On the Adjoint and the Closure of the Sum of Two Unbounded Operators

Mortad, Mohammed Hichem

doi:10.4153/cmb-2011-041-7

Cited by 13 publications

(17 citation statements)

References 23 publications

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“…Finally, it is worth mentioning that the author has similar papers about the sum of unbounded operators (see [6,7]). Since rB − I < 1, we can say that (rB − I)A is A-bounded with relative bound smaller than one.…”

Section: Introductionmentioning

confidence: 99%

“…Then BA, defined on D(BA) = D(A), is not closed as BA ⊂ I. Thus it can neither be self-adjoint nor it can be normal and yetSo, the main purpose of this paper is to give conditions that force the product of two unbounded (one of them at most is bounded) closed, selfadjoint or normal operators to be closed, self-adjoint or normal.We draw the attention of the reader to the paper [3] where some conditions implying the closedness of the product of two closed operators are established.Finally, it is worth mentioning that the author has similar papers about the sum of unbounded operators (see [6,7]). Since rB − I < 1, we can say that (rB − I)A is A-bounded with relative bound smaller than one.…”

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confidence: 99%

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On the closedness, the self-adjointness and the normality of the product of two unbounded operators

Mortad

2012

Demonstratio Mathematica

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Abstract. Let A and B be two unbounded densely defined operators on a Hilbert space H. The purpose of this work is to give simple conditions that make the product AB closed, self-adjoint and normal provided the two operators are so. IntroductionFirst, we assume the reader is very familiar with notions, definitions and results on unbounded operators. Some general references are [2,8,10].Second, we recall the following theorems which will be needed to prove some of our results. Theorem 1.[11] Let B be a closed operator. If A is B-bounded with relative bound "a" smaller than one, then A + B is closed.The analog of the previous theorem for self-adjoint operators is the KatoRellich theorem (see e.g. [9]).Theorem 2. [Kato-Rellich] Let B be an unbounded self-adjoint operator with domain D(B). If A is B-bounded with relative bound "a" smaller than one, such that A is symmetric, then A + B is self-adjoint on D(B).If A and B are self-adjoint commuting bounded operators, then it is straightforward that AB = BA is self-adjoint.If A and B are two commuting bounded normal operators, then a simple use of the Fuglede theorem yields the normality of BA = AB. In the unbounded case (at least one operator is unbounded, A say), things get a little tricky in the sense that the same hypotheses do not necessarily yield The following example will be used on several occasions throughout this paper and it shows that the product of two commuting self-adjoint (and hence it also includes the class of normal operators) operators need not be self-adjoint and it may not be even normal.Example 1. Let A be an unbounded self-adjoint operator having a trivial kernel, for example takeNow set B = A −1 (observe that both A and B are positive on their respective domains). Then BA, defined on D(BA) = D(A), is not closed as BA ⊂ I. Thus it can neither be self-adjoint nor it can be normal and yetSo, the main purpose of this paper is to give conditions that force the product of two unbounded (one of them at most is bounded) closed, selfadjoint or normal operators to be closed, self-adjoint or normal.We draw the attention of the reader to the paper [3] where some conditions implying the closedness of the product of two closed operators are established.Finally, it is worth mentioning that the author has similar papers about the sum of unbounded operators (see [6,7]). Since rB − I < 1, we can say that (rB − I)A is A-bounded with relative bound smaller than one. Since A is closed, Theorem 1 yields the closedness of rBA or BA and hence the proposition is proved. Then BA is closed.Remark. Although Condition 1 seems to be a bit strong but the power of the previous theorem lies in the fact that B is not assumed to be closed.and so BA is densely defined. Then we can write for all ϕ ∈ D(BA) rBAϕ = (rB − I)Aϕ + Aϕ.Since (rB − I)Aϕ ≤ a Aϕ + b ϕ and a < 1, this shows that (rB − I)A is A-bounded with relative bound smaller than one and hence Theorem 1 allows us to establish the closedness of rBA = (rB −I)A+A or that of BA. The self-adjointness and the normality question...

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Section: Introductionmentioning

confidence: 99%

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confidence: 99%

On the closedness, the self-adjointness and the normality of the product of two unbounded operators

Mortad

2012

Demonstratio Mathematica

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“…We digress a bit to say that the author proved in [9] results about the adjoint and the closure of the sum of two unbounded operators for which the two first assertions of the previous theorem were just a consequence.…”

Section: Any Unbounded B and If B A Is Densely Definedmentioning

confidence: 94%

On the Normality of the Sum of Two Normal Operators

Mortad

2010

Complex Anal. Oper. Theory

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“…A preprint of a paper by M. Mortad [9], dealing with when (A + B) * = A * + B * , caused me to go back to some old notes of mine [3] in which the same question was investigated but with a more specific goal: when is A + B essentially selfadjoint. In [3] A + B is a quantum mechanical Hamiltonian, e.g., A = −∆ on a core domain D(A), and B is a quantum mechanical potential.…”

Section: Introductionmentioning

confidence: 99%

“…The notes [3] contained a number of sufficient conditions, but they were not what I wanted, so were never published. Here I would like to touch on some main points from [3] to augment [9]. Recall that A + B essentially selfadjoint means that, for A selfadjoint and B a regular symmetric perturbation, i.e., D(B) ⊃ D(A), one has A + B ⊂ A + B = (A + B) * .…”

Section: Introductionmentioning

confidence: 99%