# Spatiality of Derivations of Operator Algebras in Banach Spaces

Authors: Quanyuan Chen; Xiaochun Fang
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

Suppose that𝒜 is a transitive subalgebra of B(X) and its norm closure 𝒜¯ contains a nonzero minimal left ideal ℐ. It is shown that if δ is a bounded reflexive transitive derivation from 𝒜 into B(X), then δ is spatial and implemented uniquely; that is, there exists T∈B(X) such that δ(A)=TA−AT for each A∈𝒜, and the implementation T of δ is unique only up to an additive constant. This extends a result of E. Kissin that “if 𝒜¯ contains the ideal C(H) of all compact operators in B(H), then a bounded reflexive transitive derivation from 𝒜 into B(H) is spatial and implemented uniquely.”  in an algebraic direction and provides an alternative proof of it. It is also
shown that a bounded reflexive transitive derivation from 𝒜 into B(X) is spatial and implemented uniquely, if X is a reflexive Banach space and 𝒜¯ contains a nonzero minimal right ideal ℐ.


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## Body

## 1. Introduction

Throughout this paper,X is a Banach space (and X will be replaced by H if it is a Hilbert space) and 𝒜 is a subalgebra of B(X), the Banach algebra of all bounded operators on X. Suppose that 𝒜⊆B(X) and ℳ⊆B(X) is an 𝒜-bimodule. A linear map δ from 𝒜 into ℳ is called a derivation if (1.1)δ(AB)=δ(A)B+Aδ(B)for anyA,B∈A.
Then 𝒜 is called the domain of δ and denoted by Dom(δ). The derivation δ is called inner (resp., spatial) if there exists an operator T∈ℳ (resp., T∈B(X)) such that (1.2)δ(A)=TA-ATfor anyA∈A.
If the operator T is not bounded, then δ is said to be quasispatial. More precisely, if there exists a densely defined, closed operator T:Dom(T)→X such that (1.3)A(Dom⁡(T))⊆Dom⁡(T),δ(A)x=(TA-AT)xfor anyA∈A,x∈Dom⁡(T),
then the derivation δ is called quasispatial, and the operator T is an implementation of δ. Compared to the spatiality, the quasispatiality is a slightly weaker notion.Given a bounded derivationδ on an operator algebra, the natural question is whether the derivation δ is inner (or spatial). The spatiality of derivations is a classical problem when formulated for self-adjoint algebras and non-self-adjoint reflexive operator algebras. And it has been extensively studied in the literature in a large variety of situations, and some interesting results have been obtained [1–13]. For example, every derivation of a C*-algebra is spatial [12], every derivation of a von Neumann algebra is inner [13], and so is the derivation of a nest algebra [14]. Every derivation from an atom Boolean subspace lattice algebra into its ideal is quasispatial [7]. A necessary and sufficient condition is given for a derivation on CDC algebras to be quasispatial [6]. In [10], the quasispatiality of derivations on CSL algebras is studied.As to general operator algebras, it is well known that every derivation ofB(X) is inner [9] and that every derivation of a standard operator subalgebra on a normed space X is spatial [4]. Since these operator algebras are transitive, the above question for a reflexive transitive derivation in a Banach space X is raised naturally as follows.Problem 1.
Suppose that𝒜 is a transitive subalgebra of B(X). Let δ be a bounded reflexive transitive derivation from 𝒜 into B(X). Does there always exist T∈B(X) such that δ(A)=TA-AT for each A∈𝒜? Is the implementation T unique only up to an additive constant if there is any?In the case whenX is a Hilbert space, it is Problem 2.12 of [5]. Although the problem is still open, some strong conditions have been found by Kissin to imply that such derivations are spatial and implemented uniquely (Proposition 2.11, [5]). In particular, Kissin has proved that the answer to Problem 1 is affirmative under the conditions that 𝒜 is a transitive subalgebra of B(H), and 𝒜¯ contains the ideal C(H) of all compact operators in B(H). As far as we know, there are no other solution to Problem 1.The purpose of this paper is to investigate the quasispatiality of derivations and to address the above question in a Banach spaceX. The paper is organized as follows. In Section 2, we give some preliminaries. In Section 3, we investigate the quasispatiality and spatiality of derivations. The main result (Theorem 3.1) shows that if 𝒜 is a transitive operator algebra on a Banach space X and 𝒜¯ contains a nonzero minimal left ideal ℐ, then a bounded reflexive transitive derivation δ from 𝒜 into B(X) is spatial and implemented uniquely. As an application, the quasispatiality of the adjoint of a derivation is discussed in Section 4. The main result (Theorem 4.2) in Section 4 shows that under the conditions that X is a reflexive Banach space and 𝒜¯ contains a nonzero minimal right ideal ℐ, δ is also spatial and implemented uniquely if δ is a bounded reflexive transitive derivation from 𝒜 into B(X). As another application, Proposition 2.11 (ii) of [5] can be proved by using Theorem 3.1 of this paper, which is Corollary 4.3. Since Theorem 2.5 and Proposition 2.11 of [5] hold in a Hilbert space and they are not valid in a Banach space without the approximation property, Theorems 3.1 and 4.2 of this paper extend the result of Kissin to Banach spaces in an algebraic direction. The role of “compact operators” is replaced by that of “minimal one-sided ideals”. The proof of our results relies on the quasispatiality of the derivation and Banach algebra techniques.This paper is a continuation of [5]. Some definitions and notations can be found in [5].


