Open Access
Issue
Wuhan Univ. J. Nat. Sci.
Volume 28, Number 5, October 2023
Page(s) 379 - 384
DOI https://doi.org/10.1051/wujns/2023285379
Published online 10 November 2023

© Wuhan University 2023

Licence Creative CommonsThis is an Open Access article distributed under the terms of the Creative Commons Attribution License (https://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

0 Introduction

Let A be a *-algebra over the complex field C. For A,BA, we write [A,B]=A*B-B*A,AB=AB+BA*and AB=A*B+B*A for the bi-skew Lie product, *-Jordan product and A*B+B*A product, respectively. These products have recently attracted the attention of many authors (see Refs. [1-14]).

Recall that a map ϕ:AA is said to be an additive derivation if ϕ(A+B)=ϕ(A)+ϕ(B) and ϕ(AB)=ϕ(A)B+Aϕ(B) for A,BA. Furthermore, ϕ is an additive *-derivation if it is an additive derivation and ϕ(A*)=ϕ(A)* for AA. A map ϕ:AA(without the linearity assumption) is called a nonlinear A*B+B*A derivation if ϕ(AB)=ϕ(A)B+Aϕ(B) for A,BA, where AB=A*B+B*A. Darvish et al[1] proved that every nonlinear A*B+B*A triple derivation on prime *-algebras is an additive *-derivation. A map ϕ:AA (without the linearity assumption) is called a nonlinear *-Jordan derivation if ϕ(AB)=ϕ(A)B+Aϕ(B) for A,BA. The authors of Ref. [2] introduced the concept of *-Jordan-type derivation. Suppose that n2 is a fixed positive integer. Accordingly, a nonlinear *-Jordan-type derivation is a map ϕ:AA satisfying the condition ϕ(A1A2An)=k=1nA1Ak-1ϕ(Ak)Ak+1An for all A1,A2,,AnA, where A1A2An=(((A1A2)A3)An),AiAj=AiAj+AjAi*,i,jN. Under some mild condition on a *-algebra A, they showed that ϕ is a nonlinear *-Jordan-type derivation on A if and only if ϕ is an additive *-derivation.

Motivated by the above results, we introduce the A*B+B*A type derivations. Suppose that n2 is a fixed positive integer. A nonlinear A*B+B*A type derivation is a map ϕ:AA satisfying the condition ϕ(A1A2An)=k=1nA1Ak-1ϕ(Ak)Ak+1An for all A1,A2,,AnA, where A1A2An=(((A1A2)A3)An),AiAj=Ai*Aj+Aj*Ai,i,jN. In this paper, under some mild condition on a *-algebra A, we prove that ϕ is a nonlinear A*B+B*A type derivation on A if and only if ϕ 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 PA. Suppose that A satisfies

(a) XAP=0 implies X=0 and (b) XA(I-P)=0 implies X=0.

If a map ϕ:AA satisfies ϕ(A1A2An)=k=1nA1Ak-1ϕ(Ak)Ak+1An for all A1,A2A and A3=A4==An=I2, then ϕ is an additive *-derivation.

Let P1=P and P2=I-P1. Let Aij=PiAPj,i,j=1,2; then A=i,j=12Aij. We can write every AA as A=i,j=12Aij, where Aij denotes an arbitrary element of Aij. Let ={AA:A*=A},12={P1MP2+P2MP1:M},ii=PiPi,i=1,2. Then for all M,M=M11+M12+M22, where M1212,Miiii,i=1,2.

Proof   The proof is completed by the following several claims.

Claim 1 ϕ ( 0 ) = 0 .

ϕ ( 0 ) = ϕ ( 0 0 I 2 I 2 ) = 0

Claim 2 ϕ ( I 2 ) = 0 , ϕ ( i 2 I ) = 0 , ϕ ( - i 2 I ) = 0 .

