Characterizing linear mappings through zero products or zero Jordan products

Abstract

Let ${\mathcal {A}}$ be a $*$-algebra and ${{\mathcal {M}}}$ be a $*$-${\mathcal {A}}$-bimodule. We study the local properties of $*$-derivations and $*$-Jordan derivations from ${\mathcal {A}}$ into ${{\mathcal {M}}}$ under the following orthogonality conditions on elements in ${\mathcal {A}}$: $ab^*=0$, $ab^*+b^*a=0$ and $ab^*=b^*a=0$. We characterize the mappings on zero product determined algebras and zero Jordan product determined algebras. Moreover, we give some applications on $C^*$-algebras, group algebras, matrix algebras, algebras of locally measurable operators and von Neumann algebras.

1 Introduction

Throughout this paper, let ${\mathcal {A}}$ be an associative algebra over the complex field ${\mathbb {C}}$ and ${{\mathcal {M}}}$ be an ${\mathcal {A}}$-bimodule. For each a, b in ${\mathcal {A}}$, we define the Jordan product by $a\circ b=ab+ba$. A linear mapping $\delta$ from ${\mathcal {A}}$ into ${{\mathcal {M}}}$ is called a derivation if $\delta (ab)=a\delta (b)+\delta (a)b$ for each a, b in ${\mathcal {A}}$; and $\delta$ is called a Jordan derivation if $\delta (a\circ b)=a\circ \delta (b)+\delta (a)\circ b$ for each a, b in ${\mathcal {A}}$. It follows from the results in [9, 20, 21] that every Jordan derivation from a $C^*$-algebra into its Banach bimodule is a derivation.

By an involution on an algebra ${\mathcal {A}}$ we mean a mapping $*$ from ${\mathcal {A}}$ into itself such that

$$\begin{aligned} (\lambda a+\mu b)^*=\bar{\lambda }a^*+\bar{\mu }b^*,~(ab)^*=b^*a^*~{\mathrm {and}}~(a^*)^*=a, \end{aligned}$$

whenever $a,b\in {\mathcal {A}}$, $\lambda ,\mu \in {\mathbb {C}}$ and $\bar{\lambda },\bar{\mu }$ denote the conjugate complex numbers. An algebra ${\mathcal {A}}$ equipped with an involution is called a $*$-algebra. Moreover, if ${\mathcal {A}}$ is a $*$-algebra, then an ${\mathcal {A}}$-bimodule ${{\mathcal {M}}}$ is called a $*$-${\mathcal {A}}$-bimodule if ${{\mathcal {M}}}$ is equipped with a $*$-mapping from ${{\mathcal {M}}}$ into itself such that

$$\begin{aligned} (\lambda m+\mu n)^*={\bar{\lambda }}m^*+{\bar{\mu }}n^*,~(am)^*=m^*a^*,~(ma)^*=a^*m^*~{\mathrm {and}}~(m^*)^*=m, \end{aligned}$$

whenever $a\in {\mathcal {A}}$, $m,n\in {\mathcal {M}}$ and $\lambda ,\mu \in {\mathbb {C}}$. An element a in ${\mathcal {A}}$ is called self-adjoint if $a^*=a$; an element p in ${\mathcal {A}}$ is called an idempotent if $p^2=p$; and p is called a projection if p is both a self-adjoint element and an idempotent.

In [24], A. Kishimoto studied the $*$-derivations on a $C^*$-algebra, and proved that the closure of a normal $*$-derivation of a UHF algebra satisfying a special condition is a generator of a one-parameter group of $*$-automorphisms. Let ${\mathcal {A}}$ be a $*$-algebra and ${{\mathcal {M}}}$ be a $*$-${\mathcal {A}}$-bimodule. A derivation $\delta$ from ${\mathcal {A}}$ into ${\mathcal {M}}$ is called a $*$-derivation if $\delta (a^*)=\delta (a)^*$ for every a in ${\mathcal {A}}$. Obviously, every derivation $\delta$ is a linear combination of two $*$-derivations. In fact, we can define a linear mapping $\hat{\delta }$ from ${\mathcal {A}}$ into ${\mathcal {M}}$ by $\hat{\delta }(a)=\delta (a^*)^*$ for every a in ${\mathcal {A}}$, therefore $\delta =\delta _1+i\delta _2$, where $\delta _1=\frac{1}{2}(\delta +\hat{\delta })$ and $\delta _2=\frac{1}{2i}(\delta -\hat{\delta })$. It is easy to show that $\delta _1$ and $\delta _2$ are both $*$-derivations. We can define $*$-Jordan derivations similarly.

For $*$-derivations and $*$-Jordan derivations, in [3, 13, 17, 18], the authors characterized the following two conditions on a linear mapping $\delta$ from a $*$-algebra ${\mathcal {A}}$ into its $*$-bimodule ${\mathcal {M}}$:

$$\begin{aligned}&({\mathbb {D}}_{1})~a,b\in {\mathcal {A}},~ab^*=0\ {\mathrm {implies}}\ a\delta (b)^*+\delta (a)b^*=0;\\&({\mathbb {D}}_{2})~a,b\in {\mathcal {A}},~ab^*=b^*a=0\ {\mathrm {implies}}\ a\delta (b)^*+\delta (a)b^*=\delta (b)^*a+b^*\delta (a)=0, \end{aligned}$$

where ${\mathcal {A}}$ is a $C^*$-algebra, a zero product determined algebra or a group algebra $L^1(G)$.

Let ${\mathcal {J}}$ be an ideal of ${\mathcal {A}}$. We say that ${\mathcal {J}}$ is a right separating set or left separating set of ${\mathcal {M}}$ if for every m in ${\mathcal {M}}$, ${\mathcal {J}}m=\{0\}$ implies $m=0$ or $m{\mathcal {J}}=\{0\}$ implies $m=0$, respectively. We denote by ${{\mathfrak {J}}}({\mathcal {A}})$ the subalgebra of ${\mathcal {A}}$ generated algebraically by all idempotents in ${\mathcal {A}}$.

In Sect. 2, we suppose that ${\mathcal {A}}$ is a $*$-algebra and ${\mathcal {M}}$ is a $*$-${\mathcal {A}}$-bimodule that satisfy one of the following conditions:

• (1) ${\mathcal {A}}$ is a zero product determined Banach $*$-algebra with a bounded approximate identity and ${{\mathcal {M}}}$ is an essential Banach $*$-${\mathcal {A}}$-bimodule;

• (2) ${\mathcal {A}}$ is a von Neumann algebra and ${{\mathcal {M}}}={\mathcal {A}}$;

• (3) ${\mathcal {A}}$ is a unital $*$-algebra and ${\mathcal {M}}$ is a unital $*$-${\mathcal {A}}$-bimodule with a left or right separating set ${\mathcal {J}}\subseteq {\mathfrak {J}}({\mathcal {A}})$;

and we investigate whether the linear mappings from ${\mathcal {A}}$ into ${{\mathcal {M}}}$ satisfying condition ${\mathbb {D}}_{1}$ characterize $*$-derivations. In particular, we generalize some results from [13, 17, 18].

An ${\mathcal {A}}$-bimodule ${\mathcal {M}}$ is said to have property ${\mathbb {M}}$, if there is an ideal ${\mathcal {J}}\subseteq {{\mathfrak {J}}}({\mathcal {A}})$ of ${\mathcal {A}}$ such that

$$\begin{aligned} \{m\in {\mathcal {M}}: xmx=0~{\mathrm {for~every}}~x\in {\mathcal {J}}\}=\{0\}. \end{aligned}$$

It is clear that if ${\mathcal {A}}={\mathfrak {J}}({\mathcal {A}})$, then ${\mathcal {M}}$ has property ${\mathbb {M}}$.

For $*$-Jordan derivations, we can study the following conditions on a linear mapping $\delta$ from a $*$-algebra ${\mathcal {A}}$ into its $*$-${\mathcal {A}}$-bimodule ${{\mathcal {M}}}$:

$$\begin{aligned}&({\mathbb {D}}_{3})~a,b\in {\mathcal {A}},~a\circ b^*=0\ {\mathrm {implies}}\ a\circ \delta (b)^*+\delta (a)\circ b^*=0,\\&({\mathbb {D}}_{4})~a,b\in {\mathcal {A}},~ab^*=b^*a=0\ {\mathrm {implies}}\ a\circ \delta (b)^*+\delta (a)\circ b^*=0. \end{aligned}$$

It is obvious that condition ${\mathbb {D}}_{2}$ or ${\mathbb {D}}_{3}$ implies condition ${\mathbb {D}}_{4}$.

In Sect. 3, we suppose that ${\mathcal {A}}$ is a $*$-algebra and ${\mathcal {M}}$ is a $*$-${\mathcal {A}}$-bimodule that satisfy one of the following conditions:

• (1) ${\mathcal {A}}$ is a unital zero Jordan product determined $*$-algebra and ${{\mathcal {M}}}$ is a unital $*$-${\mathcal {A}}$-bimodule;

• (2) ${\mathcal {A}}$ is a unital $*$-algebra and ${\mathcal {M}}$ is a unital $*$-${\mathcal {A}}$-bimodule such that the property ${\mathbb {M}}$;

• (3) ${\mathcal {A}}$ is a $C^*$-algebra (not necessary unital) and ${{\mathcal {M}}}$ is an essential Banach $*$-${\mathcal {A}}$-bimodule;

and we investigate whether the linear mappings from ${\mathcal {A}}$ into ${{\mathcal {M}}}$ satisfying condition ${\mathbb {D}}_{3}$ or ${\mathbb {D}}_{4}$ characterize $*$-Jordan derivations. In particular, we improve some results from [13, 17, 18].

2 $*$-derivations on some algebras

A (Banach) algebra ${\mathcal {A}}$ is said to be zero product determined if every (continuous) bilinear mapping $\phi$ from ${\mathcal {A}}\times {\mathcal {A}}$ into any (Banach) linear space $\mathcal {X}$ satisfying

$$\begin{aligned} \phi (a, b)=0~\mathrm { whenever } \ ab = 0 \end{aligned}$$

can be written as $\phi (a, b)= T(ab)$, for some (continuous) linear mapping T from ${\mathcal {A}}$ into $\mathcal X$. In [7], M. Brešar showed that if ${\mathcal {A}}={\mathfrak {J}}({\mathcal {A}})$, then ${\mathcal {A}}$ is zero product determined, and in [1], the authors proved that every $C^*$-algebra ${\mathcal {A}}$ is zero product determined.

Let ${\mathcal {A}}$ be a Banach $*$-algebra and ${{\mathcal {M}}}$ be a Banach $*$-${\mathcal {A}}$-bimodule. Denote by ${{\mathcal {M}}}^{\sharp \sharp }$ the second dual space of ${{\mathcal {M}}}$. Next, we show that ${{\mathcal {M}}}^{\sharp \sharp }$ is also a Banach $*$-${\mathcal {A}}$-bimodule.

