EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 17, No. 4, 2024, 3660-3676
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Additional Studies on Displacement Mapping with
Restrictions

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

Abstract. The theory of monotone operators is fundamental in modern optimization and various
areas of nonlinear analysis. Key classes of monotone operators include matrices with a positive
semidefinite symmetric component and subdifferential operators. In this paper, we extend our
investigation to displacement mappings. We derive formulas for set-valued and Moore-Penrose
inverses. Additionally, we conduct a thorough examination of the operators (one-half times the
identity plus T) and its inverse, providing a formula for the inverse of the operator. Our results are
illustrated through an analysis of reflected and projection operators onto closed linear subspaces.

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

Key Words and Phrases: Displacement mapping, maximally monotone operator, nonexpansive
mapping, , Moore-Penrose inverse set-valued inverse, inverse, Yosida approximation

1. Introduction

It is well known that one of important classes of monotone operators are Displace-
ment mappings of nonexpansive mappings. There are many key examples that have
proven how these mappings are highly useful in optimization problems. For example, in
2016 Heinz H. Bauschke, Warren Hare, and Walaa Moursi used displacement mappings
in analyzing the range of the Douglas–Rachford operator to derive valuable duality re-
sults, see [5]. Additionally, the asymptotic regularity results for nonexpansive mappings
were generalized in [8] to the broader context of displacement mappings. Overall, the
displacement mapping framework has emerged as a powerful tool for analyzing the be-
havior of nonexpansive mappings, with a range of important applications in optimiza-
tion and related areas. Throughout, we assume that

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

DOI: https://doi.org/10.29020/nybg.ejpam.v17i4.5504

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

https://www.ejpam.com 3660 Copyright: © 2024 The Author(s). (CC BY-NC 4.0)



S.T. Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3660-3676 3661

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

B : X ⇒ X are maximally monotone operators. The resolvent and the reflected resolvent
associated with A are

JA = (Id+A)−1 and RA = 2JA − Id, (2)

respectively. 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∥. (3)

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. (4)

Fact 1. [4, Definition 4.10] Let D be a nonempty subset of X, let T : D → X, and let β ∈ R++,
where R++ is the set of strictly positive real numbers ]0,+∞[. Then T is β-cocoercive ( or β-
inverse strongly monotone) if βT is firmly nonexpansive, i.e.,(

∀x ∈ D
)(
∀y ∈ D

)
⟨x − y, Tx − Ty⟩ ≥ β∥Tx − Ty∥2.

In optimization, we have seen the importance the displacement mappings of nonexpan-
sive mappings:

Id−R (5)

because of the nice properities that have such as monotonicity which plays a central role
in modern optimization (see [4, 11, 18, 20–23] for more details). A comprehensive anal-
ysis of the displacement mappings of nonexpansive mappings from the point of view of
monotone operator theory under the condition of isometry of finite order of R are given
in [2, Lemma] and [1, Section 3]. We refer the reder to [18, Exercise 12.16], and [4, Example
20.29], [7]. More information is in [10, 15, 17, 19].

Throughout this paper, we assume that

R : X → X is linear and nonexpansive, with D := Fix R = ker
(

Id−R
)
. (6)

In this paper, we study the displacement mapping using the assumption in (6). Our results can
be summarized as follows

• Proposition 1, Lemma 1, and Remark 1 collect some useful properities of the dis-
pdisplacment mapping and its inverse, which will be useful in our study.

• Lemma 2 provides a formula and gives nice properties of the operator T.

• We derive a formula for the inverse of the displacment mapping (see Theorem 2 (i)).
A formula for the Moore-Penrose inverse of the displacement mapping is given in
Theorem 2(ii).



S.T. Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3660-3676 3662

• Theorem 3 gives a comprehensive study of the the operators
(
1/2

)
Id+T and its

inverse. Additionaly, we derive a formula of
((

1/2
)

Id+T
)−1

and prove that is
equal to the resolvant of the operator 2T.

• We illustrates the reults by giving four examples. The first two examples are re-
lated to the projection operator to a closed linear subspace (see Example 2 and Ex-
ample 3), while the other two are related to the reflected operator to closed linear
subspace (see Example 4 and Example 5).

