aa r X i v : . [ m a t h . F A ] F e b BALANCED OPERATORS AND OPERATOR DOMAINS
KONRAD SCHM ¨UDGEN
Abstract.
We shall say that a densely defined closed operator T on a Hilbertspace is balanced if D ( T ) = D ( T ∗ ). Balanced operators are described in termsof their phase operators and their moduli. Examples of balanced operatorsare developed. A characterization of the domain equality D ( A ) = D ( B ) forpositive self-adjoint operators A and B with bounded inverses is given in termsof their spectral measures. AMS Subject Classification (2020) . 47 A 05, 47 B25.
Key words: unbounded operator, operator domain, polar decomposition, spec-tral measure 1.
Introduction
The aim of this paper is to introduce and study a class of unbounded operatorson Hilbert space, called balanced operators .Suppose T is a densely defined closed operator on a Hilbert space H with domain D ( T ) and let T ∗ denote its adjoint operator. The crucial definition is the following. Definition 1.
The operator T is called balanced if D ( T ) = D ( T ∗ ) . Obviously, any bounded operator with domain D ( T ) = H is balanced, so thisnotion is only of interest for unbounded operators. Clearly, T is balanced if andonly if T ∗ is balanced.In unbounded operator theory the bad behavior of an operator T goes often alongwith a “large” difference between the domains of T and T ∗ . A typical example arethe minimal and the maximal operators of a partial differential operator acting in L (Ω) for some domain Ω of R d .For another example, let T be a densely defined closed symmetric operator.Then we have T ⊆ T ∗ and the difference between the domains of T and T ∗ is nicelydescribed by the classical von Neumann formula D ( T ∗ ) = D ( T ) ˙+ N ( T ∗ − λI ) ˙+ N ( T ∗ − λI ) for λ ∈ C \ R . Thus, T is balanced if and only if T is self-adjoint and the balanced extensionsof T are precisely the self-adjoint extensions of T . Obviously, a formally normaloperator is balanced if and only if it is normal. Generally speaking, for a numberof classes of unbounded operators the balanced operators are precisely the “good”operators of this class. The examples in Section 3 support this statement.This paper is organized as follows. In Section 2 we develop some basic factson operator domains and balanced operators. In Section 3 we give a number ofinteresting weighted shift operators that are balanced.The main result of this paper (Theorem 19) is proved in Section 5. It containsnecessary and sufficient conditions for a densely defined closed operator T to bebalanced in terms of the spectral measure of the modulus | T | and the phase operator U appearing in the polar decomposition T = U | T | . Date : February 19, 2021.
The crucial part in the proof of Theorem 19 is a criterion for the domain equality D ( A ) = D ( B ) of positive self-adjoint operators A and B in terms of their spectralmeasures (Theorems 15 and 16). This result is proved in Section 4 and it seems tobe of interest in its own.Let us collect some notations that will be used throughout this paper. Thesymbol H refers always to a Hilbert space with scalar product h· , ·i and norm k · k .If T is an operator on H , we denote its domain by D ( T ), its kernel by N ( T ) andits the range by R ( T ). For α ∈ R , we write T + α for T + αI . The spectral measureof a self-adjoint operator T is denoted by E T .2. Preliminary facts on operator domains and balanced operators
Suppose that T is a densely defined closed operator on a Hilbert space H . It isan elementary fact of operator theory that | T | := √ T ∗ T is a positive self-adjointoperator on H such that D ( | T | ) = D ( T ). Replacing T by T ∗ , we obtain D ( | T ∗ | ) = D ( T ∗ ). Therefore, T is balanced if and only if D ( | T | ) = D ( | T ∗ | ) . Next we recall some known facts on domains of unbounded operators. For theconvenience of the reader, we include proofs of all results.
Lemma 2.
Suppose that B is a closed operator and A is a positive self-adjointoperator with bounded inverse on H . Then we have D ( A ) ⊆ D ( B ) if and only the X := BA − is a bounded operator defined on the Hilbert space H . In this case, theadjoint operator X ∗ is the closure of the operator A − B .Proof. Since A is positive and has a bounded inverse, the domain D ( A ), equippedwith the scalar product h ϕ, ψ i ′ := h Aϕ, Aψ i , is a Hilbert space H A . Since B isclosed and D ( A ) ⊆ D ( B ), B : H A
7→ H is a closed mapping. By the closed graphtheorem, B is continuous, that is, there exists a constant γ > k Bϕ k ≤ γ k Aϕ k , ϕ ∈ D ( A ) . (1)Suppose ψ ∈ H . Since A has a bounded inverse, R ( A ) = H , so there is a ϕ ∈ D ( A )such that ψ = Aϕ . Then ϕ = A − ψ and therefore k Xψ k = k BA − ψ k ≤ γ k ψ k by(1). This shows that X is bounded and defined on the whole Hilbert space H .Lez η ∈ D ( B ). Then h ψ, X ∗ η i = h Xψ, η i = h A − ψ, Bη i = h ψ, A − Bη i for ψ ∈ H . Hence X ∗ ⊇ A − B and X ∗ is the closure of A − B . (cid:3) Proposition 3.
