Nonlinear Generalized Bi-skew Jordan n-Derivations on $*$-Algebras

Abstract

Let ${\mathcal {A}}$ be a $*$-algebra over the complex field ${\mathbb {C}}.$ For any $T_1, T_2, \ldots , T_n \in {\mathcal {A}}$, define $q_1(T_1)=T_1, q_2(T_1, T_2)=T_1\diamond T_2=T_1T_2^*+T_2T_1^*$ and $q_n(T_1, T_2,\ldots , T_n)=q_{n-1}(T_1, T_2,\ldots , T_{n-1})\diamond T_n$ for all integers $n\ge 2.$ In this article, it is shown that under certain assumptions a map $D:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is a nonlinear bi-skew Jordan n-derivation (that is, D satisfies $D(q_n(T_1, T_2, \ldots , T_n))=\sum _{i=1}^{n}q_n(T_1,\ldots ,T_{i-1}, D(T_i), T_{i+1}, \ldots , T_n)$ for all $T_1, T_2, \ldots , T_n\in {\mathcal {A}}$) if and only if D is an additive $*$-derivation. Further, we introduce the notion of a nonlinear generalized bi-skew Jordan n-derivation of a $*$-algebra and prove that every nonlinear generalized bi-skew Jordan n-derivation $G:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is of the form $G(T)=ZT+D(T)$ for all $T\in {\mathcal {A}},$ where $Z\in {\mathcal {Z}}({\mathcal {A}})$ and $D:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is an additive $*$-derivation. As applications, we apply our main results to some special classes of $*$-algebras such as prime $*$-algebras, factor von Neumann algebras, von Neumann algebras with no central summonds of type $I_1$.

1 Introduction

Let ${\mathcal {A}}$ be an associative $*$-algebra over the complex field ${\mathbb {C}}.$ For any $T_1, T_2\in {\mathcal {A}},$ we denote the bi-skew Jordan product of $T_1$ and $T_2$ by $T_1\diamond T_2=T_1T_2^*+T_2T_1^*.$ Similar types of products based on the involution operation $(*)$, such as the skew Lie product ($[T_1,T_2]_{\bullet }=T_1T_2-T_2T_1^*$), the skew Jordan product ($T_1\bullet T_2=T_1T_2+T_2T_1^*$), the bi-skew Lie product ($[T_1,T_2]_{\diamond }=T_1T_2^*-T_2T_1^*$), etc., have been extensively studied on various $*$-algebras (see [8,9,10, 21,22,23,24,25, 27,28,30]). These products play an important role in some research topics, and hence, their study has attracted the attention of many authors over the past decades. For example, the skew Lie product was extensively studied because, by the fundamental theorem of Šemrl in [25], maps of the form $T\mapsto TA-AT^*$ naturally arise in the problem of representing quadratic functionals with sesquilinear functionals (see [24, 27]). Particular attention has been paid to investigating maps which preserve the aforementioned products between different kinds of $*$-algebras (see [4, 7, 9,10,11,12, 16, 29]).

Recall that an additive map $D:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is called an additive $*$-derivation if $D(T_1T_2)=D(T_1)T_2+T_1D(T_2)$ and $D(T^*)=D(T)^*$ hold for all $T, T_1, T_2\in {\mathcal {A}}.$ An additive map $G:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is said to be an additive generalized $*$-derivation if there exists an additive $*$-derivation D of ${\mathcal {A}}$ such that $G(T_1T_2)=G(T_1)T_2+T_1D(T_2)$ and $G(T^*)=G(T)^*$ hold for all $T, T_1, T_2\in {\mathcal {A}}.$ A map $D:{\mathcal {A}}\rightarrow {\mathcal {A}}$ (not necessarily additive) is called a nonlinear bi-skew Jordan derivation if $D(T_1\diamond T_2)=D(T_1)\diamond T_2+T_1 \diamond D(T_2)$ holds for all $T_1, T_2 \in {\mathcal {A}}.$ A map $G:{\mathcal {A}}\rightarrow {\mathcal {A}}$ (not necessarily additive) is said to be a nonlinear generalized bi-skew Jordan derivation if there exists a nonlinear bi-skew Jordan derivation D of ${\mathcal {A}}$ such that $G(T_1\diamond T_2)=G(T_1)\diamond T_2+T_1 \diamond D(T_2)$ holds for all $T_1, T_2 \in {\mathcal {A}}.$

Given the consideration of nonlinear bi-skew Jordan derivations and nonlinear generalized bi-skew Jordan derivations, let us introduce a more general class of mappings. Suppose that $n\ge 2$ is a fixed integer and $T_1, T_2,\ldots , T_n \in {\mathcal {A}}$. Let us define a sequence of polynomials as follows:

$$\begin{aligned} \begin{aligned} q_1(T_1)&=T_1,\\ q_2(T_1, T_2)&=q_1(T_1)\diamond T_2 =T_1\diamond T_2,\\ q_3(T_1, T_2, T_3)&=q_2(T_1, T_2)\diamond T_3=(T_1\diamond T_2)\diamond T_3,\\&\vdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots \ \ \\ q_n(T_1, T_2,\ldots , T_n)&=q_{n-1}(T_1, T_2,\ldots , T_{n-1})\diamond T_n.\\ \end{aligned} \end{aligned}$$

The element $q_n(T_1, T_2,\ldots , T_n)\ \ (n\ge 2)$ is called the bi-skew Jordan n-product of elements $T_1, T_2,\ldots , T_n \in {\mathcal {A}}$. A map $D:{\mathcal {A}}\rightarrow {\mathcal {A}}$ (not necessarily additive) is called a nonlinear bi-skew Jordan n-derivation if

$$\begin{aligned} D(q_n(T_1, T_2, \ldots , T_n))=\sum _{i=1}^{n} q_n(T_1, T_2,\ldots , T_{i-1}, D(T_i), T_{i+1}, \ldots , T_n) \end{aligned}$$

(1.1)

holds for all $T_1, T_2, \ldots , T_n\in {\mathcal {A}}.$ A map $G:{\mathcal {A}}\rightarrow {\mathcal {A}}$ (not necessarily additive) is said to be a nonlinear generalized bi-skew Jordan n-derivation if there exists a nonlinear bi-skew Jordan n-derivation D of ${\mathcal {A}}$ such that

$$\begin{aligned} G(q_n(T_1, T_2, \ldots , T_n))= & {} q_n(G(T_1),T_2, \ldots , T_n)\nonumber \\{} & {} +\sum _{i=2}^{n} q_n(T_1,T_2,\ldots , T_{i-1}, D(T_i), T_{i+1}, \ldots , T_n) \end{aligned}$$

(1.2)

holds for all $T_1, T_2, \ldots , T_n\in {\mathcal {A}}.$ By the definition, it is clear that a nonlinear bi-skew Jordan 2-derivation is a nonlinear bi-skew Jordan derivation and a nonlinear generalized bi-skew Jordan 2-derivation is a nonlinear generalized bi-skew Jordan derivation. Moreover, any nonlinear bi-skew Jordan n-derivation is an example of a nonlinear generalized bi-skew Jordan n-derivation (set $G=D$ in (1.2)).

Recently, the structure of nonlinear maps involving some new products on different kinds of $*$-algebras has been extensively studied by many authors (see [4, 5, 11, 12, 17,18,19,20, 31,32,33,34,35] and references therein). Let ${\mathcal {H}}$ be a complex Hilbert space and $\mathcal {B(H)}$ be the algebra of all bounded linear operators on ${\mathcal {H}}.$ In [18], Li et al. showed that if ${\mathcal {A}}\subseteq \mathcal {B(H)}$ is a von Neumann algebra without central summands of type $I_1,$ then a map $\delta :{\mathcal {A}}\rightarrow \mathcal {B(H)}$ is a nonlinear skew Lie (resp. Jordan) derivation if and only if $\delta$ is an additive $*$-derivation. This result has been extended to the case of nonlinear $*$-Lie (Jordan)-type derivations by Lin in [14, 15]. Kong and Zhang [13] introduced the concept of the bi-skew Lie product and proved that a map D (not necessarily additive) on a factor von Neumann algebra ${\mathcal {A}}$ satisfies $D([T_1, T_2]_\diamond )=[D(T_1), T_2]_\diamond +[T_1, D(T_2)]_\diamond$ for all $T_1, T_2\in {\mathcal {A}}$ if and only if D is an additive $*$-derivation. The authors extended the above result to the case of nonlinear bi-skew Lie n-derivation on a factor von Neumann algebra in [3]. Very recently, the authors in [2] introduced the notion of the bi-skew Jordan product and proved that each nonlinear bi-skew Jordan n-derivation of a factor von Neumann algebra is an additive $*$-derivation. The present paper is an attempt to extend the above result to the case of a nonlinear (generalized-) bi-skew Jordan n-derivation of an arbitrary unital $*$-algebras. In fact, it is shown that under certain mild assumptions every nonlinear bi-skew Jordan n-derivation is an additive $*$-derivation (Theorem 2.1), and further it is proved that every nonlinear generalized bi-skew Jordan n-derivation $G:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is of the form $G(T)=ZT+D(T)$ for all $T\in {\mathcal {A}},$ where $Z\in {\mathcal {Z}}({\mathcal {A}})$ and $D:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is an additive $*$-derivation (Theorem 3.1). As an application, we give a description of nonlinear generalized bi-skew Jordan n-derivations on classical examples of unital $*$-algebras such as unital prime $*$-algebras (Corollary 3.1), factor von Neumann algebras (Corollary 3.2) and von Neumann algebras with no central summands of type $I_1$ (Corollary 3.3). Besides the aforementioned works, the main motivation for our study actually comes from the papers [1, 6]. Benkovič [6] described the form of generalized Lie n-derivations of a unital algebra ${\mathcal {A}}$ with a nontrivial idempotent and proved that under certain assumptions every generalized Lie n-derivation $G:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is of the form $G(T)=ZT+D(T)$ for all $T\in {\mathcal {A}},$ where $Z\in {\mathcal {Z}}({\mathcal {A}})$ and $D:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is a Lie n-derivation. In [1], Ashraf and Ansari extended the above result to the case of nonlinear generalized Lie n-derivations.

