Research article Special Issues

Computational comparative analysis of fixed point approximations of generalized α-nonexpansive mappings in hyperbolic spaces

  • Received: 02 September 2022 Revised: 25 October 2022 Accepted: 31 October 2022 Published: 07 November 2022
  • MSC : 47H05, 47H10, 47J25

  • In this article, we use the Picard-Thakur hybrid iterative scheme to approximate the fixed points of generalized α-nonexpansive mappings. For generalized α-nonexpansive mappings in hyperbolic spaces, we show several weak and strong convergence results. It is proved numerically and graphically that the Picard-Thakur hybrid iterative scheme converges more faster than other well-known hybrid iterative methods for generalized α-nonexpansive mappings. We also present an application to Fredholm integral equation.

    Citation: Liliana Guran, Khushdil Ahmad, Khurram Shabbir, Monica-Felicia Bota. Computational comparative analysis of fixed point approximations of generalized α-nonexpansive mappings in hyperbolic spaces[J]. AIMS Mathematics, 2023, 8(2): 2489-2507. doi: 10.3934/math.2023129

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  • In this article, we use the Picard-Thakur hybrid iterative scheme to approximate the fixed points of generalized α-nonexpansive mappings. For generalized α-nonexpansive mappings in hyperbolic spaces, we show several weak and strong convergence results. It is proved numerically and graphically that the Picard-Thakur hybrid iterative scheme converges more faster than other well-known hybrid iterative methods for generalized α-nonexpansive mappings. We also present an application to Fredholm integral equation.



    Let A be an associative algebra. For A,BA, denote by [A,B]=ABBA the Lie product of A and B. An additive (a linear) map δ:AA is called a global Lie triple derivation if δ([[A,B],C])=[[δ(A),B],C]+[[A,δ(B)],C]+[[A,B],δ(C)] for all A,B,CA. The study of global Lie triple derivations on various algebras has attracted several authors' attention, see for example [2,11,16,17,20]. Next, let δ:AA be a map (without the additivity (linearity) assumption). δ is called a global nonlinear Lie triple derivation if δ satisfies δ([[A,B],C])=[[δ(A),B],C]+[[A,δ(B)],C]+[[A,B],δ(C)] for all A,B,CA. Ji, Liu and Zhao [4] gave the concrete form of global nonlinear Lie triple derivations on triangular algebras. Chen and Xiao [3] investigated global nonlinear Lie triple derivations on parabolic subalgebras of finite-dimensional simple Lie algebras. Very recently, Zhao and Hao [21] paid attention to non-global nonlinear Lie triple derivations. Let F:A×A×AA be a map and Q be a proper subset of A. δ is called a non-global nonlinear Lie triple derivation if δ satisfies δ([[A,B],C])=[[δ(A),B],C]+[[A,δ(B)],C]+[[A,B],δ(C)] for any A,B,CA with F(A,B,C)Q. Let M be a finite von Neumann algebra with no central summands of type I1. Zhao and Hao [21] proved that if δ:MM satisfies δ([[A,B],C])=[[δ(A),B],C]+[[A,δ(B)],C]+[[A,B],δ(C)] for any A,B,CM with ABC=0, then δ=d+τ, where d is a derivation from M into itself and τ is a nonlinear map from M into its center such that τ([[A,B],C])=0 with ABC=0.

    Let A be an associative -algebra. For A,BA, denote by [A,B]=ABBA the skew Lie product of A and B. The skew Lie product arose in representability of quadratic functionals by sesquilinear functionals [12,13]. In recent years, the study related to skew Lie product has attracted some authors' attention, see for example [1,5,6,7,8,9,10,14,15,18,19,22] and references therein. A map δ:AA (without the additivity (linearity) assumption) is called a global nonlinear skew Lie triple derivation if δ([[A,B],C])=[[δ(A),B],C]+[[A,δ(B)],C]+[[A,B],δ(C)] for all A,B,CA. A map δ:AA is called an additive -derivation if it is an additive derivation and satisfies δ(A)=δ(A) for all AA. Li, Zhao and Chen [5] proved that every global nonlinear skew Lie triple derivation on factor von Neumann algebras is an additive -derivation. Taghavi, Nouri and Darvish [15] proved that every global nonlinear skew Lie triple derivation on prime -algebras is additive. Similarly, let F:A×A×AA be a map and Q be a proper subset of A. If δ satisfies δ([[A,B],C])=[[δ(A),B],C]+[[A,δ(B)],C]+[[A,B],δ(C)] for any A,B,CA with F(A,B,C)Q, then δ is called a non-global nonlinear skew Lie triple derivation.