## 2. Preliminaries

Throughout this paper,X is a complex Banach space, and X* is the topological dual space of X, the Banach space of all continuous linear functionals on X. We denote by ℱ(X) the algebra of all finite-rank operators on X. If a subalgebra 𝒜 contains ℱ(X), then 𝒜 is called a standard operator algebra. For a bounded operator A on X, denote by LatA the lattice of all closed invariant subspaces of A and A* the adjoint operator of A. For a subalgebra 𝒜 of B(X), denote by Lat(𝒜) the lattice of all closed subspaces invariant under every operator in 𝒜. For a set ℒ of subspaces of X, denote by Algℒ the algebra of all operators in B(X) which leave all subspaces in ℒ invariant. An operator algebra 𝒜 is transitive if Lat𝒜={{0},X}; 𝒜 is reflexive if (2.1)A=Alg LatA,whereAlg LatA={T∈B(X):LatA⊂LatT}.For0≠x∈X and 0≠f∈X*, the rank-one operator x⊗f acts on X by (x⊗f)y=f(y)x for y∈X. Let A be an operator on X with Dom(A)⊆X. If x∈Dom(A) and f∈Dom(A*), then A(x⊗f)=Ax⊗fand(x⊗f)A=x⊗(A*f). Let M be a nonempty subset of X and N a nonempty subset of X*. The annihilator M⊥ of M and the preannihilator N⊥ of N are defined as follows [15]: M⊥={f∈X*:f(x)=0for allx∈M}, N⊥={x∈X:f(x)=0for allf∈N}. It is obvious that M⊥ is a weak*-closed subspace of X* and N⊥ is a norm-closed subspace of X. For a subalgebra 𝒜 and a closed, densely defined operator T with domain Dom(T), we say that Tcommutes with𝒜, if A(Dom(T))⊆Dom(T) and TAξ=ATξ for any A∈𝒜 and any ξ∈Dom(T). For a subalgebra 𝒜 of B(X), let 𝒜*={A*:A∈𝒜} in notation.A subsetℐ of an algebra 𝒜 is a left ideal of 𝒜 if 𝒜ℐ⊆ℐ, a right ideal if ℐ𝒜⊆ℐ, and a two-sided ideal if it is both a left and a right ideal. A left ideal ℐ of 𝒜 is minimal if every left ideal of 𝒜 included in ℐ is either ℐ or {0}, similarly for minimal right ideals.A derivationδ is bounded (resp., closed) if the map Dom(δ)∋A↦δ(A)∈B(X) is bounded (resp., closed) in the operator norm topology. The derivation δ is transitive if its domain Dom(δ) is a transitive operator algebra; δ is reflexive if (2.2)Aδ={Â=(Aδ(A)0A):A∈Dom⁡(δ)}
is a reflexive operator algebra on X⊕X. Denote by Imp(δ) the set of all closed, densely defined operators implementing the derivation δ as in (1.3). For a densely defined, closed operator T with domain Dom(T), we can define the derivation ΔT with domain (2.3)Dom⁡(ΔT)={A∈B(X):A(Dom⁡(T))⊆Dom⁡(T),TA-ATis bounded onDom⁡(T)}
by (2.4)ΔT(A)=Closure(TA-AT),for anyA∈Dom⁡(ΔT).If a densely defined, closed operatorT implements a derivation δ, then the derivation ΔT is an extension of the derivation δ. In fact, A∈Dom(δ) implies A∈Dom(ΔT) and ΔT(A)=δ(A) for any A∈Dom(δ) by (1.3) and (2.4). For any set Λ of derivations δλ (λ∈Λ), define the derivation δΛ=⋂λ∈Λδλ by (2.5)Dom⁡(δΛ)={A∈⋂λ∈ΛDom⁡(δλ):δλ(A)coincidefor anyλ∈Λ},δΛ(A)=δλ(A)for anyA∈Dom⁡(δΛ)and anyλ∈Λ.
