Linear maps behaving like derivations or anti-derivations at orthogonal elements on C*-algebras

TL;DR

This paper characterizes continuous linear maps on C*-algebras that behave like derivations or anti-derivations at orthogonal elements under various orthogonality conditions, with applications to von Neumann and simple C*-algebras.

Contribution

It provides a detailed structural characterization of such maps under multiple orthogonality conditions, extending understanding of derivation-like behavior in C*-algebras.

Findings

Characterization of maps acting like derivations at orthogonal elements

Application of results to von Neumann algebras

Application to unital simple C*-algebras

Abstract

Let A be a C*-algebra and d from A into A** be a continuous linear map. We assume that d acts like derivation or anti-derivation at orthogonal elements for several types of orthogonality conditions such as ab=0, ab*=0, ab=ba=0 and ab*=b*a=0. In each case, we characterize the structure of d. Then we apply our results for von Neumann algebras and unital simple C*-algebras.

Equations

$$a,b\in A,ab=z\Rightarrow \delta (ab)=a\delta (b)+\delta (a)b(⧫),$$

$$a,b\in A,ab=ba=0⟹a\delta (b)+\delta (a)b+b\delta (a)+\delta (b)a=0,$$

$$ab=0⟹a\delta (b)+\delta (a)b=0;$$

$$a{b}^{\star }=0⟹a\delta (b{)}^{\star }+\delta (a)b^{\star }=0;$$

$$a^{\star }b=0⟹a^{\star }\delta (b)+\delta (a{)}^{\star }b=0;$$

$$ab=0⟹b\delta (a)+\delta (b)a=0;$$

$$a{b}^{\star }=0⟹\delta (b{)}^{\star }a+b^{\star }\delta (a)=0;$$

$$a^{\star }b=0⟹\delta (b)a^{\star }+b\delta (a{)}^{\star }=0;$$

$$ab=ba=0⟹a\delta (b)+\delta (a)b=b\delta (a)+\delta (b)a=0;$$

$$a{b}^{\star }=b^{\star }a=0⟹a\delta (b{)}^{\star }+\delta (a)b^{\star }=\delta (b{)}^{\star }a+b^{\star }\delta (a)=0;$$

$$\mu a_i\text{xlongrightarrow}\sigma (A^{\ast \ast },A^{\ast })\mu \nu \text{and}{a}_i\mu \text{xlongrightarrow}\sigma (A^{\ast \ast },A^{\ast })\nu \mu .$$

$$ab=0⟹a\delta (b)+\delta (a)b=0(a,b\in A).$$

$$a{b}^{\star }=0⟹a\delta (b{)}^{\star }+\delta (a)b^{\star }=0(a,b\in A).$$

$$a{b}^{\star }=0⟹ad(b{)}^{\star }+d(a)b^{\star }=0(a,b\in A),$$

$$abd(c^{\star }{)}^{\star }+d(ab)c=ad(c^{\star }{b}^{\star }{)}^{\star }+d(a)bc,$$

$$u_{i}bd(c^{\star }{)}^{\star }+d(u_{i}b)c=u_{i}d(c^{\star }{b}^{\star }{)}^{\star }+d(u_i)bc.$$

$$d(a{b}^{\star })=d(a)b^{\star }+ad(b{)}^{\star },$$

$$d(u_i{b}^{\star })=d(u_i)b^{\star }+u_{i}d(b{)}^{\star },$$

$$a^{\star }b=0⟹a^{\star }\delta (b)+\delta (a{)}^{\star }b=0(a,b\in A).$$

$$ab=0⟹a\delta (b)+\delta (a)b=0(a,b\in A)$$

$$a{b}^{\star }=0⟹a\delta (b{)}^{\star }+\delta (a)b^{\star }=0(a,b\in A),$$

$$a^{\star }b=0⟹a^{\star }\delta (b)+\delta (a{)}^{\star }b=0(a,b\in A)$$

$$ab=0⟹b\delta (a)+\delta (b)a=0(a,b\in A).$$

$$a{b}^{\star }=0⟹\delta (b{)}^{\star }a+b^{\star }\delta (a)=0(a,b\in A).$$

$$c\delta (ab)+\delta (c)ab=bc\delta (a)+\delta (bc)a,$$

$$u_i\delta (ab)+\delta (u_i)ab=b{u}_i\delta (a)+\delta (b{u}_i)a.$$

$$\delta (ab)=b\delta (a)+\delta (b)a-\eta ab$$

$$bc\delta (u_i)+\delta (bc)u_i=c\delta (u_{i}b)+\delta (c)u_{i}b.$$

$$\delta (ab)=b\delta (a)+\delta (b)a-ab\eta$$

$$\delta (c^{\star }{)}^{\star }ab+c\delta (ab)=\delta (c^{\star }{b}^{\star }{)}^{\star }a+bc\delta (a),$$

Full text

MSC(2010): 46L05; 47B47; 46L57.

Keywords: derivation; anti-derivation; orthogonal elements; $C^{\star}$-algebra.

1. Introduction

Algebras and vector spaces in this paper are assumed to be those over the complex field $\mathbb{C}$. Let $A$ and $M$ be an algebra and an $A$-bimodule, respectively. Recall that a linear map $d:A\rightarrow M$ is said to be a derivation if $d(ab)=ad(b)+d(a)b$ for all $a,b\in A$. Also, $d$ is called inner derivation if for some $m\in M$, $d$ takes the form $d(a)=am-ma$ for all $a\in A$. We also call $d$ anti-derivation if $d(ab)=bd(a)+d(b)a$ for all $a,b\in A$. Derivation is an important field of research, and have been studied and applied extensively both in theory and applications. For this and related topics, see [7, 9, 10] among others.