Since ${{\mathcal {M}}}$ is a Banach $*$-${\mathcal {A}}$-bimodule, ${\mathcal {M}}^{\sharp \sharp }$ turns into a dual Banach ${\mathcal {A}}$-bimodule with the operation defined by

$$\begin{aligned} a\cdot m^{\sharp \sharp }=\lim \limits _{\mu }am_{\mu }~\mathrm {and}~m^{\sharp \sharp }\cdot a=\lim \limits _{\mu }m_{\mu }a \end{aligned}$$

for every a in ${\mathcal {A}}$ and every $m^{\sharp \sharp }$ in ${\mathcal {M}}^{\sharp \sharp }$, where $(m_{\mu })$ is a net in ${\mathcal {M}}$ with $\Vert m_{\mu }\Vert \leqslant \Vert m^{\sharp \sharp }\Vert$ and $(m_{\mu })\ {\rightarrow }\ m^{\sharp \sharp }$ in the weak$^{*}$-topology $\sigma ({\mathcal {M}}^{\sharp \sharp },{\mathcal {M}}^{\sharp })$.

We define an involution $*$ in ${{\mathcal {M}}}^{\sharp \sharp }$ by

$$\begin{aligned} (m^{\sharp \sharp })^*(\rho )=\overline{m^{\sharp \sharp }(\rho ^*)},~~~~\rho ^*(m)=\overline{\rho (m^*)}, \end{aligned}$$

where $m^{\sharp \sharp }\in {\mathcal {M}}^{\sharp \sharp }$, $\rho \in {{\mathcal {M}}}^{\sharp }$ and $m\in {{\mathcal {M}}}$. Moreover, if $(m_{\mu })$ is a net in ${\mathcal {M}}$ and $m^{\sharp \sharp }$ is an element in ${{\mathcal {M}}}^{\sharp \sharp }$ such that $m_\mu \ {\rightarrow }\ m^{\sharp \sharp }$ in $\sigma ({\mathcal {M}}^{\sharp \sharp },{\mathcal {M}}^{\sharp })$, then for every $\rho$ in ${\mathcal {M}}^{\sharp }$, we have that

$$\begin{aligned} \rho (m_\mu )=m_\mu (\rho )\ \rightarrow \ m^{\sharp \sharp }(\rho ). \end{aligned}$$

It follows that

$$\begin{aligned} (m_\mu ^*)(\rho )=\rho (m_\mu ^*)=\overline{\rho ^*(m_\mu )}\ \rightarrow \ \overline{m^{\sharp \sharp }(\rho ^*)}=(m^{\sharp \sharp })^*(\rho ) \end{aligned}$$

for every $\rho$ in ${\mathcal {M}}^{\sharp }$. This means that the involution $*$ in ${{\mathcal {M}}}^{\sharp \sharp }$ is continuous in $\sigma ({\mathcal {M}}^{\sharp \sharp }, {\mathcal {M}}^{\sharp })$. Thus, we obtain

$$\begin{aligned} (a\cdot m^{\sharp \sharp })^*=(\lim \limits _{\mu }am_{\mu })^*= \lim \limits _{\mu }m_{\mu }^*a^*=(m^{\sharp \sharp })^*\cdot a^*. \end{aligned}$$

Similarly, we can show $(m^{\sharp \sharp }\cdot a)^*=a^*\cdot (m^{\sharp \sharp })^*$. This implies that ${{\mathcal {M}}}^{\sharp \sharp }$ is a Banach $*$-${\mathcal {A}}$-bimodule.

If ${\mathcal {A}}$ is a Banach $*$-algebra, then a bounded approximate identity for ${\mathcal {A}}$ is a net $(e_{i})_{i\in \Gamma }$ of self-adjoint elements in ${\mathcal {A}}$ such that $\lim \limits _{i}\Vert ae_i-a\Vert =\lim \limits _{i}\Vert e_ia-a\Vert =0$ for every a in ${\mathcal {A}}$ and $\mathrm {sup}_{i\in \Gamma }\Vert e_{i}\Vert \le k$ for some $k>0$ .

In [18], H. Ghahramani and Z. Pan proved that if ${\mathcal {A}}$ is a unital zero product determined $*$-algebra and a linear mapping $\delta$ from ${\mathcal {A}}$ into itself satisfies the condition

$$\begin{aligned} ({\mathbb {D}}_{1})~a,b\in {\mathcal {A}},~ab^*=0\ \mathrm {implies}\ a\delta (b)^*+\delta (a)b^*=0, \end{aligned}$$

then $\delta (a)=\Delta (a)+\delta (1)a$ for every a in ${\mathcal {A}}$, where $\Delta$ is a $*$-derivation.

For general zero product determined Banach $*$-algebras with a bounded approximate identity, the following result holds.

Theorem 2.1

Suppose that ${\mathcal {A}}$ is a zero product determined Banach $*$-algebra with a bounded approximate identity, and ${{\mathcal {M}}}$ is an essential Banach $*$-${\mathcal {A}}$-bimodule. If $\delta$ is a continuous linear mapping from ${\mathcal {A}}$ into ${{\mathcal {M}}}$ such that

$$\begin{aligned} a,b\in {\mathcal {A}},~ab^*=0\ implies \ a\delta (b)^*+\delta (a)b^*=0, \end{aligned}$$

then there exists a $*$-derivation $\Delta$ from ${\mathcal {A}}$ into ${{\mathcal {M}}}^{\sharp \sharp }$ and an element $\xi$ in ${{\mathcal {M}}}^{\sharp \sharp }$ such that $\delta (a)=\Delta (a)+\xi \cdot a$ for every a in ${\mathcal {A}}$. Furthermore, $\xi$ can be chosen in ${{\mathcal {M}}}$ in each of the following cases:

$\mathrm {(1)}$:

${\mathcal {A}}$ is a unital $*$-algebra,

$\mathrm {(2)}$:

${{\mathcal {M}}}$ is a dual $*$-${\mathcal {A}}$-bimodule.

Proof

Let $(e_{i})_{i\in \Gamma }$ be a bounded approximate identity of ${\mathcal {A}}$. Since $\delta$ is continuous, the net $(\delta (e_{i}))_{i\in \Gamma }$ is bounded and we can assume that it converges to $\xi$ in ${{\mathcal {M}}}^{\sharp \sharp }$ with the topology $\sigma ({{\mathcal {M}}}^{\sharp \sharp },{{\mathcal {M}}}^{\sharp })$.

Since ${{\mathcal {M}}}$ is an essential Banach $*$-${\mathcal {A}}$-bimodule, we know that the nets $(e_i m)_{i\in \Gamma }$ and $(me_i)_{i\in \Gamma }$ converge to m with the norm topology for every m in ${{\mathcal {M}}}$. Thus, we have

$$\begin{aligned} \mathrm {Ann}_{{{\mathcal {M}}}}({\mathcal {A}})=\{m\in {{\mathcal {M}}}:amb=0~\mathrm {for~each}~a,b\in {\mathcal {A}}\}=\{0\}. \end{aligned}$$

By the hypothesis, we obtain that

$$\begin{aligned} a,b,c\in {\mathcal {A}},~ab^*=b^*c=0\ \mathrm {implies}\ a\delta (b)^*c=0. \end{aligned}$$

It follows that

$$\begin{aligned} a,b,c\in {\mathcal {A}},~ab=bc=0\ \Rightarrow \ c^*b^*=b^*a^*=0 \ \Rightarrow \ c^*\delta (b)^*a^*=0 \ \Rightarrow \ a\delta (b)c=0. \end{aligned}$$

(2.1)

By (2.1) and [1, Theorem 4.5], we know that

$$\delta (ab)=\delta (a)b+a\delta (b)-a\cdot \xi \cdot b$$

for each a, b in ${\mathcal {A}}$, and $\xi$ can be chosen in ${{\mathcal {M}}}$ if ${\mathcal {A}}$ is a unital $*$-algebra or ${{\mathcal {M}}}$ is a dual $*$-${\mathcal {A}}$-bimodule.

Define a linear mapping $\Delta$ from ${\mathcal {A}}$ into ${\mathcal {M}}$ by

$$\begin{aligned} \Delta (a)=\delta (a)-\xi \cdot a \end{aligned}$$

for every a in ${\mathcal {A}}$. It is easy to show that $\Delta$ is a norm-continuous derivation from ${\mathcal {A}}$ into ${{\mathcal {M}}}^{\sharp \sharp }$ and we only need to show that $\Delta (b^*)=\Delta (b)^*$ for every b in ${\mathcal {A}}$.

First we claim that $\Delta (e_i)=\delta (e_i)-\xi \cdot e_i$ converges to zero in ${{\mathcal {M}}}^{\sharp \sharp }$ with the topology $\sigma ({{\mathcal {M}}}^{\sharp \sharp },{{\mathcal {M}}}^{\sharp })$. In fact, since $(e_{i})_{i\in \Gamma }$ is bounded in ${\mathcal {A}}$, we assume $(e_{i})_{i\in \Gamma }$ converges to $\zeta$ in ${\mathcal {A}}^{\sharp \sharp }$ with the topology $\sigma ({\mathcal {A}}^{\sharp \sharp },{\mathcal {A}}^{\sharp })$. For every $m^{\sharp \sharp }$ in ${{\mathcal {M}}}^{\sharp \sharp }$, define

$$\begin{aligned} m^{\sharp \sharp }\cdot \zeta =\lim \limits _{i}m^{\sharp \sharp }\cdot e_{i}. \end{aligned}$$

Thus, $m\cdot \zeta =m$ for every m in ${{\mathcal {M}}}$. By [10, Proposition A.3.52], we know that the mapping $m^{\sharp \sharp }\mapsto m^{\sharp \sharp }\cdot \zeta$ from ${\mathcal {M}}^{\sharp \sharp }$ into itself is $\sigma ({{\mathcal {M}}}^{\sharp \sharp },{{\mathcal {M}}}^{\sharp })$-continuous, and by the $\sigma ({{\mathcal {M}}}^{\sharp \sharp },{{\mathcal {M}}}^{\sharp })$-denseness of ${{\mathcal {M}}}$ in ${{\mathcal {M}}}^{\sharp \sharp }$, we have

$$\begin{aligned} m^{\sharp \sharp }\cdot \zeta =m^{\sharp \sharp } \end{aligned}$$

(2.2)

for every $m^{\sharp \sharp }$ in ${{\mathcal {M}}}^{\sharp \sharp }$. Hence $\Delta (e_i)=\delta (e_i)-\xi \cdot e_i$ converges to zero in ${{\mathcal {M}}}^{\sharp \sharp }$ with the topology $\sigma ({{\mathcal {M}}}^{\sharp \sharp },{{\mathcal {M}}}^{\sharp })$.