Suppose T and S are densely defined closed operators on a Hilbertspace H . Let α > and β > . We define operators X and Y by X := ( | S | + β )( | T | + α ) − and Y := ( | T | + α )( | S | + β ) − . (2) Then the following statements are equivalent: (i) D ( T ) = D ( S ) . (ii) The operators X and Y are bounded and defined on the whole Hilbert space. (iii) The exists an operator Z ∈ B ( H ) with inverse Z − ∈ B ( H ) such that ( | T | + α ) − = ( | S | + β ) − Z. (3) Proof. (i) → (ii): We apply Lemma 2 twice, with B = | S | + β, A = | T | + α , and with A = | S | + β, B = | T | + α. (ii) → (iii): Set Z := X . Then Z ∈ B ( H ) by (ii). From (2) it follows that( | S | + β ) − Z = ( | T | + α ) − , which proves (3). Further, (2) implies that XY = Y X = I , that is, the bounded operator Y ∈ B ( H ) is the inverse of Z = X .(iii) → (i): Using (3) we derive D ( T ) = D ( | T | + α ) = R (( | T | + α ) − ) ⊆ R (( | S | + β ) − ) = D ( | S | + β ) = D ( S ) . ALANCED OPERATORS AND OPERATOR DOMAINS 3
Applying the adjoint to equation (3) yields Z ∗ ( | S | + β ) − = ( | T | + α ) − . (4)Interchanging T and S in the preceding reasoning and using (4) instead of (3) weobtain D ( S ) ⊆ D ( T ). Thus D ( T ) = D ( S ), which proves (i). (cid:3) Suppose D ( T ) = D ( S ). Then the operators X and Y defined by (2) have niceproperties: They are bounded and inverse to each other, X ∗ is a bijection of D ( S )onto D ( T ) and Y ∗ is a bijection of D ( T ) onto D ( S ).In the special case S = T ∗ , the equivalence (i) ↔ (ii) in Proposition 3 gives thefollowing criterion. Corollary 4.
Let α > and β > . A densely defined closed operator T on aHilbert space is balanced if and only if ( | T ∗ | + β )( | T | + α ) − and ( | T | + α )( | T ∗ | + β ) − . are bounded operators defined on the whole Hilbert space. Remark 5.
In the literature, domains of closed operators are often studied in termsof operator ranges. An operator range is the range R ( X ) = X H of some boundedoperator X ∈ B ( H ) . Since X and | X ∗ | = √ XX ∗ have the same range, eachoperator range is the range of a positive self-adjoint operator. The domain of adensely defined closed operator T is the operator range R (( | T | + I ) − ) .There is an extensive literature on operator ranges and operator domains. Clas-sical papers are [vN29] , [K36a] , [K36b] , [D49a] , [D49b] , [D66] , [FW71] , [F72] . Morerecent ones include [vD82] , [Sch83] , [BN93] , [K06] , [ACG13] , [AZ15] , [DM20] . A simple sufficient condition for an operator T to be balanced is the following. Proposition 6.
Suppose T is a densely defined closed operator. If D ( T ∗ T ) = D ( T T ∗ ) , then T is balanced.Proof. The assertion means that the domain equality D ( | T | ) = D ( | T ∗ | ) implies D ( | T | ) = D ( | T ∗ | ). This follows from the following result: If B and A are positiveself-adjoint operators on H such that D ( A ) ⊆ D ( B ) , then D ( A ) ⊆ D ( B ) . We include a proof of this well-known fact. Without loss of generality we canassume that A and B have bounded inverses. Since D ( A ) ⊆ D ( B ), Lemma 2implies that ( B A − ) ∗ ⊇ A − B is a bounded operator, so there is a constant γ > k A − B ϕ k ≤ γ k ϕ k for ϕ ∈ D ( B ). Hence k A − ψ k ≤ γ k B − ψ k for ψ ∈ H , so that ( A − ) ≤ ( γB − ) . Therefore A − ≤ γB − by the Heinzinequality (see e.g. [Sch90, Proposition 10.14]). Then k A − ψ k ≤ √ γ k B − ψ k for ψ ∈ H and k A − Bϕ k ≤ √ γ k ϕ k for ϕ ∈ D ( B ). The operator A − B is bounded,so ( A − B ) ∗ ⊇ BA − is bounded. Hence D ( A ) ⊆ D ( B ) again by Lemma 2. (cid:3) The following example shows that the converse of Proposition 6 does not hold.