2 Nonlinear Bi-skew Jordan n-Derivations

As preliminaries, we first introduce some notations that will play an important part in the proof of our main results. Throughout the paper, we assume that ${\mathcal {A}}$ is a $*$-algebra with identity I and a nontrivial projection $P_1$ (that is, $P_1\ne 0, I$ and ${P_1}^2=P_1={P_1}^*$), and write $P_2=I-P_1.$ Then ${\mathcal {A}}$ can be written as ${\mathcal {A}}=P_1{\mathcal {A}}P_1+P_1{\mathcal {A}}P_2+P_2{\mathcal {A}}P_1+P_2{\mathcal {A}}P_2.$ Let ${\mathcal {A}}^{s}=\{M\in {\mathcal {A}}: M^*=M\}$ and ${\mathcal {A}}^{k}=\{N\in {\mathcal {A}}: N^*=-N\}.$ Further, write ${\mathcal {A}}^{s}_{11}=P_1 {\mathcal {A}}^{s} P_1,$ ${\mathcal {A}}^{s}_{12}=P_1 {\mathcal {A}}^{s} P_2+P_2 {\mathcal {A}}^{s} P_1$ and ${\mathcal {A}}^{s}_{22}=P_2 {\mathcal {A}}^{s} P_2.$ Then, ${\mathcal {A}}^{s}$ can be expressed as ${\mathcal {A}}^{s}={\mathcal {A}}^{s}_{11}+{\mathcal {A}}^{s}_{12}+{\mathcal {A}}^{s}_{22}$ and each $M\in {\mathcal {A}}^{s}$ can be written as $M=M_{11}+M_{12}+M_{22},$ where $M_{11}\in {\mathcal {A}}^{s}_{11},$ $M_{12}\in {\mathcal {A}}^{s}_{12}$ and $M_{22}\in {\mathcal {A}}^{s}_{22}.$ Furthermore, assume that ${\mathcal {A}}$ satisfies the following property

$$\begin{aligned} T{\mathcal {A}}P_k=\{0\} \Longrightarrow T=0 \ \ (k=1,2). \end{aligned}$$

(2.1)

Some specific examples of unital $*$-algebras satisfying the property (2.1) are prime $*$-algebras, factor von Neumann algebras, von Neumann algebras with no central summands of type $I_1.$ The following lemma will be used frequently in proving the main results of the article.

Lemma 2.1

For any $T\in {\mathcal {A}}$ and integer $n\ge 2,$ we have

• (i) $q_n(T, P_1, \ldots , P_1)=2^{n-2}(P_1TP_1+P_1T^*P_1)+P_1T^*P_2+P_2TP_1,$

• (ii) $q_n(T, P_2, \ldots , P_2)=2^{n-2}(P_2TP_2+P_2T^*P_2)+P_1TP_2+P_2T^*P_1,$

• (iii) $q_n(T, I, \ldots , I)=2^{n-2}(T+T^*),$

• (iv) $q_n(T, \frac{I}{2}, \ldots , \frac{I}{2})=\frac{I}{2}(T+T^*).$

Proof

(i) We apply induction method on n. For any $T \in {\mathcal {A}},$ we see that

$$\begin{aligned} q_2(T,P_1)=T \diamond P_1=P_1TP_1+P_1T^*P_1+P_1T^*P_2+P_2T P_1. \end{aligned}$$

Thus, (i) holds true for $n=2.$ Let us assume that (i) holds for all integers $k<n,$ that is,

$$\begin{aligned} q_k(T, P_1,\ldots , P_1)=2^{k-2}(P_1TP_1+P_1T^*P_1)+P_1T^*P_2+P_2TP_1 \end{aligned}$$

for all $T \in {\mathcal {A}}.$ We now show that (i) holds true for n. Using the induction hypothesis, we have

$$\begin{aligned} q_n(T, P_1,\ldots , P_1)= & {} q_{n-1}(T,P_1,\ldots , P_1)\diamond P_1\\= & {} (2^{n-3}(P_1TP_1+P_1T^*P_1)+P_1T^*P_2+P_2TP_1)\diamond P_1\\= & {} 2^{n-2}(P_1TP_1+P_1T^*P_1)+P_1T^*P_2+P_2TP_1 \end{aligned}$$

for all $T \in {\mathcal {A}}.$ Hence, the result holds for all integers $n\ge 2.$

(iii) First, observe that $q_n(T_1, T_2,\ldots , T_n)=q_{n-1}(T_1\diamond T_2,\ldots , T_n)$ for all $T_1, T_2,\ldots , T_n\in {\mathcal {A}}$ and integer $n\ge 2.$ By a recursive calculation, we see that

$$\begin{aligned} q_n\Bigg (T, \frac{I}{2}, \ldots , \frac{I}{2}\Bigg )= & {} q_{n-1}\Bigg (\frac{I}{2}(T+T^*), \frac{I}{2},\ldots , \frac{I}{2}\Bigg ) =q_{n-2}\Bigg (\frac{I}{2}(T+T^*), \frac{I}{2},\ldots , \frac{I}{2}\Bigg )\\= & {} q_{n-3}\Bigg (\frac{I}{2}(T+T^*), \frac{I}{2},\ldots , \frac{I}{2}\Bigg )=\cdots =\frac{I}{2}(T+T^*). \end{aligned}$$

The assertions (ii) and (iv) can be obtained in a similar way. $\square$

Now, we are in a position to state the first main result of this article.

Theorem 2.1

Let ${\mathcal {A}}$ be a unital $*$-algebra with a nontrivial projection $P_1$ satisfying (2.1). Then a map $D:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is a nonlinear bi-skew Jordan n-derivation if and only if D is an additive $*$-derivation.

Proof

The ‘if’ part of the theorem is obvious. However, the ‘only if’ part of the theorem can be realized via a series of claims.

Claim 2.1

$D(0)=0.$

Taking $T_1=T_2=\cdots =T_n=0$ in (1.1), the above claim is easy to obtain.

Claim 2.2

For any $M\in {\mathcal {A}}^{s},$ we have $D(M)^*=D(M).$

Since $q_n(M, \frac{I}{2}, \ldots , \frac{I}{2})=M$ for all $M\in {\mathcal {A}}^{s},$

$$\begin{aligned} D(M)= & {} D\left( q_n \left( M, \frac{I}{2}, \ldots , \frac{I}{2}\right) \right) \\= & {} q_n\left( D(M), \frac{I}{2}, \ldots , \frac{I}{2}\right) +q_n\left( M, D\left( \frac{I}{2}\right) , \ldots , \frac{I}{2}\right) \\{} & {} +\cdots +q_n\left( M, \frac{I}{2}, \ldots , D \left( \frac{I}{2}\right) \right) \\= & {} \left( \frac{I}{2}\right) \left\{ D(M)\left( \frac{I}{2}\right) +\left( \frac{I}{2}\right) D(M)^*\right\} +(n-1)\left\{ MD\left( \frac{I}{2}\right) ^*+D\left( \frac{I}{2}\right) M\right\} . \end{aligned}$$

Thus, $D(M)^*=D(M).$

Claim 2.3

For any $M_{ii}\in {\mathcal {A}}^{s}_{ii}\ \ (i=1, 2), M_{12}\in {\mathcal {A}}^{s}_{12},$ we have $D(M_{ii}+M_{12})=D(M_{ii})+D(M_{12}).$

For any $M_{11}\in {\mathcal {A}}^{s}_{11}, M_{12}\in {\mathcal {A}}^{s}_{12},$ assume that $S=D(M_{11}+M_{12})-D(M_{11})-D(M_{12}).$ Our aim is to show $S=0.$ It is easy to see that $S^*=S$ by Claim 2.2. On the one hand, we have

$$\begin{aligned} D(q_n(M_{11}+M_{12}, P_2, \ldots , P_2))= & {} q_n(D(M_{11}+M_{12}), P_2, \ldots , P_2)\\{} & {} \quad +q_n(M_{11}+M_{12}, D(P_2), \ldots , P_2)\\{} & {} +\cdots +q_n(M_{11}+M_{12}, P_2, \ldots , D(P_2)). \end{aligned}$$

