# Generalized Derivations and Bilocal Jordan Derivations of Nest Algebras

Authors: Dangui Yan; Chengchang Zhang
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

LetH be a complex Hilbert space and B(H) the collection of all linear bounded operators, 𝔄 is the closed subspace lattice including 0 an H, then 𝔄 is a nest, accordingly alg 𝔄={T∈B(H):TN⊆N,∀N∈𝔄} is a nest algebra. It will be shown that of nest algebra, generalized derivations are generalized inner derivations, and bilocal Jordan derivations are inner derivations.


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## Body

## 1. Introduction

The concept of local derivations was introduced by Kadison [1] who showed that on a Von-nemann algebra all norm-continuous local derivations  are derivations. Larson and Sourour [2] proved that on the algebra B(X) local derivations are derivations.  M. Brešar and P. Šemrl [3, 4] generalized the results of the three authors above under a weaker condition. Shulman [5] showed that all local derivations  on C*-algebra are derivations.Based on a great deal of research works of many mathematicians, some scholars paid more interests in similar kind of problems  under more generalized conditions, such as considering local derivations on nest algebras and generalized derivation. Zhu and Xiong [6, 7] proved that local derivations of nest algebra and standard operator algebra are derivations, Zhang [8] considered the Jordan derivations of nest algebras, Lee [9] discussed generalized derivations of left faithful rings. Recently, some scholars discussed some new types of derivations, as  Li and Zhou  [10] and Majieed and Zhou  [11] investigated some new types of generalized derivations associated with Hochschild 2 cocycles, other examples are in [12–15]. In fact, under appropriate conditions, local derivations are derivations.In this paper, we will show that of nest algebra, a generalized derivation is a generalized inner derivation, and bilocal Jordan derivations are inner derivations.


## 2. Some Notations and Definitions

In what follows, some notations and basic definitions are introduced.LetH be a complex Hilbert space and B(H) the collection of all linear bounded operators on H, 𝔄 is the closed subspace lattice including 0 an H, then 𝔄 is a nest, correspondingly the Nest algebra is alg𝔄={T∈B(H):TN⊆N,∀N∈𝔄}.IfN≠0, we denote N-=∨{M∈𝔄:M⊊N} and if N≠H, denote N+=∧{M∈𝔄:M⊋N}, where ⊊ is real inclusion, and we define 0-=0, H+=H.For allN∈𝔄, P(N) represent the project operator from H to N, and N⊥={f∈H:〈x,f〉=0,∀x∈N}.Let𝔄 be a Banach algebra and 𝔄1 a subalgebra of 𝔄, we call the linear map φ:𝔄1→𝔄 a generalized inner derivation if and only if for all T∈𝔄1, there exist operators A and B in 𝔄 such that φ(T)=AT+TB; if for all T∈𝔄1, we have φ(T2)=φ(T)T+Tφ(T), then φ is called a Jordan derivation; if for all T∈𝔄1, there is a Jordan derivation φT:𝔄1→𝔄, such that φ(T)=φT(T), then φ is said to be a local Jordan derivation.Definition 2.1.
Letφ:alg𝔄→alg𝔄 be an additive mapping,  if there exists a derivation δ:alg𝔄→alg𝔄 that φ(ST)=φ(S)T+Sδ(T), for all S,T∈alg𝔄, then φ is called a generalized derivation.Definition 2.2.
We call the linear mappingφ:alg𝔄→alg𝔄 a bilocal Jordan derivation, if for every u∈H, there is a Jordan derivation δT,u:alg𝔄→alg𝔄, such that φ(T)u=δT,u(T)u.


