EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 1, Article Number 5503
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

More on the Order of Aragón Artacho–Campoy Algorithm
Operators With the Help of Douglas–Rachford Operators

Salihah Thabet Alwadani
Mathematics, Yanbu Industrial College, The Royal Comission for Jubail and Yanbu, Yanbu,
Saudi Arabia

Abstract. The Aragón Artacho–Campoy algorithm (AACA) is a new method for finding zeros
of sums of monotone operators. In this paper we complete the analysis of their algorithm by
defining their operator using Douglas Rachford operator and then study the effects of the order
of the two possible Aragón Artacho–Campoy operators.

2020 Mathematics Subject Classifications: 47H09, 47H05, 47A06, 90C25

Key Words and Phrases: Maximally monotone operator, Aragón Artacho–Campoy operators,
Affine subspace, Douglas-Rachford splitting operator, projection operator, resolvent, reflected re-
solvent

1. Introduction

Throughout, we assume that

X is a real Hilbert space with inner product ⟨·, ·⟩ : X × X → R, (1)

and induced norm ∥ · ∥ : X → R : x 7→
√
⟨x, x⟩. We also assume that A : X ⇒ X

and B : X ⇒ X are maximally monotone operators. For more details about maximally
monotone operators, we refer the reader to [3], [4], [9], [10], [11], [12], [14], [15], and the
references therein. In [3], Auslender and Teboulle provide essential tools used to study
monotone graphs. They focus on the behavior of a given subset of Rn at infinity. By
using real analysis and geometric concepts, they develop a mathematical treatment to
study the asymptotic behavior of sets. Moreover, the book by Bauschke and Combettes
[4] is one of the best sources to learn about non-linear analysis, namely, convex analy-
sis, monotone operators, and fixed point theory of operators. Additionally, [9] highlights
the importance of maximal monotone operators and describes the progress that has been
made in the field of monotone operators over the past decade. Furthermore, [10] pro-
vides a survey that discusses the developments in the theory of monotone operators. It

DOI: https://doi.org/10.29020/nybg.ejpam.v18i1.5503

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 (1) (2025), 5503 2 of 16

is well known that a prominent example of maximal monotone operators is the subd-
ifferential operator, which was investigated in section 5.1.6 of [11]. Moreover, Burachik
and Svaiter establish new connections between maximal monotone operators and convex
functions. They demonstrate that each maximal monotone operator is associated with a
family of convex functions. Their study focuses on this family, determining its extremal
elements using the concept of convex functions (see [12]). Following this, Patrick reviews
the properties of subdifferential operators as maximally monotone operators in [14], and
examines proximity operators as resolvents of these operators. Additionally, in [15], a
comprehensive treatment of monotone set-valued operators is presented, utilizing math-
ematical programming in detail. The resolvent and the reflected resolvent associated with
A are:

JA = (Id+A)−1 and RA = 2JA − Id, (2)

respectively. Suppose that

A and B are maximally monotone on X, w ∈ X, and γ ∈ ]0, 1[ . (3)

Fact 1. The resolvent averages between A, B and Nw are

Aγ : H ⇒ H : x 7→ A
(

γ−1(x − (1 − γ)w
))

+ γ−1(1 − γ
)(

x − w
)
, (4)

and
Bγ : H ⇒ H : x 7→ B

(
γ−1(x − (1 − γ)w

))
+ γ−1(1 − γ

)(
x − w

)
. (5)

Fact 2. Aγ and Bγ are maximally monotone and their resolvents are given by

JAγ
= γJA +

(
1 − γ

)
w and JBγ = γJB +

(
1 − γ

)
w, (6)

respectively. Moreover, reflected resolvents are

RAγ
= 2γJA + 2

(
1 − γ

)
w − Id, and RBγ = 2γJB + 2

(
1 − γ

)
w − Id, (7)

respectively. Then Aragón Artacho–Campoy operator [1] associated with the ordered
pair of operators

(
Aγ, Bγ

)
is

TAγ,Bγ
=

(
1 − λ

)
Id+λRBγ RAγ

. (8)

Fact 3. (Definition of the Douglas–Rachford splitting operator) The Douglas–Rachford split-
ting operator [19] associated with the ordered pair of operators

(
A, B

)
is

TA,B =
1
2
(

Id+RBRA
)
= Id−JA + JBRA. (9)

Through straightforward calculations, we can determine that

TB,A = Id+JARB − JB. (10)

In this paper, we explore the relationship between the Aragón Artacho–Campoy opera-
tors TAγ,Bγ

and TBγ,Aγ
. The key findings are summarized as follows:



S. Th. Alwadani / Eur. J. Pure Appl. Math, 18 (1) (2025), 5503 3 of 16

• Key properties of JAγ
and Aγ are presented in Proposition 1. These properties will

be valuable for our analysis.