There are a number of papers investigating the conditions under which mappings of (Banach) algebras are thoroughly determined by actions on some sets of points. We refer the reader to [2, 4, 5, 9, 11, 12, 13, 14] for a full account of the topic and a list of references. The following condition has been drawing many researchers attention working in this field:

[TABLE]

where $z\in A$ is fixed and $\delta:A\rightarrow M$ is a linear (additive) map. Bre$\check{\textrm{s}}$ar, [4] studied the derivations of rings with idempotents with $z=0$. In [4] it was demonstrated that if $A$ is a prime ring containing a non-trivial idempotent and $\delta:A\rightarrow A$ is an additive map satisfying $(\blacklozenge)$ with $z=0$, then $\delta(a)=d(a)+ca$ ($a\in A$) where $d$ is an additive derivation and $c$ is a central element of $A$. Note that the nest algebras are important operator algebras that are not prime. Jing et al. in [15] showed that, for the cases of nest algebras on a Hilbert space and standard operator algebras in a Banach space, the set of linear maps satisfying $(\blacklozenge)$ with $z=0$ and $\delta(I)=0$ coincides with the set of inner derivations. Then, many studies have been done in this case and different results were obtained, for instance, see [2, 3, 4, 5, 8, 9, 10, 19, 20, 21, 22] and the references therein. Recently, in [6] the additive maps on a prime ring acting on some orthogonality condition are described where the ring has an involution and non-trivial idempotents. Also in [3] the authors considered the subsequent condition on a continuous linear map $\delta$ from a $C^{\star}$-algebra $A$ into an essential Banach $A$-bimodule $M$:

[TABLE]

and they showed that there exist a derivation $d:A\rightarrow M$ and a bimodule homomorphism $\Phi:A\rightarrow M$ such that $\delta=d+\Phi$. Motivated by these reasons, in this paper we consider the problem of characterizing continuous linear maps on $C^{\star}$-algebras behaving like derivations or anti-derivations at orthogonal elements for several types of orthogonality conditions.

In this paper we consider the problem of characterizing continuous linear maps behaving like derivations or anti-derivations at orthogonal elements for several types of orthogonality conditions on $C^{\star}$-algebras. In particular, in this paper we consider the subsequent conditions on a continuous linear map $\delta:A\rightarrow A^{**}$ where $A$ is a $C^{\star}$-algebra:

1. (i)

derivations through one-sided orthogonality conditions

[TABLE]

[TABLE]

[TABLE] 2. (ii)

anti-derivations through one-sided orthogonality conditions

[TABLE]

[TABLE]

[TABLE] 3. (iii)

Derivations through two-sided orthogonality conditions

[TABLE]

[TABLE]

where $a,b\in A$. Our purpose is to investigate whether the above conditions characterize continuous derivations ($\star$-derivations) or continuous anti-derivations ($\star$-anti-derivations) on $C^{\star}$-algebras. Also we give applications of our results for von Neumann algebras and unital simple $C^{\star}$-algebras.

The following are the notations and terminologies which are used throughout this article.

Let $A$ be a $C^{\star}$-algebra. For $\mu,\nu\in A^{**}$, we denote by $\mu\nu$ the first Arens product. In view of the fact that any $C^{\star}$-algebra is Arens regular, the product $\mu\nu$ in $A^{**}$ coincide with the second Arens product. By [7, Theorem 2.6.17] the product $\mu\nu$ in $A^{**}$ is separately continuous with respect to the weak∗ topology ($\sigma(A^{**},A^{*})$) on $A^{**}$. The Banach algebra $A^{**}$ is unital by this product and the unity of $A^{**}$ is denoted by $e$.

If a net $(\lambda_{i})_{i\in I}$ in $A^{**}$ converges to $\lambda\in A^{**}$ with respect to the weak*∗* topology, we write it by $\lambda_{i}\xlongrightarrow{\sigma(A^{**},A^{*})}\lambda$.

Remark 1.1*.*

Let $A$ be a $C^{\star}$-algebra, and let $\mu\in A^{**}$. If $\mu A=\{0\}$ or $A\mu=\{0\}$, then $\mu=0$.

Proof.

By Goldstine Theorem there is a net $(a_{i})_{i\in I}$ in $A$ such that $a_{i}\xlongrightarrow{\sigma(A^{**},A^{*})}e$, where $e$ is the unity of $A^{**}$. Let $\mu A=\{0\}$. By separately $w^{*}$-continuity of product in $A^{**}$ we have $\mu a_{i}\xlongrightarrow{\sigma(A^{**},A^{*})}\mu e=\mu$. Hence $\mu=0$. Similarly, we can show that $A\mu=\{0\}$ implies $\mu=0$. ∎

We note that the centre of an algebra $A$ is written by $\mathcal{Z}(A)$.

Remark 1.2*.*

Let $A$ be a $C^{\star}$-algebra, and let $\mu\in A^{**}$. Suppose that $\mu a=a\mu$ for each $a\in A$. Then $\mu\in\mathcal{Z}(A^{**})$.

Proof.

