Main result

4.1 Non-Lipschitz split monotone variational inclusion problem with fixed points

4.1.1 Main result

Chapter 4

Split Monotone Variational Inclusion Problems and Fixed Point Problems

4.0.1 Introduction

In this chapter, we present our result on relaxed double inertial Tseng’s extragradient method with self-adaptive step sizes for solving split monotone variational inclusion prob- lem (SMVIP) involving non-Lipschitz operators and fixed point problem of strict pseudo- contractive mappings. Furthermore, results on generalized split feasibility problem over a solution set of monotone variational inclusion problem was studied. For each of these problems, we propose iterative algorithm with self-adaptive step size for approximating the solution and prove strong convergence theorem. We also give applications of our results and illustrate our algorithms with numerical examples.

4.1 Non-Lipschitz split monotone variational inclu-

Assumption 4.1.1. Let H₁ and H₂ be real Hilbert spaces. We make the following as- sumptions:

(1) (a)A :H₁ →H₁ and D:H₂ →H₂ are monotone and uniformly continuous.

(b) B :H1 →2^H1 and G:H2 →2^H2 are maximal monotone operators.

(c) T :H₁ →H₂ is a bounded linear operator such that T ̸= 0 and T^∗ :H₂ →H₁ is the adjoint of T.

(d) S :H1 →H1 is a k-strict pseudocontractive mapping.

(e) The solution set Υ := Fix(S)∩Ω̸=∅.

(2) {αn}^∞_n=1, {ϕn}^∞_n=1, {εn}^∞_n=1, {τn}^∞_n=1, {κn}^∞_n=1, {ϑn}^∞_n=1, {sn}^∞_n=1 and {tn}^∞_n=1 are all positive sequences which satisfy the following conditions:

(a){α_n} ⊂(0,1), lim

n→∞α_n = 0, P∞

n=1α_n=∞, lim

n→∞

ε_n

α_n = 0, lim

n→∞

τ_n

α_n = 0,0< a≤ ϕ_n≤b <1.

(b) 0 < κ < κ^′ < 1, 0 < σ < σ^′ < 1,0 < ϑ < ϑ^′ < 1, lim

n→∞κn = 0, lim

n→∞σn = 0, lim

n→∞ϑn = 0,P∞

n=1sn<+∞ and P∞

n=1tn<+∞, α∈[k,1).

Below is our proposed scheme for this paper.

Algorithm 4.1.2.

Initialization: Supposeθ, ϱ, γ₁, λ₁ >0, x₀, x₁ ∈H₁ be arbitrary and x₀ =b₀. Iterative Steps: Set n := 1. Calculate x_n+1 as follows:

Step 1. Given the iterates xn−1 and x_n (n ≥1), we choose θ_n such that 0≤θ_n ≤θ¯_n, where θ¯_n :=





 min

θ,_∥x ^εn

n−xn−1∥ , if x_n ̸=xn−1

θ, otherwise.

(4.1)

Step 2. Set

w_n=x_n+θ_n(x_n−x_n−1).

Compute

y_n = (I+γ_nG)⁻¹(I−γ_nD)T w_n, (4.2) where

γn+1 = (min

n(κn+κ)∥T w_n−y_n∥

∥DT wn−Dyn∥ , γn+sn

o

, if DT wn̸=Dyn,

γ_n+s_n, otherwise, (4.3)

and

z_n=y_n−γ_n(Dy_n−DT w_n).

Step 3. Compute

b_n=w_n+η_nT^∗(z_n−T w_n), where

ηn = (_(ϑ

n+ϑ)∥T wn−zn∥²

∥T^∗(T wn−z_n)∥² , if ∥T^∗(T w_n−z_n)∥ ̸= 0,

0, otherwise. (4.4)

Step 4. Choose ϱn such that 0≤ϱn≤ϱ¯n, where

¯ ϱ_n:=





 min

ϱ,_∥b ^τn

n−bn−1∥ , if b_n ̸=b_n−1

ϱ, otherwise.

(4.5)

Step 5 : Compute

l_n = (1−α_n) (b_n+ϱ_n(b_n−bn−1))

u_n= (I+λ_nB)⁻¹(I−λ_nA)l_n, where

λ_n+1 =

(minn_(σ

n+σ)∥un−ln∥

∥Aun−Aln∥ , λ_n+t_no

, if Au_n ̸=Al_n,

λ_n+t_n, otherwise,

(4.6) and

v_n=u_n−λ_n(Au_n−Al_n).