• We provide formulas for the Aragón Artacho–Campoy operators utilizing the Dou-
glas–Rachford splitting operator (refer to Lemma 1). For additional details on the
Douglas–Rachford splitting algorithm, see [17], [5], [8], [13], [16], and [18]. [5], [8],
[13], and [18] help to understand more about the behaviour of DRS. Paper [5] stud-
ies the range of the DRS systematically. Under the assumption that the operators
are 3∗ monotone operators. While the second one helps to understand the behav-
ior of the shadow sequence when the given functions have disjoint domains. The
main result of this paper is proving the weak and value convergence of the shadow
sequence generated by the Douglas–Rachford algorithm. Paper [13] aims to solve
convex feasibility problems by using new algorithmic structures with DRS opera-
tors. Paper [18] gives a comprehensive survey about the developments of the DRS
methods. Additionally, [17] shows an amazing connection between the alternat-
ing direction multiplier method (ADMM) and Douglas Rachford Splitting method
(DRS) for convex problems. Finally, the paper [16] shows that the proximal point
algorithm encompasses the DRS method as a specific instance, which is employed
for locating a zero of the combined sum of two monotone operators.

• With the assumption A is affine relation, we prove that RAγ
Tn

Aγ,Bγ
= Tn

Bγ,Aγ
RAγ

(see
Theorem 1).

• We demonstrate the results by providing two examples (refer to Example 1 and
Proposition 2).

• We established that the equality does not hold when substituting Aγ with Bγ in the
previous result (see Proposition 2, (vii), (viii), and (ix)).

The notation employed in this paper is standard and closely aligns with that in [2], [1],
and [4].

2. New Results

All the results in this section are new, highlighting the main ones, which are the rela-
tionships between the Aragón Artacho–Campoy operators TAγ,Bγ

and TBγ,Aγ
. Key find-

ings include the important properties of JAγ
and Aγ outlined in Proposition 1, which

support our analysis. We provide formulas for the Aragón Artacho–Campoy operators
using the Douglas–Rachford splitting operator, as detailed in Lemma 1. Under the as-
sumption that A is an affine relation, we establish the equality RAγ

Tn
Aγ,Bγ

= Tn
Bγ,Aγ

RAγ

(see Theorem 1). Our findings are further illustrated with two examples (refer to Exam-
ple 1 and Proposition 2). Additionally, we demonstrate that replacing Aγ with Bγ in this
result leads to a failure of the equality (see Proposition 2, (vii), (viii), and (ix)).



S. Th. Alwadani / Eur. J. Pure Appl. Math, 18 (1) (2025), 5503 4 of 16

Proposition 1. Let γ ∈ ]0, 1[ and assume that A is an affine relation. The following
statements are true:

(i) JAγ
is affine.

(ii) Aγ is an affine relation.

Proof. (i): From [7, Lemma 2.3] or [6, Theorem 2.1(xix)], it follows that JA is affine. Utiliz-
ing (6), we can conclude that JAγ

is also affine. Thus, JAγ
is affine. (ii): From (i), we have

that JAγ
is affine if and only if (Id+Aγ)−1 is an affine relation, which in turn is equiva-

lent to (Id+Aγ) being an affine relation, and this is also equivalent to Aγ being an affine
relation. ■

Lemma 1. Let γ ∈ ]0, 1[ and λ ∈ ]0, 1]. We derive:

RBγ RAγ
= Id+2JBγ RAγ

− 2JAγ
(11)

= Id+2γJBRAγ
− 2γJA (12)

= TA,B + (1 − 2γ)JA − JBRA + 2γJBRAγ
. (13)

Additionally,

RAγ
RBγ = Id+2JAγ

RBγ − 2JBγ (14)

= Id+2γJARBγ − 2γJB (15)

= TB,A + (1 − 2γ)JB − JARB + 2γJARBγ . (16)

Moreover,

TAγ,Bγ
= Id+2λJBγ RAγ

− 2λJAγ
(17)

= Id+2λγJBRAγ
− 2λγJA (18)

= Id+2λγ
(

JBRAγ
− JA

)
(19)

= TA,B + (1 − 2λγ)JA − JBRA + 2λγJBRAγ
(20)

= TB,A + JB − JARB + 2λγJBRAγ
− 2λγJA. (21)

Furthermore,

TBγ,Aγ
= Id+2λJAγ

RBγ − 2λJBγ (22)

= Id+2λγJARBγ − 2λγJB (23)

= Id+2λγ
(

JARBγ − JB
)

(24)

= TB,A + (1 − 2λγ)JB − JARB + 2λγJARBγ . (25)

Proof. From (7), we can conclude:

RBγ RAγ
=

(
2JBγ − Id

)
RAγ



S. Th. Alwadani / Eur. J. Pure Appl. Math, 18 (1) (2025), 5503 5 of 16

= 2JBγ RAγ
− RAγ

= 2JBγ RAγ
−

(
2JAγ

− Id
)

= Id+2JBγ RAγ
− 2JAγ

,

This establishes (11). Additionally, from (6) and (11), we have:

RBγ RAγ
= Id+2

(
γJB + (1 − γ)w

)
RAγ

− 2
(
γJA + (1 − γ)w

)
= Id+2γJBRAγ

+ 2(1 − γ)w − 2γJA − 2(1 − γ)w

= Id+2γJBRAγ
− 2γJA,

This confirms (12). Furthermore, from Fact 3, we have:

RBγ RAγ
= TA,B + JA − JBRA + 2γJBRAγ

− 2γJA

= TA,B + (1 − 2γ)JA − JBRA + 2γJBRAγ
.