    Motivated by the mentioned works, we will concentrate on characterizing a kind of non-global nonlinear skew Lie triple derivations δ on factor von Neumann algebras satisfying δ([[A,B],C])=[[δ(A),B],C]+[[A,δ(B)],C]+[[A,B],δ(C)] for any A,B,CA with ABC=0.

    As usual, C denotes the complex number field. Let H be a complex Hilbert space and B(H) be the algebra of all bounded linear operators on H. Let AB(H) be a factor von Neumann algebra (i.e., the center of A is CI, where I is the identity of A). Recall that A is prime (i.e., for any A,BA, AAB={0} implies A=0 or B=0).

    The main result is the following theorem.

    Theorem 2.1. Let A be a factor von Neumann algebra acting on a complex Hilbert space H with dimA>1. If a map δ:AA satisfies

    δ([[A,B],C])=[[δ(A),B],C]+[[A,δ(B)],C]+[[A,B],δ(C)]

    for any A,B,CA with ABC=0, then δ is an additive -derivation.

    Let P1A be a nontrivial projection. Write P2=IP1, Aij=PiAPj (i,j=1,2). Then A=A11+A12+A21+A22. For any AA, A=A11+A12+A21+A22, Aij Aij (i,j=1,2).

    Lemma 2.1. (a) δ(Pi)=δ(Pi) (i=1,2);

    (b) Piδ(Pi)Pj=Piδ(Pj)Pj (1ij2).

    Proof. (a) It is clear that δ(0)=0. For any X21A21, it follows from P1P1X21=0 and [[P1,P1],X21]=0 that

    0=δ([[P1,P1],X21])=[[δ(P1),P1],X21]+[[P1,δ(P1)],X21]+[[P1,P1],δ(X21)]=P1δ(P1)X21X21δ(P1)+X21δ(P1)P1+P1δ(P1)X21X21δ(P1)P1+X21δ(P1). (2.1)

    Multiplying (2.1) by P2 from the left and by P1 from the right, we have X21(δ(P1)P1δ(P1)P1)=0. Then by the primeness of A, we get

    P1δ(P1)P1=P1δ(P1)P1. (2.2)

    By P1P2P2=0 and [[P1,P2],P2]=0, we have

    0=δ([[P1,P2],P2])=[[δ(P1),P2],P2]+[[P1,δ(P2)],P2]+[[P1,P2],δ(P2)]=δ(P1)P2P2δ(P1)P2P2δ(P1)+P2δ(P1)P2+P1δ(P2)P2P2δ(P2)P1. (2.3)

    Multiplying (2.3) by P2 from both sides, we see that

    P2δ(P1)P2=P2δ(P1)P2. (2.4)

    From P1P1P2=0 and [[P1,P1],P2]=0, we have

    0=δ([[P1,P1],P2])=[[δ(P1),P1],P2]+[[P1,δ(P1)],P2]+[[P1,P1],δ(P2)]=P1δ(P1)P2+P2δ(P1)P1+P1δ(P1)P2P2δ(P1)P1. (2.5)

    Multiplying (2.5) by P1 from the left and by P2 from the right, then

    P1δ(P1)P2=P1δ(P1)P2. (2.6)

    Multiplying (2.5) by P2 from the left and by P1 from the right, then

    P2δ(P1)P1=P2δ(P1)P1. (2.7)

    It follows from (2.2), (2.4), (2.6) and (2.7) that δ(P1)=δ(P1). Similarly, we can obtain that δ(P2)=δ(P2).

    (b) From P2P1P2=0 and [[P2,P1],P2]=0, we have

    0=δ([[P2,P1],P2])=[[δ(P2),P1],P2]+[[P2,δ(P1)],P2]+[[P2,P1],δ(P2)]=P1δ(P2)P2+P2δ(P2)P1+P2δ(P1)P2δ(P1)P2P2δ(P1)P2+P2δ(P1). (2.8)

    Multiplying (2.8) by P1 from the left and by P2 from the right, we have P1δ(P1)P2=P1δ(P2)P2. Then P1δ(P1)P2=P1δ(P2)P2 by (a). Similarly, we can obtain that P2δ(P2)P1=P2δ(P1)P1.

    Lemma 2.2. For any AijAij (1ij2), we have

    Pjδ(Aij)Pi=0.

