Derivations, local and 2-local derivations on some algebras of operators on Hilbert C*-modules

TL;DR

This paper investigates derivations and 2-local derivations on algebras of bounded and adjointable operators over Hilbert C*-modules, proving their linearity, continuity, and inner nature under certain conditions.

Contribution

It establishes that derivations on these operator algebras are inner and that 2-local derivations are actual derivations, extending understanding of their structure.

Findings

Derivations are $ ext{A}$-linear, continuous, and inner.

2-local derivations are actual derivations.

Conditions for local derivations to be derivations are identified.

Abstract

For a commutative C*-algebra $\mathcal A$ with unit $e$ and a Hilbert~$\mathcal A$-module $\mathcal M$, denote by End$_{\mathcal A}(\mathcal M)$ the algebra of all bounded $\mathcal A$-linear mappings on $\mathcal M$, and by End$^*_{\mathcal A}(\mathcal M)$ the algebra of all adjointable mappings on $\mathcal M$. We prove that if $\mathcal M$ is full, then each derivation on End$_{\mathcal A}(\mathcal M)$ is $\mathcal A$-linear, continuous, and inner, and each 2-local derivation on End$_{\mathcal A}(\mathcal M)$ or End$^{*}_{\mathcal A}(\mathcal M)$ is a derivation. If there exist $x_0$ in $\mathcal M$ and $f_0$ in $\mathcal M^{'}$, such that $f_0(x_0)=e$, where $\mathcal M^{'}$ denotes the set of all bounded $\mathcal A$-linear mappings from $\mathcal M$ to $\mathcal A$, then each $\mathcal A$-linear local derivation on End$_{\mathcal A}(\mathcal M)$ is a derivation.

Equations

$$\sum _{i=1}^{n}A{\theta }_{x,\widehat{x_i}}{x}_i=\sum _{i=1}^n⟨x_i,x_i⟩Ax=Ax,$$

$$\sum _{i=1}^n{\theta }_{x,\widehat{x_i}}A{x}_i=\sum _{i=1}^n⟨A{x}_i,x_i⟩x.$$

$$A{\theta }_{x,f}Ax={\theta }_{Ax,f\circ A}x=f(Ax)Ax=0.$$

$$⟨⟨y,y⟩y,⟨y,y⟩y⟩=⟨y,y{⟩}^3=0.$$

$$d(T_{a}A)=d(T_a)A+T_{a}d(A)$$

$$d(A{T}_a)=Ad(T_a)+d(A)T_a,$$

$$d(aA)=d(T_{a}A)=d(T_a)A+T_{a}d(A)=T_{a}d(A)=ad(A),$$

$$d({\theta }_{x,f}{T}_n{\theta }_{y,g})=d(f(T_{n}y){\theta }_{x,g})=f(T_{n}y)d({\theta }_{x,g})\to 0,$$

$$d({\theta }_{x,f}{T}_n{\theta }_{y,g})=d({\theta }_{x,f})T_n{\theta }_{y,g}+{\theta }_{x,f}d(T_n){\theta }_{y,g}+{\theta }_{x,f}{T}_{n}d({\theta }_{y,g}).$$

$$d({\theta }_{x,f}{T}_n{\theta }_{y,g})\to {\theta }_{x,f}T{\theta }_{y,g}=f(Ty){\theta }_{x,g}.$$

$${\theta }_{ax,g}z=g(z)ax=0.$$

$$f(Ty)x=0.$$

$$Tx=\sum _{i=1}^{n}d({\theta }_{x,x_i})x_i,$$

$$TAx$$

$$=\sum _{i=1}^{n}d(A{\theta }_{x,x_i})x_i$$

$$=\sum _{i=1}^{n}d(A){\theta }_{x,x_i}{x}_i+\sum _{i=1}^{n}Ad({\theta }_{x,x_i})x_i$$

$$=d(A)\sum _{i=1}^n⟨x_i,x_i⟩x+A\sum _{i=1}^{n}d({\theta }_{x,x_i})x_i$$

$$=d(A)x+ATx.$$

$$\sum _{i=1}^n{f}_i(x_k)x_i=\sum _{i=1}^n{\theta }_{x_i,f_i}{x}_k=0$$

$$\Rightarrow$$

$$\varphi ({\theta }_{x,f}A)=\varphi ({\theta }_{x,f\circ A})=f(Ax)=\varphi ({\theta }_{Ax,f})=\varphi (A{\theta }_{x,f}).$$

$$\delta (A)S+A\delta (S)=d(A)S+Ad(S)=d(AS)=D_T(AS)=TAS-AST.$$

$$\varphi (\delta (A)S+A\delta (S))=\varphi (TAS-AST)=0,$$

$$\varphi (\delta (A+B)S)$$

$$=-\varphi (A\delta (S))-\varphi (B\delta (S))$$

$$=\varphi (\delta (A)S)+\varphi (\delta (B)S)$$

$$=\varphi ((\delta (A)+\delta (B))S).$$

$$\varphi (C{\theta }_{x,f})=f(Cx)=0\Rightarrow ⟨Cx,Cx⟩=0\Rightarrow Cx=0\Rightarrow C=0.$$