Let $\nu\in A^{**}$. We will show that $\mu\nu=\nu\mu$. By Goldstine Theorem there is a net $(a_{i})_{i\in I}$ in $A$ such that $a_{i}\xlongrightarrow{\sigma(A^{**},A^{*})}\nu$. By separately $w^{*}$-continuity of product in $A^{**}$ we have

[TABLE]

By applying assumption, $\mu\nu=\nu\mu$. ∎

Remark 1.3*.*

Let $A$ be $C^{\star}$-algebra and $(u_{i})_{i\in I}$ be a bounded approximate identity of $A$. Since $(u_{i})_{i\in I}$ is bounded, by Banach-Alaoglu Theorem we can assume that it converges to some $\mu\in A^{**}$ with respect to the weak*∗* topology. So by separately $w^{*}$-continuity of product in $A^{**}$ we have $u_{i}a\xlongrightarrow{\sigma(A^{**},A^{*})}\mu a$ for all $a\in A$. On the other hand by the fact that $(u_{i})_{i\in I}$ is an approximate identity, for each $a\in A$ we get $u_{i}a\xlongrightarrow{\sigma(A^{**},A^{*})}a$ in $A^{**}$. So $(\mu-e)A=\{0\}$ and by Remark 1.1, it follows that $\mu=e$. Therefore we can assume that the $C^{\star}$-algebra $A$ has a bounded approximate identity such that $u_{i}\xlongrightarrow{\sigma(A^{**},A^{*})}e$ in $A^{**}$.

Let $h:A\rightarrow A^{**}$ be a map. We say that $h$ is a $\star$-map whenever $h(a^{\star})=h(a)^{\star}$ for all $a\in A$.

2. Derivations and anti-derivations at orthogonality product

In this section we will consider a continuous linear map on a $C^{\star}$-algebra behaving like derivation or anti-derivation at one-sided orthogonality conditions.

In order to prove our results we need the following result for continuous bilinear maps on $C^{\star}$-algebras.

Lemma 2.1**.**

(Alaminos et al. [2]). Let $A$ be a $C^{\star}$-algebra, let $X$ be a Banach space, and let $\phi:A\times A\rightarrow X$ be a continuous bilinear map such that $\phi(a,b)=0$ whenever $a,b\in A$ are such that $ab=0$, Then $\phi(ab,c)=\phi(a,bc)$ for all $a,b,c\in A$. Also there is a continuous linear map $\Phi:A\rightarrow X$ such that $\phi(a,b)=\Phi(ab)$ for all $a,b\in A$.

It should be noted that in the above lemma if $A$ is a commutative $C^{\star}$-algebra, then $\phi$ is symmetric, that is $\phi(a,b)=\phi(b,a)$ for all $a,b\in A$.

Now, we characterize continuous linear maps on $C^{\star}$-algebras behaving like derivations through one-sided orthogonality conditions.

Theorem 2.2**.**

Let $A$ be a $C^{\star}$-algebra, and let $\delta:A\rightarrow A^{**}$ be a continuous linear map.

1. (i)

$\delta$* satisfies*

[TABLE]

if and only if there is a continuous derivation $d:A\rightarrow A^{**}$ and an element $\eta\in\mathcal{Z}(A^{**})$ such that $\delta(a)=d(a)+\eta a$ for all $a\in A$. 2. (ii)

$\delta$* satisfies*

[TABLE]

if and only if there is a continuous $\star$-derivation $d:A\rightarrow A^{**}$ and an element $\eta\in A^{**}$ such that $\delta(a)=d(a)+\eta a$ for all $a\in A$.

Proof.

$(i)$ By [2, Theorem 4.6] and Remark 1.1, there is a continuous derivation $d:A\rightarrow A^{**}$ and an element $\eta\in\mathcal{Z}(A^{**})$ such that $\delta(a)=d(a)+\eta a$ for all $a\in A$. The converse is proved easily.

$(ii)$ Suppose that $(u_{i})_{i\in I}$ is a bounded approximate identity of $A$ such that $u_{i}\xlongrightarrow{\sigma(A^{**},A^{*})}e$, where $e$ is the identity of $A^{**}$. Since the net $(\delta(u_{i}))_{i\in I}$ is bounded, we can assume that it converges to some $\eta\in A^{**}$ with respect to the weak*∗* topology. Define $d:A\rightarrow A^{**}$ by $d(a)=\delta(a)-\eta a$. Then $d$ is a continuous linear map which satisfies

[TABLE]

and $d(u_{i})\xlongrightarrow{\sigma(A^{**},A^{*})}0$. We will show that $d$ is a $\star$-derivation. In order to prove this we consider the continuous bilinear map $\phi:A\times A\rightarrow A^{**}$ by $\phi(a,b)=ad(b^{\star})^{\star}+d(a)b$. If $a,b\in A$ are such that $ab=0$, then $a(b^{\star})^{\star}=0$ and (2.1) gives $\phi(a,b)=0$. So by Lemma 2.1, we get $\phi(ab,c)=\phi(a,bc)$ for all $a,b,c\in A$. Therefore

[TABLE]

for all $a,b,c\in A$. On account of (2.2), for all $b,c\in A$ we have

[TABLE]

From continuity of $d$, we get $u_{i}d(c^{\star}b^{\star})^{\star}+d(u_{i})bc$ converges to $bd(c^{\star})^{\star}+d(b)c$ with respect to the norm topology. On the other hand, from separately $w^{*}$-continuity of product in $A^{**}$ and $d(u_{i})\xlongrightarrow{\sigma(A^{**},A^{*})}0$ it follows that $u_{i}d(c^{\star}b^{\star})^{\star}+d(u_{i})bc\xlongrightarrow{\sigma(A^{**},A^{*})}d(c^{\star}b^{\star})^{\star}$. Hence

[TABLE]

for all $a,b\in A$. Now letting $a=u_{i}$ in (2.3), we obtain

[TABLE]

for all $b\in A$. By continuity of $d$, $d(u_{i})\xlongrightarrow{\sigma(A^{**},A^{*})}0$ and using similar arguments as above it follows that $d(b^{\star})=d(b)^{\star}$ for all $b\in A$. Thus from (2.3), $d$ is a $\star$-derivation.