This confirms (13). The proof for RAγ
RBγ follows similarly to that of RBγ RAγ

. From (8),
we find:

TAγ,Bγ
= (1 − λ) Id+λRBγ RAγ

= (1 − λ) Id+λ
(

Id+2JBγ RAγ
− 2JAγ

)
(from(11))

= Id+2λJBγ RAγ
− 2λJAγ

,

This confirms (17). Utilizing (6) and (17), we derive:

TAγ,Bγ
= Id+2λ

(
γJB + (1 − γ)w

)
RAγ

− 2λ
(
γJA + (1 − γ)w

)
= Id+2λγJBRAγ

+ 2λ(1 − γ)w − 2λγJA − 2λ(1 − γ)w

= Id+2λγJBRAγ
− 2λγJA.

= Id+2λγ
(

JBRAγ
− JA

)
This confirms (18) and (19). Finally, from (18) and (9), we derive:

TAγ,Bγ
= TA,B + JA − JBRA + 2λγJBRAγ

− 2λγJA

= TA,B + (1 − 2λγ)JA − JBRA + 2λγJBRAγ
.

By merging (10) and (18), we obtain (21). The proof for TBγ,Aγ
follows a similar approach

to that of TAγ,Bγ
. ■

Example 1. Let w ∈ H, U be a closed linear subspace, γ ∈ ]0, 1[, and λ ∈ ]0, 1]. Assume
A = Id−v, where v ∈ U⊥, and B = Pa+U for some a ∈ H. From (9), we have

TA,B = Id−JA + JBRA,

and from (8), it follows that

TAγ,Bγ
= (1 − λ) Id+λ

(
RBγ RAγ

)
.

The following statements hold:



S. Th. Alwadani / Eur. J. Pure Appl. Math, 18 (1) (2025), 5503 6 of 16

(i) JA =
(
(Id+v)/2

)
and RA = v.

(ii) JAγ
= γ

(
(Id+v)/2

)
+ (1 − γ)w. Moreover,

RAγ
= γv − (1 − γ) Id+2(1 − γ)w.

(iii) JB =
(

Id− 1
2 PU

)
− PU⊥ a and RB =

(
Id−PU

)
− 2 PU⊥ a.

(iv) We have

JBγ = γ
((

Id−1
2

PU
)
− PU⊥ a

)
+ (1 − γ)w,

and
RBγ = (2γ − 1) Id−γ PU −2γ PU⊥ a + 2(1 − γ)w.

(v) TA,B =
(
(Id+v)/2

)
− PU⊥ a.

(vi) TB,A =
(
(Id+v)/2

)
.

(vii) JBRA = v − PU⊥ a.

(viii) JARB =
(
(Id+v)/2

)
− (PU /2)− PU⊥ a.

(ix) JBRAγ
= γv + (1 − γ)

((
1
2 PU − Id

)
−

(
PU −2 Id

)
w
)
− PU⊥ a.

(x) JARBγ = 1
2

(
(2γ − 1) Id−γ PU −2γ PU⊥ a + 2(1 − γ)w + v

)
.

(xi) Suppose k := λγ
[
(2γ − 1)v + 4(1 − γ)w − 2(1 − γ)PU w − 2 PU⊥ a

]
. Then

TAγ,Bγ
(x) =

(
1 − λγ(3 − 2γ)

)
x + λγ(1 − γ)PU x + k. (26)

(xii) Suppose l := λγ
[
2(1 − γ)PU⊥ a + v + 2(1 − γ)w

]
. Then

TBγ,Aγ
(x) =

(
1 − λγ(3 − 2γ)

)
x + λγ(1 − γ)PU x + l. (27)

Proof. (i): Let y ∈ H and define x = JAy. Then, we have y ∈ (Id+A)x if and only if y =
2x − v, which implies x = ((y + v)/2). This leads to JA =

(
(Id+v)/2

)
. Consequently,

we find that
RA = 2

(
(Id+v)/2

)
− Id ⇔ RB = v

by (2).
(ii): Combine (i) and (6) yields:

RAγ
(x) = 2γ

(
(x + v)/2

)
+ 2(1 − γ)w − x

= γ
(
x + v

)
+ 2

(
1 − γ

)
w − x



S. Th. Alwadani / Eur. J. Pure Appl. Math, 18 (1) (2025), 5503 7 of 16

= γv −
(
1 − γ

)
x + 2

(
1 − γ

)
w.