2. Results

Important properties of the displacement mapping (Id−R) and its inverse are given
in the next proposition.

Proposition 1. Let R be nonexpansive operator, then the following holds:
(i) 1

2 (Id−R) is firmly nonexpansive.
(ii) Id−R is nonexpansive.

(iii) Id−R and (Id−R)−1 are maximally monotone.
(iv) Id−R is 1

2 -cocoercive.
(v) (Id−R)−1 is strongly monotone*with constant 1

2 .
(vi) Id−R is 3∗ monotone.

(vii) (Id−R)−1 is 3∗ monotone
(viii) Id−R is paramonotone.

(ix)
(

Id−R
)−1 − 1

2 Id is maximally monotone.

Proof. (i): We have

R is nonexpansive ⇔ −R = 2
((

Id−R
)
/2
)

is nonexpansive

⇔
(

Id−R
)
/2 is firmly nonexpansive,

by [4, Proposition 4.4]. (ii): It follows from (i) and [4, Proposition 4.2] . (iii): See [4,
Example 25.20(v)] or [2, Theorem 7.1]. (iv): Combine (i) and Fact 1. (v): Take (x, u) ∈
gra(Id−R)−1 and (y, v) ∈ gra(Id−R)−1. Then u ∈ (Id−R)−1x ⇒ x = u − Ru and
v ∈ (Id−R)−1y ⇒ y = v − Rv.

⟨u − v, x − y⟩ ≥ 1
2
∥x − y∥2

⇔ ⟨u − v, (u − Ru)− (v − Rv)⟩ ≥ 1
2
∥(u − Ru)− (v − Rv)∥2,

∗An operator A : X ⇒ X is strongly monotone with constant β ∈ R++ if A − β Id is montone, i.e.,(
∀(x, u) ∈ gra A

)(
∀(y, v) ∈ gra A

)
⟨x − y, u − v⟩ ≥ β∥x − y∥2.



S.T. Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3660-3676 3663

which deduce from (iv) and Footnote * that (Id−R)−1 is strongly monotone with con-
stant (1/2). (vi) and (vii): It follows from (iv) that Id−R is bounded by (1/2) and
its monotone by (iii). Hence, Id−R and (Id−R)−1 are 3∗ monotone by [4, Proposi-
tion 25.16(i) & (iv)].
(viii): See [4, Example 22.9]. (ix): By (iv) and [4, Example 22.7], (Id−R)−1 is (1/2)-
strongly monotone, i.e., B := (Id−R)−1 − 1

2 Id is still monotone. If B was not maximally
monotone, then neither would be B + 1

2 Id = (Id−R)−1 which would contradict (iii).
■

Lemma 1. Set D := ker
(

Id−R
)
= Fix R. Then the following holds:

(i) D is a closed linear subspace.
(ii) Fix R∗ = D.

(iii) ran
(

Id−R
)
= ran

(
Id−R∗) = D⊥.

Proof. (i): Let x, y ∈ D such that x − Rx = 0 and y − Ry = 0. Let α, β ∈ R. Then(
Id−R

)(
αx + βy

)
=
(

Id−R
)
(αx) +

(
Id−R

)
(βy)

= α(x − Rx) + β(y − Ry)
= 0 + 0 = 0.

Therefore, αx + βy ∈ D and hence D is a linear subspace. To show that D is closed, let
(xn) be a sequence in D such that (xn) converges to x. Then

lim
n→∞

(Id−R)(x − xn) = lim
n→∞

(Id−R)x − lim
n→∞

(Id−R)xn

= (Id−R)x − (Id−R)x = 0.

Therefore, x ∈ D and hence D is closed.
(ii) and (iii): It follows from Proposition 1(iii) & (iv) that Id−R is monotone and bounded.
Hence, Fix R∗ = Fix R = D and ran

(
Id−R

)
= ran

(
Id−R∗) = D⊥ by [4, Proposi-

tion 20.17]. ■

Remark 1. Suppose that X = ℓ2(N) and that

R : X → X : (xn)n∈N 7→
(
((1 − εn)xn)

)
n∈N

, (7)

where (εn)n∈N lies in ]0, 1[ with εn → 0. Then the following holds:
(i) Id−R : (xn)n∈N 7→

(
εnxn

)
n∈N

is a compact operator.
(ii) D = Fix R = {0}.