Next we prove $\Delta (b^*)=\Delta (b)^*$ for every b in ${\mathcal {A}}$. By the definition of $\Delta$, we know that $a\Delta (b)^*+\Delta (a)b^*=0$ for each a, b in ${\mathcal {A}}$ with $ab^*=0$. Define a bilinear mapping from ${\mathcal {A}}\times {\mathcal {A}}$ into ${\mathcal {M}}^{\sharp \sharp }$ by

$$\begin{aligned} \phi (a,b)=a\Delta (b^*)^*+\Delta (a)b. \end{aligned}$$

Thus, $ab=0$ implies $\phi (a,b)=0$. Since ${\mathcal {A}}$ is a zero product determined algebra, there exists a norm-continuous linear mapping T from ${\mathcal {A}}$ into ${{\mathcal {M}}}^{\sharp \sharp }$ such that

$$\begin{aligned} T(ab)=\phi (a,b)=a\Delta (b^*)^*+\Delta (a)b \end{aligned}$$

(2.3)

for each a, b in ${\mathcal {A}}$. If $b=e_i$ in (2.3), then we obtain

$$\begin{aligned} T(ae_i)=a\Delta (e_i)^*+\Delta (a)e_i. \end{aligned}$$

By the continuity of T and (2.2), it follows that $T(a)=\Delta (a)$ for every a in ${\mathcal {A}}$. Thus,

$$\begin{aligned} T(ab)=\Delta (ab)=a\Delta (b^*)^*+\Delta (a)b. \end{aligned}$$

Since $\Delta$ is a derivation, we have $a\Delta (b^*)^*=a\Delta (b)$ and $\Delta (b^*)a^*=\Delta (b)^*a^*$. If $a=e_i$, then taking $\sigma ({{\mathcal {M}}}^{\sharp \sharp },{{\mathcal {M}}}^{\sharp })$-limits, by (2.2) it follows that $\Delta (b^*)=\Delta (b)^*$ for every b in ${\mathcal {A}}$. $\square$

Let G be a locally compact group. The group algebra and the measure convolution algebra of G are denoted by $L^{1}(G)$ and M(G), respectively. The convolution product is denoted by $\cdot$ and the involution is denoted by $*$. It is well known that M(G) is a unital Banach $*$-algebra, and $L^{1}(G)$ is a closed ideal in M(G) with a bounded approximate identity. By [3, Lemma 1.1], we know that $L^{1}(G)$ is zero product determined. By [10, Theorem 3.3.15(ii)], it follows that M(G) with respect to convolution product is the dual of $C_{0}(G)$ as a Banach M(G)-bimodule.

By [27, Corollary 1.2], we know that every continuous derivation $\Delta$ from $L^{1}(G)$ into M(G) is an inner derivation, that is, there exists $\mu$ in M(G) such that $\Delta (f)=f\cdot \mu -\mu \cdot f$ for every f in $L^{1}(G)$. Thus, by Theorem 2.1, we can prove [17, Theorem 3.1(ii)] as follows.

Corollary 2.2

Let G be a locally compact group. If $\delta$ is a continuous linear mapping from $L^{1}(G)$ into M(G) such that

$$\begin{aligned} f,g\in L^{1}(G),~f\cdot g^*=0\ implies \ f\cdot \delta (g)^*+\delta (f)\cdot g^*=0, \end{aligned}$$

then there are $\mu ,\nu$ in M(G) such that

$$\begin{aligned} \delta (f)=f\cdot \mu -\nu \cdot f \end{aligned}$$

for every f in $L^{1}(G)$ and $\mathrm {Re}\mu \in \mathcal Z(M(G))$.

Proof

By Theorem 2.1, we know that there exists a $*$-derivation $\Delta$ from $L^{1}(G)$ into M(G) and an element $\xi$ in M(G) such that $\delta (f)=\Delta (f)+\xi \cdot f$ for every f in $L^{1}(G)$. By [27, Corollary 1.2], it follows that there exists $\mu$ in M(G) such that $\Delta (f)=f\cdot \mu -\mu \cdot f$. Since $\Delta (f^*)=\Delta (f)^*$, we have that

$$\begin{aligned} f^*\cdot \mu -\mu \cdot f^*=\mu ^*\cdot f^*-f^*\cdot \mu ^* \end{aligned}$$

for every f in $L^{1}(G)$. By [3, Lemma 1.3(ii)], we know $\mathrm {Re}\mu =\frac{1}{2}(\mu +\mu ^*)\in \mathcal Z(M(G))$. If $\nu =\mu -\xi$, then from the definition of $\Delta$ we have $\delta (f)=f\cdot \mu -\nu \cdot f$ for every f in $L^{1}(G)$. $\square$

For a general $C^*$-algebra ${\mathcal {A}}$, in [13], B. Fadaee and H. Ghahramani proved that if $\delta$ is a continuous linear mapping from ${\mathcal {A}}$ into its second dual space ${\mathcal {A}}^{\sharp \sharp }$ such that condition ${\mathbb {D}}_{1}$ holds, then there exists a $*$-derivation $\Delta$ from ${\mathcal {A}}$ into ${\mathcal {A}}^{\sharp \sharp }$ and an element $\xi$ in ${\mathcal {A}}^{\sharp \sharp }$ such that $\delta (a)=\Delta (a)+\xi a$ for every a in ${\mathcal {A}}$.

In [1], the authors proved that every $C^*$-algebra ${\mathcal {A}}$ is zero product determined, and it is well known that ${\mathcal {A}}$ has a bounded approximate identity. Thus, by Theorem 2.1, we can improve the result in [13] for any essential Banach $*$-bimodule.

Corollary 2.3

Suppose that ${\mathcal {A}}$ is a $C^*$-algebra and ${{\mathcal {M}}}$ is an essential Banach $*$-${\mathcal {A}}$-bimodule. If $\delta$ is a continuous linear mapping from ${\mathcal {A}}$ into ${{\mathcal {M}}}$ such that

$$\begin{aligned} a,b\in {\mathcal {A}},~ab^*=0\ implies \ a\delta (b)^*+\delta (a)b^*=0, \end{aligned}$$

then there exists a $*$-derivation $\Delta$ from ${\mathcal {A}}$ into ${{\mathcal {M}}}^{\sharp \sharp }$ and an element $\xi$ in ${{\mathcal {M}}}^{\sharp \sharp }$ such that $\delta (a)=\Delta (a)+\xi \cdot a$ for every a in ${\mathcal {A}}$. Furthermore, $\xi$ can be chosen in ${{\mathcal {M}}}$ in each of the following cases:

$\mathrm {(1)}$:

${\mathcal {A}}$ has an identity,

$\mathrm {(2)}$:

${{\mathcal {M}}}$ is a dual $*$-${\mathcal {A}}$-bimodule.

For von Neumann algebras, we have the following result.

Theorem 2.4

Suppose that ${\mathcal {A}}$ is a von Neumann algebra. If $\delta$ is a linear mapping from ${\mathcal {A}}$ into itself such that

$$\begin{aligned} a,b\in {\mathcal {A}},~ab^*=0\ implies \ a\delta (b)^*+\delta (a)b^*=0, \end{aligned}$$

then $\delta (a)=\Delta (a)+\delta (1)a$ for every a in ${\mathcal {A}}$, where $\Delta$ is a $*$-derivation. In particular, $\delta$ is a $*$-derivation when $\delta (1)=0$.

Proof

Define a linear mapping $\Delta$ from ${\mathcal {A}}$ into ${\mathcal {M}}$ by

$$\begin{aligned} \Delta (a)=\delta (a)-\delta (1)a \end{aligned}$$

for every a in ${\mathcal {A}}$. In the following we show that $\Delta$ is a $*$-derivation. It is clear that $\Delta (1)=0$ and $ab^*=0$ imply $a\Delta (b)^*+\Delta (a)b^*=0$.

Case 1 Suppose that ${\mathcal {A}}$ is an abelian von Neumann algebra. First we show that for $\Delta$ the following holds:

$$\begin{aligned} a,b\in {\mathcal {A}},~ab=0\ \mathrm {implies}\ a\Delta (b)=0. \end{aligned}$$

It is well known that ${\mathcal {A}}\cong C(X)$, where X is a compact Hausdorff space and C(X) denotes the $C^*$-algebra of all continuous complex-valued functions on X. Thus, we have $ab=0$ if and only if $ab^*=0$ for each a, b in ${\mathcal {A}}$. Indeed, if f and g are functions in C(X) corresponding to a and b, respectively, then

$$\begin{aligned} ab^*=0\Leftrightarrow f\cdot \bar{g}=0\Leftrightarrow f\cdot g=0\Leftrightarrow ab=0. \end{aligned}$$

If a and b are in ${\mathcal {A}}$ with $ab^*=ab=0$, then $a\Delta (b)^*+\Delta (a)b^*=0$. Multiplying by a on the left side of the above equation, we obtain $a^2\Delta (b)^*=0$. If f and h are functions in C(X) corresponding to a and $\Delta (b)$, respectively, then

$$\begin{aligned} 0=f^2\bar{g}=f^2g=fg. \end{aligned}$$

This implies that $a\Delta (b)=0$. By [23, Theorem 3], the function $\Delta$ is continuous. By [19, Lemma 2.5] and $\Delta (1)=0$, we obtain $\Delta (a)=\Delta (1)a=0$ for every a in ${\mathcal {A}}$.

Case 2 Suppose ${\mathcal {A}}\cong M_n({\mathcal {B}})$, where $\mathcal B$ is also a von Neumann algebra and $n\geqslant 2$. By [6, 7] we know that ${\mathcal {A}}$ is a zero product determined algebra. Thus, by [18, Theorem 3.1] it follows that $\Delta$ is a $*$-derivation.

Case 3 Suppose that ${\mathcal {A}}$ is a von Neumann algebra without abelian direct summands. By the type decomposition theorem, we have

$$\begin{aligned} {\mathcal {A}}=\left( \sum _{n\in E}\bigoplus {\mathcal {A}}_{n}\right) \oplus {\mathcal {A}}_{\mathrm {I}_{\infty }}\oplus {\mathcal {A}}_{\mathrm {II}}\oplus {\mathcal {A}}_{\mathrm {III}}, \end{aligned}$$

where E is some set of different finite cardinal numbers and ${\mathcal {A}}_n$ is type $\mathrm {I_n}$ ($n\ge 2$).