Example 7.
Suppose A is a positive self-adjoint operator with trivial kernel and U a unitary operator on H . Then, T := U A is a densely defined closed operatoron H such that the formula T = U A is the polar decomposition of T and T ∗ T = A , T T ∗ = U A U ∗ and | T | = A, | T ∗ | = √ T T ∗ = U AU ∗ . Therefore, if U ∗ D ( A ) = D ( A ) , then D ( | T | ) = D ( | T ∗ | ) , so T is balanced. Theequality D ( T ∗ T ) = D ( T T ∗ ) holds if and only if U ∗ D ( A ) = D ( A ) .Now let A be the differential operator − ddx on L ( R ) and U the multiplicationoperator by a function ω ( x ) of modulus one. Hence, if ω is in C ( R ) , but not in C ( R ) , then T = U A is balanced, but D ( T ∗ T ) = D ( T T ∗ ) . KONRAD SCHM¨UDGEN Examples: Weighted shift operators
Let us consider the Hilbert space l ( N ). Suppose ( λ n ) n ∈ N is a complex se-quence. There is a densely defined closed linear operator T on l ( N ) defined by T ( ϕ , ϕ , ϕ , . . . ) = ( λ ϕ , λ ϕ , λ ϕ , . . . ) , (5)written shortly as T ( ϕ n ) = ( λ n +1 ϕ n +1 ), with domain D ( T ) = (cid:8) ( ϕ n ) n ∈ N ∈ l ( N ) : ( λ n +1 ϕ n +1 ) n ∈ N ∈ l ( N ) (cid:9) . (6)The adjoint operators T ∗ acts by T ∗ ( ϕ , ϕ , ϕ , . . . ) = (0 , λ ϕ , λ ϕ , λ ϕ , . . . ) , briefly T ∗ ( ϕ n ) = ( λ n ϕ n − ), and has the domain D ( T ∗ ) = (cid:8) ( ϕ n ) n ∈ N ∈ l ( N ) : ( λ n ϕ n − ) n ∈ N ∈ l ( N ) (cid:9) , where ϕ − := 0. Then | T | and | T ∗ | are diagonal operators | T | ( ϕ , ϕ , ϕ , . . . ) = (0 , | λ | ϕ , | λ | ϕ , | λ | ϕ , . . . ) , | T ∗ | ( ϕ , ϕ , ϕ , . . . ) = ( | λ | ϕ , | λ | ϕ , | λ | ϕ , . . . ) . Then, for α > , β >
0, ( | T ∗ | + β )( | T | + α ) − and ( | T | + α )( | T ∗ | + β ) − are diagonaloperators. For n ∈ N , the n -th diagonal entries are ( | λ n +1 | + β )( | λ n | + α ) − and ( | λ n | + α )( | λ n +1 | + β ) − , respectively. Therefore, both diagonal operators arebounded if and only if there are constants c > , c > , c > , c > | λ n | ≤ c | λ n +1 | + c and | λ n +1 | ≤ c | λ n | + c , n ∈ N . (7)Thus, by Corollary 4, we have proved the following. Proposition 8.
The operator T defined by (5) and (6) is balanced if and only ifcondition (7) holds. Now we mention some interesting examples of balanced weighted shift operators.
Example 9.
Set λ n = √ n for n ∈ N . Then (7) is obviously true. In this case, T is the creation operator and T ∗ is the annihilation operator of quantum mechanics.Both operators are balanced. Note that T satisfies the commutation relation T T ∗ − T ∗ T = I. (8) Example 10.
Suppose q > , q = 1 , and set λ n = q − q n − q , n ∈ N . Then (7) issatisfied. The corresponding operator T is the Fock representation (see e.g. [Sch20,Theorem 11.28(i)] ) for the q -oscillator algebra defined by the relation XX ∗ = qX ∗ X + I. (9)Proposition 8 remains valid verbatim for two-sided weighted shifts acting on theHilbert space l ( Z ). We state two examples of balanced two-sided weighted shifts. Example 11.