In particular, (2.6)δImp(δ)=⋂T∈Imp(δ)ΔT.Remark 2.1.
ΔT is a reflexive transitive derivation for any densely defined, closed operator T.
Indeed,Dom(ΔT) is a subalgebra of B(X), and ΔT is a derivation.
Ifx∈Dom(T) and h∈Dom(T*), (x⊗h)z=h(z)x∈Dom(T) for any z∈Dom(T). So (x⊗h)(Dom(T))⊆Dom(T). Since (T(x⊗h)-(x⊗h)T)(z)=(Tx⊗h)z-(x⊗T*h)z, ∥T(x⊗h)-(x⊗h)T∥≤∥Tx∥∥h∥+∥x∥∥T*h∥ so that T(x⊗h)-(x⊗h)T is bounded on Dom(T). It follows that x⊗h∈Dom(ΔT) for any x∈Dom(T) and h∈Dom(T*).
LetM∈Lat(Dom(ΔT)) with 0≠x0∈M. Then (x⊗h)(x0)=h(x0)x∈M for any x∈Dom(T) and h∈Dom(T*). Since Dom(T*) is dense, there exists h0∈Dom(T*) such that h0(x0)≠0 by Hahn-Banach Theorem. It follows that x∈M for any x∈Dom(T) so that Dom(T)⊆M. Thus X=Dom(T)¯⊆M¯=M. Therefore Dom(ΔT) is transitive.
For the derivationΔT, 𝒜ΔT={(AΔT(A)0A):A∈Dom(ΔT)}. It is easy to see that X⊕{0}∈Lat(𝒜ΔT) and G(T+tI)={(T(x)+txx):x∈Dom(T)}∈Lat(𝒜ΔT) for any t∈ℂ. Let B2=(B11B12B21B22)∈B(X⊕X) and B2∈Alg Lat(𝒜ΔT). Then Lat(𝒜ΔT)⊆Lat(B2). Since X⊕{0}∈Lat(𝒜ΔT)⊆Lat(B2), (B11B12B21B22)(x0)=(B11xB21x)∈X⊕{0} for any x∈X. It follows that B21X={0}. Thus B21=0 and B2=(B11B120B22). Since G(T+tI)∈Lat(𝒜ΔT)⊆Lat(B2), (B11B120B22)(T(x)+txx)=(B11Tx+tB11x+B12xB22x)∈G(T+tI) for any x∈Dom(T) and any t∈ℂ. Then B22x∈Dom(T) and B11Tx+tB11x+B12x=(T+tI)B22x for any x∈Dom(T) and any t∈ℂ. It follows that B11x=B22x∈Dom(T) and B11Tx+B12x=TB22x for all x∈Dom(T). Since Dom(T) is dense in X and B11,B22∈B(X), B11=B22. So B11(Dom(T))⊆Dom(T) and B12=TB11-B11T on Dom(T). Therefore, B12=TB11-B11T is bounded on Dom(T). It follows that B11=B22∈Dom(ΔT) and B12=ΔT(B11). Thus B2=(B11ΔT(B11)0B11)∈𝒜ΔT. It follows that 𝒜ΔT is a reflexive algebra. Therefore ΔT is a reflexive transitive derivation.


## 3. Main Results

In this section, we discuss the quasispatiality of bounded transitive derivations on operator algebras in a Banach space. The main result is as follows.Theorem 3.1.
Suppose that𝒜 is a transitive subalgebra of  B(X) and  𝒜¯, the norm closure of  𝒜, contains a nonzero minimal left ideal  ℐ. If  δ is a bounded reflexive transitive derivation from  𝒜 into  B(X), then  δ is spatial and implemented uniquely, or, more precisely, there exists  T∈B(X) such that  δ(A)=TA-AT for each  A∈𝒜 and the implementation  T is unique only up to an additive constant.The proof of Theorem3.1 will proceed through several lemmas, in each of which we maintain the same notation. The results similar to the following two lemmas (Lemmas 3.2 and 3.3) can be found in [5]. For the sake of completeness, we outline the proof.Lemma 3.2.
Letδ be a reflexive transitive derivation from  Dom(δ)(Dom(δ)=𝒜⊆B(X)) into  B(X). Then  δ is quasispatial.Proof.
