© Wuhan University 2023

0 Introduction

Let ${A}$ be a *-algebra over the complex field $\mathbb{C}.$ For $A,B\in {A},$ we write ${[A,B]}_{\diamond }=A^{\mathrm{*}}B-B^{\mathrm{*}}A,$$A\bullet B=AB+B{A}^{\mathrm{*}}$and $A\circ B=A^{\mathrm{*}}B+B^{\mathrm{*}}A$ for the bi-skew Lie product, *-Jordan product and $A^{\mathrm{*}}B+B^{\mathrm{*}}A$ product, respectively. These products have recently attracted the attention of many authors (see Refs. [1-14]).

Recall that a map $\varphi :{A}\to {A}$ is said to be an additive derivation if $\varphi (A+B)=\varphi (A)+\varphi (B)$ and $\varphi (AB)=\varphi (A)B+A\varphi (B)$ for $A,B\in {A}.$ Furthermore, $\varphi$ is an additive *-derivation if it is an additive derivation and $\varphi (A^{\mathrm{*}})=\varphi {(A)}^{\mathrm{*}}$ for $A\in {A}.$ A map $\varphi :{A}\to {A}$(without the linearity assumption) is called a nonlinear $A^{\mathrm{*}}B+B^{\mathrm{*}}A$ derivation if $\varphi (A\circ B)=\varphi (A)\circ B+A\circ \varphi (B)$ for $A,B\in {A},$ where $A\circ B=A^{\mathrm{*}}B+B^{\mathrm{*}}A.$ Darvish et al^[1] proved that every nonlinear $A^{\mathrm{*}}B+B^{\mathrm{*}}A$ triple derivation on prime *-algebras is an additive *-derivation. A map $\varphi :{A}\to {A}$ (without the linearity assumption) is called a nonlinear *-Jordan derivation if $\varphi (A\bullet B)=\varphi (A)\bullet B+A\bullet \varphi (B)$ for $A,B\in {A}.$ The authors of Ref. [2] introduced the concept of *-Jordan-type derivation. Suppose that $n\ge \mathrm{2}$ is a fixed positive integer. Accordingly, a nonlinear *-Jordan-type derivation is a map $\varphi :{A}\to {A}$ satisfying the condition $\varphi (A_{\mathrm{1}}\bullet A_{\mathrm{2}}\bullet \cdots \bullet A_n)={\displaystyle \sum _{k=\mathrm{1}}^n}{A}_{\mathrm{1}}\bullet \cdots \bullet A_{k-\mathrm{1}}\bullet \varphi (A_k)\bullet A_{k+\mathrm{1}}\bullet \cdots \bullet A_n$ for all $A_{\mathrm{1}},A_{\mathrm{2}},\cdots ,A_n\in {A},$ where $A_{\mathrm{1}}\bullet A_{\mathrm{2}}\bullet \cdots \bullet A_n=(\cdots ((A_{\mathrm{1}}\bullet A_{\mathrm{2}})\bullet A_{\mathrm{3}})\bullet \cdots \bullet A_n),$$A_i\bullet A_j=A_i{A}_j+A_j{A}_i^{\mathrm{*}},$$i,j\in \mathbb{N}.$ Under some mild condition on a *-algebra ${A}$, they showed that $\varphi$ is a nonlinear *-Jordan-type derivation on ${A}$ if and only if $\varphi$ is an additive *-derivation.

Motivated by the above results, we introduce the $A^{\mathrm{*}}B+B^{\mathrm{*}}A$ type derivations. Suppose that $n\ge \mathrm{2}$ is a fixed positive integer. A nonlinear $A^{\mathrm{*}}B+B^{\mathrm{*}}A$ type derivation is a map $\varphi :{A}\to {A}$ satisfying the condition $\varphi (A_{\mathrm{1}}\circ A_{\mathrm{2}}\circ \cdots \circ A_n)=$${\displaystyle \sum _{k=\mathrm{1}}^n}{A}_{\mathrm{1}}\circ \cdots \circ A_{k-\mathrm{1}}\circ \varphi (A_k)\circ A_{k+\mathrm{1}}\circ \cdots \circ A_n$ for all $A_{\mathrm{1}},A_{\mathrm{2}},\cdots ,A_n\in {A},$ where $A_{\mathrm{1}}\circ A_{\mathrm{2}}\circ \cdots \circ A_n=(\cdots ((A_{\mathrm{1}}\circ A_{\mathrm{2}})\circ A_{\mathrm{3}})\circ \cdots \circ A_n),$$A_i\circ A_j=A_i^{\mathrm{*}}{A}_j+A_j^{\mathrm{*}}{A}_i,$$i,j\in \mathbb{N}.$ In this paper, under some mild condition on a *-algebra ${A},$ we prove that $\varphi$ is a nonlinear $A^{\mathrm{*}}B+B^{\mathrm{*}}A$ type derivation on ${A}$ if and only if $\varphi$ is an additive *-derivation.

