EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 2, Article Number 6031 ISSN 1307-5543 – ejpam.com Published by New York Business Global The Composition of Modified Reflection Operators and Their Fixed Point Sets Salihah Thabet Alwadani Mathematics, Yanbu Industrial College, The Royal Comission for Jubail and Yanbu, Yanbu, Saudi Arabia Abstract. The modified reflection operator plays a crucial role in optimization, particularly in algorithms designed to solve constrained optimization problems. By effectively transforming feasible solutions while maintaining their viability within defined constraints, this operator en- ables smoother navigation through the solution space. It enhances convergence rates and stability in iterative methods, such as projected gradient descent and proximal algorithms. In this paper, we investigate the fixed point sets of the compositions of three modified reflection operators onto linear closed subspaces. We also derive formulas for the compositions under different parameters. 2020 Mathematics Subject Classifications: 47H09, 47H05, 47A06, 90C25 Key Words and Phrases: Composition, fixed point set, linear subspace, orthogonal subspace, projector, identity operator, modified reflector opeartor, reflector operator 1. Introduction The modified reflection operator is a vital component in optimization, particularly in algorithms that address constrained optimization problems. This operator facilitates the transformation of feasible solutions while ensuring they remain within the defined constraints, allowing for a more efficient exploration of the solution space. By enhancing the convergence rates and stability of iterative methods, such as projected gradient de- scent and proximal algorithms, the modified reflection operator significantly improves the performance of optimization algorithms (see [1], [2], [3], [4], [5], [6], and [7] for more information). Moreover, it is particularly effective in navigating non-convex landscapes, as it aids in the exploration of local minima, thereby reducing the risk of stagnation in suboptimal solutions. Its adaptability to various constraints further underscores its ver- satility, making it an essential tool in a wide range of applications, from machine learning DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.6031 Email address: salihah.s.alwadani@gmail.com (S. Th. Alwadani) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) S.Th. Alwadani / Eur. J. Pure Appl. Math, 18 (2) (2025), 6031 2 of 13 to engineering design. Overall, the modified reflection operator contributes to more ro- bust and efficient optimization processes, ensuring that algorithms can effectively tackle real-world