IfImp(δ) is nonempty, then the lemma is trivially true. Therefore, for the rest of the argument, we assume that Imp(δ) is empty.
First, we have that ifδ is a transitive derivation from Dom(δ) into B(X), then
(3.1)Lat(Aδ)={{0}⊕{0},X⊕{0},X⊕X,G(T):T∈Imp(δ)},
where G(T)={(T(x)x):x∈Dom(T)} is the graph of T.
Indeed, it is easy to see that{0}⊕{0}, X⊕{0}, X⊕X, and all G(T) are invariant subspaces of 𝒜δ. For the converse, suppose that M∈Lat𝒜δ such that M≠{0}⊕{0}, M≠X⊕{0}, and M≠X⊕X. If there is a vector of form (x0)∈M with x≠0, then (Ax0)∈M for any A∈𝒜. It follows that X⊕{0}⊆M by the transitivity of Dom(δ). If M contains no vector of form (0x) with x≠0, then M=X⊕{0}; otherwise, there is a vector of form (0x)∈M with x≠0, then we have that M=X⊕X. Therefore (x0)∈Mimpliesthatx=0. It follows that there is a closed operator T such that
(3.2)M=G(T)={(T(x)x):x∈Dom⁡(T)},T∈Imp(δ).
As Imp(δ)=∅, by (3.1),
(3.3)Alg LatAδ={(AB0C):A,B,C∈B(X)}≠Aδ={(Aδ(A)0A):A∈Dom⁡(δ)}.
It is a contradiction to the assumption that δ is reflexive, which shows that
(3.4)Imp(δ)≠∅.
Therefore δ is quasispatial.Lemma 3.3.
Letδ be a reflexive transitive derivation from  Dom(δ) into  B(X). Then  δ=δImp(δ).Proof.
By (3.4), Imp(δ)≠∅. And by (2.4) of [5],
(3.5)Imp(δ)=Imp(δImp(δ)).
As both δ and δImp(δ) are transitive derivations, Lat𝒜δ=Lat𝒜δImp(δ) by (3.1). Since both of these derivations are reflexive,
(3.6)Aδ=Alg LatAδ=Alg LatAδImp(δ)=AδImp(δ).
Hence
(3.7)δ=δImp(δ).Lemma 3.4.
Let𝒜 be a transitive subalgebra of  B(X). If  𝒜¯ contains a nonzero minimal left ideal  ℐ, then  𝒜¯∩ℱ(X) is also a transitive subalgebra of  B(X).Proof.
We complete the proof step by step.Step 1.
We have that(3.8)I2={B1B2:B1,B2∈I}≠{0}.
Indeed, suppose thatℐ2={0}. Let 0≠A0∈ℐ. Then 𝒜¯A0⊆ℐ. It follows that (𝒜¯A0)2⊆ℐ2 and (𝒜¯A0)2={0}. Set X0={x∈X:A(x)=0for allA∈𝒜¯}. It is obvious that X0 is an 𝒜¯-invariant closed subspace of X. Hence X0={0} since 𝒜 is a transitive operator algebra. Since (𝒜¯A0)2={0}, A0𝒜¯A0(X)⊆X0. It follows that A0𝒜¯A0(X)={0} so that A0𝒜¯A0={0}. We can choose x0∈X (x0≠0) such that A0(x0)≠0. It is obvious that 𝒜¯A0(x0) is an 𝒜¯-invariant linear manifold of X and 𝒜¯A0(x0)≠{0}. It follows that 𝒜¯A0(x0) is dense in X. However, 𝒜¯A0(x0) is contained in the null space of A0, A0=0, which is a contradiction.Step 2.
Sinceℐ2≠{0}, it follows from Lemma 2.1.5 and Corollary 2.1.6 of [16] that there exists an idempotent P in 𝒜¯ such that ℐ=𝒜¯P and P𝒜¯P is a division algebra consisting of scalar multiples of P (with identity P), that is,
(3.9)PA¯P={μP:μis a complex scalar}.
ThenP is a rank-one operator; that is, there exist x1∈X and f∈X* such that
(3.10)P=x1⊗f∈A¯,f(x1)=1.
Indeed, letx1:0≠x1∈X such that Px1=x1 and let M1=𝒜¯(x1)={A(x1):A∈𝒜¯}. It is obvious that M1≠{0} and M1 is an 𝒜¯-invariant linear manifold of X. Hence M1 is dense in X. Since PA(x1)=PAP(x1)=μP(x1)=μx1 for any A∈𝒜¯, the restriction P|M1 of P on M1 has one-dimensional range. As P is bounded, it also has one-dimensional range. Hence P=x1⊗f, where f is a continuous functional.Step 3.
