EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 17, No. 4, 2024, 3642-3659
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Compositions of Resolvents: Fixed Points Sets and Set of
Cycles

Salihah Thabet Alwadani
Mathematics, Royal Commission Yanbu Colleges and Institutes, Yanbu, Saudi Arabia

Abstract. In this paper, we investigate the cycles and fixed point sets of compositions of resol-
vents using Attouch–Théra duality. We demonstrate that the cycles defined by the resolvent op-
erators can be formulated in Hilbert space as solutions to a fixed point equation. Furthermore,
we introduce the relationship between these cycles and the fixed point sets of the compositions of
resolvents.
2020 Mathematics Subject Classifications: 47H09, 47H05, 47A06, 90C25

Key Words and Phrases: Displacement mapping, Attouch–Théra duality, maximally monotone
operator, nonexpansive mapping, , Fixed point set, resolvent operator, set-valued inverse

1. Introduction

Throughout, we assume that

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

and induced norm ∥ · ∥ : X → R : x 7→
√
⟨x, x⟩. For more details about Hilbert space, we

refere the redear to [10] and [13]. An operator T : X → X is nonexpansive if it is Lipschitz
continuous with constant 1, i.e.,(

∀x ∈ X
)(
∀y ∈ X

)
∥Tx − Ty∥ ≤ ∥x − y∥. (1)

Nonexpansive operators play a major role in optimization because the set of fixed points
Fix R := {x ∈ X | x = Rx} usually represents solutions to inclusion problems and op-
timization tasks. For more details about nonexpansive operators and the fixed point set,
we refer the reader to [1]-[6], [7]-[8], [11], [16], [17], and [2, Chapters 3 and 6]. Moreover,
T : D → X is firmly nonexpansive if(

∀x ∈ D
)(
∀y ∈ D

)
∥Tx − Ty∥2 + ∥(Id−T)x − (Id−T)y∥2 ≤ ∥x − y∥2. (2)

DOI: https://doi.org/10.29020/nybg.ejpam.v17i4.5505

Email address: salihah.s.alwadani@gmail.com (S. T. Alwadani)

https://www.ejpam.com 3642 Copyright: © 2024 The Author(s). (CC BY-NC 4.0)



S.Th.Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3642-3659 3643

Firmly nonexpansive operators are also central due to their favorable convergence prop-
erties for iterates and their correspondence with maximal monotone operators. Recall
that a set-valued operator A : X ⇒ X with graph gra A is monotone if(

∀
(
x, u

)
∈ gra A

)(
∀
(
y, v

)
∈ gra A

)
⟨x − y, u − v⟩ ≥ 0.

Furthermore, A is maximally monotone if there does not exist a monotone operator B :
X ⇒ X such that gra B properly contains gra A, i.e., for every (x, u) ∈ X × X,(

x, u
)
∈ gra A ⇔

(
∀
(
y, v

)
∈ gra A

)
⟨x − y, u − v⟩ ≥ 0.

It is well known that monotone and maximally monotone operators play central roles in
various areas of modern nonlinear analysis. See [10], [14], [15]-[22], and [20] for back-
ground material. Let A : X ⇒ X be a maximally monotone operator and denote the
associated resolvent by

JA := (Id+A)−1. (3)

In [21], Minty observed that JA is a firmly nonexpansive operator from X to X. For more
information about the relationship between firmly nonexpansive mappings and maxi-
mally monotone operators, see [12]. The Hilber product space,

X =
{

x = (xi)i∈I

∣∣∣ (∀i ∈ I) xi ∈ X
}

,

where m ∈ {2, 3, . . . } and i = {1, 2, . . . , m}. Let

Ai : X ⇒ X be maximally monotone operators, (4)

with resolvents JA1
, JA2

, . . . , JAm which we also write more simply as J1, J2, . . . , Jm. Set

A = A1 × A2 × · · · × Am. (5)

Then
JA : X → X :

(
x1, x2, . . . , xm

)
7→

(
J1x1, J2x2, . . . , Jmxm

)
. (6)

Define the circular right-shift operator

R :
(

x1, x2, . . . , xm
)
7→

(
xm, x1, x2, . . . , xm−1

)
. (7)

Define the fixed point sets of the cyclic compositions of resolvants:

F1 := Fix(J1Jm . . . J2), (8)
F2 := Fix(J2J1Jm . . . J3), (9)

... (10)
Fm := Fix(JmJm−1 . . . J1). (11)

The prospects of applying the compositions of resolvents in practical applications
are broad and significant, particularly in fields such as optimization, control theory, and
mathematical analysis. Here are some key areas where these applications are emerging:



S.Th.Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3642-3659 3644

1. In the area of Optimization and Control: Compositions of resolvents are crucial
in optimization problems, particularly in convex optimization and monotone in-
clusion problems. They provide a framework for developing algorithms that can
efficiently find solutions to complex optimization tasks. For instance, the resolvent
composition is a monotonicity-preserving operation that can be linked to proximal
compositions, which are essential in convex analysis. This relationship allows for
the relaxation of monotone inclusion problems, making it easier to solve them in
practical scenarios. See [18].