Suppose q > , q = 1 . For λ > , we set λ n = λq n/ , n ∈ Z . Thenthe operator T is also balanced and it is an irreducible representations (see e.g. [Sch20, Proposition 11.25] ) of the Hermitean quantum plane given by the relation XX ∗ = qX ∗ X. (10) Example 12.
Suppose < q < . For γ ∈ [ q, , set λ n = q q k γ − q , n ∈ Z . Thecorresponding two-sided weigthed shift operator T is balanced and it is an irreduciblenon-Fock representation [Sch20, Theorem 11.28(ii)] of the q -oscillator algebra (9). ALANCED OPERATORS AND OPERATOR DOMAINS 5
The balanced operators T in Examples 9–12 are well-behaved representations(in the sense of [Sch20, Section 11.2], see also [OS14]) of the operator relations(8)–(10). They exhaust the infinite-dimensional irreducible well-behaved represen-tations of these relations. From this fact it can be derived that all well-behavedrepresentations of the three relations (8)–(10) are given by balanced operators.4. On the equality of operator domains
Throughout this section we suppose that A and B are positive self-adjoint op-erators on H . Let A = Z + ∞ λ dE A ( λ ) and B = Z + ∞ λ dE B ( λ )be their spectral resolutions.Our aim is to characterize the domain equality D ( A ) = D ( B ) in terms of thespectral measures E A and E B . Proposition 13.
Suppose the inverse operator A − exists and is bounded. If D ( A ) ⊆ D ( B ) , then there is a number ν > such that E A ((0 , b ]) ≤ E B ((0 , bν ]) for all b > . (11) One may take any number ν such that ν > k BA − k .Proof. Since D ( A ) ⊆ D ( B ) and A − is bounded, it follows from Lemma 2 that Z δ := ( B + δ ) A − , δ ≥ , is a bounded operator defined on the whole Hilbert space H . We fix a number ν such that ν > k BA − k . Clearly, the norm of Z δ is the infimum of numbers C > k ( B + δ ) ξ k ≤ C k Aξ k for ξ ∈ H . Therefore, since A − is bounded, thereis a positive number δ such that ν > k Z δ k .Now we take a number q , 0 < q <
1, such that qν ≥ k Z δ k . Then we have k Z δ k b < q ( bν + δ ), so that k Z δ k b ( bν + δ ) − < q. (12)Fix b ∈ R , b >
0. Suppose that ϕ ∈ E A ((0 , b ]) H . We can write the vector ϕ as ϕ = ψ + η with ψ ∈ E B ((0 , bν ]) H , η ∈ E B (( bν, + ∞ )) H . Clearly, ϕ ∈ D ( A ). By the spectral calculus, k Aϕ k = Z (0 ,b ] λ d h E A ( λ ) ϕ, ϕ i ≤ Z ∞ b d h E A ( λ ) ϕ, ϕ i = b k ϕ k , k ( B + δ ) − η k = Z ( bν, + ∞ ) ( λ + δ ) − d h E B ( λ ) η, η i ≤ Z ∞ ( bν + δ ) − d h E B ( λ ) η, η i = ( bν + δ ) − k η k . Further, ( Z δ ) ∗ ξ = A − ( B + δ ) ξ for ξ ∈ D ( B ). Using the preceding facts we derive k ϕ − ψ k = k η k = |h ϕ, η i| = |h A − Aϕ, η i| = |h Aϕ, A − η i| = |h Aϕ, Z ∗ δ ( B + δ ) − η i|≤ b k ϕ k k Z δ k ( bν + δ ) − k η k = k ϕ k k η k ( k Z δ k b ( bν + δ ) − ) ≤ k ϕ k k ϕ − ψ k q, where the last inequality follows from (12). Thus, k ϕ − ψ k ≤ q k ϕ k for ϕ ∈ E A ((0 , b ]) H , ψ = E B ((0 , bν ]) E A ((0 , b ]) ϕ. (13) KONRAD SCHM¨UDGEN
Let us abbreviate P A := E A ((0 , b ]) and P B := E B ((0 , bν ]). Then (13) means k P A ϕ − P B P A ϕ k ≤ q k P A ϕ k for ϕ ∈ H . Hence k ( I − P B ) P A k ≤ q k P A k ≤ q and therefore k P A [( I − P B ) P A ] n k ≤ q n for n ∈ N . Since 0 ≤ q <
1, we conclude thatlim n →∞ k P A [( I − P B ) P A ] n k = 0 . (14)Recall that for orthogonal projections P, Q on a Hilbert space there is the well-known formula P ∧ Q = lim n →∞ P [ QP ] n in the strong operator topology (see e.g. [H67, Exercise 96]). Therefore, (14) yields P A ∧ ( I − P B ) = 0, so that P A H ∩ ( I − P B ) H = { } and hence P A H ⊆ P B H .Thus P A ≤ P B , which means that E A ((0 , b ]) ≤ E B ((0 , bν ]). (cid:3) Interchanging the role of the operators B and A in Proposition 13 gives Proposition 14.