We haveP=x1⊗f∈ℐ⊆𝒜¯. As 𝒜¯ is an algebra, the set of rank-one operators
(3.11)SP={APB:A,B∈A}={Ax1⊗B*f:A,B∈A}
lies in 𝒜¯∩ℱ(X). For each 0≠y∈X, the linear manifold
(3.12)My={∑i=1nSiy:Si∈SP,i=1,2,…,n,n=1,2,…}
is dense in X. Indeed, if My is not dense in X, by Hahn-Banach theorem, there exists 0≠g∈X* such that g(z)=0 for all z∈My. Hence
(3.13)0=g((Ax1⊗B*f)y)=g((B*f(y))Ax1)=f(By)g(Ax1)for anyA,B∈A.
If f(By)=0 for all B∈𝒜, then f vanishes on the linear manifold 𝒜y={By:B∈𝒜}. However, as 𝒜 is transitive, the manifold 𝒜y is dense in X, so that f=0. This contradiction shows that there is B1∈𝒜 such that f(B1y)≠0. Then (3.13) implies that g(Ax1)=0 for all A∈𝒜. Repeating the above argument, we obtain that g=0. This contradiction shows that any manifold My is dense in X. Therefore the algebra 𝒜¯∩ℱ(X) is transitive.Lemma 3.5.
Suppose that𝒜 is a transitive subalgebra of  B(X) and  𝒜¯ contains a nonzero minimal left ideal  ℐ. Then the operators commuting with  𝒜¯ are scalars. More precisely, if  T is a closed, densely defined operator such that  A(Dom(T))⊆Dom(T) and  ATξ=TAξ for any  A∈𝒜¯ and any  ξ∈Dom(T), then  T=μI for some complex scalar  μ.Proof.
For anyy∈X, there exists a net of operators {Aλ}⊆𝒜 such that limλAλ(x1)=y by the transitivity of 𝒜. Then 𝒜¯∋Aλ(x1⊗f)=Aλ(x1)⊗f→∥·∥y⊗f, since P=x1⊗f∈𝒜¯ as in (3.10). It follows that
(3.14)y⊗f∈A¯for anyy∈X.
Suppose thatT is a closed, densely defined operator commuting with 𝒜¯. As S=(y⊗f)∈𝒜¯ for any y∈X, then (y⊗f)(ξ)∈Dom(T)and(y⊗f)Tξ=T(y⊗f)ξfor anyξ∈Dom(T). Since Dom(T) is dense in X and f≠0, there is ξ0∈Dom(T) such that f(ξ0)≠0. So
(3.15)(y⊗f)(ξ0)=f(ξ0)y∈Dom⁡(T),f(Tξ0)y=f(ξ0)T(y),for anyy∈X.
It follows that Dom(T)=X. Set μ=f(Tξ0)/f(ξ0). Then T=μI, as required.Lemma 3.6.
Suppose that𝒜 is a subalgebra of  B(X) and  𝒜¯ is its norm closure. Then  Lat(𝒜)=Lat(𝒜¯).Proof.
Clearly,Lat(𝒜¯)⊆Lat(𝒜). Conversely, let M∈Lat(𝒜). For any B∈𝒜¯, there exists a net of operators {Aλ}⊆𝒜 such that Aλ→∥·∥B. Then Aλx∈M for any x∈M. Since Aλx→Bx and M is a closed subspace of X, Bx∈M. It follows that M∈Lat(B) so that M∈Lat(𝒜¯). Therefore Lat(𝒜)⊆Lat(𝒜¯). The Lemma follows.Let𝒜 be a subalgebra of B(X) and δ be a bounded derivation from 𝒜 (𝒜=Dom(δ)) into B(X). For any B∈𝒜¯, there exists a net of operators {Aλ}⊆𝒜 such that Aλ→∥·∥B. If {Aλ} is a uniformly convergent net in 𝒜; then {δ(Aλ)} is also a uniformly convergent net in B(X) since ∥δ(Aλ1)-δ(Aλ2)∥≤∥δ∥∥Aλ1-Aλ2∥ for any indexes λ1,λ2. Also, if two nets of operators {Aλ(1)},{Aλ(2)} in 𝒜 converge uniformly to the same limit, then {δ(Aλ(1))}, {δ(Aλ(2))} converge uniformly to the same limit, since ∥δ(Aλ(1))-δ(Aλ(2))∥≤∥δ∥∥Aλ(1)-Aλ(2)∥. Therefore a linear map δ¯ can be unambiguously defined by (3.16)Dom⁡(δ¯)=Dom⁡(δ)¯=A¯,δ¯(B)=lim⁡λδ(Aλ)
for any B∈𝒜¯ and any net {Aλ} in 𝒜 such that Aλ→∥·∥B, where δ¯(B) is the limit of δ(Aλ) in the operator norm topology. It is obvious that ∥δ¯∥=∥δ∥ if δ is bounded.Proposition 3.7.