2. In the area of Equilibrium Problems: The compositions of resolvents encapsulate
known concepts and introduce new operations that are pertinent to equilibrium
problems. This is particularly relevant in economic models and game theory, where
finding equilibria is essential. The properties established in the study of resolvent
compositions can lead to new insights and methods for analyzing these problems
[18].

3. Applications in Fluid Dynamics: In fluid dynamics, the mean resolvent operator
has been used to analyze the stability of flows and predict the behavior of turbulent
systems. The application of resolvent compositions in this context can enhance
our understanding of flow dynamics and improve control strategies for various
engineering applications [19].

4. In the area of Signal Processing and Data Analysis: Resolvent compositions can
also be applied in signal processing, particularly in filtering and data reconstruc-
tion techniques. By leveraging the mathematical properties of resolvents, engineers
can develop more effective algorithms for noise reduction and signal enhancement,
which are critical in communications and multimedia applications [18].

The compositions of resolvents hold significant promise for practical applications across
various fields, including optimization, control theory, fluid dynamics, and signal process-
ing. As research continues to explore these compositions, we can expect to see innovative
solutions and methodologies that leverage their mathematical properties to address com-
plex real-world problems.

Definition 1. [2, Definition 5.1] Let z1 ∈ F1. Set z2 := J2z1, z3 := J3z2, · · · , zm−1 :=
Jm−1zm−2, and zm := Jmzm−1. The truple z =

(
z1, z2, . . . , zm

)
∈ X is called a cycle.

The notation used in the paper is standard and follows largely, e.g., [2] and [10].

2. Aim and outline of this paper

Our main results can be summarized as follows:

• Theorem 1 and Theorem 2 sketche the relationship between the cycles and the fixed
point sets of the composition of resolvants.

• The cycles that are defined by the resolvant operators can be formulated in Hilbert
product space as a solution to a fixed point equation (see Lemma 2 and Lemma 4).



S.Th.Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3642-3659 3645

• We study the set of classical cycles that are defined by using resolvant operators
and the set of classical gap vectors see Theorem 4.

• If one of the fixed point sets of composition of resolvents is not empty, then the indi-
viduals fixed point sets are equal and their intersection is not empty (see Lemma 5).

• In Section 5, we use Attouch–Théra duality to study the cycles and the fixed point
sets of compositions of resolvents operators.

Approach of this paper is novel as it utilizes Attouch–Théra duality to conduct an in-
depth investigation of the cycles and fixed point sets related to compositions of resol-
vents. This duality offers a powerful framework for uncovering the intricate structures
and dynamics inherent in these mathematical constructs. In summary, applying Attouch–
Théra duality to analyze cycles and fixed point sets in resolvent compositions represents
a significant advancement in the field. More information about Attouch–Théra duality is
in the next section.

3. Attouch–Théra duality

Let A and B be two maximally monotone operators on X. The primal problem associ-
ated with

(
A, B

)
is to

find x ∈ X such that 0 ∈ Ax + Bx. (12)

The set of primal solutions associated with
(

A, B
)

are the solutions to the corresponding
sum problem (12) are defined as

psol(A, B) := zer(A, B) = (A, B)−1(0) =
{

x ∈ X
∣∣∣ 0 ∈ (A + B)x

}
. (13)

Now define B> := (− Id) ◦ B ◦ (− Id) and B−> := (B−1)> = (B>)−1. This allows us to
define the dual pair of (A, B):

(A, B)∗ := (A−1, B−>). (14)

Then the dual problem associated with (A, B) is defined to be the primal problem associ-
ated with the dual pair (A−1, B−>):

find y ∈ X such that 0 ∈ A−1y + B−>y = A−1y − B−1(−y). (15)

The set of of dual solutions associated with
(

A, B
)

are the solutions to the corresponding
sum problem (15):

dsol(A, B) := psol(A, B)∗ = zer (A−1 + B−>) =
{

y ∈ X
∣∣∣ 0 ∈ (A−1 + B−>)y

}
. (16)

Because (A−1)−1 = A, (A>)> = A, and (A−>)−> = A, we have

(A, B)∗∗ = (A, B). (17)



S.Th.Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3642-3659 3646

Lemma 1. Let A and B be maximally monotone on X. Let x and y in X. Then the follow-
ing holds:

(i) If psol(A, B) = {x}, then

dsol(A, B) = Ax ∩ (−Bx)

and
dsol(A, B) = Ax ∩ B>(−x).