On the other hand, using the fact that $q_n(M_{11}, P_2, \ldots , P_2)=0$ and Claim 2.1, we get

$$\begin{aligned} D(q_n(M_{11}+M_{12}, P_2, \ldots , P_2))= & {} D(q_n(M_{11}, P_2, \ldots , P_2))+D(q_n(M_{12}, P_2, \ldots , P_2))\\= & {} q_n(D(M_{11})+D(M_{12}), P_2, \ldots , P_2)\\{} & {} +\,q_n(M_{11}+M_{12}, D(P_2), \ldots , P_2)\\{} & {} +\cdots +q_n(M_{11}+M_{12}, P_2, \ldots , D(P_2)). \end{aligned}$$

Comparing the above two expressions, we get $q_n(S, P_2, \ldots , P_2)=0.$ This together with the fact that $S^*=S$ leads us to $S_{12}=S_{21}=S_{22}=0.$ Again, on the one hand,

$$\begin{aligned}{} & {} {D(q_n(M_{11}+M_{12}, P_2-P_1, P_1, \ldots , P_1))}\\{} & {} \quad = q_n(D(M_{11}+M_{12}), P_2-P_1, P_1, \ldots , P_1)\\{} & {} \qquad +\,q_n(M_{11}+M_{12}, D(P_2-P_1), P_1, \ldots , P_1)\\{} & {} \qquad +\cdots +q_n(M_{11}+M_{12}, P_2-P_1, P_1, \ldots , D(P_1)). \end{aligned}$$

On the other hand, using the fact that $q_n(M_{12}, P_2-P_1, P_1, \ldots , P_1)=0$ and Claim 2.1, we obtain

$$\begin{aligned}{} & {} {D(q_n(M_{11}+M_{12}, P_2-P_1, P_1, \ldots , P_1))}\\{} & {} \quad = D(q_n(M_{11}, P_2-P_1, P_1, \ldots , P_1))+D(q_n(M_{12}, P_2-P_1, P_1, \ldots , P_1))\\{} & {} \quad = q_n(D(M_{11})+D(M_{12}), P_2-P_1, P_1, \ldots , P_1)\\{} & {} \qquad +\,q_n(M_{11}+M_{12}, D(P_2-P_1), P_1, \ldots , P_1)\\{} & {} \qquad +\cdots +q_n(M_{11}+M_{12}, P_2-P_1, P_1, \ldots , D(P_1)). \end{aligned}$$

The last two expressions yield that $q_n(S, P_2-P_1, P_1, \ldots , P_1)=0.$ This together with the fact $S^*=S$ gives $S_{11}=0.$ Hence, $S=0,$ that is, $D(M_{11}+M_{12})=D(M_{11})+D(M_{12}).$ Symmetrically, we can prove the other case.

Claim 2.4

For any $M_{ii}\in {\mathcal {A}}^{s}_{ii}\ \ (i=1, 2), M_{12}\in {\mathcal {A}}^{s}_{12},$ we have

$$\begin{aligned} D(M_{11}+M_{12}+M_{22})=D(M_{11})+D(M_{12})+D(M_{22}). \end{aligned}$$

Set $S=D(M_{11}+M_{12}+M_{22})-D(M_{11})-D(M_{12})-D(M_{22})$, where $M_{11}\in {\mathcal {A}}^{s}_{11}, M_{12}\in {\mathcal {A}}^{s}_{12}, M_{22}\in {\mathcal {A}}^{s}_{22}.$ On the one hand, we have

$$\begin{aligned}{} & {} {D(q_n(M_{11}+M_{12}+M_{22}, P_2, \ldots , P_2))}\\{} & {} \quad = q_n(D(M_{11}+M_{12}+M_{22}), P_2, \ldots , P_2)\\{} & {} \qquad +q_n(M_{11}+M_{12}+M_{22}, D(P_2), P_2, \ldots , P_2)\\{} & {} \qquad +\cdots +q_n(M_{11}+M_{12}+M_{22}, P_2, P_2, \ldots , D(P_2)). \end{aligned}$$

On the other hand, using the fact that $q_n(M_{11}, P_2, \ldots , P_2)=0$ and Claims 2.1, 2.3, we find that

$$\begin{aligned}{} & {} {D(q_n(M_{11}+M_{12}+M_{22}, P_2, \ldots , P_2))}\\{} & {} \quad = D(q_n(M_{11}, P_2, \ldots , P_2))\\{} & {} \quad +D(q_n(M_{12}+M_{22}, P_2, \ldots , P_2))\\{} & {} \quad = q_n(D(M_{11})+D(M_{12})+D(M_{22}), P_2, \ldots , P_2)\\{} & {} \qquad +\,q_n(M_{11}+M_{12}+M_{22}, D(P_2), \ldots , P_2)\\{} & {} \qquad +\cdots +q_n(M_{11}+M_{12}+M_{22}, P_2, \ldots , D(P_2)). \end{aligned}$$

Comparing the last two expressions, we get $q_n(S, P_2, \ldots , P_2)=0.$ This together with the fact that $S^*=S$ yields $S_{12}=S_{21}=S_{22}=0.$

Again, on the one hand

$$\begin{aligned}{} & {} {D(q_{n}(M_{11}+M_{12}+M_{22}, P_1, \ldots , P_1))}\\{} & {} \quad = q_n(D(M_{11}+M_{12}+M_{22}), P_1, \ldots , P_1)\\{} & {} \qquad +\,q_n(M_{11}+M_{12}+M_{22}, D(P_1), P_1, \ldots , P_1)\\{} & {} \qquad +\cdots +q_n(M_{11}+M_{12}+M_{22}, P_1, \ldots , D(P_1)). \end{aligned}$$

On the other hand, using the fact that $q_n(M_{22}, P_1, \ldots , P_1)=0$ and Claims 2.1, 2.3, we obtain

$$\begin{aligned}{} & {} {D(q_{n}(M_{11}+M_{12}+M_{22}, P_1, \ldots , P_1))}\\{} & {} \quad = D(q_{n}(M_{11}+M_{12}, P_1, \ldots , P_1))+D(q_{n}(M_{22}, P_1, \ldots , P_1))\\{} & {} \quad = q_{n}(D(M_{11})+D(M_{12})+D(M_{22}), P_1, \ldots , P_1)\\{} & {} \qquad +\,q_{n}(M_{11}+M_{12}+M_{22}, D(P_1), \ldots , P_1)\\{} & {} \qquad +\cdots +q_{n}(M_{11}+M_{12}+M_{22}, P_1, \ldots , D(P_1)). \end{aligned}$$

Combining the above two expressions for $D(q_{n}(M_{11}+M_{12}+M_{22}, P_1, \ldots , P_1)),$ we get $q_{n}(S, P_1, \ldots , P_1)=0.$ This together with the fact that $S^*=S$ implies that $S_{11}=0.$ Hence, $S=0,$ that is,

$$\begin{aligned} D(M_{11}+M_{12}+M_{22})=D(M_{11})+D(M_{12})+D(M_{22}). \end{aligned}$$

Claim 2.5

For any $M_{12}, M'_{12}\in {\mathcal {A}}^{s}_{12},$ we have $D(M_{12}+M'_{12})=D(M_{12})+D(M'_{12}).$

For any $M_{12}, M'_{12}\in {\mathcal {A}}^{s}_{12},$ observe that

$$\begin{aligned} q_n\left( P_1+M_{12}, P_2+M'_{12}, \frac{I}{2}, \ldots , \frac{I}{2}\right) =M_{12}+M'_{12}+M_{12}M'_{12}+M'_{12}M_{12}. \end{aligned}$$

Since $M_{12}M'_{12}+M'_{12}M_{12}\in {\mathcal {A}}^{s}_{11}+{\mathcal {A}}^{s}_{22},$ write $M_{12}M'_{12}+M'_{12}M_{12}=M_{11}+M_{22}$ for some $M_{11}\in {\mathcal {A}}^{s}_{11}$ and $M_{22}\in {\mathcal {A}}^{s}_{22}.$ Invoking Claims 2.3 and 2.4, we have

$$\begin{aligned}{} & {} {D(M_{12}+M'_{12})+D(M_{11})+D(M_{22})}\\{} & {} \quad = D(M_{12}+M'_{12}+M_{11}+M_{22})\\{} & {} \quad = D\left( q_n\left( P_1+M_{12}, P_2+M'_{12}, \frac{I}{2}, \ldots , \frac{I}{2}\right) \right) \\{} & {} \quad = q_n\left( D(P_1+M_{12}), P_2+M'_{12}, \frac{I}{2}, \ldots , \frac{I}{2}\right) \\{} & {} \qquad +\,q_n\left( P_1+M_{12}, D(P_2+M'_{12}), \frac{I}{2}, \ldots , \frac{I}{2}\right) \\{} & {} \qquad +\,q_n\left( P_1+M_{12}, P_2+M'_{12}, D\left( \frac{I}{2}\right) , \ldots , \frac{I}{2}\right) \\{} & {} \qquad +\cdots +q_n\left( P_1+M_{12}, P_2+M'_{12}, \frac{I}{2}, \ldots , D\left( \frac{I}{2}\right) \right) \\{} & {} \quad = q_n\left( D(P_1)+D(M_{12}), P_2+M'_{12}, \frac{I}{2}, \ldots , \frac{I}{2}\right) \\{} & {} \qquad +\,q_n\left( P_1+M_{12}, D(P_2)+D(M'_{12}), \frac{I}{2}, \ldots , \frac{I}{2}\right) \\{} & {} \qquad +\,q_n\left( P_1+M_{12}, P_2+M'_{12}, D\left( \frac{I}{2}\right) , \ldots , \frac{I}{2}\right) \\{} & {} \qquad +\cdots +q_n\left( P_1+M_{12}, P_2+M'_{12}, \frac{I}{2}, \ldots , D\left( \frac{I}{2}\right) \right) \\{} & {} \quad = D\left( q_n \left( P_1, P_2, \frac{I}{2}, \ldots , \frac{I}{2}\right) \right) +D\left( q_n\left( M_{12}, P_2, \frac{I}{2}, \ldots , \frac{I}{2}\right) \right) \\{} & {} \qquad +D\left( q_n\left( P_1, M'_{12}, \frac{I}{2}, \ldots , \frac{I}{2}\right) \right) +D\left( q_n\left( M_{12}, M'_{12}, \frac{I}{2}, \ldots , \frac{I}{2}\right) \right) \\{} & {} \quad = D(M_{12})+D(M'_{12})+D(M_{12}M'_{12}+M'_{12}M_{12})\\{} & {} \quad = D(M_{12})+D(M'_{12})+D(M_{11}+M_{22})\\{} & {} \quad = D(M_{12})+D(M'_{12})+D(M_{11})+D(M_{22}). \end{aligned}$$