1 The Main Result and Its Proof

Theorem 1   Let ${A}$ be a unital *-algebra with the unit $I$ and a nontrivial projection $P\in {A}.$ Suppose that ${A}$ satisfies

(a) $X{A}P=\mathrm{0}$ implies $X=\mathrm{0}$ and (b) $X{A}(I-P)=\mathrm{0}$ implies $X=\mathrm{0}$.

If a map $\varphi :{A}\to {A}$ satisfies $\varphi (A_{\mathrm{1}}\circ A_{\mathrm{2}}\circ \cdots \circ A_n)=$${\displaystyle \sum _{k=\mathrm{1}}^n}{A}_{\mathrm{1}}\circ \cdots \circ A_{k-\mathrm{1}}\circ \varphi (A_k)\circ A_{k+\mathrm{1}}\circ \cdots \circ A_n$ for all $A_{\mathrm{1}},A_{\mathrm{2}}\in {A}$ and $A_{\mathrm{3}}=A_{\mathrm{4}}=\cdots =A_n=\frac{I}{\mathrm{2}},$ then $\varphi$ is an additive *-derivation.

Let $P_{\mathrm{1}}=P$ and $P_{\mathrm{2}}=I-P_{\mathrm{1}}.$ Let ${{A}}_{ij}=P_i{A}{P}_j,$$i,j=\mathrm{1,2};$ then ${A}={\displaystyle \sum _{i,j=\mathrm{1}}^{\mathrm{2}}}{{A}}_{ij}.$ We can write every $A\in {A}$ as $A={\displaystyle \sum _{i,j=\mathrm{1}}^{\mathrm{2}}}{A}_{ij},$ where $A_{ij}$ denotes an arbitrary element of ${{A}}_{ij}.$ Let $\mathrm{ℳ}=\{A\in {A}:A^{\mathrm{*}}=A\},{\mathrm{ℳ}}_{\mathrm{12}}=\{P_{\mathrm{1}}M{P}_{\mathrm{2}}+P_{\mathrm{2}}M{P}_{\mathrm{1}}:M\in \mathrm{ℳ}\},$${\mathrm{ℳ}}_{ii}=P_i\mathrm{ℳ}{P}_i,i=\mathrm{1,2}.$ Then for all $M\in \mathrm{ℳ},M=M_{\mathrm{11}}+M_{\mathrm{12}}+M_{\mathrm{22}},$ where $M_{\mathrm{12}}\in {\mathrm{ℳ}}_{\mathrm{12}},M_{ii}\in {\mathrm{ℳ}}_{ii},i=\mathrm{1,2}.$

Proof   The proof is completed by the following several claims.

Claim 1 $\varphi (\mathrm{0})=\mathrm{0}.$

$\varphi (\mathrm{0})=\varphi \left(\mathrm{0}\circ \mathrm{0}\circ \frac{I}{\mathrm{2}}\circ \dots \circ \frac{I}{\mathrm{2}}\right)=\mathrm{0}$

Claim 2 $\varphi \left(\frac{I}{\mathrm{2}}\right)=\mathrm{0},\varphi \left(\frac{\mathrm{i}}{\mathrm{2}}I\right)=\mathrm{0},\varphi \left(-\frac{\mathrm{i}}{\mathrm{2}}I\right)=\mathrm{0}.$

Using $\frac{I}{\mathrm{2}}=\frac{I}{\mathrm{2}}\circ \frac{I}{\mathrm{2}}\circ \dots \circ \frac{I}{\mathrm{2}},$ we obtain

$\begin{array}{l}\varphi \left(\frac{I}{\mathrm{2}}\right)=\varphi \left(\frac{I}{\mathrm{2}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}\right)=\varphi \left(\frac{I}{\mathrm{2}}\right)\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}+\frac{I}{\mathrm{2}}\circ \varphi \left(\frac{I}{\mathrm{2}}\right)\circ \cdots \circ \frac{I}{\mathrm{2}}+\cdots +\frac{I}{\mathrm{2}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \varphi \left(\frac{I}{\mathrm{2}}\right)\\ =\frac{n}{\mathrm{2}}\left(\varphi \left(\frac{I}{\mathrm{2}}\right)+\varphi {\left(\frac{I}{\mathrm{2}}\right)}^{\mathrm{*}}\right)\end{array}$

which implies $\varphi {\left(\frac{I}{\mathrm{2}}\right)}^{\mathrm{*}}=\varphi \left(\frac{I}{\mathrm{2}}\right),$ and then $(n-\mathrm{1})\varphi \left(\frac{I}{\mathrm{2}}\right)=\mathrm{0}.$ Since $n\ge \mathrm{2},$ we obtain