(ii) If dsol(A, B) = {y}, then

psol(A, B) = (A−1y) ∩ B−1(−y)

and
psol(A, B) = (A−1y) ∩ (−B−>(y)).

(iii) If psol(A, B) = {x} and Ax is a singelton, then dsol(A, B) = Ax.
(iv) If psol(A, B) = {x} and Bx and B>(−x) are singelton, then

dsol(A, B) = −Bx

and
dsol(A, B) = B>(−x).

(v) If dsol(A, B) = {y} and A−1y is a singelton, then psol(A, B) = A−1y.
(vi) If dsol(A, B) = {y} and B−1(−y) and (−B−>(y)) are a singelton, then

psol(A, B) = B−1(−y)

and
psol(A, B) = (−B−>(y)).

Proof. (i): From (13), it follows that

x ∈ psol(A, B) ⇔ (A + B)−1(0) ̸= ∅
⇔ ∅ ̸= Ax ∩ (−Bx)

⇔ ∅ ̸= Ax ∩
(
(− Id) ◦ B ◦ (− Id)(−x)

)
⇔ ∅ ̸= Ax ∩ B>(−x)
⇔ ∅ ̸= Ax ∩ B>(−x) ⊆ dsol(A, B)
⇔ ∅ ̸= Ax ∩ (−Bx) ⊆ dsol(A, B).

Since psol(A, B) = {x} and by using (13), it follows that Ax ∩ (−Bx) = dsol(A, B) and
Ax ∩ B>(−x) = dsol(A, B).
(ii): From (16), it follows that

y ∈ dsol(A, B) ⇔ (A−1 + B−>)−1(0) ̸= ∅



S.Th.Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3642-3659 3647

⇔ A−1(y) ∩ (−B−>(y)) ̸= ∅

⇔ A−1(y) ∩ B−1(−y) ̸= ∅

⇔ ∅ ̸= A−1(y) ∩ B−1(−y) ⊆ psol(A, B).

Since dsol(A, B) = {y} and by using (16), it follows that

A−1(y) ∩ B−1(−y) = psol(A, B).

(iii): From (i), we have Ax ∩ (−Bx) = dsol(A, B), and Ax ∩ B>(−x) = dsol(A, B). Since
Ax is a singelton then we obtain

dsol(A, B) = Ax ∩ (−Bx) = Ax,

and
dsol(A, B) = Ax ∩ B>(−x) = Ax.

(iv): From (i), we have Ax ∩ (−Bx) = dsol(A, B) and Ax ∩ B>(−x) = dsol(A, B). Since
Bx and B>(−x) are singelton, it follows that

dsol(A, B) = Ax ∩ (−Bx) = −Bx

and
dsol(A, B) = Ax ∩ B>(−x) = B>(−x).

(v): From (ii), we have (A−1y) ∩ B−1(−y) = psol(A, B) and (A−1y) ∩ (−B−>(y)) =
dsol(A, B). Since A−1y is a singelton, we obtain

psol(A, B) = (A−1y) ∩ B−1(−y) = A−1y

and
psol(A, B) = (A−1y) ∩ (−B−>(y)) = A−1y.

(vi): From (ii), we have (A−1y) ∩ B−1(−y) = psol(A, B) and (A−1y) ∩ (−B−>(y)) =
dsol(A, B). Since B−1(−y) and (−B−>(y)) are singelton, we obtain

psol(A, B) = (A−1y) ∩ B−1(−y) = B−1(−y)

and
psol(A, B) = (A−1y) ∩ (−B−>(y)) = −B−>(y).

■
For more information about the Attouch–Théra duality, we refer the reader to [5].



S.Th.Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3642-3659 3648

4. Correspondence of Properties and Results

This section presents some of our key findings regarding the cycles and fixed point
sets of resolvent compositions, beginning with an exploration of their interrelationship,
as demonstrated in Theorem 1 and Theorem 2.

Theorem 1. The following are equivalent
(i) Cycle exists.

(ii) For all 1 ≤ i ≤ m, the fixed point sets of cyclic compositions of resolvants Fi ̸= ∅.