    Proof. Let A12A12. For any X12A12, since A12X12P2=0 and [[A12,X12],P2]=0, we have

    0=δ([[A12,X12],P2])=[[δ(A12),X12],P2]+[[A12,δ(X12)],P2]+[[A12,X12],δ(P2)]=δ(A12)X12X12δ(A12)P2X12δ(A12)+P2δ(A12)X12+A12δ(X12)P2P2δ(X12)A12X12A12δ(P2)+δ(P2)A12X12. (2.9)

    Multiplying (2.9) by P2 from both sides, we have

    0=P2δ(A12)X12X12δ(A12)P2. (2.10)

    Replacing X12 with iX12 in (2.10) yields that

    0=P2δ(A12)X12+X12δ(A12)P2. (2.11)

    Combining (2.10) and (2.11), we see that P2δ(A12)X12=0. Then P2δ(A12)P1=0 by the primeness of A. Similarly, we can obtain that P1δ(A21)P2=0.

    Lemma 2.3. For any A12A12,B21A21, there exist GA12,B21A11,KA12,B21A22 such that

    δ(A12+B21)=δ(A12)+δ(B21)+GA12,B21+KA12,B21.

    Proof. Let T=δ(A12+B21)δ(A12)δ(B21). From P2(A12+B21)P2=P2A12P2=P2B21P2=0 and [[P2,B21],P2]=0, we have

    [[δ(P2),A12+B21],P2]+[[P2,δ(A12+B21)],P2]+[[P2,A12+B21],δ(P2)]=δ([[P2,A12+B21],P2])=δ([[P2,A12],P2])+δ([[P2,B21],P2])=[[δ(P2),A12+B21],P2]+[[P2,δ(A12)+δ(B21)],P2]+[[P2,A12+B21],δ(P2)],

    which implies

    [[P2,T],P2]=0. (2.12)

    Multiplying (2.12) by P1 from the left, we get T12=0. Similarly, T21=0. Let

    GA12,B21=T11,KA12,B21=T22.

    Then GA12,B21A11,KA12,B21A22, and so δ(A12+B21)=δ(A12)+δ(B21)+GA12,B21+KA12,B21.

    Lemma 2.4. (a) Pjδ(Pi)Pj=0 (1ij2);

    (b) Piδ(Pi)Pi=0 (i=1,2).

    Proof. (a) For any X12A12, since P1X12P1=0 and [[P1,X12],P1]=0, we have

    0=δ([[P1,X12],P1])=[[δ(P1),X12],P1]+[[P1,δ(X12)],P1]+[[P1,X12],δ(P1)]=X12δ(P1)P1+P1δ(P1)X12+P1δ(X12)P1δ(X12)P1P1δ(X12)P1+P1δ(X12)+X12δ(P1)δ(P1)X12. (2.13)

    Multiplying (2.13) by P1 from the left and by P2 from the right, we have

    P1δ(X12)P2+X12δ(P1)P2=0.

    It follows from Lemma 2.2 that X12δ(P1)P2=(P2δ(X12)P1)=0. Then P2δ(P1)P2=0. Similarly, P1δ(P2)P1=0.

    (b) For any X21A21, from (iX21)P1P1=0, [[iX21,P1],P1]=iX21+iX21, Lemma 2.1(a) and Lemma 2.3, there exist GiX21,iX21A11,KiX21,iX21A22 such that

    δ(iX21)+δ(iX21)+GiX21,iX21+KiX21,iX21=δ([[iX21,P1],P1])=[[δ(iX21),P1],P1]+[[iX21,δ(P1)],P1]+[[iX21,P1],δ(P1)]=δ(iX21)P1P1δ(iX21)P1P1δ(iX21)+P1δ(iX21)P1+iX21δ(P1)P1+iP1δ(P1)X21+iX21δ(P1)+iX21δ(P1)+iδ(P1)X21+iδ(P1)X21. (2.14)

    Multiplying (2.14) by P2 from the left and by P1 from the right, we have

    P2δ(iX21)P1=2iX21δ(P1)P1+iP2δ(P1)X21. (2.15)

    By (2.15), Lemma 2.2 and the fact that P2δ(P1)P2=0, we obtain X21δ(P1)P1=0. Then P1δ(P1)P1=0. Similarly, P2δ(P2)P2=0.

    Remark 2.1. Let S=P1δ(P1)P2P2δ(P1)P1. Then S=S by Lemma 2.1. We define a map Δ:AA by

    Δ(X)=δ(X)[X,S]

    for any XA. It is easy to verify that Δ is a map satisfying

    Δ([[A,B],C])=[[Δ(A),B],C]+[[A,Δ(B)],C]+[[A,B],Δ(C)]

    for any A,B,CA with ABC=0. By Lemmas 2.1–2.4, it follows that

    (a) Δ(Pi)=0 (i=1,2);

    (b) For any AijAij (1ij2), we have PjΔ(Aij)Pi=0;

    (c) For any A12A12,B21A21, there exist UA12,B21A11,VA12,B21A22 such that

    Δ(A12+B21)=Δ(A12)+Δ(B21)+UA12,B21+VA12,B21.