$\varphi \left(\frac{I}{\mathrm{2}}\right)=\mathrm{0}$(1)

From (1), we obtain

$\begin{array}{l}\mathrm{0}=\varphi \left(\frac{I}{\mathrm{2}}\right)=\varphi \left(\frac{\mathrm{i}}{\mathrm{2}}I\circ \frac{\mathrm{i}}{\mathrm{2}}I\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}\right)=\varphi \left(\frac{\mathrm{i}}{\mathrm{2}}I\right)\circ \frac{\mathrm{i}}{\mathrm{2}}I\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}+\frac{\mathrm{i}}{\mathrm{2}}I\circ \varphi \left(\frac{\mathrm{i}}{\mathrm{2}}I\right)\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}\\ =\mathrm{i}\left(\varphi {\left(\frac{\mathrm{i}}{\mathrm{2}}I\right)}^{\mathrm{*}}-\varphi \left(\frac{\mathrm{i}}{\mathrm{2}}I\right)\right)\end{array}$

i.e.,

$\varphi {\left(\frac{\mathrm{i}}{\mathrm{2}}I\right)}^{\mathrm{*}}=\varphi \left(\frac{\mathrm{i}}{\mathrm{2}}I\right)$(2)

Using (1) and (2), we have

$\mathrm{0}=\varphi \left(\frac{\mathrm{i}}{\mathrm{2}}I\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}\right)=\varphi \left(\frac{\mathrm{i}}{\mathrm{2}}I\right)\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}=\varphi \left(\frac{\mathrm{i}}{\mathrm{2}}I\right)$

In the same manner, we obtain $\varphi \left(-\frac{\mathrm{i}}{\mathrm{2}}I\right)=\mathrm{0}.$

Claim 3 For every $A\in {A},$ we have $\varphi \left(\mathrm{i}A\right)=\mathrm{i}\varphi \left(A\right).$

For all $A\in {A},$ using $\mathrm{i}A\circ \frac{I}{\mathrm{2}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}=A\circ -\frac{\mathrm{i}}{\mathrm{2}}I\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}$ and Claim 2, we obtain

$\begin{array}{l}\frac{\mathrm{1}}{\mathrm{2}}(\varphi {(\mathrm{i}A)}^{\mathrm{*}}+\varphi (\mathrm{i}A))=\varphi \left(\mathrm{i}A\right)\circ \frac{I}{\mathrm{2}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}=\varphi \left(\mathrm{i}\mathrm{A}\circ \frac{I}{\mathrm{2}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}\right)=\varphi \left(A\circ -\frac{\mathrm{i}}{\mathrm{2}}I\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}\right)\\ =\varphi \left(A\right)\circ -\frac{\mathrm{i}}{\mathrm{2}}I\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}=\frac{\mathrm{i}}{\mathrm{2}}(\varphi (A)-\varphi {(A)}^{\mathrm{*}})\end{array}$

i.e.,

$\varphi {\left(\mathrm{i}A\right)}^{\mathrm{*}}+\varphi \left(\mathrm{i}A\right)=\mathrm{i}\varphi \left(A\right)-\mathrm{i}\varphi {\left(A\right)}^{\mathrm{*}}$(3)

Using $\mathrm{i}A\circ \frac{\mathrm{i}}{\mathrm{2}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}=A\circ \frac{I}{\mathrm{2}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}$ and Claim 2, we have

$\begin{array}{l}\frac{\mathrm{i}}{\mathrm{2}}(\varphi {\left(\mathrm{i}A\right)}^{\mathrm{*}}-\varphi \left(\mathrm{i}A\right))=\varphi \left(\mathrm{i}A\right)\circ \frac{\mathrm{i}}{\mathrm{2}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}=\varphi \left(\mathrm{i}A\circ \frac{\mathrm{i}}{\mathrm{2}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}\right)=\varphi \left(A\circ \frac{I}{\mathrm{2}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}\right)\\ =\varphi \left(A\right)\circ \frac{I}{\mathrm{2}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}(\varphi (A)+\varphi {(A)}^{\mathrm{*}})\end{array}$

i.e.,

$-\varphi {\left(\mathrm{i}A\right)}^{\mathrm{*}}+\varphi \left(\mathrm{i}A\right)=\mathrm{i}\varphi \left(A\right)+\mathrm{i}\varphi {\left(A\right)}^{\mathrm{*}}$(4)