Proof. ” (i)⇒(ii)”: Let z =
(
z1, z2, . . . , zm−1, zm

)
be a cycle. Then by Definition 1, we have

z1 = J1zm, z2 = J2z1, z3 = J3z2, · · · , zm−1 = Jm−1zm−2, and zm = Jmzm−1. This gives that

z1 = J1Jm . . . J3J2z1

z2 = J2J1 . . . J4J3z2

...
zi = JiJi−1 . . . J1Jm . . . Ji+1zi

...
zm = JmJm−1 . . . J2J1zm.

Therefore, z1 ∈ F1, z2 ∈ F2, . . . , zi ∈ Fi, . . . , zm ∈ Fm by (8)-(11). This implies that F1 ̸= ∅,
F2 ̸= ∅, . . . , Fi ̸= ∅, . . . , Fm ̸= ∅.
” (ii)⇒(i)”: Let Fm ̸= ∅ and zm ∈ Fm. Then

zm ∈ Fix
(
JmJm−1 . . . J2J1

)
zm,

by (11). Therefore,
zm = JmJm−1 . . . J2J1zm.

Next, Applying J1 gives
J1zm = J1

(
JmJm−1 . . . J2

)(
J1zm

)
,

which is equivalent to J1zm ∈ Fix
(
J1JmJm−1 . . . J3J2

)
⇔ J1zm ∈ F1 ̸= ∅. Additionaly, let

z2 ∈ F2 such taht z2 = J2z1. Keep doing this gives,

zm−2 = Jm−2Jm−3 . . . J1JmJm−1zm−2.

Then,
Jm−1zm−2 = Jm−1

(
Jm−2Jm−3 . . . J1Jm

)(
Jm−1zm−2

)
,

and Jm−1zm−2 ∈ Fix
(
Jm−1Jm−2Jm−3 . . . J1Jm

)
, which is equivalent to Jm−1Jm−2 ∈ Fm−1 ̸= ∅.

Moreover, let zm−1 ∈ Fm−1 such that zm−1 = Jm−1zm−2. It follows that

zm−1 = Jm−1Jm−2Jm−3 . . . J1Jmzm−1,



S.Th.Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3642-3659 3649

and
Jmzm−1 = Jm

(
Jm−1Jm−2Jm−3 . . . J1

)
Jmzm−1.

Therefore,
Jmzm−1 ∈ Fix

(
JmJm−1Jm−2Jm−3 . . . J1

)
.

This is equivalent to Jmzm−1 ∈ Fix Fm ̸= ∅. All these together give
(
z1, z2, . . . , zm−1, zm

)
∈

X satisfying that(
z1, z2, . . . , zm−1, zm

)
=

(
J1zm, J2z1, . . . , Jm−1zm−2, Jmzm−1

)
.

■

Lemma 2. Let z ∈ X is a cycle. Then

z = JA
(
Rz

)
. (18)

Moreover, solving (18) is equivalent to solve

0 ∈ A(z) +
(
Id − R

)
(z). (19)

Proof. Given that z =
(
z1, z2, . . . , zm

)
∈ X is a cycle. Then Definition 1 gives that z1 = J1zm,

z2 = J2z1, z3 = J3z2, · · · , zm−1 = Jm−1zm−2, and zm = Jmzm−1. Hence,

z =
(
z1, z2, . . . , zm−1, zm

)
=

(
J1zm, J2z1, . . . , Jm−1zm−2, Jmzm−1

)
=

(
J1, J2, . . . , Jm−1, Jm

)(
zm, z1, . . . , zm−2, zm−1

)
= JA

(
R
(
z1, z2, . . . , zm−1, zm

))
= JA

(
Rz

)
.

Note that z = JA(Rz) ⇔ z =
(
Id + A

)−1
(Rz) by (3). Therefore, we obtain

Rz ∈ z + A(z) ⇔ 0 ∈ A(z) +
(
Id − R

)
(z).

■
Define the set of all cycles by

Z := Fix(JAR). (20)

Define
Fi :=

{
z ∈ X | z = Ji . . . J1Jm . . . Ji+1z

}
. (21)

Moreover,
Qi : X → X : z 7→ zi. (22)

The relationship between the fixed point set of composition of m resolvants Fi’s and the
set of all cycles Z are given in the following theorem.