## 3. Main Results

Next to give out the main conclusions.Theorem 3.1.
Ifφ:alg𝔄→alg𝔄 is a generalized derivation, then there are operators A and B in alg𝔄, such that φ(T)=AT+TB, for all T∈alg𝔄.Proof.
From the definition of generalized derivation, we can find a derivation:δ:alg𝔄→alg𝔄, such that φ(ST)=φ(S)T+Sδ(T), for all S,T∈alg𝔄, so when S=I, we have φ(T)=φ(I)T+δ(T), for all T∈alg𝔄, denote φ(I)=C, apparently C∈alg𝔄 and φ(T)=CT+δ(T), for all T∈alg𝔄.
Sinceδ:alg𝔄→alg𝔄 is a derivation, by [6], it is an inner derivation, namely, there exists D∈alg𝔄, such that δ(T)=DT-TD, consequently
(3.1)φ(T)=CT+DT-TD=(C+D)T-TD.
Denote A=C+D, B=-D, then φ(T)=AT+TB, for all T∈alg𝔄.Theorem 3.2.
Ifφ:alg𝔄→alg𝔄 is a local Jordan derivation, then φ is an inner derivation.Proof.
Sinceφ is a local Jordan derivation, there exists a Jordan derivation φT:alg𝔄→alg𝔄, such that φ(T)=φT(T), from Theorem  2.12 in [8], we know that the Jordan derivation of nest algebra alg𝔄 is an inner derivation, so there exists AT∈alg𝔄, such that φT(T)=TAT-ATT, by imitating the proof in [6], we can conclude that φT(T)=TA-AT, so φ(T)=TA-AT, namely, φ is an inner derivation.The following is the main result.Theorem 3.3.
Ifφ:alg𝔄→alg𝔄 is a bilocal Jordan derivation, then it is an inner derivation.Proof.
We will prove this proposition by the following three steps.
(1)φ(T)(kerT)⊆ranT, where kerT and ranT are the kernal of T and range of T, respectively. In fact, since φ(T)u=δT,u(T)u for all T∈alg𝔄, for all u∈H, and δT,u is a Jordan derivation, by Theorem  2.12 in [8], there is an AT,u∈alg𝔄, such that φ(T)u=(TAT,u-AT,uT)u, so if u∈kerT,  we have φ(T)u=TAT,uu∈ranT.
(2) For allN∈𝔄, {o}⊂N⊂H, there exists CN∈B(H) and BN∈B(H), such that φ(x⊗f)=x⊗CNf+BNx⊗f, for all x∈N, f∈N-⊥.
For arbitrary fixedf∈N-⊥, f≠0, and for all x∈N, we know that x⊗y∈alg𝔄, from step (1), we have φ(x⊗f){f}⊥⊆span{x}, so there exists a linear function λx,fover{f}⊥, such that φ(x⊗f)(u)=〈u,λx,f〉x, for all u∈{f}⊥, in succession we will prove that λx,f is independent of x. Take a z∈N which is linear independent of x, we have z⊗f∈alg𝔄, then
(3.2)φ((x+z)⊗f)(u)=φ(x⊗f)(u)+φ(z⊗f)(u)=〈u,λx,f〉x+〈u,λz,f〉z.
On the other hand, φ((x+z)⊗f)(u)=〈u,λx+z,f〉(x+z), so
(3.3)〈u,λx+z,f-λx,f〉x=〈u,λz,f-λx+z,f〉z.
Since x is linear independent of z, we know that λx,f=λx+z,f, that is, λx,f is independent of x, so λx,f can be denoted by λf, and
(3.4)φ(x⊗f)(u)=〈u,λf〉x,∀u∈{f}⊥.
Let gf be the linear continuous span on H of λf,  we define Bu,f:N→N as follows:
(3.5)Bu,f(x)=1〈u,f〉{φ(x⊗f)(u)-〈u,gf〉x},whereu∈{f}⊥.
Obviously, Bu,f is linear and φ(x⊗f)(u)=〈u,gf〉x+〈u,f〉Bu,fx, x∈N, u∈H.