The converse is proved easily. ∎

Corollary 2.3**.**

Let $A$ be a $C^{\star}$-algebra, and let $\delta:A\rightarrow A^{**}$ be a continuous linear map. Then

[TABLE]

if and only if there is a continuous $\star$-derivation $d:A\rightarrow A^{**}$ and an element $\eta\in A^{**}$ such that $\delta(a)=d(a)+a\eta$ for all $a\in A$.

Proof.

Consider the continuous linear map $\tau:A\rightarrow A^{**}$ defined by $\tau(a)=\delta(a^{\star})^{\star}$. It is easily seen that this map satisfies the conditions of Theorem 2.2-$(ii)$. So there exists a continuous $\star$-derivation $d:A\rightarrow A^{**}$ and an element $\eta_{1}\in A^{**}$ such that $\tau(a)=d(a)+\eta_{1}a$ for all $a\in A$. Then $\delta(a)=d(a)+a\eta$ for all $a\in A$, where $\eta=\eta^{\star}_{1}$.

The converse is proved easily. ∎

Note that in above corollary and part $(ii)$ of Theorem 2.2, it is not necessary true that $\eta\in\mathcal{Z}(A^{**})$. For example suppose that $\eta$ is not in $\mathcal{Z}(A^{**})$ and define $\delta:A\rightarrow A^{**}$ by $\delta(a)=\eta a$. Then $\delta$ satisfies in conditions of Theorem 2.2 and $\delta$ equals to sum of the zero derivation and $\eta a$, but $\eta$ is not belong to $\mathcal{Z}(A^{**})$.

Remark 2.4*.*

Let $A$ be a $C^{\star}$-algebra and $d:A\rightarrow A$ be an inner derivation where $d(a)=a\mu-\mu a$ for some $\mu\in A$. If $d$ is a $\star$-map, then $a^{\star}\mu-\mu a^{\star}=\mu^{\star}a^{\star}-a^{\star}\mu^{\star}$ for all $a\in A$. So $Re\mu=\dfrac{1}{2}(\mu+\mu^{\star})\in\mathcal{Z}(A)$. Conversely for $\mu\in A$ with $Re\mu\in\mathcal{Z}(A)$, the map $d:A\rightarrow A$ defined by $d(a)=a\mu-\mu a$ is a $\star$-inner derivation.

If $A$ is a von Neumann algebra or a simple $C^{\star}$-algebra with unity, then by [18, Theorem 4.1.6] and [18, Theorem 4.1.11] every derivation $d:A\rightarrow A$ is an inner derivation. In view of these results, we have the next proposition.

Proposition 2.5**.**

Let $A$ be a von Neumann algebra or a simple $C^{\star}$-algebra with unity. Suppose that $\delta:A\rightarrow A$ is a continuous linear map. Then

1. (i)

$\delta$* satisfies*

[TABLE]

if and only if there are elements $\mu,\nu\in A$ such that $\delta(a)=a\mu-\nu a$ for all $a\in A$ and $\mu-\nu\in\mathcal{Z}(A)$. 2. (ii)

$\delta$* satisfies*

[TABLE]

if and only if there are elements $\mu,\nu\in A$ such that $\delta(a)=a\mu-\nu b$ for all $a\in A$ and $Re\mu\in\mathcal{Z}(A)$. 3. (iii)

$\delta$* satisfies*

[TABLE]

if and only if there are elements $\mu,\nu\in A$ such that $\delta(a)=a\nu-\mu a$ for all $a\in A$ and $Re\mu\in\mathcal{Z}(A)$.

Proof.

$(i)$ Suppose that $\delta$ satisfies the given condition. By Theorem 2.2-$(i)$, there exists a continuous derivation $d:A\rightarrow A^{**}$ and an element $\eta\in\mathcal{Z}(A^{**})$ such that $\delta(a)=d(a)+\eta a$ for all $a\in A$. Since $d$ is an derivation and $A$ is unital, we have $d(1)=0$, so $\delta(1)=\eta\in A$ and hence $d(A)\subseteq A$, and also $\eta\in\mathcal{Z}(A)$. Since every derivation on $A$ is inner, it follows that $d(a)=a\mu-\mu a$ for all $a\in A$. Setting $\nu=\mu-\eta$. So $\delta(a)=a\mu-\nu a$ for all $a\in A$ and $\mu-\nu\in\mathcal{Z}(A)$.

The converse is proved easily.

$(ii)$ Let $\delta$ satisfies $(ii)$. By Theorem 2.2-$(ii)$, there exists a continuous $\star$-derivation $d:A\rightarrow A^{**}$ and an element $\eta\in A^{**}$ such that $\delta(a)=d(a)+\eta a$ for all $a\in A$. By using similar methods as in proof of part $(i)$, $d:A\rightarrow A$ is a $\star$-derivation, and hence it is inner. Now by Remark 2.4, there exists an element $\mu\in A$ with $Re\mu\in\mathcal{Z}(A)$ such that $d(a)=a\mu-\mu a$ for all $a\in A$, where $\mu\in A$. Setting $\nu=\mu-\eta$. So we have $\delta(a)=a\mu-\nu a$ for all $a\in A$ and $Re\mu\in\mathcal{Z}(A)$.