Suppose the inverse operator B − is bounded and D ( B ) ⊆ D ( A ) .Then there is a number ν ′ > such that E B ((0 , b ]) ≤ E A ((0 , b ν ′ ]) for all b > . (15) Any number ν ′ such that ν ′ > k AB − k can be taken. The following theorem is the main result of the section.
Theorem 15.
Suppose A and B are positive self-adjoint operators on a Hilbertspace H such that inverses A − and B − exist and are bounded operators on H .Let E A and E B denote their spectral measures. Then the following are equivalent: (i) D ( A ) = D ( B ) . (ii) There is a number µ > such that for all a, b ∈ R , < a < b,E A (( a, b ]) ≤ E B (( aµ − , bµ ]) and E B (( a, b ]) ≤ E A (( aµ − , bµ ]) . (16)(iii) There is a number µ > such that for all a, b ∈ R , < a < b,E A ([ a, b ]) ≤ E B ([ aµ − , bµ ]) and E B ([ a, b ]) ≤ E A ([ aµ − , bµ ]) . (17) Proof. (i) → (ii): Recall that E (( a, b ]) = E ((0 , b ]) − E ((0 , a ]) for any spectral measure E . Let µ be the maximum of the numbers ν and ν ′ from Propositions 13 and 14.Then E A ((0 , b ]) ≤ E B ((0 , bµ ]) and E B ((0 , aµ − ]) ≤ E A ((0 , a ]) by (11) and (15).Hence E A (( a, b ]) = E A ((0 , b ]) − E A ((0 , a ]) ≤ E B ((0 , bµ ]) − E B ((0 , aµ − ]) = E B (( aµ − , bµ ]) , which proves the first relation of (16). The second relation follows by interchangingthe role of A and B .(ii) → (i): From spectral theory we recall that a vector ϕ ∈ H belongs to D ( A )if and only if R ∞ λ d h E A ( λ ) ϕ, ϕ i < ∞ ; similarly for B .Now we suppose that (16) holds. There is no loss of generality to assume that µ >
1. We abbreviate f k := E B (( µ k , µ k +1 ]) , g k := E A (( µ k , µ k +1 ]) , e k := E A (( µ k − , µ k +2 ]) for k ∈ N . Then, f k ≤ e k by (16). Further, since ( µ k − , µ k +2 ] is the union of disjoint inter-vals ( µ k − , µ k ] , ( µ k , µ k +1 ] , ( µ k +1 , µ k +2 ], the projection e k is the sum of the threeorthogonal projections g k − , g k , g k +1 . Thus, f k ≤ e k = g k − ⊕ g k ⊕ g k +1 .Our aim is to show that D ( A ) ⊆ D ( B ). Suppose that ϕ ∈ D ( A ). Let m ∈ N .Using the preceding facts we derive Z ( µ,µ m +1 ] λ d h E B ( λ ) ϕ, ϕ i = m X k =1 Z ( µ k ,µ k +1 ] λ d h E B ( λ ) ϕ, ϕ i ALANCED OPERATORS AND OPERATOR DOMAINS 7 ≤ m X k =1 µ k +1) k f k ϕ k ≤ m X k =1 µ k +1) k e k ϕ k = m X k =1 µ k +1) k ( g k − ⊕ g k ⊕ g k +1 ) ϕ k = m X k =1 µ k +1) k g k − ϕ k + m X k =1 µ k +1) k g k ϕ k + m X k =1 µ k +1) k g k +1 ϕ k ≤ µ m +1 X j =0 µ j k g j ϕ k ≤ µ m +1 X j =0 Z ( µ j ,µ j +1 ] λ d h E A ( λ )( ϕ, ϕ i = 3 µ Z (1 ,µ m +2 ] λ d h E A ( λ ) ϕ, ϕ i≤ µ Z ∞ λ d h E A ( λ ) ϕ, ϕ i = 3 µ k Aϕ k . Passing to the limit m → ∞ and remembering that µ > Z ( µ, + ∞ ) λ d h E B ( λ ) ϕ, ϕ i ≤ µ k Aϕ k . Hence R + ∞ λ d h E B ( λ ) ϕ, ϕ i < ∞ and therefore ϕ ∈ D ( B ). We have shown that D ( A ) ⊆ D ( B ). Since condition (16) is symmetric in A and B , we also have D ( A ) ⊆ D ( B ). Thus D ( A ) = D ( B ).