$$\varphi ({\theta }_{x,\hat{y}}A)=\varphi ({\theta }_{x,\widehat{A^{\ast }y}})=⟨x,A^{\ast }y⟩=⟨Ax,y⟩=\varphi ({\theta }_{Ax,\hat{y}})=\varphi (A{\theta }_{x,\hat{y}}).$$

$$f_0(a(x_0-\lambda x))=a{f}_0(x_0-\lambda x)=a(e-\lambda f_0(x))=a{a}^{-1}=e.$$

$${\theta }_{x,f_0}={\lambda }^{-1}{\theta }_{x_0,f_0}-{\lambda }^{-1}{a}^{-1}{\theta }_{a(x_0-\lambda x),f_0}.$$

$$(a(f_0-\lambda f))(x_0)=a(e-\lambda f(x_0))=a{a}^{-1}=e.$$

$${\theta }_{x_0,f}={\lambda }^{-1}{\theta }_{x_0,f_0}-{\lambda }^{-1}{a}^{-1}{\theta }_{x_0,a(f_0-\lambda f)}.$$

Full text

Keywords:   Derivations, Hilbert C*-modules, inner derivations, local derivations, 2-local derivations

Mathematics Subject Classification(2010): 47B47; 47L10

1 Introduction and preliminaries

The structure of derivations on operator algebras is an important part of the theory of operator algebras.

Let $\mathcal{A}$ be an algebra and $\mathcal{M}$ be an $\mathcal{A}$-bimodule. Recall that a derivation is a linear mapping $d$ from $\mathcal{A}$ into $\mathcal{M}$ such that $d(xy)=d(x)y+xd(y)$, for all $x,y$ in $\mathcal{A}$. For each $m$ in $\mathcal{M}$, one can define a derivation $D_{m}$ by $D_{m}(x)=mx-xm$, for all $x$ in $\mathcal{A}$. Such derivations are called inner derivations.

It is a classical problem to identify those algebras on which all derivations are inner derivations. Several authors investigate this topic. The following two results are classical. S. Sakai [17] proves that all derivations from a W*-algebra into itself are inner derivations. E. Christensen [3] proves that all derivations from a nest algebra into itself are inner derivations.

In 1990, R. Kadison [10] and D. Larson, A. Sourour [13] independently introduce the concept of local derivation in the following sense: a linear mapping $\delta$ from $\mathcal{A}$ into $\mathcal{M}$ such that for every $a\in$ $\mathcal{A}$, there exists a derivation $d_{a}:$ $\mathcal{A}$ $\rightarrow$ $\mathcal{M}$, depending on $a$, satisfying $\delta(a)=d_{a}(a)$. In [10], R. Kadison proves that each continuous local derivation from a von Neumann algebra into its dual Banach module is a derivation. In [13], D. Larson and A. Sourour prove that each local derivation from $B(\mathcal{X})$ into itself is a derivation, where $\mathcal{X}$ is a Banach space. B. Jonson [8] proves that each local derivation from a C*-algebra into its Banach bimodule is a derivation. Z. Pan and the second author of this paper [14] prove that each local derivation from the algebra $\mathcal{M}\bigcap alg\mathcal{L}$ into $B(\mathcal{H})$ is a derivation, where $\mathcal{H}$ is a Hilbert space, $\mathcal{M}$ is a von Neumann algebra acting on $\mathcal{H}$, and $\mathcal{L}$ is a commutative subspace lattice in $\mathcal{M}$. For more information about this topic, we refer to [2, 4, 6].

In 1997, P. $\check{S}$emrl [18] introduces the concept of 2-local derivations. Recall that a mapping $\delta$ : $\mathcal{A}$ $\rightarrow$ $\mathcal{M}$ (not necessarily linear) is called a 2-local derivation if for each $a,b\in$ $\mathcal{A}$, there exists a derivation $d_{a,b}$ : $\mathcal{A}$ $\rightarrow$ $\mathcal{M}$ such that $\delta(a)=d_{a,b}(a)$ and $\delta(b)=d_{a,b}(b)$. Moreover, the author proves that every 2-local derivation on $B(\mathcal{H})$ is a derivation for a separable Hilbert space $\mathcal{H}$. J. Zhang and H. Li [19] extend the above result for arbitrary symmetric digraph matrix algebras and construct an example of 2-local derivation which is not a derivation on the algebra of all upper triangular complex $2\times 2$ matrices. S. Ayupov and K. Kudaybergenov [1] prove that each 2-local derivation on a von Neumann algebra is a derivation. For more information about this topic, we refer to [2, 7, 11].

In this paper, we study derivations, local derivations and 2-local derivations on some algebras of operators on Hilbert C*-modules. There are few results in this topic. P. Li, D. Han and W. Tang [15] prove that each derivation on End${}^{*}_{\mathcal{A}}(\mathcal{M})$ is inner, where $\mathcal{M}$ is a full Hilbert C*-module over a commutative unital C*-algebra $\mathcal{A}$. M. Moghadam, M. Miri and A. Janfada [16] prove that each $\mathcal{A}$-linear derivation on End${}_{\mathcal{A}}(\mathcal{M})$ is inner, where $\mathcal{M}$ is a full Hilbert C*-module over a commutative unital C*-algebra $\mathcal{A}$ with the property that there exist $x_{0}$ in $\mathcal{M}$ and $f_{0}$ in $\mathcal{M}^{{}^{\prime}}$ such that $f_{0}(x_{0})=e$.