Using I2=I2I2I2, we obtain

    ϕ ( I 2 ) = ϕ ( I 2 I 2 I 2 ) = ϕ ( I 2 ) I 2 I 2 + I 2 ϕ ( I 2 ) I 2 + + I 2 I 2 ϕ ( I 2 )                = n 2 ( ϕ ( I 2 ) + ϕ ( I 2 ) * )

which implies ϕ(I2)*=ϕ(I2), and then (n-1)ϕ(I2)=0. Since n2, we obtain

ϕ ( I 2 ) = 0 (1)

From (1), we obtain

0 = ϕ ( I 2 ) = ϕ ( i 2 I i 2 I I 2 I 2 ) = ϕ ( i 2 I ) i 2 I I 2 I 2 + i 2 I ϕ ( i 2 I ) I 2 I 2    = i ( ϕ ( i 2 I ) * - ϕ ( i 2 I ) )

i.e.,

ϕ ( i 2 I ) * = ϕ ( i 2 I ) (2)

Using (1) and (2), we have

0 = ϕ ( i 2 I I 2 I 2 ) = ϕ ( i 2 I ) I 2 I 2 = ϕ ( i 2 I )

In the same manner, we obtain ϕ(-i2I)=0.

Claim 3 For every AA, we have ϕ(iA)=iϕ(A).

For all AA, using iAI2I2I2=A-i2II2I2 and Claim 2, we obtain

   1 2 ( ϕ ( i A ) * + ϕ ( i A ) ) = ϕ ( i A ) I 2 I 2 I 2 = ϕ ( i A I 2 I 2 I 2 ) = ϕ ( A - i 2 I I 2 I 2 )    = ϕ ( A ) - i 2 I I 2 I 2 = i 2 ( ϕ ( A ) - ϕ ( A ) * )

i.e.,

ϕ ( i A ) * + ϕ ( i A ) = i ϕ ( A ) - i ϕ ( A ) * (3)

Using iAi2I2I2=AI2I2I2 and Claim 2, we have

   i 2 ( ϕ ( i A ) * - ϕ ( i A ) ) = ϕ ( i A ) i 2 I 2 I 2 = ϕ ( i A i 2 I 2 I 2 ) = ϕ ( A I 2 I 2 I 2 ) = ϕ ( A ) I 2 I 2 I 2 = 1 2 ( ϕ ( A ) + ϕ ( A ) * )

i.e.,

- ϕ ( i A ) * + ϕ ( i A ) = i ϕ ( A ) + i ϕ ( A ) * (4)

From (3) and (4), we obtain ϕ(iA)=iϕ(A).

Claim 4 For every A, we have ϕ(A)*=ϕ(A).

For all A, we have

ϕ ( A ) = ϕ ( A I 2 I 2 I 2 ) = ϕ ( A ) I 2 I 2 I 2 = 1 2 ( ϕ ( A ) + ϕ ( A ) * )

which indicates ϕ(A)*=ϕ(A).

Claim 5 For every M1111,M1212,M2222, we have (i) ϕ(M11+M12)=ϕ(M11)+ϕ(M12); (ii) ϕ(M12+M22)=ϕ(M12)+ϕ(M22).

Setting T=ϕ(M11+M12)-ϕ(M11)-ϕ(M12), let us prove that T=0. Based on Claim 4, we obtain T*=T. Since P2M11I2I2=0, it follows from Claim 1 and Claim 2 that

    ϕ ( P 2 ) ( M 11 + M 22 ) I 2 I 2 + P 2 ϕ ( M 11 + M 22 ) I 2 I 2 = ϕ ( P 2 ( M 11 + M 22 ) I 2 I 2 ) = ϕ ( P 2 M 11 I 2 I 2 ) + ϕ ( P 2 M 22 I 2 I 2 ) = ϕ ( P 2 ) ( M 11 + M 22 ) I 2 I 2 + P 2 ( ϕ ( M 11 ) + ϕ ( M 22 ) ) I 2 I 2

i.e., P2TI2I2=0. This together with T*=T shows that P1TP2=P2TP1=P2TP2=0. Using (P1-P2)M12I2I2=0, Claim 1 and Claim 2, we obtain

    ϕ ( P 1 - P 2 ) ( M 11 + M 12 ) I 2 I 2 + ( P 1 - P 2 ) ϕ ( M 11 + M 12 ) I 2 I 2 = ϕ ( ( P 1 - P 2 ) ( M 11 + M 12 ) I 2 I 2 ) = ϕ ( ( P 1 - P 2 ) M 11 I 2 I 2 ) + ϕ ( ( P 1 - P 2 ) M 12 I 2 I 2 ) = ϕ ( P 1 - P 2 ) ( M 11 + M 12 ) I 2 I 2 + ( P 1 - P 2 ) ( ϕ ( M 11 ) + ϕ ( M 12 ) ) I 2 I 2

i.e., (P1-P2)TI2I2=0. This together with T*=T shows that P1TP1=0. And then T=0.