(ii) ↔ (iii): Suppose that the first relation of (16) holds. Then, for 0 < a ′ < a , E A ([ a, b ]) ≤ E A (( a ′ , b ]) ≤ E B (( a ′ µ − , bµ ]) . Letting a ′ ↑ a , we get E A ([ a, b ]) ≤ E B ([ aµ − , bµ ]), which is the first relation of (17).Conversely, assume the first relation of (17). For 0 < a < a ′ , we have E A (( a, b ]) ≤ E B ([ a ′ µ − , bµ ]) ≤ E B ([ a ′ µ − , bµ ]) . Taking a ′ ↓ a yields E A (( a, b ]) ≤ E B (( aµ − , bµ ]). This is the first relation of (16).The equivalence of the second relations in (16) and (17) follows by interchangingthe role of A and B . (cid:3) The preceding proof shows that the implications (iii) ↔ (ii) → (i) in Theorem 15are valid without the assumption that A and B have bounded inverses. For theimplication (i) → (ii) this is not true, as we will see in Example 17 below.From its proof it follows that the assertion of Proposition 13 remains true if weonly assume that D ( A ) ⊆ D ( B ) and the inverse A − exists and BA − is bounded.(Since D ( A ) ⊆ D ( B ), the domain of BA − is the range of A , so it is dense if weassume that N ( A ) = { } .) Similarly, if we assume D ( A ) = D ( B ) together with theexistence of A − and B − and the boundedness of BA − and AB − , the implication(i) → (ii) of Theorem 15 holds.If we omit the boundedness assumption for the inverses, then Theorem 15 hasthe following reformulation. Theorem 16.
Suppose A and B are positive self-adjoint operators on a Hilbertspace with spectral measures E A and E B . Let ǫ > and δ > . Then the followingstatements are equivalent: (i) D ( A ) = D ( B ) . KONRAD SCHM¨UDGEN (ii)
There is a number µ > such that for all a, b ∈ R , < a < b,E A (( a + ε, b + ε ]) ≤ E B (( aµ − + δ, bµ + δ ]) , E B (( a + δ, b + δ ]) ≤ E A (( aµ − + ε, bµ + ε ]) . (iii) There is a number µ > such that for all a, b ∈ R , < a < b,E A ([ a + ε, b + ε ]) ≤ E B ([ aµ − + δ, bµ + δ ]) , E B ([ a + δ, b + δ ]) ≤ E A ([ aµ − + ε, bµ + ε ]) . Proof.
The self-adjoint operators A ′ := A + ε and B ′ := B + δ have trivial kernelsand bounded inverses and satisfy D ( A ) = D ( A ′ ), D ( B ) = D ( B ′ ), so Theorem15 applies to theses operators. Using the formulas E A ′ ( M ) = E A ( M + ε ) and E B ′ ( M ) = E B ( M + δ ) this yields the assertion. (cid:3) We illustrate the preceding by a simple example dealing with multiplicationoperators.
Example 17.
