International Journal of Analysis and Applications Volume 18, Number 4 (2020), 644-662 URL: https://doi.org/10.28924/2291-8639 DOI: 10.28924/2291-8639-18-2020-644 Received April 1st, 2020; accepted April 20th, 2020; published May 28th, 2020. 2010 Mathematics Subject Classification. 47L15. Key words and phrases. commutative; compact; operation; W*-algebras. ©2020 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 644 A COMMUTATIVE AND COMPACT DERIVATIONS FOR W* ALGEBRAS ABDELGABAR ADAM HASSAN1,2,*, MOHAMMAD JAWED1 1Jouf University, College of Science and Arts in Tabrjal, Department of Mathematics, Kingdom of Saudi Arabia 2University of Nyala, Department of Mathematics, Sudan *Corresponding Author: aahassan@ju.edu.sa ABSTRACT. In this paper, we study the compact derivations on W* algebras. Let M be W*-algebra, let ( )LS M be algebra of all measurable operators with M , it is show that the results in the maximum set of orthogonal predictions. We have found that W* algebra A contains the Center of a W* algebra ß and is either a commutative operation or properly infinite. We have considered derivations from W* algebra two-sided ideals. 1. INTRODUCTION Let M be a W*-algebra and let ( )Z M be the center of M . Fix a M and consider the inner derivation a  on M generated by the component a , which is  :( ) ,a a =  . The norm closing two sided ideal ( )f B generated by the finite projections of a W* algebra B behaves somewhat similar to the idealized compact operators of ( )B H (see [11],[8],[9]). Therefore, it is natural to ask about any sub-algebras d of B that is any derivation from  into ( )f B implemented from an element of ( )y B . https://doi.org/10.28924/2291-8639 https://doi.org/10.28924/2291-8639-18-2020-644 Int. J. Anal. Appl. 18 (4) (2020) 645 We perform two main difficulties: the presence of the center of B and the fact that the main characteristic in [8] proof (that is, if n Q , is a sequence of mutually orthogonal projections and ( )T B H hence 0n nQ TQ   for all n implies that T is not compact) failure to generalize to the case in which g is of Type II . Finally, we have considered derivations from d at the two-sided ( ) ( )( )11 , , 1 1C B B L B      + + =   +   to obtain faithful finite normal trace  on B . 2. NOTATIONS PRELIMINARY Lemma (1). Let B be a semi-finite algebra, let ( )0Q p B and 0 0x Q be such that 0x , is a faithful trace on 0Q B . Assume there are ( )nQ p B , ( )nF p and nU B for 1, n ,...i in n += , such that the projections n Q are mutually orthogonal and * *, n n n o n n n Q U U Q F U U= = for all n (i.e., 0n n Q Q F ). Let 0n n n x U F x= . Then n JRW x O→ . Proof. Assume that i n in n Q n  = = . Let  be a faithful semi-finite normal ( )fsn trace on B + to be agreed on 0Q B with 0x  . Then for all nQ B B +  we have ( ) ( ) ( ) ( ) ( ) ( ) 0 * * * * * * . n n n n n n n n n o n n n n o x n n n n x B U U BU U U U U BU Q F U BU F Q F U BU F B       = = = = = Let ( )P p B be any semi-finite projection. Then by [11] there is a central decomposition of the identity ( )1, , 0E E p E E     =  = for   such that ( )PE   for all   . Then ( ) ( ) ( ) 1 1 n n n n x n n n PE Q PE Q Q PE Q        =  = = =   2 1 n n PE x   =   Int. J. Anal. Appl. 18 (4) (2020) 646 whence 0 n PE x   for all   . Let 0  and let   be a finite index set such that 2 0 E x     . Then for all n , 2 2 0 2 0 2 0 n n n n n PE x PE U F x PU F E x E x              = =       Hence from 22 n n Px PE x     + where 0nPx → , to completes the proof. Lemma (2). Let ( )T f P , then there is an 0  and ( )0 E p  such that for every ( )0 F p  with F E we have ( )TF  . Proof. Let ( )12 0T =  and let G be the sum of a maximal family of mutually orthogonal central projections G  such that ( )TG  . Then ( ) ( )supTG TG   =  , hence 1G  . Let E Z G= − and let ( )0 F P  with F E . Since 0FG = , by the maximally of the family we have ( )TF  . 