Step 6. Compute

x_n+1 = (1−ϕ_n)l_n+ϕ_nS_αv_n, (4.7) where

S_α =αI + (1−α)S.

Set n :=n+ 1 and go back to Step 1.

Remark 4.1.3. Observe from (4.1) and Assumption 4.1.1(2)(a) that

n→∞lim θ_n∥x_n−xn−1∥= 0 and lim

n→∞

θ_n αn

∥x_n−xn−1∥= 0.

Similarly, we obtain from (4.5) and Assumption 4.1.1 (2)(a) that

n→∞lim ϱ_n∥b_n−b_n−1∥= 0 and lim

n→∞

ϱ_n αn

∥b_n−b_n−1∥= 0.

Lemma 4.1.4. The sequences {γ_n} and {λ_n} of step sizes generated by Algorithm 4.1.2 under Assumption 4.1.1 are well-defined and bounded.

Proof. Since D is uniformly continuous, using (2.55), we have that for any given ϵ > 0, there existsK <+∞such that∥DT w_n−Dy_n∥ ≤K∥T w_n−y_n∥+ϵ.Furthermore, whenever DT wn−Dyn̸= 0 for all n≥1,we obtain

(κ_n+κ)∥T w_n−y_n∥

∥DT w_n−Dy_n∥ ≥ (κ_n+κ)∥T w_n−y_n∥

K∥T w_n−y_n∥+ϵ = (κ_n+κ)∥T w_n−y_n∥

(K+ϵ₁)∥T w_n−y_n∥ = κ_n+κ

N ≥ κ

N, whereϵ=ϵ₁∥T w_n−y_n∥for someϵ₁ ∈(0,1) andN =K+ϵ₁.It follows from the definition of γ_n+1 that the sequence {γ_n}has lower bound min{_N^κ, γ₁} and has upper boundγ₁+ Ψ.

By Lemma 2.5.23, the limit lim

n→∞γ_n exists and lim

n→∞γ_n = γ. Obviously, we have that γ ∈

min{_N^κ, γ₁}, γ₁+ Ψ

; where Ψ =P∞ n=1s_n. Similarly, we obtain lim

n→∞λ_n = λ ∈

min{_M^σ, λ₁}, λ₁+ Φ

; for some M > 0 where Φ = P∞

n=1t_n.

Lemma 4.1.5. Suppose ∥T^∗(T w_n−z_n)∥ ̸= 0, then the sequence {η_n} generated by Algo- rithm 4.1.2 has a positive lower bound.

Proof. Let ∥T^∗(T w_n−z_n)∥ ̸= 0,it follows that

η_n= (ϑ_n+ϑ)∥T w_n−z_n∥²

∥T^∗(T w_n−z_n)∥² . Since T is a bounded linear operator,ϑ∈(0,1) and lim

n→∞ϑn= 0,we have (ϑ_n+ϑ)∥T w_n−z_n∥²

∥T^∗(T w_n−z_n)∥² ≥ (ϑ_n+ϑ)∥T w_n−z_n∥²

∥T∥²∥(T w_n−z_n)∥² = (ϑ_n+ϑ)

∥T∥² ≥ ϑ

∥T∥². Thus, _∥T^ϑ_∥2 is a lower bound of {η_n}.

Lemma 4.1.6. Let Assumption 4.1.1 (2) hold. Then, there exists a positive integer N such that ϑ_n+ϑ,^γn(κ_γ^n+κ)

n+1 , ^λn_λ^(σn+σ)

n+1 ∈(0,1) ∀ n ≥N.

Proof. Since 0< ϑ < ϑ^′ <1 and lim

n→∞ϑn= 0,there exists a positive integer N1 such that 0< ϑ+ϑ_n ≤ϑ^′ <1, ∀n ≥N₁.

Also, since 0 < κ < κ^′ <1, lim

n→∞κ_n = 0 and lim

n→∞γ_n =γ, we obtain

n→∞lim

1− γ_n(κ_n+κ) γ_n+1

= 1−κ >1−κ^′ >0.

Thus, there exists a positive integer N₂ such that 1− γ_n(κ_n+κ)

γ_n+1 >0, ∀ n ≥N₂.