From (3) and (4), we obtain $\varphi \left(\mathrm{i}A\right)=\mathrm{i}\varphi \left(A\right).$

Claim 4 For every $A\in \mathrm{ℳ},$ we have $\varphi {\left(A\right)}^{\mathrm{*}}=\varphi \left(A\right).$

For all $A\in \mathrm{ℳ},$ we have

$\varphi \left(A\right)=\varphi \left(A\circ \frac{I}{\mathrm{2}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}\right)=\varphi \left(A\right)\circ \frac{I}{\mathrm{2}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}(\varphi (A)+\varphi {(A)}^{\mathrm{*}})$

which indicates $\varphi {\left(A\right)}^{\mathrm{*}}=\varphi \left(A\right).$

Claim 5 For every $M_{\mathrm{11}}\in {\mathrm{ℳ}}_{\mathrm{11}},M_{\mathrm{12}}\in {\mathrm{ℳ}}_{\mathrm{12}},M_{\mathrm{22}}\in {\mathrm{ℳ}}_{\mathrm{22}},$ we have (i) $\varphi \left(M_{\mathrm{11}}+M_{\mathrm{12}}\right)=\varphi \left(M_{\mathrm{11}}\right)+\varphi \left(M_{\mathrm{12}}\right);$ (ii) $\varphi \left(M_{\mathrm{12}}+M_{\mathrm{22}}\right)=\varphi \left(M_{\mathrm{12}}\right)+\varphi \left(M_{\mathrm{22}}\right).$

Setting $T=\varphi \left(M_{\mathrm{11}}+M_{\mathrm{12}}\right)-\varphi \left(M_{\mathrm{11}}\right)-\varphi \left(M_{\mathrm{12}}\right),$ let us prove that $T=\mathrm{0}.$ Based on Claim 4, we obtain $T^{\mathrm{*}}=T.$ Since $P_{\mathrm{2}}\circ M_{\mathrm{11}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}=\mathrm{0},$ it follows from Claim 1 and Claim 2 that

$\begin{array}{l}\varphi \left(P_{\mathrm{2}}\right)\circ (M_{\mathrm{11}}+M_{\mathrm{22}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}+P_{\mathrm{2}}\circ \varphi (M_{\mathrm{11}}+M_{\mathrm{22}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}=\varphi (P_{\mathrm{2}}\circ (M_{\mathrm{11}}+M_{\mathrm{22}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}})\\ =\varphi (P_{\mathrm{2}}\circ M_{\mathrm{11}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}})+\varphi (P_{\mathrm{2}}\circ M_{\mathrm{22}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}})\\ =\varphi \left(P_{\mathrm{2}}\right)\circ (M_{\mathrm{11}}+M_{\mathrm{22}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}+P_{\mathrm{2}}\circ (\varphi (M_{\mathrm{11}})+\varphi (M_{\mathrm{22}}))\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}\end{array}$

i.e., $P_{\mathrm{2}}\circ T\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}=\mathrm{0}.$ This together with $T^{\mathrm{*}}=T$ shows that $P_{\mathrm{1}}T{P}_{\mathrm{2}}=P_{\mathrm{2}}T{P}_{\mathrm{1}}=P_{\mathrm{2}}T{P}_{\mathrm{2}}=\mathrm{0}.$ Using $(P_{\mathrm{1}}-P_{\mathrm{2}})\circ M_{\mathrm{12}}\circ \frac{I}{\mathrm{2}}$$\circ \cdots \circ \frac{I}{\mathrm{2}}=\mathrm{0},$ Claim 1 and Claim 2, we obtain

$\begin{array}{l}\varphi \left(P_{\mathrm{1}}-P_{\mathrm{2}}\right)\circ (M_{\mathrm{11}}+M_{\mathrm{12}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}+\left(P_{\mathrm{1}}-P_{\mathrm{2}}\right)\circ \varphi (M_{\mathrm{11}}+M_{\mathrm{12}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}\\ =\varphi (\left(P_{\mathrm{1}}-P_{\mathrm{2}}\right)\circ (M_{\mathrm{11}}+M_{\mathrm{12}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}})\\ =\varphi (\left(P_{\mathrm{1}}-P_{\mathrm{2}}\right)\circ M_{\mathrm{11}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}})+\varphi (\left(P_{\mathrm{1}}-P_{\mathrm{2}}\right)\circ M_{\mathrm{12}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}})\\ =\varphi \left(P_{\mathrm{1}}-P_{\mathrm{2}}\right)\circ (M_{\mathrm{11}}+M_{\mathrm{12}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}+\left(P_{\mathrm{1}}-P_{\mathrm{2}}\right)\circ (\varphi (M_{\mathrm{11}})+\varphi (M_{\mathrm{12}}))\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}\end{array}$

i.e., $(P_{\mathrm{1}}-P_{\mathrm{2}})\circ T\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}=\mathrm{0}.$ This together with $T^{\mathrm{*}}=T$ shows that $P_{\mathrm{1}}T{P}_{\mathrm{1}}=\mathrm{0}.$ And then $T=\mathrm{0}.$

In the second case, we can similarly prove that the conclusion is valid.