By [22, Theorem 6.6.5], we know that ${\mathcal {A}}_{n}$ is $*$-isomorphic to $M_n(\mathcal Z)$, where $\mathcal Z$ is the center of ${\mathcal {A}}_{n}$. Since ${\mathcal {A}}_{\mathrm {I}_{\infty }}$ is a properly infinite von Neumann algebra and (${\mathcal {A}}_{\mathrm {II}}\oplus {\mathcal {A}}_{\mathrm {III}}$) is a continuous von Neumann algebra, by [22, Lemma 6.3.3] and [26, Theorem 6.8.3], we know that there are two equivalent projections in (${\mathcal {A}}_{\mathrm {I}_{\infty }}\oplus {\mathcal {A}}_{\mathrm {II}}\oplus {\mathcal {A}}_{\mathrm {III}}$) with sum the unit element of (${\mathcal {A}}_{\mathrm {I}_{\infty }}\oplus {\mathcal {A}}_{\mathrm {II}}\oplus {\mathcal {A}}_{\mathrm {III}}$). By [22, Lemmas 6.6.3 and 6.6.4], it follows that (${\mathcal {A}}_{\mathrm {I}_{\infty }}\oplus {\mathcal {A}}_{\mathrm {II}}\oplus {\mathcal {A}}_{\mathrm {III}}$) is $*$-isomorphic to $M_2(\mathcal B)$ for some von Neumann algebra $\mathcal B$.

Hence, for a general von Neumann algebra ${\mathcal {A}}$, we have ${\mathcal {A}}\cong \sum _{i=1}^{n}\bigoplus {\mathcal {A}}_i$ (n is a finite integer or infinite), where each ${\mathcal {A}}_i$ coincides with either Case 1 or Case 2. Denote the unit element of ${\mathcal {A}}_i$ by $1_i$ and the restriction of $\Delta$ in ${\mathcal {A}}_i$ by $\Delta _i$. Since $1_i(1-1_i)=0$ and $\Delta (1)=0$, we have

$$1_i\Delta (1-1_i)^*+\Delta (1_i)(1-1_i)=0,$$

therefore

$$\begin{aligned} -1_i\Delta (1_i)^*+\Delta (1_i)-\Delta (1_i)1_i=0. \end{aligned}$$

(2.4)

Multiplying by $1_i$ on the left side of (2.4) and using $1_i\Delta (1_i)=\Delta (1_i)1_i$, we obtain $1_i\Delta (1_i)^*=0$. This implies $\Delta (1_i)=0$. For every a in ${\mathcal {A}}$, we write $a=\sum _{i=1}^{n}a_i$ with $a_i$ in ${\mathcal {A}}_i$. Since $a_i(1-1_i)=0$, we have $\Delta (a_i)(1-1_i)=0$, which means $\Delta (a_i)\in {\mathcal {A}}_i$. If $a_i, b_i$ are in ${\mathcal {A}}_i$ with $a_ib_i^*=0$, then

$$\begin{aligned} \Delta (a_i)b_i^*+a_i\Delta (b_i)^*=\Delta _i(a_i)b_i^*+a_i\Delta _i(b_i)^*=0. \end{aligned}$$

By Cases 1 and 2, we know that every $\Delta _i$ is a $*$-derivation. Thus, $\Delta$ is a $*$-derivation. $\square$

In what follows, we characterize the linear mappings $\delta$ that satisfy condition ${\mathbb {D}}_{1}$ from a unital $*$-algebra into a unital $*$-${\mathcal {A}}$-bimodule with a right or left separating set ${\mathcal {J}}\subseteq {\mathfrak {J}}({\mathcal {A}})$.

Lemma 2.5

([7, Theorem 4.1]) Suppose that ${\mathcal {A}}$ is a unital algebra and $\mathcal X$ is a linear space. If $\phi$ is a bilinear mapping from ${\mathcal {A}}\times {\mathcal {A}}$ into $\mathcal {X}$ such that

$$\begin{aligned} a,b\in {\mathcal {A}},~ab=0\ implies \ \phi (a,b)=0, \end{aligned}$$

then

$$\begin{aligned} \phi (a,x)=\phi (ax,1)~\mathrm {and}~\phi (x,a)=\phi (1,xa) \end{aligned}$$

for every a in ${\mathcal {A}}$ and every x in ${\mathfrak {J}}({\mathcal {A}})$.

Theorem 2.6

Suppose that ${\mathcal {A}}$ is a unital $*$-algebra and ${{\mathcal {M}}}$ is a unital $*$-${\mathcal {A}}$-bimodule with a right or left separating set ${\mathcal {J}}\subseteq {\mathfrak {J}}({\mathcal {A}})$. If $\delta$ is a linear mapping from ${\mathcal {A}}$ into ${\mathcal {M}}$ such that

$$\begin{aligned} a,b\in {\mathcal {A}},~ab^*=0\ implies \ a\delta (b)^*+\delta (a)b^*=0, \end{aligned}$$

then $\delta (a)=\Delta (a)+\delta (1)a$ for every a in ${\mathcal {A}}$, where $\Delta$ is a $*$-derivation. In particular, $\delta$ is a $*$-derivation when $\delta (1)=0$.

Proof

Since ${\mathcal {A}}$ is a unital $*$-algebra and ${{\mathcal {M}}}$ is a unital $*$-${\mathcal {A}}$-bimodule, we know that ${\mathcal {J}}\subseteq {\mathfrak {J}}({\mathcal {A}})$ is a right separating set of ${\mathcal {M}}$ if and only if ${\mathcal {J}}^*=\{x^*:x\in {\mathcal {J}}\}\subseteq {\mathfrak {J}}({\mathcal {A}})$ is a left separating set of ${\mathcal {M}}$. Thus, without loss of generality, we can assume that ${\mathcal {J}}$ is a left separating set of ${\mathcal {A}}$, otherwise, we replace ${\mathcal {J}}$ by ${\mathcal {J}}^*$.

Define a linear mapping $\Delta$ from ${\mathcal {A}}$ into ${\mathcal {M}}$ by

$$\begin{aligned} \Delta (a)=\delta (a)-\delta (1)a \end{aligned}$$

for every a in ${\mathcal {A}}$. In what follows, we show that $\Delta$ is a $*$-derivation.

It is clear that $\Delta (1)=0$ and $ab^*=0$ imply $a\Delta (b)^*+\Delta (a)b^*=0$. Define a bilinear mapping $\phi$ from ${\mathcal {A}}\times {\mathcal {A}}$ into ${\mathcal {M}}$ by

$$\begin{aligned} \phi (a,b)=a\Delta (b^*)^*+\Delta (a)b \end{aligned}$$

for each a and b in ${\mathcal {A}}$. By the assumption, $ab=0$ implies $\phi (a,b)=0$.

Let a, b be in ${\mathcal {A}}$ and x be in ${\mathcal {J}}$. By Lemma 2.5, we obtain

$$\begin{aligned} \phi (x,1)=\phi (1,x)~\mathrm {and}~\phi (a,x)=\phi (ax,1). \end{aligned}$$

Hence, the following two identities hold:

$$\begin{aligned} x\Delta (1)^*+\Delta (x)=\Delta (x^*)^*+\Delta (1)x \end{aligned}$$

(2.5)

and

$$\begin{aligned} a\Delta (x^*)^*+\Delta (a)x=ax\Delta (1)^*+\Delta (ax). \end{aligned}$$

(2.6)

By (2.5) and $\Delta (1)=0$, we obtain $\Delta (x)^*=\Delta (x^*)$. Thus, by (2.6), this implies

$$\begin{aligned} \Delta (ax)=a\Delta (x)+\Delta (a)x. \end{aligned}$$

Similarly to the proof of [4, Theorem 2.3], we obtain $\Delta (ab)=a\Delta (b)+\Delta (a)b$ for each a and b in ${\mathcal {A}}$.

It remains to show that $\Delta (a)^*=\Delta (a^*)$ holds for every a in ${\mathcal {A}}$. Indeed, for every a in ${\mathcal {A}}$ and every x in ${\mathcal {J}}$, we have $\Delta (ax)^*=\Delta ((ax)^*)$. This implies

$$\begin{aligned} (\Delta (a)x+a\Delta (x))^*=\Delta (x^*)a^*+x^*\Delta (a^*). \end{aligned}$$

Thus, we obtain $x^*(\Delta (a)^*-\Delta (a^*))=0$, hence $(\Delta (a)-\Delta (a^*)^*)x=0$. Therefore $\Delta (a)^*=\Delta (a^*)$ for every a in ${\mathcal {A}}$. $\square$

Remark 1

Let ${\mathcal {A}}$ be a $*$-algebra, ${{\mathcal {M}}}$ a $*$-${\mathcal {A}}$-bimodule, and $\delta$ a linear mapping from ${\mathcal {A}}$ into ${\mathcal {M}}$. Similarly to condition ${\mathbb {D}}_{1}$ which we have characterized in Sect. 2 as follows:

$$\begin{aligned} ({\mathbb {D}}_{1})~a,b\in {\mathcal {A}},~ab^*=0\ {\mathrm {implies}}\ a\delta (b)^*+\delta (a)b^*=0, \end{aligned}$$

we can consider condition ${\mathbb {D}}_{1}'$:

$$\begin{aligned} ({\mathbb {D}}_{1}')~a,b\in {\mathcal {A}},~a^*b=0\ {\mathrm {implies}}\ a^*\delta (b)+\delta (a)^*b=0. \end{aligned}$$

Through minor modifications, we can obtain the corresponding results.

Remark 2

A linear mapping $\delta$ from ${\mathcal {A}}$ into ${\mathcal {M}}$ is called a local derivation if, for every a in ${\mathcal {A}}$, there exists a derivation $\delta _{a}$ (depending on a) from ${\mathcal {A}}$ into ${\mathcal {M}}$ such that $\delta (a)=\delta _{a}(a)$. It is clear that every local derivation satisfies the following condition:

$$\begin{aligned} (\mathbb {H})~a,b,c\in {\mathcal {A}},~ab=bc=0\ \mathrm {implies}\ a\delta (b)c=0. \end{aligned}$$

In [1], the authors proved that every continuous linear mapping from a unital $C^*$-algebra into its unital Banach bimodule such that condition $\mathbb {H}$ holds and $\delta (1)=0$ is a derivation.

Let ${\mathcal {A}}$ be a $*$-algebra and ${\mathcal {M}}$ a $*$-${\mathcal {A}}$-bimodule. The natural way to translate condition $\mathbb {H}$ to the context of $*$-derivations is to consider the following condition:

$$\begin{aligned} (\mathbb {H}')~a,b,c\in {\mathcal {A}},~ab^*=b^*c=0\ \mathrm {implies}\ a\delta (b)^*c=0. \end{aligned}$$

However, conditions $\mathbb {H}'$ and $\mathbb {H}$ are equivalent. Indeed, if condition $\mathbb {H}'$ holds, then

$$\begin{aligned} a,b,c\in {\mathcal {A}},~ab=bc=0\ \Rightarrow \ c^*b^*=b^*a^*=0\ \Rightarrow \ c^*\delta (b)^*a^*=0\ \Rightarrow \ a\delta (b)c=0, \end{aligned}$$

and if condition $\mathbb {H}$ holds, then

$$\begin{aligned} a,b,c\in {\mathcal {A}},~ab^*=b^*c=0\ \Rightarrow \ c^*b=ba^*=0\ \Rightarrow \ c^*\delta (b)a^*=0\ \Rightarrow \ a\delta (b)^*c=0. \end{aligned}$$

This means that condition $\mathbb {H}'$ and $\delta (1)=0$ do not imply that $\delta$ is a $*$-derivation.