3. RELATIVELY COMPACT DERIVATION Let M be a W  -algebra and let ( )Z M be the center of M . Fix a M and consider the inner derivation a  on M generated by the element a , that is  :( ) ,a a  =  . Obviously, a there is a linear bounded operator on ( ), M M  , where M  is a C  -norm on M . It is known that there exists ( )c Z M such that the following estimate holds: a Ma c  − . In view of this result, it is natural to ask whether there exists is an element y M with 1y  and ( )c Z M such that  , a y a c − . Definition (3). A linear subspace I in the W* algebra M equipped with a norm I  is said to be a symmetric operator ideal if Int. J. Anal. Appl. 18 (4) (2020) 647 (i) , I S S for all S I  (ii) , II S S for all S I  =  (iii) , , I I ASB A S B for all S I A B M   . Observe, that every symmetric operator ideal I is a two-sided ideal in M , and therefore by [13], it follows from 0 S T  and T I that S I and I I S T . Corollary (4). Let M be a W  -algebra and let I be an ideal in M . Let : M I → be a derivation. Then there exists an element a I , such that  ,a a = =  . Proof. Since  is a derivation on a W  -algebra, it is necessarily inner [8]. Thus, there exists an element d M , such that ( ) ( )  · ,d d  = =  . It follows from the hypothesis that  ,d M I . Using [22] (or [20]), we obtain  , ,d M d M I I    = −  =  and  , , 1, 2kd M I k = , where 1 2 , k d d id dk d M  = + =  , for 1, 2k = . It follows now, that there exist ( )1 2, c c Z M and ( )1 2,u u U M , such that  , 1 / 2k k k kd u d c − for 1, 2k = . Again applying [20], we obtain k k d c I−  , for 1, 2k = . Setting ( ) ( )1 1 2 2:a d c i d c= − + − , we deduce that a I and  ,a =  . Corollary (5). Let be a semi-finite W* -algebra and let be a symmetric operator space. Fix and consider inner derivation on the algebra given by . If , then there exists satisfying the inequality and such that . Proof. The existence of such that . Now, if , then . Hence, if , then , where and for , and so , that is . M E ( )a a S M=  a = ( )LS M ( )   ( ), , x a x x LS M =  ( )M E  d E E M E d  →  ( )  ,x d x = d E ( )  ,x d x = ( )u U M ( ) 2 E E E EE u du ud du ud d = −  + =  1 : 1x M x M x =   4 1 i ii x u = = ( )iu U M 1i  1, 2, 3, 4i = ( ) ( ) 4 1 8 i i EiE E x u d   =   8M E Ed →    Int. J. Anal. Appl. 18 (4) (2020) 648 4. A COMMUTATIVE OPERATION ON W* SUB-ALGEBRAS When A a commutative operation is is crucial because it provides the following explicit way to find an operator T B implementing the derivation. For the rest of this section let  be any a commutative operation sub-algebras of B and : B  → be any derivation. Let u be the unitary group of  and M be a given invariant mean on u , i.e., a linear functional on the algebra of bounded complex-valued functions on u such that (i) For all real ( )  ( ) , f inf f U U u Mf sup f U U u    (ii) For all ( ) ( ), , U UU u Mf MS where f V f UV for V u = =  . Thus M is bounded and ( )  f U uMf sup U  for all f (see [8] for the existence and properties of M ). For each B   the map ( )( )M U U  → is linear