(iii): Let y ∈ H and define x = JBy. Our goal is to determine x. We have:

y ∈ (Id+Pa+U)x ⇔ y = x + a + PU(x − a)
⇔ y = x + (Id−PU)a + PU x
⇔ y = x + PU⊥ a + PU x.

Hence,
y = x + PU⊥ a + x∗, where x∗ = PU x. (28)

Applying PU to (28) results in:

PU y = PU x + PU PU⊥ a + x∗ ⇔ PU y = 2x∗ ⇔ x∗ =
1
2

PU y. (29)

Inserting (29) back into (28) results in:

y = x + PU⊥ a +
1
2

PU y ⇔ x =
(

Id−1
2

PU

)
y − PU⊥ a.

Therefore,

JB = (Id−1
2

PU)− PU⊥ a,

and

RB = 2
(

Id−1
2

PU
)
− 2 PU⊥ a − Id = (Id−PU)− 2 PU⊥ a,

by (2).
(iv): From (6) and (iii), it can be concluded that:

JBγ = γJB + 2(1 − γ)w

= γ
((

Id−1
2

PU

)
− PU⊥ a

)
+ (1 − γ)w.

According to (7), we obtain:

RBγ = 2γ
((

Id−1
2

PU

)
− PU⊥ a

)
+ 2(1 − γ)w − Id

= (2γ − 1) Id−γ PU −2γ PU⊥ a + 2(1 − γ)w.

(v): Utilizing (iii), (i), and (9) yields:

TA,B(x) = x −
( x + v

2

)
+

(
Id−1

2
PU −PU⊥ a

)
RAx

=
x
2
− v

2
+ RAx − 1

2
PU RAx − PU⊥ a



S. Th. Alwadani / Eur. J. Pure Appl. Math, 18 (1) (2025), 5503 8 of 16

=
( x + v

2

)
− PU⊥ a.

(vi): Based on (iii), (ii), and (9), we find:

TB,A(x) =
1
2

x +
1
2

RARBx

=
1
2

x +
1
2
(v)

((
x − PU x

)
− 2 PU⊥ a

)
=

1
2
(

x + v
)
.

(vii): By employing (iii) and (i), we derive:

JBRAx =
(

Id−1
2

PU −PU⊥ a
)

RAx

= RAx − 1
2

PU RAx − PU⊥ a

= v − PU⊥ a.

(viii): Applying (i) and (iii) results in:

JARBx =
( Id+v

2

)
RBx

=
1
2

RBx +
1
2

v

=
1
2

(
x − PU x − 2 PU⊥ a

)
+

1
2

v

=
( x + v

2

)
− 1

2
PU x − PU⊥ a.

(ix): Through the application of (ii) and (iii), we obtain:

JBRAγ
x =

(
Id−1

2
PU −PU⊥ a

)
RAγ

x

= RAγ
x − 1

2
PU RAγ

x − PU⊥ a

= γv − (1 − γ)x + 2(1 − γ)w − 1
2

PU

(
2(1 − γ)w − (1 − γ)x

)
− PU⊥ a

= γv + (1 − γ)
((1

2
PU − Id

)
x −

(
PU −2 Id

)
w
)
− PU⊥ a.

(x): Based on (i) and (iv), we derive:

JARBγ x =
( Id+v

2

)(
(2γ − 1)x − γ PU x − 2γ PU⊥ a + 2(1 − γ)w

)
=

1
2

(
(2γ − 1)x − γ PU x − 2γ PU⊥ a + 2(1 − γ)w + v

)
.



S. Th. Alwadani / Eur. J. Pure Appl. Math, 18 (1) (2025), 5503 9 of 16

(xi): Merging (vii), (ix), (v), and (20) results in:

TAγ,Bγ
x = TA,B + (1 − 2λγ)JA − JBRA + 2λγJBRAγ

=
( x + v

2

)
− PU⊥ a + (1 − 2λγ)

( x + v
2

)
− JBRA + 2λγJBRAγ

= (x + v)− PU⊥ a − λγ(x + v)− JBRA + 2λγJBRAγ

= (x + v)− λγ(x + v)− v + 2λγJBRAγ

= x − λγ(x + v) + 2λγJBRAγ

= x − λγ(x + v) + 2λγ
[
γv +

(1 − γ)

2
PU x − (1 − γ)(x + PU w − 2w)− PU⊥ a

]
=

(
1 − λγ(3 − 2γ)

)
x + λγ(1 − γ)PU x + λγ

(
(2γ − 1)v

+ 4(1 − γ)w − 2(1 − γ)PU w − 2 PU⊥ a
)

.