Hilbert C*-modules provide a natural generalization of Hilbert spaces by replacing the complex field $\mathbb{C}$ with an arbitrary C*-algebra. The theory of Hilbert C*-modules plays an important role in the theory of operator algebras, as it can be applied in many fields, such as index theory of elliptic operators, K- and K K-theory, noncommutative geometry, and so on.

In the following, we would firstly review some properties of Hilbert C*-modules [12].

Let $\mathcal{A}$ be a C*-algebra and $\mathcal{M}$ be a left $\mathcal{A}$-module.

$\mathcal{M}$ is called a Pre-Hilbert $\mathcal{A}$-module if there exists a mapping $\langle\cdot,\cdot\rangle:\mathcal{M}\times\mathcal{M}\longrightarrow\mathcal{A}$ with the following properties: for each $\lambda\in\mathbb{C},a\in\mathcal{A},x,y,z\in\mathcal{M}$,

(1) $\langle x,x\rangle\geq 0$, and $\langle x,x\rangle=0$ implies that $x=0$,

(2) $\langle\lambda x+y,z\rangle=\lambda\langle x,z\rangle+\langle y,z\rangle$,

(3) $\langle ax,y\rangle=a\langle x,y\rangle$,

(4) $\langle x,y\rangle=\langle y,x\rangle^{*}$.

The mapping $\langle\cdot,\cdot\rangle$ is called an $\mathcal{A}$-valued inner product. The inner product induces a norm on $\mathcal{M}$: $\|x\|=\|\langle x,x\rangle\|^{1/2}$. $\mathcal{M}$ is called a Hilbert $\mathcal{A}$-module(or more exactly, a Hilbert C*-module over $\mathcal{A}$), if it is complete with respect to this norm.

We denote by $\langle\mathcal{M},\mathcal{M}\rangle$ the closure of the linear span of all the elements of the form $\langle x,y\rangle,x,y\in\mathcal{M}$. $\mathcal{M}$ is called a full Hilbert $\mathcal{A}$-module if $\langle\mathcal{M},\mathcal{M}\rangle=\mathcal{A}$.

For a full Hilbert $\mathcal{A}$-module $\mathcal{M}$, we have the following lemma.

Lemma 1.1**.**

Let $\mathcal{A}$ be a C-algebra with unit $e$ and $\mathcal{M}$ be a full Hilbert $\mathcal{A}$-module. There exists a sequence $\{x_{i}\}_{i=1}^{n}\subseteq\mathcal{M}$, such that $\sum_{i=1}^{n}\langle x_{i},x_{i}\rangle=e$.*

A linear mapping $T$ from $\mathcal{M}$ into itself is said to be $\mathcal{A}$-linear if $T(ax)=aT(x)$ for each $a\in\mathcal{A}$ and $x\in\mathcal{M}$. A bounded $\mathcal{A}$-linear mapping from $\mathcal{M}$ into itself is called an operator on $\mathcal{M}$. Denote by End${}_{\mathcal{A}}(\mathcal{M})$ all operators on $\mathcal{M}$. End${}_{\mathcal{A}}(\mathcal{M})$ is a Banach algebra.

A mapping $T$ from $\mathcal{M}$ into itself is said to be adjointable if there exists a mapping $T^{*}$ such that $\langle Tx,y\rangle=\langle x,T^{*}y\rangle$, for all $x,y\in\mathcal{M}$. Notice that each adjointable mapping must be an operator. Denote by End${}^{*}_{\mathcal{A}}(\mathcal{M})$ all adjointable operators on $\mathcal{M}$. End${}^{*}_{\mathcal{A}}(\mathcal{M})$ is a C*-algebra.

Similarly, a linear mapping $f$ from $\mathcal{M}$ into $\mathcal{A}$ is said to be $\mathcal{A}$-linear if $f(ax)=af(x)$ for each $a\in\mathcal{A}$ and $x\in\mathcal{M}$. The set of all bounded $\mathcal{A}$-linear mappings from $\mathcal{M}$ to $\mathcal{A}$ is denoted by $\mathcal{M}^{{}^{\prime}}$.

For each $x$ in $\mathcal{M}$, one can define a mapping $\hat{x}$ from $\mathcal{M}$ to $\mathcal{A}$ by follows: $\hat{x}(y)=\langle y,x\rangle$, for all $y\in\mathcal{M}$. Obviously, $\hat{x}\in\mathcal{M}^{{}^{\prime}}$.

For each $x$ in $\mathcal{M}$ and $f$ in $\mathcal{M}^{{}^{\prime}}$, one can define a mapping $\theta_{x,f}$ from $\mathcal{M}$ into itself by follows: $\theta_{x,f}y=f(y)x$, for all $y\in\mathcal{M}$. Obviously, $\theta_{x,f}\in$ End${}_{\mathcal{A}}(\mathcal{M})$.

In particular, for each $x,y$ in $\mathcal{M}$, we have $\theta_{x,\hat{y}}z=\hat{y}(z)x=\langle z,y\rangle x$, for all $z\in\mathcal{M}$.

For the operators of the above forms, we have the following lemmas.

Lemma 1.2**.**

Let $\mathcal{M}$ be a Hilbert C-module over a C*-algebra $\mathcal{A}$.