Let J be an interval and H the Hilbert space L ( J ) , with respect tothe Lebesgue measure, let f and g be nonnegative Borel functions on J such thatthe sets { t ∈ J : f ( t ) = 0 } and { t ∈ J : g ( t ) = 0 } have Lebesgue measure zero. Let A and B denote the multiplication operators by f and g , respectively. Then A and B are positive self-adjoint operators on H with trivial kernels.We have D ( A ) ⊆ D ( B ) if and only if there are constants C ≥ , c ≥ such that g ( t ) ≤ C ( f ( t ) + c ) a.e. on J. (18) Clearly, A − is bounded if and only if there is an ε > such that g ( t ) ≥ ε a.e. on J . For a Borel set M , the spectral projection E A ( M ) is the multiplication operatorby the characteristic function of f − ( M ) := { t ∈ J : f ( t ) ∈ M } ; similarly for B .Using these facts one can easily construct interesting cases. For instance, set J = ( α, + ∞ ) and f ( x ) = x , g ( x ) = x β , with α ≥ , β > .First we discuss Proposition 13 and condition (11). We note that D ( A ) ⊆ D ( B ) if and only if β ≤ , and A − is bounded if and only if α > . For b > , ν > ,the spectral projections E A ((0 , b ]) and E B ([0 , bν ]) are multiplication operators bythe characteristic functions of { t ∈ J : t ≤ b } and { t ∈ J : t ≤ ( νb ) /β } .Suppose a > . Then (11) holds if and only if β ≤ , or equivalently, D ( A ) ⊆D ( B ) . In this case, A − is bounded.Now let a = 0 . Then condition (11) is fulfilled if and only if β = 1 , that is, A = B . For β < , we have D ( A ) ⊆ D ( B ) , but (11) is not valid. Thus, theassertion of Proposition 13 does not necessarily hold if the operator A − is notbounded.Next we consider Theorem 15 and condition (16). Let f ( x ) = x and g ( x ) = x if x ≥ , g ( x ) = √ x if x ≤ . Then, obviously, D ( A ) = D ( B ) , but condition (16)is not satisfied. That is, the implication (i) → (ii) in Theorem 15 is not true if theassumption that the operators A − and B − are bounded is omitted. Polar decomposition and balanced operators
Suppose that T is a densely defined closed operator on H . By the theorem onthe polar decomposition (see e.g. [Sch90, Theorem 7.2] or [K66, Chapter VII, § U with initial space K and final space K , where K : = N ( T ) ⊥ = R ( T ∗ ) = N ( | T | ) ⊥ = R ( | T | ) = U ∗ ( R ( T ) ) , (19) K : = R ( T ) = N ( T ∗ ) ⊥ = N ( | T ∗ | ) ⊥ = R ( | T ∗ | ) = U ( R ( T ∗ ) ) , (20)such that T = U | T | . The formula T = U | T | is called the polar decomposition of T ,the partial isometry U is the phase operator and | T | is the modulus of T . We have U ∗ U = P K , U U ∗ = P K , (21) ALANCED OPERATORS AND OPERATOR DOMAINS 9 where P K and P K are the projections onto K and K , respectively. Moreover, T = U | T | = | T ∗ | U, T ∗ = U ∗ | T ∗ | = | T | U ∗ (22)In fact, T ∗ = U ∗ | T ∗ | is the polar decomposition of the operator T ∗ .Let E | T | and E | T ∗ | denote the spectral measures of the positive self-adjoint op-erators | T | and | T ∗ | , respectively. Next we relate these spectral measures.Since U | T | = | T ∗ | U and U ∗ | T ∗ | = | T | U ∗ , it follows from [Sch90, Proposition5.15] that for any Borel set M ⊆ [0 , + ∞ ), U E | T | ( M ) = E | T ∗ | ( M ) U, E | T | ( M ) U ∗ = U ∗ E | T ∗ | ( M ) . (23)Further, since N ( | T | ) = E | T | ( { } ) H and N ( | T ∗ | ) = E | T ∗ | ( { } ) H , we have N ( U ) = E | T | ( { } ) H , N ( U ∗ ) = E | T ∗ | ( { } ) H . (24)Now we multiply the first equality of (23) by U ∗ and the second by U from theright. Using (21) we then obtain for each Borel M ⊆ (0 , + ∞ ), U E | T | ( M ) U ∗ = E | T ∗ | ( M ) , E | T | ( M ) = U ∗ E | T ∗ | ( M ) U. (25)(Note that the second equalities in (23) and (25) follow also from the first ones byapplying the adjoint operation.)Conversely, suppose that E is a spectral measure on [0 , + ∞ ) and U a partialisometry on H with kernel E ( { } ) H .The spectral integral A := Z [0 , + ∞ ) λdE ( λ )is a positive self-adjoint operator on H . Set T := U A . Since D ( T ) = D ( A ), theoperator T is densely defined. We show that T is closed. For let ( ϕ n ) be a sequenceof vectors ϕ n ∈ D ( T ) such that ϕ n → T ϕ n → ψ for some vector ψ ∈ H .Using that U is zero on N ( A ) = E ( { } ) H and isometric on the complement, wederive k Aϕ k − Aϕ n k = (cid:13)(cid:13)(cid:13) Z (0 , + ∞ ) λdE ( λ ) ϕ k − Z (0 , + ∞ ) λdE ( λ ) ϕ k (cid:13)(cid:13)(cid:13) = (cid:13)(cid:13)(cid:13) U Z (0 , + ∞ ) λdE ( λ ) ϕ k − U Z (0 , + ∞ ) λdE ( λ ) ϕ k (cid:13)(cid:13)(cid:13) = (cid:13)(cid:13)(cid:13) U Z [0 , + ∞ ) λdE ( λ ) ϕ k − U Z [0 , + ∞ ) λdE ( λ ) ϕ k (cid:13)(cid:13)(cid:13) = k T ϕ k − T ϕ n k . Therefore, since A is closed, there is a vector ϕ ∈ D ( A ) such that Aϕ n → Aϕ .Hence T ϕ n = U Aϕ n → U Aϕ = T ϕ = ψ , which proves that T is closed.From the properties of U it follows that | T | = A and T = U A is the polardecomposition of T . By the preceding we have proved the following proposition. Proposition 18.