Thus, $D(M_{12}+M'_{12})=D(M_{12})+D(M'_{12})$ for all $M_{12}, M'_{12}\in {\mathcal {A}}^{s}.$

Claim 2.6

For any $M_{ii}, M'_{ii}\in {\mathcal {A}}^{s}_{11} (i=1,2),$ we have $D(M_{ii}+M'_{ii})=D(M_{ii})+D(M'_{ii}).$

Let $S=D(M_{11}+M'_{11})-D(M_{11})-D(M'_{11}),$ where $M_{11}, M'_{11}\in {\mathcal {A}}^{s}_{11}.$ On the one hand, we have

$$\begin{aligned} D(q_n(M_{11}+M'_{11}, P_2, \ldots , P_2))= & {} q_n(D(M_{11}+M'_{11}), P_2, \ldots , P_2))\\{} & {} +\,q_n(M_{11}+M'_{11}, D(P_2), P_2, \ldots , P_2)\\{} & {} +\cdots +q_n(M_{11}+M'_{11}, P_2, \ldots , D(P_2)). \end{aligned}$$

On the other hand, using the fact that $q_n(M_{11}, P_2, \ldots , P_2)=q_n(M'_{11}, P_2, \ldots , P_2)=0$ and Claim 2.1, we obtain

$$\begin{aligned} D(q_n(M_{11}+M'_{11}, P_2, \ldots , P_2))= & {} D(q_n(M_{11}, P_2, \ldots , P_2))+D(q_n(M'_{11}, P_2, \ldots , P_2))\\= & {} q_n(D(M_{11})+D(M'_{11}), P_2, \ldots , P_2)\\{} & {} +\,q_n(M_{11}+M'_{11}, D(P_2), \ldots , P_2)\\{} & {} +\cdots +q_n(M_{11}+M'_{11}, P_2, \ldots , D(P_2)). \end{aligned}$$

Comparing the last two expressions, we have $q_n(S, P_2, \ldots , P_2)=0.$ This together with the fact that $S^*=S$ leads us to $S_{12}=S_{21}=S_{22}=0.$

Next, we show that $S_{11}=0.$ For any $T\in {\mathcal {A}},$ let $M_{12}=P_1TP_2+(P_1TP_2)^{*}.$ Then

$$\begin{aligned} q_n\left( M_{11}, M_{12}, \frac{I}{2}, \ldots , \frac{I}{2}\right) , q_n\left( M'_{11}, M_{12}, \frac{I}{2}, \ldots , \frac{I}{2}\right) \in {\mathcal {A}}^{s}_{12}. \end{aligned}$$

On the one hand, we have

$$\begin{aligned}{} & {} {D\left( q_n\left( M_{11}+M'_{11}, M_{12}, \frac{I}{2}, \ldots , \frac{I}{2}\right) \right) }\\{} & {} \quad = q_n\left( D(M_{11}+M'_{11}), M_{12}, \frac{I}{2}, \ldots , \frac{I}{2}\right) +q_n\left( M_{11}+M'_{11}, D(M_{12}), \frac{I}{2}, \ldots , \frac{I}{2}\right) \\{} & {} \qquad +\,q_n\left( M_{11}+M'_{11}, M_{12}, D\left( \frac{I}{2}\right) , \ldots , \frac{I}{2}\right) \\{} & {} \qquad +\cdots +q_n\left( M_{11}+M'_{11}, M_{12}, \frac{I}{2}, \ldots , D\left( \frac{I}{2}\right) \right) . \end{aligned}$$

On the other hand, Claim 2.5 gives

$$\begin{aligned}{} & {} {D\left( q_n\left( M_{11}+M'_{11}, M_{12}, \frac{I}{2}, \ldots , \frac{I}{2}\right) \right) }\\{} & {} \quad = D\left( q_n\left( M_{11}, M_{12}, \frac{I}{2}, \ldots , \frac{I}{2}\right) \right) +D\left( q_n\left( M'_{11}, M_{12}, \frac{I}{2}, \ldots , \frac{I}{2}\right) \right) \\{} & {} \quad = q_n\left( D(M_{11})+D(M'_{11}), M_{12}, \frac{I}{2}, \ldots , \frac{I}{2}\right) \\{} & {} \qquad +\,q_n\left( M_{11}+M'_{11}, D(M_{12}), \frac{I}{2}, \ldots , \frac{I}{2}\right) \\{} & {} \qquad +\,q_n\left( M_{11}+M'_{11}, M_{12}, D\left( \frac{I}{2}\right) , \ldots , \frac{I}{2}\right) \\{} & {} \qquad +\cdots +q_n\left( M_{11}+M'_{11}, M_{12}, \frac{I}{2}, \ldots , D\left( \frac{I}{2}\right) \right) . \end{aligned}$$

Combining the last two expressions, we get $q_n\big (S, M_{12}, \frac{I}{2}, \ldots , \frac{I}{2}\big )=0,$ which in turn implies that

$$\begin{aligned} SM_{12}+SM_{12}=0. \end{aligned}$$

Using the facts that $S_{12}=S_{21}=S_{22}=0$ and $M_{12}=P_1TP_2+(P_1TP_2)^{*},$ we obtain

$$\begin{aligned} S_{11}TP_2+P_2T^{*}S_{11}=0\ \ \text {for all}\ \ T\in {\mathcal {A}}. \end{aligned}$$

Multiplying the above equation by $P_1$ from the left and by $P_2$ from the right, we have $S_{11}TP_2=0$ for all $T\in {\mathcal {A}}.$ By (2.1), we get $S_{11}=0.$ Thus, $S=0,$ that is, $D(M_{11}+M'_{11})=D(M_{11})+D(M'_{11}).$ Symmetrically, one can prove that $D(M_{22}+M'_{22})=D(M_{22})+D(M'_{22})$ for all $M_{22}, M'_{22}\in {\mathcal {A}}^{s}_{22}.$

Using Claims 2.4, 2.5 and 2.6, one can readily verify the following claim.

Claim 2.7

D is additive on ${\mathcal {A}}^{s}.$

Claim 2.8

For any $N\in {\mathcal {A}}^{k},$ we have $D(N)^*=-D(N).$

Let us first show that $D\left( \frac{I}{2}\right) =0.$ Since $q_n\left( \frac{I}{2}, \frac{I}{2}, \ldots , \frac{I}{2}\right) =\frac{I}{2}$ and $D\left( \frac{I}{2}\right) ^*=D\left( \frac{I}{2}\right)$ (by Claim 2.2), we have

$$\begin{aligned} D\left( \frac{I}{2}\right)= & {} q_n\left( D\left( \frac{I}{2}\right) , \frac{I}{2}, \ldots , \frac{I}{2}\right) +q_n\left( \frac{I}{2}, D\left( \frac{I}{2}\right) , \ldots , \frac{I}{2}\right) \\{} & {} +\cdots +q_n\left( \frac{I}{2}, \frac{I}{2}, \ldots , D\left( \frac{I}{2}\right) \right) \\= & {} n D\left( \frac{I}{2}\right) . \end{aligned}$$

Hence $D\left( \frac{I}{2}\right) =0.$ Now, using Claim 2.1 and the fact that $q_n\left( N, \frac{I}{2}, \ldots , \frac{I}{2}\right) =0$ for all $N\in {\mathcal {A}}^{k},$ we get

$$\begin{aligned} 0= & {} D\left( q_n\left( N, \frac{I}{2}, \ldots , \frac{I}{2}\right) \right) \\= & {} q_n\left( D(N), \frac{I}{2}, \ldots , \frac{I}{2}\right) +q_n\left( N, D\left( \frac{I}{2}\right) , \ldots , \frac{I}{2}\right) +\cdots +q_n\left( N, \frac{I}{2}, \ldots , D\left( \frac{I}{2}\right) \right) \\= & {} \frac{I}{2}\bigg \{D(N)\frac{I}{2}+\frac{I}{2}D(N)^*\bigg \}. \end{aligned}$$