For all $a\in\mathcal{A},x,y\in\mathcal{M},f,g\in\mathcal{M}^{{}^{\prime}},A\in$ End${}_{\mathcal{A}}(\mathcal{M})$, we have

(1) $\theta_{x,f}A=\theta_{x,f\circ A}$,

(2) $A\theta_{x,f}=\theta_{Ax,f}$,

(3) if in addition, $\mathcal{A}$ is commutative, then $\theta_{x,f}\theta_{y,g}=f(y)\theta_{x,g}$, $\theta_{ax,f}=a\theta_{x,f}$.*

Lemma 1.3**.**

Let $\mathcal{M}$ be a Hilbert C-module over a C*-algebra $\mathcal{A}$.

For all $a\in\mathcal{A},x,y,z,w\in\mathcal{M},A\in$ End*${}_{\mathcal{A}}(\mathcal{M})$, we have

(1) $\theta_{x,\hat{y}}\in$ End*${}_{\mathcal{A}}(\mathcal{M})$, and $\theta^{*}_{x,\hat{y}}=\theta_{y,\hat{x}}$,

(2) $\theta_{x,\hat{y}}A=\theta_{x,\hat{y}\circ A}=\theta_{x,\widehat{A^{*}y}}$,

(3) $A\theta_{x,\hat{y}}=\theta_{Ax,\hat{y}}$,

(4) if in addition, $\mathcal{A}$ is commutative, then $\theta_{x,\hat{y}}\theta_{z,\hat{w}}=\langle z,y\rangle\theta_{x,\hat{w}}$, $\theta_{ax,\hat{y}}=a\theta_{x,\hat{y}}=\theta_{x,\widehat{a^{*}y}}$.*

For a commutative C*-algebra $\mathcal{A}$, for each $a$ in $\mathcal{A}$, one can define a mapping $T_{a}$ from $\mathcal{M}$ into itself by follows: $T_{a}x=ax$, for all $x\in\mathcal{M}$. Obviously, $T_{a}\in$ End${}_{\mathcal{A}}(\mathcal{M})$. It is worthwhile to notice that if $\mathcal{A}$ is not commutative, then $T_{a}$ is not $\mathcal{A}$-linear. In this case, $T_{a}$ is not in End${}_{\mathcal{A}}(\mathcal{M})$.

Lemma 1.4**.**

Let $\mathcal{A}$ be a commutative C-algebra with unit $e$ and $\mathcal{M}$ be a full Hilbert $\mathcal{A}$-module. Then $\mathcal{Z}($End${}_{\mathcal{A}}(\mathcal{M}))=\{T_{a}:~{}a\in\mathcal{A}\}$.*

Proof.

For each $A$ in End${}_{\mathcal{A}}(\mathcal{M})$ and $x$ in $\mathcal{M}$, since $AT_{a}x=A(ax)=aAx=T_{a}Ax$, we have $AT_{a}=T_{a}A$. It is to say $T_{a}\in\mathcal{Z}($End${}_{\mathcal{A}}(\mathcal{M}))$.

On the other hand, assume $A\in\mathcal{Z}($End${}_{\mathcal{A}}(\mathcal{M}))$. By Lemma 1.1, there exists a sequence $\{x_{i}\}_{i=1}^{n}\subseteq\mathcal{M}$, such that $\sum_{i=1}^{n}\langle x_{i},x_{i}\rangle=e$. Thus we have

[TABLE]

and

[TABLE]

Let $\sum_{i=1}^{n}\langle Ax_{i},x_{i}\rangle=a$. Then we have $A=T_{a}$. The proof is complete. ∎

For an algebra $\mathcal{A}$, if for each $a$ in $\mathcal{A}$, $a\mathcal{A}a=0$ implies that $a=0$, then it is said to be semi-prime.

Lemma 1.5**.**

Let $\mathcal{A}$ be a C-algebra and $\mathcal{M}$ be a Hilbert $\mathcal{A}$-module. Then End${}_{\mathcal{A}}(\mathcal{M})$ is a semi-prime Banach algebra.*

Proof.

Let $A$ be in End${}_{\mathcal{A}}(\mathcal{M})$. Assume that $ABA=0$ for each $B$ in End${}_{\mathcal{A}}(\mathcal{M})$. In particular, for each $x\in\mathcal{M}$ and $f\in\mathcal{M}^{{}^{\prime}}$, we have

[TABLE]

By taking $y=Ax$ and $f=\hat{y}$, we have $\langle y,y\rangle y=0$. It follows that

[TABLE]

Since $\langle y,y\rangle$ is a self-adjoint element, we have $\langle y,y\rangle=0$, and $y=0$. Hence $A=0$, and End${}_{\mathcal{A}}(\mathcal{M})$ is semi-prime. The proof is complete. ∎

2 Derivations on End${}_{\mathcal{A}}(\mathcal{M})$

In this section, we study derivations on End${}_{\mathcal{A}}(\mathcal{M})$. We begin with several lemmas.

Lemma 2.1**.**

Let $\mathcal{A}$ be a commutative unital C-algebra and $\mathcal{M}$ be a full Hilbert $\mathcal{A}$-module. Then each derivation on $End_{\mathcal{A}}(\mathcal{M})$ is $\mathcal{A}$-linear, i.e. $d(aA)=ad(A),$ for each $a\in\mathcal{A}$ and $A\in End_{\mathcal{A}}(\mathcal{M})$.*

Proof.