The converse is proved easily.

$(iii)$ By using Corollary 2.3 and a similar proof as that of part $(ii)$, we can obtain this result. ∎

Let $A$ be an algebra and $M$ be an $A$-bimodule. Recall that a linear map $d:A\rightarrow M$ is said to be a Jordan derivation if $d(a^{2})=ad(a)+d(a)a$ for all $a\in A$. Clearly, each derivation is a Jordan derivation. The converse is not true, in general. Johnson in [17] has shown that any continuous Jordan derivation from a $C^{\star}$-algebra $A$ into any Banach $A$-bimodule is a derivation.

In the next theorem we characterize anti-derivations through one-sided orthogonality conditions.

Theorem 2.6**.**

Let $A$ be a $C^{\star}$-algebra, and let $\delta:A\rightarrow A^{**}$ be a continuous linear map.

1. (i)

Assume that

[TABLE]

Then there is a continuous derivation $d:A\rightarrow A^{**}$ and an element $\eta\in\mathcal{Z}(A^{**})$ such that $\delta(a)=d(a)+\eta a$ for all $a\in A$. 2. (ii)

Assume that

[TABLE]

Then there is a continuous derivation $d:A\rightarrow A^{**}$ and an element $\eta\in A^{**}$ such that $\delta(a)=d(a)+a\eta$ for all $a\in A$.

Proof.

Suppose that $(u_{i})_{i\in I}$ is a bounded approximate identity of $A$ such that $u_{i}\xlongrightarrow{\sigma(A^{**},A^{*})}e$, where $e$ is the identity of $A^{**}$.

$(i)$ Define a continuous bilinear map $\phi:A\times A\rightarrow A^{**}$ by $\phi(a,b)=b\delta(a)+\delta(b)a$. Then $\phi(a,b)=0$ for all $a,b\in A$ with $ab=0$. By applying Lemma 2.1, we obtain $\phi(ab,c)=\phi(a,bc)$ for all $a,b,c\in A$. So

[TABLE]

for all $a,b,c\in A$. Since the net $(\delta(u_{i}))_{i\in I}$ is bounded, we can assume that it converge to some $\eta\in A^{**}$ with respect to the weak*∗* topology. On account of (2.4), for all $a,b\in A$ we have

[TABLE]

From continuity of $\delta$, we get $bu_{i}\delta(a)+\delta(bu_{i})a$ converges to $b\delta(a)+\delta(b)a$ with respect to the norm topology. On the other hand, by separately $w^{*}$-continuity of product in $A^{**}$, it follows that $u_{i}\delta(ab)+\delta(u_{i})ab$ converges to $\delta(ab)+\eta ab$ with respect to the weak*∗* topology. Hence

[TABLE]

for all $a,b\in A$. Now letting $a=u_{i}$ in (2.4), we obtain

[TABLE]

By this identity and using similar arguments as above it follows that

[TABLE]

for all $a,b\in A$. Hence from (2.5) and (2.6), for each $a,b\in A$, we find that $\mu ab=ab\eta$. So by the fact that $A^{2}=A$ and Remark 1.2, it follows that $\eta\in\mathcal{Z}(A^{**})$. Define $d:A\rightarrow A^{**}$ by $d(a)=\delta(a)-\eta a$. The linear map $d$ is continuous and by (2.5) and the fact that $\eta\in\mathcal{Z}(A^{**})$, it follows that $d$ is an Jordan derivation. From [17], $d$ is a derivation.

$(ii)$ In order to prove this we consider the continuous bilinear map $\phi:A\times A\rightarrow A^{**}$ defined by $\phi(a,b)=\delta(b^{\star})^{\star}a+b\delta(a)$. If $a,b\in A$ are such that $ab=0$, then $\phi(a,b)=0$. So by Lemma 2.1, we get $\phi(ab,c)=\phi(a,bc)$ for all $a,b,c\in A$. Therefore

[TABLE]

for all $a,b,c\in A$. Setting $a=u_{i}$ in (2.7) and by using similar methods as in part $(i)$, we get

[TABLE]

for all $b,c\in A$, where $\eta\in A^{**}$ and $\delta(u_{i})\xlongrightarrow{\sigma(A^{**},A^{*})}\eta$. By (2.8) we have

[TABLE]

for all $b,c\in A$. Letting $c^{\star}=u_{i}$, we arrive at

[TABLE]

for all $b\in A$. Hence

[TABLE]

for all $b\in A$. From (2.9) we have

[TABLE]

for all $a,b\in A$. Define $d:A\rightarrow A^{**}$ by $d(a)=\delta(a)-a\eta$. The linear map $d$ is continuous and from (2.10), it follows that $d$ is a $\star$-map. Now from (2.11) we have

[TABLE]

for all $a\in A$. Thus $d$ is a continuous $\star$-derivation and by [17], $d$ is a $\star$-derivation. ∎

Corollary 2.7**.**

Let $A$ be a $C^{\star}$-algebra, and let $\delta:A\rightarrow A^{**}$ be a continuous linear map. Suppose that

[TABLE]

Then there is a continuous $\star$-derivation $d:A\rightarrow A^{**}$ and an element $\eta\in A^{**}$ such that $\delta(a)=d(a)+\eta a$ for all $a\in A$.

Proof.

