Remarks on Fuglede-Putnam Theorem for Normal Operators Modulo the Hilbert-Schmidt Class

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1 International Mathematical Forum, Vol. 9, 2014, no. 29, HIKARI Ltd, Remarks on Fuglede-Putnam Theorem for Normal Operators Modulo the Hilbert-Schmidt Class Vasile Lauric Department Mathematics Florida A&M University Tallahassee, FL USA Copyright c 2014 Vasile Lauric. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, reproduction in any medium, provided the original work is properly cited. Abstract In a recent note, we proved a Fuglede-Putnam commutativity theorem for almost normal operators with finite modulus of C 2 -quasitriangularity modulo the Hilbert- Schmidt class. In this note we show how our proof can be adjusted to the case of normal operators to obtain an optimal norm estimate obtained by G. Weiss. The result is also reviewed in the case of almost normal operators with zero modulus of C 2 -quasitriangularity for the operator its adjoint an example is provided. Mathematics Subject Classification: 47B20 Keywords: Hilbert-Schmidt trace-class operators, C 2 -quasitriangularity modulus Let H be a separable, infinite dimensional, complex Hilbert space, denote by L(H) the algebra of all bounded linear operators on H by C 1 (H) C 2 (H) the trace class Hilbert-Schmidt class, respectively. For an operator T L(H), let D T denote its self-commutator, that is T T T T. An operator T L(H) for which D T C 1 is called almost normal. We will denote the class of almost normal operators on H by AN(H). The classical Fuglede-Putnam theorem states if M, N L(H) are normal operators X L(H) is such that MX = XN, then M X = XN. G.

2 1390 Vasile Lauric Weiss (see [8,9]) extended this result by proving that if M, N L(H) are normal operators X L(H) such that MX XN C 2 (H), then M X XN C 2 (H) M X XN 2 = MX XN 2. T. Furuta ([1]) F. Kittaneh ([2]) extended the above result to subnormal operators (a subnormal operator is the restriction of a normal operator to an invariant subspace) by proving that if M, N L(H) are subnormal operators X L(H) such that MX XN C 2 (H), then M X XN C 2 (H) M X XN 2 MX XN 2. Kittaneh [2] provided some result that suggested that it would be interesting to investigate Fuglede-Putnam generalizations in which operators M, N are almost normal. For instance, he proved that if S L(H) is an almost normal subnormal operator SX XS C 2 (H) for some X L(H), then S X XS C 2 (H). In [3] we extended Fuglede-Putnam theorem to operators that are almost normal operators that have finite modulus of C 2 -quasitriangularity modulo Hilbert-Schmidt class. Let P(H), (or simply P) denote the set of all finite rank orthogonal projections on H. Recall (see [4]) that the modulus of C 2 -quasitriangularity of an operator T L(H) is q 2 (T ) = lim inf P P (I P )T P 2, where lim inf is with respect to the natural order on P. Recall that if m(t ) denotes the rational cyclicity of T, then ([6]) q 2 (T ) (m(t )) 1 2 T, that is, operators of finite rational cyclicity have finite modulus of C 2 -quasitriangularity. According to [7], if T AN(H) q 2 (T ) <, then q 2 (T ) < there exists a sequence P n P, n 1, with P n I strongly, so that lim (I P n)t P n 2 = q 2 (T ) lim (I P n )T P n 2 = q 2 (T ), n n q 2 2(T ) = q 2 2(T ) + Tr(D T ). We will use the following idea that was first used by T. Furuta [1], namely if T 1, T 2 L(H), then the operator T1,T 2 (X) = T 1 X XT 2 defined on the Hilbert space C 2 (H) (with inner product X, Y 2 = Tr (XY )) has its adjoint