(iii) ran (Id−R) is not closed.
(iv) ran R is a closed subspace.

Proof. (i) and (ii): See [12, PropositionII.4.6]. (iii): It follows from [16, Proposition 3.4.6]
that ran (Id−R) is closed if and only if ran (Id−R) is finite-dimensional. On the other
hand, X = D⊥ = ran (Id−R), i.e., the range of Id−R is dense in the infinite-dimensional
space X. Altogether,

ran
(

Id−R
)

is not closed.

(iv): See [16, Lemma 3.4.20]. ■



S.T. Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3660-3676 3664

Lemma 2. Suppose that ran (Id−R) is closed; equivalently,

ran (Id−R) = D⊥.

Set

T := PD⊥(Id−R)−1 PD⊥ −
1
2

PD⊥ . (8)

Then,
(i) ran (Id−R)∗ = D⊥.

(ii) T is a linear and continuous.
(iii) T is monotone.
(iv) T is maximally monotone.
(v) ran T ⊆ D⊥, where D = ker(Id−R).

(vi) PD⊥ T = T PD⊥ = T.

Proof. (i): By using the closeness of ran (Id−R) and ...

ran (Id−R)∗ = ran (Id−R∗)

= ran (Id−R)

= D⊥.

(ii): This is clear because T is defined using PD⊥ , which is a linear and continuous oper-
ator. (iii): See [4, Example 20.12]. (iv): Combine (ii), (iii) and [4, Corollary 20.28]. (v): It
follows directly from (8). (vi): Since ran T ⊆ D⊥ by using (v), we obtain

PD⊥ T = T.

Moreover, both T and PD⊥ commute and so

T PD⊥ = PD⊥ T = T.

■

Remark 2. It is well known that ran (Id−R) is closed if and only if there exists α > 0 such that(
∀y ∈ (ker(Id−R))⊥ = D⊥

)
∥y − Ry∥ ≥ α∥y∥; (9)

Proof. See [13, Theorem 8.18]. ■

Proposition 2. Suppose that (9) holds, then the operator

PD⊥(Id−R)−1 : D⊥ → D⊥, (10)

(i) is a linear selection of T−1.
(ii) is continuous and its norm is bounded above by 1/α.



S.T. Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3660-3676 3665

Proof. (i): It follows from (8) and Lemma 2. (ii): Clear from (9). ■

Theorem 1. Suppose that ran (Id−R) is closed. Set

A := (Id−R)−1 − 1
2

Id, (11)

and defined
QA : dom A → X : y 7→ PAy y. (12)

Set
B := Pdom A QA Pdom A . (13)

Then the following holds;
(i) dom A = D⊥ and is closed.

(ii) A is linear relation.
(iii) A is maximally monotone.
(iv) we have

(∀y ∈ dom A) QAy = PD⊥(Id−R)−1y − 1
2

PD⊥ y.

(v) B is maximally monotone, linear and continuous.
(vi) A = ND⊥ + B.

(vii) B = T.
(viii) B|dom A is a selection of A|dom A.

Proof. (i): From (11) dom A = ran (Id−R) = D⊥, which is closed by the assumption.
(ii): It is clear that A is a linear relation, i.e., gra A is a linear subspace, that A0 = D, and
by (i) the dom A = D⊥ is closed.
(iii): It follows directly from Proposition 1(ix).
(iv): By [9, Proposition 6.2], we have

(
∀y ∈ dom A

)
QAy = P(A0)⊥(Ay) ∈ Ay. Hence,

(
∀y ∈ D⊥) QAy = PD⊥(Ay) = PD⊥

(
(Id−R)−1y − 1

2
y
)

= PD⊥(Id−R)−1y − 1
2

PD⊥ y.

(v): See [9, Example 6.4(i)]. (vi): Combining (i) and [9, Example 6.4(iii)] gives

A = Ndom A + B = ND⊥ + B.