Thus, we obtain $D(N)^*=-D(N)$ for all $N\in {\mathcal {A}}^{k}.$

Claim 2.9

$D(I)=0,$ $D(\text {i}I)=0.$

Using Claim 2.7 and the fact that $D\left( \frac{I}{2}\right) =0,$ we see that $D(I)=D\left( \frac{I}{2}+\frac{I}{2}\right) =D\left( \frac{I}{2}\right) +D\left( \frac{I}{2}\right) =0.$ Next we show that $D(\text {i}I)=0.$ Since $D(I)=0,$ using Claim 2.7, we find that

$$\begin{aligned} 0= & {} 2^{n-1}D(I)\\= & {} D(2^{n-1}I)\\= & {} D(q_n(\text {i}I, \text {i}I, I, \ldots , I))\\= & {} q_n(D(\text {i}I), \text {i}I, I, \ldots , I)+q_n(\text {i}I, D(\text {i}I), I, \ldots , I)\\= & {} q_{n-1}(-2\text {i}D(\text {i}I)+2\text {i}D(\text {i}I)^*, I,\ldots , I)+q_{n-1}(2\text {i}D(\text {i}I)^*-2\text {i}D(\text {i}I), I,\ldots , I)\\= & {} -2^{n-1}\text {i}\{D(\text {i}I)^*-D(\text {i}I)\}. \end{aligned}$$

Thus, $D(\text {i}I)^*=D(\text {i}I).$ Moreover, by Claim 2.8, we have $D(\text {i}I)^*=-D(\text {i}I).$ Therefore, $D(\text {i}I)=0.$

Claim 2.10

For any $M\in {\mathcal {A}}^{s},$ we have $D(\text {i}M)=\text {i}D(M).$

Applying Claims 2.7, 2.8 and 2.9, we obtain

$$\begin{aligned} 2^{n-1}D(M)= & {} D(2^{n-1}M)\\= & {} D(q_n(\text {i}M, \text {i}I, I, \ldots , I))\\= & {} q_n(D(\text {i}M), \text {i}I, I, \ldots , I)\\= & {} q_{n-1}(-2\text {i}D(\text {i}M), I, \ldots , I)\\= & {} -2^{n-1}\text {i}D(\text {i}M). \end{aligned}$$

That is, $D(\text {i}M)=\text {i}D(M)$ for all $M\in {\mathcal {A}}^{s}.$

Claim 2.11

For any $T\in {\mathcal {A}},$ we have $D(T^*)=D(T)^*.$

For any $M_1, M_2\in {\mathcal {A}}^{s},$ in view of the fact that $D(I)=0,$ we have

$$\begin{aligned} D(q_n(M_1+\text {i}M_2, I, \ldots , I))= & {} q_n(D(M_1+\text {i}M_2), I, \ldots , I)\\= & {} 2^{n-2}\{D(M_1+\text {i}M_2)+D(M_1+\text {i}M_2)^*\}. \end{aligned}$$

On the other hand, using the fact that $q_n(\text {i}M, I, \ldots , I)=0$ for all $M \in {\mathcal {A}}^{s},$ and Claims 2.1, 2.10, we get

$$\begin{aligned} D(q_n(M_1+\text {i}M_2, I, \ldots , I))= & {} D(q_n(M_1, I, \ldots , I))+D(q_n(\text {i}M_2, I, \ldots , I))\\= & {} q_n(D(M_1), I, \ldots , I))+q_n(\text {i}D(M_2), I, \ldots , I)\\= & {} q_n(D(M_1)+\text {i}D(M_2), I, \ldots , I)\\= & {} 2^{n-1}D(M_1). \end{aligned}$$

From the above two expressions for $D(q_n(M_1+\text {i}M_2, I, \ldots , I)),$ we obtain

$$\begin{aligned} D(M_1+\text {i}M_2)+D(M_1+\text {i}M_2)^*=2D(M_1) \end{aligned}$$

(2.2)

for all $M_1, M_2\in {\mathcal {A}}^{s}.$ Again, invoking Claim 2.9, we have

$$\begin{aligned} D(q_n(M_1+\text {i}M_2, \text {i}I, I, \ldots , I))= & {} q_n(D(M_1+\text {i}M_2), \text {i}I, I, \ldots , I)\\= & {} q_{n-1}(-\text {i}D(M_1+\text {i}M_2)+\text {i}D(M_1+\text {i}M_2)^*, I, \ldots , I)\\= & {} -2^{n-2}\text {i}\{D(M_1+\text {i}M_2)-D(M_1+\text {i}M_2)^*. \end{aligned}$$

On the other hand, using the fact $q_n(M, \text {i}I, I, \ldots , I)=0$ and Claims 2.1, 2.9 and 2.10, we get

$$\begin{aligned} D(q_n(M_1+\text {i}M_2, \text {i}I, I, \ldots , I))= & {} D(q_n(M_1, \text {i}I, I, \ldots , I))+D(q_n(\text {i}M_2, \text {i}I, I, \ldots , I))\\= & {} q_n(D(M_1)+\text {i}D(M_2), \text {i}I, I, \ldots , I)\\= & {} q_{n-1}(2D(M_2), I, \ldots , I )\\= & {} 2^{n-1}D(M_2). \end{aligned}$$

Comparing the last two expressions for $D(q_n(M_1+\text {i}M_2, \text {i}I, I, \ldots , I)),$ we obtain

$$\begin{aligned} D(M_1+\text {i}M_2)-D(M_1+\text {i}M_2)^*=2\text {i}D(M_2) \end{aligned}$$

(2.3)

for all $M_1, M_2\in {\mathcal {A}}^{s}.$ Addition of (2.2) and (2.3) yields

$$\begin{aligned} D(M_1+\text {i}M_2)=D(M_1)+\text {i}D(M_2) \end{aligned}$$

(2.4)

for all $M_1, M_2\in {\mathcal {A}}^{s}.$

Now, we show that $D(T^*)=D(T)^*$ for all $T\in {\mathcal {A}}.$ Let $T\in {\mathcal {A}}$ be an arbitrary element. Observe that any $T\in {\mathcal {A}}$ can be written as $T=\frac{T+T^*}{2}+\text {i}\frac{T-T^*}{2\text {i}},$ where $\frac{T+T^*}{2},\frac{T-T^*}{2\text {i}}\in {\mathcal {A}}^{s}.$ Hence, without loss of generality, we assume that $T=M_1+\text {i}M_2,$ where $M_1, M_2\in {\mathcal {A}}^{s}.$ Using Claim 2.2, Eq. (2.4) and the fact that $D(-M)=-D(M)$ for all $M\in {\mathcal {A}},$ we see that

$$\begin{aligned} D(T)^*= & {} D(M_1+\text {i}M_2)^*=D(M_1)^*-\text {i}D(M_2)^*\\= & {} D(M_1)+\text {i}D(-M_2)=D(M_1-\text {i}M_2)=D(T^*) \end{aligned}$$

for all $T\in {\mathcal {A}}.$

Claim 2.12

D is additive on ${\mathcal {A}}.$

Let $T_1=M_1+\text {i}M_2, T_2=M'_1+\text {i}M'_2\in {\mathcal {A}},$ where $M_1, M_2, M'_1, M'_2\in {\mathcal {A}}^{s}.$ Using (2.4) and Claim 2.7, we have

$$\begin{aligned} D(T_1+T_2)= & {} D((M_1+M'_1)+\text {i}(M_2+M'_2))\\= & {} D(M_1+M'_1)+\text {i}D(M_2+M'_2)\\= & {} D(M_1)+\text {i}D(M_2)+D(M'_1)+\text {i}D(M'_2)\\= & {} D(M_1+\text {i}M_2)+D(M'_1+\text {i}M'_2)\\= & {} D(T_1)+D(T_2). \end{aligned}$$

Claim 2.13

For any $T\in {\mathcal {A}},$ we have $D(\text {i}T)=\text {i}D(T).$

Let $T=M_1+\text {i}M_2,$ where $M_1, M_2\in {\mathcal {A}}^{s}.$ Using Claims 2.10 and 2.12, we obtain

$$\begin{aligned} D(\text {i}T)= & {} D(\text {i}M_1-M_2)\\= & {} D(\text {i}M_1)-D(M_2)\\= & {} \text {i}D(M_1)-D(M_2)\\= & {} \text {i}\{D(M_1)+\text {i}D(M_2)\}\\= & {} \text {i}D(M_1+\text {i}M_2)\\= & {} \text {i}D(T). \end{aligned}$$