3 Remarks on Fuglede-Putnam theorem 1391 given by T 1,T 2 (X) = T1 X XT2 the self-commutator of 12 := T1,T 2 is D 12 (X) = D T1 X XD T2 = DT1,D T2 (X). Using operator theoretic concepts, we proved in [3] the following. Theorem 1. Let S AN(H) with q 2 (S) < let X L(H) such that R := SX XS C 2 (H). Then Q := S X XS C 2 (H). The proof of Theorem 1 provided the estimate Q R X 2 ( 8 D S q 2 2(S) + 26 q 2 2(S ) ), which implies that for a normal operator S L(H), Q 2 2 R 2, (since for a normal operator S, q 2 (S) = q 2 (S ) = D S 1 = 0.) The purpose of this note is to review the proof given in [3] show that in the case of a normal operator S L(H), it gives the optimal estimate, namely S X XS 2 = SX XS 2. The proof depends only operator theoretic concepts rather than using generating functions defined by G. Weiss. Proposition 2. Let S L(H) be a normal operator let X L(H) such that R := SX XS C 2 (H). Then Q := S X XS C 2 (H) Q 2 R 2. Proof. Let P n P, n 1, such that P n I lim (I P n)sp n 2 = q 2 (S) lim (I P n )S P n 2 = q 2 (S ) <. n n It is known now ([5]) that for a normal operator S, q 2 (S) = q 2 (S ) = 0. Then QP n 2 2 = Tr (P n Q QP n ) Tr (Q Q). We need to prove that the sequence { QP n 2 } n is bounded above. Write ( ) ( ) S1n S S = 2n X1n X X = 2n, S 3n S 4n X 3n X 4n relative to the decomposition of H as P n H (I P n )H. Thus ( ) ( ) S Q = 1n X 1n + S3nX 3n X 1n S1n X 2n S2n Q11 + Q S2nX 1n + S4nX 3n X 3n S1n X 4n S2n = 32, Q 41 + Q 22 where Q 11 = S 1nX 1n X 1n S 1n, Q 32 = S 3nX 3n X 2n S 2n, Q 41 = S 4nX 3n X 3n S 1n, Q 22 = S 2nX 1n X 4n S 2n. Consequently, QP n 2 2 = Q 11 + Q Q 41 + Q

4 1392 Vasile Lauric Next we estimate the Hilbert-Schmidt norm of each of the Q s. We begin with Q 32, namely Q = S 3nX 3n X 2n S 2n 2 2 ( X S 3n 2 + X S 2n 2 ) 2 2 X 2 ( S 3n S 2n 2 2) 2 X 2 ( q 2 2(S) + q 2 2(S ) ), Q , (1) (recall that all Q s depend on n.) Denote 11 (X) = S 1n X XS 1n defined on C 2 (P n H); therefore Q 11 = 11(X 1n ) = S 1nX 1n X 1n S 1n, then [ 11, 11 ](X 1n ) = D S1n X 1n X 1n D S1n, 11 (X 1n ) (X 1n ) 2 2 = [ 11, 11 ](X 1n ), X 1n 2 = = Tr (D S1n X 1n X 1n) Tr (X 1n D S1n X 1n) = = Tr (X 1nD S1n X 1n ) Tr (X 1n D S1n X 1n). A matrix calculation shows that P n D S P n = D S1n + S 3nS 3n S 2n S 2n, X 1nD S1n X 1n = X 1nD S X 1n X 1nS 3nS 3n X 1n + X 1nS 2n S 2nX 1n X 1n D S1n X 1n = X 1n D S X 1n X 1n S 3nS 3n X 1n + X 1n S 2n S 2nX 1n, respectively. Therefore 11(X 1n ) (X 1n ) 2 2 = Tr (X 1nD S1n X 1n ) + Tr (X 1n D S1n X 1n) = Tr (X 1n D S X 1n) Tr (X 1n S 3nS 3n X 1n) + Tr (X 1n S 2n S 2nX 1n) Tr (X 1nD S X 1n ) + Tr (X 1nS 3nS 3n X 1n ) Tr (X 1nS 2n S 2nX 1n ) Tr (X 1n D S X 1n) Tr (X 1nD S X 1n )+ +Tr (X 1n S 2n S 2nX 1n) + Tr (X 1nS 3nS 3n X 1n ) 2 X 2 D S 1 + S 2nX 1n S 3n X 1n 2 2 X 2 ( 2 D S 1 + q 2 2(S) + q 2 2(S ) ), consequently Q 11 2 R 11 2, (2) where R 11 = S 1n X 1n X 1n S 1n = 11 (X 1n ), means that the difference between right h side left h side is bounded above by a non-negative sequence that converges to 0. Combining (1) (2), one obtains Q 11 + Q 32 2 Q Q 32 2 R (3)