S.Th.Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3642-3659 3650

Theorem 2. For every 1 ≤ i, j ≤ m, the following hold:
(i) Fi are closed and convex. Moreover,

Fm =
(
JmJm−1 . . . J3J2

)(
F1
)
=

(
JmJm−1 . . . J3

)(
F2
)
= · · · = JmJm−1

(
Fm−2

)
= Jm

(
Fm−1

)
. (23)

Fm−1 =
(
Jm−1 . . . J3J2J1

)(
Fm

)
=

(
Jm−1 . . . J3J2

)(
F1
)
= · · · = Jm−1

(
Fm−2

)
. (24)

... (25)

F2 =
(
J2J1Jm . . . J4

)(
F3
)
=

(
J2J1Jm . . . J5

)(
F4
)
= · · · = J2J1

(
Fm

)
= J2

(
F1
)
. (26)

F1 =
(
J1Jm . . . J4J3

)(
F2
)
=

(
J1Jm . . . J4

)(
F3
)
= · · · =

(
J1Jm

)(
Fm−1

)
= J1

(
Fm

)
. (27)

(ii) ∩m
i=1 Fix Ji ⊆ ∩m

i=1Fi. If Fi = ∅, then ∩m
i=1 Fix Ji = ∅.

(iii) For 1 ≤ i ≤ m − 1, Ji+1
(

Fi
)
= Fi+1 and J1

(
Fm

)
= F1. This implies that

JAR
(

F1 × F2 × · · · × Fm
)
= F1 × F2 × · · · × Fm. (28)

(iv) Fi ̸= ∅ if and only if Fj ̸= ∅ if and only if Z = ∅.
(v) Z is closed and convex, and Z ⊆ F1 × F2 × · · · × Fm.

(vi) The mapping Qi|Z : Z → Fi is bijective and Qi
(
Z
)
= Fi.

Proof. (i): Since each Ji is firmly nonexpansive, it follows that Ji is nonexpansive. There-
fore, by [16, Lemma 2.1.12 (ii)], the composition

Ji . . . J1Jm . . . Ji+1

is also nonexpansive. As a result, Fi is closed and convex by [16, Proposition 2.1.11].
Let x ∈ Fm ⇔ x ∈ Fix

(
JmJm−1 . . . J3J2J1

)
⇔ x = JmJm−1 . . . J3J2J1x. Then,

J1x = J1
(
JmJm−1 . . . J3J2J1

)
x =

(
J1JmJm−1 . . . J3J2

)(
J1x

)
.

Therefore, J1x ∈ Fix
(
J1JmJm−1 . . . J3J2

)
⇔ J1x ∈ F1. It follows that

J1
(

Fm
)
⊆ F1. (29)

Moreover,(
J2J1

)(
Fm

)
=

(
J2J1

)(
Fix

(
JmJm−1 . . . J2J1

))
⊆ J2

(
Fix

(
J1JmJm−1 . . . J3J2

))
(30)

⊆ Fix
(
J2J1JmJm−1 . . . J4J3

)
= F2, (31)

hence (
J3J2J1

)(
Fm

)
=

(
J3J2J1

)(
Fix

(
JmJm−1 . . . J2J1

))
⊆

(
J3J2

)(
Fix

(
J1JmJm−1 . . . J3J2

))
(32)

⊆ J3
(

Fix
(
J2J1JmJm−1 . . . J4J3

))
(33)

⊆ Fix
(
J3J2J1JmJm−1 . . . J5J4

)
= F3, (34)

until finally

Fm = Fix
(
JmJm−1 . . . J2J1 =

(
JmJm−1 . . . J2J1

)(
Fix

(
JmJm−1 . . . J2J1

))
(35)



S.Th.Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3642-3659 3651

⊆
(
JmJm−1 . . . J2

)(
Fix

(
J1Jm . . . J3J2

))
(36)

... (37)

⊆ Jm
(

Fix
(
Jm−1Jm−2 . . . J2J1Jm

))
= Jm

(
Fm−1

)
(38)

⊆ Fix
(
JmJm−1 . . . J2J1

)
= Fm. (39)

Hence, equality holds throughout (35) to (39), and we are done. The same approach will
verify (24)-(27).
(ii): It is well known that ∩m

i=1 Fix Ji ⊆ F1, ∩m
i=1 Fix Ji ⊆ F2, · · · , ∩m

i=1 Fix Ji ⊆ Fm. Hence

m⋂
i=1

Fix Ji ⊆
m⋂

i=1

Fi.

This also implies that ∩m
i=1 Fix Ji = ∅ if Fi = ∅. (iii): From (i), we have

J1
(

Fm
)
= F1, J2

(
F1
)
= F2, . . . , Jm

(
Fm−1

)
= Fm. (40)

Using (6), (7) and (40), we obtain

JAR
(

F1 × F2 × · · · × Fm−1 × Fm
)
= JA

(
Fm × F1 × F2 × · · · × Fm−1

)
=

(
J1, J2, · · · , Jm

)(
Fm × F1 × F2 × · · · × Fm−1

)
= F1 × F2 × · · · × Fm−1 × Fm.