Claim 2.14

D is an additive $*$-derivation on ${\mathcal {A}}.$

In view of Claims 2.11 and 2.12, it suffices to show that D is a derivation on ${\mathcal {A}}.$ First, we show that D is a derivation on ${\mathcal {A}}^{s},$ that is, $D(MM')=D(M)M'+MD(M')$ for all $M, M'\in {\mathcal {A}}^{s}.$ Using Claims 2.2, 2.9, 2.11 and 2.13, we see that

$$\begin{aligned}{} & {} {2^{n-2}D(MM'+M'M)}\\{} & {} \quad = D(2^{n-2}(MM'+M'M))\\{} & {} \quad = D(q_n(M, M', I, \ldots , I))\\{} & {} \quad = q_n(D(M), M', I, \ldots , I)+q_n(M, D(M'), I, \ldots , I)\\{} & {} \quad = q_{n-1}(D(M)M'+M'D(M), I, \ldots , I)+q_{n-1}(MD(M')+D(M')M, I, \ldots , I)\\{} & {} \quad = 2^{n-2}\{D(M)M'+M'D(M)\}+2^{n-2}\{MD(M')+D(M')M\} \end{aligned}$$

and

$$\begin{aligned}{} & {} {2^{n-2}\text {i}D(MM'-M'M)}\\{} & {} \quad = D(2^{n-2}(\text {i}MM'-\text {i}M'M))\\{} & {} \quad = D(q_n(\text {i}M, M', I, \ldots , I))\\{} & {} \quad = q_n(D(\text {i}M), M', I, \ldots , I)+q_n(\text {i}M, D(M'), I, \ldots , I)\\{} & {} \quad = q_{n-1}(\text {i}D(M)M'-\text {i}M'D(M), I, \ldots , I)+q_{n-1}(\text {i}MD(M')-\text {i}D(M')M, I, \ldots , I)\\{} & {} \quad = 2^{n-2}\text {i}\{D(M)M'-M'D(M)\}+2^{n-2}\text {i}\{MD(M')-D(M')M\}. \end{aligned}$$

It follows from the above two equations that $D(MM')=D(M)M'+MD(M')$ for all $M, M'\in {\mathcal {A}}^{s}.$ Take any two arbitrary elements $T_1=M_1+\text {i}M_2, T_2=M'_1+\text {i}M'_2 \in {\mathcal {A}},$ where $M_1, M_2, M'_1, M'_2\in {\mathcal {A}}^{s}.$ Using Claim 2.13, we get

$$\begin{aligned} D(T_1T_2)= & {} D((M_1M'_1-M_2M'_2)+\text {i}(M_1M'_2+M_2M'_1))\\= & {} D(M_1M'_1-M_2M'_2)+\text {i}D(M_1M'_2+M_2M'_1)\\= & {} D(M_1M'_1)-D(M_2M'_2)+\text {i}D(M_1M'_2)+\text {i}D(M_2M'_1)\\= & {} D(M_1)M'_1+M_1D(M'_1)-D(M_2)M'_2-M_2D(M'_2)+\text {i}D(M_1)M'_2\\{} & {} +\text {i}M_1D(M'_2)+\text {i}D(M_2)M'_1+\text {i}M_2D(M'_1) \end{aligned}$$

and

$$\begin{aligned} D(T_1)T_2+T_1D(T_2)= & {} D(M_1+\text {i}M_2)(M'_1+\text {i}M'_2)+(M_1+\text {i}M_2)D(M'_1+\text {i}M'_2)\\= & {} D(M_1)M'_1+M_1D(M'_1)-D(M_2)M'_2-M_2D(M'_2)+\text {i}D(M_1)M'_2\\{} & {} +\text {i}M_1D(M'_2)+\text {i}D(M_2)M'_1+\text {i}M_2D(M'_1). \end{aligned}$$

Thus, $D(T_1T_2)=D(T_1)T_2+T_1D(T_2)$ for all $T_1, T_2\in {\mathcal {A}}.$ Consequently, D is an additive $*$-derivation and the proof of the theorem is completed. $\square$

Now we apply Theorem 2.1 to certain special classes of $*$-algebras, namely prime $*$-algebras, factor von Neumann algebras and von Neumann algebras with no central summands of type $I_{1}.$

Recall that an algebra ${\mathcal {A}}$ is prime if for any $T_1, T_2\in {\mathcal {A}},$ $T_1{\mathcal {A}}T_2=\{0\}$ implies that either $T_1=0$ or $T_2=0.$ It is easy to verify that every prime $*$-algebra with a nontrivial projection satisfies (2.1). Therefore, as a direct consequence of Theorem 2.1, we have:

Corollary 2.1

Let ${\mathcal {A}}$ be a unital prime $*$-algebra with a nontrivial projection. Then, every nonlinear bi-skew Jordan n-derivation of ${\mathcal {A}}$ is an additive $*$-derivation.

Let ${\mathcal {H}}$ be a complex Hilbert space and $\mathcal {B(H)}$ be the algebra of all bounded linear operators on ${\mathcal {H}}$. Let $\mathcal {F(H)}\subseteq \mathcal {B(H)}$ denote the subalgebra of all bounded finite rank operators. A subalgebra ${\mathcal {A}}\subseteq \mathcal {B(H)}$ is called a standard operator algebra if it contains $\mathcal {F(H)}.$ Now we have the following result:

Corollary 2.2

Let ${\mathcal {H}}$ be an infinite dimensional complex Hilbert space and ${\mathcal {A}}$ be a standard operator algebra on ${\mathcal {H}}$ containing the identity operator I. Suppose that ${\mathcal {A}}$ is closed under the adjoint operation. Then every nonlinear bi-skew Jordan n-derivation $D:{\mathcal {A}}\rightarrow \mathcal {B(H)}$ is a linear $*$-derivation. Moreover, there exists an operator $T\in \mathcal {B(H)}$ satisfying $T+T^{*}=0$ such that $D(A)=AT-TA$ for all $A\in {\mathcal {A}},$ that is, D is inner.

Proof

Since ${\mathcal {A}}$ is a unital prime $*$-algebra containing nontrivial projections, by Corollary 2.1, we see that D is an additive $*$-derivation. It follows from [26] that D is an linear inner derivation; that is, there exists an operator $S\in \mathcal {B(H)}$ such that $D(A)=AS-SA$ for all $A\in {\mathcal {A}},$ Using the fact that $D(A^{*})={D(A)}^{*},$ we have

$$\begin{aligned} A^{*}S-SA^{*}=D(A^{*})={D(A)}^{*}=S^{*}A^{*}-A^{*}S^{*} \end{aligned}$$

for any $A\in {\mathcal {A}},$ This leads to $A^{*}(S+S^{*}) = (S+S^{*})A^{*}.$ Hence, $S+S^{*}=\lambda I$ for some $\lambda \in {\mathbb {R}}.$ Letting $T = S-\frac{1}{2}\lambda I,$ one can check that $T + T^{*} = 0$ and $D(A)=AT-TA$ for all $A\in {\mathcal {A}}.$ $\square$

A von Neumann algebra ${\mathcal {A}}$ is a weakly closed, self-adjoint subalgebra of $\mathcal {B(H)}$ containing the identity operator I. A von Neumann algebra ${\mathcal {A}}$ is a factor if its center is ${\mathbb {C}}I$. It is well known that a factor von Neumann algebra is prime. Hence, as an immediate consequence of Corollary 2.1, we get:

Corollary 2.3

[2, Theorem 2.1] Let ${\mathcal {A}}$ be a factor von Neumann algebra with dim$({\mathcal {A}})\ge 2.$ Then, every nonlinear bi-skew Jordan n-derivation of ${\mathcal {A}}$ is an additive $*$-derivation.

Further, it is well known that every von Neumann algebra with no central summands of type $I_1$ satisfies (2.1) (see [7, 18] for details). Therefore, applying Theorem 2.1, we obtain:

Corollary 2.4

Let ${\mathcal {A}}$ be a von Neumann algebra with no central summands of type $I_1.$ Then, every nonlinear bi-skew Jordan n-derivation of ${\mathcal {A}}$ is an additive $*$-derivation.

3 Nonlinear Generalized Bi-skew Jordan n-Derivations

In this section, we present the second main result of this article which gives a description of nonlinear generalized bi-skew Jordan n-derivations of unital $*$-algebras with nontrivial projections. Observe that if $G:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is a nonlinear generalized bi-skew Jordan n-derivation with associated nonlinear bi-skew Jordan n-derivation $D:{\mathcal {A}}\rightarrow {\mathcal {A}},$ then it follows from (1.1) and (1.2) that a map $\delta =G-D$ satisfies

$$\begin{aligned} \delta (p_n(T_1, T_2, \ldots , T_n))=p_n(\delta (T_1), T_2, \ldots , T_n) \end{aligned}$$

(3.1)

for all $T_1, T_2, \ldots , T_n\in {\mathcal {A}}.$ Therefore, to obtain a description of the map G it suffices to consider a map $\delta$ with property (3.1). The following proposition provides a description of such maps on unital $*$-algebras containing nontrivial projections.

Proposition 3.1

Let ${\mathcal {A}}$ be a unital $*$-algebra with a nontrivial projection. Then, a map $\delta :{\mathcal {A}}\rightarrow {\mathcal {A}}$ satisfying $\delta (q_n(T_1,T_2, \ldots , T_n))=q_n(\delta (T_1),T_2, \ldots , T_n)$ for all $T_1,T_2, \ldots , T_n\in {\mathcal {A}}$ is of the form $\delta (T)=ZT$ for all $T\in {\mathcal {A}},$ where $Z\in {\mathcal {Z}}({\mathcal {A}}).$

Proof

In the following, we prove that the proposition holds by checking several claims.