Claim 6 For every $M_{\mathrm{11}}\in {\mathrm{ℳ}}_{\mathrm{11}},M_{\mathrm{12}}\in {\mathrm{ℳ}}_{\mathrm{12}},M_{\mathrm{22}}\in {\mathrm{ℳ}}_{\mathrm{22}},$ we have $\varphi \left(M_{\mathrm{11}}+M_{\mathrm{12}}+M_{\mathrm{22}}\right)=\varphi (M_{\mathrm{11}})$+$\varphi \left(M_{\mathrm{12}}\right)+\varphi \left(M_{\mathrm{22}}\right).$

Setting $T=\varphi \left(M_{\mathrm{11}}+M_{\mathrm{12}}+M_{\mathrm{22}}\right)-\varphi \left(M_{\mathrm{11}}\right)-\varphi \left(M_{\mathrm{12}}\right)-\varphi \left(M_{\mathrm{22}}\right),$ since $P_{\mathrm{1}}\circ M_{\mathrm{22}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}=\mathrm{0},$ applying Claim 1, Claim 2 and Claim 5(i), we obtain

$\begin{array}{l}\varphi \left(P_{\mathrm{1}}\right)\circ (M_{\mathrm{11}}+M_{\mathrm{12}}+M_{\mathrm{22}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}+P_{\mathrm{1}}\circ \varphi (M_{\mathrm{11}}+M_{\mathrm{12}}+M_{\mathrm{22}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}\\ =\varphi (P_{\mathrm{1}}\circ (M_{\mathrm{11}}+M_{\mathrm{12}}+M_{\mathrm{22}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}})\\ =\varphi (P_{\mathrm{1}}\circ (M_{\mathrm{11}}+M_{\mathrm{12}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}})+\varphi (P_{\mathrm{1}}\circ M_{\mathrm{22}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}})\\ =\varphi \left(P_{\mathrm{1}}\right)\circ (M_{\mathrm{11}}+M_{\mathrm{12}}+M_{\mathrm{22}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}+P_{\mathrm{1}}\circ (\varphi (M_{\mathrm{11}})+\varphi (M_{\mathrm{12}})+\varphi (M_{\mathrm{22}}))\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}\end{array}$

i.e., $P_{\mathrm{1}}\circ T\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}=\mathrm{0}.$ This together with $T^{\mathrm{*}}=T$ shows that $P_{\mathrm{1}}T{P}_{\mathrm{1}}=P_{\mathrm{1}}T{P}_{\mathrm{2}}=P_{\mathrm{2}}T{P}_{\mathrm{1}}=\mathrm{0}.$ In the same manner, by applying the above proof for $P_{\mathrm{2}}$ instead of $P_{\mathrm{1}}$ and Claim 5(ii) instead of Claim 5(i), we have $P_{\mathrm{2}}T{P}_{\mathrm{2}}=\mathrm{0}.$

Claim 7 For every $M_{\mathrm{12}},B_{\mathrm{12}}\in {\mathrm{ℳ}}_{\mathrm{12}},$ we have $\varphi \left(M_{\mathrm{12}}+B_{\mathrm{12}}\right)=\varphi (M_{\mathrm{12}})+\varphi (B_{\mathrm{12}}).$

Let $M_{\mathrm{12}},B_{\mathrm{12}}\in {\mathrm{ℳ}}_{\mathrm{12}},$ we obtain $M_{\mathrm{12}}=U_{\mathrm{12}}+U_{\mathrm{12}}^{\mathrm{*}},B_{\mathrm{12}}=V_{\mathrm{12}}+V_{\mathrm{12}}^{\mathrm{*}},$where $U_{\mathrm{12}},V_{\mathrm{12}}\in {\mathrm{ℳ}}_{\mathrm{12}}.$ Since $M_{\mathrm{12}}{B}_{\mathrm{12}}^{\mathrm{*}}+B_{\mathrm{12}}{M}_{\mathrm{12}}^{\mathrm{*}}=U_{\mathrm{12}}{V}_{\mathrm{12}}^{\mathrm{*}}+V_{\mathrm{12}}{U}_{\mathrm{12}}^{\mathrm{*}}+U_{\mathrm{12}}^{\mathrm{*}}{V}_{\mathrm{12}}+V_{\mathrm{12}}^{\mathrm{*}}{U}_{\mathrm{12}},$ we set $U_{\mathrm{12}}{V}_{\mathrm{12}}^{\mathrm{*}}+V_{\mathrm{12}}{U}_{\mathrm{12}}^{\mathrm{*}}=M_{\mathrm{11}}$$\in {\mathrm{ℳ}}_{\mathrm{11}},$$U_{\mathrm{12}}^{\mathrm{*}}{V}_{\mathrm{12}}+V_{\mathrm{12}}^{\mathrm{*}}{U}_{\mathrm{12}}=M_{\mathrm{22}}$$\in {\mathrm{ℳ}}_{\mathrm{22}},$ then