Letδ be a bounded transitive derivation from  𝒜(𝒜=Dom(δ)) into  B(X). Then  δ¯ is a bounded transitive derivation from  𝒜¯ into  B(X) and
(3.17)Imp(δ¯)=Imp(δ).Proof.
Since𝒜 is transitive, 𝒜¯ is also transitive. It is obvious that δ¯ is a linear map. For any B1,B2∈𝒜¯, there are two nets of operators {Aλ(1)},{Aλ(2)} in 𝒜 such that Aλ(i)→∥·∥Bi(i=1,2). Then Aλ(1)Aλ(2)→∥·∥B1B2 and
(3.18)δ¯(B1B2)=lim⁡λδ(Aλ(1)Aλ(2))=lim⁡λδ(Aλ(1))Aλ(2)+lim⁡λAλ(1)δ(Aλ(2))=δ¯(B1)B2+B1δ¯(B2).
It follows that the linear map δ¯ is a transitive derivation from 𝒜¯ into B(X).
By (2.2), 𝒜δ={(Aδ(A)0A):A∈𝒜} and 𝒜δ¯={(Bδ¯(B)0B):B∈𝒜¯}. So that 𝒜δ¯=𝒜δ¯ by (3.16) and Lat(𝒜δ¯)=Lat(𝒜δ) by Lemma 3.6. By (3.1),
(3.19)Lat(Aδ)={{0}⊕{0},X⊕{0},X⊕X,G(F):F∈Imp(δ)},Lat(Aδ¯)={{0}⊕{0},X⊕{0},X⊕X,G(F1):F1∈Imp(δ¯)}.
It follows that Imp(δ¯)=Imp(δ).Proof of Theorem3.1.
By (3.4), Imp(δ)≠∅. Suppose that T∈Imp(δ). Then T is a closed, densely defined operator on X and
(3.20){(T+μI):μis a complex scalar}⊆Imp(δ).
Letδ0=0 be the derivation (with the same domain 𝒜 of δ) from 𝒜 into B(X), that is, δ0(A)=0 for any A∈𝒜. If T0∈Imp(δ0) with domain Dom(T0), then T0∈Imp(δ0¯) by (3.17), and
(3.21)A(Dom⁡(T0))⊆Dom⁡(T0),δ0¯(A)x=(T0A-AT0)xfor anyA∈A¯,x∈Dom⁡(T0),
that is, T0Ax=AT0x for any x∈Dom(T0). Then T0=μI for some scalar μ by Lemma 3.5. It follows that
(3.22)Imp(δ0)={μI:μis a complex scalar}.
LetT1∈Imp(δ) be any closed, densely defined operators implementing δ. Then T1∈Imp(δ¯) by (3.17). Then Dom(T1) is a nonzero Dom(δ¯)-invariant linear manifold of X by (1.3); that is, Dom(T1) is a nonzero 𝒜¯-invariant linear manifold of X. Therefore A(Dom(T1))⊆Dom(T1) for any A∈𝒜¯. If y∈X, y⊗f∈𝒜¯ by (3.14) and (y⊗f)ξ=f(ξ)y∈Dom(T1) for any ξ∈Dom(T1). Since Dom(T1) is dense in X and f≠0, there is ξ0∈Dom(T1) such that f(ξ0)≠0. So y∈Dom(T1). Since y is arbitrary chosen, it follows that Dom(T1)=X, the whole space. Then Dom(T)=X also holds since T1 is an arbitrary implementation of δ. It is obvious that Dom(T1-T)=(Dom(T)∩Dom(T1))=X. Therefore the operator T1-T is everywhere defined.
SetT2=T1-T. Then T2 is closable.