(vii): Using (13), (iv) and (i) gives

B = Pdom A QA Pdom A

= PD⊥

(
PD⊥(Id−R)−1 − 1

2
PD⊥

)
PD⊥

= PD⊥(Id−R)−1 PD⊥ −
1
2

PD⊥



S.T. Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3660-3676 3666

= T (from (8)).

(viii): Using (i) gives

A|dom A = (ND⊥ + B)|D⊥ (from (vi))
= (ND⊥ + T)|D⊥ (from (vii))

=
(

ND⊥ + PD⊥(Id−R)−1 PD⊥ −
1
2

PD⊥

)∣∣∣∣
D⊥

(from (8))

≡ D + D⊥ (because ND⊥ |D⊥ ≡ D),

and

B|dom A = T|D⊥ (from (i) and (vii))

=
(

PD⊥(Id−R)−1 PD⊥ −
1
2

PD⊥

)∣∣∣∣
D⊥

(from (8))

= D⊥.

Hence, B|dom A is a selection of A|dom A. ■
In the next theorem we derive formulas for the inverse and Moore-Penrose inverse of

the operator (Id−R).

Theorem 2. Recall from (8) and (11) that

T := PD⊥(Id−R)−1 PD⊥ −
1
2

PD⊥ ,

and
A := (Id−R)−1 − 1

2
Id,

respectively.Then the following holds;
(i) The set-valued inverse of Id−R is

(Id−R)−1 =
1
2

Id+T + ND⊥ . (14)

(ii) The Moore-Penrose inverse of Id−R is

(Id−R)† = T +
1
2

PD⊥ . (15)

Proof. (i): Combining Theorem 1(vi) & (vii) and (11) gives

(Id−R)−1 − 1
2

Id = A

= ND⊥ + T,



S.T. Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3660-3676 3667

Hence,

(Id−R)−1 = ND⊥ + T +
1
2

Id .

(ii): By using [6, Proposition 2.1] and we obtain

(Id−R)† = P(Id−R)∗ ◦(Id−R)−1 ◦ Pran (Id−R)

= PD⊥ ◦(Id−R)−1 ◦ PD⊥ (from Lemma 2(i))

= PD⊥ ◦
(1

2
Id+T + ND⊥

)
◦ PD⊥ (from (i))

= PD⊥ ◦
(1

2
PD⊥ +T PD⊥ +D

)
(Because ND⊥ |D⊥ ≡ D)

=
1
2

PD⊥ +PD⊥ T PD⊥ +0

=
1
2

PD⊥ +T (from Lemma 2(vi)),

which verified (15). ■

Proposition 3 (uniqueness of T). Let T◦ : X → X be such that

(Id−R)−1 =
1
2

Id+T◦ + ND⊥ , (16)

and
PD⊥ T◦ PD⊥ = T◦. (17)

Then T◦ = T.

Proof. By using (8), we have

T = PD⊥(Id−R)−1 PD⊥ −
1
2

PD⊥

= PD⊥

(1
2

Id+T◦ + ND⊥

)
PD⊥ −

1
2

PD⊥ (from (16))

= PD⊥ T◦ PD⊥

= T◦ (from (17)),

as claimed. ■

Theorem 3. Recall from (8) that

T = PD⊥(Id−R)−1 PD⊥ −
1
2

PD⊥ .

Then the following holds;
(i) (1/2) Id+T is 1

2 -strongly monotone.
(ii)

(
(1/2) Id+T

)−1
= 2J2T.



S.T. Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3660-3676 3668

(iii) 2T + Id = 2 PD⊥(Id−R)−1 PD⊥ +PD.
(iv) J2T = PD + 1

2 (Id−R)PD⊥ .
(v) 2J2T = (Id−R)PD⊥ +2 PD.

(vi) (Id−R)PD⊥ +2 PD = Id−R + 2 PD.
(vii) We have (1

2
Id+T

)−1
= 2J2T = (Id−R)PD⊥ +2 PD = Id−R + 2 PD . (18)

(viii)
(

1
2 Id+T

)−1
∣∣∣∣

D⊥
= Id−R.