problems (see [8] for more information). Throughout, we assume that H is a real Hilbert space with inner product ⟨·, ·⟩ : H×H → R, (1) and induced norm ∥ · ∥ : H → R : x 7→ √ ⟨x, x⟩. Let A : H ⇒ H be an arbitrary set valued operator, i.e., Ax ⊆ H ( ∀x ∈ H ) . The graph of A, donoted by Then gra A, is defined as gra A = { (x, y) ∈ H×H | y ∈ Ax } . A set-valued operator A : H ⇒ H is a monotone if( ∀ ( x, u ) ∈ gra A ) ( ∀ ( y, v ) ∈ gra A ) ⟨x − y, u − v⟩ ≥ 0. A monotone operator A is a maximally monotone if there exists no monotone operator B such that gra A ⊂ gra B. That is, for every (x, u) ∈ H×H,(( x, u ) ∈ gra A ) ⇔ ( ∀ ( y, v ) ∈ gra A ) ⟨x − y, u − v⟩ ≥ 0. The Id is the identity operator defines as Id : H → H : x → x and satisfies (8). Definition 1. [9, Definition 3.28] Let A be a monotone operator from H ⇒ H and denote the associated resolvent by JA = (Id+A)−1. (2) The reflected resolvent of A is denoted by RA and defined by RA = 2JA − Id . (3) Example 1. Let A = Id. Then JA = 1 2 Id and RA = 0. To show that let y ∈ H and set x = JAy. Then y ∈ ( Id+A ) x. This implies that y = x + x ⇔ y = 2x ⇔ x = 1 2 y. Therefore, JA = 1 2 Id . Using (3) gives RA = 2(1/2) Id− Id = 0. Definition 2. [9, Definition 4.1] Let U be a nonempty subset of H. A mapping T : U → H is nonexpansive or Lipschitz continuous with constant 1, i.e.,( ∀x ∈ U ) ( ∀y ∈ U ) ∥Tx − Ty∥ ≤ ∥x − y∥. (4) Moreover, T : U → H is firmly nonexpansive if( ∀x ∈ U )( ∀y ∈ U ) ∥Tx − Ty∥2 + ∥(Id−T)x − (Id−T)y∥2 ≤ ∥x − y∥2. (5) The Fix T is the set of fixed points of T defined as Fix T := {x ∈ H | x = Tx}. (6) S.Th. Alwadani / Eur. J. Pure Appl. Math, 18 (2) (2025), 6031 3 of 13 Definition 3. Let U be a nonempty closed and convex subset of H and let z ∈ H. The projection operator (this is also known as the closet point mapping) of z onto U is the unique point in U denoted by PU z that satisfies ∥z − x∥ = inf∥u − z∥, where x = PU z. Let U be closed linear subspace. Then, RU := 2 PU − Id . (7) and Id := PU +PU⊥ . (8) Example 2. Let U be a nonempty closed convex subset of H and let RU = 2 PU − Id be a nonexpansive operator on U. Then Fix RU = C. To show that let x ∈ H. Then x = RUx. Using (3) gives x = 2 PU x − x. This implies that PU x = x ⇔ x ∈ U. Example 3. Let A = a + PU , where U is a closed linear subspace of H and a ∈ H. Then JA = ( Id− 1 2 PU ) + ( 1 2 PU − Id ) a and RA = ( Id−PU ) + ( PU −2 Id ) a. Proof. See [9, Lemma 4.3 (i) and (ii)]. ■ For more details about the composition of reflectors, see [10], [9], [11], [2], [12], and [13]. A comprehensive analysis of nonexpansive mappings under the condition of isom- etry of finite order of R was provided in [9, Lemma] and [14, Section 3]. We