Indeed, suppose thatξn∈X, ξn→0 and T2(ξn)→η. Since T,T1∈Imp(δ¯), δ¯(A)x=(T1A-AT1)x and δ¯(A)x=(TA-AT)x for any A∈Dom(δ¯) and any x∈X. So T2Aξn=AT2ξn for any A∈𝒜¯. As P=x1⊗f∈𝒜¯ by (3.10) and 𝒜¯ is an algebra, T2PAξn=PAT2ξnfor anyA∈𝒜¯. Then
(3.23)f(Aη)x1=(x1⊗f)Aη=PAη=lim⁡n→∞PAT2ξn=lim⁡n→∞T2PAξn=lim⁡n→∞T2(x1⊗f)Aξn=lim⁡n→∞f(Aξn)T2x1=0
for any A∈𝒜¯. Since x1≠0, f(Aη)=0 for any A∈𝒜¯. Thus f vanishes on the linear manifold 𝒜¯η={Aη:A∈𝒜¯}. The transitivity of 𝒜¯ assures that 𝒜¯η={0} or 𝒜¯η is dense in X. Since f≠0, 𝒜¯η={0} and η=0.
It follows thatT2 is a closable operator with domain Dom(T2)=X, the whole space. So that T2 is a closed operator. Clearly, T2=T1-T∈Imp(δ0). Then T1-T=μI for some scalar μ by (3.22), that is, T1=T+μI. It follows that
(3.24)Imp(δ)={(T+μI):μis a complex scalar}.
Since ΔT=ΔT+μI, by (3.7),
(3.25)δ=δImp(δ)=ΔT.
Since T is closed and Dom(T)=X, T is bounded by the closed graph theorem. It follows from (2.4), (3.24), and (3.25) that δ(A)=TA-AT for each A∈𝒜 and the implementation T is unique only up to an additive constant.


## 4. Applications

Suppose that𝒜 is a subalgebra of B(X) and δ is a derivation from 𝒜 (𝒜=Dom(δ)) into B(X). We can define the adjoint δ* of δ by (4.1)Dom⁡(δ*)=(A)*={A*:A∈A},δ*(A*)=(δ(A))*for anyA∈A.Lemma 4.1.
Letδ be a derivation from  𝒜(𝒜=Dom(δ)) into  B(X).  (1)
The adjointδ* of δ is a derivation from 𝒜* into B(X*).(2)
Furthermore, suppose thatX is a reflexive Banach space and 𝒜 is a subalgebra of B(X). Then δ* is a reflexive (resp., transitive) derivation from 𝒜* into B(X*) if and only if δ is a reflexive (resp., transitive) derivation from 𝒜 into B(X). And Imp(δ*)={-T*:T∈Imp(δ)}.Proof.
(1) It is obvious thatδ* is a linear map from 𝒜* into B(X*). If A1*,A2*∈Dom(δ*), then
(4.2)δ*(A1*A2*)=δ*((A2A1)*)=(δ(A2A1))*=(δ(A2)A1+A2δ(A1))*=A1*(δ(A2))*+(δ(A1))*A2*=A1*(δ*(A2*))+(δ*(A1*))A2*.
It follows that δ* is a derivation.
(2) Suppose thatδ is transitive. Then 𝒜 is transitive and Lat𝒜={X,{0}}. Since M∈LatAif and only ifM⊥∈Lat(A*)for anyA∈B(X),
(4.3)Lat(Dom⁡(δ*))=Lat(A*)={M⊥:M∈LatA}={{0}⊥,X⊥}={X*,{0}}.
We have that Dom(δ*) is transitive, that is, δ* is transitive.
Conversely, suppose thatδ* is transitive. We obtain, as above, that δ** is transitive. Since X is a reflexive Banach space, A**=A for any bounded operator A∈B(X). It follows from (4.1) that δ=δ** is transitive by the definition of δ*.
SetJ=(0II0), where I is the identity operator on X* and 0 is the zero operator on X*. Since
(4.4)Aδ*={(A*δ*(A*)0A*):A∈A}={J(A*0δ(A)*A*)J:A∈A}={J(Aδ(A)0A)*J:A∈A}=J(Aδ)*J=J-1(Aδ)*J,
the two operator algebras 𝒜δ* and (𝒜δ)* are similar. We have that
(4.5)Lat(Aδ*)=J(Lat((Aδ)*))=J(Lat(Aδ))⊥.
It is obvious that
(4.6)(X⊕X)⊥={0}⊕{0},({0}⊕{0})⊥=X*⊕X*,(X⊕{0})⊥={0}⊕X*,J({0}⊕X*)=X*⊕{0}.
If T∈Imp(δ), (g-T*(g))∈(G(T))⊥ for any g∈Dom(T*), since
(4.7)G(T)={(T(x)x):x∈Dom⁡(T)},g(T(x))+(-T*(g))(x)=0for anyx∈Dom⁡(T).
Conversely, if g,h∈X* such that (gh)∈(G(T))⊥, then
(4.8)g(Tx)+h(x)=0for anyx∈Dom⁡(T).