(iv): It is clear from the definitions of Fi, Fj and Z.
(v): Since JAR is nonexpansive and Z = Fix JAR, it follows that Z is closed and convex by
[16, Proposition 2.1.11]. Moreover, let z =

(
z1, z2, . . . , zm

)
∈ Fix JAR ⇔ z = JARz. This

implies

z1 = J1Jm . . . J2z1,
...

zi = JiJi−1 . . . J1Jm . . . Ji+1zi,
...

zm = JmJm−1 . . . J1zm.

Hence, z =
(
z1, z2, . . . , zm

)
∈ F1 × F2 × · · · × Fm−1 × Fm. Since this is true for all z ∈ Z, it

follows that
Z ⊆ F1 × F2 × · · · × Fm−1 × Fm.

(vi): It is clear from (i) that Qi : Z → Fi is surjective. To show Qi is injective, suppose
z =

(
z1, z2, . . . , zm

)
, z̃ =

(
z̃1, z̃2, . . . , z̃m

)
∈ Z and Qi

(
z
)
= Qi

(
z̃
)
. This implies that zi = z̃i.

Because z and z̃ are cycles, it follows that

zi+1 = Ji+1zi = Ji+1z̃i = z̃i+1 (41)



S.Th.Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3642-3659 3652

... (42)
zm = Jmzm−1 = Jm z̃m−1 = z̃m (43)
z1 = J1zm = J1z̃m = z̃1 (44)
z2 = J2z1 = J2z̃1 = z̃2 (45)

... (46)
zi−1 = Ji−1zi−2 = Ji−1z̃i−2 = z̃i−1. (47)

From (41)-(47), we have
z = z̃.

■

Lemma 3. Let ∩m
i=1 Fix Ji ̸= ∅ := D. Then the following holds:

(i) For all i such that 1 ≤ i ≤ m, it holds that Fi = D.
(ii) Z = {

(
z, z, · · · , z

)
| z ∈ D} = Dm ∩ ∆.

Proof. (i): Since Ji is firmly nonexpansive for every 1 ≤ i ≤ m, then by [10, Corollary 4.51],
we have (

∀(1 ≤ i ≤ m)
)
, Fix

(
JiJi−1 · · · J1Jm · · · Ji+1

)
= D.

(ii): Let z =
(
z1, z2, · · · , zm

)
∈ Z. Then,

z1 = J1JmJm−1 · · · J2z1 ⇔ z1 ∈ F1 = D
z2 = J2J1Jm · · · J3z2 ⇔ z2 ∈ F2 = D
...
zm = JmJm−1Jm−2 · · · J1zm ⇔ zm ∈ Fm = D.

Therefore,

z =
(
z1, z2, · · · , zm

)
∈ F1 × F2 × · · · × Fm = D × U × D × · · · × D = Dm

and
z =

(
z1, z2, · · · , zm

)
=

(
z, z, · · · , z

)
∈ D.

Therefore,
z ∈ Dm ∩ D.

■

Remark 1. When m = 2, we have J1
(

F2
)
= F1 and J2

(
F1
)
= F2.

Proof. Let z ∈ F1. This implies that z = J1J2z and J2z = J2J1(J2z). Therefore,

J2
(

F1
)
⊆ F2. (48)



S.Th.Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3642-3659 3653

Let z̃ ∈ F2. It follows that z̃ = J2J1z̃ and J1z̃ = J1J2(J1z̃). Thus,

J1
(

F2
)
⊆ F1 (49)

Now, applying J1 and J2 to (48) and (49), respectively, we obtain

F1 ⊆ J1
(

F2
)

and F2 ⊆ J2
(

F1
)
.

Hence, F1 = J1
(

F2
)

and F2 = J2
(

F1
)
. ■

Lemma 4. Recall from (20) that Z = Fix JAR. Then it follows that

Z = Fix JAR = Fix J 1
2 A

( Id + R
2

)
.

Proof. Let x ∈ Fix
(
JAR

)
. Then

x = JARx ⇔ Rx ∈ x + Ax ⇔ 0 ∈
(
x − Rx

)
+ A(x)

⇔ 0 ∈
(
x − Rx

)
2

+
A(x)

2

⇔ 0 ∈ x −
( Id + R

2

)
x +

A(x)
2

adding and subtracting
x
2

⇔
( Id + R

2

)
x ∈

(
Id +

1
2

A
)(

x
)

⇔ x = J 1
2 A

( Id + R
2

)(
x
)

⇔ x ∈ Fix J 1
2 A

( Id + R
2

)
.