$\begin{array}{l}(P_{\mathrm{1}}+U_{\mathrm{12}}+U_{\mathrm{12}}^{\mathrm{*}})\circ (P_{\mathrm{2}}+V_{\mathrm{12}}+V_{\mathrm{12}}^{\mathrm{*}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}\\ =(U_{\mathrm{12}}+U_{\mathrm{12}}^{\mathrm{*}})+(V_{\mathrm{12}}+V_{\mathrm{12}}^{\mathrm{*}})+(U_{\mathrm{12}}{V}_{\mathrm{12}}^{\mathrm{*}}+V_{\mathrm{12}}{U}_{\mathrm{12}}^{\mathrm{*}}+U_{\mathrm{12}}^{\mathrm{*}}{V}_{\mathrm{12}}+V_{\mathrm{12}}^{\mathrm{*}}{U}_{\mathrm{12}})\\ =M_{\mathrm{12}}+B_{\mathrm{12}}+M_{\mathrm{12}}{B}_{\mathrm{12}}^{\mathrm{*}}+B_{\mathrm{12}}{M}_{\mathrm{12}}^{\mathrm{*}}=M_{\mathrm{12}}+B_{\mathrm{12}}+M_{\mathrm{11}}+M_{\mathrm{22}}\end{array}$

Using $U_{\mathrm{12}}+U_{\mathrm{12}}^{\mathrm{*}},V_{\mathrm{12}}+V_{\mathrm{12}}^{\mathrm{*}}$$\in {\mathrm{ℳ}}_{\mathrm{12}}$ and Claim 6, we obtain

$\varphi (M_{\mathrm{12}}+B_{\mathrm{12}})+\varphi (M_{\mathrm{11}})+\varphi (M_{\mathrm{22}})=\varphi (M_{\mathrm{12}}+B_{\mathrm{12}}+M_{\mathrm{12}}{B}_{\mathrm{12}}^{\mathrm{*}}+B_{\mathrm{12}}{M}_{\mathrm{12}}^{\mathrm{*}})$

$=\varphi ((P_{\mathrm{1}}+U_{\mathrm{12}}+U_{\mathrm{12}}^{\mathrm{*}})\circ (P_{\mathrm{2}}+V_{\mathrm{12}}+V_{\mathrm{12}}^{\mathrm{*}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}})=(\varphi (P_{\mathrm{1}})+\varphi (U_{\mathrm{12}}+U_{\mathrm{12}}^{\mathrm{*}}))\circ (P_{\mathrm{2}}+V_{\mathrm{12}}+V_{\mathrm{12}}^{\mathrm{*}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}$

$\begin{array}{l}+(P_{\mathrm{1}}+U_{\mathrm{12}}+U_{\mathrm{12}}^{\mathrm{*}})\circ (\varphi (P_{\mathrm{2}})+\varphi (V_{\mathrm{12}}+V_{\mathrm{12}}^{\mathrm{*}}))\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}+\dots +(P_{\mathrm{1}}+U_{\mathrm{12}}+U_{\mathrm{12}}^{\mathrm{*}})\circ (P_{\mathrm{2}}+V_{\mathrm{12}}+V_{\mathrm{12}}^{\mathrm{*}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \varphi (\frac{I}{\mathrm{2}})\\ =\varphi (P_{\mathrm{1}}\circ P_{\mathrm{2}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}})+\varphi (P_{\mathrm{1}}\circ (V_{\mathrm{12}}+V_{\mathrm{12}}^{\mathrm{*}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}})\\ +\varphi ((U_{\mathrm{12}}+U_{\mathrm{12}}^{\mathrm{*}})\circ P_{\mathrm{2}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}})+\varphi ((U_{\mathrm{12}}+U_{\mathrm{12}}^{\mathrm{*}})\circ (V_{\mathrm{12}}+V_{\mathrm{12}}^{\mathrm{*}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}})\\ =\varphi (M_{\mathrm{12}})+\varphi (B_{\mathrm{12}})+\varphi (M_{\mathrm{12}}{B}_{\mathrm{12}}^{\mathrm{*}}+B_{\mathrm{12}}{M}_{\mathrm{12}}^{\mathrm{*}})=\varphi (M_{\mathrm{12}})+\varphi (B_{\mathrm{12}})+\varphi (M_{\mathrm{11}})+\varphi (M_{\mathrm{22}})\end{array}$

