisimilar class p- wA(s, t) operators are unitarily equivalent; and (iv) Fuglede-Putnam type theorem holds for a class p-wA(s, t) operator T for 0 < s, t, s + t = 1 and 0 < p ≤ 1 if T satisfies a kernel condition ker(T ) ⊂ ker(T ∗). 2020 Mathematics Subject Classifications: 47A10, 47A11, 47B20 Key Words and Phrases: Quasinormal, Class A(s, t) operators, Class p-(A(s, t) operators, Fuglede-Putnam theorem 1. Introduction On a complex Hilbert space H, let B(H) be the algebra of all bounded linear opera- tors. Aluthge [2] investigated the p-hyponormal operator T , which is defined as (T ∗T )p ≥ (TT ∗)p with 0 ≤ p ≤ 1 using the Furuta inequality [14]. When p = 1, T is said to be hyponormal. As a result, p-hyponormality is a broadening of hyponormality. Following [2], several authors are looking towards novel hyponormal operator generalizations. It is known that p-hyponormal operators have many interesting properties as hyponor- mal operators, for example, Putnam’s inequality, Fuglede-Putnam type theorem, Bishop’s property (β), Weyl’s theorem and polaroid. Let T ∈ B(H) and |T | = (T ∗T ) 1 2 . By taking U |T |x = Tx for x ∈ H and Ux = 0 for x ∈ ker |T |, T has a unique polar decomposition T = U |T | with condition kerU = ker |T |. We say that T = U |T | is the polar decompo- sition of T . In [2], Aluthge extended the class of hyponormal operators by introducing p-hyponormal operators and obtained some properties with the help of the transformation ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v15i3.4412 Email addresses: malik okasha@yahoo.com (M.H.M. Rashid), naltawil@ut.edu.sa (N. H. Altaweel) https://www.ejpam.com 1067 © 2022 EJPAM All rights reserved. M.H.M.Rashid, N. H. Altaweel / Eur. J. Pure Appl. Math, 15 (3) (2022), 1067-1089 1068 T (12 , 1 2) = |T | 1 2U |T | 1 2 , which now known as the Aluthge transform. The introduction of these operators by Aluthge has inspired many researchers not only to expose some impor- tant properties of p-hyponormal operators but also to introduce the number of extensions ([1, 7, 8, 13]). The Aluthge transform, and more broadly, the generalized Aluthge transform defined as T (s, t) = |T |sU |T |t with s, t > 0, have proven to be useful tools in this attempt. The generalized Aluthge transform is used to analyze class p-wA(s, t) operators in this article. Definition 1. Let T = U |T | be the polar decomposition of an operator T ∈ B(H). Then the generalized Aluthge transform T (s, t) of T is defined as follows: T (s, t) = |T |sU |T |t. Moreover, for each nonnegative integer n, the n-th generalized Aluthge transform ∆n(T (s, t)) of T (s, t) is defined as follows: ∆n(T (s, t)) = ∆(∆n−1(T (s, t))),∆0(T (s, t)) = T (s, t). Definition 2. Let 0 < s, t, and 0 < p ≤ 1. An operator T is said to be a class (i) p-wA(s, t) if (|T ∗|t|T |2s|T ∗|t) tp s+t ≥ |T ∗|2tp and |T |2sp ≥ (|T |s|T ∗|2t|T |s) sp s+t . (ii) p-A(s, t) if (|T ∗|t|T |2s|T ∗|t) tp s+t ≥ |T ∗|2tp. (iii) p-A if |T 2|p ≥ |T |2p. (iv) (s, p)-w-hyponormal if |T (s, s)|p ≥ |T |2sp ≥ |(T (s, s)∗|p. It is known that p-hyponormal operators and log-hyponormal operators are class 1- wA(s, t) for any 0 < s, t. Class p-wA(s, s) is called class (s, p)-w-hyponormal, class 1- wA(1, 1) is called class A and class 1-wA(12 , 1 2) is called w-hyponormal [13, 15, 18, 19, 33]. Hence class p-wA(s, t) operator is a generalization of class (s, p)-w-hyponormal, class A and w-hyponormal operators. C. Yang and J. Yuan [34–36] studied class wF (p, r, q) operator T , i.e., ( |T ∗|r|T |2p|T ∗|r ) 1 q ≥ |T ∗| 2(p+r) q and |T |2(p+r)(1− 1 q ) ≥ ( |T |p|T ∗|2r|T |p )1− 1 q where 0 < p, 0 < r, 1 ≤ q. If we take small p1 such that 0 < p1 ≤ p+r qr and p1 ≤ (p+r)(q−1) pq , then T is class p1-wA(p, r). Hence class p1-wA(p, r) is a generalization of class wF (p, r, q). We will use this property frequently. M.H.M.Rashid, N. H. Altaweel / Eur. J. Pure Appl. Math, 15 (3) (2022), 1067-1089 1069 It is known that T = U |T | is class p-wA(s, t) if and only if |T (s, t)| 2tp s+t ≥ |T |2tp, |T |2sp ≥ |T (s, t)∗| 2sp s+t by [26]. Hence |T (s, t)| 2rp s+t ≥ |T |2rp ≥ |T (s, t)∗| 2rp s+t and T (s, t) is rp-hyponormal for all r ∈ (0,min{s, t}]. The following is a breakdown of the paper’s structure: In section 2, we prove that if T is a class of p-wA(s, t) operators and its Aluthge transform T (s, t) is quasinormal (respectively, normal), then T is also quasinormal (resp., normal). The normal parts of quasisimilar class p-wA(s, t) operators are unitarily equivalent in section 3. The major goal of Section 4 is to demonstrate that the Fuglede-Putnam theorem holds for a class p-wA(s, t) operator T with 0 < s, t, s+t = 1 and 0 < p ≤ 1 if T fulfills the kernel condition ker(T ) ⊂ ker(T ∗). 