3 $*$-Jordan derivations on some algebras

A (Banach) algebra ${\mathcal {A}}$ is said to be zero Jordan product determined if every (continuous) bilinear mapping $\phi$ from ${\mathcal {A}}\times {\mathcal {A}}$ into any (Banach) linear space $\mathcal {X}$ satisfying

$$\begin{aligned} \phi (a, b)=0~\mathrm { whenever } \ a\circ b = 0 \end{aligned}$$

can be written as $\phi (a, b)= T(a\circ b)$ for some (continuous) linear mapping T from ${\mathcal {A}}$ into $\mathcal X$. In [5], we showed that if ${\mathcal {A}}$ is a unital algebra with ${\mathcal {A}}={\mathfrak {J}}({\mathcal {A}})$, then ${\mathcal {A}}$ is a zero Jordan product determined algebra.

Theorem 3.1

Suppose that ${\mathcal {A}}$ is a unital zero Jordan product determined $*$-algebra, and ${{\mathcal {M}}}$ is a unital $*$-${\mathcal {A}}$-bimodule. If $\delta$ is a linear mapping from ${\mathcal {A}}$ into ${{\mathcal {M}}}$ such that

$$\begin{aligned} a,b\in {\mathcal {A}},~a\circ b^*=0\ implies \ a\circ \delta (b)^*+\delta (a)\circ b^*=0~\mathrm {and}~\delta (1)a=a\delta (1), \end{aligned}$$

then $\delta (a)=\Delta (a)+\delta (1)a$ for every a in ${\mathcal {A}}$, where $\Delta$ is a $*$-Jordan derivation. In particular, $\delta$ is a $*$-Jordan derivation when $\delta (1)=0$.

Proof

Define a linear mapping $\Delta$ from ${\mathcal {A}}$ into ${\mathcal {M}}$ by $\Delta (a)=\delta (a)-\delta (1)a$ for every a in ${\mathcal {A}}$. It is sufficient to show that $\Delta$ is a $*$-Jordan derivation.

It is clear that $\Delta (1)=0$, and by $\delta (1)a=a\delta (1)$ we have

$$\begin{aligned} a,b\in {\mathcal {A}},~a\circ b^*=0\ \mathrm {implies}\ a\circ \Delta (b)^*+\Delta (a)\circ b^*=0. \end{aligned}$$

Define a bilinear mapping from ${\mathcal {A}}\times {\mathcal {A}}$ into ${\mathcal {M}}$ by

$$\begin{aligned} \phi (a,b)=a\circ \Delta (b^*)^*+\Delta (a)\circ b. \end{aligned}$$

Thus, $a\circ b=0$ implies $\phi (a,b)=0$. Since ${\mathcal {A}}$ is a zero Jordan product determined algebra, there exists a linear mapping T from ${\mathcal {A}}$ into ${{\mathcal {M}}}$ such that

$$\begin{aligned} T(a\circ b)=\phi (a,b)=a\circ \Delta (b^*)^*+\Delta (a)\circ b \end{aligned}$$

(3.1)

for each a, b in ${\mathcal {A}}$. Let $a=1$ and $b=1$ in (3.1). By $\Delta (1)=0$, we obtain

$$\begin{aligned} T(a)=\Delta (a)~\mathrm {and}~T(b)=\Delta (b^*)^*. \end{aligned}$$

It follows that $\Delta (a^*)=\Delta (a)^*$ holds for every a in ${\mathcal {A}}$. By (3.1),

$$\begin{aligned} T(a\circ b)=\Delta (a\circ b)=\phi (a,b)=a\circ \Delta (b)+\Delta (a)\circ b. \end{aligned}$$

This means that $\Delta$ is a $*$-Jordan derivation. $\square$

In [5], we proved that the matrix algebra $M_n(\mathcal B)$ for $n\ge 2$ is zero Jordan product determined, where $\mathcal B$ is a unital algebra. In [16], H. Ghahramani showed that every Jordan derivation from $M_n(\mathcal B)$ with $n\ge 2$ into its unital bimodule ${{\mathcal {M}}}$ is a derivation. Hence we have the following result.

Corollary 3.2

Suppose that $\mathcal B$ is a unital $*$-algebra, $M_n(\mathcal B)$ is a matrix algebra with $n\ge 2$, and ${{\mathcal {M}}}$ is a unital $*$-$M_n(\mathcal B)$-bimodule. If $\delta$ is a linear mapping from $M_n(\mathcal B)$ into ${{\mathcal {M}}}$ such that

$$\begin{aligned} a,b\in M_n(\mathcal B),~a\circ b^*=0\ implies \ a\circ \delta (b)^*+\delta (a)\circ b^*=0~\mathrm {and}~\delta (1)a=a\delta (1), \end{aligned}$$

then $\delta (a)=\Delta (a)+\delta (1)a$ for every a in $M_n(\mathcal B)$, where $\Delta$ is a $*$-derivation. In particular, $\delta$ is a $*$-derivation when $\delta (1)=0$.

Let $\mathcal H$ be a complex Hilbert space and $B(\mathcal H)$ be the algebra of all bounded linear operators on $\mathcal H$. Suppose that ${\mathcal {A}}$ is a von Neumann algebra on $\mathcal {H}$ and $LS({\mathcal {A}})$ is the set of all locally measurable operators affiliated with the von Neumann algebra ${\mathcal {A}}$.

In [28], M. Muratov and V. Chilin proved that $LS({\mathcal {A}})$ is a unital $*$-algebra and ${\mathcal {A}}\subset LS({\mathcal {A}})$. By [25, Proposition 21.20, Exercise 21.18], we know that if ${\mathcal {A}}$ is a von Neumann algebra without abelian direct summands, and $\mathcal B$ is a $*$-algebra with ${\mathcal {A}}\subseteq \mathcal B\subseteq LS({\mathcal {A}})$, then $\mathcal B\cong \sum _{i=1}^{k}\bigoplus M_{n_i}(\mathcal B_i)$ (k is a finite integer or infinite), where $\mathcal B_i$ is a unital algebra. By Theorem 3.1, we have the following result.

Corollary 3.3

Suppose that ${\mathcal {A}}$ is a von Neumann algebra without abelian direct summands, and $\mathcal B$ is a $*$-algebra with ${\mathcal {A}}\subseteq \mathcal B\subseteq LS({\mathcal {A}})$. If $\delta$ is a linear mapping from $\mathcal B$ into $LS({\mathcal {A}})$ such that

$$\begin{aligned} a,b\in \mathcal B,~a\circ b^*=0\ implies \ a\circ \delta (b)^*+\delta (a)\circ b^*=0~\mathrm {and}~\delta (1)a=a\delta (1), \end{aligned}$$

then $\delta (a)=\Delta (a)+\delta (1)a$ for every a in $\mathcal B$, where $\Delta$ is a $*$-Jordan derivation. In particular, $\delta$ is a $*$-Jordan derivation when $\delta (1)=0$.

For von Neumann algebras, by Corollary 3.2 and similarly to the proof of Theorem 2.4, we can easily obtain the following result and we omit the proof.

Corollary 3.4

Suppose that ${\mathcal {A}}$ is a von Neumann algebra. If $\delta$ is a linear mapping from ${\mathcal {A}}$ into itself such that

$$\begin{aligned} a,b\in {\mathcal {A}},~a\circ b^*=0\ implies \ a\circ \delta (b)^*+\delta (a)\circ b^*=0~\mathrm {and}~\delta (1)a=a\delta (1), \end{aligned}$$

then $\delta (a)=\Delta (a)+\delta (1)a$ for every a in ${\mathcal {A}}$, where $\Delta$ is a $*$-derivation. In particular, $\delta$ is a $*$-derivation when $\delta (1)=0$.

Lemma 3.5

([5, Theorem 2.1]) Suppose that ${\mathcal {A}}$ is a unital algebra and $\mathcal X$ is a linear space. If $\phi$ is a bilinear mapping from ${\mathcal {A}}\times {\mathcal {A}}$ into $\mathcal {X}$ such that

$$\begin{aligned} a,b\in {\mathcal {A}},~a\circ b=0\ implies \ \phi (a,b)=0, \end{aligned}$$

then

$$\begin{aligned} \phi (a,x)=\frac{1}{2}\phi (ax,1)+\frac{1}{2}\phi (xa,1) \end{aligned}$$

for every a in ${\mathcal {A}}$ and every x in ${\mathfrak {J}}({\mathcal {A}})$.

Suppose that ${\mathcal {A}}$ is a unital algebra and ${{\mathcal {M}}}$ is a unital ${\mathcal {A}}$-bimodule satisfying

$$\begin{aligned} \{m\in {\mathcal {M}}: xmx=0~\mathrm {for~every}~x\in {\mathcal {J}}\}=\{0\}, \end{aligned}$$

where ${\mathcal {J}}$ is an ideal of ${\mathcal {A}}$ linear generated by idempotents in ${\mathcal {A}}$. In [15, Theorem 4.3], H. Ghahramani studied the linear mapping $\delta$ from ${\mathcal {A}}$ into ${{\mathcal {M}}}$ that satisfies

$$\begin{aligned} a,b\in {\mathcal {A}},~a\circ b=0\ \mathrm {implies}\ a\circ \delta (b)+\delta (a)\circ b=0, \end{aligned}$$

and showed that $\delta$ is a generalized Jordan derivation. In what follows, we suppose that ${\mathcal {J}}$ is an ideal of ${\mathcal {A}}$ generated algebraically by all idempotents in ${\mathcal {A}}$, and have the following result.

Theorem 3.6

Suppose that ${\mathcal {A}}$ is a unital $*$-algebra, ${{\mathcal {M}}}$ is a unital $*$-${\mathcal {A}}$-bimodule, and ${\mathcal {J}}\subseteq {\mathfrak {J}}({\mathcal {A}})$ is an ideal of ${\mathcal {A}}$ such that

$$\begin{aligned} \{m\in {\mathcal {M}}: xmx=0~\mathrm {for~every}~x\in {\mathcal {J}}\}=\{0\}. \end{aligned}$$

If $\delta$ is a linear mapping from ${\mathcal {A}}$ into ${\mathcal {M}}$ such that

$$\begin{aligned} a,b\in {\mathcal {A}},~a\circ b^*=0\ implies \ a\circ \delta (b)^*+\delta (a)\circ b^*=0~ and ~\delta (1)a=a\delta (1), \end{aligned}$$

then $\delta (a)=\Delta (a)+\delta (1)a$ for every a in ${\mathcal {A}}$, where $\Delta$ is a $*$-Jordan derivation. In particular, $\delta$ is a $*$-Jordan derivation when $\delta (1)=0$.

