EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 17, No. 4, 2024, 3399-3414 ISSN 1307-5543 – ejpam.com Published by New York Business Global Nonlinear Mixed λ-Jordan Triple Derivation on ∗-algebras Amal S. Alali1, Junaid Nisar2,∗, Nadeem ur Rehman3, Hafedh M. Alnoghashi4 1 Department of Mathematical Sciences, College of Science, Princess Nourah bint Abdulrahman University, P. O. Box 84428, Riyadh 11671, Saudi Arabia 2 Department of Applied Sciences, Symbiosis Institute of Technology, Symbiosis International (Deemed) University, Lavale, Pune,India 3 Department of Mathematics, Aligarh Muslim University, Aligarh-202002 India 4 Department of Computer Science, College of Engineering and Information Technology, Amran University, Amran, Yemen. Abstract. Let A be a ∗-algebra with unit I and P1 and P2 = I − P1 includes a non-trivial projections, and let λ ∈ C \ {0,−1}. In this paper, we aim to study the characterization of nonlinear mixed λ-Jordan triple derivation on ∗-algebras. As an application, we can also apply our results on prime ∗-algebras, factor von-Neumann algebras and standard operator algebras. 2020 Mathematics Subject Classifications: 16W10, 47B47, 46K15. Key Words and Phrases: λ-Mixed Jordan triple derivation, ∗-derivation, ∗- algebra. 1. Introduction Consider an ∗-algebra A defined over the complex field C. Introducing the λ-Jordan product U♢λV = UV + λV U and the skew Lie product [U, V ]∗ = UV − V U∗ for nonzero scalar λ, these algebraic structures have gained significant attention in various research do- mains, as evidenced by studies such as [1–5, 7, 8, 12]. In the context of additive mappings, an additive derivation is characterized by Π(UV ) = Π(U)V + UΠ(V ) for all U, V ∈ A. If the additional condition Π(U∗) = Π(U)∗ holds for all U ∈ A, then Π is termed an additive ∗-derivation. Now, let Π : A → A be a map without assuming additivity. The concept of a nonlinear skew Lie derivation is introduced, defined by the relation Π([U, V ]∗) = [Π(U), V ]∗ + [U,Π(V )]∗ for all U, V ∈ A. Notably, Kong and Zhang [3] established the result that every nonlinear skew Lie derivation is, in fact, an additive ∗-derivation. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v17i4.5390 Email addresses: asalali@pnu.edu.sa (A. Alali), junaidnisar73@gmail.com (J. Nisar), nu.rehman.mm@amu.ac.in (N. Rehman), halnoghashi@gmail.com (H. Alnoghashi) https://www.ejpam.com 3399 Copyright: © 2024 The Author(s). (CC BY-NC 4.0) . Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3400 Similarly, a map Π : A → A is termed a nonlinear skew Lie triple derivation if it satisfies the equation Π([[U, V ]∗,W ]∗) = [[Π(U), V ]∗,W ]∗ + [[U,Π(V )]∗,W ]∗ + [[U, V ]∗,Π(W )]∗ for all U, V,W ∈ A. Several recent studies have delved into the exploration of derivations and isomorphisms associated with innovative products resulting from the combination of Lie and skew Lie products, as evidenced by works such as [6, 9, 11]. Notably, Zhou et al. [13] established the result that every nonlinear mixed Lie triple derivation on a prime ∗-algebra is, in fact, an additive ∗-derivation. Additionally, Pang et al. [10] demonstrated that every second nonlinear mixed Jordan triple derivable