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Maximality of Linear Operators

We present maximality results in the setting of non necessarily bounded operators. In particular, we discuss and establish results showing when the"inclusion"between operators becomes a full equality.


Introduction
In the theory of non necessarily bounded linear operators on a complex Hilbert space H, we say that an operator T with domain D(T ) ⊂ H is an extension of S with domain D(S) ⊂ H when: D(S) ⊂ D(T ) and Sx = T x for all x ∈ D(S). We then write S ⊂ T . We say that S is closed if it possess a closed graph in H ⊕ H.
The product of S and T is defined for each x on the natural domain where I is the identity operator on H. It is known that the product ST is closed if for instance S is closed and T ∈ B(H), or if S −1 ∈ B(H) and T 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. 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).
A densely defined operator S is said to be symmetric if S ⊂ S * . It is called self-adjoint if S = S * . We say that S is normal if S is densely defined, closed and SS * = S * S. Recall that the previous is equivalent to Sx = S * x for all x ∈ D(S) = D(S * ). We say that S is formally normal if Sx = S * x for all x ∈ D(S) ⊂ D(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. [2] or [17]). We do recall the celebrated Fuglede-Putnam Theorem though: Theorem 1.1. (for a proof, see e.g. [2]) Let T ∈ B(H) and let M, N be two normal non necessarily bounded operators. Then One of the main aims of this work is to seek conditions which transform S ⊂ T into S = T (which we call a maximality condition) for some classes of operators, and also in the case of a product of two operators. This type of results is a powerful tool when proving 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. [12]). For other uses, see e.g. [6] or [9].
Let us now list some known (see e.g. [12] or [13]) maximality results: 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: (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).
In fact, Statements (2) to (4) of the preceding result are all simple consequences of the following readily verified result: Let us now say a few words about "double maximality". A known property (Theorem 5.31, [17]) states that if S is a symmetric operator such that S ⊂ R and S ⊂ T where R, T are self-adjoint and D(R) ⊂ D(T ), then T = R. Observe that the assumption S symmetric is tacitly assumed in S ⊂ R so there was no need to assume it. What is more, is that the assumption S being symmetric is not used in the proof of the previous result. So, we restate this result as (cf. Proposition 3.1): Closely related to what has just been said, we have: [9], cf. [16]) Let R, S, T be three densely defined operators on a Hilbert space H with respective domains D(R), D(S) and D(T ). Assume that Assume further that R and S are self-adjoint. Let D ⊂ D(T ) (⊂ D(R) ∩ D(S)) be dense. Let D be a core, for instance, for S. Then R = S.
Finally, we recall results on the case when we have a product on one side of the "inclusion": Theorem 1.6. Let R, S, T, A, B, C be operators such that T ⊂ RS and AB ⊂ C. Then: (1) T = RS if all R, S, T are self-adjoint (see [3]). [10]).
Remark. As observed in [4], the first statement in the previous theorem does not extend to normal operators. Indeed, just in the naive case of unitary operators, we have that a product of any two unitary operators is always unitary even when the two factors of the product do not commute. This observation motivates the investigation in the case where one operator is normal.
Remark. Another natural question may pop up. In [3], the authors before showing that T = RS, they first showed that R and S commute strongly (i.e. the corresponding spectral measures commute). So what if we have T ⊂ ABC, do we still expect T = ABC when all of T, A, B, C are self-adjoint? The answer is negative (at least as far as the idea of their proof is concerned). Indeed, we can have a self-adjoint product of three self-adjoint operators which do not commute pairwise.
In R 2 , consider the following self-adjoint matrices: is self-adjoint. Nevertheless, none of the products AB, AC and BC is self-adjoint, that is, AB = BA, AC = CA and BC = CB.

Some Results on Normality
The normality of unbounded products of normal operators has been studied recently. See e.g. [5] and the references therein. We recall The chosen idea of proof of the following result is via the Fuglede-Putnam Theorem (for a different proof, we may proceed as in [1]). Proof. Since BA ⊂ AB, Fuglede Theorem yields BA * ⊂ A * B. Hence (since also AB is densely defined), Since AB is closed, it follows that (AB) * AB is self-adjoint, and by the boundedness of B * B, we get and this marks the end of the proof of the normality of AB.
To show that BA is normal, we first observe that The first part of the proof leads to the normality of A * B * because both A * and B * are normal. Accordingly, The following result is known to most readers (a proof based on the spectral theorem may be found in [1]). We can equally regard it as a consequence of the preceding theorem: Proof. Since BA ⊂ AB, and A and B are self-adjoint, the previous theorem yields the normality of BA. But i.e. BA is symmetric as well. Therefore, BA is self-adjoint. Accordingly, AB = (BA) * = BA, and so AB is also self-adjoint, as required.