Consider the continuous linear map $\tau:A\rightarrow A^{**}$ defined by $\tau(a)=\delta(a^{\star})^{\star}$. It is easily seen that the map satisfies in conditions of Theorem 2.6-$(ii)$. So there exists a continuous $\star$-derivation $d:A\rightarrow A^{**}$ and an element $\eta_{1}\in A^{**}$ such that $\tau(a)=d(a)+a\eta_{1}$ for all $a\in A$. Then $\delta(a)=d(a)+\eta a$ for all $a\in A$, where $\eta=\eta_{1}^{\star}$. ∎

Note that in part (ii) of Theorem 2.6 and above corollary, it is not necessary true that $\eta\in\mathcal{Z}(A^{**})$. For example suppose that $\eta$ is not in $\mathcal{Z}(A^{**})$ and $\eta^{\star}[a,b]+[a,b]\eta=0$. Define $\delta:A\rightarrow A^{**}$ by $\delta(a)=a\eta$. Then $\delta$ satisfies in condition of Theorem 2.6-$(ii)$, but $\delta$ equals to sum of the zero derivation and $a\eta$, while $\eta$ is not belong to $\mathcal{Z}(A^{**})$.

Proposition 2.8**.**

Let $A$ be a von Neumann algebra or a simple $C^{\star}$-algebra with unity. Suppose that $\delta:A\rightarrow A$ is a continuous linear map. Then

1. (i)

$\delta$* satisfies*

[TABLE]

if and only if there are elements $\mu,\nu\in A$ such that $\delta(a)=a\mu-\nu a$, where $\mu-\nu\in\mathcal{Z}(A)$ and

[TABLE]

for all $a,b\in A$. 2. (ii)

$\delta$* satisfies*

[TABLE]

if and only if there are elements $\mu,\nu\in A$ such that $\delta(a)=a\nu-\mu a$ for all $a\in A$ and $Re\mu\in\mathcal{Z}(A)$ and

[TABLE]

for all $a,b\in A$ . 3. (iii)

$\delta$* satisfies*

[TABLE]

if and only if there are elements $\mu,\nu\in A$ such that $\delta(a)=a\mu-\nu a$ for all $a\in A$ and $Re\mu\in\mathcal{Z}(A)$ and

[TABLE]

for all $a,b\in A$ .

Proof.

$(i)$ Let $\delta$ satisfies $(i)$. By Theorem 2.6-$(i)$, there is a continuous derivation $d:A\rightarrow A^{**}$ and an element $\eta\in\mathcal{Z}(A^{**})$ such that $\delta(a)=d(a)+\eta a$ for all $a\in A$. Since $d(1)=0$, we have $\eta\in\mathcal{Z}(A)$, and $d$ is a derivation on $A$. By the fact that every derivation on $A$ is inner, it follows that $d(a)=a\mu-\mu a$ for all $a\in A$, where $\mu\in A$. Setting $\nu=\mu-\eta$. So $\delta(a)=a\mu-\nu a$ for all $a\in A$ and $\mu-\nu\in\mathcal{Z}(A)$.

Now by (2.6) and the fact that $d$ is a derivation we see that

[TABLE]

for all $a,b\in A$. So

[TABLE]

and hence

[TABLE]

for all $a,b\in A$. Therefore

[TABLE]

for all $a,b\in A$.

Conversely, suppose that there are elements $\mu,\nu\in A$ such that $\delta(a)=a\mu-\nu a$, where $\mu-\nu\in\mathcal{Z}(A)$ and

[TABLE]

for all $a,b\in A$. From this equation for all $a,b\in A$ with $ab=0$, we have

[TABLE]

Since $\mu-\nu\in\mathcal{Z}(A)$, it follows that

[TABLE]

Therefore

[TABLE]

for all $a,b\in A$ with $ab=0$.

$(ii)$ By Theorem 2.6-(ii), there is a continuous $\star$-derivation $d:A\rightarrow A^{**}$ and an element $\eta\in A^{**}$ such that $\delta(a)=d(a)+a\eta$ for all $a\in A$, and by using similar arguments as above, it follows that $\eta\in A$ and $d$ is a derivation on $A$. The derivation $d$ is an inner derivation and by Remark 2.4, there is an element $\mu\in A$ with $Re\mu\in\mathcal{Z}(A)$, such that $d(a)=a\mu-\mu a$ for all $a\in A$.

Now by (2.11) and the fact that $d$ is a $\star$-derivation, we have

[TABLE]

for all $a,b\in A$. So from the fact that $\delta(a)=d(a)+a\eta=a\mu-\mu a+a\eta$ for all $a\in A$, we have

[TABLE]

and hence

[TABLE]

for all $a,b\in A$. By setting $\nu=\mu+\eta$, we have $\delta(a)=a\nu-\mu a$ and

[TABLE]

for all $a,b\in A$, where $Re\mu\in\mathcal{Z}(A)$.

Conversely, suppose that there are elements $\mu,\nu\in A$ such that $\delta(a)=a\nu-\mu a$ for all $a\in A$ and $Re\mu\in\mathcal{Z}(A)$ and

[TABLE]

for all $a,b\in A$ . By this equation for all $a,b\in A$ with $ab^{\star}=0$, we have

[TABLE]

So by $Re\mu\in\mathcal{Z}(A)$ we get

[TABLE]

for all $a,b\in A$ with $ab^{\star}=0$.