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

Claim 6 For every M1111,M1212,M2222, we have ϕ(M11+M12+M22)=ϕ(M11)+ϕ(M12)+ϕ(M22).

Setting T=ϕ(M11+M12+M22)-ϕ(M11)-ϕ(M12)-ϕ(M22), since P1M22I2I2=0, applying Claim 1, Claim 2 and Claim 5(i), we obtain

     ϕ ( P 1 ) ( M 11 + M 12 + M 22 ) I 2 I 2 + P 1 ϕ ( M 11 + M 12 + M 22 ) I 2 I 2 = ϕ ( P 1 ( M 11 + M 12 + M 22 ) I 2 I 2 ) = ϕ ( P 1 ( M 11 + M 12 ) I 2 I 2 ) + ϕ ( P 1 M 22 I 2 I 2 ) = ϕ ( P 1 ) ( M 11 + M 12 + M 22 ) I 2 I 2 + P 1 ( ϕ ( M 11 ) + ϕ ( M 12 ) + ϕ ( M 22 ) ) I 2 I 2

i.e., P1TI2I2=0. This together with T*=T shows that P1TP1=P1TP2=P2TP1=0. In the same manner, by applying the above proof for P2 instead of P1 and Claim 5(ii) instead of Claim 5(i), we have P2TP2=0.

Claim 7 For every M12,B1212, we have ϕ(M12+B12)=ϕ(M12)+ϕ(B12).

Let M12,B1212, we obtain M12=U12+U12*,B12=V12+V12*,where U12,V1212. Since M12B12*+B12M12*=U12V12*+V12U12*+U12*V12+V12*U12, we set U12V12*+V12U12*=M1111,U12*V12+V12*U12=M2222, then

    ( P 1 + U 12 + U 12 * ) ( P 2 + V 12 + V 12 * ) I 2 I 2 = ( U 12 + U 12 * ) + ( V 12 + V 12 * ) + ( U 12 V 12 * + V 12 U 12 * + U 12 * V 12 + V 12 * U 12 ) = M 12 + B 12 + M 12 B 12 * + B 12 M 12 * = M 12 + B 12 + M 11 + M 22

Using U12+U12*,V12+V12*12 and Claim 6, we obtain

   ϕ ( M 12 + B 12 ) + ϕ ( M 11 ) + ϕ ( M 22 ) = ϕ ( M 12 + B 12 + M 12 B 12 * + B 12 M 12 * )

= ϕ ( ( P 1 + U 12 + U 12 * ) ( P 2 + V 12 + V 12 * ) I 2 I 2 ) = ( ϕ ( P 1 ) + ϕ ( U 12 + U 12 * ) ) ( P 2 + V 12 + V 12 * ) I 2 I 2

          + ( P 1 + U 12 + U 12 * ) ( ϕ ( P 2 ) + ϕ ( V 12 + V 12 * ) ) I 2 I 2 + + ( P 1 + U 12 + U 12 * ) ( P 2 + V 12 + V 12 * ) I 2 ϕ ( I 2 )          = ϕ ( P 1 P 2 I 2 I 2 ) + ϕ ( P 1 ( V 12 + V 12 * ) I 2 I 2 )           + ϕ ( ( U 12 + U 12 * ) P 2 I 2 I 2 ) + ϕ ( ( U 12 + U 12 * ) ( V 12 + V 12 * ) I 2 I 2 )          = ϕ ( M 12 ) + ϕ ( B 12 ) + ϕ ( M 12 B 12 * + B 12 M 12 * ) = ϕ ( M 12 ) + ϕ ( B 12 ) + ϕ ( M 11 ) + ϕ ( M 22 )

i.e., ϕ(M12+B12)=ϕ(M12)+ϕ(B12).

Claim 8 For each Cii,Diiii,i=1,2, we have (i) ϕ(C11+D11)=ϕ(C11)+ϕ(D11); (ii) ϕ(C22+D22)=ϕ(C22)+ϕ(D22).