refer the reader to [15, Exercise 12.16], [2, Example 20.29], and [16] In this paper, we study the composition of three modified reflection operators and their fixed point sets.. Our results can be summarized as follows: • Lemma 1 and Lemma 2 provide key properties concerning the fixed point set of the composition of two modified reflection operators. • Theorem 1, Theorem 2, and Theorem 3 show that the sequence in which three mod- ified reflection operators are applied affects the fixed point set of their composition. These theorems offer valuable insights into the fixed point set resulting from the composition of three modified reflection operators. • Under different parameters γ, β, α ∈ (0, 1], we derive formulas for the composition of two modified reflection operators (see Lemma 1 and Lemma 2). Additionally, formulas for the composition of three modified reflection operators are given in Theorem 1, Theorem 2, and Theorem 3. The notation employed in this paper is standard and closely aligns with that in [9, 17], and [2]. S.Th. Alwadani / Eur. J. Pure Appl. Math, 18 (2) (2025), 6031 4 of 13 2. Results In this section, we will present significant new results regarding the composition of two and three modified reflection operators, as well as their corresponding fixed point sets. We will explore the properties and interactions of these operators, highlighting how their compositions influence the structure of the fixed point sets. Lemma 1. Let U be a closed linear subspace of H, and let U⊥ denote the orthogonal complement of U. Let β ∈]0, 1]. Recall from (7) that RU := 2 PU − Id, where PU is defined in Definition 3. The following results hold: (i) RU⊥RU = − Id = RU RU⊥ . (ii) ( 2β PU − Id ) = βRU + (1 − β)(− Id). (iii) ( 2β PU⊥ − Id ) = βRU⊥ + (1 − β)(− Id). (iv) − ( 2β PU − Id ) = βRU⊥ + (1 − β) Id. (v) ( 2β PU − Id ) ◦ (− Id) = − ( 2β PU − Id ) . (vi) Fix (( 2β PU − Id ) ◦ (− Id) ) = Fix ( − ( 2β PU − Id )) = Fix ( βRU +(1− β)(− Id) ) = U⊥. Proof. (i): See [9, Lemma 6.2 (i)]. (ii): Using (7) gives ( 2β PU − Id ) = 2β PU − Id+β Id−β Id = ( 2β PU −β Id ) + (1 − β)(− Id) = βRU + (1 − β)(− Id). (iii): Applying (7) yields( 2β PU⊥ − Id ) = 2β PU⊥ − Id+β Id−β Id = ( 2β PU⊥ −β Id ) + (1 − β)(− Id) = βRU⊥ + (1 − β)(− Id). (iv): Using (7), (8), and (ii), we obtain − ( 2β PU − Id ) = − ( βRU + (1 − β)(− Id) ) = −βRU + (1 − β) Id = −β ( 2 PU − Id ) + (1 − β) Id = −β ( PU − ( Id−PU )) + (1 − β) Id = −β ( PU −PU⊥ ) + (1 − β) Id = −β ( PU −PU⊥ +PU⊥ −PU⊥ ) + (1 − β) Id S.Th. Alwadani / Eur. J. Pure Appl. Math, 18 (2) (2025), 6031 5 of 13 = −β ( PU +PU⊥ −2 PU⊥ ) + (1 − β) Id = −β ( Id−2 PU⊥ ) + (1 − β) Id = βRU⊥ + (1 − β) Id, as required. (v): Let x ∈ H. By using (ii) and (iv), we have( 2β PU − Id ) ◦ (− Id)(x) = ( 2β PU − Id ) (−x) = 2β PU(−x)− (−x) = −2β PU(x) + x = − ( 2β PU x − x ) = βRU⊥x + (1 − β)x (vi): It follows from (ii), (iv), and (v) that Fix (( 2β PU − Id ) ◦ (− Id) ) = Fix ( − ( 2β PU − Id )) = Fix ( βRU + (1 − β)(− Id) ) . Let x ∈ H. Then x = βRU⊥(x) + (1 − β)x = βRU⊥(x) + x − βx, therefore, x − x = βRU⊥(x)− βx 0 = β ( 2 PU⊥ x − x ) − βx 0 = 2β PU⊥ x − 2βx, and 2βx = 2β PU⊥ x. Hence, x = PU⊥ x ⇔ Fix (( 2β PU − Id ) ◦ (− Id) ) = U⊥. ■ Lemma 2. Let U be a closed linear subspace of H, and let γ, β ∈]0, 1]. Then the following holds: ( 2β PU⊥ − Id )( 2γ PU − Id ) =  − Id, for β = γ = 1 0, for β = γ = 1/2 Id−2 ( β PU⊥ +γ PU ) , for β, γ ̸= 1 = ( 2γ PU − Id )( 2β PU⊥ − Id ) . S.Th. Alwadani / Eur. J. Pure Appl. Math, 18 (2) (2025), 6031 6 of 13 Proof. Using Lemma 1, (i), (ii), and (iii) yields( 2β PU⊥ − Id )( 2γ PU − Id ) = ( βRU⊥ + (1 − β)(− Id) )( γRU + (1 − γ)(− Id) ) = −βγ Id+β(1 − γ)RU − (1 − β)γRU + (1 − β)(1 − γ) Id = βRU − γRU − (β + γ) Id+ Id = Id−2 ( β PU⊥ +γ PU ) . There are 3 cases: Case 1: If β = γ = 1, then( 2β PU⊥ − Id )( 2γ PU − Id ) = Id−2 ( PU⊥ +PU ) = Id−2 Id = − Id, by (8). Case 2: If β = γ = 1/2, then( PU⊥ − Id )( PU − Id ) = Id− ( PU⊥ +PU ) = Id− Id = 0, Case 3: If β, γ ̸= 1 and β, γ ̸= 1/2,then( 2β PU⊥ − Id )( 2γ PU − Id ) = Id−2 ( β PU⊥ +γ PU ) . Applying Lemma 1, (i), (ii), and (iii), the same strategy can be appied to show that ( 2γ PU − Id )( 2β PU⊥ − Id ) =  − Id, for γ = β = 1 0, for γ = β = 1/2 Id−2 ( γ PU +β PU⊥ ) , for γ, β ̸= 1 Therefore, ( 2β PU⊥ − Id )( 2γ PU − Id ) = ( 2γ PU − Id )( 2β PU⊥ − Id ) . ■ Example 4. Let X = R2 and suppose that U = R× {0} and x = (2, 2). Then U⊥ = {0} ×R, and by Lemma 2, we obtain the following: Case 1. If β = γ = 1, then( 2 P{0}×R − Id )( 2 PR×{0}(2, 2)− (2, 2) ) = ( 2 PR×{0} − Id )( 2 P{0}×R(2, 2)− (2, 2) ) = (−2,−2). S.Th. Alwadani / Eur. J. Pure Appl. Math, 18 (2) (2025), 6031 7 of 13 Case 2. If β = γ = 1/2, then( P{0}×R − Id )( PR×{0}(2, 2)− (2, 2) ) = ( PR×{0} − Id )( P{0}×R(2, 2)− (2, 2) ) = ( 0, 0 ) . Case 3. If β, γ ̸= 1/2 and β, γ ̸= 1, then( 2β P{0}×R − Id )( 2γ PR×{0}(2, 2)− (2, 2) ) = ( 2γ PR×{0} − Id )( 2β P{0}×R(2, 2)− (2, 2) ) = (2, 2)− 2 ( β(0, 2) + γ(2, 0) ) Note that when β, γ → 0, then( 2β P{0}×R − Id )( 2γ PR×{0}(2, 2)− (2, 2) ) → (2, 2). Additionally, when β, γ → 1, then( 2β P{0}×R − Id )( 2γ PR×{0}(2, 2)− (2, 2) ) → (−2,−2), which satisfies the first case. From now and on, deffine the modified reflector operators; RU,γ := 2γ PU − Id, (9) RU⊥,β := 2β PU⊥ − Id, (10) RV,α := 2α PV − Id . (11) Theorem 1. Let U and V be claosed linear subspaces of H. Suppose that β, γ, α ∈]0, 1] and recall from (9), (10), and (11) the