Suppose $d$ is a derivation on End${}_{\mathcal{A}}(\mathcal{M})$.

By Lemma 1.4, we have $\mathcal{Z}($End${}_{\mathcal{A}}(\mathcal{M}))=\{T_{a}:~{}a\in\mathcal{A}\}$. For each $A$ in End${}_{\mathcal{A}}(\mathcal{M})$, By

[TABLE]

and

[TABLE]

we obtain $d(T_{a})A=Ad(T_{a})$. Hence $d(T_{a})\in\mathcal{Z}($End${}_{\mathcal{A}}(\mathcal{M}))$, and $d(\mathcal{Z}($End${}_{\mathcal{A}}(\mathcal{M})))\subseteq\mathcal{Z}($End${}_{\mathcal{A}}(\mathcal{M}))$.

Since $\mathcal{Z}($End${}_{\mathcal{A}}(\mathcal{M}))=\{T_{a}:~{}a\in\mathcal{A}\}$ is a commutative C*-algebra, and every derivation on a commutative C*-algebra is zero, we have $d(T_{a})=0$.

It follows that

[TABLE]

which means that $d$ is $\mathcal{A}$-linear. The proof is complete. ∎

Lemma 2.2**.**

Let $\mathcal{A}$ be a commutative unital C-algebra and $\mathcal{M}$ be a full Hilbert $\mathcal{A}$-module. Then each derivation on End${}_{\mathcal{A}}(\mathcal{M})$ is continuous.*

Proof.

Suppose $d$ is a derivation on End${}_{\mathcal{A}}(\mathcal{M})$. Assume that $\{T_{n}\}$ is a sequence converging to zero in End${}_{\mathcal{A}}(\mathcal{M})$, and $\{d(T_{n})\}$ converges to $T$.

According to the closed graph theorem, to show $d$ is continuous, it is sufficient to prove that $T=0$.

By Lemma 2.1, we know $d$ is $\mathcal{A}$-linear. For $x,y\in\mathcal{M}$, $f,g\in\mathcal{M}^{{}^{\prime}}$, we have

[TABLE]

and

[TABLE]

Since $\{T_{n}\}$ converges to zero and $\{d(T_{n})\}$ converges to $T$, we have

[TABLE]

It follows that $f(Ty)\theta_{x,g}=0.$

Let $a=f(Ty)$, then we have $a\theta_{x,g}=\theta_{ax,g}=0$. For each $z\in\mathcal{M}$, we have

[TABLE]

By taking $g=\widehat{ax}$ and $z=ax$ in (2.1), we can obtain $ax=0$, i.e.

[TABLE]

By taking $f=\widehat{Ty}$ and $x=Ty$ in (2.2), we can obtain $Ty=0$. i.e. $T=0$. The proof is complete. ∎

Now we can prove our main theorem in this section.

Theorem 2.3**.**

Let $\mathcal{A}$ be a commutative C-algebra with unit $e$ and $\mathcal{M}$ be a full Hilbert $\mathcal{A}$-module. Then each derivation on End${}_{\mathcal{A}}(\mathcal{M})$ is an inner derivation.*

Proof.

Suppose $d$ is a derivation on End${}_{\mathcal{A}}(\mathcal{M})$ and $\{x_{i}\}_{i=1}^{n}$ is a sequence in $\mathcal{M}$ such that $\sum_{i=1}^{n}\langle x_{i},x_{i}\rangle=e$.

Define a mapping $T$ from $\mathcal{M}$ into itself by follows:

[TABLE]

for all $x\in\mathcal{M}.$

By Lemmas 2.1 and 2.2, $d$ is $\mathcal{A}$-linear and continuous, thus $T$ is also $\mathcal{A}$-linear and continuous. That is to say $T\in$ End${}_{\mathcal{A}}(\mathcal{M})$.

Now it is sufficient to show that $d(A)=TA-AT$, for each $A\in$ End${}_{\mathcal{A}}(\mathcal{M})$.

For each $x\in\mathcal{M}$, we have

[TABLE]

It implies that $d(A)=TA-AT$. Hence $d$ is an inner derivation. The proof is complete. ∎

3 2-Local derivations on End${}_{\mathcal{A}}(\mathcal{M})$ and End${}^{*}_{\mathcal{A}}(\mathcal{M})$

In this section, we characterize 2-local derivations on End${}_{\mathcal{A}}(\mathcal{M})$ and End${}^{*}_{\mathcal{A}}(\mathcal{M})$. Firstly, we show the following lemma.

Lemma 3.1**.**

Let $\mathcal{A}$ be a commutative unital C-algebra and $\mathcal{M}$ be a Hilbert $\mathcal{A}$-module. For $x_{i}\in\mathcal{M}$ and $f_{i}\in\mathcal{M}^{{}^{\prime}}$, if  $\sum_{i=1}^{n}\theta_{x_{i},f_{i}}=0$, then $\sum_{i=1}^{n}f_{i}(x_{i})=0$.*

Proof.

Let $a_{i,j}=f_{j}(x_{i})\in\mathcal{A}$ and $\Lambda=(a_{i,j})_{n\times n}\in M_{n}(\mathcal{A})$.

