International Journal of Analysis and Applications

Volume 18, Number 4 (2020), 624-632

URL: https://doi.org/10.28924/2291-8639

DOI: 10.28924/2291-8639-18-2020-624

ON THE NORMALITY OF THE PRODUCT OF TOW OPERATORS IN HILBERT

SPACE

MOHAMMED MEZIANE1,∗, ABDELKADER BENALI2 AND MOHAMMED HICHEM

MORTAD3

1Higher school of economics, B.P 65 CH2 PTT ACHABA Hnifi, Bir El Djir, Oran,Laboratoire d’analyse

mathematiques et applications (LAMA), Algeria

2Faculty of The Exact sciences And computer, Mathematics Department, University of Hassiba Benbouali,

Chlef Algeria. B.P. 151 Hay Essalem, chlef 02000, Algeria

3Department of Mathematics, University of Oran, B.P. 1524, El Menouar, Oran 31000, Algeria

∗Corresponding author: benali4848@gmail.com

Abstract. In this paper we present the results of the maximality of operators not necessarily bounded. For

that, we will see the results obtained by operators in situation of extension. Regarding the normal product

of normal operators we seem to be the key to maximality.

1. Introduction

First, we assume that all operators operators are non necessarily bounded on a complex Hilbert space

H, Let us, however, recall some notations that will be met below. If A and B are two operators with

dense domains D(A) and D(B) respectively, then B is called an extension of A, and we write A ⊂ B, if

D(A) ⊂ D(B) and if A and B coincide on D(A). The product AB of two operators is definded by

AB(x) = A(Bx) for x ∈ D(AB)

Received February 20th, 2020; accepted April 1st, 2020; published May 19th, 2020.

2010 Mathematics Subject Classification. Primary 47A05, Secondary 47A10, 47B20, 47B25.

Key words and phrases. normal; self-adjoint; symmetric operators; commutativity; maximality of operators.

©2020 Authors retain the copyrights

of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

624

https://doi.org/10.28924/2291-8639
https://doi.org/10.28924/2291-8639-18-2020-624


Int. J. Anal. Appl. 18 (4) (2020) 625

wehere

D(AB) = {x ∈ D(B) : Bx ∈ D(A)}.

Recall too that the unbounded operator A, defined on D(A), is said to be invertible if there exists an

everywhere defined (i.e. on the whole of H) bounded operator B, which then will be designated by A−1,

such that

A−1A ⊂ AA−1 = I

where I is the indentity operator on H. An operator A is said to be closed if its graph is closed in H ⊕H.

The closing of the domain D(A) of A implies the closing of A if A is bounded on D(A). It is known that the

product operators AB is closed if for instance A is closed and B ∈ B(H), or A−1 ∈ B(H) and B is closed.

We also recall that an operator S is said to be densely defined if its domain D(S) is dense in H. It is known

that in such case its adjoint S∗ exists and is unique. If T ⊂ S, then S∗ ⊂ T∗. Notice that if S, T and ST

are all densely defined, then we are only sure of

T∗S∗ ⊂ (ST)∗,

and a full equality occurring if e.g. T−1 ∈ B(H) or S ∈ B(H). The bounded operator T ∈ B(H) is said to

be unitary if TT∗ = T∗T = I. A densely defined operator S is said to be symmetric if S ⊂ S∗. It is called

self-adjoint if S = S∗. S is called essential self-adjoint if the closure of S is self-adjoint (i.e. (S)∗ = S). We

say that S is normal if S is densely defined, and ‖Sx‖ = ‖S∗x‖ for all x ∈ D(S) = D(S∗) (hence from known

facts normal operators are automatically closed). Recall that the previous is equivalent to S is closed and

SS∗ = S∗S. Other classes of operators are defined in the usual fashion. Let us also agree that any operator

is linear and non necessarily bounded unless we specify that it belongs to B(H). We also assume the basic

theory of operators (see e.g. [1] or [20]). We do recall the celebrated Fuglede-Putnam Theorem though:

Theorem 1.1. (for a proof, see e.g. [1]) Let T ∈ B(H) and let M,N be two normal non necessarily bounded

operators. Then

TN ⊂ MT =⇒ TN∗ ⊂ M∗T.

One of the main objectives of this work is to impose conditions to obtain other results, starting from an

extension. The following theorem and corollary result are a powerful tool to prove results on unbounded

operators. For instance, Statement (3) of the next theorem is used in the proof of the ”unbounded” version

of the spectral theorem of normal operators (see e.g. [15]). For other uses, see e.g. [6] or [10].

Let us now list some known (see e.g. [15] or [16]) maximality results:

Theorem 1.2. Let S,T be two operators with (dense when necessary) domains D(S) and D(T) respectively

such that S ⊂ T . Then S = T when one of the following occurs:



Int. J. Anal. Appl. 18 (4) (2020) 626

(1) S is surjective and T is injective.

(2) T is symmetric and S is self-adjoint (resp. normal). We then say that self-adjoint (resp. normal)

operators are maximally symmetric.