■

Lemma 5. Suppose that Fix Ji ̸= ∅ for each 1 ≤ i ≤ m. Then the following are equivalent
:

(i) ∩m
i=1 Fix Ji ̸= ∅.

(ii) F1 = F2 = · · · = Fm ̸= ∅.

Proof. (i): Let ∩m
i=1 Fix Ji ̸= ∅ ⇒ F1 ̸= ∅, F2 ̸= ∅, · · · , Fm ̸= ∅ and from Lemma 3(i), it

follows that
F1 = F2 = · · · = Fm = ∩m

i=1 Fix Ji.

(ii): Let F1 = F2 = · · · = Fm ̸= ∅. Applying Theorem 2 (iv) gives Z ̸= ∅. ■

5. Consequences of Attouch-Théra duality

Recall (5) that
A = A1 × A2 × · · · × Am.



S.Th.Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3642-3659 3654

From now on, suppose that

A is maximally monotone on X, (50)

and
C := zer A is not empty. (51)

Recall (20), which states that
Z := Fix(JAR).

Proposition 1. The following holds:
(i) Z = psol

(
Id − R

)
.

(ii) A + Id − R is maximally monotone.
(iii) Z is closed and convex.

Proof. (i): Combine Lemma 2 and (13). (ii): Note that Id − R is linear, full domain,
and maximally monotone by [2, Theorem 7.1]. Moreover, A is maximally monotone by
assumption. Therefore, the sum is maximally monotone by [10, Corollary 25.5 (i)] (iii): It
follows directly from (i) and Theorem 2 (v). ■

Theorem 3. Recall from Lemma 2, the primal (Attouch-Théra) problem:

0 ∈ A(z) + (Id − R)(z),

for the pair (A, Id − R). The Attouch-Théra dual problem is

0 ∈ A−1(y) + (Id − R)−1(y) (52)

or

0 ∈
(
A−1 + ND⊥

)
(y) +

(1
2

Id + T
)
(y). (53)

Moreover,

dsol(A, Id − R) = zer
(

A−1 + ND⊥ +
1
2

Id + T
)

. (54)

Proof. The dual pair of
(
A,

(
Id − R

))
is(

A,
(
Id − R

))∗
=

(
A−1,

(
Id − R

)−>).

Because of the linearity of R, it follows that(
Id − R

)−>
=

(
− Id

)
◦
(
Id − R

)−1 ◦
(
−Id

)
=

(
Id − R

)−1.

Hence, Attouch-Théra dual problem simplifies to

0 ∈ A−1(y
)
+

(
Id − R

)−1(y
)
.



S.Th.Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3642-3659 3655

From [3, Theorem 2.8 (i)], we obtain

0 ∈ A−1(y
)
+

(
Id − R

)−1(y
)
⇔ 0 ∈ A−1(y

)
+

(
ND⊥ +

1
2

Id + T
)(

y
)

⇔ 0 ∈
(
A−1 + ND⊥

)
(y) +

(1
2

Id + T
)
(y),

which verifies (53). Next, applying (16), (52), and (53) yields

dsol(A, Id − R) = zer
(

A−1 +
(
Id − R

)−1
)

= zer
(

A−1 + ND⊥ +
1
2

Id + T
)

.

■

Proposition 2. The solution set of (52) is at most a sigleton and possibly empty.

Proof. [3, Theorem 2.8 (i)] gives
(
Id − R

)−1
= ND⊥ + 1

2 Id + T. Id − R is (1/2)-cocoercive

because R is nonexpansive by [10, Proposition 4.11]. Hence,
(
Id − R

)−1
= ND⊥ + 1

2 Id +
T is (1/2)-strongly monotone by [2, Lemma 7.8(iv)]. Then

A−1 +
(
Id − R

)−1
=

1
2

Id +
(
ND⊥ + T + A−1)

is strongly monotone. Hence, it follows that

zer
(
A−1 +

(
Id − R

)−1)
=

(
A−1 +

(
Id − R

)−1
)−1

(0)

is at most a singleton by [10, Proposition 23.35]. ■

Theorem 4. Let psol(A, Id − R) = Z and recall (54), which states that

dsol(A, Id − R) = zer
(

A−1 + ND⊥ +
1
2

Id + T
)

.