We have

[TABLE]

Since $\mathcal{A}$ is a commutative unital C*-algebra, it is well known that $\mathcal{A}$ is $*$-isomorphic to $C(\mathcal{S})$ for some compact Hausdorff space $\mathcal{S}$. Without loss of generality, we can assume $\mathcal{A}=C(\mathcal{S})$.

Then for each $t\in\mathcal{S}$, we have $a_{i,j}(t)\in\mathbb{C}$ and $\Lambda(t),\Lambda^{2}(t)\in M_{n}(\mathbb{C})$.

Recall that for a matrix $A$ in $M_{n}(\mathbb{C})$, $A^{2}=0$ implies that $tr(A)=0$, where $tr(A)$ denotes the trace of $A$, i.e. the sum of all the diagonal elements.

Hence $\Lambda^{2}(t)=0$ implies that $tr(\Lambda(t))=0$. It follows that $tr(\Lambda)=0$, that is to say $\sum_{i=1}^{n}f_{i}(x_{i})=0$. The proof is complete. ∎

Theorem 3.2**.**

Let $\mathcal{A}$ be a commutative unital C-algebra and $\mathcal{M}$ be a full Hilbert $\mathcal{A}$-module. Then each 2-local derivation on End${}_{\mathcal{A}}(\mathcal{M})$ is a derivation.*

Proof.

Denote by $\Gamma(\mathcal{M})$ the linear span of the set $\{\theta_{x,f}:x\in\mathcal{M},f\in\mathcal{M}^{{}^{\prime}}\}$. By Lemma 1.2, $\Gamma(\mathcal{M})$ is a two-side ideal of End${}_{\mathcal{A}}(\mathcal{M})$.

For each $S=\sum_{i=1}^{n}\theta_{x_{i},f_{i}}\in\Gamma(\mathcal{M})$, define $\phi(S)=\sum_{i=1}^{n}f_{i}(x_{i})$.

One can verify that $\phi$ is well defined by Lemma 3.1. And obviously, $\phi$ is $\mathcal{A}$-linear. Moreover, for each $A\in$ End${}_{\mathcal{A}}(\mathcal{M})$, we have

[TABLE]

It follows that $\phi(SA)=\phi(AS)$ for each $A\in$ End${}_{\mathcal{A}}(\mathcal{M})$ and $S\in\Gamma(\mathcal{M})$.

Suppose $\delta$ is a 2-local derivation on End${}_{\mathcal{A}}(\mathcal{M})$. By the definition of 2-local derivation, there exists a derivation $d$ on End${}_{\mathcal{A}}(\mathcal{M})$ such that $\delta(A)=d(A)$ and $\delta(S)=d(S)$. By Theorem 2.3, $d$ is an inner derivation, i.e. there exists an element $T\in$ End${}_{\mathcal{A}}(\mathcal{M})$ such that $d=D_{T}$.

Thus we have

[TABLE]

Since $\Gamma(\mathcal{M})$ is a two-side ideal of End${}_{\mathcal{A}}(\mathcal{M})$, we know that $AS\in\Gamma(\mathcal{M})$.

Hence

[TABLE]

which follows that $\phi(\delta(A)S)=-\phi(A\delta(S))$.

Now, for each $A,B\in$ End${}_{\mathcal{A}}(\mathcal{M})$ and $S\in\Gamma(\mathcal{M})$, we have

[TABLE]

Let $C=\delta(A+B)-\delta(A)-\delta(B)$, we obtain $\phi(CS)=0$.

By taking $S=\theta_{x,f}$, we have

[TABLE]

It means that $\delta(A+B)=\delta(A)+\delta(B)$. That is to say $\delta$ is an additive mapping. In addition, by the definition of 2-local derivation, it is easy to show that $\delta$ is homogeneous and $\delta(A^{2})=A\delta(A)+\delta(A)A$ for each $A\in$ End${}_{\mathcal{A}}(\mathcal{M})$. Hence $\delta$ is a Jordan derivation.

By Lemma 1.5, End${}_{\mathcal{A}}(\mathcal{M})$ is a semi-prime Banach algebra. According to the classical result that every Jordan derivation on a semi-prime Banach algebra is a derivation [5], we obtain that $\delta$ is a derivation. The proof is complete. ∎

Theorem 3.3**.**

Let $\mathcal{A}$ be a commutative unital C-algebra and $\mathcal{M}$ be a full Hilbert $\mathcal{A}$-module. Then each 2-local derivation on End*${}_{\mathcal{A}}(\mathcal{M})$ is a derivation.*

Proof.

Denote by $\Gamma^{*}(\mathcal{M})$ the linear span of the set $\{\theta_{x,\hat{y}}:x,y\in\mathcal{M}\}$. By Lemma 1.3, $\Gamma^{*}(\mathcal{M})$ is a two-side ideal of End${}^{*}_{\mathcal{A}}(\mathcal{M})$.

For each $S=\sum_{i=1}^{n}\theta_{x_{i},\widehat{y_{i}}}\in\Gamma^{*}(\mathcal{M})$, define $\phi(S)=\sum_{i=1}^{n}\langle x_{i},y_{i}\rangle$.