$(iii)$ Define the map $d:A\rightarrow A$ by $d(a)=\delta(a^{\star})^{\star}$. It is easily seen that the map $d$ satisfies in conditions of part $(ii)$. So, there exists $\mu_{1},\nu_{1}\in A$ such that $d(a)=a\nu_{1}-\mu_{1}a$ for all $a\in A$ with $Re\mu_{1}\in\mathcal{Z}(A)$ and

[TABLE]

for all $a,b\in A$. Then $\delta(a)=a\mu-\nu a$ for all $a\in A$, where $\nu=-\nu_{1}^{\star}$, $\mu=-\mu_{1}^{\star}$ with $Re\mu\in\mathcal{Z}(A)$ and

[TABLE]

for all $a,b\in A$.

The converse is proved by a similar method as in part $(ii)$. ∎

3. Derivations through two-sided orthogonality conditions

In this section we will consider a linear map behaving like derivation at two-sided orthogonality conditions. In order to prove our results we need the following result.

Lemma 3.1**.**

Let $A$ be a unital $C^{\star}$-algebra, let $X$ be a Banach space, and let $\phi:A\times A\rightarrow X$ be a continuous bilinear map satisfying

[TABLE]

Then for all $a,c\in\mathcal{Z}(A)$ and $b\in A$ we have

[TABLE]

Proof.

Let $a,c\in\mathcal{Z}(A)$, and $b$ be a self-adjoint element i $A$. Let $I$ be a compact interval of $\mathbb{R}$ containing the spectrum of $b$. Define $\psi:C(I)\times C(I)\rightarrow X$ by

[TABLE]

If $f,g\in C(I)$ are such that $fg=0$, then $af(b)g(b)c=g(b)caf(b)=g(b)f(b)ca=0$ and so $\psi(f,g)=0$. On account of Lemma 2.1, $\psi$ is symmetric, i.e. $\psi(f,g)=\psi(g,f)$ for all $f,g\in C(I)$, hence

[TABLE]

Now, for $f(s)=1$ and $g(s)=s$ we obtain $\phi(a,bc)=\phi(ab,c)$, which readily implies the desired conclusion (since $b$ is self-adjoint). ∎

In continue we give the main results of this section.

Proposition 3.2**.**

Let $A$ be a $C^{\star}$-algebra, and let $\delta:A\rightarrow A^{**}$ be a continuous linear map. Assume that

[TABLE]

Then there is a continuous derivation $d:A\rightarrow A^{**}$ and an element $\eta\in\mathcal{Z}(A^{**})$ such that $\delta(a)=d(a)+\eta a$ for all $a\in A$.

Proof.

For all $a,b\in A$ with $ab=ba=0$, we have

[TABLE]

So by [3, Theorem 4.1], there exist a continuous derivation $d:A\rightarrow A^{**}$ and a bimodule homomorphism $\Phi:A\rightarrow A^{**}$ such that $\delta=d+\Phi$.

Suppose that $(u_{i})_{i\in I}$ is a bounded approximate identity of $A$, By [16], $\Phi$ is continuous and so the net $\{\Phi(u_{i})\}_{i\in I}$ in $A^{**}$ is bounded. Hence we can assume that $\Phi(u_{i})\xlongrightarrow{\sigma(A^{**},A^{*})}\eta$ for some $\eta\in A^{**}$. Now, for all $a\in A$ we have $u_{i}a\rightarrow a$ and $au_{i}\rightarrow a$. Thus $\Phi(u_{i}a)\rightarrow\Phi(a)$ and $\Phi(au_{i})\rightarrow\Phi(a)$. On the other hand by separately $w^{*}$-continuity of product in $A^{**}$, we see that

[TABLE]

for all $a\in A$. Since $\Phi$ is a bimodule homomorphism, it follows that

[TABLE]

and so

[TABLE]

for all $a\in A$. Also from Remark 1.2, $\eta\in\mathcal{Z}(A^{**})$. It is clear that $d$ is continuous. ∎

Theorem 3.3**.**

Let $A$ be a unital $C^{\star}$-algebra, and let $\delta:A\rightarrow A^{**}$ be a continuous linear map. Assume that

[TABLE]

Then there are continuous $\star$-derivations $d_{1},d_{2}:A\rightarrow A^{**}$ and an element $\eta\in A^{**}$ with $Re\eta\in\mathcal{Z}(A)$ such that $\delta(a)=d_{1}(a)+\eta a=d_{2}(a)+a\eta$ for all $a\in A$.

Proof.

Define continuous bilinear maps $\phi,\psi:A\times A\rightarrow A^{**}$ by

[TABLE]

It is easily seen that $\phi(a,b)=0$ and $\psi(a,b)=0$, whenever $a,b\in A$ are such that $ab=ba=0$. By Lemma 3.1 we have

[TABLE]

[TABLE]

for all $a,c\in\mathcal{Z}(A)$ and $b\in A$. Now letting $a=1$ and $\eta=\delta(1)$ in (3.1) and (3.2), we obtain

[TABLE]

[TABLE]

for all $c\in\mathcal{Z}(A)$ and $b\in A$. By applying the $\star$ on above equations, and setting $c^{\star}=1$, we arrive at

[TABLE]

and

[TABLE]

for all $b\in A$. By (3.3) and (3.4), we get $(\eta+\eta^{\star})b^{\star}=b^{\star}(\eta+\eta^{\star})$ for all $b\in A$. Therefore by Remark 1.2, $Re\eta\in\mathcal{Z}(A^{**})$. Define the map $d_{1}:A\rightarrow A^{**}$ by $d_{1}(a)=\delta(a)-\eta a$. Then $d_{1}$ is a continuous linear map, and by (3.3) we have $d_{1}(a^{\star})=d_{1}(a)^{\star}$ for all $a\in A$. So $d$ is a $\star$-map. If $ab=ba=0$, then by hypothesis, definition of $d_{1}$ and the fact that $d_{1}$ is a $\star$-map and $Re\eta\in\mathcal{Z}(A^{**})$, we have

