Characterization of Non-Linear Bi-Skew Jordan n-Derivations on Prime ∗-Algebras

Abstract

Let $\mathfrak{A}$ be a prime *-algebra. A product defined as $U•V=U{V}^{\ast }+V{U}^{\ast }$ for any $U,V\in \mathfrak{A}$, is called a bi-skew Jordan product. A map $\xi :\mathfrak{A}\to \mathfrak{A}$, defined as $\xi \left(p_n\left(U_1,U_2,\cdots ,U_n\right)\right)={\sum }_{k=1}^n{p}_n\left(U_1,U_2,...,U_{k-1},\xi \left(U_k\right),U_{k+1},\cdots ,U_n\right)$ for all $U_1,U_2,...,U_n\in \mathfrak{A}$, is called a non-linear bi-skew Jordan n-derivation. In this article, it is shown that $\xi$ is an additive ∗-derivation.

1. Introduction

Let $\mathfrak{A}$ be an associative ∗-algebra. Recall that a map $\xi :\mathfrak{A}\to \mathfrak{A}$, is called an additive derivation if $\xi (U+V)=\xi \left(U\right)+\xi \left(V\right)$ and $\xi \left(UV\right)=\xi \left(U\right)V+U\xi \left(V\right)$ for all $U,V\in \mathfrak{A}$. Let $U\ast V=UV+V{U}^{\ast }$ and ${[U,V]}_{\ast }=UV-V{U}^{\ast }$ denote the skew Jordan product and skew Lie product of elements $U,V\in \mathfrak{A}$, respectively. These products are also called ∗-Jordan product and ∗-Lie product, respectively. The difficulty of the representability of quadratic functionals by sesqui-linear functionals on left-modules over ∗-algebras is greatly impacted by the existence of such Jordan bracket-based products in regard to the so-called Jordan ∗-derivations (see [1,2,3]). We say a map $\xi :\mathfrak{A}\to \mathfrak{A}$, without considering the linearity assumption, is called a multiplicative skew (or ∗)-Jordan derivation if

$$\xi (U\ast V)=\xi \left(U\right)\ast V+U\ast \xi \left(V\right)$$

for all $U,V\in \mathfrak{A}$. Furthermore, without the linearity assumption, a map $\xi :\mathfrak{A}\to \mathfrak{A}$ is called a multiplicative skew or ∗-Jordan triple derivation if it satisfies

$$\xi (U\ast V\ast W)=\xi \left(U\right)\ast V\ast W+U\ast \xi \left(V\right)\ast W+U\ast V\ast \xi \left(W\right)$$

for all $U,V,W\in \mathfrak{A}$. A map $\xi :\mathfrak{A}\to \mathfrak{A}$ is said to be an additive ∗-derivation if it is an additive derivation and satisfies $\xi \left(U^{\ast }\right)=\xi {\left(U\right)}^{\ast }$ for all $U\in \mathfrak{A}$. Many authors investigated the structure of skew Jordan derivations and skew Jordan triple derivations on different algebras, see, e.g., [2,3,4,5,6]. For instance, Taghavi et al. [5] showed that a non-linear ∗-Jordan derivation on a factor von Neumann algebra is an additive ∗-derivation. Zhao and Li [6] proved that every non-linear ∗-Jordan triple derivation on a von Neumann algebra with no central summands of type $I_1$ is an additive ∗-derivation. A lot of work was also carried out by considering Lie product ($[U,V]=UV-VU$) and ∗-Lie product (${[U,V]}_{\ast }=UV-V{U}^{\ast }$) (see [7,8,9,10,11,12,13,14,15,16]). In [15], Yu and Zhang proved that every non-linear Lie derivation on triangular algebras has the standard form, i.e., it is a sum of an additive derivation and a central valued map. Furthermore, the authors of [7,8], respectively, established that a non-linear Lie triple derivation on triangular algebras and a non-linear Lie type derivation on von Neumann algebras have the standard form. The structure of non-linear ∗-Lie derivation on factor von Neumann algebra was also explored by Yu and Zhang [16], and they proved that such a map is an additive ∗-derivation. On similar grounds, the characterization of non-linear skew Lie triple derivations on factor von Neumann algebras [11], non-linear ∗-Lie derivations on standard operator algebras [9], non-linear ∗-Lie-type derivations on von Neumann algebras [12] and non-linear ∗-Lie type derivations on standard operator algebras [13] is performed, and they are proven to be additive ∗-derivations.

Let us recall the definition of a prime ∗-algebra. A prime ∗-algebra is an algebra $\mathfrak{A}$ with involution ∗, in which $U\mathfrak{A}V$ equates to $\left(0\right)$, gives either $U=0$ or $V=0$. The class of prime ∗-algebras is very important and has numerous applications in various disciplines. In the context of operator theory and quantum mechanics, prime ∗-algebras are used to study the behavior of operators on Hilbert spaces and provide insights into the nature of physical observables and symmetries in quantum systems. Prime ∗-algebras are a larger class containing factor von Neumann algebras and standard operator algebras. Therefore, it would be of great importance to characterize a map on prime ∗-algebras. In recent years, some mathematicians focus to explore the structure of ∗-Jordan type derivations on prime ∗-algebras, see [17,18]. Inspired by skew Jordan product, very recently, Kong and Li [19] introduced a new product, namely, bi-skew Jordan product, as $U•V=U{V}^{\ast }+V{U}^{\ast }$ for all $U,V\in \mathfrak{A}$. They proved that every non-linear/multiplicative bi-skew Jordan derivation, i.e., a map $\xi$ from $\mathfrak{A}$ to itself, (where $\mathfrak{A}$ is a prime ∗-algebra) satisfying $\xi (U•V)=\xi \left(U\right)•V+U•\xi \left(V\right)$ for all $U,V\in \mathfrak{A}$, is an additive ∗-derivation on $\mathfrak{A}$ provided $dim\left(\mathfrak{A}\right)\ge 2$. Later, Khan and Alhazmi [20] extended the results of Kong and Li [19] to multiplicative bi-skew Jordan triple derivation and proved that every multiplicative bi-skew Jordan triple derivation, i.e., a map $\xi :\mathfrak{A}\to \mathfrak{A}$ satisfying $\xi (U•V•W)=\xi \left(U\right)•V•W+U•\xi \left(V\right)•W+U•V•\xi \left(W\right)$ for all $U,V,W\in \mathfrak{A}$, is an additive ∗-derivation. We can naturally develop them further when bi-skew Jordan derivations and bi-skew Jordan triple derivations are taken into account. Let’s assume that $n\ge 2$ is a fixed positive integer and see the list of polynomials with involution.