    Lemma 2.5. Δ(Aii)Aii (i=1,2).

    Proof. Let A11A11. From A11P2P2=0, [[A11,P2],P2]=0 and Δ(P2)=0, we have

    0=Δ([[A11,P2],P2])=[[Δ(A11),P2],P2]=Δ(A11)P2P2Δ(A11)P2P2Δ(A11)+P2Δ(A11)P2. (2.16)

    Multiplying (2.16) by P1 from the left, we get P1Δ(A11)P2=0. Since P2A11P1=0, [[P2,A11],P1]=0 and Δ(P1)=Δ(P2)=0, we have

    0=Δ([[P2,A11],P1])=[[P2,Δ(A11)],P1]=P2Δ(A11)P1P1Δ(A11)P2. (2.17)

    Multiplying (2.17) by P2 from the left, we get P2Δ(A11)P1=0. For any X12A12, from X12A11P2=0, [[X12,A11],P2]=0 and Δ(P2)=0, we have

    0=Δ([[X12,A11],P2])=[[Δ(X12),A11],P2]+[[X12,Δ(A11)],P2]=A11Δ(X12)P2+P2Δ(X12)A11+X12Δ(A11)P2P2Δ(A11)X12. (2.18)

    Multiplying (2.18) by P1 from the left, we get A11Δ(X12)P2+X12Δ(A11)P2=0. It follows from Remark 2.1(b) that X12Δ(A11)P2=A11(P2Δ(X12)P1)=0. Then P2Δ(A11)P2=0. Hence Δ(A11)A11. Similarly, Δ(A22)A22.

    Lemma 2.6. Δ(Aij)Aij (1ij2).

    Proof. Let A12A12. Then P2Δ(A12)P1=0 by Remark 2.1(b). For any X12A12, from X12A12P1=0 and Δ(P1)=0, we have

    Δ(A12X12+X12A12)=Δ([[X12,A12],P1])=[[Δ(X12),A12],P1]+[[X12,Δ(A12)],P1]=A12Δ(X12)P1+P1Δ(X12)A12+X12Δ(A12)P1Δ(A12)X12P1Δ(A12)X12+X12Δ(A12). (2.19)

    Multiplying (2.19) by P2 from the left and by P1 from the right, then by Lemma 2.5, we get P2Δ(A12)X12=0. Hence P2Δ(A12)P2=0. Since A12X12P2=0, [[A12,X12],P2]=0 and Δ(P2)=0, we have

    0=Δ([[A12,X12],P2])=[[Δ(A12),X12],P2]+[[A12,Δ(X12)],P2]=Δ(A12)X12X12Δ(A12)P2X12Δ(A12)+P2Δ(A12)X12+A12Δ(X12)P2P2Δ(X12)A12. (2.20)

    Multiplying (2.20) by P1 from the left and by P2 from the right, then by P2Δ(A12)P2=P2Δ(X12)P2=0, we have P1Δ(A12)X12=0. It follows that P1Δ(A12)P1=0. Therefore Δ(A12)A12. Similarly, Δ(A21)A21.

    Lemma 2.7. For any AiiAii,BijAij,BjiAji (1ij2), we have

    (a) Δ(Aii+Bij)=Δ(Aii)+Δ(Bij);

    (b) Δ(Aii+Bji)=Δ(Aii)+Δ(Bji).

    Proof. (a) Let T=Δ(Aii+Bij)Δ(Aii)Δ(Bij). Since (iPj)I(Aii+Bij)=(iPj)IAii=(iPj)IBij=0 and [[iPj,I],Aii]=0, we have

    [[Δ(iPj),I],Aii+Bij]+[[iPj,Δ(I)],Aii+Bij]+[[iPj,I],Δ(Aii+Bij)]=Δ([[iPj,I],Aii+Bij])=Δ([[iPj,I],Aii])+Δ([[iPj,I],Bij])=[[Δ(iPj),I],Aii+Bij]+[[iPj,Δ(I)],Aii+Bij]+[[iPj,I],Δ(Aii)+Δ(Bij)],

    which implies

    [[iPj,I],T]=0. (2.21)

    Multiplying (2.21) by Pi from the left, by Pi from the right, by Pj from both sides, respectively, we get Tij=Tji=Tjj=0. Hence

    Δ(Aii+Bij)=Δ(Aii)+Δ(Bij)+Tii. (2.22)