i.e., $\varphi \left(M_{\mathrm{12}}+B_{\mathrm{12}}\right)=\varphi (M_{\mathrm{12}})+\varphi (B_{\mathrm{12}}).$

Claim 8 For each $C_{ii},D_{ii}\in {\mathrm{ℳ}}_{ii},i=\mathrm{1,2},$ we have (i) $\varphi \left(C_{\mathrm{11}}+D_{\mathrm{11}}\right)=\varphi \left(C_{\mathrm{11}}\right)+\varphi \left(D_{\mathrm{11}}\right);$ (ii)　$\varphi \left(C_{\mathrm{22}}+D_{\mathrm{22}}\right)=\varphi \left(C_{\mathrm{22}}\right)+\varphi \left(D_{\mathrm{22}}\right).$

Setting $T=\varphi \left(C_{\mathrm{11}}+D_{\mathrm{11}}\right)-\varphi \left(C_{\mathrm{11}}\right)-\varphi \left(D_{\mathrm{11}}\right),$ we obtain

$\begin{array}{l}\varphi \left(P_{\mathrm{2}}\right)\circ (C_{\mathrm{11}}+D_{\mathrm{11}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}+P_{\mathrm{2}}\circ \varphi (C_{\mathrm{11}}+D_{\mathrm{11}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}\\ =\varphi (P_{\mathrm{2}}\circ (C_{\mathrm{11}}+D_{\mathrm{11}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}})=\varphi (P_{\mathrm{2}}\circ C_{\mathrm{11}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}})+\varphi (P_{\mathrm{2}}\circ D_{\mathrm{11}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}})\\ =\varphi \left(P_{\mathrm{2}}\right)\circ (C_{\mathrm{11}}+D_{\mathrm{11}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}+P_{\mathrm{2}}\circ (\varphi (C_{\mathrm{11}})+\varphi (D_{\mathrm{11}}))\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}\end{array}$

i.e., $P_{\mathrm{2}}\circ T\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}=\mathrm{0}.$ This together with the fact $T^{\mathrm{*}}=T$ shows that $P_{\mathrm{1}}T{P}_{\mathrm{2}}=P_{\mathrm{2}}T{P}_{\mathrm{1}}=P_{\mathrm{2}}T{P}_{\mathrm{2}}=\mathrm{0}.$ For all $A_{\mathrm{12}}\in {{A}}_{\mathrm{12}},$ take $M=A_{\mathrm{12}}+A_{\mathrm{12}}^{\mathrm{*}}.$ Then $M\circ C_{\mathrm{11}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}\in {\mathrm{ℳ}}_{\mathrm{12}},M\circ D_{\mathrm{11}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}\in {\mathrm{ℳ}}_{\mathrm{12}}.$ We get from Claim 7 that

$\begin{array}{l}\varphi \left(M\right)\circ (C_{\mathrm{11}}+D_{\mathrm{11}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}+M\circ \varphi (C_{\mathrm{11}}+D_{\mathrm{11}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}\\ =\varphi (M\circ (C_{\mathrm{11}}+D_{\mathrm{11}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}})=\varphi (M\circ C_{\mathrm{11}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}})+\varphi (M\circ D_{\mathrm{11}}\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}})\\ =\varphi \left(M\right)\circ (C_{\mathrm{11}}+D_{\mathrm{11}})\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}+M\circ (\varphi (C_{\mathrm{11}})+\varphi (D_{\mathrm{11}}))\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}\end{array}$

i.e., $M\circ T\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}=\mathrm{0}.$ Then $P_{\mathrm{1}}T{A}_{\mathrm{12}}+A_{\mathrm{12}}^{\mathrm{*}}T{P}_{\mathrm{1}}=\mathrm{0}$ for all $A_{\mathrm{12}}\in {{A}}_{\mathrm{12}}.$ Hence $P_{\mathrm{1}}T{P}_{\mathrm{1}}A{P}_{\mathrm{2}}=\mathrm{0}$ for all $A\in {A}.$ From (b), we obtain $P_{\mathrm{1}}T{P}_{\mathrm{1}}=\mathrm{0}$ and then $T=\mathrm{0}.$

In the second case, we can similarly prove that the conclusion is valid.