Main Results on Maximality
The same idea of proof of (Theorem 5.31, [17], discussed above) may lead to the following result which seems to have escaped notice up to now. Since D(S) is dense, it follows that T x = T * x for all x ∈ D(T ) = D(T * ), that is, T is self-adjoint.
Corollary 3.2. Let S be a densely defined operator such that S ⊂ T and S ⊂ T * . If T is normal, then it is self-adjoint.
The next result is perhaps known: Particularly, if C is invertible, then Proof. Since S is closed and B ∈ B(H), the general theory says that SB is closed. This combined with SB ⊂ I completes the proof. Proof. Since, by assumption, A := B −1 is self-adjoint and unbounded, we see that D(A 2 ) D(A) (by applying Lemma A.1 in [14] to R = |B|)). Then, take a (necessarily nonzero) vector e ∈ D(A) \ D(A 2 ). It follows from Lemma 3.2 of [14], that M := D(A)⊖ A < e > is a vector subspace of D(A) which is dense in H, where ⊖ designates the orthogonal difference with respect to the graph inner product of A (cf. [14]) and < e >= C · e. Set S = A| M . Since M is a closed vector subspace of D(A) with respect to the graph norm of A, we see that the operator S is closed, densely defined and injective. Then clearly The following gives more information about Theorem 1.6 is: Theorem 3.6. Let A, B, T be non necessarily bounded operators such that A is self-adjoint, B is symmetric with B −1 ∈ B(H) (hence B is self-adjoint) and T is symmetric. Assume further that AB ⊂ T . Then: (1) AB ⊂ BA.
(4) T is essentially self-adjoint. If T is also closed, then BA is self-adjoint and T = BA and T = AB.

Proof.
• Since T is densely defined, so is AB and so since also B −1 ∈ B(H) and A and B are self-adjoint. Since T is symmetric, we obtain AB ⊂ T ⊂ T * ⊂ BA. Lemma 2.1 (or else) then yields the normality of BA. Now, since T * ⊂ BA, we get (BA) * ⊂ T * * = T . Because BA is normal, so is (BA) * . But, normal operators are maximally symmetric. Therefore, we finally infer that (BA) * = T , i.e. T is essentially self-adjoint (for T is normal and symmetric).  We also have: In particular, if we further assume that T is closed, then we obtain T = AB.
The Fugelde-Putnam Theorem then gives Reasoning as in the proof of Theorem 2.2, we may prove i.e. AB is normal. Hence (AB) * too is normal. Since normal operators are maximally symmetric, we get Since normal operators are maximally symmetric, we obtain as needed.
From the proof of Theorem 3.8, we have: Proof. We have already obtained: BA ⊂ A * B and BA * ⊂ AB.
These two inequalities allow us to establish the normality of both BA and A * B (cf. Theorem 2.2). Therefore, BA = A * B Corollary 3.11. Let A, B, T be non necessarily bounded operators such that A is self-adjoint, B is symmetric with B −1 ∈ B(H) and T is normal. Then: Proof. Obviously, where we used the Fuglede-Putnam Theorem in the lase implication. As in the preceding corollary, we may show the normality of T B −1 . This, combined with the self-adjointness of A and A ⊂ T B −1 lead finally to A = T B −1 , as needed.
Remark. We already observed in the remark just above Theorem 3.8 that if A, B and T are as in the previous corollary, then we must not have T = AB. The same counterexample may be reused here.
The following is also worth stating.

Corollary 3.12. Let A, B, T be operators such that A is normal, B is bounded and self-adjoint and T is self-adjoint. Then
T ⊂ AB =⇒ T = AB.
Proof. As in the proofs above, we can easily show that AB is normal. Then Theorem 1.2 does the remaining job.
Theorem 3.13. Let A, B and T be non-necessarily bounded operators. Assume that B is normal, that A is symmetric and invertible (hence A is self-adjoint) and that T is self-adjoint. Then Proof. We claim that AB is normal. First we have: Hence (AB) * AB ⊃ B * BA 2 or (AB) * AB ⊂ A 2 B * B as (AB) * AB is self-adjoint since AB is closed because also A −1 ∈ B(H) . Therefore, by Theorem 1.6. Similarly, we may prove that Accordingly, AB is normal. In the end, since self-adjoint operators are maximally normal, we obtain T ⊂ AB =⇒ T = AB, as required.

A Conjecture
Unfortunately, if we switch the roles of A and B in Corollary 3.12, then we have not been able so far to find a complete answer. Indeed, we need a version of Fuglede-Putnam Theorem which is not available in the literature yet. Even with help from Bent Fuglede himself, we have only got as far as the following (we have chosen not to include the proof in this paper):  Proof. Just apply Theorem 4.1 to the functions f, g : z → z (so that g • f becomes the identity map on C). Related to what has just been discussed, we propose the following conjecture: Remark. What makes the previous conjecture interesting is that it is known to hold if A ∈ B(H) (Fuglede-Putnam Theorem), and as it is posed, it is covered by none of the known (unbounded) generalizations of Fuglede-Putnam Theorem (see e.g. [7], [11] and [15]).