In the same vein, there exists a positive integer N₃ such that 1− λ_n(σ_n+σ)

λ_n+1 >0, ∀ n≥N3. Now, setting N = max{N₁, N₂, N₃}, we have ϑ_n+ϑ,^γn(κ_γ^n+κ)

n+1 , ^λn(σ_λ ^n+σ)

n+1 ∈ (0,1) ∀ n ≥ N.

Lemma 4.1.7. Suppose Assumptions 4.1.1 holds and let {z_n} be a sequence generated by Algorithm 4.1.2. Then, for all x¯∈Υ, we have

∥z_n−Tx∥¯ ² ≤ ∥T w_n−Tx∥¯ ²−

1− γ_n²(κ_n+κ)² γ_n+1²

∥T w_n−y_n∥², (4.8) and

∥z_n−y_n∥ ≤ γ_n(κ_n+κ) γn+1

∥T w_n−y_n∥. (4.9)

Proof. From (4.3), we obtain

∥DT w_n−Dy_n∥ ≤ (κ_n+κ)

γ_n+1 ∥T w_n−y_n∥, ∀n∈N. (4.10) Observe that (4.10) holds for both DT w_n =Dy_n and DT w_n ̸=Dy_n. Let ¯x∈ Υ, then we have Tx¯∈(D+G)⁻¹(0).From Step 2 and by applying (4.10) and Lemma2.5.18 (i), we obtain

∥z_n−Tx∥¯ ² =∥y_n−γ_n(Dy_n−DT w_n)−Tx∥¯ ²

=∥yn−Tx∥¯ ²+γ_n²∥Dyn−DT wn∥²−2γn⟨yn−Tx, Dy¯ n−DT wn⟩

=∥y_n−T w_n∥²+∥T w_n−Tx∥¯ ²+ 2⟨y_n−T w_n, T w_n−Tx⟩¯ +γ_n²∥Dy_n−DT w_n∥²−2γ_n⟨y_n−Tx, Dy¯ _n−DT w_n⟩

=∥T w_n−Tx∥¯ ²+∥y_n−T w_n∥²−2⟨y_n−T w_n, y_n−T w_n⟩

+ 2⟨y_n−T w_n, y_n−Tx⟩¯ +γ_n²∥Dy_n−DT w_n∥²−2γ_n⟨y_n−Tx, Dy¯ _n−DT w_n⟩

=∥T w_n−Tx∥¯ ²− ∥y_n−T w_n∥²+ 2⟨y_n−T w_n, y_n−Tx⟩¯ +γ_n²∥Dy_n−DT w_n∥²−2γ_n⟨y_n−Tx, Dy¯ _n−DT w_n⟩

=∥T wn−Tx∥¯ ²− ∥yn−T wn∥²−2⟨T wn−yn−γn(DT wn−Dyn), yn−Tx⟩¯ +γ_n²∥Dy_n−DT w_n∥²

≤ ∥T wn−Tx∥¯ ²−

1− γ_n²(κ_n+κ)² γ_n+1²

∥T wn−yn∥²

−2⟨T wn−yn−γn(DT wn−Dyn), yn−Tx⟩.¯ (4.11) Now, recall from Step 2 that y_n = (I +γ_nG)⁻¹(I −γ_nD)T w_n, which implies that (I − γnD)T wn∈(I+γnG)yn. Since G is maximal monotone, there existshn ∈Gyn such that

T w_n−γ_nDT w_n =y_n+γ_nh_n,

which implies that

h_n = 1 γn

(T w_n−γ_nDT w_n−y_n). (4.12) Furthermore, sinceTx¯∈(D+G)⁻¹(0),we have 0 ∈(D+G)T(¯x) andDy_n+h_n∈(D+G)y_n. In addition, since D+G is maximal monotone, we obtain

⟨Dy_n+h_n, y_n−Tx⟩ ≥¯ 0. (4.13) Using (4.12) and (4.13), we have

1

γ_n⟨T w_n−γ_nDT w_n−y_n+γ_nDy_n, y_n−Tx⟩ ≥¯ 0, which implies that

⟨T w_n−y_n−γ_n(DT w_n−Dy_n), y_n−Tx⟩ ≥¯ 0. (4.14) Therefore, applying (4.14) in (4.11), we obtain

∥zn−Tx∥¯ ² ≤ ∥T wn−Tx∥¯ ²−

1− γ_n²(κ_n+κ)² γ_n+1²

∥T wn−yn∥². (4.15) Additionally, from the definition of zn and inequality (4.10), we have

∥z_n−y_n∥ ≤ γ_n(κ_n+κ) γn+1

∥T w_n−y_n∥.