Proof. (i): Showing that 1
2 Id+T is (1/2)-strongly monotone ⇔ 1

2 Id+T − 1
2 Id = T

is montone, which is verified by Lemma 2(iii). (ii): From Lemma 2(ii) & (iv) and [14,
Lemma 2], we have (1

2
Id+T

)−1
=
(1

2
(Id+2T)

)−1

= 2
(

Id+2T)−1

= 2J2T.

(iii): By using (8) and Lemma 2(ii), we obtain

2T = 2
(

PD⊥(Id−R)−1 PD⊥ −
1
2

PD⊥

)
= 2 PD⊥(Id−R)−1 PD⊥ −PD⊥ ,

hence

2T + Id = 2 PD⊥(Id−R)−1 PD⊥ −PD⊥ + Id

= 2 PD⊥(Id−R)−1 PD⊥ +PD .

(iv): From (iii), we obtain 2T + Id = 2 PD⊥(Id−R)−1 PD⊥ +PD. Put differently,

2T + Id : D ⊕ D⊥ → D ⊕ D⊥ : d ⊕ d⊥ 7→ d + 2 PD⊥(Id−R)−1d⊥.

For two vectors d⊥, e⊥ in D⊥, we have the equivalences,

e⊥ = 2 PD⊥(Id−R)−1d⊥ ⇔ d⊥ =
(

2 PD⊥(Id−R)−1
)−1

e⊥,

and therefore,

d⊥ =
(

2 PD⊥(Id−R)−1
)−1

e⊥

=
(

2
(

PD⊥(Id−R)−1))−1
e⊥



S.T. Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3660-3676 3669

=
1
2

(
PD⊥(Id−R)−1

)−1
e⊥

=
1
2
(

Id−R
)

P−1
D⊥ e⊥

=
1
2
(

Id−R
)
e⊥.

Hence,

(2T + Id)−1 : D ⊕ D⊥ → D ⊕ D⊥ : d ⊕ d⊥ 7→ d +
1
2
(Id−R)d⊥;

equivalently,

J2T = (2T + Id)−1 : z 7→ PD z +
1
2
(Id−R)PD⊥ z.

(v): It follows directly from (iv). (vi): Because ker(Id−R) = D, we have (Id−R)PD ≡ 0.
Therefore,

(Id−R)PD⊥ +2 PD = Id−R + 2 PD .

(vii): Combine (ii), (v), and (vi). (viii): From (v), we obtain(1
2

Id+T
)−1

∣∣∣∣
D⊥

=
(

Id−R + 2 PD
)∣∣

D⊥ = Id−R.

■

Proposition 4. Let m ∈ {2, 3, . . . } and assume that Rm = Id, i.e., R is an isometry of finite
rank m. Assume that X = Rm and recall from [2, Lemma] that

PD =
1
m

m−1

∑
k=0

Rk and PD⊥ = Id− 1
m

m−1

∑
k=0

Rk, (19)

where D = Fix R. Then

1
2

PD⊥
(

R + R∗)PD⊥ =
1
m

(
− Id−

m−2

∑
k=2

Rk +
max{1, m − 2}

2
(

R + Rm−1)). (20)

Proof. Noted that R is an isometry ⇒ R∗R = RR∗ = Id, so R−1 = R∗. But also R has rank
m, hence Rm−1 = R−1 = R∗. By using these facts, we obtain

PD⊥(R + R∗)PD⊥ = PD⊥
(

R + R−1)PD⊥

= PD⊥
(

R + R−1)( Id− 1
m

m−1

∑
k=0

Rk
)

(from (19))

= PD⊥

((
R + R−1)− 1

m

m−1

∑
k=0

(
R + R−1)Rk

)
= PD⊥

((
R + R−1)− 1

m

m−1

∑
k=0

(
Rk+1 + Rk−1)).