By Lemma 3.1, $\phi$ is well defined. For each $A\in$ End${}^{*}_{\mathcal{A}}(\mathcal{M})$, we have

[TABLE]

It follows that $\phi(SA)=\phi(AS)$ for each $A\in$ End${}^{*}_{\mathcal{A}}(\mathcal{M})$ and $S\in\Gamma^{*}(\mathcal{M})$.

In [15], the authors prove that for a commutative unital C*-algebra $\mathcal{A}$ and a full Hilbert $\mathcal{A}$-module $\mathcal{M}$, each derivation on End${}^{*}_{\mathcal{A}}(\mathcal{M})$ is an inner derivation.

The rest of the proof is similar to Theorem 3.2, so we omit it. ∎

4 Local derivations on End${}_{\mathcal{A}}(\mathcal{M})$

In this section, we discuss local derivations on End${}_{\mathcal{A}}(\mathcal{M})$. Through this section, we assume that $\mathcal{A}$ is a commutative C*-algebra with unit $e$, and $\mathcal{M}$ is a Hilbert $\mathcal{A}$-module, and moreover, there exist $x_{0}$ in $\mathcal{M}$ and $f_{0}$ in $\mathcal{M}^{{}^{\prime}}$ such that $f_{0}(x_{0})=e$. Denote the unit of End${}_{\mathcal{A}}(\mathcal{M})$ by $I$. Define $\mathcal{L}=span\{\theta_{x,f_{0}}:x\in\mathcal{M}\}$, and $\mathcal{R}=span\{\theta_{x_{0},f}:f\in\mathcal{M}^{{}^{\prime}}\}.$

Lemma 4.1**.**

*(1) $\theta_{x_{0},f_{0}}$ is an idempotent;

(2) each element in $\mathcal{L}$ is an $\mathcal{A}$-linear combination of some idempotents in $\mathcal{L}$, and each element in $\mathcal{R}$ is an $\mathcal{A}$-linear combination of some idempotents in $\mathcal{R}$;

(3) $\mathcal{L}$ is a left ideal of $End_{\mathcal{A}}(\mathcal{M})$, and $\mathcal{R}$ is a right ideal of $End_{\mathcal{A}}(\mathcal{M})$;

(4) $\mathcal{L}$ is a left separating set of End${}_{\mathcal{A}}(\mathcal{M})$, i.e. for each $A$ in End${}_{\mathcal{A}}(\mathcal{M})$, $A\mathcal{L}=0$ implies that $A=0$, and $\mathcal{R}$ is a right separating set of End${}_{\mathcal{A}}(\mathcal{M})$, i.e. for each $A$ in End${}_{\mathcal{A}}(\mathcal{M})$, $\mathcal{R}A=0$ implies that $A=0$*

Proof.

(1) $\theta_{x_{0},f_{0}}\theta_{x_{0},f_{0}}=f_{0}(x_{0})\theta_{x_{0},f_{0}}=\theta_{x_{0},f_{0}}$.

(2) For each $x\in\mathcal{M}$, there exists a non-zero complex number $\lambda\in\mathbb{C}$, such that $e-\lambda f_{0}(x)$ is invertible in $\mathcal{A}$. Denote $e-\lambda f_{0}(x)$ by $a^{-1}$, then we have

[TABLE]

By (1), we know that $\theta_{a(x_{0}-\lambda x),f_{0}}$ is an idempotent.

Thus we have

[TABLE]

That is to say $\theta_{x,f_{0}}$ is an $\mathcal{A}$-linear combination of idempotents in $\mathcal{L}$.

Similarly, for each $f\in\mathcal{M}^{{}^{\prime}}$, there exists a non-zero complex number $\lambda\in\mathbb{C}$, such that $e-\lambda f(x_{0})$ is invertible in $\mathcal{A}$. Denote $e-\lambda f(x_{0})$ by $a^{-1}$, then we have

[TABLE]

Again by (1), we know that $\theta_{x_{0},a(f_{0}-\lambda f)}$ is an idempotent.

Thus we have

[TABLE]

(3) For each $A\in$ End${}_{\mathcal{A}}(\mathcal{M}),$ since $A\theta_{x,f_{0}}=\theta_{Ax,f_{0}}$, we know that $\mathcal{L}$ is a left ideal of End${}_{\mathcal{A}}(\mathcal{M})$. Similarly, $\mathcal{R}$ is a right ideal of End${}_{\mathcal{A}}(\mathcal{M})$ since $\theta_{x_{0},f}A=\theta_{x_{0},f\circ A}$.

(4) Suppose $A\in$ End${}_{\mathcal{A}}(\mathcal{M}),\theta_{x,f_{0}}\in\mathcal{L},\theta_{x_{0},f}\in\mathcal{R}$.

If $A\theta_{x,f_{0}}=0$, then

[TABLE]

i.e. $A=0$.

If $\theta_{x_{0},f}A=0$, then

[TABLE]

The proof is complete. ∎

Let $\mathcal{J}$ be a left $\mathcal{A}$-module, and $\phi$ be a bilinear mapping from End${}_{\mathcal{A}}(\mathcal{M})\times$ End${}_{\mathcal{A}}(\mathcal{M})$ into $\mathcal{J}$.

We say that $\phi$ is $\mathcal{A}$-bilinear if $\phi(aA,B)=\phi(A,aB)=a\phi(A,B)$ for each $A,B\in$ End${}_{\mathcal{A}}(\mathcal{M})$ and $a\in\mathcal{A}$.