[TABLE]

and

[TABLE]

So $d_{1}$ satisfies in conditions of Proposition 3.2 and hence there exist a continuous derivation $d:A\rightarrow A^{**}$ and an element $\eta\in\mathcal{Z}(A^{**})$ such that

[TABLE]

By definition of $d_{1}$, we have $d_{1}(1)=0$, on the other hand $d$ is a derivation, so $d(1)=0$. Hence $\eta=0$ and $d=d_{1}$. Hence $d_{1}$ is a $\star$-derivation and so

[TABLE]

with $Re\eta\in\mathcal{Z}(A^{**})$. By defining $d_{2}(a)=\delta(a)-a\eta$ and by using similar arguments as above, it follows that $d_{2}$ is a continuous $\star$-derivation. ∎

If $\delta(1)\in\mathcal{Z}(A^{**})$, it is obvious that $d_{1}=d_{2}$ in above theorem. Indeed, in this case there is a derivation $d:A\rightarrow A^{**}$ such that $\delta(a)=d(a)+\delta(1)a$ for all $a\in A$. Note that in this theorem, it is not necessary true that $\eta\in\mathcal{Z}(A^{**})$. For example suppose that $\eta\in A^{**}$, $\eta$ is not in $\mathcal{Z}(A^{**})$ and $Re\eta\in\mathcal{Z}(A^{**})$. Define $\delta:A\rightarrow A^{**}$ by $\delta(a)=\eta a$. Then $\delta$ satisfies in conditions of above theorem, but $\delta$ equals to sum of the zero derivation and $\eta a$, while $\eta$ is not belong to $\mathcal{Z}(A^{**})$.

In the next proposition we consider von Neumann algebras or simple $C^{\star}$-algebras with unity.

Proposition 3.4**.**

Let $A$ be a von Neumann algebra or a simple $C^{\star}$-algebra with unity. Suppose that $\delta:A\rightarrow A$ is a continuous linear map. Then

1. (i)

$\delta$* satisfies*

[TABLE]

if and only if there are elements $\mu,\nu\in A^{**}$ such that $\delta(a)=a\mu-\nu a$ for all $a\in A$, where $\mu-\nu\in\mathcal{Z}(A^{**})$. 2. (ii)

$\delta$* satisfies*

[TABLE]

if and only if there are elements $\mu,\nu\in A^{**}$ such that $\delta(a)=a\mu-\nu a$ for all $a\in A$, where $Re\mu\in\mathcal{Z}(A^{**})$ and $Re(\mu-\nu)\in\mathcal{Z}(A^{**})$.

Proof.

$(i)$ By Proposition 3.2, there exist a continuous derivation $d:A\rightarrow A^{**}$ and an element $\eta\in\mathcal{Z}(A^{**})$ such that

[TABLE]

By using similar arguments as above, it follows that $\eta\in\mathcal{Z}(A)$ and $d:A\rightarrow A$ is a derivation. So $d$ is inner, and $d(a)=a\mu-\mu a$ for all $a\in A$ whenever $\mu\in A$. Setting $\nu=\mu-\eta$. Thus $\delta(a)=a\mu-\nu a$ for all $a\in A$ and $\mu-\nu\in\mathcal{Z}(A)$.

The converse is proved easily.

$(ii)$ By Theorem 3.3, there exist a continuous $\star$-derivation $d_{1}:A\rightarrow A^{**}$ and an element $\eta\in\mathcal{Z}(A^{**})$ such that $\delta(a)=d_{1}(a)+\eta a$ for all $a\in A$. By using similar arguments as above, it follows that $\eta\in A$ whenever $Re\eta\in\mathcal{Z}(A)$ and $d_{1}:A\rightarrow A$ is a $\star$-derivation. Therefore $d_{1}$ is inner, and there is an element $\mu\in A$ with $Re\mu\in\mathcal{Z}(A)$ such that $d_{1}(a)=a\mu-\mu a$ for all $a\in A$. Taking $\nu=\mu-\eta$. Thus $\delta(a)=a\mu-\nu a$ for all $a\in A$ whenever $Re\mu\in\mathcal{Z}(A)$ and $Re(\mu-\nu)\in\mathcal{Z}(A)$.

Conversely, suppose that there are elements $\mu,\nu\in A^{**}$ such that $\delta(a)=a\mu-\nu a$ for all $a\in A$, where $Re\mu\in\mathcal{Z}(A^{**})$ and $Re(\mu-\nu)\in\mathcal{Z}(A^{**})$.

By $Re\mu\in\mathcal{Z}(A^{**})$, for $a,b\in A$ with $ab^{\star}=b^{\star}a=0$ we have

[TABLE]

Also for $a,b\in A$ with $ab^{\star}=b^{\star}a=0$, by $Re\mu\in\mathcal{Z}(A^{**})$ and $Re(\mu-\nu)\in\mathcal{Z}(A^{**})$ we arrive at

[TABLE]

∎

Note that in Proposition 3.4 the converses of Results 3.2 and 3.3 hold, but we do not know the converses of these results are true or not, in general.

Acknowledgment

The author thanks the referee for careful reading of the manuscript and for helpful suggestions.