This completes the proof.

Lemma 4.1.8. Let {xn} be a sequence generated by Algorithm 4.1.2 under Assumption 4.1.1. Then, {x_n} is bounded.

Proof. Let ¯x∈Υ.FromStep 1and Assumption4.1.1(2), we obtain thatθ_n∥x_n−xn−1∥ ≤ ε_n ∀ n∈N, and by implication we have

θ_n

α_n∥x_n−xn−1∥ ≤ ε_n

α_n →0, asn → ∞.

Thus, there exists M₁ >0 such that θ_n

α_n∥x_n−xn−1∥ ≤M₁ ∀n ∈N. Thus, from Step 2, we have

∥w_n−x∥ ≤ ∥x¯ _n−x∥¯ +θ_n∥x_n−xn−1∥

=∥x_n−x∥¯ +α_nθn

α_n∥x_n−xn−1∥ (4.16)

≤ ∥x_n−x∥¯ +α_nM₁. (4.17)

Furthermore, by Lemma 4.1.6there existsN ∈N such that

1− ^γn2(κ_γ2^n+κ)2 n+1

>0∀ n≥N.

Therefore, from (4.8), we have

∥z_n−Tx∥¯ ² ≤ ∥T w_n−x∥¯ ², ∀n ≥N. (4.18) Using Lemma 2.5.18, (4.18) and the definition of b_n, we have

∥b_n−x∥¯ ² =∥w_n+η_nT^∗(z_n−T w_n)−x∥¯ ²

=∥w_n−x∥¯ ² +η²_n∥T^∗(z_n−T w_n)∥²+ 2η_n⟨w_n−x, T¯ ^∗(z_n−T w_n)⟩ (4.19)

=∥w_n−x∥¯ ² +η²_n∥T^∗(z_n−T w_n)∥²+ 2η_n⟨T w_n−Tx, z¯ _n−T w_n⟩

=∥w_n−x∥¯ ² +η²_n∥T^∗(z_n−T w_n)∥² +η_n

∥z_n−Tx∥¯ ²− ∥T w_n−Tx∥¯ ²− ∥z_n−T w_n∥²

≤ ∥w_n−x∥¯ ²+η_n²∥T^∗(z_n−T w_n)∥²−η_n∥z_n−T w_n∥²

=∥w_n−x∥¯ ² −η_n[∥z_n−T w_n∥²−η_n∥T^∗(z_n−T w_n)∥²]. (4.20) If ∥T^∗(z_n−T w_n)∥²∥ ̸= 0, we obtain from (4.4) that

∥z_n−T w_n∥²−η_n∥T^∗(z_n−T w_n)∥² = (1−(ϑ_n+ϑ))∥T w_n−z_n∥² ≥0. (4.21) It follows from (4.17) and (4.20) that

∥b_n−x∥ ≤ ∥w¯ _n−x∥ ≤ ∥x¯ _n−x∥¯ +α_nM₁, ∀ n ≥N. (4.22) On the other hand, if ∥T^∗(z_n−T w_n)∥² = 0,then from (4.19),∥b_n−x∥ ≤ ∥w¯ _n−x∥¯ clearly holds.

Next, observe that from (4.6) we have

∥Au_n−Al_n∥ ≤ (σ_n+σ)

λ_n+1 ∥u_n−l_n∥, ∀n ∈N. (4.23) Observe that (4.23) holds for both Au_n =Al_n and Au_n ̸=Al_n. Now, from the definition of v_n and by applying (4.23) we have

∥vn−x∥¯ ² =∥un−λn(Aun−Aln)−x∥¯ ²

=∥u_n−x∥¯ ²+λ²_n∥Au_n−Al_n∥²−2λ_n⟨u_n−x, Au¯ _n−Al_n⟩

=∥u_n−l_n∥²+∥l_n−x∥¯ ²+ 2⟨u_n−l_n, l_n−x⟩¯ +λ²_n∥Au_n−Al_n∥²

−2λ_n⟨u_n−x, Au¯ _n−Al_n⟩

=∥l_n−x∥¯ ²+∥u_n−l_n∥²−2⟨u_n−l_n, u_n−l_n⟩+ 2⟨u_n−l_n, u_n−x⟩¯ +λ²_n∥Au_n−Al_n∥²−2λ_n⟨u_n−x, Au¯ _n−Al_n⟩