2. Quasinormality Let T = U |T | be the polar decomposition of T ∈ B(H) . T is said to be quasinormal if |T |U = U |T | , or equivalently, TT ∗T = T ∗TT . S. M. Patel, K. Tanahashi, A. Uchiyama and M. Yanagida [27] proved that if T is class A(s, t) and T (s, t) is quasinormal, then T is quasinormal and T = T (s, t) if s+ t = 1. The following is a generalization of this result. Theorem 1. Let T be a class p-wA(s, t) operator with the polar decomposition T = U |T |. If T (s, t) = |T |sU |T |t is quasinormal, then T is also quasinormal. Hence T coincides with its generalized Aluthge transform T (s, t). Proof. Since T is a class p-A(s, t) operator, |T (s, t)| 2rp s+t ≥ |T |2rp ≥ |(T (s, t))∗| 2rp s+t (1) for all r ∈ (0,min{s, t}) by [19, Theorem 3] and Löwner-Heinz inequality. Then Douglas’s theorem [11] implies ran(T (s, t)) = ran((|T (s, t))∗|) ⊂ ran(|T |) = ran(|T (s, t)|) where M denotes the norm closure of M . Let T (s, t) = W |T (s, t)| be the polar decom- position of T (s, t) . Then E := W ∗W = U∗U ≥ WW ∗ =: F . Put |(T (s, t))∗| 1 s+t = ( X 0 0 0 ) ,W = ( W1 W2 0 0 ) on H = ran(T (s, t))⊕ ker((T (s, t))∗). Then X is injective and has a dense range. Since T (s, t) is quasinormal, W commutes with |T (s, t)| and |T (s, t)| 2rp s+t = W ∗W |T (s, t)| 2rp s+t = W ∗|T (s, t)| 2rp s+tW M.H.M.Rashid, N. H. Altaweel / Eur. J. Pure Appl. Math, 15 (3) (2022), 1067-1089 1070 ≥ W ∗|T |2rpW ≥ W ∗|(T (s, t))∗| 2rp s+tW = |T (s, t)| 2rp s+t . Hence |T (s, t)| 2rp s+t = W ∗|T (s, t)| 2rp s+tW = W ∗|T |2rpW, and |(T (s, t))∗| 2rp s+t = W |T (s, t)| 2rp s+tW ∗ = WW ∗|T (s, t)| 2rp s+tWW ∗ (2) = WW ∗|T |2rpWW ∗ = ( X2rp 0 0 0 ) . (3) Since WW ∗ = ( 1 0 0 0 ) , (1), (2) and (3) imply that |T (s, t)| 2rp s+t and |T |2rp are of the forms |T (s, t)| 2rp s+t = ( X2rp 0 0 Y 2rp ) ≥ |T |2rp = ( X2rp 0 0 Z2rp ) , (4) where ran(Y ) = ran(Z) = ran(|T |)⊖ ran(T (s, t)) = ker((T (s, t))∗)⊖ ker(T ). Since W commutes with |T (s, t)| ,( W1 W2 0 0 )( X 0 0 Y ) = ( X 0 0 Y )( W1 W2 0 0 ) . So W1X = XW1 and W2Y = XW2 , and hence ran(W1) and ran(W2) are reducing subspaces of X . Since W ∗W |T (s, t)| = |T (s, t)| , we have W ∗ 1W1 = 1 and Xk = W ∗ 1W1X k = W ∗ 1X kW1, Y k = W ∗ 2W2Y k = W ∗ 2X kW2, for k = 1, 2, · · · . Put U = ( U11 U12 U21 U22 ) . Then T (s, t) = |T |sU |T |t = W |T (s, t)| implies ( Xs 0 0 Zs )( U11 U12 U21 U22 )( Xt 0 0 Zt ) = ( W1 W2 0 0 )( Xs+t 0 0 Y s+t ) . Hence XsU11X t = W1X s+t = XsW1X t, XsU12Z t = W2Y s+t = Xs+tW2 and Xs(U11 −W1)X t = 0, Xs(U12Z t −XtW2) = 0. M.H.M.Rashid, N. H. Altaweel / Eur. J. Pure Appl. Math, 15 (3) (2022), 1067-1089 1071 Since X is injective and has a dense range, U11 = W1 is isometry and U12Z t = XtW2. Then U∗U = ( U∗ 11U11 + U∗ 21U21 U∗ 11Ul2 + U∗ 21U22 U∗ 12U11 + U∗ 22U21 U∗ 12U12 + U∗ 22U22 ) onH = ran(T (s, t))⊕ker((T (s, t))∗) is the orthogonal projection onto ran(|T |) ⊃ ran(T (s, t)), we have U21 = 0 and U∗U = ( 1 0 0 U∗ 12U12 + U∗ 22U22 ) . Since U12Z t = XtW2 , we have Z2t ≥ ZtU∗ 12U12Z t = W ∗ 2X 2tW2 = Y 2t, and Z2rp ≥ (ZtU∗ 12U12Z t) rp t = (W ∗ 2X tW2) rp t = Y 2rp ≥ Z2rp by Löwner-Heinz inequality and (4). Hence (ZtU∗ 12U12Z t) rp t = Z2rp = Y 2rp, so Z = Y and |T (s, t)| = |T |s+t . Since Z2t = ZtU∗ 12U12Z t ≤ ZtU∗ 12U12Z t + ZtU∗ 22U22Z t ≤ Z2t ZtU∗ 22U22Z t = 0 and U22Z t = 0 . This implies ran(U∗ 22) ⊂ ker(Z). Since ran(U∗ 12U12 + U∗ 22U22) ⊂ ran(Z) and U∗ 22U22 ≤ U∗ 12U12 + U∗ 22U22 , we have ran(U∗ 22) ⊂ ran(Z) . Hence U22 = 0, U = ( W1 U12 0 0 ) and ran(U) ⊂ ran(T (s, t)) ⊂ ℜ(|T |) = ran(E). Since W commutes with |T (s, t)| = |T |s+t, W commutes with |T | and |T |s(W − U)|T |t = W |T |s|T |t − |T |sU |T |t = W |T (s, t)| − T (s, t) = 0. Hence E(W − U)E = 0 and U = UE = EUE = EWE = WE = W. Thus U = W commutes with |T | and T is quasinormal. Corollary 1. Let T = U |T | be a class p-wA(s, t) operator. If T (s, t) = |T |sU |T |t is normal, then T is also normal. M.H.M.Rashid, N. H. Altaweel / Eur. J. Pure Appl. Math, 15 (3) (2022), 1067-1089 1072 Proof. Since T (s, t) is normal, T is quasinormal by Theorem 1. Hence T (s, t) = |T |sU |T |t = U |T |s+t and (T (s, t))∗ = |T |s+tU∗. Hence |T |2(s+t) = |T (s, t)|2 = |(T (s, t))∗|2 = |T ∗|2(s+t). This implies |T | = |T ∗| and T is normal. Theorem 2. [25] Let s1 > 0, s2 > 0, t1 > 0, t2 > 0 and 0 < p ≤ 1. If T belongs to class p1-wA(s1, t1) for 0 < p1 ≤ p and T ∗ belongs to class p2-wA(s2, t2) for 0 < p2 ≤ p , then T is normal. To prove Theorem 2, we need the following results. Lemma 1. ([21]) If T is class p-wA(s, t) and 0 < s ≤ s1, 0 < t ≤ t1, 0 < p1 ≤ p < 1, then T is class p1-wA(s1, t1). Theorem 3 (Furuta theorem [14]). If A ≥ B ≥ 0, then for each r ≥ 0, (i) (B r 2ApB r 2 ) 1 q ≥ B r+p q and (ii) A r+p q ≥ (A r 2BpA r 2 ) 1 q hold for p ≥ 0 and q ≥ 1 with (1 + r)q ≥ p+ r. Proposition 1. ([19]) Let A ≥ 0 and B ≥ 0. If B 1 2AB 1 2 ≥ B2 and A 1 2BA 1 2 ≥ A2, (5) then A = B. Proof. [Proof of Theorem 2] Let r = max{s1, s2, t1, t2} and let q = min{p1, p2}. Firstly, if T belongs to class p1-wA(s1, t1), then T belongs to class q-wA(r, r) by Lemma 1. Hence we have (|T ∗|r|T |2r|T ∗|r) q 2 ≥ |T ∗|2rq and |T |2rq ≥ (|T |r|T ∗|2r|T |r) q 2 (6) Secondly, if T ∗ belongs to class p2-wA(s2, t2), then T ∗ belongs to class q-wA(r, r) by Lemma 