(xii): Using (iii), (vi), (viii), and (x), we derive:

TBγ,Aγ
x = TB,Ax + (1 − 2λγ)JBx − JARBx + 2λγJARBγ x

=
( x + v

2

)
+ (1 − 2λγ)JBx −

( x + v
2

)
+

1
2

PU x + PU⊥ a + 2λγJARBγ x

= (1 − 2λγ)
(

x − 1
2

PU x − PU⊥ a
)
+

1
2

PU x + PU⊥ a + 2λγJARBγ x

= (1 − 2λγ)x + 2λγ PU⊥ a + λγ PU x + 2λγJARBγ x

=
(

1 − λγ(3 − 2γ)
)

x + λγ(1 − γ)PU x + λγ
[
2(1 − γ)PU⊥ a + v + 2(1 − γ)w

]
,

which verifies (xii). ■

Lemma 2. Let A : H ⇒ H be a maximally monotone and γ ∈ ]0, 1[. If JA is affine, then:

JAγ
RAγ

= RAγ
JAγ

. (30)

Proof. Combine Proposition 1(i) and [7, Lemma 2.4 (i)]. ■

Lemma 3. Assuming that A is an affine relation, we can conclude:

RAγ
TAγ,Bγ

− TBγ,Aγ
RAγ

= 2
(

JAγ
TAγ,Bγ

− (1 − λ)JAγ
− λJAγ

RBγ RAγ

)
. (31)

= 2γ
(

JATA,B − (1 − λ)JA − λJARBγ RAγ

)
. (32)

Proof. From Proposition 1(ii), it follows that Aγ is an affine relation. Therefore, applying
(7) and (17), we derive:

RAγ
TAγ,Bγ

− TBγ,Aγ
RAγ

=
(
2JAγ

− Id
)
TAγ,Bγ

− TBγ,Aγ
RAγ

= 2JAγ
TAγ,Bγ

− TAγ,Bγ
− TBγ,Aγ

RAγ

= 2JAγ
TAγ,Bγ

− TAγ,Bγ
−

(
Id+2λJAγ

RBγ − 2λJBγ

)
RAγ



S. Th. Alwadani / Eur. J. Pure Appl. Math, 18 (1) (2025), 5503 10 of 16

= 2JAγ
TAγ,Bγ

− TAγ,Bγ
− RAγ

− 2λ
(

JAγ
RBγ RAγ

− JBγ RAγ

)
= 2JAγ

TAγ,Bγ
−

(
Id+2λJBγ RAγ

− 2λJAγ

)
− RAγ

− 2λ
(

JAγ
RBγ RAγ

− JBγ RAγ

)
= 2JAγ

TAγ,Bγ
− Id+2λJAγ

− RAγ
− 2λJAγ

RBγ RAγ

= 2JAγ
TAγ,Bγ

− 2(1 − λ)JAγ
− 2λJAγ

RBγ RAγ
,

this confirms (31). Subsequently, utilizing (6) and (31) gives

RAγ
TAγ,Bγ

− TBγ,Aγ
RAγ

= 2JAγ
TAγ,Bγ

− 2(1 − λ)JAγ
− 2λJAγ

RBγ RAγ

= 2
(
γJA + (1 − γ)w

)
TAγ,Bγ

− 2(1 − λ)
(
γJA + (1 − γ)w

)
− 2λ

(
γJA + (1 − γ)w

)
RBγ RAγ

= 2γJATAγ,Bγ
− 2γ(1 − λ)JA − 2λγJARBγ RAγ

.

■

Theorem 1. Let γ ∈ ]0, 1[ and λ ∈ ]0, 1], and suppose that A is an affine realtion. Then:

RAγ
Tn

Aγ,Bγ
= Tn

Bγ,Aγ
RAγ

. (33)

Proof. We will demonstrate by induction that RAγ
Tn

Aγ,Bγ
= Tn

Bγ,Aγ
RAγ

. Starting with n = 1,
we can use (32) to derive:

RAγ
TAγ,Bγ

− TBγ,Aγ
RAγ

= 2γJATAγ,Bγ
− 2γ(1 − λ)JA − 2λγJARBγ RAγ

= JA

(
2γ

(
TAγ,Bγ

− (1 − λ) Id
))

− 2λγJARBγ RAγ

= JA

(
2γ

(
(1 − λ) Id+λRBγ RAγ

)
− 2γ(1 − λ) Id

)
− 2λγJARBγ RAγ

= 2λγJARBγ RAγ
− 2λγJARBγ RAγ

= 0.

Hypothesis assumption: when n = k;

RAγ
Tk

Aγ,Bγ
− Tk

Bγ,Aγ
RAγ

= 0. (34)

For n = k + 1, and utilizing (34), we obtain:

RAγ
Tk+1

Aγ,Bγ
− Tk+1

Bγ,Aγ
RAγ

= RAγ
Tk

Aγ,Bγ
TAγ,Bγ

− Tk
Bγ,Aγ

TBγ,Aγ
RAγ

= RAγ
Tk

Aγ,Bγ
TAγ,Bγ

− Tk
Bγ,Aγ

RAγ
TAγ,Bγ

= RAγ
Tk

Aγ,Bγ
TAγ,Bγ

− RAγ
Tk

Aγ,Bγ
TAγ,Bγ

= 0.