    For any XijAij, from (Aii+Bij)XijPj=AiiXijPj=BijXijPj=0, [[Bij,Xij],Pj]=0 and (2.22), we have

    [[Δ(Aii)+Δ(Bij)+Tii,Xij],Pj]+[[Aii+Bij,Δ(Xij)],Pj]+[[Aii+Bij,Xij],Δ(Pj)]=[[Δ(Aii+Bij),Xij],Pj]+[[Aii+Bij,Δ(Xij)],Pj]+[[Aii+Bij,Xij],Δ(Pj)]=Δ([[Aii+Bij,Xij],Pj])=Δ([[Aii,Xij],Pj])+Δ([[Bij,Xij],Pj])=[[Δ(Aii)+Δ(Bij),Xij],Pj]+[[Aii+Bij,Δ(Xij)],Pj]+[[Aii+Bij,Xij],Δ(Pj)].

    This implies

    [[Tii,Xij],Pj]=0. (2.23)

    Multiplying (2.23) by Pj from the right, we see that TiiXij=0. Hence Tii=0, and so we obtain (a).

    Similarly, we can show that (b) holds.

    Lemma 2.8. For any Aij,BijAij (1ij2), we have

    Δ(Aij+Bij)=Δ(Aij)+Δ(Bij).

    Proof. For any A12,B12A12, it follows that

    [[P1+A12,P2+B12],P2]=A12+B12A12B12. (2.24)

    Then by (2.24) and Remark 2.1(c), there exist UA12+B12,A12B12A11, VA12+B12,A12B12 A22 such that

    Δ([[P1+A12,P2+B12],P2])=Δ(A12+B12)+Δ(A12B12)+UA12+B12,A12B12+VA12+B12,A12B12. (2.25)

    From (P1+A12)(P2+B12)P2=0, Δ(P1)=Δ(P2)=0, (2.25), Lemmas 2.6 and 2.7, we have

    Δ(A12+B12)+Δ(A12B12)+UA12+B12,A12B12+VA12+B12,A12B12=Δ([[P1+A12,P2+B12],P2])=[[Δ(A12),P2+B12],P2]+[[P1+A12,Δ(B12)],P2]=Δ(A12)+Δ(B12)Δ(A12)Δ(B12). (2.26)

    Multiplying (2.26) by P1 from the left and by P2 from the right, then by Lemma 2.6 and the fact that UA12+B12,A12B12A11,VA12+B12,A12B12A22, we see that Δ(A12+B12)=Δ(A12)+Δ(B12). Similarly, we can show that Δ(A21+B21)=Δ(A21)+Δ(B21).

    Lemma 2.9. For any Aii,BiiAii (i=1,2), we have

    Δ(Aii+Bii)=Δ(Aii)+Δ(Bii).

    Proof. For any A11,B11A11,B12A12, from A11B12P2=0, Δ(P2)=0, [[A11,B12],P2]=A11B12B12A11, Lemmas 2.5, 2.6 and 2.8, we have

    Δ(A11B12)+Δ(B12A11)=Δ([[A11,B12],P2])=[[Δ(A11),B12],P2]+[[A11,Δ(B12)],P2]=Δ(A11)B12+A11Δ(B12)B12Δ(A11)Δ(B12)A11. (2.27)

    Multiplying (2.27) by P1 from the left and by P2 from the right, we have

    Δ(A11B12)=Δ(A11)B12+A11Δ(B12). (2.28)

    Similarly, we can show that

    Δ(A22B21)=Δ(A22)B21+A22Δ(B21). (2.29)

    For any X12A12, it follows from Lemma 2.8 and (2.28) that

    Δ(A11+B11)X12+(A11+B11)Δ(X12)=Δ((A11+B11)X12)=Δ(A11X12)+Δ(B11X12)=Δ(A11)X12+A11Δ(X12)+Δ(B11)X12+B11Δ(X12).

    It follows that (Δ(A11+B11)Δ(A11)Δ(B11))X12=0. Then Δ(A11+B11)=Δ(A11)+Δ(B11). Similarly, we can show that Δ(A22+B22)=Δ(A22)+Δ(B22).

    Lemma 2.10. For any A12A12,B21A21, we have

    Δ(A12+B21)=Δ(A12)+Δ(B21).