1. Hence we have (|T |r|T ∗|2r|T |r) q 2 ≥ |T |2rq and |T ∗|2rq ≥ (|T ∗|r|T |2r|T ∗|r) q 2 (7) Therefore |T ∗|r|T |2r|T ∗|r = |T ∗|4r and |T |4r = |T |r|T ∗|2r|T |r hold by (6) and (7), and then |T | = |T ∗| by Proposition 1. The following result is very important in the sequal M.H.M.Rashid, N. H. Altaweel / Eur. J. Pure Appl. Math, 15 (3) (2022), 1067-1089 1073 Theorem 4. [17, Jensen’s Operator Inequality (JOI)] Suppose that f is a continuous function defined on an interval I. Then f is operator convex on an interval I containing 0 with f(0) ≤ 0 if and only if f(a∗xa) ≤ a∗f(x)a for every self-adjoint x with spectrum in I and every contraction a. Theorem 5. ([11]) Let A and B be bounded linear operators on a Hilbert space H. Then the following are equivalent: (i) ran(A) ⊆ ran(B); (ii) AA∗ ≤ λ2BB∗ for some λ ≥ 0; and (i) there exists a bounded linear operator C on H so that A = BC. Lemma 2. Let A,B and C be positive operators. Then the following assertions hold for each p ≥ 0, r ∈ [0, 1] and 0 < q ≤ 1: (i) If (Br/2ApBr/2) rq p+r ≥ Brq and B ≥ C, then (Cr/2ApCr/2) rq p+r ≥ Crq. (ii) If A ≥ B, Brq ≥ (Br/2CpBr/2) rq p+r and the condition if lim n→∞ B1/2xn = 0 and lim n→∞ A1/2xn exists, then lim n→∞ A1/2xn = 0 for any sequence of vectors {xn} (8) hold, then Arq ≥ (Ar/2CpAr/2) rq p+r . Lemma 2 can be obtained as an application of the following results. Theorem 6. ([11]) Let A and B be bounded linear operators on a Hilbert space H. Then the following are equivalent: (i) ran(A) ⊆ ran(B); (ii) AA∗ ≤ λ2BB∗ for some λ ≥ 0; and (iii) there exists a bounded linear operator C on H so that A = BC. Moreover, if (i), (ii) and (iii) are valid, then there exists a unique operator C so that (a) ∥C∥2 = inf{µ : AA∗ ≤ µBB∗}; (b) ker(A) = ker(C); and (c) ran(C) ⊆ ran(B∗). Theorem 7. ([16]) Let X and A be bounded linear operator on a Hilbert space H. We suppose that A ≥ 0 and ∥X∥ ≤ 1. If f is an operator monotone function defined on [0,∞), then X∗f(A)X ≤ f(X∗AX). M.H.M.Rashid, N. H. Altaweel / Eur. J. Pure Appl. Math, 15 (3) (2022), 1067-1089 1074 We remark that the condition (c) of Theorem 6 is equivalent to (c′): ran(C) ⊆ ran(B∗). Here we consider when the equality of (c′) holds. Lemma 3. ([33]) Let A and B be operators which satisfy (i), (ii) and (iii) of Theorem 6 and C be the operator which is given in (iii) and determined uniquely by (a), (b) and (c) of Theorem 6. Then the following assertions are mutually equivalent: (i) ran(C) = ran(B∗). (ii) If lim n→∞ A∗xn = 0 and lim n→∞ B∗xn exists, then lim n→∞ B∗xn = 0 for any sequence of vectors {xn}. We also prepare the following lemma in order to give a proof of Lemma 2. Lemma 4. ([33]) Let S be a positive operator and 0 < q ≤ 1. If lim n→∞ Sxn = 0 and lim n→∞ Sqxn exists, then lim n→∞ Sqxn = 0 for any sequence of vectors {xn}. Proof. [Proof of Lemma 2] (i) The hypothesis B ≥ C ensures then Bt ≥ Ct for each t ∈ (0, 1] by Löwner-Heinz theorem. By Theorem 6, there exists an operator X with ∥X∥ ≤ 1 such that B t 2X = X∗B t 2 = C t 2 . (9) Then we have (Cr/2ApCr/2) rq p+r = (X∗Br/2ApBr/2X) rq p+r ≥ X∗(Br/2ApBr/2) rq p+rX (by Theorem 7) ≥ X∗BrqX (by the hypothesis) = X∗(Br)qX ≥ (X∗B r 2B r 2X)q (by Theorem 4) = (C r 2C r 2 )q = Crq (by Equation (9)). (ii) The hypothesis A ≥ B ensures As ≥ Bs for s ∈ (0, 1] by Löwner-Heinz theorem. By Theorem 6, there exists an operator X with ∥X∥ ≤ 1 such that As/2X = X∗As/2 = Bs/2. (10) Then we have X∗(Ar/2CpAr/2) rq p+rX ≤ (X∗Ar/2CpAr/2X) rq p+r (by Theorem 7) = (Br/2CpBr/2) rq p+r ≤ Brq (by the hypothesis) = (Br)q = (X∗A r 2A r 2X)q ≤ X∗ArqX (by Theorem 4) so that Arq ≥ (Ar/2CpAr/2) rq p+r holds on ran(X). On the other hand, the hypothesis (8) implies the following (11) If lim n→∞ Br/2xn = 0 and lim n→∞ Ar/2xn exists, M.H.M.Rashid, N. H. Altaweel / Eur. J. Pure Appl. Math, 15 (3) (2022), 1067-1089 1075 then lim n→∞ Ar/2xn = 0 for any sequence of vectors {xn}. (11) since lim n→∞ Br/2xn = 0 and lim n→∞ Ar/2xn exists, then lim n→∞ B1/2xn = B(1−r)/2( lim n→∞ Br/2xn) = 0 and lim n→∞ A1/2xn = A(1−r)/2( lim n→∞ Ar/2xn) ex- ists, so that lim n→∞ A1/2xn = 0 by (8), hence lim n→∞ Ar/2xn = 0 by Lemma 4. (11) ensures ran(X) = ran(Ar/2) by Lemma 3, hence we have ker((Ar/2CpAr/2) rq p+r ) = ker(Ar/2CpAr/2) ⊇ ker(Ar/2) = ker(Ar) = ker(Aqr) = ker(X∗), so that Aqr = (Ar/2CpAr/2) rq p+r = 0 holds on ker(X∗). Consequently the proof is complete since H = ran(X)⊕ ker(X∗). Lemma 5. ([26]) Let T = U |T | ∈ B(H) be the polar decomposition of T . Then T is class p-wA(s, t) if and only if |T (s, t)| 2tp s+t ≥ |T |2tp and |T |2sp ≥ |(T (s, t))∗| 2sp s+t . Lemma 6. Let 0 < s, t, s + t ≤ 1 and 0 < p ≤ 1. Let T ∈ B(H) be class p-wA(s, t) and let M an invariant subspace of T . Then the restriction T |M is also class p-wA(s, t). Proof. Let T = ( T1 S 0 T2 ) on H = M ⊕ M⊥ and P the orthogonal projection onto M. Let T0 := TP = PTP = ( T1 0 0 0 ) . Then |T0|2t = (P |T |2P )t ≥ P |T |2tP for each 0 < t ≤ 1 by Hansen’s inequality, and |T ∗|2 = TT ∗ ≥ TPT ∗ = |T ∗ 0 |2. Hence T is class p-A(s, t) ⇐⇒ |T ∗|2tp ≤ (|T ∗|t|T |2s|T ∗|t) tp s+t =⇒ |T ∗ 0 |2tp ≤ (|T ∗ 0 |t|T |2s|T ∗ 0 |t) tp s+t (by Lemma 2) =⇒ |T ∗ 0 |2tp ≤ (|T ∗ 0 |t|T0|2s|T ∗ 0 |t) tp s+t (since |T ∗ 0 |t = |T ∗ 0 |tP = P |T ∗ 0 |t for every 0 < t ≤ 1). Now |T0| = P |T̃ |P ≥ P |T |P ≥ P |(T̃ )∗|P = |T ∗ 0 |. Then by Theorem 3 it follows that |T0|2sp ≥ (|T0|s|T ∗ 0 |2t||T0|s|) ps s+t . Therefore, T |M is class p-A(s, t) operator. The following example shows that there exists a class p-wA(s, t) operator T such that T |M is quasinormal but M does not reduce T . M.H.M.Rashid, N. H. Altaweel / Eur. J. Pure Appl. Math, 15 (3) (2022), 1067-1089 1076 Example 1. Let T be a bilateral shift on ℓ2(Z) defined by Ten = en+1 and M = ∨ n≥0 Cen. Then T is unitary and T |M is isometry. However, M does not reduce T. Lemma 7. Let 0 < s, t, s+t = 1 and 0 < p ≤ 1. Let T ∈ B(H) be class p-wA(s, t) operator, let M be an invariant subspace for T and a reducing subspace for T (s, t) such that T (s, t)|M the restriction of T (s, t) to M is an injective normal operator, then T |M = T (s, t)|M and M reduces T. Proof. Let T (s, t) = ( T0 0 0 A ) , T = ( S B 0 D ) on H = M⊕M⊥. Since T is class p-wA(s, t) we have |T (s, t)|2rp ≥ |T |2rp ≥ |(T (s, t))∗|2rp for r ∈ min{s, t}. Let P be the orthogonal projection onto M. Then |T0| = P |T (s, t)|P ≥ P |T |P ≥ P |(T (s, t))∗|P = |T ∗ 0 |. By Löwner-Heinz theorem we get |T0|2rp = P |T (s, t)|2rpP ≥ P |T |2rpP ≥ P |(T (s, t))∗|2rpP = |T ∗ 0 |2rp. Since |T |sT = T (s, t)|T |s and P |T |sP = |T0|s, we deduce that |T0|sS = T0|T0|s. We have T0 is an injective normal operator, then S = T |M = T0 = T (s, t)|M, consequently T = ( T0 B 0 D ) on H = M⊕M⊥. Hence T ∗T = ( T ∗ 0 T0 T ∗ 0B B∗T0 B∗B +D∗D ) on H = M⊕M⊥. So we can write |T |rp = ( |T0|rp X X∗ Y ) on H = M⊕M⊥. Since P |T |pr|T |prP = |T0|2rp, then |T0|2rp = |T0|2rp +XX∗, and thus X = 0. It follows that |T |rp = |T0|rp⊕Y 2 implying |T |2rp = |T0|2rp⊕Y 4. Consequently we get B∗B = 0 it follows that B = 0 and hence M reduces T . The next lemma is a simple consequence of the preceding one. M.H.M.Rashid, N. H. Altaweel / Eur. J. Pure Appl. Math, 15 (3) (2022), 1067-1089 1077 Lemma 8. Let 0 < s, t, s + t = 1 and 0 < p ≤ 1. Let T ∈ B(H) be a class p-wA(s, t) operator with ker(T ) ⊂ ker(T ∗). Then T = T1 ⊕ T2 on H = H1 ⊕H2 where T1 is normal, ker(T2) = {0} and T2 is pure class p-wA(s, t) i.e., T2 has no non-zero invariant subspace M such that T2|M is normal. Lemma 9. Let 0 < s, t, s+ t = 1 and 0 < p ≤ 1. Let T = U |T | ∈ B(H) be class p-wA(s, t) and ker(T ) ⊂ ker(T ∗). Suppose T (s, t) = |T |sU |T |t be of the form N⊕T ′ on H = M⊕M⊥, where N is a normal operator on M. Then T = N ⊕ T1 and U = U11 ⊕ U22 where T1 is class p-wA(s, t) with ker(T1) ⊂ ker(T ∗ 1 ) and N = U11|N | is the polar decomposition of N . Proof. Since |T (s, t)|2rp ≥ |T |2rp ≥ |(T (s, t))∗|2rp for r ∈ min{s, t}, we have |N |2rp ⊕ |T ′|2rp ≥ |T |2rp ≥ |N |2rp ⊕ |T ′∗|2rp by assumption. This implies that |T | is of the form |N | ⊕L for some positive operator L. Let U = ( U11 U12 U21 U22 ) be 2×2 matrix representation of U with respect to the decomposition H = M⊕M⊥. Then the definition T (s, t) means( N 0 0 T ′ ) = ( |N |s 0 0 Ls )( U11 U12 U21 U22 )( |N |t 0 0 Lt ) Hence, we have N = |N |sU11|N |t, |N |sU12L t = 0 and LsU21|N |t = 0. Since ker(T ) ⊂ ker(T ∗), ran(U) = ran(T ) = ker(T ∗)⊥ ⊂ ker(T )⊥ = ran(|T |). Let Nx = 0 for x ∈ M. Then x ∈ ker(|T |) = ker(U), and Ux = ( U11 U12 U21 U22 )( x 0 ) = ( U11x U21x ) = 0. Hence ker(N) ⊂ ker(U11) ∩ ker(U21). Let x ∈ M. Then U ( x 0 ) = ( U11x U21x ) ∈ ran(|T |) = ran(|N | ⊕ L). Hence ran(U11) ⊂ ran(|N |), ran(U21) ⊂ ran(L). M.H.M.Rashid, N. H. Altaweel / Eur. J. Pure Appl. Math, 15 (3) (2022), 1067-1089 1078 Similarly ran(U12) ⊂ ran(|N |), ran(U22) ⊂ ran(L). Let Lx = 0 for x ∈ M⊥. Then x ∈ ker(|T |) = ker(U) and U ( 0 x ) = ( U12x U22x ) = 0 Hence ker(L) ⊂ ker(U12) ∩ ker(U22). Let N = V |N | be the polar decomposition of N . Then (V |N |s − |N |sU11)|N |t = 0. Hence V |N |s − |N |sU11 = 0 on ran(|N |). Since ker(N) ⊂ ker(U11), this implies 0 = V |N |s − |N |sU11 = |N |s(V − U11). Hence ran(V − U11) ⊂ ker(|N |) ∩ ran(|N |) = {0}. Hence V = U11 and N = U11|N | is the polar decomposition of N . Since |N |sU12L t = 0, ran(U11L t) ⊂ ker(|N |) ∩ ran(|N |) = {0}. Hence U12L t and U12 = 0. Similarly we have U21 = 0 by LsU21|N |t = 0. Hence U = U11 ⊕ U22. So we obtain T = U |T | = U11|N | ⊕ U22L = N ⊕ T1, where T1 = U22L. 3. Quasisimilarity An operator X ∈ B(K,H) is called quasiaffinity if X is both injective and has a dense range. For T ∈ B(H) and S ∈ B(K), if there exist quasiaffinities X ∈ B(K,H) and Y ∈ B(H,K) such that TX = XS and Y T = SY, then we say that T and S are quasisimilar. The operator T ∈ B(H) is said to be pure if there exists no non- trivial reducing subspace M of H such that the restriction of T to M is normal and is completely hyponormal if it is pure. Recall that every operator T ∈ B(H) has a direct sum decomposition T = T1⊕T2, where T1 and T2 are normal and pure parts, respectively. Of course in the sum decomposition, either T1 or T2 may be absent. The following lemma is due to Williams [32, Lemma 1.1]. Lemma 10. Let T ∈ B(H) and S ∈ B(K) be normal operators. It there exist injective operators X ∈ B(K,H) and Y ∈ B(H,K) such that TX = XS and Y T = SY , then T and S are unitarily equivalent. M.H.M.Rashid, N. H. Altaweel / Eur. J. Pure Appl. Math, 15 (3) (2022), 1067-1089 1079 Corollary 2. Let T ∈ B(H) be class p-wA(s, t)operator for 0 < s, t, s + t = 1 and 0 < p ≤ 1. Then T = T1 ⊕ T2 on the space H = H1 ⊕H2, where T1 is normal and T2 is pure and class p-wA(s, t), i.e., T2 has no invariant subspace M such that T2|M is normal. The next result was proved for dominant operators in [28, Theorem 1], for p-hyponormal operators in [20] and for w-hyponormal operators in [22, Lemma 2.12]. Proposition 2. Let T ∈ B(H) be class p-wA(s, t)operator for 0 < s, t, s + t = 1 and 0 < p ≤ 1 such that ker(T ) ⊂ ker(T ∗) and let S ∈ B(K) be a normal operator. If there exists a quasiaffinity X ∈ B(K,H) with dense range such that TX = XS, then T is normal. To prove Proposition 2, we need the following lemmas. Lemma 11. [9] If N is a normal operator on H, then we have⋂ λ∈C (N − λ)H = {0}. Lemma 12. ([10]) Let T ∈ B(H), D ∈ B(H) with 0 ≤ D ≤ M(T − λ)(T − λ)∗ for all λ ∈ C, where M is a positive real number. Then for every x ∈ D 1 2H there exists a bounded function f : C −→ H such that (T − λ)f(λ) ≡ x. Proof. [Proof of Proposition 2] ker(T ) ⊂ ker(T ∗) implies ker(T ) reduces T . Also ker(S) reduces S since S is normal. Using the orthogonal decompositions H = ran(|T |)⊕ ker(T ) and H = ran(S) ⊕ ker(S), we can represent T and S as follows: T = ( T1 0 0 0 ) , S = ( S1 0 0 0 ) , where T1 is an injective class p-wA(s, t) operator on ran(|T |) and S1 is injective normal on ran(S). The assumption TX = XS asserts that X maps ran(S) to ran(T ) ⊂ ran(|T |) and ker(S) to ker(T ), hence X is the form: X = ( X1 0 0 X2 ) , where X1 ∈ B(ran(S), ran(|T |), X2 ∈ B(ker(S), ker(T )). Since TX = XS, we have that T1X1 = X1S1. Since X is injective with dense range, X1 is also injective with dense range. Put W1 = |T1|sX1, then W1 is also injective with dense range and satisfies T (s, t)W1 = W1S. Put Wn = |∆n(T (s, t))|sWn−1, then Wn is also injective with dense range and satisfies ∆n(T (s, t))Wn = WnS. From [26, Corollary 2.7] and [6], if there exists an integer m such that ∆m(T (s, t)) is a hyponormal operator, then ∆n(T (s, t)) is a hyponormal operator for n ≥ m. It follows from Lemma 12 that there exists a bounded function f : C −→ H such that (∆n(T1(s, t)) ∗ − λ)f(λ) ≡ x, for every x ∈ (∆n(T1(s, t)) ∗∆n(T1(s, t)− ∆n(T1(s, t)(∆ n(T1(s, t)) ∗) 1 2H. Hence W ∗ nx = W ∗ n(∆ n(T1(s, t)) ∗ − λ)f(λ) = (S∗ 1 − λ)W ∗ nf(λ) ∈ ran(S∗ 1 − λ) for all λ ∈ C. M.H.M.Rashid, N. H. Altaweel / Eur. J. Pure Appl. Math, 15 (3) (2022), 1067-1089 1080 By Lemma 11, we have W ∗ nx = 0, and hence x = 0 because W ∗ n is injective. This implies that ∆n(T1(s, t) is normal. By Corollary 1, T1 is normal and therefore T = T1 ⊕ 0 is also normal. Theorem 8. Let T and S∗ be class p-wA(s, t) operators with 0 < s, t, s + t = 1 and 0 < p ≤ 1 such that ker(T ) ⊂ ker(T ∗) and ker(S∗) ⊂ ker(S). If there exist a quasiaffinity X such that TX = XS, then T and S are unitarily equivalent normal operators. Proof. First decompose T and S∗ into their normal and pure parts by T = T1 ⊕ T2 on H = H1 ⊕ H2 and S∗ = S∗ 1 ⊕ S∗ 2 on K = K1 ⊕ K2, where T1, S1 are normal and T2, S ∗ 2 are pure. Let X = [Xij ] 2 i,j=1. Then TX = XS implies that T2X21 = X21S1 and T2X22 = X22S2. Let T2 = U2|T2|, S∗ 2 = V ∗ 2 |S∗ 2 | be the polar decompositions of T2 and S∗ 2 , respectively and T2(s, t) = |T2|sU2|T2|t, S∗ 2(s, t) = |S∗ 2 |sV ∗ 2 |S∗ 2 |t, W = |T2|sX22|S∗ 2 |s. Then T2(s, t)W = |T2|sT2X22|S∗ 2 |s = |T2|sX22S2|S∗ 2 |s = W (S∗ 2(s, t)) ∗. Since ran(W ) reduces T2(s, t) and ker(W )⊥ reduces S∗ 2(s, t) and T2(s, t)|ran(W ) and S∗ 2(s, t)|ker(W )⊥ are unitarily equivalent normal operators, and since T2, S ∗ 2 are injective class p-wA(s, t) operators, we have T2|ran(W ) = T2(s, t)|ran(W ) and S∗ 2 |ker(W )⊥ = S∗ 2(s, t)|ker(W )⊥ by Lemma 9. Since T2, S ∗ 2 are pure, it implies W = |T2|sX22|S∗ 2 |s = 0. Hence X22 = 0. Similarly X12 = 0, X21 = 0. Hence X = X11 and S, T are unitarily equivalent normal operators. The following lemma is due to Williams [32, Lemma 1.1] Lemma 13. Let N1 ∈ B(H) and N2 ∈ B(K) be normal. If X ∈ B(K,H) and Y ∈ B(H,K) are injective such that N1X = XN2 and Y N1 = N2Y , then N1 and N2 are unitarily equivalent. Stampfli and Wadhwa [28] proved that the normal parts of quasisimilar dominant operators are unitarily equivalent. This result was generalized to classes of p-hyponormal operators in [12]. We prove that theses results hold for class p-wA(s, t) operators. Theorem 9. Suppose that 0 < s, t, s + t = 1 and ) < p ≤ 1. For