Therefore, (33) has been verified. ■

Lemma 4. Suppose both A and B are affine relations. Then the following holds:



S. Th. Alwadani / Eur. J. Pure Appl. Math, 18 (1) (2025), 5503 11 of 16

(i) The operators TAγ,Bγ
and TBγ,Aγ

are affine.

(ii) The equation TAγ,Bγ
RBγ RAγ

= RBγ RAγ
TAγ,Bγ

is satisfied.

(iii) We have

λ−2(TAγ,Bγ
TBγ,Aγ

− TBγ,Aγ
TAγ,Bγ

)
= RBγ R2

Aγ
RBγ − RAγ

R2
Bγ

RAγ
.

(iv) The equality
TAγ,Bγ

TBγ,Aγ
= TBγ,Aγ

TAγ,Bγ

holds if and only if RBγ R2
Aγ

RBγ = RAγ
R2

Bγ
RAγ

.

(v) If R2
Aγ

= R2
Bγ

, then it follows that TAγ,Bγ
TBγ,Aγ

= TBγ,Aγ
TAγ,Bγ

.

Proof. (i): Clear. (ii): From (i), we conclude that:

TAγ,Bγ
RBγ RAγ

= TAγ,Bγ

(
λ−1TAγ,Bγ

− λ−1(1 − λ) Id
)

= λ−1T2
(Aγ,Bγ)

− λ−1(1 − λ)TAγ,Bγ

=
(
λ−1TAγ,Bγ

− λ−1(1 − λ) Id
)
TAγ,Bγ

= RBγ RAγ
TAγ,Bγ

.

(iii): Utilizing (8), we find that:

λ−2(TAγ,Bγ
TBγ,Aγ

)
= λ−2((1 − λ) Id+λRBγ RAγ

)(
(1 − λ) Id+λRAγ

RBγ

)
Hence,

λ−2(TAγ,Bγ
TBγ,Aγ

)
= λ−2((1 − λ)2 Id+λ(1 − λ)RAγ

RBγ + λ(1 − λ)RBγ RAγ

+ λ2RBγ R2
Aγ

RBγ

)
. (35)

Moreover,

λ−2(TBγ,Aγ
TAγ,Bγ

)
= λ−2((1 − λ) Id+λRAγ

RBγ

)(
(1 − λ) Id+λRBγ RAγ

)
Hence,

λ−2(TBγ,Aγ
TAγ,Bγ

)
= λ−2((1 − λ)2 Id+λ(1 − λ)RBγ RAγ

+ λ(1 − λ)RAγ
RBγ

+ λ2RAγ
R2

Bγ
RAγ

)
. (36)

Taking the difference of (35) and (36) yields:

λ−2(TAγ,Bγ
TBγ,Aγ

− TBγ,Aγ
TAγ,Bγ

)
= RBγ R2

Aγ
RBγ − RAγ

R2
Bγ

RAγ
.

(iv) and (v): They are derived from (iii). ■



S. Th. Alwadani / Eur. J. Pure Appl. Math, 18 (1) (2025), 5503 12 of 16

Proposition 2. Let U be a closed linear subspace, and define A = Id+v with v ∈ U⊥. Further-
more, let B = Pa+U , where a ∈ U⊥ and a ̸= v. The following statements are true:

(i) We have

TAγ,Bγ
(x) =

(
1 − λγ(3 − 2γ)

)
x + λγ(1 − γ)PU x + k,

where
k = λγ

(
(2γ − 1)v + 4(1 − γ)w − 2(1 − γ)PU w − 2a

)
.

(ii) We have

TBγ,Aγ
(x) =

(
1 − λγ(3 − 2γ)

)
x + λγ(1 − γ)PU x + l,

where

l = λγ
(
2(1 − γ)a + v + 2(1 − γ)w

)
.

(iii) We have

RAγ
TAγ,Bγ

(x) = TBγ,Aγ
RAγ

(x)

=
(
1 − γ

)((
λγ

(
3 − 2γ

)
− 1

)
x − λγ

(
1 − γ

)
PU x

)
+ h,

where

h = γ
[
λγ

(
2γ − 3

)
+ 1 + λ

]
v + 2

(
1 − γ

)[(
1 − 2λγ

(
1 − γ

))
w

+ λγ
(
1 − γ

)
PU w

]
+ 2λγ

(
1 − γ

)
a.