and bounded and hence defines an element ( )T B    . Explicitly, ( ) ( )( ) T M U U for all B    →  The same easy computation as in [8] shows that a T =  . Notice that for all A B the map ( ) ( )( )M U BU E B  → = defines an element ( )E B which clearly belongs to B . Moreover it is easy to see that E is a conditional expectation (i.e., a projection of norm one) from B onto B (see [6]). Theorem (6). Let  be a commutative operation W* sub-algebras of B containing the center of B . For every derivation ( ): f B  → there is a ( )T f B such that a T =  . We have seen that given an invariant mean M on u there is a unique T B such that a T =  and ( ) 0E T = . We are going to show that ( )T B . Reasoning by contradiction assume that ( )T B . We proof requires several reductions to the restricted derivation Int. J. Anal. Appl. 18 (4) (2020) 649 ( ):E E f B  → for some ( )0 E p  . To simplify notations we shall assume each time that 1E = . Let us start by noticing that if ( )iQ p  for 1, 1, , 0n ni n n Q Q += + = and 1n nP Q Q += + , then ( ) ( ) 1 1 1 n i i n n n n i n PTP Q TQ Q Q Q Q  + + + = = + + hence ( ) ( ) ( ) 1 max n i i i i i i n PTP Q TQ Q TQ   + = = + Definition (7). For every ( )Q p  define    , Q Q = to be the central projection. Set ( )   1P P p P=   = . Thus P p iff ( ) ( )PTPG TG = for all ( )G p . We collect several properties of  Q . Corollary (8). Let B be a semi-finite W* algebra with a trace  , let  be a properly infinite W* sub-algebras of B and let 1 1  +   . Then for every derivation ( )1: ,C B +→ there is ( )1 ,a T C B + such that a T =  . In the notations introduced there, it is easy to see that ( )( ) ( )1 1, ,C B C B   + += ), where 0   =  and 0  is the usual trace on ( )0B H . We can actually simplify the proof by choosing n I  = since the condition  is no longer required. Corollary (9). Let 1n n P Q Q + = + . Then there is a largest central projection  1,n nQ Q + such that for every ( )G p with  1,n nG Q Q + , we have ( ) ( )1 1Q TQ G PTPG = . Proof. Let ( ) ( ) ( ) i i iG G p Q TQ G PTPG =  = and ( )G p = +  if ( )G p and 0  then nG G . Since ( ) ( )maxi i iPTPG Q TQ G = for all ( )G p , we see that ( )1n nG G p+ = . Notice that  is hereditary (i.e., G −  and ( ), FF p G   + imply F  ). Int. J. Anal. Appl. 18 (4) (2020) 650 Let  1, supn nQ Q + =  . We have only to show that  1,n nQ Q +  . Let ( )G G  + = + be the sum of a maximal collection of mutually orthogonal projections ( )G  +  . Then for every F  we have   ( )( )1, 0n nQ Q G F+ − + = because of the maximal of the collection of  . Then  1,n nQ Q G + = + . Consider now any ( ), 0G p   , then ( )G G G  = + and since ( ) ( )G G G    +  +  , we have ( )( ) ( )( )n nQ TQ G G PTPG G    + = + for all  . Since ( ) ( )( ). n nQ TQ G resp PTPG  is the direct sum of then ( )( ) ( )( )( ). n nQ TQ G G resp PTPG G    + + , then we have ( ) ( )( ) ( )( ) ( ) sup sup n n n n Q TQ G Q TQ G G PTPG G PTPG           += = = + whence n G G . Since 0  is arbitrary, we have  1,n nG Q Q ++ =  which completes the proof. Corollary (10). (i) If 1 0 n n Q Q + = with ( )iQ p  then  11 ,n nQ Q +−  1,n nQ Q + . (ii) If 1n n Q Q +  with ( )iQ p  then    1n nQ Q + . (iii) If 0  with ( ), QQ p p  +  then    , Q Q = and    1 Q −  If ( ) 0TG  for all ( )0 E p  then the following hold: (iv) If ( )E p then  E E= . (v) If ( )Q p  then   ( )Q Qc , where ( )c Q is the central support of Q . Proof. We have to show that for every ( )  1, 1 , n nG p G Q Q +  − we have 1nG G + . Let E + be the sum E  of a maximal collection of mutually orthogonal projections of 1nG + that are majored by G . Then Int. J. Anal. Appl. 18 (4) (2020) 651 ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 1 1 1 1 1 sup sup n n n n n n n n n n n n Q TQ F Q TQ F TQ Q Q Q Q Q Q Q F T F         + + + + + + + + + + = + + = = whence 1n E G + +  . By the maximalist of the collection, ( )0 G E  − + does not majority any nonzero projection of 1n G + and since ( ) 1n np G G +=  , any central projection ( )G G E   − + must be in n G . By definition of  , this implies that ( )G E − +  whence ( )  1,n nG E Q Q +− +  . So, ( )  11 ,n nG E G Q Q +− +   − and hence 1nG E G += +  which completes the proof. (ii) Let ( )G p and  nG Q . Then ( ) ( ) ( )1 1n n n nTG Q TQ G Q TQ G   + +=  ( )TG whence equality holds and    1n nQ Q + by the maximalist of  1nQ + . (iii)  , Q  is maximal under the condition: if ( )G p and  , G Q  then ( ) ( ) ( )( ) ( )QTQG Q T Q G TG     + + = which is the same condition defining    ,Q I Q Q− = . Thus    , Q Q = . Applying this to  we have    ,Q = and thus by (i) we have      1 , 1Q Q  − = − . (ii) Let ( ),E E p+  then ( )( ) ( )( )ETE E TE E   + = + . This implies that if 0  , then  E E+  so  E E and if    E E E E+ = −  then ( )( ) ( )( )0 ETE E T E   = + = + whence  E E= . (v) Follows at once from (ii) and (iv). The condition that ( ) 0TE  for all ( )0 E p  is of course meaningless unless B is properly infinite. Hence, we may assume without loss of generality that: B is properly infinite and semi-finite. There is an 0  such that ( )TE  for all ( )0 E p  . Int. J. Anal. Appl. 18 (4) (2020) 652 Lemma (11). Let P p and  ) ( 1, , ,n PTP n PTPR X R X +=  = − − , where ( )PTPX denotes the spectral measure of the self-adjoint operator PTP . Then there is an ( )nE p , with n E I E= − such that i i R E are properly infinite and ( ) i i jc R E E= for , n 1i n= + . Proof. Let  )1 ,n n PTPR R R X += + =  and let 0F  be any central projection. If RF were finite, we would have ( ) ( ) ( )( ) ( )( ) 1 1 TF PTPF PTP R F PTP R F      = = − = −  Thus RF is infinite and nonzero. Hence R is properly infinite and ( )c R n= . Now let 1E be the maximal central projection majored by ( )nc R , such that n nR F is properly infinite. Then ( ), n n nc R E E= and ( )n nR n E− is finite, hence ( ) 1 1n n n nR n E R E+ + +− = is properly infinite and ( )1 1 1, n n nc R E E+ + += . End of the Proof of Theorem (6). Take any ( )00 Q p B  such that 0Q B has a faithful trace 0x  with 0 0x Q H and assume 0 1x = . Let ,P p   be the not decreasing to zero. We are going to construct inductively a sequence ( ) ( ), , ,n n n nF p Q p B U    partial isometrics in B, n x H such that (a) * * 0 0 , , . ., n n n n n n n n U U Q U U Q F i e Q Q F= = (b) 0n n n n x U F x Q H=  (c) 0 n m Q Q for n m=  (d) ( ) n mn m hence P P for n m       (e) nn Q p   (f) 1 1 n nn p x  +  (g) 2,n nTx x  . Int. J. Anal. Appl. 18 (4) (2020) 653 The induction can be started with an arbitrary P  ; assume we have the construction for 1n − . Let us apply Lemma(11) to n P P  = and obtain ( ) ( ), = , 1i iE p R p B for i n n  + as defined there. Then 2 2 2 0 1 0 1 o n n x E x E x + = = + Let n F be (any of) the projection n E or 1n E + for which 2 1 0 2i E x  and let i be the corresponding index. Then i n R F is properly infinite and has central support n F . Now 0 Q is finite having a finite faithful trace 0x  , hence so is 0 0 1 1 