5 Remarks on Fuglede-Putnam theorem 1393 The operator Q 41 = S4nX 3n X 3n S1n 41 is the adjoint of can be viewed as 41(X 3n ), where 41 : C 2 (P n H, (I P n )H) C 2 (P n H, (I P n )H) defined by 41 (Y ) = S 4n Y Y S 1n. Denoting 41 (X 3n ) by R 41, we have R Q = 41 (X 3n ) (X 3n ) 2 2 = = [ 41, 41 ](X 3n ), X 3n 2 = Tr (D S4n X 3n X 3n) Tr (X 3n D S1n X 3n) = = Tr (X 3nD S4n X 3n ) Tr (X 3n D S1n X 3n). Since X 3n = (I P n )XP n, X 3n = P n X (I P n ), using again the matrix representation of D S relative to the decomposition of H into P n H (I P n )H, that is P n D S P n = D S1n + S 3nS 3n S 2n S 2n, we obtain X 3n D S1n X 3n = X 3n (D S S 3nS 3n + S 2n S 2n)X 3n = = X 3n D S X 3n X 3n S 3nS 3n X 3n + X 3n S 2n S 2nX 3n. On other h (I P n )D S (I P n ) = D S4n + S 2nS 2n S 3n S 3n, which implies Thus X 3nD S4n X 3n = X 3n(D S S 2nS 2n + S 3n S 3n)X 3n = = X 3nD S X 3n X 3nS 2nS 2n X 3n + X 3nS 3n S 3nX 3n. Q R = Tr (X 3nD S4n X 3n ) + Tr (X 3n D S1n X 3n) = = Tr (X 3n D S X 3n) Tr (X 3n S 3nS 3n X 3n) + Tr (X 3n S 2n S 2nX 3n)+ Tr (X 3nD S X 3n ) + Tr (X 3nS 2nS 2n X 3n ) Tr (X 3nS 3n S 3nX 3n ) Tr (X 3n D S X 3n) Tr (X 3nD S X 3n )+ +Tr (X 3n S 2n S 2nX 3n) + Tr (X 3nS 2nS 2n X 3n ) 2 X 2 D S 1 + S 2nX 3n S 2n X 3n 2 2 X 2 ( 2 D S q 2 2(S ) ), so Q 41 2 R (4) Finally, Q 22 = S 2nX 1n X 4n S 2n can be hled similarly to Q 32, namely Q 22 2 S 2n 2 X 1n + X 4n S 2n 2 2 X S 2n 2, According to (4) (5), Q (5) Q 41 + Q 22 2 Q Q 22 2 R 41 2, (6)

6 1394 Vasile Lauric consequently, QP n 2 R R (7) The proof will be finished after we establish that R R 41 2 is bounded. Since R C 2 (H), RP n 2 2 Tr (R R), RP n 2 2 is bounded above by R 2 2. The representation of R relative to the decomposition of H into P n H (I P n )H is ( R11 + R 23 ), R 41 + R 33 where R 23 = S 2n X 3n X 2n S 3n R 33 = S 3n X 1n X 4n S 3n, therefore On other h, Similarly therefore According to (9) (10), RP n 2 2 = R 11 + R R 41 + R (8) R 11 2 R 11 + R R 23 2 R X 2 ( q 2 2(S) + q 2 2(S ) ) 0, R 11 2 R 11 + R (9) R 41 2 R 41 + R R 33 2 R , R 41 2 R 41 + R (10) R R 41 2 R 11 + R R 41 + R 33 2, (11) Finally, according to (7) (12), R R 41 2 RP n 2. (12) QP n 2 RP n 2, which ends the proof. Q 2 R 2, Applying Proposition 2 to S, one obtains the following.