each i = 1, 2, let Ti ∈ B(Hi) be class p-wA(s, t) operators such that ker(Tj) ⊂ ker(T ∗ j ) and let Ti = Ni ⊕ Vi on Hi = Hi1 ⊕ Hi2, where Ni and Vi are the normal and pure parts, respectively of Ti. If T1 and T2 are quasisimilar, then N1 and N2 are unitarily equivalent and there exist X∗ ∈ B(H22,H12) and Y∗ ∈ B(H12,H22) having dense range such that V1X∗ = X∗V2 and Y∗V1 = V2Y∗. M.H.M.Rashid, N. H. Altaweel / Eur. J. Pure Appl. Math, 15 (3) (2022), 1067-1089 1081 Proof. By hypothesis there exist quasiaffinities X ∈ B(H2,H1) and Y ∈ B(H1,H2) such that T1X = XT2 and Y T1 = T2Y . Let X = ( X1 X2 X3 X4 ) and Y = ( Y1 Y2 Y3 Y4 ) with respect to H2 = H21 ⊕ H22 and H1 = H11 ⊕ H12, respectively. A simple matrix calculation shows that V1X3 = X3N2 and V2Y3 = Y3N1. We claim that X3 = Y3 = 0. Let M = ran(X3). Then M is a non-trivial invariant subspace of V1. Since V ∗ 1 X3 = X3N ∗ 2 by Proposition 2, M is an invariant subspace of V ∗ 1 . Hence M reduces V1, σ(V1|M) ⊂ σ(V1) and V1|M is invertible. Let V ′ 1 = V1|M and define an operator X ′ 3 : H12 −→ M by X ′ 3x = X3x for each x ∈ H12. Then V ′ 1 is class p-wA(s, t) by Lemma 6, so that X ′ 3 has dense range and satisfies V ′ 1X ′ 3 = X ′ 3N2. Hence V ′ 1 is normal by Propsition 2. Since V1 is pure, this implies that M = {0} and X3 = 0. Similarly, we have Y3 = 0. Hence X1 and Y1 are injective. SInce N1X1 = X1N2 and Y1N1 = N2Y1, N1 and N2 are unitarily equivalent, by Lemma 13. Also, X4 and Y4 have dense ranges. Hence V1X4 = X4V2 and Y4V1 = V2Y4, so the proof is complete. Corollary 3. Let T1 ∈ B(H1) and T2 ∈ B(H2) be quasisimilar class p-wA(s, t) operators for 0 < s, t, s+ t = 1 and 0 < p ≤ 1. If T1 is pure, then T2 is also pure. Corollary 4. Let T1 ∈ B(H1) be class p-wA(s, t) operators for 0 < s, t, s + t = 1 and 0 < p ≤ 1 and T2 ∈ B(H2) be normal. If T1 and T2 are quasisimilar, then T1 and T2 are unitarily equivalent normal operators. 4. The Fuglede-Putnam Theorem We offer various results related to the Fuglede-Putnam theorem in this section. If T ∗X = XS∗ whenever TX = XS for every X ∈ B(K,H), a pair (T, S) is said to have the Fuglede-Putnam property. In operator theory, the Fuglede-Putnam theorem is well- known. It claims that the pair (T, S) possesses the Fuglede-Putnam property for any normal operators T and S. There are several generalizations of this theorem, the majority of which loosen the normality of T and S; see, for example, [22–24, 27, 28], and some references therein and for more details (see [3],[5],[4]). The Fuglede-Putnam theorem is the subject of the next lemma, which we will require in the future. Lemma 14. ([29]) Let T ∈ B(H) and S ∈ B(K). Then the following assertions equivalent. (i) The pair (T, S) has the Fuglede-Putnam property. (ii) If TX = XS, then ran(X) reduces T , ker(X)⊥ reduces S, and T | ran(X) , S|ker(X)⊥ are unitarily equivalent normal operators. M.H.M.Rashid, N. H. Altaweel / Eur. J. Pure Appl. Math, 15 (3) (2022), 1067-1089 1082 Remark 1. A necessary condition for the pair (T, T ∗) to satisfy Fuglede-Putnam’s theorem is ker(T ) ⊂ ker(T ∗). Since for a class p-wA(s, t) operator this is not always true, class p-wA(s, t) operator do not Fuglede-Putnam’s theorem. For example, if P is the orthogonal projection onto ker(T ), with T is class p-wA(s, t), then TP = PT ∗ but T ∗P ̸= PT. The following result (Corollary 6) prove that if T ∗, S are p-class A(s, t) operators for 0 < s, t, s+ t = 1 and 0 < p ≤ 1 such that ker(T ∗) reduces T ∗ and ker(S) reduces S, then the pair (T, S) satisfy Fuglede-Putnam’s theorem. Theorem 10. Let T ∈ B(H) be class p-wA(s, t) operator for 0 < s, t, s + t = 1 and 0 < p ≤ 1 and ker(T ) ⊂ ker(T ∗). If L is self-adjoint and TL = LT ∗, then T ∗L = LT. Proof. Since ker(T ) ⊂ ker(T ∗) and TL = LT ∗, ker(T ) reduces T and L. Hence T = T1 ⊕ 0, L = L1 ⊕ L2 on H = ran(T ∗)⊕ ker(T ), T1L1 = L1T ∗ and {0} = ker(T1) ⊂ ker(T ∗ 1 ). Since ran(L1) is invariant under T1 and reduces L1, T = ( T11 S 0 T22 ) , L1 = L11 ⊕ 0 on H = ran(T ∗) = ran(L1)⊕ ker(L1). T11 is an injective class p-wA(s, t) operator by Lemma 6 and L11 is an injective self-adjoint operator (hence it has dense range) such that T11L11 = L11T ∗ 11. Let T11 = V11|T11| be the polar decomposition of T11 and T11(s, t) = |T11|sV11|T11|t, W = |T11|sL11|T11|s. Then T11(s, t)W = |T11|sV11|T11|t|T11|sL11|T11|s = |T11|sT11L11|T11|s = |T11|sL11T ∗ 11|T11|s = |T11|sL11|T11|s|T11|tV ∗ 11|T11|s = W (T11(s, t)) ∗. Since T11(s, t) is min{sp, tp}-hyponormal and ran(W ) is dense (because ker(W ) = {0}), T11(s, t) is normal by [12, Theorem 7]. Hence T11 is normal and T11 = T11(s, t) by Corollary 1. Then ran(L1) reduces T1 by Lemma 7 and T ∗ 11L11 = L11T11 by Lemma 14. Hence T = T11 ⊕ T22 ⊕ 0, L = L11 ⊕ 0⊕ L2 and T ∗L = T ∗ 11L11 ⊕ 0⊕ 0 = L11T11 ⊕ 0⊕ 0 = LT. Example 2. Let H = ∞⊕ n=0 C2 and define an operator R on H by R(· · · ⊕ x−2 ⊕ x−1 ⊕ x (0) 0 ⊕ x1 ⊕ · · · ) = · · · ⊕Ax−2 ⊕Ax (0) −1 ⊕Bx0 ⊕Bx1 ⊕ · · · , M.H.M.Rashid, N. H. Altaweel / Eur. J. Pure