S.T. Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3660-3676 3670

Since R has rank m, the following holds:

m−1

∑
k=0

Rk+1 =
m−1

∑
k=0

Rk−1 =
m−1

∑
k=0

Rk. (21)

Moreover,

Rl
m−1

∑
k=0

Rk =
m−1

∑
k=0

Rl+k =
m−1

∑
k=0

Rk (22)

Thus, (
R + R−1)( 1

m

m−1

∑
k=0

Rk
)
=

1
m

m−1

∑
k=0

(
Rk+1 + Rk−1) = 2

m

m−1

∑
k=0

Rk. (23)

Therefore,

PD⊥
(

R + R∗)PD⊥ = PD⊥

((
R + R−1)− 1

m

m−1

∑
k=0

(
Rk+1 + Rk−1))

= PD⊥

((
R + R−1)− 2

m

m−1

∑
k=0

Rk

)

=

(
Id− 1

m

m−1

∑
k=0

Rk

)((
R + R−1)− 2

m

m−1

∑
k=0

Rk

)

=

((
R + R−1)− 2

m

m−1

∑
k=0

Rk

)
−
(

1
m

m−1

∑
k=0

Rk

)((
R + R−1)− 2

m

m−1

∑
k=0

Rk

)

=

((
R + R−1)− 2

m

m−1

∑
k=0

Rk

)
−
(

2
m

m−1

∑
k=0

Rk − 2
m2

m−1

∑
l=0

Rl
m−1

∑
k=0

Rk

)

=

((
R + R−1)− 2

m

m−1

∑
k=0

Rk

)
−
(

2
m

m−1

∑
k=0

Rk − 2
m2

m−1

∑
l=0

m−1

∑
k=0

Rk

)

=

((
R + R−1)− 2

m

m−1

∑
k=0

Rk

)
−
(

2
m

m−1

∑
k=0

Rk − 2m
m2

m−1

∑
k=0

Rk

)

=

((
R + R−1)− 2

m

m−1

∑
k=0

Rk

)
−
(

2
m

m−1

∑
k=0

Rk − 2
m

m−1

∑
k=0

Rk

)

=

((
R + R−1)− 2

m

m−1

∑
k=0

Rk

)
.

First: assume that m > 2. Therefore, max{1, m − 2} = m − 2. Then

PD⊥
(

R + R∗)PD⊥ =
(

R + R−1)− 2
m

m−1

∑
k=0

Rk



S.T. Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3660-3676 3671

=
2
m

(
m
2
(

R + R−1)− m−1

∑
k=0

Rk

)

=
2
m

((
m
2
− 1

)(
R + R−1)− Id−

m−2

∑
k=2

Rk

)

=
2
m

(
− Id+

m − 2
2

(
R + Rm−1)− m−2

∑
k=2

Rk

)
,

which prove (20) when m > 2.
Next, assume that m = 2. Then max{1, m − 1} = 1 and R−1 = R2−1 = R. Therefore,

PD⊥
(

R + R∗)PD⊥ =
(

R + R−1)− 2
m

m−1

∑
k=0

Rk

= 2R − 2
2
(

Id+R
)

= 2R − Id−R
= R − Id .

On the other hand,

2
m

(
− Id+

max{1, m − 2}
2

(
R + Rm−1)− m−2

∑
k=2

Rk

)
=

2
2

(
− Id+

1
2
(

R + R
)
−

0

∑
k=2

Rk

)

= − Id+
1
2
(2R)− 0

= − Id+R,

so equality holds when m = 2. ■

3. Examples

Example 1 (isometry of finite rank). Let m ∈ {2, 3, . . . } and assume that

Rm = Id . (24)

Then the results in Section 2 were derived already in [1]. Moreover, the work there based on
exploiting (24) yielded to (19) and

T =
1

2m

m−1

∑
k=1

(
m − 2k

)
Rk = −T∗, (25)

which is always skew right-shift operator, T is symmetric only when m = 2.



S.T. Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3660-3676 3672

Example 2. Let U be a closed subspace of X and suppose that

R = PU . (26)

Then
(i) D = U.

(ii) Id−R = PU⊥ .
(iii) ran (Id−R) = D⊥ is closed.
(iv)

(
Id−R

)−1
= Id+NU .

(v) T = 1
2 PU⊥ = T∗.

(vi) T is always symmetric, but skew only when U = X.

Proof. (i): D = Fix R = Fix PU = {x ∈ X | x = PU x} = U. (ii): Id−R = Id−PU = PU⊥ .
(iii): By using (ii), we obtain ran (Id−R) = ran (Id−PU) = U⊥ = D⊥. (iv): From [4,
Example 1], we have

(
Id−R

)−1
=
(

Id−PU
)−1

= P−1
U⊥ = Id+NU⊥ .