Then

dsol
(
Id − R

)
=

(
R − Id

)
Z =


{

J2(A−1+ND⊥+T)(0)
}

, i f Z ̸= ∅

∅, i f Z = ∅.
(55)

Moreover, if y∗ := J2(A−1+ND⊥+T)(0) exists, then the following holds:

(i) y∗ ∈ D⊥.
(ii) y∗ is the only vector that makes A−1y ∩−(ND⊥y + 1

2 y + Ty) non-empty.
(iii) Z = A−1y∗ ∩ (− 1

2 y∗ − Ty∗ − D).



S.Th.Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3642-3659 3656

Proof. By using (16), we have y ∈ dsol
(
A, Id − R

)
. This implies that

0 ∈ A−1(y) + (Id − R)−1(y) (∀z ∈ Z)

⇔ z ∈ A−1(y) and − z ∈ (Id − R)−1(y) (∀z ∈ Z)
⇔ y ∈ A(z) and y = (Id − R)(−z) (∀z ∈ Z).

Hence, for all z ∈ Z, it follows that

y = Rz − z

and

dsol
(
Id − R

)
= ∪z∈Z{Rz − z | z ∈ Z}
=

(
R − Id

)
Z.

Additionally, if Z ̸= ∅, then using [3, Theorem 2.4] and (3) gives

0 ∈ A−1(y) + (Id − R)−1(y)

⇔ 0 ∈ A−1(y) +
(1

2
Id + T + ND⊥

)
(y)

⇔ 0 ∈
(

Id + 2
(

A−1 + T + ND⊥

))
(y)

⇔ y =
(

Id + 2
(

A−1 + T + ND⊥

))−1
(0)

⇔ y = J2(A−1+T+ND⊥ )(0).

However, if Z = ∅, then using [9, Proposition 2.4 (v)]

∅ = dsol
(
A, Id − R

)
= dsol

(
Id − R

)
.

(i): By [3, Theorem 2.7], we have dom(Id − R)−1 = D⊥. This implies that

y ∈ D⊥.

(ii): Combine Proposition 2 and [9, Proposition 2.4].
(iii): Combine (i), (ii), and [2, Proposition 9.3 (i)] where N−1

C is replaced by A−1. ■

Lemma 6. Denote by y∗ =
(
y1, y2, · · · , ym

)
the unique solution of (53). Then the follow-

ing holds:
(i) The mapping J1 : Fm → F1 is bijective on Fm and it is given by J1

(
z
)
= z − y1.

Moreover, for 1 ≤ i ≤ m − 1 the mapping Ji+1 : Fi → Fi+1 is bijective and is given
by Ji+1

(
z
)
= z − yi+1.

(ii) The fixed point sets F1 = Fm − y1 and Fi+1 = Fi − yi+1.



REFERENCES 3657

Proof. (i): Let z and z̃ be in Fm satisfying that J1z = J1z̃. Our goal is to show that J1 is
injective on Fm. Then, we have

z = JmJm−1 · · · JiJi−1 · · · J1z

and
z̃ = JmJm−1 · · · JiJi−1 · · · J1z̃.

Since J1z = J1z̃, then we obtain z = z̃. Thus, J1 is an injective mapping on Fm. Moreover,
Remark 1 shows that J1 is a surjective mapping on Fm. Therefore, J1 is a bijective mapping
on Fm. For every z ∈ Fm, we have

z = JmJm−1 · · · JiJi−1 · · · J1z. (56)

Set z1 = J1z, z2 = J2z1, · · · , zm−1 = Jm−1zm−2, zm = Jmzm−1. Therefore, using (56), we
have z = Jmzm−1 and z =

(
z1, z2, · · · , zm−1, z

)
. Therefore, Theorem 4 gives(

y1, y2, · · · , ym
)
=

(
z, z1, z2, · · · , zm−1

)
−

(
z1, z2, · · · , zm−1, zm

)
and therefore, y1 = z − z1 ⇒ z1 = z − y1 ⇒ J1z = z − y1. The proof of Ji is the same as J1.
(ii): It follows from (i) that for every z ∈ Fm, we obtain J1z = z − y1. Then by Theo-
rem 2(iii) we have F1 = Fm − y1. The proof for Fi+1 = Fi − yi+1 is the same as F1. ■

Acknowledgements

The author expresses gratitude to the reviewers for their insightful comments and
constructive feedback, which greatly contributed to enhancing the quality of the work.

6. Clarification

Please note that a preprint has been published on arXiv and is referenced in [4]. There
is no conflict of interest, and no data were used to support this study.

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