Claim 3.1

For any $T_1, \ldots T_n\in {\mathcal {A}},$ $\delta (q_n(T_1, T_2, T_3, \ldots , T_n))=q_n(\delta (T_1), T_2, T_3, \ldots , T_n)$ if and only if $\delta (q_n(T_1, T_2, T_3, \ldots , T_n))=q_n(T_1, \delta (T_2), T_3, \ldots , T_n).$

Suppose that

$$\begin{aligned} \delta (q_n(T_1, T_2, T_3, \ldots , T_n))=q_n(\delta (T_1), T_2, T_3, \ldots , T_n) \end{aligned}$$

for all $T_1, \ldots , T_n\in {\mathcal {A}}.$ Using the fact that $q_n(T_1, T_2, T_3, \ldots , T_n)=q_{n-1}(T_1\diamond T_2, T_3, \ldots , T_n),$ we get

$$\begin{aligned} \delta (q_{n-1}(T_1\diamond T_2, T_3, \ldots , T_n))=q_{n-1}(\delta (T_1)\diamond T_2, T_3, \ldots , T_n) \end{aligned}$$

for all $T_1, \ldots , T_n\in {\mathcal {A}}.$ Interchanging $T_1$ and $T_2$ in the above expression, we obtain

$$\begin{aligned} \delta (q_{n-1}(T_2\diamond T_1, T_3, \ldots , T_n))=q_n(\delta (T_2)\diamond T_1, T_3, \ldots , T_n) \end{aligned}$$

for all $T_1, \ldots , T_n\in {\mathcal {A}}.$ Since $T_1\diamond T_2=T_2\diamond T_1$ for all $T_1,T_2\in {\mathcal {A}},$ we have

$$\begin{aligned} \delta (q_{n-1}(T_1\diamond T_2, X_3, \ldots , T_n))=q_{n-1}(T_1\diamond \delta (T_2), T_3, \ldots , T_n) \end{aligned}$$

for all $T_1, \ldots , T_n\in {\mathcal {A}}.$ This implies that

$$\begin{aligned} \delta (q_n(T_1, T_2, T_3, \ldots , T_n))=q_n(T_1, \delta (T_2), T_3, \ldots , T_n) \end{aligned}$$

for all $T_1, \ldots , T_n\in {\mathcal {A}}.$ Similarly, one can verify the ‘only if’ part.

Claim 3.2

$\delta$ is additive.

For any $T_1, T_2\in {\mathcal {A}},$ set $S=\delta (T_1+T_2)-\delta (T_1)-\delta (T_2).$ Our aim is to show $S=0.$ For any $W\in {\mathcal {A}},$ using Claim 3.1, we get

$$\begin{aligned} q_n(\delta (T_1+T_2), W, I, \ldots , I)= & {} q_n(T_1+T_2, \delta (W), I, \ldots , I)\\= & {} q_n(T_1, \delta (W), I, \ldots , I)+q_n(T_2, \delta (W), I, \ldots , I)\\= & {} q_n(\delta (T_1), W, I, \ldots , I)+q_n(\delta (T_2), W, I, \ldots , I)\\= & {} q_n(\delta (T_1)+\delta (T_2), W, I, \ldots , I). \end{aligned}$$

This implies that

$$\begin{aligned} q_n(S, W, I, \ldots , I)=0 \ \ \text {for all}\ \ W\in {\mathcal {A}}. \end{aligned}$$

(3.2)

Choosing $W=I$ in (3.2) and simplifying, we get $S^*=-S.$ Similarly, taking $W=\text {i}I$ in (3.2) and simplifying, we obtain $S^*=S.$ Thus, $S=0$ and hence $\delta$ is additive.

Claim 3.3

We have

(i):

$\delta (iT)=i\delta (T)$ for all $T\in {\mathcal {A}},$

(ii):

$\delta (T^*)=\delta (T)^*$ for all $T\in {\mathcal {A}}.$

(i):

First we show that $\delta (iM)=i\delta (M)$ for all $M\in {\mathcal {A}}^{s}.$ Since $\delta$ is additive, we have $\delta (0)=0.$ Let $T=M_1+\text {i}M_2$ for some $M_1, M_2\in {\mathcal {A}}^{s}.$ As $q_n(\text {i}M_2, I, \ldots , I)=0,$ using Claim 3.2, we have

$$\begin{aligned} 2^{n-1}\delta (M_1)= & {} \delta (2^{n-1}M_1)\\= & {} \delta (q_n(M_1+\text {i}M_2, I, \ldots , I))\\= & {} q_n(\delta (M_1)+\delta (\text {i}M_2), I, \ldots , I)\\= & {} 2^{n-2}\{\delta (M_1)+\delta (\text {i}M_2)+\delta (M_1)^*+\delta (\text {i}M_2)^*\}. \end{aligned}$$

That is,

$$\begin{aligned} 2\delta (M_1)=\{\delta (M_1)+\delta (\text {i}M_2)+\delta (M_1)^*+\delta (\text {i}M_2)^*\}. \end{aligned}$$

(3.3)

Similarly, as $q_n(M_1, \text {i}I, I, \ldots , I)=0,$ we have

$$\begin{aligned} 2^{n-1}\delta (M_2)= & {} \delta (2^{n-1}M_2)\\= & {} \delta (q_n(M_1+\text {i}M_2, \text {i}I, I, \ldots , I))\\= & {} q_n(\delta (M_1)+\delta (\text {i}M_2), \text {i}I, I, \ldots , I)\\= & {} q_{n-1}(\text {i}(-\delta (M_1)-\delta (\text {i}M_2)+\delta (M_1)^*+\delta (\text {i}M_2)), I, \ldots , I)\\= & {} 2^{n-2}\text {i}\{-\delta (M_1)-\delta (\text {i}M_2)+\delta (M_1)^*+\delta (\text {i}M_2)\}. \end{aligned}$$

That is,

$$\begin{aligned} 2\text {i}\delta (M_2)=\{\delta (M_1)+\delta (\text {i}M_2)-\delta (M_1)^*-\delta (\text {i}M_2)^*\}. \end{aligned}$$

(3.4)

Addition of (3.3) and (3.4) yields $\delta (\text {i}M_2)=\text {i}\delta (M_2)$ for all $M_2\in {\mathcal {A}}^{s}.$ That is, $\delta (\text {i}M)=\text {i}\delta (M)$ for all $M\in {\mathcal {A}}^{s}.$ Now, let $T=M_1+\text {i}M_2,$ where $M_1, M_2\in {\mathcal {A}}^{s}.$ Using and Claim 3.2, we obtain

$$\begin{aligned} \delta (\text {i}T)= & {} \delta (\text {i}M_1-M_2)\\= & {} \delta (\text {i}M_1)-\delta (M_2)\\= & {} \text {i}\delta (M_1)-\delta (M_2)\\= & {} \text {i}\{\delta (M_1)+\text {i}\delta (M_2)\}\\= & {} \text {i}\delta (M_1+\text {i}M_2)\\= & {} \text {i}\delta (T). \end{aligned}$$

(ii):

First we show that $\delta (M)^*=\delta (M)$ for all $M\in {\mathcal {A}}^{s}.$ Observe that $q_n\big (M, \frac{I}{2}, \ldots , \frac{I}{2}\big )=M$ for all $M\in {\mathcal {A}}^{s}.$ Thus, we see that

$$\begin{aligned} \delta (M)=\delta \bigg (q_n\bigg (M, \frac{I}{2}, \ldots , \frac{I}{2}\bigg )\bigg ) =q_n\bigg (\delta (M), \frac{I}{2}, \ldots , \frac{I}{2}\bigg ) =\frac{\delta (M)}{2}+\frac{\delta (M)^*}{2} \end{aligned}$$

Therefore, $\delta (M)^*=\delta (M)$ for all $M\in {\mathcal {A}}^{s}.$ Now, let $T=M_1+\text {i}M_2,$ where $M_1, M_2\in {\mathcal {A}}^{s}.$ Then, using part (i) and Claim 3.2, we find that

$$\begin{aligned} \delta (T)^*=\delta (M_1)^*-\text {i}\delta (M_2)^*=\delta (M_1^*)+\delta (-\text {i}M_2^*) =\delta ((M_1-\text {i}M_2)^*)=\delta (T^*) \end{aligned}$$

for all $T\in {\mathcal {A}}.$

Claim 3.4

$\delta (T_1T_2)=\delta (T_1)T_2=T_1\delta (T_2)$ for all $T_1, T_2\in {\mathcal {A}}.$

First we show that $\delta (M_1M_2)=\delta (M_1)M_2=M_1\delta (M_2)$ for all $M_1, M_2\in {\mathcal {A}}^{s}.$ Invoking Claims 3.2 and 3.3, we find that

$$\begin{aligned} 2^{n-2}\{\delta (M_1M_2)+\delta (M_2M_1)\}= & {} \delta (q_n(M_1, M_2, I, \ldots , I))\\= & {} q_n(\delta (M_1), M_2, I, \ldots , I)\\= & {} q_{n-1}(\delta (M_1)M_2+M_2\delta (M_1), I, \ldots , I)\\= & {} 2^{n-2}\{\delta (M_1)M_2+M_2\delta (M_1)\}. \end{aligned}$$

That is,

$$\begin{aligned} \delta (M_1M_2)+\delta (M_2M_1)=\delta (M_1)M_2+M_2\delta (M_1) \ \ \text {for all} \ \ M_1, M_2\in {\mathcal {A}}^{s}. \end{aligned}$$