So that the functional x↦g(Tx)(=-h(x)) is a bounded functional on Dom(T). Therefore g∈Dom(T*) and h=(-T*(g)) holds on Dom(T) by (4.8). As (-T*(g)) and h are bounded functionals on X and Dom(T) is dense in X, h=(-T*(g)) on X. It follows that
(4.9)(G(T))⊥={(g-T*(g)):g∈Dom⁡(T*)},J(G(T))⊥={(-T*(g)g):g∈Dom⁡(T*)}=G(-T*).
Hence(4.10)Lat(Aδ*)={{0}⊕{0},X*⊕{0},X*⊕X*,G(-T*):T∈Imp(δ)}.
It follows from (3.4) that
(4.11)Imp(δ*)={-T*:T∈Imp(δ)}.
Suppose thatδ is reflexive. Then 𝒜δ is a reflexive operator algebra on X⊕X. Since 𝒜δ*=J-1(𝒜δ)*J, the two operator algebras 𝒜δ* and (𝒜δ)* are similar. It follows that 𝒜δ* is reflexive if and only if (𝒜δ)* is reflexive. Since 𝒜δ is reflexive, (𝒜δ)* is also reflexive by the reflexivity of the Banach space X⊕X. Thus 𝒜δ* is reflexive, that is, δ* is reflexive.
Conversely, suppose thatδ* is reflexive. We obtain, as above, that δ** is reflexive. It follows that δ=δ** is reflexive.Theorem 4.2.
Suppose thatX is a reflexive Banach space. Let  𝒜 be a transitive subalgebra of  B(X) and let  𝒜¯ contain a nonzero minimal right ideal  ℐ. If  δ is a bounded reflexive transitive derivation from  𝒜 into  B(X), then  δ is spatial and implemented uniquely, or more precisely, there exists  T∈B(X) such that  δ(A)=TA-AT for each  A∈𝒜, and the implementation  T is unique only up to an additive constant.Proof.
Consider the derivationδ*. δ* is a bounded reflexive transitive derivation from 𝒜*={A*:A∈𝒜} into B(X*) by Lemma 4.1.
Setℐ*={B*:B∈ℐ}. Since ℐ is a right ideal of 𝒜¯, BA∈ℐ for any B∈ℐ and A∈𝒜¯. For any A*∈(𝒜¯)*, B*∈ℐ*, that is, A∈𝒜¯, B∈ℐ, we have that BA∈ℐ. Then A*B*=(BA)*∈ℐ*. It follows that ℐ* is a left ideal of (𝒜¯)*. If there exists a nonzero left ideal ℐ1 of (𝒜¯)* included in ℐ*, then ℐ1* is a nonzero right ideal of 𝒜¯ included in ℐ. Thus ℐ1*=ℐ and ℐ1=ℐ*. It follows that ℐ* is a minimal left ideal of (𝒜¯)*=𝒜*¯.
By (3.24), there exists a bounded operator T1∈B(X*) such that
(4.12)Imp(δ*)={(T1+μI):μis a complex scalar},
where I is the identity operator on X*. By (4.11),
(4.13)Imp(δ)=Imp(δ**)={(-T1*-μ¯I):μis a complex scalar},
where I is the identity operator on X. Set T=-T1*. Then T is bounded with Dom(T)=X. By (3.7) and (4.13), δ=δImp(δ)=ΔT. It follows from (2.4) that δ(A)=TA-AT for each A∈𝒜, and the implementation T is unique only up to an additive constant.Corollary 4.3 (Proposition 2.11, [5]).
Suppose that𝒜 is a transitive subalgebra of  B(H). Let  δ be a bounded reflexive transitive derivation from  𝒜 into  B(H). If  C(H)⊆𝒜¯, then there exists  T∈B(H) such that  δ(A)=TA-AT for each  A∈𝒜, and the implementation  T is unique only up to an additive constant.Proof.
SinceC(H)⊆𝒜¯, for a fixed vector f∈H,
(4.14)I={x⊗f∣x∈H}
is a minimal left ideal of 𝒜¯, where x⊗f is the rank-one operator on H with (x⊗f)h=(h,f)x for h∈H. By Theorem 3.1, there exists T∈B(H) such that δ(A)=TA-AT for each A∈𝒜, and the implementation T is unique only up to an additive constant.Remark 4.4.
In a Banach space without the approximation property (there exists such space as this, e.g., [17]), not all compact operators can be approximated by finite-rank operators in the norm operator topology. Therefore, Theorems 3.1 and 4.2 of this paper improve [5, Proposition 2.11].



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