j j Q F Q Q for j n   − and ( ) 1 1 n j nj Q F − = . Let ( )  1 1 , 1 n n i n j nj S inf R F Q F − = = − . By the parallelogram law (see [2]) applied to n F we have that ( ) 1 1 1 1 inf , 1 n n i n n j n j n i n j j R F S Q F Q F R F − − = =       − − −             whence i n n R F S− is finite and hence n S is properly infinite and ( )n nc S F= . Since 0 nQ F is finite and ( )0 n nc Q F F we have 0 n nQ F S , i.e., there is a partial isometry nU B and a ( ),n n nQ p B Q S  such that (a) holds. Let nx be defined by (b) and choose 1n n + so that (d) and (f) hold. -Since nn i Q R P    we have (e), since ( ) 1 1 1 n n j nj Q Q F − =  − we have (c). Finally nn i n n x R x P x  = = hence (g) follows from ( ) ( ) ( ) ( ) 2 2 0 1 2 , , , , . n n n n n n n n i n i n i n i n n n Tx x P TP x x P TP R x R x R x R x x F x         = =  = =  Let now 1nn n n y x P x  + = − . B is semi-finite, hence we can apply Lemma (1) to obtain that 0 n BEW x → . Since 1 0 n n P x  + → we thus have 0 n BEW y → and n n y P H , where Int. J. Anal. Appl. 18 (4) (2020) 654 ( ) 1n nn P P P p d   + = −  and are mutually orthogonal by (d). Clearly for n large enough, ( ) ( ) 14, nn n yTy y T =  . Since ( ) ( )( )n ny yT U UM    = , by the properties of the invariant mean mentioned, we have that ( )( )  14sup ny U U U u      . Thus we can find for every n , a unitary n V u such that ( )( ) 14,n n n nV V y y    . Let 1 n nn A V P  = = , then A d and ( ) ( )( ) ( )( ) ( )( ) ( )( ) 1 4 , , , , , n n n n n n n n n n n n n n n n n n n n A A y y P A A P y y P A AT A TA P y y P V V P y y V V y y            = = − = = = for all n . Therefore ( ) 0nA y → . But because of (Π), we have ( ) ( )A f B  , which completes the proof. 5. THE PROPERTY OF INFINITE W* SUB-ALGEBRA Lemma (12). Let be a properly infinite projection and . Let projection be finite or properly infinite, and . Let . For every we denote by such a projection that is the largest central projection, for which holds. We have and for the following relations and . Moreover, if all projections are finite then is a finite projection as well. Proof. Since, we have ( ) ( ) ( )0 , 1; 0,azb Z M s b e  =  ( )( )0, 1azc e  = ( )q P M ( ) 1c q = ( )0,azq e  0n  n nz 1 nz− ( ) ( )1 1 an n zz q z e−  − ( ),nb + 1n nz  ( )1 1 1 1 1 : n n n n d z z z b   + + =    −  = +    ( ) 1: , , 0 a z hold q e d d b+   ( ) 1s d = ( ), , 1az ne b n +  ( ), a z e d + ( ) ( )1 , , a a z n z n e b e b  + +  + Int. J. Anal. Appl. 18 (4) (2020) 655 . Hence, for every . In addition, and is properly infinite projection. Hence, in the case when is finite projection, it follows that . Let us consider the case when is a properly infinite projection with and such that . In this case, with and deduce . All other statements follow from the form of element . Since, and for every . Observe also that . Finally, let all projections be finite. Since , we have for every . There projections standing on the right-hand sides are finite. Hence, is finite projection as a sum of the left-hand sides [22]. We shall use a following well-known implication . We supply here a straightforward argument. Let be such that . Then and therefore . This means . As in [6] we can use Theorem (6) to extend the result to the properly infinite case. Theorem (13). Let  be a properly infinite W* sub-algebra of B containing the center of B . For every derivation ( ): f B  → there is a ( )T f B such that a T =  . Before we start the proof let us recall