mapping on factor von Neumann algebras also an additive ∗-derivation. Motivated by these previous works, our paper introduces the λ-Jordan product defined as U♢λV = UV +λV U . We specifically focus on the derivation corresponding to the novel product obtained by combining the skew Lie product and the λ-Jordan product. In this context, we define a map Π : A → A as a mixed λ-Jordan triple derivation if it satisfies the equation Π([U, V ]∗♢λW ) = [Π(U), V ]∗ ⋄Λ W + [U,Π(V )]∗♢λW + [U, V ]∗♢λΠ(W ) for all U, V,W ∈ A. Our main result establishes that Π is a nonlinear mixed λ-Jordan triple derivation on ∗-algebras if and only if Π is an additive ∗-derivation. 2. Main Result Theorem 2.1. Let A be a unital ∗-algebra with unity I containing a non-trivial projection P satisfies XAP = 0 =⇒ X = 0 (▲) and XA(I − P ) = 0 =⇒ X = 0. (▼) Define a map Π : A → A such that Π([U, V ]∗♢λW ) = [Π(U), V ]∗♢λW ) + [U,Π(V )]∗♢λW + [U, V ]∗♢λΠ(W ). Then Π is an additive ∗-derivation. Consider a non-trivial projection P = P1 in the algebra A, and let P2 = I −P1, where I is the unity element of the algebra. Utilizing the Peirce decomposition of A, we express A as the direct sum A = P1AP1⊕P1AP2⊕P2AP1⊕P2AP2. Denoting the corresponding subspaces as A11 = P1AP1, A12 = P1AP2, A21 = P2AP1, and A22 = P2AP2, we can represent any element U ∈ A as the sum U = U11 + U12 + U21 + U22, where Uij ∈ Aij and U∗ ij ∈ Aji for i, j = 1, 2. Before proving Theorem 2.1, we need several lemmas and remarks. . Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3401 Lemma 2.1. Π(0) = 0. Proof. It is obvious that Π(0) = Π([0, 0]∗♢λ0) = [Π(0), 0]∗♢λ0 + [0,Π(0)]∗♢λ0 + [0, 0]∗♢λΠ(0) = 0. Lemma 2.2. Let U12 ∈ A12 and U21 ∈ A21. Then Π(U12 + U21) = Π(U12) + Π(U21). Proof. Let T = Π(U12 + U21) − Π(U12) − Π(U21). Since [U12, P1]∗♢λP2 = 0 and by using Lemma 2.1, we have Π([U12 + V21, P1]∗♢λP2) = Π([U12, P1]∗♢λP2) + Π([V21, P1]∗♢λP2) = [Π(U12), P1]∗♢λP2 + [U12,Π(P1)]∗♢λP2 + [U12, P1]∗♢λΠ(P2) +[Π(V21), P1]∗♢λP2 + [V21,Π(P1)]∗♢λP2 + [V21, P1]∗♢λΠ(P2). On the other hand, we find Π([U12 + V21, P1]∗♢λP2 = [Π(U12 + V21), P1]∗♢λP2 + [U12 + V21,Π(P1)]∗♢λP2 +[U12 + V21, P1]∗♢λΠ(P2). From the above two equations, we get [T, P1]∗♢λP2 = 0. That means−P1T ∗P2+λP2TP1 = 0. Multiplying by P2 from the left and since λ ̸= 0, we get P2TP1 = 0. Similarly, one can show that P1TP2 = 0. Now, for every X21 ∈ A21, it follows from [X21, U12]∗♢λP1 = 0 and using Lemma 2.1 that Π([X21, U12 + V21]∗♢λP1) = Π([X21, U12]∗♢λP1) + Π([X21, V21]∗♢λP1) = [Π(X21), U12]∗♢λP1 + [X21,Π(U12)]∗♢λP1 +[X21, U12]∗♢λΠ(P1) + [Π(X21), V21]∗♢λP1 +[X21,Π(V21)]∗♢λP1 + [X21, V21]∗♢λΠ(P1). On the other hand, we have Π([X21, U12 + V21]∗♢λP1) = [Π(X21), U12 + V21]∗♢λP1 + [X21,Π(U12 + V21)]∗♢λP1 +[X21, U12 + V21]∗♢λΠ(P1). From the above expressions, we find that [X21, T ]∗♢λP1 = 0. That means X21TP1 − λP1TX ∗ 21 = 0. Multiplying both sides by P1 from right, we get X21TP1 = 0. By using (▲) and (▼), we have P1TP1 = 0. Similarly, we can show that P2TP2 = 0. Hence, T = 0 i.e., Π(U12 + U21) = Π(U12) + Π(U21). Lemma 2.3. For any Uij ∈ Aij , 1 ≤ i, j ≤ 2, we have Π( 2∑ i,j=1 Uij) = 2∑ i,j=1 Π(Uij). . Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3402 Proof. Let T = Π(U11 + U12 + U21 + U22)−Π(U11)−Π(U12)−Π(U21)−Π(U22). For every X12 ∈ A12, also [P1, U11]∗♢λX12 = [P1, U22]∗♢λX12 = 0 and using Lemmas 2.1 and 2.2, we get Π([P1, U11 + U12 + U21 + U22]∗♢λX12) = Π([P1, U11]∗♢λX12) + Π([P1, U12]∗♢λX12) +Π([P1, U21]∗♢λX12) + Π([P1, U22]∗♢λX12) = [Π(P1), U11]∗♢λX12 + [P1,Π(U11)]∗♢λX12 +[P1, U11]∗♢λΠ(X12) + [Π(P1), U12]∗♢λX12 +[P1,Π(U12)]∗♢λX12 + [P1, U12]∗♢λΠ(X12) +[Π(P1), U21]∗♢λX12 + [P1,Π(U21)]∗♢λX12 +[P1, U21]∗♢λΠ(X12) + [Π(P1), U22]∗♢λX12 +[P1,Π(U22)]∗♢λX12 + [P1, U22]∗♢λΠ(X12). On the other hand, we have Π([P1, U11 + U12 + U21 + U22]∗♢λX12) = [Π(P1), U11 + U12 + U21 + U22]∗♢λX12 +[P1,Π(U11 + U12 + U21 + U22)]∗♢λX12 +[P1, U11 + U12 + U21 + U22]∗♢λΠ(X12). By comparing the above two equations, we get [P1,M ]∗♢λX12 = 0 from which we obtain P1TX12 − TX12 − λX12TP1 = 0. Multiplying P2 from left and right, we get P2TX12 = 0. By using (▲) and (▼), we have P2TP1 = 0. Similarly, we can show that P1TP2 = 0. Again forX12 ∈ A12, it follows from [X12, U11]∗♢λP2 = [X12, U12]∗♢λP2 = [X12, U12]∗♢λP2 = 0 that Π([X12, U11 + U12 + U21 + U22]∗♢λP2) = Π([X12, U11]∗♢λP2) + Π([X12, U12]∗♢λP2) +Π([X12, U21]∗♢λP2) + Π([X12, U22]∗♢λP2) = [Π(X12), U11]∗♢λP2 + [X12,Π(U11)]∗♢λP2 +[X12, U11]∗♢λΠ(P2) + [Π(X12), U12]∗♢λP2 +[X12,Π(U12)]∗♢λP2 + [X12, U12]∗♢λΠ(P2) +[Π(X12), U21]∗♢λP2 + [X12,Π(U21)]∗♢λP2 +[X12, U21]∗♢λΠ(P2) + [Π(X12), U22]∗♢λP2 +[X12,Π(U22)]∗♢λP2 + [X12, U22]∗♢λΠ(P2). On the other hand, we get Π([X12, U11 + U12 + U21 + U22]∗♢λP2) = [Π(X12), U11 + U12 + U21 + U22]∗♢λP2 +[X12,Π(U11 + U12 + U21 + U22)]∗♢λP2 +[X12, U11 + U12 + U21 + U22]∗♢λΠ(P2). By the above two equations, we get [X12, T ]∗♢λP2 = 0. That means that X12TP2 − λP2TX ∗ 12 = 0. When we multiply both sides by P1 on the left, the result is X12TP2 = 0. . Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3403 By using (▲) and (▼), we have P2TP2 = 0. Similarly, P1TP1 = 0. Hence, T = 0 i.e., Π(U11 + U12 + U21 + U22) = Π(U11) + Π(U12) + Π(U21) + Π(U22). Lemma 2.4. For any Uij , Vij ∈ Aij with (1 ≤ i ̸= j ≤ 2), we have Π(Uij + Vij) = Π(Uij) + Π(Vij). Proof. Initially, we establish the result for i = 1 and j = 2. Let T = Π(U12 + V12) − Π(U12)−Π(V12). Since [X12, U12]∗♢λP2 = 0, and using Lemma 2.1, we get Π([X12, U12 + V12]∗♢λP2) = Π([X12, U12]∗♢λP2) + Π([X12, V12]∗♢λP2) = [Π(X12), U12]∗♢λP2 + [X12,Π(U12)]∗♢λP2 +[X12, U12]∗♢λΠ(P2) + [Π(X12), V12]∗♢λP2 +[X12,Π(V12)]∗♢λP2 + [X12, V12]∗♢λΠ(P2). On the other hand, we have Π([X12, U12 + V12]∗♢λP2) = [Π(X12), U12 + V12]∗♢λP2 + [X12,Π(U12 + V12)]∗♢λP2 +[X12, U12 + V12]∗♢λΠ(P2). By comparing the last two expressions, we get [X12, T ]∗♢λP2 = 0. That means X12TP2− P2TX ∗ 12 = 0. By left-multiplying both sides of the preceding equation by P1 and utilizing (▲) and (▼), we obtain P2TP2 = 0. Similarly, we can show that P1TP1 = 0. Now, again for any X12 ∈ A12. Since [P1, U12]∗♢λX12 = 0 and using Lemma 2.1, we have Π([P1, U12 + V12]∗♢λX12) = Π([P1, U12]∗♢λX12) + Π([P1, V12]∗♢λX12) = [Π(P1), U12]∗♢λX12 + [P1,Π(U12)]∗♢λX12 +[P1, U12]∗♢λΠ(X12) + [Π(P1), V12]∗♢λX12 +[P1,Π(V12)]∗♢λX12 + [P1, U12]∗♢λΠ(X12). On the other hand, we find Π([P1, U12 + V12]∗♢λX12) = [Π(P1), U12 + V12]∗♢λX12 + [P1,Π(U12 + V12)]∗♢λX12 +[P1, U12 + V12]∗♢λΠ(X12). From the last two expressions, we find [P1, T ]∗♢λX12 = 0. That means P1TX12−TX12− λX12TP1 = 0. Multiplying both sides by P1 from right and since λ ̸= 0, we haveX12TP1 = 0. Thus, P2TP1 = 0 follows from (▲) and (▼). Similarly, we can show that P1TP2 = 0. Hence, T = 0 i.e., Π(U12 + V12) = Π(U12) + Π(V12). By using the same technique as above, one can show that Π(U21 + V21) = Π(U21) + Π(V21). . Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3404 Lemma 2.5. For any U11, V11 ∈ A11 and U22, V22 ∈ A22, we have (i) Π(U11 + V11) = Π(U11) + Π(V11). (ii) Π(U22 + V22) = Π(U22) + Π(V22). Proof. Let T = Π(U11 + V11)−Π(U11)−Π(V11). On the one hand, we have Π([P2, U11 + V11]∗♢λP1) = [Π(P2), U11 + V11]∗♢λP1 + [P2,Π(U11 + V11)]∗♢λP1 +[P2, U11 + V11]∗♢λΠ(P1). On the other hand, it follows from [P2, U11]∗♢λP1 = 0 that Π([P2, U11 + V11]∗♢λP1) = Π([P2, U11]∗♢λP1) + Π([P2, V11]∗♢λP1) = [Π(P2), U11]∗♢λP1 + [P2,Π(U11)]∗♢λP1 + [P2, U11]∗♢λΠ(P1) +[Π(P2), V11]∗♢λP1 + [P2,Π(V11)]∗♢λP1 + [P2, V11]∗♢λΠ(P1). By comparing the last two equations, we find [P2, T ]∗♢λP1 = 0. This gives P2TP1 − λP1TP2 = 0 and hence, P2TP1 = P1TP2 = 0. Again for any X12 ∈ A12 and since [X12, U11]∗♢λP2 = 0, we find Π([X12, U11 + V11]∗♢λP2) = Π([X12, U11]∗♢λP2) + Π([X12, V11]∗♢λP2) = [Π(X12), U11]∗♢λP2 + [X12,Π(U11)]∗♢λP2 +[X12, U11]∗♢λΠ(P2) + [Π(X12), V11]∗♢λP2 +[X12,Π(V11)]∗♢λP2 + [X12, V11]∗♢λΠ(P2). From the other side, we get Π([X12, U11 + V11]∗♢λP2) = [Π(X12), U11 + V11]∗♢λP2 + [X12,Π(U11 + V11)]∗♢λP2 +[X12, U11 + V11]∗♢λΠ(P2). From the last two equations, we get [X12, T ]∗♢λP2 = 0. That means X12TP2−λP2TX ∗ 12 = 0. Thus, X12TP2 = 0. By using (▲) and (▼), we get P2TP2 = 0. Similarly, we can show that P1TP1 = 0. Hence, T = 0 i,e. Π(U11 + V11) = Π(U11) + Π(V11). (ii). By using the same argument as in (i), one can show that Π(U22 + V22) = Π(U22) + Π(V22). Lemma 2.6. Π is an additive map. . Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3405 Proof. For any U, V ∈ A, we write U = U11 + U12 + U21 + U22 and V = V11 + V12 + V21 + V22. By using Lemmas 2.3 - 2.5, we get Π(U + V ) = Π(U11 + U12 + U21 + U22 + V11 + V12 + V21 + V22) = Π(U11 + V11) + Π(U12 + V12) + Π(U21 + V21) + Π(U22 + V22) = Π(U11) + Π(V11) + Π(U12) + Π(V12) + Π(U21) + Π(V21) + Π(U22) + Π(V22) = Π(U11 + U12 + U21 + U22) + Π(V11 + V12 + V21 + V22) = Π(U) + Π(V ). Lemma 2.7. (i) Π(P1) ∗ = Π(P1), Π(P2) ∗ = Π(P2). (ii) P1Π(P1)P2 = −P1Π(P2)P2. (iii) P2Π(P2)P1 = −P2Π(P1)P1. Proof. (i) In the view of [P1, P1]∗♢λI = 0 and using Lemma 2.1, we have 0 = Π([P1, P1]∗♢λI) = [Π(P1), P1]∗♢λI + [P1,Π(P1)]∗♢λI + [P1, P1]∗♢λΠ(I) = (1 + λ)(P1Π(P1) ∗ − P1Π(P1)). Since λ ̸= −1, we have P1Π(P1) ∗ = P1Π(P1). That means P1Π(P1) ∗P1 = P1Π(P1)P1. (2.1) P1Π(P1) ∗P2 = P1Π(P1)P2 (2.2) P2Π(P1) ∗P1 = P2Π(P1)P1 (2.3) Also, [P1, P2]∗♢λI = 0 and Π(0) = 0, we have 0 = Π([P1, P2]∗♢λI = 0) = [Π(P1), P2]∗♢λI + [P1,Π(P2)]∗♢λI + [P1, P2]∗♢λΠ(I) = (1 + λ)(P2Π(P1)P2 − P2Π(P1) ∗P2). That is, P2Π(P1) ∗P2 = P2Π(P1)P2. (2.4) From equations (2.1)-(2.4), we conclude Π(P1) ∗ = Π(P1). Similarly, by using the same technique as above, we get Π(P2) ∗ = Π(P2). (ii) Again [P1, P2]∗♢λP2 = 0 and Π(0) = 0, we find 0 = Π([P1, P2]∗♢λP2) . Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3406 = [Π(P1), P2]∗♢λP2 + [P1,Π(P2)]∗♢λP2 + [P1, P2]∗♢λΠ(P2) = Π(P1)P2 − P2Π(P1) ∗P2 + λP2Π(P1)P2 − λP2Π(P1) ∗ + P1Π(P2)P2 − λP2Π(P2)P1. Multiplying above equation by P1 from left, we get P1Π(P1)P2 = −P1Π(P2)P2. (iii) By using the same technique as in (ii), we can show that P2Π(P2)P1 = −P2Π(P1)P1. Lemma 2.8. For every Uij ∈ Aij(1 ≤ i ̸= j ≤ 2), we have PjΠ(λUij)Pi = 0. Proof. To begin, we establish the result for i = 1 and j = 2. For any U12 ∈ A12, we get Π(λU12) = Π([P1, λU12]∗♢λP2) = [Π(P1), λU12]∗♢λP2 + [P1,Π(λU12)]∗♢λP2 + [P1, λU12]∗♢λΠ(P2) = λΠ(P1)U12 − λU12Π(P1)P2 + λ2P2Π(P1)U12 + P1Π(λU12)P2 −λP2Π(λU12)P1 + λU12Π(P2) + λ2Π(P2)U12. By left-multiplying the above equation with P2 and right-multiplying with P1, we obtain (1 + λ)P2Π(λU12)P1 = 0. (2.5) Since λ ̸= −1, we find P2Π(λU12)P1 = 0. Similarly, by using the same technique for i = 2, j = 1, one can show that P1Π(λU21)P2 = 0. Lemma 2.9. (i) P1Π(P2)P1 = P2Π(P1)P2 = 0. (ii) P1Π(P1)P1 = P2Π(P2)P2 = 0. Proof. (i) For every X12 ∈ A12, it follows from [X12, P1]∗♢λP1 = 0 that 0 = Π([X12, P1]∗♢λP1) = [Π(X12), P1]∗♢λP1 + [X12,Π(P1)]∗♢λP1 + [X12, P1]∗♢λΠ(P1) = Π(X12)P1 − P1Π(X12) ∗P1 + λP1Π(X12)P1 − λP1Π(X12) ∗ +X12Π(P1)P1 −Π(P1)X ∗ 12 + λX12Π(P1)− λP1Π(P1)X ∗ 12. By left-multiplying the preceding equation by P1 and right-multiplying by P2, and con- sidering the fact that λ ̸= 0, we obtain X12Π(P1)P2 = P1Π(X12) ∗P2. (2.6) That means, P2Π(P1)X ∗ 12 = P2Π(X12)P1. . Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3407 Also, [P1, X12]∗♢λX12 = 0 and using Lemma 2.1, we get 0 = Π([P1, X12]∗♢λX12) = [Π(P1), X12]∗♢λX12 + [P1,Π(X12)]∗♢λX12 + [P1, X12]∗♢λΠ(X12) = −X12Π(P1) ∗X12 + λX12Π(P1)X12 + P1Π(X12)X12 −Π(X12)X12 − λX12Π(X12)P1 +X12Π(X12) + λX12Π(X12). Multiplying the above equation by P1 from left and right, we find X12Π(X12)P1 = 0. By using (▲) and (▼), we get P2Π(X12)P1 = 0. That means P1Π(X12) ∗P2 = 0. From Equation (2.6),(▲) and (▼), we have P2Π(P1)P2 = 0. Similarly, we can show that P1Π(P2)P1 = 0. (ii) For any X21 ∈ A21 and using Lemma 2.6, we have Π([X21, P1]∗♢λP1) = Π(X21)−Π(λX∗ 21). On the other hand, we have Π([X21, P1]∗♢λP1) = [Π(X21), P1]∗♢λP1 + [X21,Π(P1)]∗♢λP1 + [X21, P1]∗♢λΠ(P1) = Π(X21)P1 − P1Π(X21) ∗P1 + λP1Π(X21)P1 − λP1Π(X21) ∗ +X21Π(P1)P1 − λP1Π(P1)X ∗ 21 +X21Π(P1) + λΠ(P1)X21. By comparing the aforementioned two equations and subsequently left-multiplying by P2 and right-multiplying by P1, we obtain 2X21Π(P1)P1 + λP2Π(P1)X21 + P2Π(λX ∗ 21)P1 = 0 Now, by using Lemma 2.8 and Lemma 2.9(i), we getX21Π(P1)P1 = 0. Hence, P1Π(P1)P1 = 0. Similarly, we can show that P2Π(P2)P2 = 0. Now, let M = P1Π(P1)P2 − P2Π(P1)P1. Then M = −M∗. We define a mapping ∆ : A → A as ∆(U) = Π(U)−(UM−MU) for all U ∈ A. It can be easily verified that for all U, V,W ∈ A, ∆([U, V ]∗♢λW ) = [∆(U), V ]∗♢λW )+ [U,∆(V )]∗♢λW +[U, V ]∗♢λ∆(W ). Remark 2.1. The mapping ∆ possesses the following properties: (i) ∆ is additive. (ii) ∆(P1) = ∆(P2) = 0. (iii) ∆(I) = 0 (iv) For every Uij ∈ Aij(1 ≤ i ̸= j ≤ 2), we have Pj∆(λUij)Pi = 0 . Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3408 (v) ∆ is a ∗-derivation if and only if Π is an ∗ -derivation. Proof. (i) Since [U,M ] is additive and also using Lemma 2.6, it is clear that ∆ is additve. (ii) By using Lemma 2.9, we have ∆(P1) = Π(P1)− P1Π(P1)P2 − P2Π(P1)P1 = P1Π(P1)P2 + P2Π(P1)P1 − P1Π(P1)P2 − P2Π(P1)P1 = 0. Similarly, we can show that ∆(P2) = 0. (iii) By using additivity of ∆, we have ∆(I) = ∆(P1 + P2) = ∆(P1) + ∆(P2) = 0. (iv) For i = 1 and j = 2, it can be inferred from Lemma 2.8 that P2∆(λU12)P1 = P2(Π(λU12)− λU12M +MλU12)P1 = 0. In the same way, one can show for i = 2, j = 1, i.e., P1∆(λU21)P2 = 0. (v) Since [U,M ] = UM − MU is an additive ∗-derivation. Therefore, ∆ qualifies as a ∗-derivation if and only if Π is a ∗-derivation. Lemma 2.10. ∆(Uij) ⊆ Uij , i, j = 1, 2. Proof. First, we prove for i = 1, j = 1. For every U11 ∈ A11, it follows from Remark 2.1 that 0 = ∆([P1, U11]∗♢λP1) = [P1,∆(U11)]∗♢λP1 = P1∆(U11)P1 −∆(U11)P1 + λP1∆(U11)− λP1∆(U11)P1. Left multiplying the above equation by P2, we find P2∆(U11)P1 = 0. (2.7) Similarly, again by using Remark 2.1, we have 0 = ∆([P2, U11]∗♢λP1) = [P2,∆(U11)]∗♢λP1 = P2∆(U11)P1 − λP1∆(U11)P2. Multiplying P2 on the right and since λ ̸= 0, we get P1∆(U11)P2 = 0 (2.8) . Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3409 Now, for every X12 ∈ A12 and ∆(P2) = 0, it follows that 0 = ∆([X12, U11]∗♢λP2) = [∆(X12), U11]∗♢λP2 + [X12,∆(U11)]∗♢λP2 = −U11∆(X12) ∗P2 + λP2∆(X12)U11 +X12∆(U11)P2 −λP2∆(U11)X ∗ 12. Multiplying P1 from left and P2 from right and using Lemma 2.8, we get X12∆(U11)P2 = U11∆(X12) ∗P2 = U11(P2∆(X12)P1) ∗ = 0. Thus, X12∆(U11)P2 = 0. By using (▲) and (▼), we have P2∆(U11)P2 = 0. (2.9) From Equations (2.7)-(2.9), we have ∆(U11) ⊆ U11. Similarly, we can show that ∆(U22) ⊆ U22. Next, we establish the result for i = 1, j = 2. Additionally, for any U12 ∈ A12 and ∆(P2) = 0, we have ∆(U12) = ∆([P1, U12]∗♢λP2) = [P1,∆(U12)]∗♢λP2 = P1∆(U12)P2 − λP2∆(U12)P1. By left and right multiplying the above equation by P1, we obtain P1∆(U12)P1 = 0. (2.10) Additionally, by left and right-multiplying by P2, we find P2∆(U12)P2 = 0. (2.11) Similarly, multiplying P2 from left and P1 from right and since λ ̸= −1, we get P2∆(U12)P1 = 0 (2.12) From equations (2.10)- (2.12), we get ∆(U12) ⊆ U12. Similarly, by using the same technique as above, one can prove that ∆(U21) ⊆ U21. Lemma 2.11. For any Ui,j , Vi,j ∈ Aij , 1 ≤ i, j ≤ 2, we have (i) ∆(U11V12) = ∆(U11)V12 + U11∆(V12) and ∆(U22V21) = ∆(U22)V21 + U22∆(V21). (ii) ∆(U12V21) = ∆(U12)V21 + U12∆(V21) and ∆(U21V12) = ∆(U21)V12 + U21∆(V12). (iii) ∆(U11V11) = ∆(U11)V11 + U11∆(V11) and ∆(U22V22) = ∆(U22)V22 + U22∆(V22). . Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3410 (iv) ∆(U12V22) = ∆(U12)V22 + U12∆(V22) and ∆(U21V11) = ∆(U21)V11 + U21∆(V11). Proof. (i) Using Lemma 2.10 and ∆(P2) = 0, we get ∆([U11, V12]∗♢λP2) = [∆(U11), V12]∗♢λP2 + [U11,∆(V12)]∗♢λP2 = ∆(U11)V12 + U11∆(V12). On the other side, we get ∆([U11, V12]∗♢λP2) = ∆(U11V12). By comparing the above two equations, we get ∆(U11V12) = ∆(U11)V12 + U11∆(V12). Similarly, we can show that ∆(U22V21) = ∆(U22)V21 + U22∆(V21). (ii) For any X12 ∈ A12 and by using Lemma 2.11 (1), we have ∆([U12, V21]∗♢λX12) = ∆(U12V21X21) = ∆(U12V21)X12 + U12V21∆(X12). On the other hand, we have ∆([U12, V21]∗♢λX12) = [∆(U12), V21]∗♢λX12 + [U12,∆(V21)]∗♢λX12 +[U12, V21]∗♢λ∆(X12) = ∆(U12)V21X12 + U12∆(V21)X12 + U12V21∆(X12). From