$$\begin{array}{ccc}\hfill p_1\left(U_1\right)& =& U_1,\hfill \\ \hfill p_2(U_1,U_2)& =& p_1\left(U_1\right)•U_2=U_1•U_2=U_1{U}_2^{\ast }+U_2{U}_1^{\ast },\hfill \\ \hfill p_3(U_1,U_2,U_3)& =& p_2(U_1,U_2)•U_3=U_1•U_2•U_3,\hfill \\ \hfill p_4(U_1,U_2,U_3,U_4)& =& p_3(U_1,U_2,U_3)•U_4=U_1•U_2•U_3•U_4\hfill \\ & & ...,\hfill \\ \hfill p_n(U_1,U_2,U_3...,U_n)& =& p_{n-1}(U_1,U_2,...,U_{n-1})•U_n\hfill \\ & =& U_1•U_2•U_3•...U_{n-1}•U_n\hfill \end{array}$$

Accordingly, a multiplicative bi-skew Jordan n-derivation is a mapping $\xi :\mathfrak{A}\to \mathfrak{A}$, satisfying the condition

$$\begin{array}{ccc}\hfill \xi \left(p_n\left(U_1,U_2,\cdots ,U_n\right)\right)& =& \sum _{k=1}^n{p}_n\left(U_1,\cdots ,U_{k-1},\xi \left(U_k\right),U_{k+1},\cdots ,U_n\right),\hfill \end{array}$$

for all $U_1,U_2,\cdots ,U_n\in \mathfrak{A}$. This is the best way to define multiplicative bi-skew Jordan n-derivations, using this notion. Every multiplicative bi-skew Jordan derivation is a multiplicative bi-skew Jordan 2-derivation according to the definition, and every multiplicative bi-skew Jordan triple derivation is a multiplicative bi-skew Jordan 3-derivation. One can easily check that every multiplicative bi-skew Jordan derivation on any ∗-algebra is a multiplicative bi-skew Jordan triple derivation but the converse is not true, in general. Multiplicative bi-skew Jordan-type derivations refer to the multiplicative bi-skew Jordan 2-, multiplicative bi-skew Jordan 3- and multiplicative bi-skew Jordan n-derivations. Inspired by the above mentioned work in this article, we focus our study on multiplicative bi-skew Jordan type derivations on prime ∗-algebras.

2. Preliminaries

We need to give some preliminaries in order to state and prove our main theorem. Throughout the work, $\mathfrak{A}$ represents a prime ∗-algebra and $\mathbb{C}$ denotes the field of complex numbers. Let H be a complex Hilbert space. We denote by $\mathcal{B}\left(H\right)$ the algebra of all bounded linear operators on H. An operator $P\in \mathcal{B}\left(H\right)$ is called a projection provided $P^{\ast }=P$ and $P^2=P$. Any operator $U\in \mathcal{B}\left(H\right)$ can be expressed as $U=\mathfrak{R}U+i\mathfrak{T}U$, where i is the imaginary unit, $\mathfrak{R}U=\frac{U+U^{\ast }}{2}$ and $\mathfrak{T}U=\frac{U-U^{\ast }}{2i}$. Note that both $\mathfrak{R}U$ and $\mathfrak{T}U$ are self-adjoint. Let $P=P_1\in \mathfrak{A}$ be a projection. Write $P_2=I-P_1$ and ${\mathfrak{A}}_{ij}=P_i\mathfrak{A}{P}_j$. Then, $\mathfrak{A}={\mathfrak{A}}_{11}+{\mathfrak{A}}_{12}+{\mathfrak{A}}_{21}+{\mathfrak{A}}_{22}$. Let $\mathcal{M}=\{M\in \mathfrak{A}|M^{\ast }=M\}$ and $\mathcal{N}=\{N\in \mathfrak{A}|N^{\ast }=-N\}$, ${\mathcal{M}}_{12}=\{P_{1}M{P}_2+P_{2}M{P}_1|M\in \mathcal{M}\}$ and ${\mathcal{M}}_{ii}=P_i\mathcal{M}{P}_i$$(i=1,2)$. Thus, for every $M\in \mathcal{M},M=M_{11}+M_{12}+M_{22}$ for every $M_{12}\in {\mathcal{M}}_{12}$ and $M_{ii}\in {\mathcal{M}}_{ii}$$(i=1,2)$.

In proving our main theorem, we frequently use the following lemma and remark.

Lemma 1. 

For any $U\in \mathfrak{A},p_n\left(U,\frac{1}{2}I,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)=\frac{1}{2}(U+U^{\ast })$.

Proof. 

By doing the recursive calculation, we obtain

$$\begin{array}{ccc}\hfill p_n\left(U,\frac{1}{2}I,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)& =& p_{n-1}\left(\frac{1}{2}(U+U^{\ast }),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & =& p_{n-2}\left(\frac{1}{2}(U+U^{\ast }),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & =& ...=p_2\left(\frac{1}{2}(U+U^{\ast }),\frac{1}{2}I\right)\hfill \\ & =& \frac{1}{2}(U+U^{\ast }).\hfill \end{array}$$

□

Remark 1. 

If $U\in \mathcal{M}$, i.e., $U^{\ast }=U$, then

$$p_n\left(U,\frac{1}{2}I,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)=U.$$

3. Main Result

Theorem 1. 

Let $\mathfrak{A}$ be a prime ∗-algebra with $dim\left(\mathfrak{A}\right)\ge 2$, containing the identity element I and a non-trivial projection P. A map $\xi :\mathfrak{A}\to \mathfrak{A}$ is a multiplicative bi-skew Jordan-type derivation if and only if it is an additive ∗-derivation.

Only the necessity needs to be established. The proof of the theorem is demonstrated in a series of claims, which are as follows.

Claim 1. 

$\xi \left(0\right)=0$.

Proof. 

It follows that

$$\begin{array}{ccc}\hfill \xi \left(0\right)& =& \xi \left(p_n\left(0,0,\cdots ,0\right)\right)=p_n\left(\xi \left(0\right),0,\cdots ,0,\cdots ,0\right)+p_n\left(0,\xi \left(0\right),\cdots ,0,\cdots ,0\right)\hfill \\ & +& ...+p_n\left(0,0,\cdots ,\xi \left(0\right),\cdots ,0\right)+...+p_n\left(0,0,\cdots ,0,\cdots ,\xi \left(0\right)\right)\hfill \\ & =& 0.\hfill \end{array}$$

□

Claim 2. 

$\xi {\left(M\right)}^{\ast }=\xi \left(M\right)$ for every $M\in \mathcal{M}$.

Proof. 