    Proof. For any X12A12, by X12(A12+B21)P1=X12A12P1=X12B21P1=0,

    [[X12,A12+B21],P1]=[[X12,A12],P1]+[[X12,B21],P1]A11,

    Remark 2.1(c) and Lemma 2.9, there exist UA12,B21A11,VA12,B21A22 such that

    [[Δ(X12),A12+B21],P1]+[[X12,Δ(A12)+Δ(B21)+UA12,B21+VA12,B21],P1]+[[X12,A12+B21],Δ(P1)]=Δ([[X12,A12+B21],P1])=Δ([[X12,A12],P1])+Δ([[X12,B21],P1])=[[Δ(X12),A12+B21],P1]+[[X12,Δ(A12)+Δ(B21)],P1]+[[X12,A12+B21],Δ(P1)].

    Then

    0=[[X12,UA12,B21+VA12,B21],P1]=VA12,B21X12+X12VA12,B21. (2.30)

    Multiplying (2.30) by P1 from the right, we get VA12,B21X12=0. Hence VA12,B21=0. Then by Remark 2.1(c), we get

    Δ(A12+B21)=Δ(A12)+Δ(B21)+UA12,B21. (2.31)

    For any X21A21, from X21(A12+B21)P2=X21A12P2=X21B21P2=0,

    [[X21,A12+B21],P2]=[[X21,A12],P2]+[[X21,B21],P2]A22,

    Lemma 2.9 and (2.31), we have

    [[Δ(X21),A12+B21],P2]+[[X21,Δ(A12)+Δ(B21)+UA12,B21],P2]+[[X21,A12+B21],Δ(P2)]=Δ([[X21,A12+B21],P2])=Δ([[X21,A12],P2])+Δ([[X12,B21],P2])=[[Δ(X21),A12+B21],P2]+[[X21,Δ(A12)+Δ(B21)],P2]+[[X21,A12+B21],Δ(P2)],

    which implies

    0=[[X21,UA12,B21],P2]=UA12,B21X21+X21UA12,B21. (2.32)

    Multiplying (2.32) by P2 from the right, we obtain UA12,B21X21=0. Then UA12,B21=0. Hence we obtain the desired result.

    Lemma 2.11. For any A11A11,B12A12,C21A21,D22A22, we have

    (a) Δ(A11+B12+C21)=Δ(A11)+Δ(B12)+Δ(C21);

    (b) Δ(B12+C21+D22)=Δ(B12)+Δ(C21)+Δ(D22).

    Proof. (a) Let T=Δ(A11+B12+C21)Δ(A11)Δ(B12)Δ(C21). From P2(A11+B12+C21)P2=P2A11P2=P2B12P2=P2C21P2=0 and [[P2,A11],P2]=[[P2,C21],P2]=0, we have

    [[Δ(P2),A11+B12+C21],P2]+[[P2,Δ(A11+B12+C21)],P2]+[[P2,A11+B12+C21],Δ(P2)]=Δ([[P2,A11+B12+C21],P2])=Δ([[P2,A11],P2])+Δ([[P2,B12],P2])+Δ([[P2,C21],P2])=[[Δ(P2),A11+B12+C21],P2]+[[P2,Δ(A11)+Δ(B12)+Δ(C21)],P2]+[[P2,A11+B12+C21],Δ(P2)].

    This implies

    [[P2,T],P2]=0. (2.33)

    Multiplying (2.33) by P1 from the left, we have T12=0. For any X12A12, from P1X12(A11+B12+C21)=P1X12A11=P1X12B12=P1X12C21=0, [[P1,X12],A11+B12+C21]=X12C21B12X12 and Lemma 2.9, we have

    [[Δ(P1),X12],A11+B12+C21]+[[P1,Δ(X12)],A11+B12+C21]+[[P1,X12],Δ(A11+B12+C21)]=Δ([[P1,X12],A11+B12+C21])=Δ(X12C21)+Δ(B12X12)=Δ([[P1,X12],A11])+Δ([[P1,X12],B12])+Δ([[P1,X12],C21])=[[Δ(P1),X12],A11+B12+C21]+[[P1,Δ(X12)],A11+B12+C21]+[[P1,X12],Δ(A11)+Δ(B12)+Δ(C21)],

    which implies

    [[P1,X12],T]=0. (2.34)

    Multiplying (2.34) by P1 from both sides, we obtain X12TP1P1TX12=0. Then X12TP1=0 by T12=0. Hence T21=0. Multiplying (2.34) by P2 from the right, we have X12TP2=0 and so T22=0. Let SA11,B12,C21=T11. Then SA11,B12,C21A11 and

    Δ(A11+B12+C21)=Δ(A11)+Δ(B12)+Δ(C21)+SA11,B12,C21.