=∥ln−x∥¯ ²− ∥un−ln∥²+ 2⟨un−ln, un−x⟩¯ +λ²_n∥Aun−Aln∥²

−2λ_n⟨u_n−x, Au¯ _n−Al_n⟩

=∥l_n−x∥¯ ²− ∥u_n−l_n∥²−2⟨l_n−u_n−λ_n(Al_n−Au_n), u_n−x⟩¯ +λ²_n∥Au_n−Al_n∥²

≤ ∥l_n−x∥¯ ²−

1− λ²_n(σn+σ)² λ²_n+1

∥u_n−l_n∥²

−2⟨l_n−u_n−λ_n(Al_n−Au_n), u_n−x⟩.¯ (4.24)

From u_n = (I +λ_nB)⁻¹(I −λ_nA)l_n, we have (I −λ_nA)l_n ∈ (I +λ_nB)u_n. Since B is maximal monotone, there exists d_n ∈Bu_n such that

(I−λ_nA)l_n=u_n+λ_nd_n, which implies that

d_n= 1

λ_n(l_n−u_n−λ_nAl_n). (4.25) Since ¯x∈ Υ, we have 0 ∈(A+B)¯x and Au_n+d_n∈ (A+B)u_n. Since A+B is maximal monotone, we obtain

⟨Au_n+d_n, u_n−x⟩ ≥¯ 0. (4.26) Substituting (4.25) into (4.26), we obtain

1

λ_n⟨l_n−u_n−λ_nAl_n+λ_nAu_n, u_n−x⟩ ≥¯ 0, which gives

⟨l_n−u_n−λ_n(Al_n−Au_n), u_n−x⟩ ≥¯ 0. (4.27) Applying (4.27) in (4.24), we have

∥v_n−x∥¯ ² ≤ ∥l_n−x∥¯ ²−

1−λ²_n(σ_n+σ)² λ²_n+1

∥u_n−l_n∥². (4.28)

By the definition of v_n and applying (4.23), we obtain

∥vn−un∥ ≤ λ_n(σ_n+σ)

λ_n+1 ∥un−ln∥. (4.29)

Now, from Step 4 and Assumption4.1.1 (2), we have ϱ_n

αn

∥b_n−b_n−1∥ ≤ τ_n αn

→0, as n→ ∞. (4.30)

By the definition of l_n, we have

∥l_n−x∥¯ =∥(1−α_n)(b_n+ϱ_n(b_n−bn−1))−x∥¯

=∥(1−α_n)(b_n−x) + (1¯ −α_n)ϱ_n(b_n−bn−1)−α_nx∥¯

≤(1−α_n)∥b_n−x∥¯ + (1−α_n)ϱ_n∥b_n−bn−1∥+α_n∥¯x∥

= (1−αn)∥bn−x∥¯ +αn

(1−αn)ϱ_n

α_n∥bn−bn−1∥+∥¯x∥

, (4.31)

By (4.30), we obtain

n→∞lim

(1−α_n)ϱ_n αn

∥b_n−b_n−1∥+∥¯x∥

=∥¯x∥.

Thus, there exists M₂ >0 such that (1−α_n)ϱn

α_n∥b_n−bn−1∥+∥¯x∥ ≤M₂, ∀n ∈N. (4.32) Combining (4.31) and (4.32), we have

∥l_n−x∥ ≤¯ (1−α_n)∥b_n−x∥¯ +α_nM₂, ∀n ≥N. (4.33) By Lemma 4.1.6 there exists N ∈N such that

1−^λ2n(σ_λ2^n+σ)2 n+1

>0 ∀ n ≥ N. Thus, from (4.28) we obtain

∥v_n−x∥¯ ² ≤ ∥l_n−x∥¯ ², ∀ n ≥N. (4.34) Using the last inequality together with (4.22) and (4.33) we obtain

∥v_n−x∥ ≤ ∥l¯ _n−x∥ ≤¯ (1−α_n)∥w_n−x∥¯ +α_nM₂ (4.35)

≤(1−α_n)(∥x_n−x∥¯ + +α_nM₁) +α_nM₂

≤(1−α_n)∥x_n−x∥¯ +α_n(M₁+M₂).