the above two expressions, we get (∆(U12V21)−∆(U12)V21 − U12∆(V21))X12 = 0. Thus, by (▲) and (▼), ∆(U12V21) = ∆(U12)V21 + U12∆(V21). Similarly, we can show that ∆(U21V12) = ∆(U21)V12 + U21∆(V12). (iii) For any X12 ∈ A12, it follows from Lemma 2.11(i) that ∆(U11V11X12) = ∆(U11V11)X12 + U11V11∆(X12). Again using Lemma 2.11 from the other side, we have ∆(U11V11X12) = ∆(U11)V11X12 + U11∆(V11X12) = ∆(U11)V11X12 + U11∆(V11)X12 + U11V11∆(X12). By comparing the aforementioned two equations, we obtain (∆(U11V11) − ∆(U11)V11 − U11∆(V11))X12 = 0. Therefore, utilizing (▲) and (▼), we conclude that ∆(U11V11) = ∆(U11)V11 + U11∆(V11). Similarly, one can show that ∆(U22V22) = ∆(U22)V22 + U22∆(V22). (iv) For any X21 ∈ A21. It follows from Lemma 2.11(2) that ∆(U12V22X21) = ∆(U12V22)X21 + U12V22∆(X21). . Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3411 Again on the other side, it follows from Lemma 2.11(i) and Lemma 2.11(ii) that ∆(U12V22X21) = ∆(U12)V22X21 + U12∆(V22X21) = ∆(U12)V22X21 + U12∆(V22)X21 + U12V22∆(X21). From the above two equations and using (▲) and (▼), we find ∆(U12V22) = ∆(U12)V22 + U12∆(V22). Similarly, one can show that ∆(U21V11) = ∆(U21)V11 + U21∆(V11). Lemma 2.12. ∆(U∗) = ∆(U)∗ for all U ∈ A. Proof. For any X12 ∈ A12, it follows from Remark 2.1 and Lemma 2.11(i) that ∆([U11, P1]∗♢λX12) = ∆(U11X12)−∆(U∗ 11X12) = ∆(U11)X12 + U11∆(X12)−∆(U∗ 11)X12 − U∗ 11∆(X12). Alternatively, it can be deduced from ∆(P1) = 0 that ∆([U11, P1]∗♢λX12) = [∆(U11), P1]∗♢λX12 + [U11, P1]∗♢λ∆(X12) = ∆(U11)X12 −∆(U11) ∗X12 + U11∆(X12)− U∗ 11∆(X12). From the above two equations, we have (∆(U11) ∗ −∆(U∗ 11))X12 = 0. Now, by using (▲) and (▼), we get ∆(U11) ∗ = ∆(U∗ 11). (2.13) Similarly, by using the same technique, one can show that ∆(U22) ∗ = ∆(U∗ 22). (2.14) Again, it follows from Lemma 2.10 and ∆(P2) = 0 that ∆([U12, P2]∗♢λX12) = [∆(U12), P2]∗♢λX12 + [U12, P2]∗♢λ∆(X12) = −∆(U12) ∗X12 − λX12∆(U12) ∗ − U∗ 12∆(X12)− λ∆(X12)U ∗ 12. On the other hand, using Lemma 2.11, we have ∆([U12, P2]∗♢λX12) = ∆(−U∗ 12X12 − λX12U ∗ 12) = −∆(U∗ 12X12)−∆(X12λU ∗ 12) = −∆(U∗ 12)X12 − U∗ 12∆(X12)− λ∆(X12)U ∗ 12 −X12∆(λU∗ 12). By comparing the above two equations, we get (∆(U12) ∗ −∆(U∗ 12))X12 + λX12∆(U12) ∗ −X12∆(λU∗ 12) = 0. By left-multiplying both sides by P2 and utilizing (▲) and (▼), we obtain ∆(U12) ∗ = ∆(U∗ 12). (2.15) . Nisar et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 3399-3414 3412 Similarly, we can show that ∆(U21) ∗ = ∆(U∗ 21). (2.16) From equations (2.13)-(2.16) and using additivity of ∆, we get ∆(U∗) = ∆(U)∗. Proof of Theorem 2.1 For every U, V ∈ A, we can write U = U11+U12+U21+U22 and V = V11 + V12 + V21 + V22. Since, ∆ is additive and using Lemma 2.11, we get ∆(UV ) = ∆(U11V11 + U11V12 + U12V21 + U12V22 +U21V11 + U21V12 + U22V21 + U22V22) = ∆(U11V11) + ∆(U11V12) + ∆(U12V21) + ∆(U12V22) +∆(U21V11) + ∆(U21V12) + ∆(U22V21) + ∆(U22V22) = ∆(U11 + U12 + U21 + U22)(V11 + V12 + V21 + V22) +(U11 + U12 + U21 + U22) ∆(V11 + V12 + V21 + V22) = ∆(U)V + U∆(V ). So, ∆ is a derivation. By using Lemma 2.12, ∆ is an additive ∗-derivation. Hence, by Remark 2.1, Π is an additive ∗-derivation. This completes the proof of Theorem 2.1. The corollaries following directly from