    Similarly, there exists a RB12,C21,D22A22 such that

    Δ(B12+C21+D22)=Δ(B12)+Δ(C21)+Δ(D22)+RB12,C21,D22. (2.35)

    For any X21A21, by [[P2,X21],A11+B12+C21]=A11X21+X21A11+X21B12C21X21 and (2.35), there exist a RA11X21,X21A11,X21B12C21X21A22 such that

    Δ([[P2,X21],A11+B12+C21])=Δ(A11X21)+Δ(X21A11)+Δ(X21B12C21X21)+RA11X21,X21A11,X21B12C21X21. (2.36)

    From P2X21(A11+B12+C21)=P2X21A11=P2X21B12=P2X21C21=0, (2.36), Lemmas 2.9 and 2.10, we have

    [[Δ(P2),X21],A11+B12+C21]+[[P2,Δ(X21)],A11+B12+C21]+[[P2,X21],Δ(A11+B12+C21)]=Δ([[P2,X21],A11+B12+C21])=Δ(A11X21)+Δ(X21A11)+Δ(X21B12C21X21)+RA11X21,X21A11,X21B12C21X21=Δ(A11X21+X21A11)+Δ(X21B12)+Δ(C21X21)+RA11X21,X21A11,X21B12C21X21=Δ([[P2,X21],A11])+Δ([[P2,X21],B12])+Δ([[P2,X21],C21])+RA11X21,X21A11,X21B12C21X21=[[Δ(P2),X21],A11+B12+C21]+[[P2,Δ(X21)],A11+B12+C21]+[[P2,X21],Δ(A11)+Δ(B12)+Δ(C21)]+RA11X21,X21A11,X21B12C21X21.

    It follows that

    [[P2,X21],T]=RA11X21,X21A11,X21B12C21X21. (2.37)

    Multiplying (2.37) by P1 from the right, then by RA11X21,X21A11,X21B12C21X21A22, we obtain X21TP1=0. Hence SA11,B12,C21=T11=0, and so Δ(A11+B12+C21)=Δ(A11)+Δ(B12)+Δ(C21).

    Similarly, we can show that (b) holds.

    Lemma 2.12. For any A11A11,B12A12,C21A21,D22A22, we have

    Δ(A11+B12+C21+D22)=Δ(A11)+Δ(B12)+Δ(C21)+Δ(D22).

    Proof. Let T=Δ(A11+B12+C21+D22)Δ(A11)Δ(B12)Δ(C21)Δ(D22). From (A11+B12+C21+D22)P1P2=A11P1P2=B12P1P2=C21P1P2=D22P1P2=0 and [[A11+B12+C21+D22,P1],P2]=C21+C21, we have

    [[Δ(A11+B12+C21+D22),P1],P2]+[[A11+B12+C21+D22,Δ(P1)],P2]+[[A11+B12+C21+D22,P1],Δ(P2)]=Δ([[A11+B12+C21+D22,P1],P2])=Δ([[A11,P1],P2])+Δ([[B12,P1],P2])+Δ([[C21,P1],P2])+Δ([[D22,P1],P2])=[[Δ(A11)+Δ(B12)+Δ(C21)+Δ(D22),P1],P2]+[[A11+B12+C21+D22,Δ(P1)],P2]+[[A11+B12+C21+D22,P1],Δ(P2)].

    This implies

    [[T,P1],P2]=0. (2.38)

    Multiplying (2.38) by P2 from the left, we have T21=0. Similarly, T12=0. For any X12A12, from P1X12(A11+B12+C21+D22)=P1X12A11=P1X12B12=P1X12C21=P1X12D22=0, [[P1,X12],A11+B12+C21+D22]=X12C21B12X12+X12D22D22X12, Lemmas 2.9 and 2.12, we have

    [[Δ(P1),X12],A11+B12+C21+D22]+[[P1,Δ(X12)],A11+B12+C21+D22]+[[P1,X12],Δ(A11+B12+C21+D22)]=Δ([[P1,X12],A11+B12+C21+D22])=Δ(X12C21)+Δ(B12X12)+Δ(X12D22D22X12)=Δ([[P1,X12],A11])+Δ([[P1,X12],B12])+Δ([[P1,X12],C21])+Δ([[P1,X12],D22])=[[Δ(P1),X12],A11+B12+C21+D22]+[[P1,Δ(X12)],A11+B12+C21+D22]+[[P1,X12],Δ(A11)+Δ(B12)+Δ(C21)+Δ(D22)].

    This implies

    [[P1,X12],T]=0. (2.39)

    Multiplying (2.39) by P2 from the right, we obtain X12TP2=0. Then T22=0. Similarly, T11=0. Hence we obtain the desired result.