For any $M\in \mathcal{M}$, observe that $M=p_n\left(M,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)$. Thus,

$$\begin{array}{ccc}\hfill \xi \left(M\right)& =& \xi \left(p_n\left(M,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)\hfill \\ & =& p_n\left(\xi \left(M\right),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(M,\xi \left(\frac{1}{2}I\right),\cdots ,\frac{1}{2}I\right)\hfill \\ & +& ...+p_n\left(M,\frac{1}{2}I,\cdots ,\xi \left(\frac{1}{2}I\right)\right)\hfill \\ & =& p_{n-1}\left(\frac{1}{2}(\xi \left(M\right)+\xi {\left(M\right)}^{\ast }),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_{n-1}\left(M\xi {\left(\frac{1}{2}I\right)}^{\ast }+\xi \left(\frac{1}{2}I\right)M,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& ...+p_{n-1}\left(M,\frac{1}{2}I,\cdots ,\xi \left(\frac{1}{2}I\right)\right)\hfill \\ & =& \frac{1}{2}\left(\xi \left(M\right)+\xi {\left(M\right)}^{\ast }\right)+(n-1)\left(M\xi {\left(\frac{1}{2}I\right)}^{\ast }+\xi \left(\frac{1}{2}I\right)M\right).\hfill \end{array}$$

This implies that

$$\begin{array}{ccc}\hfill \xi \left(M\right)& =& \xi {\left(M\right)}^{\ast }+2(n-1)\left(M\xi {\left(\frac{1}{2}I\right)}^{\ast }+\xi \left(\frac{1}{2}I\right)M\right).\hfill \end{array}$$

(1)

It follows that

$$\begin{array}{ccc}\hfill \xi {\left(M\right)}^{\ast }& =& \xi \left(M\right)+2(n-1)\left(\xi \left(\frac{1}{2}I\right)M+M\xi {\left(\frac{1}{2}I\right)}^{\ast }\right).\hfill \end{array}$$

(2)

Combining (1) and (2), we obtain $\xi {\left(M\right)}^{\ast }=\xi \left(M\right)$. This completes the proof. □

Claim 3. 

For any $U_{11}\in {\mathcal{M}}_{11},V_{12}\in {\mathcal{M}}_{12}$ and $W_{22}\in {\mathcal{M}}_{22}$, we have

(i)

$\xi (U_{11}+V_{12})=\xi \left(U_{11}\right)+\xi \left(V_{12}\right)$;

(ii)

$\xi (V_{12}+W_{22})=\xi \left(V_{12}\right)+\xi \left(W_{22}\right)$.

Proof. 

$\left(i\right)$ Let $T=\xi (U_{11}+V_{12})-\xi \left(U_{11}\right)-\xi \left(V_{12}\right)$. It is obvious from Claim 2 that $T\in \mathcal{M}$, i.e., $T^{\ast }=T$. Our aim is to show that $T=T_{11}+T_{12}+T_{22}=0$. In view of $p_n\left(P_2,U_{11},P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)=0$ and Claim 1, we have

$$\begin{array}{ccc}& & \xi \left(p_n\left(P_2,U_{11}+V_{12},P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)\hfill \\ & =& \xi \left(p_n\left(P_2,U_{11},P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)+\xi \left(p_n\left(P_2,V_{12},P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)\hfill \\ & =& p_n\left(\xi \left(P_2\right),U_{11},P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_2,\xi \left(U_{11}\right),P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(P_2,U_{11},\xi \left(P_1\right),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_2,U_{11},P_1,\xi \left(\frac{1}{2}I\right),\cdots ,\frac{1}{2}I\right)\hfill \\ & +& ...+p_n\left(P_2,U_{11},P_1,\frac{1}{2}I,\cdots ,\xi \left(\frac{1}{2}I\right)\right)+p_n\left(\xi \left(P_2\right),V_{12},P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(P_2,\xi \left(V_{12}\right),P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_2,V_{12},\xi \left(P_1\right),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(P_2,V_{12},P_1,\xi \left(\frac{1}{2}I\right),\cdots ,\frac{1}{2}I\right)+...+p_n\left(P_2,V_{12},P_1,\frac{1}{2}I,\cdots ,\xi \left(\frac{1}{2}I\right)\right)\hfill \\ & =& p_n\left(\xi \left(P_2\right),U_{11}+V_{12},P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_2,\xi \left(U_{11}\right)+\xi \left(V_{12}\right),P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(P_2,U_{11}+V_{12},\xi \left(P_1\right),\frac{1}{2}I,...\frac{1}{2}I\right)+p_n\left(P_2,U_{11}+V_{12},P_1,\xi \left(\frac{1}{2}I\right),\cdots ,\frac{1}{2}I\right)\hfill \\ & +& ...+p_n\left(P_2,U_{11}+V_{12},P_1,\frac{1}{2}I,\cdots ,\xi \left(\frac{1}{2}I\right)\right).\hfill \end{array}$$

Furthermore, we have

$$\begin{array}{ccc}& & \xi \left(p_n\left(P_2,U_{11}+V_{12},P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)\hfill \\ & =& p_n\left(\xi \left(P_2\right),U_{11}+V_{12},P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_2,\xi (U_{11}+V_{12}),P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(P_2,U_{11}+V_{12},\xi \left(P_1\right),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_2,U_{11}+V_{12},P_1,\xi \left(\frac{1}{2}I\right),\cdots ,\frac{1}{2}I\right)\hfill \\ & +& ...+p_n\left(P_2,U_{11}+V_{12},P_1,\frac{1}{2}I,\cdots ,\xi \left(\frac{1}{2}I\right)\right).\hfill \end{array}$$

From the last two expressions, we conclude that $p_n\left(P_2,T,P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)=0$. Using the primeness of $\mathfrak{A}$, we obtain $T_{12}=0$. Furthermore, since $p_n\left(P_2-P_1,V_{12},\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)=0$, we can write