(v): By using (8), we have

T = PD⊥(Id−R)−1 PD⊥ −
1
2

PD⊥

= PU⊥(Id+NU⊥)PU⊥ −
1
2

PU⊥

=
1
2

PU⊥

= T∗.

(vi): Follows from (v). ■

Example 3. Let U be a closed subspace of X and suppose that

R = −PU . (27)

Then
(i) D = {0}.

(ii) Id−R = Id+PU .
(iii) ran (Id−R) = X .
(iv)

(
Id−R

)−1
= 1

2 Id+ 1
2 PU⊥ .

(v) T = 1
2 PU .

Proof. (i): D = Fix R = Fix(−PU) = {x ∈ X | x = −PU x} = {0}. (ii): Id−R = Id+PU .
(iii): By [4, Minty Theorem], Id + PU has full range D = X. (iv):

(
Id−R

)−1
= JPU =

1
2 PU +PU⊥ = 1

2 Id+ 1
2 PU⊥ . (v): We have

T = PD⊥(Id−R)−1 PD⊥ −
1
2

PD⊥

=
1
2

Id+
1
2

Id−1
2

PU⊥ −
1
2

Id



S.T. Alwadani / Eur. J. Pure Appl. Math, 17 (4) (2024), 3660-3676 3673

=
1
2

PU .

■

Example 4. Let U be a closed subspace of X and suppose that

R = RU . (28)

Then
(i) D = U.

(ii) Id−R = 2 PU⊥ .
(iii) ran (Id−R) = D⊥ is closed.
(iv)

(
Id−R

)−1
= 1

2 Id+NU .
(v) T = 0.

Proof. (i): D = Fix R = Fix(RU) = {x ∈ X | x = RUx} = {x ∈ X | 2x = 2 PU} = U.
(ii): Id−R = Id−RU =

(
PU +PU⊥

)
−
(

PU −PU⊥
)

= 2 PU⊥ . (iii): ran (Id−R) =

ran (2 PU⊥) = D⊥ is closed. (iv):
(

Id−R
)−1

=
(
2(Id−PU)

)−1
= 1

2 Id+NU⊥ . (v): We
have

T = PD⊥(Id−R)−1 PD⊥ −
1
2

PD⊥

= PD⊥

(1
2

Id+NU⊥

)
PU⊥ −

1
2

PU⊥

=
1
2

PU⊥ −
1
2

PU⊥

= 0.

■

Example 5. Let U be a closed subspace of X and suppose that

R = −RU . (29)

Then
(i) D = Fix

(
− RU

)
= U⊥.

(ii) Id−R = 2 PU .
(iii) ran (Id−R) = U is closed.
(iv)

(
Id−R

)−1
= 1

2 Id+NU .
(v) T = 0.

Proof. (i): Note that −RU = RU⊥ and we learn from Example 4 that D = Fix R = U⊥. (ii):
Id−R = Id−RU⊥ =

(
PU +PU⊥

)
−
(
2 PU⊥ − Id

)
=
(

PU +PU⊥
)
−
(

PU⊥ −PU
)
= 2 PU .

(iii): By using (ii), we have ran
(

Id−R
)
= ran

(
2 PU

)
= D = U. (iv):

(
Id−R

)−1
=



REFERENCES 3674(
Id−(RU⊥)

)−1
=
(

Id−(2 PU⊥ − Id)
)−1

=
(
2(Id−PU⊥)

)−1
= 1

2 Id+NU by [4, Exam-
ple]. (v): By using (8), we have

T = PD⊥(Id−R)−1 PD⊥ −
1
2

PD⊥

= PD⊥

(1
2

Id+NU

)
PU⊥ −

1
2

PU⊥

=
1
2

PU⊥ −
1
2

PU⊥

= 0.

■

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.

Clarification

Please note that a preprint has previously been published in arXiv and available in
[3]. There is no conflict of interest and there is no data were used to support this study.