(3) T and S are normal (we say that normal operators are maximally normal). Hence, self-adjoint (resp.

normal) operators are maximally normal (resp. self-adjoint).

Commutativity of operators must be handled with care. First, recall the definition of two strongly

commuting (normal) operators (see e.g. [16]):

Definition 1.1. Let A and B be two normal operators. We say that A and B strongly commute if all the

projections in their associated projection-valued measures commute.

Now, let us recall results obtained by Devinatz-Nussbaum (and von Neumann) on strong commutativity:

Theorem 1.3. (Devinatz-Nussbaum-von Neumann, [2] and cf. [13]). If there exists a self-adjoint operator

A such that A ⊆ BC, where B and C are self-adjoint, then B and C strongly commute.

Corollary 1.1. Let A, B and C be self-adjoint operators. Then

A ⊆ BC =⇒ A = BC

2. Main Results

The normality of unbounded products of normal operators has been studied recently. See e.g. [5] and the

references therein. We recall

Theorem 2.1. (for a proof, see e.g. [11]) Let A,B be normal operators with B ∈ B(H). If BA ⊂ AB, then

AB and BA are both normal (and so AB = BA).

Theorem 2.2. Let T,A,B be non necessarily bounded operators such that T and B are self-adjoint with

B ∈ B(H) and A is normal. Assume further that BA ⊂ T . Then

BA = T.

Proof. We have:

BA ⊂ T =⇒ BA ⊂ T ⊂ A∗B

=⇒ BA∗ ⊂ AB (by Fuglede-Putnam Theorem).

It is clear that BA is closable and densely defined. Let’s show now that BA is normal. Indeed

(BA)∗BA = (BA)∗(BA)∗∗ = A∗B(A∗B)∗ ⊃ A∗BBA ⊃ BABA ⊃ B2A∗A.



Int. J. Anal. Appl. 18 (4) (2020) 627

Since the operators B2,A∗A, (BA)∗BA are self-adjoint with B2 ∈ B(H), then

(BA)∗BA ⊂ A∗AB2,

by corollary 1.1, we obtain

(BA)∗BA = A∗AB2.

Similarly, we obtain

BA(BA)∗ = A∗AB2,

establishing the normality of BA. Theorem 1.2 gives us

BA = T.

�

Corollary 2.1. Let T,B,A be non necessarily bounded operators such that T is normal and B symmetric

and invertible (hence B is self-adjoint) and that A is self-adjoint, then

T ⊂ BA =⇒ A = B−1T.

Proof. Clearly,

T ⊂ BA =⇒ B−1T ⊂ A,

by theorem 2.2, we obtain

A = B−1T.

�

Proposition 2.1. Let A,B and T be operators where B ∈ B(H). Assume that T∗ is symmetric, B is

self-adjoint and A is normal. If T ⊂ AB, then BA is essential self-adjoint.

Proof. Since AB is closed, we have

T ⊂ AB =⇒ T ⊂ AB,

and

T ⊂ AB =⇒ BA∗ ⊂ T∗ ⊂ T∗∗ = T ⊂ AB

=⇒ BA ⊂ A∗B (by Fuglede-Putnam Theorem)

=⇒ BA ⊂ A∗B (because A∗B is closed ).

We can show the normality of BA. We have

(BA)∗BA = A∗B(A∗B)∗ ⊃ A∗BBA ⊃ B2A∗A.



Int. J. Anal. Appl. 18 (4) (2020) 628

Since B2, A∗A are self-adjoint with B2 ∈ B(H), then

(BA)∗BA ⊂ A∗AB2,

by corollary 1.1, we obtain

(BA)∗BA = A∗AB2.

Similarly,

BA(BA)∗ = A∗AB2,

i.e. BA is normal. Since (BA)∗ too is normal and normal operators are maximally normal, we get

BA = (BA)∗ = A∗B,

i.e. BA is essentially self-adjoint. �

Proposition 2.2. Let T,B,A be non necessarily bounded operators such that T is sels-adjoint and B sym-

metric and invertible (hence B is self-adjoint) and that A∗ is symmetric. Then

AB ⊂ T =⇒ TB−1 = B−1T = A.

Proof. Clearly,

AB ⊂ T =⇒ A ⊂ TB−1

=⇒ B−1T ⊂ A∗ ⊂ A∗∗ = A ⊂ TB−1 = TB−1.

From theorem 2.1, we have TB−1 and B−1T are normal. Hence

TB−1 = B−1T = A.

�

Proposition 2.3. Let T,B,A be non necessarily bounded operators such that T and B are self-adjoint with

B ∈ B(H) and invertible and A∗ is symmetric. If BA ⊂ T , then BA = AB = T .

Proof. We have:

BA ⊂ T =⇒ A ⊂ B−1T

=⇒ TB−1 ⊂ A∗ ⊂ A∗∗ = A ⊂ B−1T = B−1T.

Left and right multiplying by B give

BT ⊂ BAB ⊂ TB.