Proof

Let $\widehat{\mathcal {J}}$ be an algebra generated algebraically by ${\mathcal {J}}$ and ${\mathcal {J}}^*$. Since ${\mathcal {J}}\subseteq {\mathfrak {J}}({\mathcal {A}})$ is an ideal of ${\mathcal {A}}$, it is easy to show that $\widehat{\mathcal {J}}\subseteq {\mathfrak {J}}({\mathcal {A}})$ is also an ideal of ${\mathcal {A}}$, and also

$$\begin{aligned} \{m\in {\mathcal {M}}: xmx=0~\mathrm {for~every}~x\in \widehat{\mathcal {J}}\}=\{0\}. \end{aligned}$$

Thus, without loss of generality, we can assume that ${\mathcal {J}}$ is a self-adjoint ideal of ${\mathcal {A}}$, otherwise we may replace ${\mathcal {J}}$ by $\widehat{\mathcal {J}}$.

Define a linear mapping $\Delta$ from ${\mathcal {A}}$ into ${\mathcal {M}}$ by

$$\begin{aligned} \Delta (a)=\delta (a)-\delta (1)a \end{aligned}$$

for every a in ${\mathcal {A}}$. Next, we show that $\Delta$ is a $*$-derivation.

It is clear that $\Delta (1)=0$ holds and, by $\delta (1)a=a\delta (1)$, the equation $a\circ b^*=0$ implies $a\circ \Delta (b)^*+\Delta (a)\circ b^*=0$.

Define a bilinear mapping $\phi$ from ${\mathcal {A}}\times {\mathcal {A}}$ into ${\mathcal {M}}$ by

$$\begin{aligned} \phi (a,b)=a\circ \Delta (b^*)^*+\Delta (a)\circ b \end{aligned}$$

for each a and b in ${\mathcal {A}}$. By the assumption, $a\circ b=0$ implies $\phi (a,b)=0$.

Let a, b be in ${\mathcal {A}}$ and x be in ${\mathcal {J}}$. By Lemma 3.5, we obtain

$$\begin{aligned} \phi (x,1)=\phi (1,x), \end{aligned}$$

hence

$$\begin{aligned} x\circ \Delta (1)^*+\Delta (x)\circ 1=1\circ \Delta (x^*)^*+\Delta (1)\circ x. \end{aligned}$$

(3.2)

By (3.2) and $\Delta (1)=0$, we know that $\Delta (x)^*=\Delta (x^*)$. Again by Lemma 3.5, it follows that

$$\begin{aligned} a\circ \Delta (x^*)^*+\Delta (a)\circ x=\frac{1}{2}[\Delta (ax)\circ 1+\Delta (xa)\circ 1]. \end{aligned}$$

(3.3)

By (3.3) and $\Delta (x)^*=\Delta (x^*)$, it is easy to show that

$$\begin{aligned} \Delta (a\circ x)=a\circ \Delta (x)+\Delta (a)\circ x. \end{aligned}$$

(3.4)

Next, we prove that $\Delta$ is a Jordan derivation.

Define $\{a,m,b\}=amb+bma$ and $\{a,b,m\}=\{m,b,a\}=abm+mba$ for each a, b in ${\mathcal {A}}$ and every m in ${\mathcal {M}}$. Let a be in ${\mathcal {A}}$ and x, y be in ${\mathcal {M}}$.

By the technique of the proof of [15, Theorem 4.3] and (3.4), we obtain the following two identities:

$$\begin{aligned} \Delta \{x,a,y\}=\{\Delta (x),a,y\}+\{x,\Delta (a),y\}+\{x,a,\Delta (y)\}, \end{aligned}$$

(3.5)

and

$$\begin{aligned} \Delta \{x,a^{2},y\}=\{\Delta (x),a^{2},y\}+\{x,a\circ \Delta (a),y\}+\{x,a^{2},\Delta (y)\}. \end{aligned}$$

(3.6)

On the other hand, by (3.5),

$$\begin{aligned} \Delta \{x,a^{2},x\}=\{\Delta (x),a^{2},x\}+\{x,\Delta (a^{2}),x\}+\{x,a^{2},\Delta (x)\}. \end{aligned}$$

(3.7)

By comparing (3.6) and (3.7), it follows that $\{x,\Delta (a^{2}),x\}=\{x,a\circ \Delta (a),x\}$ holds. That is, $x(\Delta (a^{2})-a\circ \Delta (a))x=0.$ By the assumption, this implies that $\Delta (a^{2})-a\circ \Delta (a)=0$ is true for every a in ${\mathcal {A}}$.

It remains to show that $\Delta (a)^*=\Delta (a^*)$ holds for every a in ${\mathcal {A}}$. Indeed, for every a in ${\mathcal {A}}$ and every x in ${\mathcal {J}}$, we have $\Delta (xax)^*=\Delta ((xax)^*)$. Since $\Delta$ is a Jordan derivation, this implies

$$\begin{aligned} (\Delta (x)ax+x\Delta (a)x+xa\Delta (x))^*=\Delta (x^*)a^*x^*+x^*\Delta (a^*)x^*+x^*a^*\Delta (x^*). \end{aligned}$$

Thus, we can obtain $x^*(\Delta (a)^*-\Delta (a^*))x^*=0$. Since ${\mathcal {J}}$ is a self-adjoint ideal of ${\mathcal {A}}$, the equation $\Delta (a)^*=\Delta (a^*)$ follows. $\square$

Let ${\mathcal {A}}$ be a $C^*$-algebra and ${{\mathcal {M}}}$ a Banach $*$-${\mathcal {A}}$-bimodule. Denote by ${\mathcal {A}}^{\sharp \sharp }$ and ${{\mathcal {M}}}^{\sharp \sharp }$ the second dual space of ${\mathcal {A}}$ and ${{\mathcal {M}}}$, respectively. By [11, p. 26], we can define a product $\diamond$ in ${\mathcal {A}}^{\sharp \sharp }$ by

$$\begin{aligned} a^{\sharp \sharp }\diamond b^{\sharp \sharp }=\lim \limits _{\lambda }\lim \limits _{\mu }\alpha _{\lambda }\beta _{\mu } \end{aligned}$$

for each $a^{\sharp \sharp }$, $b^{\sharp \sharp }\in {\mathcal {A}}^{\sharp \sharp }$, where $(\alpha _{\lambda })$ and $(\beta _{\mu })$ are two nets in ${\mathcal {A}}$ with $\Vert \alpha _{\lambda }\Vert \leqslant \Vert a^{\sharp \sharp }\Vert$ and $\Vert \beta _{\mu }\Vert \leqslant \Vert b^{\sharp \sharp }\Vert$, such that $\alpha _{\lambda }\rightarrow a^{\sharp \sharp }$ and $\beta _{\mu }\rightarrow b^{\sharp \sharp }$ in the weak$^{*}$-topology $\sigma ({\mathcal {A}}^{\sharp \sharp }, {\mathcal {A}}^{\sharp })$. Moreover, we can define an involution $*$ in ${\mathcal {A}}^{\sharp \sharp }$ by

$$\begin{aligned} (a^{\sharp \sharp })^*(\rho )=\overline{a^{\sharp \sharp }(\rho ^*)},~~~~\rho ^*(a)=\overline{\rho (a^*)}, \end{aligned}$$

where $a^{\sharp \sharp }\in {\mathcal {A}}^{\sharp \sharp }$, $\rho \in A^{\sharp }$ and $a\in {\mathcal {A}}$. By [22, p. 726], we deduce that ${\mathcal {A}}^{\sharp \sharp }$ is a von Neumann algebra with the product $\diamond$ and the involution $*$.

Since ${{\mathcal {M}}}$ is a Banach ${\mathcal {A}}$-bimodule, ${\mathcal {M}}^{\sharp \sharp }$ turns into a dual Banach $({\mathcal {A}}^{\sharp \sharp },\diamond )$-bimodule with the operation defined by

$$\begin{aligned} a^{\sharp \sharp }\cdot m^{\sharp \sharp }=\lim \limits _{\lambda }\lim \limits _{\mu }a_{\lambda }m_{\mu }~\mathrm {and}~m^{\sharp \sharp }\cdot a^{\sharp \sharp }=\lim \limits _{\mu }\lim \limits _{\lambda }m_{\mu }a_{\lambda } \end{aligned}$$

for every $a^{\sharp \sharp }$ in ${\mathcal {A}}^{\sharp \sharp }$ and every $m^{\sharp \sharp }$ in ${\mathcal {M}}^{\sharp \sharp }$, where $(a_{\lambda })$ is a net in ${\mathcal {A}}$ with $\Vert a_{\lambda }\Vert \leqslant \Vert a^{\sharp \sharp }\Vert$ and $(a_{\lambda })\rightarrow a^{\sharp \sharp }$ in $\sigma ({\mathcal {A}}^{\sharp \sharp },{\mathcal {A}}^{\sharp })$, $(m_{\mu })$ is a net in ${\mathcal {M}}$ with $\Vert m_{\mu }\Vert \leqslant \Vert m^{\sharp \sharp }\Vert$ and $(m_{\mu })\rightarrow m^{\sharp \sharp }$ in $\sigma ({\mathcal {M}}^{\sharp \sharp },{\mathcal {M}}^{\sharp })$.