7 Remarks on Fuglede-Putnam theorem 1395 Corollary 3. If S L(H) is a normal operator X L(H) is such that R := SX XS C 2 (H), then Q := S X XS C 2 (H) Q 2 = R 2. It is an elementary step to extend the above Corollary to two normal operators to deduce the Hilbert-Schmidt norm equality obtained by G. Weiss. A careful reviewing of the proof of Proposition 2 leads to the following. Theorem 4. Let S AN(H) satisfy q 2 (S) = q 2 (S ) = 0 let X L(H) such that R := SX XS C 2 (H). Then Q := S X XS C 2 (H) Q 2 R 2 + c X D S 1 2 1, where c > 0. This implies, by a stard argument, the following. Theorem 5. Let S 1, S 2 AN(H) satisfy q 2 (S 1 ) = q 2 (S 2 ) = q 2 (S 1) = q 2 (S 2) = 0 let X L(H) such that R := S 1 X XS 2 C 2 (H). Then Q := S 1X XS 2 C 2 (H) Q 2 R 2 + c X ( D S1 1 + D S2 1 ) 1 2, where c > 0. Proof. Set S = S 1 S 2 observe that S AN(H) q 2 (S) = 0 since q 2 2(S) q 2 2(S 1 ) + q 2 2(S 2 ) Tr (D S ) = Tr (D S1 ) + Tr (D S2 ) = 0, q 2 (S ) = 0, apply Theorem 4 to get the desired conclusion. The following is an example of a subnormal operator that is almost normal with modulus of C 2 -quasitriangularity the trace of its self-commutator equal zero, such that the Hilbert-Schmidt norm of R Q are not equal. Example 6. Let H = l 2 (N), S k be the weighted (unilateral) shift with weights 1, 1,..., let S be a block-weighted shift defined on k k+1 l2 (N) ( ) in which the nonzero blocks are S 1, S 2,.... Let X L(l 2 (N) ( ) ) be the also a block-shift with all nonzero blocks equal to the identity, that is X = S 0 I, where S 0 is the stard (unilateral) shift. It is an entertaining exercise is left to the reader to see that the operator S satisfies the above properties that SX XS 2 S X XS 2. References [1] T. Furuta, An extension of the Fuglede-Putnam theorem to subnormal operators using a Hilbert-Schmidt norm inequality, Proc. Amer. Math. Soc., 81 (1981), [2] F. Kittaneh, On generalized Fuglede-Putnam theorems of Hilbert-Schmidt type, Proc. Amer.Math. Soc., 88 (1983),

8 1396 Vasile Lauric [3] F. Kittaneh, A Fuglede-Putnam theorem for almost normal operators with finite modulus of C -quasitriangularity modulo the Hilbert-Schmidt class, Submitted. [4] D. V. Voiculescu, Some extensions of quasitriangularity, Rev. Roumaine Math. Pures Appl., 18 (1973), [5] D. V. Voiculeascu, Some results on norm-ideal perturbation of Hilbert space operators, J. Operator Theory, 2 (1979), [6] D. V. Voiculescu, A note on quasitriangularity trace-class selfcommutators, Acta Sci. Math. (Szeged), 42 (1980), [7] D. V. Voiculescu, Remarks on Hilbert-Schmidt perturbations of almost normal operators, Topics in Modern Operator Theory; Operator Theory: Advances Applications-Birkhäuser, 2 (1981), [8] G. Weiss, Fuglede s commutativity theorem modulo the Hilbert-Schmidt class generating functions for matrix operators. II, J. Operator Theory, 5 (1981), [9] G. Weiss, The Fuglede commutativity theorem modulo operator ideals, Proc. Amer.Math. Soc., 83 (1981), Received: September 7, 2014