modified reflector operators. If β, γ = 1, then the following are holds true: (i) RV,αRU⊥,βRU,γ = αRV⊥ + (1 − α) Id (ii) RV,αRU,γRU⊥,β = αRV⊥ + (1 − α) Id (iii) RU⊥,βRU,γRV,α = αRV⊥ + (1 − α) Id (iv) RU,γRU⊥,βRV,α = αRV⊥ + (1 − α) Id (v) Fix ( RV,αRU⊥,βRU,γ ) = V⊥ (vi) Fix ( RV,αRU,γRU⊥,β ) = V⊥ (vii) Fix ( RU⊥,βRU,γRV,α ) = V⊥ (viii) Fix ( RU,γRU⊥,βRV,α ) = V⊥ Additionally, if β, γ ̸= 1 and α = 1, then the following hold true: S.Th. Alwadani / Eur. J. Pure Appl. Math, 18 (2) (2025), 6031 8 of 13 (ix) RV,αRU⊥,βRU,γ = 2 PV +2γ PU +2β PU⊥ −4γ PV PU −4β PV PU⊥ − Id. (x) RV,αRU,γRU⊥,β = RV,αRU⊥,βRU,γ (xi) RU⊥,βRU,γRV,α = 2 PV +2γ PU +2β PU⊥ −4γ PU PV −4β PU⊥ PV − Id (xii) RU,γRU⊥,βRV,α = RU⊥,βRU,γRV,α. Moreover, if β, γ, α ̸= 1, then the following hold true: (xiii) RV,αRU⊥,βRU,γ = 2α PV +2γ PU +2β PU⊥ −4αγ PV PU −4αβ PV PU⊥ − Id. (xiv) RV,αRU,γRU⊥,β = RV,αRU⊥,βRU,γ (xv) RU⊥,βRU,γRV,α = 2α PV +2γ PU +2β PU⊥ −4γα PU PV −4βα PU⊥ PV − Id (xvi) RU,γRU⊥,βRV,α = RU⊥,βRU,γRV,α. Proof. (i): Using Lemma 1, (i), (iv), and (v) gives RV,αRU⊥,1RU,1 = RV,α ( − Id ) = −RV,α = αRV⊥ + (1 − α) Id (ii), (iii), (iv): The proof follows a similar approach as in (i). (v): It follows from (i) that RV,αRU⊥,βRU,γ = αRV⊥ + (1− α) Id, and using (6) and (i) gives Fix ( RV,αRU⊥,βRU,γ ) = Fix ( αRV⊥ + (1 − α) Id ) . Next, applying Lemma 1 (vi) yields, x ∈ Fix ( RV,αRU⊥,βRU,γ ) ⇔ x ∈ V⊥. (vi): The proof follows a similar approach as in statement (v), combining (6), (ii), and Lemma 1 with (vi). (vii): The proof adopts a similar method to that used in statement (v), combining (6), (iii) and Lemma 1 (vi). (viii): The proof follows a similar approach as in statement (v), combining (6), (iv) and Lemma 1 with (vi). (ix): Using Lemma 2 and (3) gives RV,αRU⊥,βRU,γ = RV,1RU⊥,βRU,γ = RV,1 − RV,12γ PU +RV,12β PU⊥ = ( 2 PV − Id ) − ( 2 PV − Id ) 2γ PU + ( 2 PV − Id ) 2β PU⊥ = 2 PV +2γ PU +2β PU⊥ −4γ PV PU −4β PV PU⊥ − Id . (x): Following the same approach as in statement (ix) gives RV,αRU,γRU⊥,β = RV,1RU,γRU⊥ = 2 PV +2γ PU +2β PU⊥ −4γ PV PU −4β PV PU⊥ − Id . Combining this result with (x) illustrates that when α = 1 RV,αRU,γRU⊥,β = RV,αRU,γRU⊥,β. (xi): Using Lemma 2 and (3) yields RU⊥,βRU,γRV,α = RU⊥,βRU,γRV,1 = RV,1 − (2γ PU)RV,1 + (2β PU⊥)RV,1 S.Th. Alwadani / Eur. J. Pure Appl. Math, 18 (2) (2025), 6031 9 of 13 = ( 2 PV − Id ) − 2γ PU ( 2 PV − Id ) + 2β PU⊥ ( 2 PV − Id ) = 2 PV +2γ PU +2β PU⊥ −4γ PU PV −4β PU⊥ PV − Id . (xii): Applying the same method as in statement (xi) gives RU,γRU⊥,βRV,α = RU,γRU⊥RV,1 = 2 PV +2γ PU +2β PU⊥ −4γ PU PV −4β PU⊥ PV − Id . Combining this result with (xi) illustrates that when α = 1 RU,γRU⊥,βRV,α = RU,γRU⊥,βRV,α. (xiii): Using Lemma 2 and (11) gives RV,αRU⊥,βRU,γ = RV,α − RV,α2γ PU +RV,α2β PU⊥ = ( 2α PV − Id ) − ( 2α PV − Id ) 2γ PU + ( 2α PV − Id ) 2β PU⊥ = 2α PV +2γ PU +2β PU⊥ −4αγ PV PU −4αβ PV PU⊥ − Id . (xiv): Following the same approach as in statement (xiii) gives RV,αRU,γRU⊥,β = 2α PV +2γ PU +2β PU⊥ −4αγ PV PU −4αβ PV PU⊥ − Id . Combining this result with (xiii) illustrates that when α ̸= 1 RV,αRU,γRU⊥,β = RV,αRU,γRU⊥,β. (xv): Using Lemma 2 and (11) yields RU⊥,βRU,γRV,α = RV,α − (2γ PU)RV,α + (2β PU⊥)RV,α = ( 2α PV − Id ) − 2γ PU ( 2α PV − Id ) + 2β PU⊥ ( 2α PV − Id ) = 2α PV +2γ PU +2β PU⊥ −4γα PU PV −4βα PU⊥ PV − Id . (xvi): Applying the same method as in statement (xv) gives RU,γRU⊥,βRV,α = 2α PV +2γ PU +2β PU⊥ −4γα PU PV −4βα PU⊥ PV − Id . Combining this result with (xv) illustrates that when α ̸= 1 RU,γRU⊥,βRV,α = RU,γRU⊥,βRV,α. ■ Remark 1. Regarding Theorem 1, if β = γ = α = 1, then the following holds: (i) RV,αRU⊥,βRU,γ = RV⊥ (ii) RV,αRU,γRU⊥,β = RV⊥ (iii) RU⊥,βRU,γRV,α = RV⊥ S.Th. Alwadani / Eur. J. Pure Appl. Math, 18 (2) (2025), 6031 10 of 13 (iv) RU,γRU⊥,βRV,α = RV⊥ (v) Fix ( RV,αRU⊥,βRU,γ ) = V⊥ (vi) Fix ( RV,αRU,γRU⊥,β ) = V⊥ (vii) Fix ( RU⊥,βRU,γRV,α ) = V⊥ (viii) Fix ( RU,γRU⊥,βRV,α ) = V⊥ Proof. (i), (ii), (iii), (iv), (v), (vi), (vii), and (viii): See [9, Proposition 6.3 (i), (ii), (iii), (iv), (v), (vi), (vii), and (viii)] for the proof. ■ Theorem 2. Let U and V be claosed linear subspaces of H. Suppose that β, γ, α ∈]0, 1] and recall from (9), (10), and (11) the modified reflector operators. If β = γ = α = 1, then the following holds: (i) RU,γRV,αRU⊥,β = Id+2 PV +8 PU PV PU⊥ −4 PU PV −4 PV PU⊥ (ii) RU⊥,βRV,αRU,γ = Id+2 PV +8 PU⊥ PV PU −4 PU⊥ PV −4 PV PU (iii) Fix ( RU,γRV,αRU⊥,β ) = RU ( V⊥) (iv) Fix ( RU⊥,βRV,αRU,γ ) = RU ( V⊥). Additionally, if β, γ = 1, then the following holds: (v) RU,γRV,αRU⊥,β = Id+2α PV +8α PU PV PU⊥ −4α PU PV −4α PV PU⊥ (vi) RU⊥,βRV,αRU,γ = Id+2α PV +8α PU⊥ PV PU −4α PU⊥ PV −4α PV PU (vii) Fix ( RU,γRV,αRU⊥,β ) = RU,1 ( V⊥) (viii) Fix ( RU⊥,βRV,αRU,γ ) = RU,1 ( V⊥) Moreover, if β, γ ̸= 1 and α = 1, then the following holds: (ix) We obtain RU,γRV,αRU⊥,β = 8γβ PU PV PU⊥ −4γ PU PV −4β PV PU⊥ +2 ( γ PU +β PU⊥ +PV ) − Id (x) Also, RU⊥,βRV,αRU,γ = 8βγ PU⊥ PV PU −4β PU⊥ PV −4γ PV PU +2 ( β PU⊥ +γ PU +PV ) − Id Proof. (i): From (3), we have RU,γRV,αRU⊥,β = RU,1RV,1RU⊥,1 = ( 2 PU − Id )( 4 PV PU⊥ −2 PV −2 PU⊥ + Id ) = 8 PU PV PU⊥ −4 PU PV +2 PU −4 PV PU⊥ +2 PV +2 PU⊥ − Id = 8 PU PV PU⊥ −4 PU PV +2 ( PU +PU⊥ ) − Id−4 PV PU⊥ +2 PV = 8 PU PV PU⊥ −4 PU PV + Id−4 PV PU⊥ +2 PV = Id+2 PV +8 PU PV PU⊥ −4 PU PV −4 PV PU⊥ (ii): The proof follows a similar approach as in (i). (iii), (iv): See [9, Proposition 6.5 (i) and (ii)]. (v): Using (3) yields RU,γRV,αRU⊥,β = RU,1RV,αRU⊥,1 