Setting T=ϕ(C11+D11)-ϕ(C11)-ϕ(D11), we obtain

    ϕ ( P 2 ) ( C 11 + D 11 ) I 2 I 2 + P 2 ϕ ( C 11 + D 11 ) I 2 I 2 = ϕ ( P 2 ( C 11 + D 11 ) I 2 I 2 ) = ϕ ( P 2 C 11 I 2 I 2 ) + ϕ ( P 2 D 11 I 2 I 2 ) = ϕ ( P 2 ) ( C 11 + D 11 ) I 2 I 2 + P 2 ( ϕ ( C 11 ) + ϕ ( D 11 ) ) I 2 I 2

i.e., P2TI2I2=0. This together with the fact T*=T shows that P1TP2=P2TP1=P2TP2=0. For all A12A12, take M=A12+A12*. Then MC11I2I212,MD11I2I212. We get from Claim 7 that

ϕ ( M ) ( C 11 + D 11 ) I 2 I 2 + M ϕ ( C 11 + D 11 ) I 2 I 2 = ϕ ( M ( C 11 + D 11 ) I 2 I 2 ) = ϕ ( M C 11 I 2 I 2 ) + ϕ ( M D 11 I 2 I 2 ) = ϕ ( M ) ( C 11 + D 11 ) I 2 I 2 + M ( ϕ ( C 11 ) + ϕ ( D 11 ) ) I 2 I 2

i.e., MTI2I2=0. Then P1TA12+A12*TP1=0 for all A12A12. Hence P1TP1AP2=0 for all AA. From (b), we obtain P1TP1=0 and then T=0.

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

Claim 9 ϕ is additive on .

By Claims 6-8, ϕ is additive on .

Claim 10 ϕ is additive on A and ϕ(A*)=ϕ(A)* for all AA.

For all H,K, from Claim 2 we obtain

ϕ ( H ) = ϕ ( ( H + i K ) I 2 I 2 ) = ϕ ( H + i K ) I 2 I 2 = 1 2 ( ϕ ( H + i K ) + ϕ ( H + i K ) * ) (5)

On the other hand, from Claim 2, we have

ϕ ( K ) = ϕ ( ( H + i K ) i 2 I I 2 I 2 ) = ϕ ( H + i K ) i 2 I I 2 I 2 = - i 2 ( ϕ ( H + i K ) - ϕ ( H + i K ) * )

i.e.,

i ϕ ( K ) = 1 2 ( ϕ ( H + i K ) - ϕ ( H + i K ) * ) (6)

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

ϕ ( H + i K ) = ϕ ( H ) + i ϕ ( K ) (7)

For all AA, we have A=A1+iA2 with A1,A2. From (7), Claim 4 and Claim 9, we obtain

ϕ ( A ) * = ϕ ( A 1 + i A 2 ) * = ( ϕ ( A 1 ) + i ϕ ( A 2 ) ) * = ϕ ( A 1 ) - i ϕ ( A 2 ) = ϕ ( A 1 ) + i ϕ ( - A 2 ) = ϕ ( A 1 - i A 2 ) = ϕ ( A * )

For all A,BA, we have A=A1+iA2,B=B1+iB2 with A1,A2,B1,B2. From (7) and Claim 9, we obtain

    ϕ ( A + B ) = ϕ ( ( A 1 + B 1 ) + i ( A 2 + B 2 ) ) = ϕ ( A 1 + B 1 ) + i ϕ ( A 2 + B 2 ) = ϕ ( A 1 ) + ϕ ( B 1 ) + i ( ϕ ( A 2 ) + ϕ ( B 2 ) ) = ( ϕ ( A 1 ) + i ϕ ( A 2 ) ) + ( ϕ ( B 1 ) + i ϕ ( B 2 ) ) = ϕ ( A ) + ϕ ( B )

Claim 11 ϕ is an additive *-derivation on A.