NextBu,f is independent of u, which reduce to show (i) Bau,f=Bu,f, a∈C; (ii) Bu,f=Bv,f, where v≠u, v∈H.
In fact, (i) is evident. For (ii), since for allx∈N, φ(x⊗f)(v)=〈v,gf〉x+〈v,f〉Bv,fx and φ(x⊗f)(u+v)=〈u+v,gf〉x+〈u+v,f〉Bu+v,fx=〈u+v,gf〉x+〈v,f〉Bv,fx+〈u,f〉Bu,fx, we have 〈u,f〉Bu+v,f+〈v,f〉Bu+v,f=〈u,f〉Bu,f+〈v,f〉Bv,f, namely, 〈u,f〉(Bu+v,f-Bu,f)=〈v,f〉(Bv,f-Bu+v,f), on account of u≠v, so Bu+v,f=Bu,f, that is, Bu,f is independent of u, so we can mark Bu,f by Bf, as a result, we have
(3.6)φ(x⊗f)(u)=〈u,gf〉x+〈u,f〉Bfx=x⊗gf(u)+Bfx⊗f(u),∀u∈H.
Consequently
(3.7)φ(x⊗f)=x⊗gf+Bfx⊗f.
DefineCN:N-⊥→N-⊥:CNf→gf, now we will show that CN is a linear bounded operator. Because gf is a continuous linear function, so gf is bounded, consequently CN is bounded, according to (3.4), we know
(3.8)φ(x⊗af)(u)=〈u,λaf〉x=a̅φ(x⊗f)(u)=a̅〈u,λf〉x=〈u,aλf〉x,
so λa,f=aλf; on the other hand,
(3.9)φ(x⊗(f1+f2))(u)=〈u,λf1+f2〉x=φ(x⊗f1)(u)+φ(x⊗f2)(u)=〈u,λf1〉x+〈u,λf2〉x=〈u,λf1+λf2〉x,
so λf1+f2=λf1+λf2, this is enough to show that gaf=agf, gf1+f2=gf1+gf2, so when af1+f2∈N-⊥,
(3.10)CN(af1+f2)=gaf1+f2=gaf1+gf2=aCNf1+CNf2.
That is to say CN is linear, so (3.7) has the form of
(3.11)φ(x⊗f)=x⊗CNf+Bfx⊗f.
In succession we will prove that Bf is independent of f, arbitrarily choose y∈N-⊥, where y is linear independent of f, then y+f∈N-⊥ and
(3.12)φ(x⊗(f+y))=x⊗CN(f+y)+Bf+yx⊗(f+y)=x⊗CNf+x⊗CNy+Bf+yx⊗f+Bf+yx⊗y.
On the other hand,
(3.13)φ(x⊗(f+y))=φ(x×⊗f)+φ(x⊗y)=x⊗CNf+Bfx⊗f+x⊗CNy+Byx⊗y.
So (Bf+y-Bf)x⊗f=(By-By+f)x⊗f, then Bf=Bf+y=By, that is, Bf is independent of f, which can be marked by BN, so (3.11) has  the form of
(3.14)φ(x⊗f)=x⊗CNf+BNx⊗f.We proved thatBN is bounded, because ∥BNx⊗f∥≤∥φ(x⊗f)∥+∥x⊗CNf∥≤M+∥CN∥∥x∥∥f∥.
On account of the boundary ofCN and φ(x⊗f)∈algN⊆B(H), we know that BN is bounded, namely, BN∈B(N), CN∈B(N-⊥).
(3) For arbitraryx⊗y∈alg𝔄, there is φ(x⊗f)=x⊗Cf+Bx⊗f.
For allM,N∈𝔄, {0}⊂N⊂M⊂H, select x∈N, f∈M-⊥, then x∈M, f∈N-⊥ and x⊗f∈alg𝔄, from the result of step (2), it is easy to know that
(3.15)φ(x⊗f)=x⊗CNf+BNx⊗f,φ(x⊗f)=x⊗CMf+BMx⊗f.
Consequently x⊗(CN-CM)f=(BM-BN)x⊗f, so there exists a scalar λ(N,M), such that
(3.16)(CN-CM)|M-⊥=λP(M-⊥),(BN-BM)|N=-λP(N).
By imitating Lemma  2 mentioned in [6], we can prove that
(3.17)φ(x⊗f)=x⊗Cf+Bx⊗f,x⊗y∈algN.
Since the collection of all rank one operators is dense inalg𝔄, so for every T∈alg𝔄, we have  φ(T)=TC*+BT, let T=I, then φ(I)=B+C*, considering φ to be a bilocal Jordan derivation, namely, φ(I)(u)=δI,u(I2)u=(IδI,u(I)+δI,u(I)I)u=2δI,u(I)u, we can conclude that δI,u(I)=0, so φ(I)=0, thereby B+C*=0 and φ(T)=BT-TB, which shows that φ is an inner derivation.



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*Source: 748159-2011-02-08.xml*