$$\begin{array}{ccc}& & p_n\left(\xi (P_2-P_1),U_{11}+V_{12},\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_2-P_1,\xi (U_{11}+V_{12}),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(P_2-P_1,U_{11}+V_{12},\xi \left(\frac{1}{2}I\right),\cdots ,\frac{1}{2}I\right)+...+p_n\left(P_2-P_1,U_{11}+V_{12},\frac{1}{2}I,\cdots ,\xi \left(\frac{1}{2}I\right)\right)\hfill \\ & =& \xi \left(p_n\left(P_2-P_1,U_{11}+V_{12},\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)\hfill \\ & =& \xi \left(p_n\left(P_2-P_1,U_{11},\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)+\xi \left(p_n\left(P_2-P_1,V_{12},\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)\hfill \\ & =& p_n\left(\xi (P_2-P_1),U_{11},\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_2-P_1,\xi \left(U_{11}\right),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(P_2-P_1,U_{11},\xi \left(\frac{1}{2}I\right),\cdots ,\frac{1}{2}I\right)+...+p_n\left(P_2-P_1,U_{11},\frac{1}{2}I,\cdots ,\xi \left(\frac{1}{2}I\right)\right)\hfill \\ & +& p_n\left(\xi (P_2-P_1),V_{12},\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_2-P_1,\xi \left(V_{12}\right),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(P_2-P_1,V_{12},\xi \left(\frac{1}{2}I\right),\cdots ,\frac{1}{2}I\right)+...+p_n\left(P_2-P_1,V_{12},\frac{1}{2}I,\cdots ,\xi \left(\frac{1}{2}I\right)\right)\hfill \\ & =& p_n\left(\xi (P_2-P_1),U_{11}+V_{12},\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_2-P_1,\xi \left(U_{11}\right)+\xi \left(V_{12}\right),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(P_2-P_1,U_{11}+V_{12},\xi \left(\frac{1}{2}I\right),\cdots ,\frac{1}{2}I\right)+...+p_n\left(P_2-P_1,U_{11}+V_{12},\frac{1}{2}I,\cdots ,\xi \left(\frac{1}{2}I\right)\right).\hfill \end{array}$$

From this, we obtain $p_n(P_2-P_1,T,\frac{1}{2}I,\cdots ,\frac{1}{2}I)=0$. Using Claim 2, we obtain $P_{2}T+T{P}_2-P_{1}T-T{P}_1=0$. Multiplying this equation by $P_1$ and $P_2$, respectively, on both sides, we obtain $T_{11}=T_{22}=0$. Therefore, $T=0$. In a similar manner, we can establish $\left(ii\right)$. Thereby the proof is completed. □

Claim 4. 

For any $U_{11}\in {\mathcal{M}}_{11},V_{12}\in {\mathcal{M}}_{12}$ and $W_{22}\in {\mathcal{M}}_{22}$, we have

$$\xi (U_{11}+V_{12}+W_{22})=\xi \left(U_{11}\right)+\xi \left(V_{12}\right)+\xi \left(W_{22}\right).$$

Proof. 

We show that $T=\xi (U_{11}+V_{12}+W_{22})-\xi \left(U_{11}\right)-\xi \left(V_{12}\right)-\xi \left(W_{22}\right)=0$. In view of Claim 3 and $p_n(P_1,W_{22},P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I)=0$, we have

$$\begin{array}{ccc}& & \xi \left(p_n\left(P_1,U_{11}+V_{12}+W_{22},P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)\hfill \end{array}$$

$$\begin{array}{ccc}& =& \xi \left(p_n\left(P_1,U_{11}+V_{12},P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)+\xi \left(p_n\left(P_1,W_{22},P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)\hfill \\ & =& p_n\left(\xi \left(P_1\right),U_{11}+V_{12},P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_1,\xi (U_{11}+V_{12}),P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(P_1,U_{11}+V_{12},\xi \left(P_1\right),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_1,U_{11}+V_{12},P_1,\xi \left(\frac{1}{2}I\right),\cdots ,\frac{1}{2}I\right)\hfill \\ & +& ...+p_n\left(P_1,U_{11}+V_{12},P_1,\frac{1}{2}I,\cdots ,\xi \left(\frac{1}{2}I\right)\right)+p_n\left(\xi \left(P_1\right),W_{22},P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(P_1,\xi \left(W_{22}\right),P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_1,W_{22},\xi \left(P_1\right),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(P_1,W_{22},P_1,\xi \left(\frac{1}{2}I\right),\cdots ,\frac{1}{2}I\right)+...+p_n\left(P_1,W_{22},P_1,\frac{1}{2}I,\cdots ,\xi \left(\frac{1}{2}I\right)\right)\hfill \\ & =& p_n\left(\xi \left(P_1\right),U_{11}+V_{12},P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_1,\xi \left(U_{11}\right)+\xi \left(V_{12}\right),P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(P_1,U_{11}+V_{12},\xi \left(P_1\right),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_1,U_{11}+V_{12},P_1,\xi \left(\frac{1}{2}I\right),\cdots ,\frac{1}{2}I\right)\hfill \\ & +& ...+p_n\left(P_1,U_{11}+V_{12},P_1,\frac{1}{2}I,\cdots ,\xi \left(\frac{1}{2}I\right)\right)+p_n\left(\xi \left(P_1\right),W_{22},P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(P_1,\xi \left(W_{22}\right),P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_1,W_{22},\xi \left(P_1\right),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(P_1,W_{22},P_1,\xi \left(\frac{1}{2}I\right),\cdots ,\frac{1}{2}I\right)+...+p_n\left(P_1,W_{22},P_1,\frac{1}{2}I,\cdots ,\xi \left(\frac{1}{2}I\right)\right)\hfill \\ & =& p_n\left(\xi \left(P_1\right),U_{11}+V_{12}+W_{22},P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n(P_1,\xi \left(U_{11}\right)+\xi \left(V_{12}\right)+\xi \left(W_{22}\right),\hfill \\ & & P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I)+p_n\left(P_1,U_{11}+V_{12}+W_{22},\xi \left(P_1\right),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(P_1,U_{11}+V_{12}+W_{22},P_1,\xi \left(\frac{1}{2}I\right),\cdots ,\frac{1}{2}I\right)+...\hfill \\ & +& p_n\left(P_1,U_{11}+V_{12}+W_{22},P_1,\frac{1}{2}I,\cdots ,\xi \left(\frac{1}{2}I\right)\right).\hfill \end{array}$$

Furthermore, we can write

$$\begin{array}{ccc}& & \xi \left(p_n\left(P_1,U_{11}+V_{12}+W_{22},P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)\hfill \\ & =& p_n\left(\xi \left(P_1\right),U_{11}+V_{12}+W_{22},P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_1,\xi (U_{11}+V_{12}+W_{22}),P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(P_1,U_{11}+V_{12}+W_{22},\xi \left(P_1\right),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_1,U_{11}+V_{12}+W_{22},P_1,\xi \left(\frac{1}{2}I\right),\cdots ,\frac{1}{2}I\right)\hfill \\ & +& ...+p_n\left(P_1,U_{11}+V_{12}+W_{22},P_1,\frac{1}{2}I,\cdots ,\xi \left(\frac{1}{2}I\right)\right).\hfill \end{array}$$

Equating the above two relations, we have $p_n(P_1,T,P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I)=0$. The primeness of $\mathfrak{A}$ and $T^{\ast }=T$ imply that $T_{11}=T_{12}=0$. It remains to show that $T_{22}=0$. Observe that $p_n(P_2,U_{11},P_2,\frac{1}{2}I,\cdots ,\frac{1}{2}I)=0$. Reasoning as above, we obtain $T_{22}=0$ and, hence, $T=0$. □

Claim 5. 