S.Th. Alwadani / Eur. J. Pure Appl. Math, 18 (2) (2025), 6031 11 of 13 = ( 2 PU − Id )( 4α PV PU⊥ −2α PV −2 PU⊥ + Id ) = 8α PU PV PU⊥ −4α PU PV +2 PU −4α PV PU⊥ +2α PV +2 PU⊥ − Id = 8α PU PV PU⊥ −4α PU PV +2 ( PU +PU⊥ ) − Id−4α PV PU⊥ +2α PV = 8α PU PV PU⊥ −4α PU PV + Id−4α PV PU⊥ +2α PV = Id+2α PV +8α PU PV PU⊥ −4α PU PV −4α PV PU⊥ (vi): The proof adopts a similar method to that used in statement (v). (vii): Using [9, Lemma 6.4] and Lemma 1 (vi) gives Fix ( RU,γRV,αRU⊥,β ) = Fix ( RU,1RV,αRU⊥,1 ) = RU,1 ( Fix RV,αRU⊥,1RU,1 ) = RU,1 ( Fix ( RV,α(− Id) )) = RU,1 ( Fix RV⊥,α ) = RU,1 ( V⊥), as required. (viii): The proof follows a similar approach as in (vii). (ix): Using (9), (10), and (11) gives RU,γRV,αRU⊥,β = RU,γRV,1RU⊥,β = RU,γ ( 4β PV PU⊥ −2 PV −2β PU⊥ + Id ) = ( 2γ PU − Id )( 4β PV PU⊥ −2 PV −2β PU⊥ + Id ) = 8γβ PU PV PU⊥ −4γ PU PV −4β PV PU⊥ +2 ( γ PU +β PU⊥ +PV ) − Id, as required. (x): The proof follows a similar approach as in (ix). ■ Theorem 3. let X, Y, and Z be closed subspaces of H. Suppose that β, γ, α ∈]0, 1] and recall from (9) the modified reflector operator. Then Fix(RX,αRY,βRZ,γ) = Fix ( α PX +β PY +γ PZ +4αβγ PX PY PZ −2αβ PX PY − 2αγ PX PZ −2βγ PY PZ ) and Fix(RX,αRY,βRZ,γ) ⊆ X + Y + Z (12) Proof. Using (9) implies that RY,βRZ,γ = ( 2β PY − Id )( 2γ PZ − Id ) = 4βγ PY PZ −2β PY −2γ PZ + Id . Next, RX,α ( 4βγ PY PZ −2β PY −2γ PZ + Id ) = ( 2α PX − Id )( 4βγ PY PZ −2β PY −2γ PZ + Id ) S.Th. Alwadani / Eur. J. Pure Appl. Math, 18 (2) (2025), 6031 12 of 13 = 2α PX +2β PY +2γ PZ +8αβγ PX PY PZ −4αβ PX PY − 4αγ PX PZ −4βγ PY PZ − Id . Therefore, RX,αRY,βRZ,γ = 2α PX +2β PY +2γ PZ +8αβγ PX PY PZ −4αβ PX PY −4αγ PX PZ − 4βγ PY PZ − Id . Let x ∈ Fix ( RX,αRY,βRZ,γ ) . Then we obtain x = RX,αRY,βRZ,γ(x) ⇔ 2x = 2α PX(x) + 2β PY(x) + 2γ PZ(x) + 8αβγ PX PY PZ(x)− 4αβ PX PY(x) − 4αγ PX PZ(x)− 4βγ PY PZ(x) ⇔ x = α PX(x) + β PY(x) + γ PZ(x) + 4αβγ PX PY PZ(x)− 2αβ PX PY(x) − 2αγ PX PZ(x)− 2βγ PY PZ(x) As a result, Fix(RX,αRY,βRZ,γ) = Fix ( α PX +β PY +γ PZ +4αβγ PX PY PZ −2αβ PX PY − 2αγ PX PZ −2βγ PY PZ ) Consequently, Fix(RX,αRY,βRZ,γ) ⊆ X + Y + Z. ■ References [1] Oluwatayomi Rereloluwa Adegboye, Afi Kekeli Feda, Opeoluwa Seun Ojekemi, Ephraim Bonah Agyekum, Abdelazim G Hussien, and Salah Kamel. Chaotic oppo- sition learning with mirror reflection and worst individual disturbance grey wolf op- timizer for continuous global numerical optimization. Scientific Reports, 14(1):4660, 2024. [2] Heinz H Bauschke, Patrick L Combettes, Heinz H Bauschke, and Patrick L Com- bettes. Correction to: convex analysis and monotone operator theory in Hilbert spaces. Springer, 2017. [3] Andrzej Cegielski. 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