For all H, K, from Claim 2 and Claim 4, we have

     ϕ ( H K + K H ) = ϕ ( H K I 2 I 2 ) = ϕ ( H ) K I 2 I 2 + H ϕ ( K ) I 2 I 2 = ϕ ( H ) K + K ϕ ( H ) + H ϕ ( K ) + ϕ ( K ) H

On the other hand, from (7), Claim 2 and Claim 4, we obtain

    i ϕ ( H K - K H ) = ϕ ( i ( H K - K H ) ) = ϕ ( H i K I 2 I 2 ) = ϕ ( H ) i K I 2 I 2 + H ϕ ( i K ) I 2 I 2 = i ( ϕ ( H ) K - K ϕ ( H ) + H ϕ ( K ) - ϕ ( K ) H )

So we have ϕ(HK)=ϕ(H)K+Hϕ(K).

For all A,BA, we have A=A1+iA2,B=B1+iB2, where A1,A2,B1,B2. From Claim 10 and (7), we obtain

    ϕ ( A B ) = ϕ ( A 1 B 1 + i A 1 B 2 + i A 2 B 1 - A 2 B 2 ) = ϕ ( A 1 B 1 ) + ϕ ( i A 1 B 2 ) + ϕ ( i A 2 B 1 ) - ϕ ( A 2 B 2 ) = ϕ ( A 1 ) B 1 + A 1 ϕ ( B 1 ) + i ϕ ( A 1 ) B 2 + i A 1 ϕ ( B 2 ) + i ϕ ( A 2 ) B 1 + i A 2 ϕ ( B 1 ) - ϕ ( A 2 ) B 2 - A 2 ϕ ( B 2 ) = ϕ ( A 1 ) B 1 + A 1 ϕ ( B 1 ) + ϕ ( A 1 ) i B 2 + A 1 ϕ ( i B 2 ) + ϕ ( i A 2 ) B 1 + i A 2 ϕ ( B 1 ) + ϕ ( i A 2 ) i B 2 + i A 2 ϕ ( i B 2 ) = ( ϕ ( A 1 ) + ϕ ( i A 2 ) ) ( B 1 + i B 2 ) + ( A 1 + i A 2 ) ( ϕ ( B 1 ) + ϕ ( i B 2 ) ) = ϕ ( A ) B + A ϕ ( B )

From this and Claim 10, we have proved that ϕ is an additive *-derivation. This completes the proof.

2 Corollaries

Now we give some applications of Theorem 1 to operator algebras. We say that A is prime when for A,BA, if AAB={0}, then A=0 or B=0. It is easy to show that prime *-algebras satisfy (a) and (b), and the following corollary is immediate.

Corollary 1   Let A be a prime *-algebra with unit I and a nontrivial projection. Then ϕ is a nonlinear A*B+B*A type derivation on A if and only if  ϕ is an additive *-derivation.

Recall that a von Neumann algebra A is weakly closed, self-adjoint algebra of operators on a Hilbert space containing the identity operator I. By Ref.[3], if a von Neumann algebra has no central summands of type I1, then A satisfies (a) and (b). So the following corollary is obvious.

Corollary 2   Let A be a von Neumann algebra with no central summands of type I1. Then ϕ is a nonlinear A*B+B*A type derivation on A if and only if ϕ is an additive *-derivation.

A is a factor von Neumann algebra means that its center only contains the scalar operators. Clearly, A is prime. So the following corollary is obvious from Corollary 1.

Corollary 3   Let A be a factor von Neumann algebra. Then ϕ is a nonlinear A*B+B*A type derivation on A if and only if ϕ is an additive *-derivation.

( ) is the algebra of all bounded linear operators on a complex Hilbert space . ()() is all bounded finite rank operators. A() is said to a standard operator algebra if it contains (). When A is a standard operator algebra, a more concrete form is achieved.

Corollary 4   Let be an infinite dimensional complex Hilbert space and A be a standard operator algebra on containing the identity operator I. Assume that A is closed under the adjoint operation. Let ϕ:AA be a nonlinear A*B+B*A type derivation. Then there exists T() satisfying T+T*=0 such that ϕ(A)=AT-TA for all AA.

Proof   Since A is prime, from Corollary 1, we obtain ϕ is an additive *-derivation. According to the result of Ref.[4], ϕ is linear, and then it is inner. Thus there exists S() such that ϕ(A)=AS-SA. Hence A*S-SA*=ϕ(A*)=ϕ(A)*=S*A*-A*S* for all AA. This indicates that S+S*=λI for certain λR. Take T=S-12λI, then T+T*=0 and ϕ(A)=AT-TA for all AA.

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