For any $U_{12},V_{12}\in {\mathcal{M}}_{12}$, we have

$$\xi (U_{12}+V_{12})=\xi \left(U_{12}\right)+\xi \left(V_{12}\right).$$

Proof. 

For any $A_{12},B_{12}\in {\mathfrak{A}}_{12}$, assume that $U_{12}=A_{12}+A_{12}^{\ast }\in {\mathcal{M}}_{12}$ and $V_{12}=B_{12}+B_{12}^{\ast }\in {\mathcal{M}}_{12}$. Thus,

$$\begin{array}{ccc}& & p_n(P_1+A_{12}+A_{12}^{\ast },P_2+B_{12}+B_{12}^{\ast },\frac{1}{2}I,\cdots ,\frac{1}{2}I)\hfill \\ & =& (A_{12}+A_{12}^{\ast })+(B_{12}+B_{12}^{\ast })+(A_{12}{B}_{12}^{\ast }+A_{12}^{\ast }{B}_{12}+B_{12}{A}_{12}^{\ast }+B_{12}^{\ast }{A}_{12})\hfill \\ \hfill & =& U_{12}+V_{12}+U_{12}{V}_{12}^{\ast }+V_{12}{U}_{12}^{\ast }.\hfill \end{array}$$

Note that $U_{12}{V}_{12}^{\ast }+V_{12}{U}_{12}^{\ast }=A_{12}{B}_{12}^{\ast }+B_{12}{A}_{12}^{\ast }+A_{12}^{\ast }{B}_{12}+B_{12}^{\ast }{A}_{12}=W_{11}+X_{22}$, where $W_{11}=A_{12}{B}_{12}^{\ast }+B_{12}{A}_{12}^{\ast }\in {\mathcal{M}}_{11}$ and $X_{22}=A_{12}^{\ast }{B}_{12}+B_{12}^{\ast }{A}_{12}\in {\mathcal{M}}_{22}$. Since $A_{12}+A_{12}^{\ast },B_{12}+B_{12}^{\ast }\in {\mathcal{M}}_{12}$, it follows from Claims 3 and 4 that

$$\begin{array}{ccc}& & \xi (U_{12}+V_{12})+\xi \left(W_{11}\right)+\xi \left(X_{22}\right)\hfill \\ & =& \xi (U_{12}+V_{12}+W_{11}+X_{22})\hfill \\ & =& \xi (U_{12}+V_{12}+U_{12}{V}_{12}^{\ast }+V_{12}{U}_{12}^{\ast })\hfill \\ & =& \xi \left(p_n\left(P_1+A_{12}+A_{12}^{\ast },P_2+B_{12}+B_{12}^{\ast },\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)\hfill \\ & =& p_n\left(\xi \left(P_1\right)+\xi (A_{12}+A_{12}^{\ast }),P_2+B_{12}+B_{12}^{\ast },\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(P_1+A_{12}+A_{12}^{\ast },\xi \left(P_2\right)+\xi (B_{12}+B_{12}^{\ast }),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(P_1+A_{12}+A_{12}^{\ast },P_2+B_{12}+B_{12}^{\ast },\xi \left(\frac{1}{2}I\right),\cdots ,\frac{1}{2}I\right)+...\hfill \\ & +& p_n\left(P_1+A_{12}+A_{12}^{\ast },P_2+B_{12}+B_{12}^{\ast },\frac{1}{2}I,\cdots ,\xi \left(\frac{1}{2}I\right)\right)\hfill \\ & =& \xi \left(p_n\left(P_1,P_2,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)+\xi \left(p_n\left(P_1,B_{12}+B_{12}^{\ast },\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)\hfill \\ & +& \xi \left(p_n\left(A_{12}+A_{12}^{\ast },P_2,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)+\xi \left(p_n\left(A_{12}+A_{12}^{\ast },B_{12}+B_{12}^{\ast },\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)\hfill \\ & =& \xi \left(U_{12}\right)+\xi \left(V_{12}\right)+\xi (U_{12}{V}_{12}^{\ast }+V_{12}{U}_{12}^{\ast })\hfill \\ & =& \xi \left(U_{12}\right)+\xi \left(V_{12}\right)+\xi \left(W_{11}\right)+\xi \left(X_{22}\right).\hfill \end{array}$$

Therefore, we have

$$\xi (U_{12}+V_{12})=\xi \left(U_{12}\right)+\xi \left(V_{12}\right).$$

□

Claim 6. 

For every $U_{ii},V_{ii}\in {\mathcal{M}}_{ii}$$(i=1,2)$, we have

$$\xi (U_{ii}+V_{ii})=\xi \left(U_{ii}\right)+\xi \left(V_{ii}\right).$$

Proof. 

We will prove for $i=1$, the other case can be proven analogously. To prove this, we show that $T=\xi (U_{11}+V_{11})-\xi \left(U_{11}\right)-\xi \left(V_{11}\right)=0$. We have

$$\begin{array}{ccc}& & \xi \left(p_n\left(P_2,U_{11}+V_{11},P_2,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)\hfill \\ & =& \xi \left(p_n\left(P_2,U_{11},P_2,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)+\xi \left(p_n\left(P_2,V_{11},P_2,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)\hfill \\ & =& p_n\left(\xi \left(P_2\right),U_{11},P_2,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_2,\xi \left(U_{11}\right),P_2,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(P_2,U_{11},\xi \left(P_2\right),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_2,U_{11},P_2,\xi \left(\frac{1}{2}I\right),\cdots ,\frac{1}{2}I\right)\hfill \\ & +& ...+p_n\left(P_2,U_{11},P_2,\frac{1}{2}I,\cdots ,\xi \left(\frac{1}{2}I\right)\right)+p_n\left(\xi \left(P_2\right),V_{11},P_2,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(P_2,\xi \left(V_{11}\right),P_2,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_2,V_{11},\xi \left(P_2\right),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(P_2,V_{11},P_2,\xi \left(\frac{1}{2}I\right),\cdots ,\frac{1}{2}I\right)+...+p_n\left(P_2,V_{11},P_2,\frac{1}{2}I,\cdots ,\xi \left(\frac{1}{2}I\right)\right)\hfill \\ & =& p_n\left(\xi \left(P_2\right),U_{11}+V_{11},P_2,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_2,\xi \left(U_{11}\right)+\xi \left(V_{11}\right),P_2,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(P_2,U_{11}+V_{11},\xi \left(P_2\right),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_2,U_{11}+V_{11},P_2,\xi \left(\frac{1}{2}I\right),\cdots ,\frac{1}{2}I\right)\hfill \\ & +& ...+p_n\left(P_2,U_{11}+V_{11},P_2,\frac{1}{2}I,\cdots ,\xi \left(\frac{1}{2}I\right)\right).\hfill \end{array}$$