(3.5)

Similarly,

$$\begin{aligned} 2^{n-2}\text {i}\{\delta (M_1M_2)-\delta (M_2M_1)\}= & {} \delta (q_n(\text {i}M_1, M_2, I, \ldots , I))\\= & {} q_n(\delta (\text {i}M_1), M_2, I, \ldots , I)\\= & {} q_n(\text {i}\delta (M_1), M_2, I, \ldots , I)\\= & {} q_{n-1}(\text {i}\delta (M_1)M_2-\text {i}M_2\delta (M_1), I, \ldots , I)\\= & {} 2^{n-2}\text {i}\{\delta (M_1)M_2-M_2\delta (M_1)\}. \end{aligned}$$

That is,

$$\begin{aligned} \delta (M_1M_2)-\delta (M_2M_1)=\delta (M_1)M_2-M_2\delta (M_1) \ \ \text {for all} \ \ M_1, M_2\in {\mathcal {A}}^{s}. \end{aligned}$$

(3.6)

Addition of (3.5) and (3.6) yields $\delta (M_1M_2)=\delta (M_1)M_2$ for all $M_1, M_2\in {\mathcal {A}}^{s}.$ Similarly, one can obtain that $\delta (M_1M_2)=\delta (M_1)M_2$ for all $M_1, M_2\in {\mathcal {A}}^{s}.$ Now, taking two arbitrary elements $T_1=M_1+\text {i}M_2, T_2=M'_1+\text {i}M'_2\in {\mathcal {A}},$ where $M_1, M'_1,M_2, M'_2\in {\mathcal {A}}^{s},$ and using Claims 3.2 and 3.3, we see that

$$\begin{aligned} \delta (T_1T_2)= & {} \delta ((M_1M'_1-M_2M'_2)+\text {i}(M_1M'_2+M_2M'_1))\\= & {} \delta (M_1M'_1)-\delta (M_2M'_2)+\text {i}\delta (M_1M'_2)+\text {i}\delta (M_2M'_1)\\= & {} \delta (M_1)M'_1-\delta (M_2)M'_2+\text {i}\delta (M_1)M'_2+\text {i}\delta (M_2)M'_1. \end{aligned}$$

Also,

$$\begin{aligned} \delta (T_1)T_2= & {} \delta (M_1+\text {i}M_2)(M'_1+\text {i}M'_2)\\= & {} (\delta (M_1)+\text {i}\delta (M_2))(M'_1+\text {i}M'_2)\\= & {} \delta (M_1)M'_1-\delta (M_2)M'_2+\text {i}\delta (M_1)M'_2+\text {i}\delta (M_2)M'_1. \end{aligned}$$

Therefore, $\delta (T_1T_2)=\delta (T_1)T_2$ for all $T_1, T_2\in {\mathcal {A}}.$ Symmetrically, one can obtain that $\delta (T_1T_2)=T_1 \delta (T_2)$ for all $T_1, T_2\in {\mathcal {A}}.$

Claim 3.5

$\delta (T)=ZT$ for all $T\in {\mathcal {A}},$ where $Z\in {\mathcal {Z}}({\mathcal {A}}).$

Let T be an arbitrary element in ${\mathcal {A}}.$ Using Claim 3.4, we have

$$\begin{aligned} \delta (T)=\delta (IT)=\delta (I)T \ \ \text {and} \ \ \delta (T)=\delta (TI)=T\delta (I). \end{aligned}$$

This implies that $\delta (I)\in {\mathcal {Z}}({\mathcal {A}}).$ Thus, $\delta (T)=ZT$ for all $T\in {\mathcal {A}},$ where $Z=\delta (I)\in {\mathcal {Z}}({\mathcal {A}}).$ $\square$

The following proposition provides the structure of nonlinear generalized bi-skew Jordan n-derivations on unital $*$-algebras with nontrivial projection (not necessarily satisfying (2.1)).

Proposition 3.2

Let ${\mathcal {A}}$ be a unital $*$-algebra with a nontrivial projection. Then, every nonlinear generalized bi-skew Jordan n-derivation $G:{\mathcal {A}}\rightarrow {\mathcal {A}}$ with associated nonlinear bi-skew Jordan n-derivation $D:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is of the form $G(T)=ZT+D(T)$ for all $T\in {\mathcal {A}},$ where $Z\in {\mathcal {Z}}({\mathcal {A}}).$

Proof

Let $G:{\mathcal {A}}\rightarrow {\mathcal {A}}$ be a nonlinear generalized bi-skew Jordan n-derivation with associated nonlinear bi-skew Jordan n-derivation $D:{\mathcal {A}}\rightarrow {\mathcal {A}}.$ Then, the map $\delta =G-D$ satisfies

$$\begin{aligned} \delta (p_n(T_1, T_2, \ldots , T_n))=p_n(\delta (T_1), T_2, \ldots , T_n) \end{aligned}$$

for all $T_1, T_2, \ldots , T_n\in {\mathcal {A}}.$ Hence, it follows from Proposition 3.1 that $\delta (T)=Z T$ for all $T\in {\mathcal {A}},$ where $Z\in {\mathcal {Z}}({\mathcal {A}}).$ Consequently, $G(T)=ZT+D(T)$ for all $T\in {\mathcal {A}},$ where $Z\in {\mathcal {Z}}({\mathcal {A}}).$ This completes the proof. $\square$

Now, we are in a position to state and prove the second main result of the article.

Theorem 3.1

Let ${\mathcal {A}}$ be a unital $*$-algebra with a nontrivial projection $P_1$ satisfying (2.1). Then every nonlinear generalized bi-skew Jordan n-derivation $G:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is of the form $G(T)=ZT+D(T)$ for all $T\in {\mathcal {A}},$ where $Z\in {\mathcal {Z}}({\mathcal {A}})$ and $D:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is an additive $*$-derivation. Moreover, $G:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is an additive generalized $*$-derivation.

Proof

Let $G:{\mathcal {A}}\rightarrow {\mathcal {A}}$ be a nonlinear generalized bi-skew Jordan n-derivation with associated nonlinear bi-skew Jordan n-derivation $D:{\mathcal {A}}\rightarrow {\mathcal {A}}.$ Then, it follow from Proposition 3.2 that G is of the form $G(T)=Z T+D(T)$ for all $T\in {\mathcal {A}},$ where $Z\in {\mathcal {Z}}({\mathcal {A}}).$ Moreover, in view of Theorem 2.1, D is an additive $*$-derivation. We only need to show that G is an additive generalized $*$-derivation. We assert that G is an additive generalized $*$-derivation with associated additive $*$-derivation D. Clearly, G is additive. For any $T_1, T_2\in {\mathcal {A}},$ we have

$$\begin{aligned} G(T_1)T_2+T_1D(T_2)= & {} (ZT_1+D(T_1))T_2+T_1D(T_2)\\= & {} (ZT_1)T_2+D(T_1)T_2+T_1D(T_2)\\= & {} Z(T_1T_2)+D(T_1T_2)\\= & {} G(T_1T_2). \end{aligned}$$

Now, define a map $\delta =G-D.$ Then, $\delta$ satisfies (3.1). Hence, by Claim 3.3, $\delta (T^*)={\delta (T)}^*$ for all $T\in {\mathcal {A}}.$ Thus, for any $T\in {\mathcal {A}},$ we get

$$\begin{aligned} G(T^*)=\delta (T^*)-D(T^*)=\delta (T)^*-D(T)^*=(\delta (T)-D(T))^*=G(T)^*. \end{aligned}$$

Consequently, G is an additive generalized $*$-derivation with associated additive $*$-derivation D. $\square$

As we have discussed earlier that unital prime $*$-algebras with nontrivial projections, factor von Neumann algebras and von Neumann algebras with no central summands of type $I_1$ satisfy (2.1). Therefore, as a direct consequence of Theorem 3.1, we have the following results.

Corollary 3.1

Let ${\mathcal {A}}$ be a unital prime $*$-algebra with a nontrivial projection. Then, every nonlinear generalized bi-skew Jordan n-derivation $G:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is of the form $G(T)=ZT+D(T)$ for all $T\in {\mathcal {A}},$ where $Z\in {\mathcal {Z}}({\mathcal {A}})$ and $D:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is an additive $*$-derivation. Moreover, $G:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is an additive generalized $*$-derivation.

Corollary 3.2

Let ${\mathcal {A}}$ be a factor von Neumann algebra with dim$({\mathcal {A}})\ge 2.$ Then, every nonlinear generalized bi-skew Jordan n-derivation $G:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is of the form $G(T)=ZT+D(T)$ for all $T\in {\mathcal {A}},$ where $Z\in {\mathcal {Z}}({\mathcal {A}})$ and $D:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is an additive $*$-derivation. Moreover, $G:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is an additive generalized $*$-derivation.

Corollary 3.3

Let ${\mathcal {A}}$ be a von Neumann algebra having no central summands of type $I_1.$ Then, every nonlinear generalized bi-skew Jordan n-derivation $G:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is of the form $G(T)=ZT+D(T)$ for all $T\in {\mathcal {A}},$ where $Z\in {\mathcal {Z}}({\mathcal {A}})$ and $D:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is an additive $*$-derivation. Moreover, $G:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is an additive generalized $*$-derivation.