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GLASNIK MATEMATI CKI Vol. 46(66)(2011), 433 – 438

AN IMPLICIT DIVISION OF BOUNDED AND UNBOUNDED LINEAR OPERATORS WHICH PRESERVES THEIR PROPERTIES

Mohammed Hichem Mortad Universit´e d’Oran, Algeria

Abstract. We give an answer to the following problem: Given two linear operators A and B such that BA and A verify some property P , then when does B verify the same property P ? Of course, we have to assume that B satisﬁes some condition Q independent of (or weaker than) P . This problem is solved in the setting of both bounded and unbounded operators on a Hilbert space. Some interesting counterexamples are al so given.

1. Introduction

A widely known property is that every self-adjoint operator is normal whether it is bounded or not. Conversely, a normal operator is self-adjoint if it has a real spectrum. An unbounded normal symmetric operato r is self-

adjoint. Another question is if an operator is a normal product of two self- adjoint operators, then when is it self-adjoint? This has been completely

solved in the setting of bounded and unbounded

In this paper we consider in some sense the converse of the previous question for bounded and unbounded operators and its analog for diﬀerent classes of operators. To be more precise, the problem considered here is the following: given two linear operators (bounded or not) A and B such that BA satisﬁes a property ” P ” such that A (or B ) also satisﬁes the same property ” P ”, then when does B (or A) satisfy the property ” P ”? Of course one will have to assume that B satisﬁes some property ” Q ” which is either independent

operators, see [1, 7, 8, 10].

2010 Mathematics Subject Classiﬁcation. 47A05. Key words and phrases. Products of operators, bounded and unbounded operators, self-adjoint, closed and normal operators. Partially supported by ”Laboratoire de Math´ematiques, G´eom´etrie, Analyse, Contrˆole & Applications”.

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of ” P ” or weaker than it, e.g. if BA and A are self-adjoint, then when is B self-adjoint? Similarly if both BA and A are nonnegative, then when is B

nonnegative?

answer for positive real numbers (if ab > 0 and b > 0, then a > 0) and this is in fact what has made us choose the title of the paper. We assume that the reader is familiar with notions and results about linear operators (both bounded and unbounded) and we refer the reader to the references [2, 11] for further details. The question considered for bounded operators is standard enough so that we include it in the introduction. In general, if BA and A are self-adjoint, then B need not remain self- adjoint. For consider the matrices

etc. This last question in particular has obviously a po sitive

A =

1

0

0

2

, B = 2 1

1 1 . Then BA =

1

2

2

2

.

Thus BA is self-adjoint while B is not (it is not even normal). Observe that

A and B are both invertible. If, however, A and B are two commuting bounded operators and if BA

and A are both self-adjoint such that A is invertible, then B is self-adjoint. The commutativity of A and B in the previous result can be dropped at the cost of adding the assumption ” B normal” and we can say that if BA and

A are two self-adjoint operators such that 0 / W (A) (the numerical range of

A). Then B is self-adjoint if it is normal. The proof is a mere consequence of

a result by M. R. Embry (). As for normal operators, one can easily prove that if A and B are two commuting operators and if BA is normal and A is unitary, then B is normal. The proof is elementary and can be easily established by doing some straightforward arithmetic. We also have a similar result for positive operators. The proof is again standard.

Proposition 1.1 . Assume that A and B are bounded and commute. If BA and A are positive such that A is invertible, then B is positive.

2. The Mixed Bounded-Unbounded Case

We recall the Fuglede theorem (. Cf. ) for the reader’s convenie nce.

Theorem 2.1 . If A is a bounded operator and N is an unbounded normal operator, then

AN NA =AN N A.

Assume that A is an

unbounded closed operator commuting with B , i.e., B A AB . If BA

is normal, then so is A.

Theorem 2.2 . Let B be a unitary operator.

IMPLICIT DIVISION OF OPERATORS

Proof. We have

435

B A AB =BB A = B (BA) (BA)B

and since BA is normal, the Fuglede theorem yields

(2.1)

B (BA) (BA) B =B A B A B B .

Now we have

AA = B BAA B B = B (BA)(BA) B

= (by the normality of BA)

B (BA) (BA)B

= B A B BAB

A B B B AB (by (2.1)) = A B AB I

A AB B (by the commutativity hypothesis)

= A A

Since A is closed, AA is self-adjoint. But self-adjoint operators are maximally symmetric (see e.g. ). Thus

AA = A A or A is normal. Theorem 2.3 . Let A be a unitary operator. Assume that B is closed and commutes with A (i.e., A B BA ). If BA is normal, then so is B .

Proof. The idea of proof is similar to the preceding one. However, some details are diﬀerent and need to be emphasized. We have

A B

BA =A BA BA A = BAA

(to simplify the proof a bit we do not replace AA = A A by I at present). Since BA is normal, the Fuglede theorem then implies that

(2.2)

We have on the one hand

A (BA) (BA) A .

 (BA) ∗ BA = AA ∗ (BA) ∗ BA ⊂ A(BA) ∗ A ∗ BA (by (2.2)) ⊂ A(BA) ∗ BA ∗ A (by the commutativity hypothesis) ⊂ (BAA ∗ ) ∗ B = B ∗ B.

436

On the other hand

 BB ∗ = BAA ∗ (BAA ∗ ) ∗ ⊂ BAA ∗ (A ∗ BA) ∗ (by the commutativity hypothesis) = BAA ∗ (BA) ∗ A ⊂ BA(BA) ∗ A ∗ A (by the Fuglede theorem) = BA(BA) ∗ .