References

[1] Salihah Alwadani, Heinz H Bauschke, Julian P Revalski, and Xianfu Wang. Re-
solvents and yosida approximations of displacement mappings of isometries. Set-
Valued and Variational Analysis, 29:721–733, 2021.

[2] Salihah Thabet Alwadani. On the behaviour of algorithms featuring compositions of
projectors and proximal mappings with no solutions. PhD thesis, University of British
Columbia, 2021.

[3] Salihah Thabet Alwadani. Additional studies on displacement mapping with re-
strictions. arXiv preprint arXiv:2405.13510, 2024.

[4] 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.

[5] Heinz H Bauschke, Warren L Hare, and Walaa M Moursi. On the range of the
douglas–rachford operator. Mathematics of Operations Research, 41(3):884–897, 2016.



REFERENCES 3675

[6] Heinz H Bauschke, Victoria Martı́n-Márquez, Sarah M Moffat, and Xianfu Wang.
Compositions and convex combinations of asymptotically regular firmly nonexpan-
sive mappings are also asymptotically regular. Fixed Point Theory and Applications,
2012:1–11, 2012.

[7] Heinz H Bauschke and Walaa M Moursi. On the order of the operators in the
douglas–rachford algorithm. Optimization Letters, 10:447–455, 2016.

[8] Heinz H Bauschke and Walaa M Moursi. The magnitude of the minimal displace-
ment vector for compositions and convex combinations of firmly nonexpansive
mappings. Optimization Letters, 12:1465–1474, 2018.

[9] Heinz H Bauschke, Xianfu Wang, and Liangjin Yao. On borwein–wiersma decompo-
sitions of monotone linear relations. SIAM Journal on Optimization, 20(5):2636–2652,
2010.

[10] Imtiyaz Ahmad Bhat, Lakshmi Narayan Mishra, Vishnu Narayan Mishra, and Cemil
Tunç. Analysis of efficient discretization technique for nonlinear integral equations
of hammerstein type. International Journal of Numerical Methods for Heat & Fluid Flow,
2024.

[11] Regina S Burachik, Alfredo N Iusem, Regina S Burachik, and Alfredo N Iusem. En-
largements of monotone operators. Springer, 2008.

[12] John B Conway. A course in functional analysis, volume 96. Springer, 2019.

[13] Frank Deutsch and F Deutsch. Best approximation in inner product spaces, volume 7.
Springer, 2001.

[14] Jonathan Eckstein and Dimitri P Bertsekas. On the douglas—rachford splitting
method and the proximal point algorithm for maximal monotone operators. Mathe-
matical programming, 55:293–318, 1992.

[15] Mohammed Sumebo Hogeme, Mesfin Mekuria Woldaregay, Laxmi Rathour, and
Vishnu Narayan Mishra. A stable numerical method for singularly perturbed fred-
holm integro differential equation using exponentially fitted difference method.
Journal of Computational and Applied Mathematics, 441:115709, 2024.

[16] Robert E Megginson. A course in functional analysis.

[17] Naol Tufa Negero, Gemechis File Duressa, Laxmi Rathour, and Vishnu Narayan
Mishra. A novel fitted numerical scheme for singularly perturbed delay parabolic
problems with two small parameters. Partial Differential Equations in Applied Mathe-
matics, 8:100546, 2023.

[18] R Tyrrell Rockafellar and Roger J-B Wets. Springer-Verlag, corrected 3rd printing, 2009.



REFERENCES 3676

[19] MK Sharma, Nitesh Dhiman, Shubham Kumar, Laxmi Rathour, Vishnu Narayan
Mishra, et al. Neutrosophic monte carlo simulation approach for decision making
in medical diagnostic process under uncertain environment. Int J Neutrosophic Sci,
22(1):08–16, 2023.

[20] Stephen Simons. Minimax and monotonicity. Springer, 2006.

[21] Stephen Simons and F Takens. From Hahn-Banach to Monotonicity, volume 1693.
Springer, 2008.

[22] Eberhard Zeidler. Nonlinear functional analysis and its applications: II/B: nonlinear
monotone operators. Springer Science & Business Media, 2013.

[23] Eberhard Zeidler. Nonlinear functional analysis and its applications: III: variational meth-
ods and optimization. Springer Science & Business Media, 2013.