Claim 9 $\varphi$ is additive on $\mathrm{ℳ}.$

By Claims 6-8, $\varphi$ is additive on $\mathrm{ℳ}.$

Claim 10 $\varphi$ is additive on ${A}$ and $\varphi (A^{\mathrm{*}})=\varphi {(A)}^{\mathrm{*}}$ for all $A\in {A}.$

For all $H,K\in \mathrm{ℳ},$ from Claim 2 we obtain

$\varphi (H)=\varphi ((H+\mathrm{i}K)\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}})=\varphi (H+\mathrm{i}K)\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}(\varphi (H+\mathrm{i}K)+\varphi {(H+\mathrm{i}K)}^{\mathrm{*}})$(5)

On the other hand, from Claim 2, we have

$\varphi (K)=\varphi ((H+\mathrm{i}K)\circ \frac{\mathrm{i}}{\mathrm{2}}I\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}})=\varphi (H+\mathrm{i}K)\circ \frac{\mathrm{i}}{\mathrm{2}}I\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}=-\frac{\mathrm{i}}{\mathrm{2}}(\varphi (H+\mathrm{i}K)-\varphi {(H+\mathrm{i}K)}^{\mathrm{*}})$

i.e.,

$\mathrm{i}\varphi (K)=\frac{\mathrm{1}}{\mathrm{2}}(\varphi (H+\mathrm{i}K)-\varphi {(H+\mathrm{i}K)}^{\mathrm{*}})$(6)

By adding (5) and (6), from Claim 3, we obtain

$\varphi (H+\mathrm{i}K)=\varphi (H)+\mathrm{i}\varphi (K)$(7)

For all $A\in {A}$, we have $A=A_{\mathrm{1}}+\mathrm{i}{A}_{\mathrm{2}}$ with $A_{\mathrm{1}},A_{\mathrm{2}}\in \mathrm{ℳ}.$ From (7), Claim 4 and Claim 9, we obtain

$\varphi {(A)}^{\mathrm{*}}=\varphi (A_{\mathrm{1}}+\mathrm{i}{A}_{\mathrm{2}}{)}^{\mathrm{*}}=(\varphi (A_{\mathrm{1}})+\mathrm{i}\varphi (A_{\mathrm{2}}{))}^{\mathrm{*}}=\varphi (A_{\mathrm{1}})-\mathrm{i}\varphi (A_{\mathrm{2}})=\varphi (A_{\mathrm{1}})+\mathrm{i}\varphi (-A_{\mathrm{2}})=\varphi (A_{\mathrm{1}}-\mathrm{i}{A}_{\mathrm{2}})=\varphi (A^{\mathrm{*}})$

For all $A,B\in {A}$, we have $A=A_{\mathrm{1}}+\mathrm{i}{A}_{\mathrm{2}},B=B_{\mathrm{1}}+\mathrm{i}{B}_{\mathrm{2}}$ with $A_{\mathrm{1}},A_{\mathrm{2}},B_{\mathrm{1}},B_{\mathrm{2}}\in \mathrm{ℳ}.$ From (7) and Claim 9, we obtain

$\begin{array}{l}\varphi (A+B)=\varphi ((A_{\mathrm{1}}+B_{\mathrm{1}})+\mathrm{i}(A_{\mathrm{2}}+B_{\mathrm{2}}))=\varphi (A_{\mathrm{1}}+B_{\mathrm{1}})+\mathrm{i}\varphi (A_{\mathrm{2}}+B_{\mathrm{2}})\\ =\varphi (A_{\mathrm{1}})+\varphi (B_{\mathrm{1}})+\mathrm{i}(\varphi (A_{\mathrm{2}})+\varphi (B_{\mathrm{2}}))=(\varphi (A_{\mathrm{1}})+\mathrm{i}\varphi (A_{\mathrm{2}}))+(\varphi (B_{\mathrm{1}})+\mathrm{i}\varphi (B_{\mathrm{2}}))=\varphi (A)+\varphi (B)\end{array}$

Claim 11 $\varphi$ is an additive *-derivation on ${A}.$

For all $H,K\in \mathrm{ℳ},$ from Claim 2 and Claim 4, we have

$\begin{array}{l}\varphi (HK+KH)=\varphi (H\circ K\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}})\\ =\varphi (H)\circ K\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}+H\circ \varphi (K)\circ \frac{I}{\mathrm{2}}\circ \cdots \circ \frac{I}{\mathrm{2}}=\varphi (H)K+K\varphi (H)+H\varphi (K)+\varphi (K)H\end{array}$