We say that $\phi$ preserves zero product if $AB=0$ implies that $\phi(A,B)=0$ for each $A,B\in$ End${}_{\mathcal{A}}(\mathcal{M})$.

Lemma 4.2**.**

Let $\mathcal{J}$ be a left $\mathcal{A}$-module, and $\phi:End_{\mathcal{A}}(\mathcal{M})\times End_{\mathcal{A}}(\mathcal{M})\rightarrow\mathcal{J}$ be an $\mathcal{A}$-bilinear mapping preserving zero product. Then for each $A,B\in End_{\mathcal{A}}(\mathcal{M}),L\in\mathcal{L}$, and $R\in\mathcal{R}$, we have:

[TABLE]

and

[TABLE]

Proof.

Suppose $P$ is an idempotent in End${}_{\mathcal{A}}(\mathcal{M})$. Let $Q=I-P$.

Since $\phi$ preserves zero product, we have

[TABLE]

By Lemma 4.1(2), each element in $\mathcal{L}$ is an $\mathcal{A}$-linear combination of idempotents in $\mathcal{L}$. Considering $\phi$ is $\mathcal{A}$-bilinear, we obtain that $\phi(A,LB)=\phi(AL,B)$.

By Lemma 4.1(3), $\mathcal{L}$ is a left ideal, so $AL\in\mathcal{L}$. Hence $\phi(AL,B)=\phi(I,ALB)$.

Similarly, we can show the equation (4.2) is true. ∎

For an algebra $\mathcal{A}$ with unit $e$, a linear mapping $\delta$ on $\mathcal{A}$ is said to be a generalized derivation if $\delta(ab)=a\delta(b)+\delta(a)b-a\delta(e)b$, for all $a,b$ in $\mathcal{A}$.

Theorem 4.3**.**

Suppose that $\mathcal{A}$ is a commutative C-algebra with unit $e$, and $\mathcal{M}$ is a Hilbert $\mathcal{A}$-module, and moreover, there exist $x_{0}$ in $\mathcal{M}$ and $f_{0}$ in $\mathcal{M}^{{}^{\prime}}$ such that $f_{0}(x_{0})=e$. If $\delta$ is an $\mathcal{A}$-linear mapping from $End_{\mathcal{A}}(\mathcal{M})$ into itself such that: for each $A,B,C$ in End${}_{\mathcal{A}}(\mathcal{M})$, $AB=BC=0$ implies that $A\delta(B)C=0$, then $\delta$ is a generalized derivation. In particular, if $\delta(I)=0$, where $I$ is the unit of End${}_{\mathcal{A}}(\mathcal{M})$, then $\delta$ is a derivation.*

Proof.

Suppose $A,B,X,Y,A_{0},B_{0}$ are arbitrary elements in End${}_{\mathcal{A}}(\mathcal{M})$, where $A_{0}B_{0}=0$, $L$ and $R$ are arbitrary elements in $\mathcal{L}$ and $\mathcal{R}$, respectively.

Define a bilinear mapping $\phi_{1}$: $\phi_{1}(X,Y)=X\delta(YA_{0})B_{0}$. Then $\phi_{1}$ is an $\mathcal{A}$-bilinear mapping preserving zero product.

By Lemma 4.2, we have

[TABLE]

i.e.

[TABLE]

Since $\mathcal{R}$ is a right separating set of End${}_{\mathcal{A}}(\mathcal{M})$, we have

[TABLE]

Now define a bilinear mapping $\phi_{2}$: $\phi_{2}(X,Y)=\delta(AX)Y-A\delta(X)Y$. Then $\phi_{2}$ is also an $\mathcal{A}$-bilinear mapping preserving zero product.

Again by Lemma 4.2, we have

[TABLE]

i.e.

[TABLE]

Since $\mathcal{L}$ is a left separating set of End${}_{\mathcal{A}}(\mathcal{M})$, we obtain that

[TABLE]

That is to say $\delta$ is a generalized derivation. The proof is complete. ∎

Applying the above Theorem, we can get the following corollary immediately.

Corollary 4.4**.**

Suppose $\mathcal{A}$ is a commutative C-algebra with unit $e$, $\mathcal{M}$ is a Hilbert $\mathcal{A}$-module, and moreover, there exist $x_{0}$ in $\mathcal{M}$ and $f_{0}$ in $\mathcal{M}^{{}^{\prime}}$ such that $f_{0}(x_{0})=e$. Then each $\mathcal{A}$-linear local derivation $\delta$ on $End_{\mathcal{A}}(\mathcal{M})$ is a derivation.*

Proof.

For each $A,B,C$ in End${}_{\mathcal{A}}(\mathcal{M})$, if $AB=BC=0$, by the definition of local derivation, there exists a derivation $\delta_{B}$ such that $\delta_{B}(B)=\delta(B)$. Thus we have

[TABLE]

Let $I$ be the unit of End${}_{\mathcal{A}}(\mathcal{M})$, by the definition of local derivation, there exists a derivation $\delta_{I}$ such that $\delta_{I}(I)=\delta(I)=0$.

By Theorem 4.3, $\delta$ is a derivation. The proof is complete. ∎

Acknowledgements. This paper was partially supported by National Natural Science Foundation of China(Grant No. 11371136).