Apparently, we can have

$$\begin{array}{ccc}& & \xi \left(p_n\left(P_2,U_{11}+V_{11},P_2,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)\hfill \\ & =& p_n\left(\xi \left(P_2\right),U_{11}+V_{11},P_2,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_2,\xi (U_{11}+V_{11}),P_2,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(P_2,U_{11}+V_{11},\xi \left(P_2\right),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_2,U_{11}+V_{11},P_2,\xi \left(\frac{1}{2}I\right),\cdots ,\frac{1}{2}I\right)\hfill \\ & +& ...+p_n\left(P_2,U_{11}+V_{11},P_2,\frac{1}{2}I,\cdots ,\xi \left(\frac{1}{2}I\right)\right).\hfill \end{array}$$

From the last two expressions, we have $p_n(P_2,T,P_2,\frac{1}{2}I,\cdots ,\frac{1}{2}I)=0$, and thus, the primeness of $\mathfrak{A}$ gives that $T_{12}=T_{22}=0$. Now, to show that $T_{11}=0$, assume that $W=A_{12}+A_{12}^{\ast }\in {\mathcal{M}}_{12}$ for $A_{12}\in {\mathfrak{A}}_{12}$. Then $p_n\left(W,U_{11},\frac{1}{2}I,\cdots ,\frac{1}{2}I\right),p_n\left(W,V_{11},\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\in {\mathcal{M}}_{12}$. Therefore, from Claim 5, we can write

$$\begin{array}{ccc}& & p_n\left(\xi \left(W\right),U_{11}+V_{11},\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(W,\xi (U_{11}+V_{11}),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(W,U_{11}+V_{11},\xi \left(\frac{1}{2}I\right),\cdots ,\frac{1}{2}I\right)+...+p_n\left(W,U_{11}+V_{11},\frac{1}{2}I,\cdots ,\xi \left(\frac{1}{2}I\right)\right)\hfill \\ & =& \xi \left(p_n\left(W,U_{11}+V_{11},\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)\hfill \\ & =& \xi \left(p_n\left(W,U_{11},\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)+\xi \left(p_n\left(W,V_{11},\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)\hfill \\ & =& p_n\left(\xi \left(W\right),U_{11}+V_{11},\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(W,\xi \left(U_{11}\right)+\xi \left(V_{11}\right),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\hfill \\ & +& p_n\left(W,U_{11}+V_{11},\xi \left(\frac{1}{2}I\right),\cdots ,\frac{1}{2}I\right)+...+p_n\left(W,U_{11}+V_{11},\frac{1}{2}I,\cdots ,\xi \left(\frac{1}{2}I\right)\right).\hfill \end{array}$$

Thus, we obtain $p_n\left(W,T,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)=0$. This gives $T_{11}=0$. Hence, the proof is completed. □

Remark 2. 

It follows from Claims 3–6 that ξ is additive on $\mathcal{M}$.

Claim 7. 

$\xi \left(I\right)=0$.

Proof. 

In view of Claim 2 and Remark 2, we have

$$\begin{array}{ccc}\hfill \xi \left(P_1\right)& =& \xi \left(p_n\left(P_1,\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)\right)\hfill \\ & =& p_n\left(\xi \left(P_1\right),\frac{1}{2}I,\cdots ,\frac{1}{2}I\right)+p_n\left(P_1,\xi \left(\frac{1}{2}I\right),\cdots ,\frac{1}{2}I\right)\hfill \\ & +& ...+p_n\left(P_1,\frac{1}{2}I,\cdots ,\xi \left(\frac{1}{2}I\right)\right)\hfill \\ & =& \xi \left(P_1\right)+(n-1)\left(P_1\xi \left(\frac{1}{2}I\right)+\xi \left(\frac{1}{2}I\right)P_1\right)\hfill \end{array}$$

This implies

$$P_1\xi \left(\frac{1}{2}I\right)+\xi \left(\frac{1}{2}I\right)P_1=0.$$

Multiplying the above equation by $P_2$ from left, right and by $P_1$ on both sides, respectively, we obtain $P_2\xi \left(\frac{1}{2}I\right)P_1=0,P_1\xi \left(\frac{1}{2}I\right)P_2=0$ and $P_1\xi \left(\frac{1}{2}I\right)P_1=0$. By replacing $P_1$ with $P_2$ in the above calculation, we can obtain $P_2\xi \left(\frac{1}{2}I\right)P_2=0$. Therefore, we obtain $\xi \left(\frac{1}{2}I\right)=0$, and thus, using Remark 2, we obtain $\xi \left(I\right)=0$. □

Claim 8. 

$\xi {\left(N\right)}^{\ast }=-\xi \left(N\right)$, for every $N\in \mathcal{N}$.

Proof. 

Observe that $p_n\left(N,I,\cdots ,I\right)=0$ for any $N\in \mathcal{N}$; therefore, from Claim 7, we have

$$\begin{array}{ccc}\hfill 0& =& \xi \left(p_n\left(N,I,\cdots ,I\right)\right)=p_n(\xi \left(N\right),I,\cdots ,I)\hfill \\ & =& 2^{n-2}\left(\xi \left(N\right)+\xi {\left(N\right)}^{\ast }\right).\hfill \end{array}$$

Thus, $\xi {\left(N\right)}^{\ast }=-\xi \left(N\right)$ for all $N\in \mathcal{N}$. □

Claim 9. 

$\xi \left(iI\right)\in \mathcal{Z}\left(\mathfrak{A}\right)$.

Proof. 

Let $M\in \mathcal{M}$. Then, from Claim 8, we have

$$\begin{array}{c}\hfill 0=\xi \left(p_n\left(M,iI,I,\cdots ,I\right)\right)=2^{n-2}\left(\xi \left(iI\right)M-M\xi \left(iI\right)\right).\end{array}$$

This gives $\xi \left(iI\right)M=M\xi \left(iI\right)$. Since for any $U\in \mathfrak{A}$, $U=M_1+i{M}_2$ for $M_1,M_2\in \mathcal{M}$. Therefore, $U\xi \left(iI\right)=\xi \left(iI\right)U$ for all $U\in \mathfrak{A}$, and, hence, $\xi \left(iI\right)\in \mathcal{Z}\left(\mathfrak{A}\right)$. □

Claim 10. 

For any $N\in \mathcal{N}$, $\xi \left(iN\right)=i\xi \left(N\right)+\xi \left(iI\right)N$.