(iv) We have

RBγ TAγ,Bγ
=

(
1 − 2γ

)((
λγ(3 − 2γ)− 1

)
x − λγ(1 − γ)PU x

)
+ m,

where

m =
(
1 − 2γ

)
λγ

[(
1 − 2γ

)
v + 2

(
1 − γ

)
PU w − 4

(
1 − γ

)
w
]

+ 2
[(

1 − γ
)
w + γ

(
λ − 1 − 2λγ

)
a
]
.

(v) We have

TBγ,Aγ
RBγ =

(
1 − 2γ

)((
λγ(3 − 2γ)− 1

)
x − γ

(
1 + λ − λγ

(
5 − 3γ

))
PU x + s,

where

s = λγv + 2
(

λγ2 −
(
2λ + 1

)
γ + 1

)
w

− 2γ
(

λ
(
3γ + 4

)
+ 1

)
a + 2λγ

(
1 − γ

)2 PU w.



S. Th. Alwadani / Eur. J. Pure Appl. Math, 18 (1) (2025), 5503 13 of 16

(vi) RBγ TAγ,Bγ
̸= TBγ,Aγ

RBγ .

(vii) We have

RBγ TBγ,Aγ
=

(
2γ − 1

)(
1 − λγ

(
3 − 2γ

))
x + γ

(
λγ

(
5 − 3γ

)
− λ − 1

)
PU x + b,

where

b = −2γ
(

λγ
(
2γ − 3

)
+ λ + 1

)
a − 2

(
γ
(

λγ
(
2γ − 3

)
+ 1

)
− 1

)
w + λγ

(
2γ − 1

)
v

− 2λγ2(1 − γ
)

PU w.

(viii) We have

TAγ,Bγ
RBγ =

(
2γ − 1

)(
1 − λγ

(
3 − 2γ

))
x + γ

(
λγ

(
5 − 3γ

)
− λ − 1

)
PU x + c,

where

c = −2γ
(

λγ
(
2γ − 3

)
+ λ + 1

)
a − 2

(
γ
(

λγ
(
2γ − 3

)
+ λ + 1

)
− 1

)
w + λγ

(
2γ − 1

)
v

− 2λγ2(1 − γ
)

PU w.

(ix) RBγ TBγ,Aγ
̸= TAγ,Bγ

RBγ .

Proof. (i): This is derived from Example 1 (xi). (ii) : This is derived from Example 1 (xii).
(iii) : Utilizing (i), (ii), and Example 1(ii), we find that:

RAγ
TAγ,Bγ

(x) =
(

γv −
(
1 − γ

)
Id+2

(
1 − γ

)
w
)

TAγ,Bγ
(x)

= γv + 2
(
1 − γ

)
w −

(
1 − γ

)
TAγ,Bγ

(x)

= γv −
(
1 − γ

)
λγ

(
2γ − 1

)
v + 2

(
1 − γ

)
w − 4λγ

(
1 − γ

)2w

−
(
1 − γ

)((
1 − λγ

(
3 − 2γ

))
x + λγ

(
1 − γ

)
PU x + λγ

(
− 2

(
1 − γ

)
PU w − 2a

))
=

(
1 − γ

)((
λγ

(
3 − 2γ

)
− 1

)
x − λγ

(
1 − γ

)
PU x

)
+ γ

[
λγ

(
2γ − 3

)
+ 1 + λ

]
v + 2

(
1 − γ

)[(
1 − 2λγ

(
1 − γ

))
w

+ λγ
(
1 − γ

)
PU w

]
+ 2λγ

(
1 − γ

)
a.

Moreover,

TBγ,Aγ
RAγ

(x) =
(
1 − λγ

(
3 − 2γ

))
RAγ

(x) + λγ
(
1 − γ

)
PU RAγ

(x)

+ λγ
(
2
(
1 − γ

)
a + v + 2

(
1 − γ

)
w
)
.

=
(
1 − γ

)((
λγ

(
3 − 2γ

)
− 1

)
x − λγ

(
1 − γ

)
PU x

)
+ γ

[
λγ

(
2γ − 3

)
+ 1 + λ

]
v + 2

(
1 − γ

)[(
1 − 2λγ

(
1 − γ

))
w



S. Th. Alwadani / Eur. J. Pure Appl. Math, 18 (1) (2025), 5503 14 of 16

+ λγ
(
1 − γ

)
PU w

]
+ 2λγ

(
1 − γ

)
a.

Therefore,
RAγ

TAγ,Bγ
(x) = TBγ,Aγ

RAγ
(x).

and (iii) is verified.
(iv): Using Example 1(iv) and (i) gives

RBγ TAγ,Bγ
(x) =

((
2γ − 1

)
Id−γ PU −2γa + 2

(
1 − γ

)
w
)

TAγ,Bγ
(x)

= 2
(
1 − γ

)
w − 2γa −

(
1 − 2γ

)
TAγ,Bγ

(x)− γ PU TAγ,Bγ
(x)

= 2
(
1 − γ

)
w − 2γa −

(
1 − 2γ

)[(
1 − λγ

(
3 − 2γ

))
x + λγ

(
1 − γ

)
PU x + k

]
=

(
1 − 2γ

)[(
λγ

(
3 − 2γ

)
− 1

)
x − λγ

(
1 − γ

)
PU x

]
+ λγ

(
1 − 2γ

)[(
1 − 2γ

)
v + 2

(
1 − γ

)
PU w − 4

(
1 − γ

)
w
]

+ 2
[(

1 − γ
)
w + γ

(
λ − 1 − 2λγ

)
a
]
.