Appl. Math, 15 (3) (2022), 1067-1089 1083 where A = 1 4 ( 1 2 1 2 1 2 1 2 ) and B = ( 1 0 0 0 ) . Then R is a class p-wA(s, t). Moreover, ran(E) = ker(R), E is not a self-adjoint and ker(R) ̸= ker(R∗), where E is the Riesz idempotent with respect to 0, see [31, Example 13]. Let T = R and L = P be the orthogonal projection onto ker(T ). Then T is a class p-wA(s, t) operator and TL = 0 = LT ∗, but T ∗L ̸= LT. Hence the kernel condition ker(T ) ⊂ ker(T ∗) is necessary for Theorem 10. Corollary 5. Let T ∈ B(H) be a class p-wA(s, t) operator for 0 < s, t, s + t = 1 and 0 < p ≤ 1 and ker(T ) ⊂ ker(T ∗). If TX = XT ∗ for some X ∈ B(H) then T ∗X = XT. Proof. Let X = L+ iJ be the Cartesian decomposition of X. Then we have TL = LT ∗ and TJ = JT ∗ by the assumption. By Theorem 10, we have T ∗L = LT and T ∗J = JT. This implies that T ∗X = XT . If we use the 2× 2 matrix trick, we easily deduce the following result. Corollary 6. Suppose that 0 < s, t, s + t = 1 and 0 < p ≤ 1. Let T ∗ ∈ B(H) be a class p-wA(s, t) operator and S ∈ B(K) be a class p-wA(s, t) operator with ker(T ∗) ⊂ ker(T ) and ker(S) ⊂ ker(S∗). If X ∈ B(H,K) and XT = SX, then XT ∗ = S∗X. Proof. Put A = ( T ∗ 0 0 S ) and B = ( 0 0 X 0 ) on H⊕K. Then A is a class p-wA(s, t) operator on H ⊕ K that satisfies BA∗ = AB and ker(A) ⊂ ker(A∗). Hence we have BA = A∗B, by Corollary 5, and so XT ∗ = S∗X. Example 3. Let S = T ∗ = R as in Example 2 and X = P be the orthogonal projection onto ker(S). Then SX = 0 = XT, but S∗X ̸= XT ∗. Hence the kernel condition is necessary for Corollary 6. As an application of Corollary 6, we establish the following result. Corollary 7. Suppose that 0 < s, t, s + t = 1. Let T ∈ B(H) and S∗ ∈ B(K) be class p-wA(s, t) and ker(T ) ⊂ ker(T ∗), ker(S∗) ⊂ ker(S). Let TX = XS for some operator X ∈ B(K,H). Then ran(X) reduces T , ker(S)⊥ reduces S and T | ran(X) , S|ker(X)⊥ are unitarily equivalent normal operators. Proof. By Corollary 6, T ∗X = XS∗. Therefore T ∗TX = XS∗S and so |T |X = X|S|. Let T = U |T |, S = V |S| be the polar decomposition. Then UX|S| = U |T |X = TX = XS = XV |S|. Let x ∈ ker(|S|). Then V x = 0 and TXx = XSx = 0. Hence Xx ∈ ker(T ) = ker(U) and UXx = 0. Hence UX = XV . Since ker(U) = ker(T ) ⊂ ker(T ∗) = ker(U∗), UU∗ ≤ U∗U . Hence U∗UU = U∗UUU∗U = UU∗U = U . This implies U and V ∗ are quasinormal. Hence U∗X = XV ∗, ran(X) reduces U , |T |, ker(X)⊥ reduces V , |S|. We may assume t < s. Then T, S∗ are class p-wA(s, s) operators with reducing kernels. M.H.M.Rashid, N. H. Altaweel / Eur. J. Pure Appl. Math, 15 (3) (2022), 1067-1089 1084 Let T (s, s) = |T |sU |T |s, S(s, s) = |S|sV |S|s. Then T (s, s), S∗(s, s) = |S∗|sV ∗|S∗|s = V S(s, s)∗V ∗ are p 2 -hyponormal. Also, since |S(s, s)∗| − |S(s, s)| = V ∗(|S∗(s, s)| − |S∗(s, s)∗|)V ≥ 0, S(s, s)∗ is p 2 -hyponormal, too. Then T (s, s)X = |T |sU |T |sX = |T |sUX|S|s = |T |sXV |S|s = XS(s, s), hence T (s, s)∗X = XS(s, s)∗, ran(X) reduces T (s, s), ker(X)⊥ reduces S(s, s) and T | ran(X) (s, s) = T (s, s)| ran(X) ≃ S(s, s)|ker(X)⊥ = S|ker(X)⊥(s, s) are unitarily equivalent normal operators. Hence T | ran(X) , S|ker(X)⊥ are normal by Corol- lary 1, and that they are unitarily equivalent follows from the fact that if N = U |N | and M = W |M | are normal operators, then for a unitary operator V , N = V ∗MV if and only if U = V ∗WV and |N |s = V ∗|M |sV for any s > 0. Theorem 11. Suppose that 0 < s, t, s + t = 1. Let T ∈ B(H) be class p-wA(s, t) and N a normal operator. Let TX = XN . Then the following assertions hold. (i) If the range ran(X) is dense, then T is normal. (ii) If ker(X∗) ⊂ ker(T ∗), then T is quasinormal. Proof. Let Z = |T |sX. Then T (s, t)Z = |T |sU |T |t|T |sX = |T |sTX = |T |sXN = ZN. Since T (s, t) is min{sp, tp}-hyponormal, we have T (s, t)∗Z = ZN∗ by [30]. Hence (T (s, t)∗T (s, t)− T (s, t)T (s, t)∗)|T |sX = T (s, t)∗T (s, t)Z − T (s, t)T (s, t)∗Z = T (s, t)∗ZN − T (s, t)ZN∗ = ZN∗N − ZNN∗ = 0. (i) If ran(X) is dense, then (T (s, t)∗T (s, t)− T (s, t)T (s, t)∗)|T |s = 0. Since ker(|T |s) ⊂ ker(T (s, t)) ∩ ker(T (s, t)∗), M.H.M.Rashid, N. H. Altaweel / Eur. J. Pure Appl. Math, 15 (3) (2022), 1067-1089 1085 this implies T (s, t) is normal. Hence T is normal by Corollary 1. (ii) Let X∗|T |sx = 0. Then |T |sx ∈ ker(X∗) ⊂ ker(T ∗) = ker(U∗) and T (s, t)∗x = |T |tU∗|T |sx = 0. Hence ker(X∗|T |s) ⊂ ker(T (s, t)∗) and ran(T (s, t)) ⊂ ran(|T |sX). Hence (T (s, t)∗T (s, t)− T (s, t)T (s, t)∗)T (s, t) = 0 by (i). This implies T (s, t) is quasinormal, and T is quasinormal by Theorem 1. Theorem 12. Suppose that 0 < s, t, s + t = 1 and 0 < q ≤ 1. Let T ∈ B(H) be such that T ∗ is p-hyponormal or log-hyponormal. Let S ∈ B(K) be class q-wA(s, t) with ker(S) ⊂ ker(S∗). If XT = SX, for some X ∈ B(H,K). Then XT ∗ = S∗X. Proof. Let T ∗ be a p-hyponormal operator for p ≥ 1 2 and let T = U |T | be the polar decomposition of T . Then the generalized Aluthge transform T ∗(s, t) of T ∗ is hyponormal and satisfies |T ∗(s, t)|2 ≥ |T |2 ≥ |(T ∗(s, t))∗|2, (12) X ′T (s, t) = SX ′ (13) whereX ′ = XU |T |t. Using the decompositionsH = ker(X ′)⊥⊕ker(X ′) and K = ran(X ′)⊕ ran(X ′)⊥, we see that T (s, t), S and X ′ are of the form T ∗(s, t) = ( T1 0 T2 T3 ) , S = ( S1 S2 0 S3 ) , X ′ = ( X1 0 0 0 ) where T ∗ 1 is hyponormal, S1 is class q-wA(s, t) with ker(S1) ⊂ ker(S∗ 1) and X1 is a one-one operator with dense range. Since X ′T (s, t) = SX ′, we have X1T1 = S1X1. (14) Hence T1 and S1 are normal by Corollary 6, so that T2 = 0, by Lemma 12 of [30] and S2 = 0 by Lemma 7. Then |T | = |T1|⊕P, for some positive operator P, by (12) and U = ( U1 U2 0 U3 ) by Lemma 13 of [30]. Let X = ( X11 X12 X21 X22 ) be a 2 × 2 matrix representation of X with respect to the decomposition H = ker(X ′)⊥⊕ker(X ′) and K = ran(X ′)⊕ ran(X ′)⊥. Then X ′ = XU |T |t implies that X1 = X11U1|T1|t and hence ker(T1) ⊂ ker(X1) = {0}. This shows that T1 is one-one and hence it has dense range, so that U2 = 0 and T = T1 ⊕ T4 for some hyponormal operator T ∗ 4 by [30, Lemma 13]. Since( X1 0 0 0 ) = X ′ = XU |T |t = ( X11 X12 X21 X22 )( U1|T1|t 0 0 U3|T4|t ) we deduce the following assertions. X12U2|T4|t = 0; hence X12T3 = 0 because T4 = U3|T4|. M.H.M.Rashid, N. H. Altaweel / Eur. J. Pure Appl. Math, 15 (3) (2022), 1067-1089 1086 X21U1|T1|t; hence X12 = 0 because U1|T1| 1 2 has dense range. X22U3|T4|t = 0; hence X22T3 = 0. The assumption XT = SX tell us that, X11T1 = S1X11 X12T4 = S1X12 = 0, X22T4 = S3X22 = 0. Since T1 and S1 are normal, we have X11T ∗ 1 = S∗ 1X11, by Fuglede-Putnam theorem. The p-hyponormality of T ∗ 4 shows that ran(T ∗ 4 ) ⊂ ran(T4). Also, we have ker(S3) ⊂ ker(S∗ 3). Hence, we also have X12T ∗ 4 = S∗ 1X12 = 0 and X22T ∗ 4 S ∗ 3X22 = 0. This implies that XT ∗ = X11T ∗ 1 ⊕ 0 = S∗ 1X11 ⊕ 0 = S∗X. Next, we prove the case where T ∗ is p-hyponormal for 0 < p ≤ 1 2 . Let X ′ be as above. Then T ∗(s, t) is (p+ 1 2)-hyponormal and satisfies X ′T (s, t) = SX ′. Use the same argument as above. We obtain T (s, t) = T1⊕T3 on H = ker(X ′)⊥⊕ker(X ′) and S = S1⊕S3, where T1 is an injective normal operator and S1 is also normal. Hence we have T = T1 ⊕ T4 for some p-hyponormal T ∗ 4 , by Lemma 13 of [30]. Again using the same argument as above, we obtain X21 = 0, X11T ∗ 1 = S∗ 1X11, X12T ∗ 4 = S∗ 1X12 = 0 and X22T ∗ 4 = S∗ 3X22 = 0. Hence we have XT ∗ = S∗X. Finally, we assume that T ∗ is log-hyponormal. Let T (s, t) and X ′ be as above. Then X ′T (s, t) = SX ′ and T ∗(s, t) is semi-hyponormal and satisfies |T ∗(s, t)| ≥ |T ∗| ≥ |(T ∗(s, t)∗|. By the same argument as above, we have T (s, t) = T1⊕T3 on H = ker(X ′)⊥⊕ker(X ′) and S = S1 ⊕ S3 on K = ran(X ′)⊕ ran(X ′)⊥, where T1 is an injective normal operator, S1 is normal, T ∗ 3 is invertible semi-hyponormal and S3 is class q-wA(s, t) with ker(S3) ⊂ ker(S∗ 3). By Lemma 13 of [30], we have that T is of the form T = T1⊕T4, for some log-hyponormal T ∗ 4 . Let X = ( X11 X12 X21 X22 ) . Then X ′ = XU |T |t implies that X12 = 0, X21 = 0 and X22 = 0. The assumption XT = SX implies that X11T1 = S1X11, hence X11T ∗ 1 ⊕ 0 = S∗ 1X11 by Fuglede-Putnam theorem. Thus we have XT ∗ = X11T ∗ 1 ⊕ 0 = S∗ 1X11 ⊕ 0 = S∗X. Therefore, the proof of the theorem is achieved. Example 4. Let R be an operator such that ker(R) does not reduce R and let P be the orthogonal projection onto ker(R). Then P does not commute with T ; otherwise ran(R) = ker(R) reduce T . Hence PR ̸= 0 = RP. It is easy to see that RP = PR∗ = 0 but R∗P ̸= PR(̸= 0) because ran(R∗P ) ⊂ ran(R∗) ⊂ ker(R⊥) = I − P. If we put T = R, then the assertion of Theorem 10 does not hold for such T . Also, if we put T = R∗, S = I −P and X = P, then XT = PR∗ = 0 = (I − P )P = SX. However, XT ∗ = PR ̸= 0 = (I − P )P = S∗X. Hence the assertion of Theorem 12 does not hold for such T . REFERENCES 1087 Theorem 13. Let T ∈ B(H) be such that T ∗ is an injective class p-wA(s, t) for 0 < s, t, s + t = and 0 < p ≤ 1. Let S ∈ B(K) be dominant. If XT = SX, for some X ∈ B(H,K). Then XT ∗ = S∗X. Proof. Assume that T ∗ is an injective p-w-hyponormal and let T = U |T | be the polar decomposition of T . Let T (s, t) be the aluthge transform of T and X ′ = XU |T |t. Then X ′T (s, t) = SX ′ and T ∗(s, t) is rp-hyponormal and satisfies |T ∗(s, t)|2rp ≥ |T ∗|2rp ≥ |(T ∗(s, t))∗|2rp for r ∈ min{s, t}. By the same argument in the proof of Theorem 12, we conclude that T ∗(s, t) = T1 ⊕ T3 on H = ker(X ′)⊥ ⊕ ker(X ′) and S = S1 ⊕ S3, where T1 is an injective normal operator and S1 is also normal, T ∗ 3 is invertible class p-wA(s, t) and S3 is dominant. Hence by Lemma 7, we have that T is of the form T = T1 ⊕ T4 for some class p-wA(s, t) T ∗ 4 . Let X = ( X11 X12 X21 X22 ) . Then X ′ = XU |T |t implies that X12 = 0, X21 = 0 and X22 = 0. The assumption XT = SX implies that X11T1 = S1X11, hence X11T ∗ 1 = S∗ 1X11 by Fuglede-Putnam theorem. Thus we have XT ∗ = X11T ∗ 1 ⊕ 0 = S∗ 1X11 ⊕ 0 = S∗X. Therefore, the proof of the theorem is achieved. 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