Theorem 2.1 are as follows: Corollary 2.1. Let A be a standard operator algebra on an infinite dimensional complex Hilbert space H containing identity operator I. Suppose that A is closed under adjoint operation. Define Π : A → A such that Π([U, V ]∗♢λW ) = [Π(U), V ]∗♢λW ) + [U,Π(V )]∗♢λW + [U, V ]∗♢λΠ(W ) for all U, V,W ∈ A . Then Π is an additive ∗-derivation. Corollary 2.2. Let A ba a factor von Neumann algebra with dimM ≥ 2. Define Π : M → M such that Π([U, V ]∗♢λW ) = [Π(U), V ]∗♢λW ) + [U,Π(V )]∗♢λW + [U, V ]∗♢λΠ(W ) for all U, V,W ∈ A . Then Π is an additive ∗-derivation. Corollary 2.3. Let A be a prime ∗-algebra with unit I containing non-trivial projection P . A map Π : A → A satisfies Π([U, V ]∗♢λW ) = [Π(U), V ]∗♢λW ) + [U,Π(V )]∗♢λW + [U, V ]∗♢λΠ(W ) for all U, V,W ∈ A . Then Π is an additive ∗-derivation. REFERENCES 3413 Acknowledgements The authors extended their appreciation to Princess Nourah bint Abdulrahman Uni- versity for funding this research under Researchers Supporting Project number (PNURSP2024R231), Princess Nourah bint Abdulrahman University, Riyadh Saudi Arabia. Conflicts of Interest: The authors declare no conflict of interest. References [1] M. Ashraf, Md. Shamim Akhter, and M. Ansari. Nonlinear bi-skew jordan-type derivations on factor von neumann algebras. Filomat, 37(17):5591–5599, 2023. [2] D. Huo, B. Zheng, J. Xu, and H. Liu. Nonlinear mappings preserving jordan multiple– product on factor von neumann algebras. Linear and Multilinear Algebra, 63(5):1026– 1036, 2015. [3] L. Kong and J. Zhang. Nonlinear skew lie derivations on prime-rings. Indian Journal of Pure and Applied Mathematics, 54(2):475–484, 2023. [4] C. Li, Q. Chen, and T. Wang. Nonlinear maps preserving the jordan triple*- product on factor von neumann algebras. Chinese Annals of Mathematics, Series B, 39(4):633–642, 2018. [5] C. Li and F. Lu. Nonlinear maps preserving the jordan triple 1-*-product on von neumann algebras. Complex Analysis and Operator Theory, 11:109–117, 2017. [6] C. Li and D. Zhang. Nonlinear mixed jordan triple-derivations on-algebras. Siberian Mathematical Journal, 63(4):735–742, 2022. [7] C. Li, F. Zhao, and Q. Chen. Nonlinear skew lie triple derivations between factors. Acta Mathematica Sinica, English Series, 32(7):821–830, 2016. [8] C. Li, Y. Zhao, and F. Zhao. Nonlinear*-jordan-type derivations on*-algebras. Rocky Mountain Journal of Mathematics, 51(2):601–612, 2021. [9] Y. Liang and J. Zhang. Nonlinear mixed lie triple derivations on factor von neumann algebras. Acta Math Sci Chinese Series, 62:1–13, 2019. [10] Y. Pang, D. Zhang, and D. Ma. The second nonlinear mixed jordan triple deriv- able mapping on factor von neumann algebras. Bulletin of the Iranian Mathematical Society, 48(3):951–962, 2022. [11] N. Rehman, J. Nisar, and M. Nazim. A note on nonlinear mixed jordan triple deriva- tion on*-algebras. Communications in Algebra, 51(4):1334–1343, 2023. REFERENCES 3414 [12] F. Zhang. Nonlinear η-jordan triple ∗-derivation on prime ∗-algebras. Rocky Mountain J. Math, 52:323–333, 2022. [13] Z. Zhou, Y. Zhujun, and Z Jianhua. Nonlinear mixed lie triple derivations on prime *-algebras. Communications in Algebra, 47(11):4791–4796, 2019.