Proof. 

It follows from Claims 2, 7 and 8 that

$$\begin{array}{ccc}\hfill \xi \left(p_n\left(N,iI,I,\cdots ,I\right)\right)& =& p_n\left(\xi \left(N\right),iI,I,\cdots ,I\right)\hfill \\ \hfill & +& p_n\left(N,\xi \left(iI\right),\cdots ,I\right)\hfill \\ \hfill & =& -2^{n-1}\left(\xi \left(iI\right)N+i\xi \left(N\right)\right).\hfill \end{array}$$

(3)

Furthermore, from Remark 2, we have

$$\begin{array}{c}\hfill \xi \left(p_n\left(N,iI,I,\cdots ,I\right)\right)=-2^{n-1}\left(\xi \left(iN\right)\right)\end{array}$$

(4)

Equations (3) and (4) lead to

$$\xi \left(iN\right)=i\xi \left(N\right)+\xi \left(iI\right)N.$$

□

Claim 11. 

$\xi$ is additive on $\mathcal{N}$.

Proof. 

Let $N_1$, $N_2\in \mathcal{N}$. Then, from Claim 10 and Remark 2, we have

$$\begin{array}{ccc}& & i\xi (N_1+N_2)+\xi \left(iI\right)(N_1+N_2)\hfill \\ & & =\xi (i(N_1+N_2)=\xi \left(i{N}_1\right)+\xi \left(i{N}_2\right)\hfill \\ & & =i(\xi \left(N_1\right)+\xi \left(N_2\right))+\xi \left(iI\right)(N_1+N_2).\hfill \end{array}$$

This gives

$$\xi (N_1+N_2)=\xi \left(N_1\right)+\xi \left(N_2\right).$$

□

Claim 12. 

$\xi$ is additive on $\mathfrak{A}$.

Proof. 

Let $N,N^{\prime }\in \mathcal{N}$. In view of Remark 2, Claims 7, 9 and 10, we have

$$\begin{array}{ccc}\hfill & & 2^{n-1}\left(i\xi \left(N^{\prime }\right)+\xi \left(iI\right)N^{\prime }\right)=2^{n-1}\xi \left(i{N}^{\prime }\right)=\xi \left(2^{n-1}i{N}^{\prime }\right)\hfill \\ \hfill & & =\xi \left(p_n\left(I,I,\cdots ,(N+i{N}^{\prime })\right)\right)\hfill \\ \hfill & & =2^{n-2}\xi {(N+i{N}^{\prime })}^{\ast }+2^{n-2}\xi (N+i{N}^{\prime })\hfill \end{array}$$

(5)

and

$$\begin{array}{ccc}\hfill & & -2^{n-1}\left(i\xi \left(N\right)+\xi \left(iI\right)N\right)=-2^{n-1}\xi \left(iN\right)=\xi (-2^{n-1}iN)\hfill \\ \hfill & =& \xi \left((N+i{N}^{\prime }),iI,I,\cdots ,I\right)\hfill \\ \hfill & =& -2^{n-2}i\xi (N+i{N}^{\prime })+2^{n-2}i\xi {(N+i{N}^{\prime })}^{\ast }-2^{n-1}\xi \left(iI\right)N.\hfill \end{array}$$

(6)

So, we have from Equations (5) and (6) that

$$\begin{array}{c}\hfill \xi (N+i{N}^{\prime })=\xi \left(N\right)+i\xi \left(N^{\prime }\right)+\xi \left(iI\right)N^{\prime }\end{array}$$

(7)

for all N, $N^{\prime }\in \mathcal{N}$. Now let $U,V\in \mathfrak{A}$ such that $U=U_1+i{U}_2$ and $V=V_1+i{V}_2$ for all $U_1,U_2,V_1,V_2\in \mathcal{N}$. Using Equation (7) and Claim 11, we have

$$\begin{array}{cc}\hfill \xi (U+V)& =\xi \left((U_1+V_1)+i(U_2+V_2)\right)\hfill \\ \hfill & =\xi (U_1+V_1)+i\xi (U_2+V_2)+\xi \left(iI\right)(U_2+V_2)\hfill \\ \hfill & =\left(\xi \left(U_1\right)+i\xi \left(U_2\right)+\xi \left(iI\right)U_2\right)\hfill \\ \hfill & +\left(\xi \left(V_1\right)+i\xi \left(V_2\right)+\xi \left(iI\right)V_2\right)\hfill \\ \hfill & =\xi (U_1+i{U}_2)+\xi (V_1+i{V}_2)\hfill \\ \hfill & =\xi \left(U\right)+\xi \left(V\right).\hfill \end{array}$$

□

Claim 13. 

$\xi \left(U^{\ast }\right)=\xi {\left(U\right)}^{\ast }$ for all $U\in \mathfrak{A}$.

Proof. 

We know that any element $U\in \mathfrak{A}$ can be expressed as $U=U_1+i{U}_2$ for $U_1,U_2\in \mathcal{N}$, so it follows from Equation (7) and Claim 8 that

$$\begin{array}{ccc}\hfill \xi {\left(U\right)}^{\ast }& =& \xi {(U_1+i{U}_2)}^{\ast }\hfill \\ \hfill & =& {\left(\xi \left(U_1\right)+i\xi \left(U_2\right)+\xi \left(iI\right)U_2\right)}^{\ast }\hfill \\ \hfill & =& \xi (-U_1)+i\xi \left(U_2\right)+\xi \left(iI\right)U_2.\hfill \end{array}$$

(8)

On the other hand,

$$\begin{array}{c}\hfill \xi \left(U^{\ast }\right)=\xi (-U_1+i{U}_2)=\xi (-U_1)+i\xi \left(U_2\right)+\xi \left(iI\right)U_2.\end{array}$$

(9)

From Equations (8) and (9), we obtain $\xi \left(U^{\ast }\right)=\xi {\left(U\right)}^{\ast }$. □

Claim 14. 

$\xi$ is a derivation on $\mathcal{N}$.

Proof. 