But BA is normal and hence

BB (BA)(BA) = (BA) BA

B B =BB B B.

Since B is closed, similar arguments to those of the foregoing proof apply and yield the normality of B . The proof is over.

Corollary 2.4 . Assume B is a bounded and self-adjoint operator verifying B 2 = I . Assume that A is a closed and symmetric operator commuting with B . If BA is self-adjoint, then so is A.

Proof. Since BA is self-adjoint, it is normal. Theorem 2.2 then implies that A is normal. But A is assumed to be symmetric and hence A is self- adjoint. Theorem 2.3 can be used analogously to establish the following result.

Corollary 2.5 . Assume A is a bounded and self-adjoint operator verifying A 2 = I . Assume that B is a closed and symmetric operator commuting with A. If BA is self-adjoint, then B is also self-adjoint.

As done in the bounded case we can drop the commutativity hypothe sis on A and B and the invertibility of A and replace it by diﬀerent conditions. The proof uses the following result

Theorem 2.6 (). Assume H is an unbounded normal operator. Also assume that A is a bounded operator for which 0 W (A) and such that

 AH ⊂ H ∗ A. Then H = H ∗ . We have

Corollary 2.7 . Assume B is an unbounded normal operator and A is a bounded operator such that 0 W (A). If BA and A are both self-adjoint, then so is B .

Proof. As A and BA are self-adjoint, then

AB (BA) = BA.

Since B is normal and 0 W (A), Theorem 2.6 applies and yields the self- adjointness of B . IMPLICIT DIVISION OF OPERATORS

437

3. The Unbounded Case

Theorem 3.1 . There exist an unbounded closed (self-adjoint or normal) operator A and a non-closed (non self-adjoint or non-normal) unbounded operator B such that BA is closed (self-adjoint or normal). Hence ” BA closed (self-adjoint or normal), A closed (self-adjoint or normal) B closed (self- adjoint or normal)”.

Proof. We give an explicit counterexample. Consider the following two operators deﬁned by

Af (x) = e 2x f (x) and Bf (x) = (e x + 1)f (x)

on their respective domains

D (A) = { f L 2 (R ) : e 2x f L 2 (R )}

and

D (B ) = { f L 2 (R ) : e 2x f, e x f L 2 (R )}

(think of B as a sum of two operators S = e 2x + e x + 1 and

on their maximal domains). Then obviously A is closed as it is self-adjoint while B is not closed since it has a closure B deﬁned by Bf (x) = (e x +1)f (x) on D (B ) = { f L 2 (R ) : e x f L 2 (R )} . Now the operator BA is deﬁned by BAf (x) = (e 2x + e x )f (x) on

T = e 2x , say,

D (BA) = { f

D (A) : Af D (B )} = { f L 2 (R ) : e 2x f, e x f L 2 (R )}

and one can see that BA is closed on this domain as it is self-adjoint. One can give some positive results using known results from the litera ture. An instance of that can be found in  where it is proved, among othe r things, that if HK is a normal product of two self-adjoint operators (H is unbounded and σ (K ) σ (K ) ⊆ { 0 } ), then HK is self-adjoint. This can be used to prove the following

Corollary 3.2 . Let B be a densely deﬁned normal operator. Assume that A is a self-adjoint, densely deﬁned (or bounded) and boundedly invertible operator with inverse A 1 . One assumes further that σ (A 1 ) σ (A 1 ) ⊆ { 0 } . If BA is self-adjoint, then B is self-adjoint.

Proof. Since A is invertible, AA 1 = I and hence

B = B (AA 1 ) = (BA)A 1 .

Now both BA and A 1 are self-adjoint and hence B is self-adjoint as it is normal and since σ (A 1 ) σ (A 1 ) ⊆ { 0 } . 438

References

 E. Albrecht and P. G. Spain, When products of selfadjoints are normal, Proc. Amer. Math. Soc. 128 (2000), 2509–2511.  J. B. Conway, A course in functional analysis, Springer, New York, 1990.

 M. R. Embry, Similarities involving normal operators on Hilbert space , Pacif. J. Math.

35 (1970), 331–336.

 B. Fuglede, A commutativity theorem for normal operators, Proc. Nati. Acad. Sci. 36 (1950), 35–40.

 T. Furuta, Invitation to linear operators. From matrices to bounded linear operators on a Hilbert space, Taylor & Francis, Ltd., London, 2001.  M. H. Mortad, An All-Unbounded-Operator Version of the Fuglede-Putnam Theorem, Complex Anal. Oper. Theory (to appear), DOI: 10.1007/s11785-011-0133-6.  M. H. Mortad, An application of the Putnam-Fuglede theorem to normal products of self-adjoint operators, Proc. Amer. Math. Soc. 131 (2003), 3135–3141.  M. H. Mortad, On some product of two unbounded self-adjoint operators, Integral Equations Operator Theory 64 (2009), 399–407.  M. H. Mortad, Similarities involving unbounded normal operators, Tsukuba J. Math.

34 (2010), 129–136.

 W. Rehder, On the product of self-adjoint operators, Internat. J. Math. and Math. Sci. 5 (1982), 813–816.  W. Rudin, Functional Analysis, McGraw-Hill, New York, 1991 (2nd edition).

M. H. Mortad D´epartement de Math´ematiques Universit´e d’Oran (Es-senia) B.P. 1524, El Menouar, Oran 31000 Algeria Mailing address:
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