We remarked in the discussion preceding Theorem 2.1 that ${\mathcal {M}}^{\sharp \sharp }$ has an involution $*$ and it is continuous in $\sigma ({\mathcal {M}}^{\sharp \sharp }, {\mathcal {M}}^{\sharp })$. By [1, p. 553], we know that every continuous bilinear map $\varphi$ from ${\mathcal {A}}\times {\mathcal {M}}$ into ${\mathcal {M}}$ is Arens regular, which means that

$$\begin{aligned} \lim \limits _{\lambda }\lim \limits _{\mu }\varphi (a_{\lambda },m_{\mu })=\lim \limits _{\mu }\lim \limits _{\lambda }\varphi (a_{\lambda },m_{\mu }) \end{aligned}$$

holds for every $\sigma ({\mathcal {A}}^{\sharp \sharp }, {\mathcal {A}}^{\sharp })$-convergent net $(a_{\lambda })$ in ${\mathcal {A}}$ and every $\sigma ({\mathcal {M}}^{\sharp \sharp }, {\mathcal {M}}^{\sharp })$-convergent net $(m_{\mu })$ in ${\mathcal {M}}$. Thus, we obtain

$$\begin{aligned} (a^{\sharp \sharp }\cdot m^{\sharp \sharp })^*=(\lim \limits _{\lambda }\lim \limits _{\mu }a_{\lambda }m_{\mu })^*= \lim \limits _{\lambda }\lim \limits _{\mu }m_{\mu }^*a_{\lambda }^*=\lim \limits _{\mu }\lim \limits _{\lambda }m_{\mu }^*a_{\lambda }^*=(m^{\sharp \sharp })^*\cdot (a^{\sharp \sharp })^*, \end{aligned}$$

where $(a_{\lambda })$ is a net in ${\mathcal {A}}$ with $(a_{\lambda })\rightarrow a^{\sharp \sharp }$ in $\sigma ({\mathcal {A}}^{\sharp \sharp },{\mathcal {A}}^{\sharp })$ and $(m_{\mu })$ is a net in ${\mathcal {M}}$ with $(m_{\mu })\rightarrow m^{\sharp \sharp }$ in $\sigma ({\mathcal {M}}^{\sharp \sharp },{\mathcal {M}}^{\sharp })$. Similarly, we can show $(m^{\sharp \sharp }\cdot a^{\sharp \sharp })^*=(a^{\sharp \sharp })^*\cdot (m^{\sharp \sharp })^*$. This implies that ${{\mathcal {M}}}^{\sharp \sharp }$ is a Banach $*$-${\mathcal {A}}^{\sharp \sharp }$-bimodule.

A projection p in ${\mathcal {A}}^{\sharp \sharp }$ is called open if there exists an increasing net $(a_{\alpha })$ of positive elements in ${\mathcal {A}}$ such that $p=\lim \limits _{\alpha }a_{\alpha }$ in the weak$^*$-topology of ${\mathcal {A}}^{\sharp \sharp }$. If p is open, then we say that the projection $1-p$ is closed.

For a unital $C^*$-algebra, the following result holds.

Theorem 3.7

Suppose that ${\mathcal {A}}$ is a unital $C^*$-algebra and ${{\mathcal {M}}}$ is a unital Banach $*$-${\mathcal {A}}$-bimodule. If $\delta$ is a continuous linear mapping from ${\mathcal {A}}$ into ${{\mathcal {M}}}$ such that $\delta (1)a=a\delta (1)$ holds for every a in ${\mathcal {A}}$, then the following three statements are equivalent:

• (1) $a,b\in {\mathcal {A}},~a\circ b^*=0\ implies \ a\circ \delta (b)^*+\delta (a)\circ b^*=0;$

• (2) $a,b\in {\mathcal {A}},~ab^*=b^*a=0\ implies \ a\circ \delta (b)^*+\delta (a)\circ b^*=0;$

• (3) $\delta (a)=\Delta (a)+\delta (1)a$ holds for every a in ${\mathcal {A}}$, where $\Delta$ is a $*$-derivation from ${\mathcal {A}}$ into ${{\mathcal {M}}}$.

Proof

It is clear that $(1)\ \mathrm {implies}\ (2)$ and $(3)\ \mathrm {implies}\ (1)$. It is sufficient the prove that $(2)\ \mathrm {implies}\ (3)$.

Define a linear mapping $\Delta$ from ${\mathcal {A}}$ into ${{\mathcal {M}}}$ by $\Delta (a)=\delta (a)-\delta (1)a$ for every a in ${\mathcal {A}}$. It is sufficient to show that $\Delta$ is a $*$-derivation. First we prove $\Delta (a^*)=\Delta (a)^*$ for every a in ${\mathcal {A}}$.

By assumption, we can easily to show that

$$\begin{aligned} a,b\in {\mathcal {A}},~ab^*=b^*a=0\ \mathrm {implies}\ a\circ \Delta (b)^*+\Delta (a)\circ b^*=0~\mathrm {and}~\Delta (1)=0. \end{aligned}$$

Next, we verify $\Delta (b)=\Delta (b)^*$ for every self-adjoint element b in ${\mathcal {A}}$.

Since $\Delta$ is a norm-continuous linear mapping form ${\mathcal {A}}$ into ${{\mathcal {M}}}$, we know that $\Delta ^{\sharp \sharp }:({\mathcal {A}}^{\sharp \sharp }, \diamond )\rightarrow {\mathcal {M}}^{\sharp \sharp }$ is the weak$^{*}$-continuous extension of $\Delta$ to the double duals of ${\mathcal {A}}$ and ${\mathcal {M}}$.

Let b be a nonzero self-adjoint element in ${\mathcal {A}}$, $\sigma (b)\subseteq [-\Vert b\Vert ,\Vert b\Vert ]$ the spectrum of b and $r(b)\in {\mathcal {A}}^{\sharp \sharp }$ the range projection of b.

Denote by ${\mathcal {A}}_b$ the $C^*$-subalgebra of ${\mathcal {A}}$ generated by b, and by $C(\sigma (b))$ the $C^*$-algebra of all continuous complex-valued functions on $\sigma (b)$. By Gelfand theory we know that there is an isometric $*$ isomorphism between ${\mathcal {A}}_b$ and $C(\sigma (b))$.

For every n in $\mathbb {N}$, let $p_n$ be the projection in ${\mathcal {A}}_b^{\sharp \sharp }\subseteq {\mathcal {A}}^{\sharp \sharp }$ corresponding to the characteristic function $\chi _{([-\Vert b\Vert ,-\frac{1}{n}]\cup [\frac{1}{n},\Vert b\Vert ])\cap \sigma (b)}$ in $C(\sigma (b))$, and let $b_n$ be in ${\mathcal {A}}_b$ such that

$$\begin{aligned} b_np_n=p_nb_n=b_n=b_n^*~\mathrm {and}~\Vert b_n-b\Vert <\frac{1}{n}. \end{aligned}$$

By [29, Section 1.8], we know that $(p_n)$ converges to r(b) in the strong$^*$-topology of ${\mathcal {A}}^{\sharp \sharp }$, and hence in the weak$^*$-topology.

It is well known that $p_n$ is a closed projection in ${\mathcal {A}}_b^{\sharp \sharp }\subseteq {\mathcal {A}}^{\sharp \sharp }$ and $1-p_n$ is an open projection in ${\mathcal {A}}_b^{\sharp \sharp }$. Thus, there exists an increasing net $(z_\lambda )$ of positive elements in $((1-p_n){\mathcal {A}}^{\sharp \sharp }(1-p_n))\cap {\mathcal {A}}$ such that

$$\begin{aligned} 0\le z_\lambda \le 1-p_n \end{aligned}$$

and $(z_\lambda )$ converges to $1-p_n$ in the weak$^*$-topology of ${\mathcal {A}}^{\sharp \sharp }$. Since

$$\begin{aligned} 0\le ((1-p_n)-z_\lambda )^2\le (1-p_n)-z_\lambda \le (1-p_n), \end{aligned}$$

the net $(z_\lambda )$ also converges to $1-p_n$ in the strong$^*$-topology of ${\mathcal {A}}^{\sharp \sharp }$.

By $b_n=b_n^*$ and $z_{\lambda }b_n=b_nz_{\lambda }=0$, it follows that

$$\begin{aligned} z_{\lambda }\circ \Delta ^{\sharp \sharp }(b_n)^*+\Delta ^{\sharp \sharp }(z_{\lambda })\circ b_n=0. \end{aligned}$$

(3.8)

Taking weak$^*$-limits in (3.8) and since $\Delta ^{\sharp \sharp }$ is weak$^*$-continuous, we deduce

$$\begin{aligned} (1-p_n)\circ \Delta ^{\sharp \sharp }(b_n)^*+\Delta ^{\sharp \sharp }((1-p_n))\circ b_n=0. \end{aligned}$$

(3.9)

Since $(p_n)$ converges to r(b) in the weak$^*$-topology of ${\mathcal {A}}^{\sharp \sharp }$ and $(b_n)$ converges to b in the norm-topology of ${\mathcal {A}}$, by (3.9), we have that

$$\begin{aligned} (1-r(b))\circ \Delta ^{\sharp \sharp }(b)^*+\Delta ^{\sharp \sharp }(1-r(b))\circ b=0. \end{aligned}$$

(3.10)

Now the range projection of every power $b^m$ with $m\in \mathbb {N}$ coincides with the r(b), and by (3.10), hence

$$\begin{aligned} (1-r(b))\circ \Delta ^{\sharp \sharp }(b^m)^*+\Delta ^{\sharp \sharp }(1-r(b))\circ b^m=0 \end{aligned}$$

holds for every $m\in \mathbb {N}$, and by the linearity and norm continuity of the product we obtain

$$\begin{aligned} (1-r(b))\circ \Delta ^{\sharp \sharp }(z)^*+\Delta ^{\sharp \sharp }(1-r(b))\circ z=0 \end{aligned}$$

for every $z=z^*$ in ${\mathcal {A}}_b$. A standard argument involving the weak$^*$-continuity of $\Delta ^{\sharp \sharp }$ gives

$$\begin{aligned} (1-r(b))\circ \Delta ^{\sharp \sharp }(r(b))^*+\Delta ^{\sharp \sharp }(1-r(b))\circ r(b)=0. \end{aligned}$$

(3.11)

By (3.11), we obtain

$$\begin{aligned} (\Delta ^{\sharp \sharp }(r(b))^*+\Delta ^{\sharp \sharp }(r(b))-\Delta ^{\sharp \sharp }(1))\circ r(b)=2\Delta ^{\sharp \sharp }(r(b))^*. \end{aligned}$$

By $\Delta (1)=0$, the equality $\Delta ^{\sharp \sharp }(1)=0$ holds, hence

$$\begin{aligned} \Delta ^{\sharp \sharp }(r(b))^*=\Delta ^{\sharp \sharp }(r(b)). \end{aligned}$$

(3.12)

It is clear that every characteristic function

$$\begin{aligned} p=\chi _{([-\Vert b\Vert ,-\alpha ]\cup [\alpha ,\Vert b\Vert ])\cap \sigma (b)} \end{aligned}$$

(3.13)

in $C_0(\sigma (b))^{\sharp \sharp }$ with $0<\alpha <\Vert b\Vert$ is the range projection of a function in $C(\sigma (b))$. Moreover, every projection of the form

$$\begin{aligned} q=\chi _{([-\beta ,-\alpha ]\cup [\alpha ,\beta ])\cap \sigma (b)} \end{aligned}$$

(3.14)

in $C_0(\sigma (b))^{\sharp \sharp }$ with $0<\alpha<\beta <\Vert b\Vert$ can be written as the difference of two projections of the type in (3.13).