By theorem 2.1, we obtain TB,BT are normal and

TB = BT = BAB,



Int. J. Anal. Appl. 18 (4) (2020) 629

hence

BA = AB = T.

�

Theorem 2.3. Let T,A,B be non necessarily bounded operators such that A and B are normal with B ∈

B(H) and T is unitary. Assume further that TA est normal. If BA ⊂ TAB and , then AB and BA are

normal. Also, if A and T commute, then TAB is normal.

Proof. Obviously,

BA ⊂ TAB =⇒ B∗A∗T∗ ⊂ A∗B∗

=⇒ B∗TA ⊂ AB∗ (by Fuglede-Putnam Theorem).

=⇒ BA∗ ⊂ A∗T∗B.

It is clear that AB is closed and we have:

(AB)∗AB ⊃ B∗A∗AB ⊃ B∗A∗BA ⊃ B∗BA∗A.

Since (AB)∗AB,B∗B,A∗A are self-adjoint with B∗B ∈ B(H), then

(AB)∗AB ⊂ A∗AB∗B,

and by corollary 1.1, we obtain

(AB)∗AB = A∗AB∗B.

We also have,

AB(AB)∗ ⊃ ABB∗A∗ ⊃ AB∗BA∗ ⊃ B∗TABA∗ ⊃ B∗TT∗BAA∗ = B∗BAA∗.

Similarly, we obtain

AB(AB)∗ = A∗AB∗B (because A and B are normal).

and this marks the end of the proof of the normality of AB. Let’s show now that BA is normal. Indeed

(BA)∗BA = A∗B∗(A∗B∗)∗ ⊃ A∗B∗BA ⊃ B∗A∗T∗BA ⊃ B∗BA∗A,

i.e.

(BA)∗BA = A∗AB∗B

Similarly,

BA(BA)∗ = A∗AB∗B,



Int. J. Anal. Appl. 18 (4) (2020) 630

that is, BA is normal.

Let’s show now that TAB is normal. We have TAB is closed because T is invertible and AB is closed. we

Also have

TA ⊂ AT =⇒ TA∗ ⊂ A∗T (by Fuglede-Putnam Theorem).

=⇒ T∗A ⊂ AT∗

=⇒ T∗A∗ ⊂ A∗T∗ (by Fuglede-Putnam Theorem).

Indeed

TAB(TAB)∗ ⊃ B∗A∗T∗TAB = B∗A∗AB ⊃ B∗A∗T∗BA ⊃ B∗BA∗A,

since TAB(TAB)∗,B∗B,A∗A are self-adjoint with B∗B ∈ B(H), we get

TAB(TAB)∗ ⊂ A∗AB∗B.

By corollary 1.1, we obtain

TAB(TAB)∗ = A∗AB∗B.

Similarly,

(TAB)∗TAB = TAB(TAB)∗ = A∗AB∗B,

and this marks the end of the proof of the normality of TAB. �

The folowing result is already seen in ( [12]), we can consider it as a consequence of the prceding theorem.

Also for T = I (where I is the indentity operator on H) we will get the theorem 2.1.

Corollary 2.2. Let A,B be normal operators with B ∈ B(H). Assume that BA ⊂ λAB where λ ∈ C. Then

AB and BA are both normal if |λ| = 1 (and so AB = λBA).

Proof. For T = λI where I is the indentity operator on H, we obtain T∗ = λI, i.e. T is unitary (because

|λ| = 1). Theorem 2.3 yiels the normality of AB, BA and λAB. Since AB is closed, we may also write

BA ⊂ λAB =⇒ BA ⊂ λAB

But, normal operators are maximally normal, therefore, we finally infer that

BA = λAB.

�

Closely related to the previous results, we have another proof for the closure of bounded operators on a

domain.

Proposition 2.4. Let T is a bounded operator on D(T). Then T is closed if D(T) is closed on H.



Int. J. Anal. Appl. 18 (4) (2020) 631

The proof requires the following lemma whose proof is very akin to the one in [15].

Lemma 2.1. Let f : X −→ Y is continous such that Y is Hausdorff space. Then the graph of f is closed

on X ×Y .

Now we prove proposition 2.4

Proof. We denote the graphe of T by Gr(T). By lemma 2.1, we obtain Gr(T) is closed on D(T) ×H, i.e.

(Gr(T))
C is open. We may write

(Gr(T))
C = ∪

(i,j)∈I×J
U1i ×U

2
j

where I, J are arbitrary and U1i , U
2
j are open on D(T), H respectively. Hence

Gr(T) = ∩
(i,j)∈I×J

(F1i ∩D(T)) ×F
2
j ,

with F1i , F
2
j closed sets on H. Therefore Gr(T) becommes closed on H ×H when D(T) is closed on H (It’s

the induced topology). �

Remark 2.1. In the previous proof, we did not use the linearity of T , we used only topological notions.

Conflicts of Interest: The author(s) declare that there are no conflicts of interest regarding the publication

of this paper.

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	1. Introduction
	2. Main Results
	References