that if  is properly infinite there is an infinite countable decomposition of the identity into mutually orthogonal projections of  , all ( ) ( ) ( ) ( ) ( )1 1 1 11 1 , 1 , a a a z n n z n n z n e z q z e b z e b  + + + + −  − +  − + 1n n z z +  n ( ) ( ), 0,a az n n ze b e +  + ( )0, a z e + q 1 n n z  q ( ) 1c q = ( )0,azq e  ( ) ( ), 0, , ,a az n z np q q e q e b= = + = + ( )1 1nn z c q  =  = d 1 1 1 ,z d z b= ( ) ( )1 1 1n n n n nz z z z b+ + +− = − ( ), a n n z n z q z e b + n ( ) ( ) ( )( )1 11 1n nns d s b z z z  += = + − = ( ), , 1az ne b n +  ( )1 1 1, n ndz b d z z += − = ( )1 1n n nb z z + + − ( ) ( )1 1 1, , , a a z z e d z e b z+ = + ( )( ) ( )( )1 1 1, , a a z n n z n n n e d z z e b z z + + + + − = + − n ( ),aze d + ( )( ) ( ) ( ), , 0p q zp zq z P Z M z c p c q      ( )z z Z M  ( ) ( ) ( ) ( )( )0 z c pz c qz z c p c q    ( ) ( )z c p c q   ( ) ( )z zp z p z q z zq   = = zp zq Int. J. Anal. Appl. 18 (4) (2020) 656 equivalent in  to I, and thus a fortify equivalent in B to 1 [8]. Therefore there is a spatial isomorphism ( )0: B B B B H → =  with ( )10 n H l + = and ( ) ( )0B H  =  =  [5]. Recall also that the elements B of B (or  ) are represented by bounded matrices , , ij B i j    with entries in B (or  ) by the formula ( ) ( )ij kl jk ilI E T I E T E  =  where ij E is the canonical matrix unit of ( )0B H . In particular if ,  are the maximal a commutative operation subalgebras of ( )0B H of Laurent (resp. diagonal) matrices, then ( ).B B resp B B    iff ijB   is a Laurent matrix with entries in B , i.e., ij i jB B −= ,where k B , denotes the entry along the kth diagonal( ). ij ij iiresp B B= for all ,i j  . Proof. Let 1    − = then ( )( ) ( ): d f B f B → = is a relative compact derivation. Let us define the following W* algebras: ( )1 1, , ,n n n nB B  − + =   =   =   =  , and 2n n+  =  . First, let us notice that ( ) ( ) ( )( ) ( ) ( ) ( )   0 0 n f B B B H f B B f B   =     =   = by [22]. Therefore ( ) ( ) ( ) ( )( )  1 1 0n n nf B A f B f B − −    =  =   = because  is spatial Now ( )( ) ( )0 . n B B H B I I =    =      Int. J. Anal. Appl. 18 (4) (2020) 657 Thus we can apply Theorem(6) to the derivation  restricted to the a commutative operation sub-algebra n  of B and we obtain a ( )nT f B such that n na T = −  vanishes on n . Now 1n n B +       =  . Therefore, for all n n A  and 1 1n n A + +  we have ( ) ( ) ( ) ( )1 1 1 1n n n n n n n n n n n nA A A A A A A A   + + + += = = i.e., ( )1n nA + and nA commute and hence ( ) ( )  1 0n n n f B +    = Thus n  also vanishes on 1n+  . Now n  is a commutative operation and hence so are n and 2n+  . Moreover, n B     Implies ( )1 n B − =      and hence 2n n I I B + =          Thus we can apply again Theorem(6) to the relative compact derivation n  restricted to 2n+  . Let ( )1nT f B+  be such that n agrees with ad 1nT + on 2n+ . Since 1n n I I +       =  and n  vanishes on 1n+  , we see that ad 1nT + vanishes on n I  , i.e., ( ) ( ) ( )( ) ( )1 0n n nT I f B B H fg B+      =    Then for all ( )1, , n niji j T +   and ( ) ( ) ( ) ( )2 1nn ni n jnijT E I E T I E f B+ =    Int. J. Anal. Appl. 18 (4) (2020) 658 whence by Lemma(12)(a) ( ) ( )1n ijT f B+  . But we saw that ( )  0nd f B  = , hence ( )1 0n ijT + = for all ,i j  , so 1 0 n T + = . Therefore n  vanishes also on 2n+  and hence on I . Now and  generate ( )0B H , whence 1n+ =  and I  generate  . Thus by the  -weak continuity of n  (see [6]) we see that 0, . ., n n n a T i e a T  − = = =  . Clearly ( )1 nad T  − = and ( ) ( )1 nT B −  . Let us assume in this part that B is semi-finite and let  be a fsn trace on it. Beside the closed ideal ( )f B we can also consider the (non closed) two-sided norm-ideals ( )1 , 1 1C B for  +  +   defined by ( ) ( )  ( ) ( ) 1 1 1 1 1 11 , , . C B B B B B B for B C B          + + + + ++ =    =  Obviously, ( ) ( )11 , , ,C B B L B     + + =  where the latter is the non commutative 1 L + -space of B relative to  (see [14]). Recall the following facts about ( )1L M+ spaces in the case of a general W* algebra M and 1 1  +   ( ( )L M is identified with M ): ( )1L M+ is a Banach space, its dual is isomorphic to ( )1L M  + (with 111 1    + + + = ), and the duality is established by the functional tr on ( ) 1 L M , where if ( ) ( )11 ,A L M B L M   ++  we have ( ), nAB BA L M and ( ) ( ) ( ) 1 1 ,tr AB tr BA tr AB A B    + + =  , ( ) ( )  1 1 1 1 1 1 max , 1A tr A trAB B L M B       + + + + + = =   (see [14]). Of course, if M B= we can identify ( )1L M+ with ( )1 ,L B + and tr with  . The following inequality will be used here only in the semi-finite case and in the context of 1 C + - ideals, but since the same proof holds for 1 L + -spaces, we shall consider the general case. Int. J. Anal. Appl. 18 (4) (2020) 659 Corollary (14). Let M be a W* algebra, ( )10, A L M +  and ( )1 1 1, , 0, 1n n n n n nQ Q p M Q Q Q Q+ + + = + = . Then 1 1 1 1 1 1 1 1n n n n A Q AQ Q AQ      + + + + + + + +  + Proof. Let us first note that 1 1 1 1n n i i i i i n i n Q AQ Q AQ   + + + + = = =  And 1 1 1 1 1 1 i i n n i i i n i n Q AQ Q AQ     + + + + + = =+ =  Consider first 1 n+ = and take the polar decomposition's , , 1 i i i i i Q AQ U Q AQ i n n= = + . Then i i U U  and i i U U  are majored by i Q and hence i U commutes with j Q . Therefore ( ) * 1n n B U U + = + commutes with i Q and 1B = . Then 1 1 1 1 . n i i i n n i i i n n i i i n n A trAB tr Q BAQ tr Q AQ Q AQ + = + = + =    =       =     =    Consider now 0  . Let ( )1B L M  + be such that 1 1B  +  and 1 1 1 . n n i i i i i n i n Q A Q tr Q A Q B  + + = =+    =         Take the polar decomposition's A U A= and B V B= , then VU are in M and ,A B are in ( ) ( )1 1,L M L M   + + , respectively. Let Int. J. Anal. Appl. 18 (4) (2020) 660 ( ) ( ) ( ) 1 1 1 1 . n z z i i i n f z tr Q U A Q V B    +   + −  +  =   =      Then by standard arguments, it is easy to see that f is analytic on 0 Re z n  and continuous and bounded on 0 Re z n  . Then by the three-line theorem (see [4]) we have ( ) ( ) ( ) 1 1 11 1 1 t t if Ma t itx f Max f     + + +   + Now 1 1 1 1 n i ii n Q A Qf  + = +   =  +   and by Holder’s inequality ( ) ( ) ( ) 1 1 1 1 1 1 1 1 1 n i t i t j j j n n i t i t j j j n f it tr Q U A Q V B B Q U A Q V B B           +   + −   + +  = +   + −   + +  =   =        ( )( )1 11 1 max . i t i t j j j Q U A Q V B B n        + −   ++  +   Again by Holder’s inequality applied twice and by the result already obtained in the 0 = case, ( ) ( ) ( ) ( ) ( ) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 n i t i t j j j n n i t i t j j j n i t i t f it tr Q U A A Q V B Q U A A Q V B U A A U A A A               +   + + −   +  = +   + + −   +  = + + + + + +   + =           Thus ( )11 1f A + + whence by the second equality in this proof, 1 1 1 1 1 1 1 1 n n i i i i i n i n A Q AQ Q AQ       + + + + + + + = =+  =  Data Availability No data were used to support this study. 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