Now, from Step 6, (4.34),Lemma2.5.13 and Lemma 2.5.18(iii), we obtain

∥x_n+1−x∥¯ ² =∥(1−ϕ_n)l_n+ϕ_nS_αv_n−x∥¯ ²

=∥(1−ϕ_n)(l_n−x) +¯ ϕ_n(S_αv_n−x)∥¯ ²

= (1−ϕn)∥ln−x∥¯ ²+ϕn∥Sαvn−x∥¯ ²−(1−ϕn)ϕn∥Sαvn−ln∥²

≤(1−ϕ_n)∥l_n−x∥¯ ²+ϕ_n∥v_n−x∥¯ ²−(1−ϕ_n)ϕ_n∥S_αv_n−l_n∥²

≤ ∥l_n−x∥¯ ²−(1−ϕ_n)ϕ_n∥S_αv_n−l_n∥², (4.36) which follows from (4.35) that

∥x_n+1−x∥ ≤ ∥l¯ _n−x∥¯

≤(1−α_n)∥x_n−x∥¯ +α_n(M₁+M₂)

= (1−α_n)∥x_n−x∥¯ +α_nK

≤max{∥xn−x∥, K}¯ ...

≤max{∥x_N −x∥, K}.¯

where K =M₁+M₂.Therefore, the sequence {x_n} is bounded. Consequently, {wn}, {yn}, {zn}, {bn}, {ln}, {un}, {vn} are all bounded.

Lemma 4.1.9. Suppose Assumption 4.1.1 hold; lim

j→∞∥b_nj −w_nj∥ = 0 = lim

j→∞∥l_nj −v_nj∥ and {x_n} is a sequence generated by Algorithm 4.1.2. If there exists a subsequence {x_nj} of {x_n} which converges weakly to an elementxˆ∈H₁, then xˆ∈Ω.

Proof. Let ¯x∈Ω. Then, we obtain from (4.20) and (4.21) that

∥b_nj −x∥¯ ² ≤ ∥w_nj−x∥¯ ²−η_nj

1−(ϑ_nj +ϑ)

∥T w_nj −z_nj∥², (4.37) which implies that

η_nj

1−(ϑ_nj+ϑ)

∥T w_nj−z_nj∥² ≤ ∥w_nj−x∥¯ ²− ∥b_nj −x∥¯ ²

≤ ∥wnj−bnj∥²+ 2∥wnj−bnj∥∥bnj −x∥.¯ (4.38) Since {b_nj} is bounded, then by the hypothesis that lim

n→∞∥b_nj −w_nj∥= 0, from (4.38) we obtain

j→∞lim η_nj

1−(ϑ_nj +ϑ)

∥T w_nj−z_nj∥² = 0, which by the conditions on the control parameters implies that

j→∞lim ∥T w_nj −z_nj∥= 0. (4.39) From (4.8), we obtain

∥z_nj−Tx∥¯ ² ≤ ∥T w_nj−Tx∥¯ ²− 1− γ_n²_j(κ_nj+κ)² γ_n²_j+1

!

∥T w_nj −y_nj∥², which implies that

1−γ_n²

j(κ_nj +κ)² γ_n²_j+1

!

∥T w_nj−y_nj∥² ≤ ∥T w_nj −Tx∥¯ ²− ∥z_nj −Tx∥¯ ²

=∥T w_nj −z_nj∥²+∥z_nj−Tx∥¯ ²

+ 2⟨T w_nj −z_nj, z_nj−Tx⟩ − ∥z¯ _nj−Tx∥¯ ²

≤ ∥T w_nj−z_nj∥²+ 2∥T w_nj −z_nj∥∥z_nj−Tx∥.¯ It follows from (4.39) and Assumption 4.1.1(2) that

j→∞lim ∥T wnj−ynj∥= 0. (4.40) Now, let (v, w) ∈ Graph(D +G), then, w− Dv ∈ Gv. Since y_nj = (I +γ_njG)⁻¹(I − γ_njD)T w_nj, we have