Since for any $N_1$, $N_2\in \mathcal{N}$, $N_1{N}_2-N_2{N}_1\in \mathcal{N}$, it follows from Claim 10 that

$$\begin{array}{ccc}\hfill -2^{n-2}\xi \left(N_1{N}_2+N_2{N}_1\right)& =& \xi \left(p_n\left(N_1,N_2,I,\cdots ,I\right)\right)\hfill \\ \hfill & =& p_n\left(\xi \left(N_1\right),N_2,I,\cdots ,I\right)\hfill \\ \hfill & +& p_n\left(N_1,\xi \left(N_2\right),I,\cdots ,I\right)\hfill \\ \hfill & =& -2^{n-2}\xi \left(N_1\right)N_2-2^{n-2}{N}_2\xi \left(N_1\right)\hfill \\ \hfill & -& 2^{n-2}{N}_1\xi \left(N_2\right)-2^{n-2}\xi \left(N_2\right)N_1.\hfill \end{array}$$

(10)

Moreover,

$$\begin{array}{ccc}\hfill & & 2^{n-2}i\xi \left(N_1{N}_2-N_2{N}_1\right)+2^{n-2}\xi \left(iI\right)\left(N_1{N}_2-N_2{N}_1\right)\hfill \\ \hfill & =& \xi \left(2^{n-2}i(N_1{N}_2-N_2{N}_1)\right)\hfill \\ \hfill & =& \xi \left(p_n\left(N_1,i{N}_2,I,\cdots ,I\right)\right)\hfill \\ \hfill & =& p_n\left(\xi \left(N_1\right),i{N}_2,I,\cdots ,I\right)+p_n\left(N_1,\xi \left(i{N}_2\right),I,\cdots ,I\right)\hfill \\ \hfill & =& 2^{n-2}i\xi \left(N_1\right)N_2-2^{n-2}i{N}_2\xi \left(N_1\right)+2^{n-2}i{N}_1\xi \left(N_2\right)\hfill \\ \hfill & -& 2^{n-2}i\xi \left(N_2\right)N_1+2^{n-2}\xi \left(iI\right)(N_1{N}_2-N_2{N}_1)\hfill \end{array}$$

(11)

for all $N_1$, $N_2\in \mathcal{N}$. Equations (10) and (11) conclude that

$$\xi \left(N_1{N}_2\right)=\xi \left(N_1\right)N_2+N_1\xi \left(N_2\right)$$

for all $N_1$, $N_2\in \mathcal{N}$. Hence, the proof. □

Claim 15. 

$\xi \left(iI\right)=0$.

Proof. 

We know from Claim 7 that $\xi \left(I\right)=0$. Thus, by Remark 2 and Claim 14, we have

$$0=\xi (-I)=\xi \left(\left(iI\right)\left(iI\right)\right)=\xi \left(iI\right)\left(iI\right)+\left(iI\right)\xi \left(iI\right)=2i\xi \left(iI\right).$$

Thus, $\xi \left(iI\right)=0$. □

Claim 16. 

$\xi \left(iU\right)=i\xi \left(U\right)$ for all $U\in \mathfrak{A}$.

Proof. 

From Claims 10 and 15, we obtain $\xi \left(iN\right)=i\xi \left(N\right)$ for all $N\in \mathcal{N}$. Since for any $U\in \mathfrak{A}$, we can write $U=N_1+i{N}_2$ for $N_1,N_2\in \mathcal{N}$. It follows from Claim 12 that

$$\begin{array}{cc}& \xi \left(iU\right)=\xi \left(i(N_1+i{N}_2)\right)=i\left(\xi \left(N_1\right)+i\xi \left(N_2\right)\right)=i\xi \left(U\right).\end{array}$$

Hence, the result. □

Proof of Theorem 1. 

By Claims 12 and 13, $\xi$ is additive with $\xi \left(U^{\ast }\right)=\xi {\left(U\right)}^{\ast }$. The final step in the proof is to demonstrate that $\xi$ is a derivation on $\mathfrak{A}$.

For any $U,V\in \mathfrak{A}$, assume that $U=U_1+i{U}_2$ and $V=V_1+i{V}_2$ for all $U_1,U_2,V_1,V_2\in \mathcal{N}$. Thus, it follows from Claims 14–16 that

$$\begin{array}{ccc}\hfill \xi \left(UV\right)& =& \xi \left((U_1+i{U}_2)(V_1+i{V}_2)\right)\hfill \\ \hfill & =& \xi (U_1{V}_1+i{U}_1{V}_2+i{U}_2{V}_1-U_2{V}_2)\hfill \\ \hfill & =& \xi \left(U_1\right)V_1+U_1\xi \left(V_1\right)+i\xi \left(U_1\right)V_2\hfill \\ \hfill & +& i{U}_1\xi \left(V_2\right)+i\xi \left(U_2\right)V_1+i{U}_2\xi \left(V_1\right)\hfill \\ \hfill & -& \xi \left(U_2\right)V_2-U_2\xi \left(V_2\right)\hfill \end{array}$$

(12)

On the other hand,

$$\begin{array}{ccc}\hfill \xi \left(U\right)V+U\xi \left(V\right)& =& \xi (U_1+i{U}_2)(V_1+i{V}_2)\hfill \\ \hfill & +& (U_1+i{U}_2)\xi (V_1+i{V}_2)\hfill \\ \hfill & =& (\xi \left(U_1\right)+i\xi \left(U_2\right))(V_1+i{V}_2)\hfill \\ \hfill & +& (U_1+i{U}_2)(\xi \left(V_1\right)+i\xi \left(V_2\right))\hfill \\ \hfill & =& \xi \left(U_1\right)V_1+U_1\xi \left(V_1\right)+i\xi \left(U_1\right)V_2\hfill \\ \hfill & +& i{U}_1\xi \left(V_2\right)+i\xi \left(U_2\right)V_1+i{U}_2\xi \left(V_1\right)\hfill \\ \hfill & -& \xi \left(U_2\right)V_2-U_2\xi \left(V_2\right)\hfill \end{array}$$

(13)

Comparing Equations (12) and (13), we conclude that $\xi$ is a derivation on $\mathfrak{A}$. This completes the theorem’s proof. □

4. Discussion

Previously, the authors studied the structures of multiplicative/non-linear bi-skew Jordan (i.e., $n=2$) and Jordan triple (i.e., $n=3$) derivations on prime ∗-algebras. In this article, we have given a characterization of multiplicative/non-linear bi-skew Jordan n-derivations (i.e., for any $n\ge 2$) on prime ∗-algebras. Therefore, our result is more general. In particular, one can easily obtain the result for $n=2$ (respectively, for $n=3$) easily in the case of multiplicative bi-skew Jordan (respectively Jordan triple) derivations on prime ∗-algebras.

5. Conclusions

In this article we explored the structure of non-linear bi-skew Jordan n-derivation ($\xi$) acting on a prime ∗-algebra $\mathfrak{A}$. Indeed, we proved that such a map is additive derivation preserving the ∗-structure of algebra $\mathfrak{A}$, i.e., $\xi \left(U^{\ast }\right)=\xi {\left(U\right)}^{\ast }$ for all $U\in \mathfrak{A}$. One can further investigate the structure of non-linear bi-skew Jordan n-derivations on different algebras such as triangular algebras, generalized matrix algebras, incidence algebras, etc.