(v): Utilizing (ii) and Example 1(iv), we obtain

TBγ,Aγ
RBγ =

((
1 − λγ

(
3 − 2γ

)
Id

)
+ λγ

(
1 − γ

)
PU +l

)
RBγ(x)

=
(

1 − λγ
(
3 − 2γ

)
RBγ(x) + λγ

(
1 − γ

)
PU

(
RBγ(x)

)
+ l

(
RBγ(x)

)
=

(
1 − λγ

(
3 − 2γ

))(
2γ − 1

)
x − γ

(
1 + λ − λγ

(
5 − 3γ

))
PU x

+ 2λγ
(
1 − γ

)(
a + w

)
+

(
1 − λγ

(
3 − 2γ

))(
2
(
1 − γ

)
w − 2γa

)
+ 2λγ

(
1 − γ

)2 PU w + λγv

=
(
1 − 2γ

)((
λγ(3 − 2γ)− 1

)
x − γ

(
1 + λ − λγ

(
5 − 3γ

))
PU x

+ λγv + 2
(

λγ2 −
(
2λ + 1

)
γ + 1

)
w

− 2γ
(

λ
(
3γ + 4

)
+ 1

)
a + 2λγ

(
1 − γ

)2 PU w.

(vi): This is derived from (iv) and (v).
(vii): By using Example 1(iv) and (ii) we have

RBγ TBγ,Aγ
(x) =

((
2γ − 1

)
Id−γ PU −2γa + 2

(
1 − γ

)
w
)

TBγ,Aγ
(x)

=
(
2γ − 1

)
TBγ,Aγ

(x)− γ PU
(
TBγ,Aγ

(x)
)
− 2γa + 2

(
1 − γ

)
w

=
(
2γ − 1

)[(
1 − λγ

(
3 − 2γ

))
x + λγ

(
1 − γ

)
PU x + l

]
− γ PU

[(
1 − λγ

(
3 − 2γ

))
x + λγ

(
1 − γ

)
PU x + l

]
− 2γa + 2

(
1 − γ

)
w



S. Th. Alwadani / Eur. J. Pure Appl. Math, 18 (1) (2025), 5503 15 of 16

=
(
2γ − 1

)[(
1 − λγ

(
3 − 2γ

))
x + λγ

(
1 − γ

)
PU x

]
− γ

[(
1 − λγ

(
3 − 2γ

))
PU x + λγ

(
1 − γ

)
PU x

]
+

(
2γ − 1

)
l − γ PU l − 2γa + 2

(
1 − γ

)
w

=
(
2γ − 1

)(
1 − λγ

(
3 − 2γ

))
x + γ

(
λγ

(
5 − 3γ

)
− λ − 1

)
PU x

− 2γ
(

λγ
(
2γ − 3

)
+ λ + 1

)
a − 2

(
γ
(

λγ
(
2γ − 3

)
+ 1

)
− 1

)
w + λγ

(
2γ − 1

)
v

− 2λγ2(1 − γ
)

PU w.

(viii): Utilizing Example 1(iv) and (ii) yields

TAγ,Bγ
RBγ(x) =

(
1 − λγ

(
3 − 2γ

)
Id+λγ

(
1 − γ

)
PU +k

)
RBγ(x)

=
(

1 − λγ
(
3 − 2γ

))
RBγ(x) + λγ

(
1 − γ

)
PU RBγ(x) + k

=
(

1 − λγ
(
3 − 2γ

))((
2γ − 1

)
x − γ PU x

)
+

(
1 − λγ

(
3 − 2γ

))(
2
(
1 − γ

)
w − 2γa

)
+ λγ

(
1 − γ

)
PU

((
2γ − 1

)
x − γ PU x

)
+ λγ

(
1 − γ

)
PU

(
2
(
1 − γ

)
w − 2γa

)
+ k

=
(
2γ − 1

)(
1 − λγ

(
3 − 2γ

))
x + γ

(
λγ

(
5 − 3γ

)
− λ − 1

)
PU x

− 2γ
(

λγ
(
2γ − 3

)
+ λ + 1

)
a − 2

(
γ
(

λγ
(
2γ − 3

)
+ λ + 1

)
− 1

)
w

+ λγ
(
2γ − 1

)
v − 2λγ2(1 − γ

)
PU w.

(ix): It follows from (vii) and (viii). ■

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