Since ${\mathcal {A}}_b$ and $C(\sigma (b))$ are isometric $*$-isomorph and $\Delta ^{\sharp \sharp }(r(b))^*=\Delta ^{\sharp \sharp }(r(b))$ holds for the range projection of b in ${\mathcal {A}}^{\sharp \sharp }$, we infer $\Delta ^{\sharp \sharp }(p)^*=\Delta ^{\sharp \sharp }(p)$ for every projection p of the type in (3.13). It follows that $\Delta ^{\sharp \sharp }(q)^*=\Delta ^{\sharp \sharp }(q)$ holds for every projection q of the type in (3.14).

It is well known that b can be approximated in norm by finite linear combinations of mutually orthogonal projections $q_j$ of the type in (3.14). Therefore, using the continuity of $\Delta$, we obtain $\Delta (b)^*=\Delta (b)$. Thus, $\Delta (a)^*=\Delta (a)$ for every a in ${\mathcal {A}}$.

By the assumption, it follows that

$$\begin{aligned} a,b\in {\mathcal {A}},~ab=ba=0\ \mathrm {implies}\ a\circ \Delta (b)+\Delta (a)\circ b=0. \end{aligned}$$

By [2, Theorem 4.1], we infer that $\Delta$ is a $*$-derivation. $\square$

Next, we consider general $C^*$-algebras ${\mathcal {A}}$. If $(e_{i})_{i\in \Gamma }$ is a bounded approximate identity of ${\mathcal {A}}$, ${{\mathcal {M}}}$ is an essential Banach $*$-${\mathcal {A}}$-bimodule, and $\delta$ is a continuous linear mapping from ${\mathcal {A}}$ into ${{\mathcal {M}}}$, then $(\delta (e_{i}))_{i\in \Gamma }$ is bounded and we can assume that it converges to $\xi$ in ${{\mathcal {M}}}^{\sharp \sharp }$ in the topology $\sigma ({{\mathcal {M}}}^{\sharp \sharp },{{\mathcal {M}}}^{\sharp })$.

Theorem 3.8

Suppose that ${\mathcal {A}}$ is a $C^*$-algebra (not necessary unital) and ${{\mathcal {M}}}$ is an essential Banach $*$-${\mathcal {A}}$-bimodule. If $\delta$ is a continuous linear mapping from ${\mathcal {A}}$ into ${{\mathcal {M}}}$ such that $\xi \cdot a=a\cdot \xi$ for every a in ${\mathcal {A}}$, then the following three statements are equivalent:

• (1) $a,b\in {\mathcal {A}},~a\circ b^*=0\ implies \ a\circ \delta (b)^*+\delta (a)\circ b^*=0;$

• (2) $a,b\in {\mathcal {A}},~ab^*=b^*a=0\ implies \ a\circ \delta (b)^*+\delta (a)\circ b^*=0;$

• (3) $\delta (a)=\Delta (a)+\xi \cdot a$ for every a in ${\mathcal {A}}$, where $\Delta$ is a $*$-derivation from ${\mathcal {A}}$ into ${{\mathcal {M}}}^{\sharp \sharp }$.

Proof

It is clear that $(1)\ \mathrm {implies}\ (2)$ and $(3)\ \mathrm {implies}\ (1)$. We only need to prove that $(2)\ \mathrm {implies}\ (3)$.

Define a linear mapping $\Delta$ from ${\mathcal {A}}$ into ${{\mathcal {M}}}^{\sharp \sharp }$ by

$$\begin{aligned} \Delta (a)=\delta (a)-\xi \cdot a \end{aligned}$$

for every a in ${\mathcal {A}}$. It is sufficient to show that $\Delta$ is a $*$-derivation.

By the definition of $\Delta$ and $\xi \cdot a=a\cdot \xi$ for every a in ${\mathcal {A}}$, we can easily to show that

$$\begin{aligned} a,b\in {\mathcal {A}},~ab^*=b^*a=0\ \mathrm {implies}\ a\circ \Delta (b)^*+\Delta (a)\circ b^*=0. \end{aligned}$$

By [10, Proposition 2.9.16], we know that $(e_{i})_{i\in \Gamma }$ converges to the identity 1 in ${\mathcal {A}}^{\sharp \sharp }$ with the topology $\sigma ({\mathcal {A}}^{\sharp \sharp },{\mathcal {A}}^{\sharp })$. By the proof of Theorem 2.1, we infer that $\Delta (e_i)=\delta (e_i)-e_i\cdot \xi$ converges to zero in ${{\mathcal {M}}}^{\sharp \sharp }$ in the topology $\sigma ({{\mathcal {M}}}^{\sharp \sharp },{{\mathcal {M}}}^{\sharp })$, and we obtain

$$\begin{aligned} m^{\sharp \sharp }\cdot 1=m^{\sharp \sharp } \end{aligned}$$

for every $m^{\sharp \sharp }$ in ${\mathcal {M}}^{\sharp \sharp }$. Since ${\mathcal {M}}^{\sharp \sharp }$ is a Banach $*$-${\mathcal {A}}^{\sharp \sharp }$-bimodule,

$$\begin{aligned} 1\cdot m^{\sharp \sharp }=m^{\sharp \sharp } \end{aligned}$$

holds for every $m^{\sharp \sharp }$ in ${\mathcal {M}}^{\sharp \sharp }$. Since $\Delta$ is a norm-continuous linear mapping form ${\mathcal {A}}$ into ${{\mathcal {M}}}^{\sharp \sharp }$, the mapping $\Delta ^{\sharp \sharp }:({\mathcal {A}}^{\sharp \sharp }, \diamond )\rightarrow {\mathcal {M}}^{\sharp \sharp \sharp \sharp }$ is the weak$^{*}$-continuous extension of $\Delta$ to the double duals of ${\mathcal {A}}$ and ${\mathcal {M}}^{\sharp \sharp }$ such that $\Delta ^{\sharp \sharp }(1)=0$.

By [10, Proposition A.3.52], we know that the mapping $m^{\sharp \sharp \sharp \sharp }\mapsto m^{\sharp \sharp \sharp \sharp }\cdot 1$ from ${\mathcal {M}}^{\sharp \sharp \sharp \sharp }$ into itself is $\sigma ({{\mathcal {M}}}^{\sharp \sharp \sharp \sharp },{{\mathcal {M}}}^{\sharp \sharp \sharp })$-continuous and, by the $\sigma ({{\mathcal {M}}}^{\sharp \sharp \sharp \sharp },{{\mathcal {M}}}^{\sharp \sharp \sharp })$-denseness of ${{\mathcal {M}}}^{\sharp \sharp }$ in ${{\mathcal {M}}}^{\sharp \sharp \sharp \sharp }$, the equality

$$\begin{aligned} m^{\sharp \sharp \sharp \sharp }\cdot 1=m^{\sharp \sharp \sharp \sharp } \end{aligned}$$

holds for every $m^{\sharp \sharp \sharp \sharp }$ in ${{\mathcal {M}}}^{\sharp \sharp \sharp \sharp }$. Since ${\mathcal {M}}^{\sharp \sharp \sharp \sharp }$ is a Banach $*$-${\mathcal {A}}^{\sharp \sharp }$-bimodule,

$$\begin{aligned} 1\cdot m^{\sharp \sharp \sharp \sharp }=m^{\sharp \sharp \sharp \sharp } \end{aligned}$$

holds for every $m^{\sharp \sharp \sharp \sharp }$ in ${\mathcal {M}}^{\sharp \sharp \sharp \sharp }$.

Finally, we use the proof of Theorem 3.7 to show that $\Delta$ is a $*$-derivation from ${\mathcal {A}}$ into ${{\mathcal {M}}}^{\sharp \sharp }$. $\square$

Remark 3

In [12], A. Essaleh and A. Peralta investigated the concept of triple derivation on $C^*$-algebras. Suppose that ${\mathcal {A}}$ is a $C^*$-algebra. If a, b and c be in ${\mathcal {A}}$, define the ternary product by $\{a,b,c\}=\frac{1}{2}(ab^*c+cb^*a)$. A linear mapping $\delta$ from ${\mathcal {A}}$ into itself is called a triple derivation if

$$\begin{aligned} \delta \{a,b,c\}=\{\delta (a),b,c\}+\{a,\delta (b),c\}+\{a,b,\delta (c)\} \end{aligned}$$

holds for each a, b and c in ${\mathcal {A}}$. If z is an element in ${\mathcal {A}}$, then $\delta$ is called a triple derivation at z if

$$\begin{aligned} a,b,c\in {\mathcal {A}},~\{a,b,c\}=z\ \mathrm {implies}\ \delta (z)=\{\delta (a),b,c\}+\{a,\delta (b),c\}+\{a,b,\delta (c)\}. \end{aligned}$$

In [12], A. Essaleh and A. Peralta proved that every continuous linear mapping $\delta$ which is a triple derivation at zero from a unital $C^*$-algebra into itself with $\delta (1)=0$ is a $*$-derivation.

On the other hand, it is easy to show that if $\delta$ is a triple derivation at zero, then

$$\begin{aligned} a,b\in {\mathcal {A}},~ab^*=b^*a=0\ \mathrm {implies}\ a\circ \delta (b)^*+\delta (a)\circ b^*=0. \end{aligned}$$

Thus, Theorem 3.7 generalizes [12, Corollary 2.10].

Remark 4

In [8], M. Brešar and J. Vukman introduced left derivations and Jordan left derivations. A linear mapping $\delta$ from an algebra ${\mathcal {A}}$ into its bimodule ${\mathcal {M}}$ is called a left derivation if $\delta (ab)=a\delta (b)+b\delta (a)$ holds for each a, b in ${\mathcal {A}}$; and $\delta$ is called a Jordan left derivation if $\delta (a\circ b)=2a\delta (b)+2b\delta (a)$ holds for each a, b in ${\mathcal {A}}$.

Let ${\mathcal {A}}$ be a $*$-algebra and ${{\mathcal {M}}}$ a $*$-${\mathcal {A}}$-bimodule. A left derivation (Jordan left derivation) $\delta$ from ${\mathcal {A}}$ into ${\mathcal {M}}$ is called a $*$-left derivation ($*$-Jordan left derivation) if $\delta (a^*)=\delta (a)^*$ for every a in ${\mathcal {A}}$.

We also can investigate the following conditions on a linear mapping $\delta$ from ${\mathcal {A}}$ into ${\mathcal {M}}$:

$$\begin{aligned}&(\mathbb {J}_{1})~a,b\in {\mathcal {A}},~ab^*=0\ \mathrm {implies}\ a\delta (b)^*+b^*\delta (a)=0;\\&(\mathbb {J}_{2})~a,b\in {\mathcal {A}},~a\circ b^*=0\ \mathrm {implies}\ a\delta (b)^*+b^*\delta (a)=0;\\&(\mathbb {J}_{3})~a,b\in {\mathcal {A}},~ab^*=b^*a=0\ \mathrm {implies}\ a\delta (b)^*+b^*\delta (a)=0. \end{aligned}$$