The densely defined closed operators T on H are in one-to-onecorrespondence, given by | T | = R [0 , + ∞ ) dE ( λ ) and T = U | T | , with pairs of spectralmeasures E on [0 , + ∞ ) and partial isometries U on H with kernels E ( { } ) H . Now we are ready to characterize a balanced operator T in terms of the spectralmeasure E | T | of its modulus | T | and its phase operator U . Theorem 19.
Suppose T is a densely defined closed operator on H with polardecomposition T = U | T | . Let ε > and δ > be fixed numbers. Then the followingstatements are equivalent: (i) T is balanced. (ii) There exists a number µ > such that for all a, b ∈ R , < a < b,E | T | (( a + ε, b + ε ]) ≤ U E | T | (( aµ − + δ, bµ + δ ]) U ∗ , (26) U E | T | (( a + δ, b + δ ]) U ∗ ≤ E | T | (( aµ − + ε, bµ + ε ]) . (27)(iii) There is a number µ > such that for all a, b ∈ R , < a < b,E | T | ([ a + ε, b + ε ]) ≤ U E | T | ([ aµ − + δ, bµ + δ ]) ,U E | T | ([ a + δ, b + δ ]) U ∗ ≤ E | T | ([ aµ − + ε, bµ + ε ]) . Proof.
The result is derived from Theorem 16, applied with B = | T ∗ | , A = | T | andcombined with the first equality of (25).Indeed, by (25), the inequality of (26) means that E A (( a + ε, b + ε ]) ≤ E B (( aµ − + δ, bµ + δ ])and the inequality of (27) is E B (( a + δ, b + δ ]) ≤ E A (( aµ − + ε, bµ + ε ]) . (Since all these intervals are contained in (0 , + ∞ ), (25) applies.) Recall that,as noted above, T is balanced if and only if D ( | T | ) = D ( | T ∗ | ). Therefore, theequivalence of (i) and (ii) in Theorem 16 gives the equivalence of assertions (i) and(ii) of Theorem 19. The proof of the equivalence of (i) and (iii) is similar. (cid:3) References [ACG13] Arias, M.L., Corach, G. and M.C. Gonzalez, Additivity properties of operator ranges,Linear Algebra Appl. (2013), 3581–3590.[AZ15] Arlinskii, Yu. and V.A. Zagrebnov, Around the van Daele-Schm¨udgen theorem, IntegralEqu. Operator Theory (2015), 53–95.[BN93] Brasche, J.F. and H. Neidhardt, Has every symmetric operator a closed symmetric re-striction whose square has a trivial domain?, Acta Sci. Math. (Szeged) (1993), 425–430.[DM20] Dehimi, S. and M.M. Mortad, Chernoff-like counterexamples related to unbounded oper-ators, Kyushu J. Math. (2020), 105–108.[D49a] Dixmier, J., Sur les varietes J d’un espace de Hilbert, J. Math. Pures Appl. (1949),321–358.[D49b] Dixmier, J., Etude sur les varietes et les operateurs de Julia, Bull. Soc. Math. France (1949), 11–101.[D66] Douglas, R.G., On majorization, factorization, and range inclusion of operators on Hilbertspace, Proc. Amer. Math. Soc. (1966), 413–416.[FW71] Fillmore, P.A. and J.P. Williams, On operator ranges, Adv. Math. (1971), 254–281.[F72] Foias, C., Invariant para-closed subspaces, Indiana Univ. Math. J. (1972), 887-906.[H67] Halmos, P.R., A Hilbert Space Problem Book , van Nostrand, Princeton, 1967.[K66] Kato, T.,
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