    Lemma 2.13. For any Aii,BiiAii,Aij,BijAij,BjiAji,BjjAjj (1ij2), we have

    (a) Δ(AiiBij)=Δ(Aii)Bij+AiiΔ(Bij);

    (b) Δ(AijBjj)=Δ(Aij)Bjj+AijΔ(Bjj);

    (c) Δ(AiiBii)=Δ(Aii)Bii+AiiΔ(Bii);

    (d) Δ(AijBji)=Δ(Aij)Bji+AijΔ(Bji).

    Proof. (a) It follows from (2.28) and (2.29) that (a) holds.

    (b) Let A12A12,B22A22. From A12B22P2=0, Δ(P2)=0, [[A12,B22],P2]=A12B22B22A12, Lemmas 2.5, 2.6 and 2.12, we have

    Δ(A12B22)+Δ(B22A12)=Δ([[A12,B22],P2])=[[Δ(A12),B22],P2])+[[A12,Δ(B22)],P2]=Δ(A12)B22+A12Δ(B22)B22Δ(A12)Δ(B22)A12. (2.40)

    Multiplying (2.40) by P1 from the left and by P2 from the right, we have Δ(A12B22)=Δ(A12)B22+A12Δ(B22). Similarly, Δ(A21B11)=Δ(A21)B11+A21Δ(B11).

    (c) Let A11,B11A11,X12A12. It follows from (a) that

    Δ(A11B11)X12+A11B11Δ(X12)=Δ(A11B11X12)=Δ(A11)B11X12+A11Δ(B11X12)=Δ(A11)B11X12+A11Δ(B11)X12+A11B11Δ(X12).

    It follows that (Δ(A11B11)Δ(A11)B11A11Δ(B11))X12=0. Hence Δ(A11B11)=Δ(A11)B11+A11Δ(B11). Similarly, Δ(A22B22)=Δ(A22)B22+A22Δ(B22).

    (d) Let A12A12,B21A21. From B21P1A12=0, Δ(P1)=0, [[B21,P1],A12]=B21A12+A12B21, Lemmas 2.6 and 2.12, we have

    Δ(B21A12)+Δ(A12B21)=Δ([[B21,P1],A12])=[[Δ(B21),P1],A12]+[[B21,P1],Δ(A12)]=Δ(B21)A12+A12Δ(B21)+B21Δ(A12)+Δ(A12)B21. (2.41)

    Multiplying (2.41) by P1 from both sides, we have Δ(A12B21)=Δ(A12)B21+A12Δ(B21). Similarly, Δ(A21B12)=Δ(A21)B12+A21Δ(B12).

    Now, we give the proof of Theorem 2.1 in the following.

    Proof of Theorem 2.1. By Lemmas 2.5, 2.6, 2.8, 2.9, 2.12 and 2.13, it is easy to verify that Δ is an additive derivation on A. Let AijAij (1ij2). By AijPjPj=0, Δ(Pj)=0 and Lemma 2.6, we have

    Δ(Aij)Δ(Aij)=Δ([[Aij,Pj],Pj])=[[Δ(Aij),Pj],Pj]=Δ(Aij)Δ(Aij).

    It follows that

    Δ(Aij)=Δ(Aij). (2.42)

    Let AiiAii, XjiAji (1ij2). Since AiiPiXji=0, Δ(Pi)=0, [[Aii,Pi],Xji]=XjiAiiXjiAii, Lemmas 2.5, 2.6 and 2.13(b), we have

    Δ(Xji)Aii+XjiΔ(Aii)Δ(Xji)AiiXjiΔ(Aii)=Δ(XjiAii)Δ(XjiAii)=Δ([[Aii,Pi],Xji])=[[Δ(Aii),Pi],Xji]+[[Aii,Pi],Δ(Xji)]=Δ(Xji)Aii+XjiΔ(Aii)Δ(Xji)AiiXjiΔ(Aii).

    It follows that Xji(Δ(Aii)Δ(Aii))=0. Then

    Δ(Aii)=Δ(Aii). (2.43)

    For any AA, we have A=2i,j=1Aij. By (2.42), (2.43) and the additivity of Δ on A, it follows that

    Δ(A)=2i,j=1Δ(Aij)=2i,j=1Δ(Aij)=Δ(A).

    Hence Δ is an additive -derivation. Therefore, δ is an additive -derivation on A by Remark 2.1.

    In this paper, we gave the characterization of a kind of non-global nonlinear skew Lie triple derivations on factor von Neumann algebras.

    This research is supported by Scientific Research Project of Shangluo University (21SKY104).

    The authors declare that there are no conflicts of interest.



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