T wnj −γnjDT wnj ∈ynj+γnjGynj, which implies that

1

γ_nj(T w_nj −y_nj−γ_njDT w_nj)∈Gy_nj. Thanks to the maximal monotonicity of G, which guarantees that

⟨v −y_nj, w−Dv− 1

γ_nj T w_nj −y_nj −γ_njDT w_nj

⟩ ≥0. (4.41)

Using (4.41) and the monotonicity ofD, we have

⟨v−y_nj, w⟩ ≥ ⟨v−y_nj, Dv+ 1

γ_nj(T w_nj −y_nj−γ_njDT w_nj)⟩

=⟨v−ynj, Dv−Dynj⟩+⟨v−ynj, Dynj−DT wnj⟩ +⟨v−y_nj, 1

γ_nj(T w_nj −y_nj)⟩

≥ ⟨v−y_nj, Dy_nj−DT w_nj⟩+⟨v−y_nj, 1

γ_nj(T w_nj −y_nj)⟩. (4.42) Moreover, we obtain from the definition of w_nj and θ_nj and by Remark5.2.4 that

j→∞lim ∥w_nj −x_nj∥= lim

j→∞θ_nj∥x_nj −x_nj−1∥= 0. (4.43) Furthermore, since x_nj ⇀xˆ∈H₁, then it follows that there exists a subsequence {w_nj} of {w_n}such thatw_nj ⇀x. Also, sinceˆ T is a bounded linear operator, we haveT w_nj ⇀ Tx.ˆ It follows from (4.40) and the continuity of D that

y_nj ⇀ Txˆand lim

j→∞∥DT w_nj−Dy_nj∥= 0. (4.44) Thus, passing limit as j → ∞ in (4.42), we obtain

⟨v −Tx, w⟩ ≥ˆ 0, ∀(v, w)∈Graph(D+G). (4.45) Thus, by the maximal monotonicity of D+G, we conclude that Txˆ∈(D+G)⁻¹(0).

Also, by applying similar argument used in obtaining (4.40), (4.42), (4.43), we have

j→∞lim ∥l_nj −u_nj∥= 0, (4.46)

j→∞lim ∥l_nj −b_nj∥= 0, (4.47)

⟨ˆv −u_nj,w⟩ ≥ˆ 0, ∀ (ˆv,w)ˆ ∈Graph(A+B). (4.48) In the same vein, sincexnj ⇀xˆ∈H1, we obtain using (4.43) and our hypothesis lim

j→∞∥bnj− wnj∥= 0 together with (4.46)-(4.48) that{wnj},{bnj},{lnj}and{unj}all converge weakly to ˆx.Hence, we have

⟨ˆv−x,ˆ w⟩ˆ = lim

j→∞⟨ˆv−unj,w⟩ ≥ˆ 0, ∀ ( ˆw,ˆv)∈Graph(A+B).

By the maximal monotonicity of A+B, we conclude that ˆx ∈ (A+B)⁻¹(0). Therefore, ˆ

x∈Ω.

Theorem 4.1.10. Let {x_n} be an iterative sequence generated by Algorithm 4.1.2 such that Assumptions 4.1.1 are satisfied. Then, {x_n} converges strongly to an element x¯∈Υ, where ∥¯x∥= min{∥p∥:p∈Υ}.

Proof. Let ¯x ∈ Υ. Since ¯x = min{∥p∥ : p ∈ Υ}, we have ¯x = PΥ(0). We obtain using Cauchy-Schwartz inequality, Step 2, Lemma2.5.18 and (4.16) that

∥w_n−x∥¯ ² =∥x_n+θ_n(x_n−xn−1)−x∥¯ ²

=∥x_n−x∥¯ ²+ 2θ_n⟨x_n−x, x¯ _n−x_n−1⟩+θ_n²∥x_n−x_n−1∥²

≤ ∥xn−x∥¯ ²+ 2θn∥xn−x∥.∥x¯ n−xn−1∥+θ²_n∥xn−xn−1∥²

=∥xn−x∥¯ ²+θn∥xn−xn−1∥(2∥xn−x∥¯ +αn

θ_n

α_n∥xn−xn−1∥)

≤ ∥x_n−x∥¯ ²+θ_n∥x_n−xn−1∥[α_nM₁+ 2∥x_n−x∥].¯ (4.49) Now, since {xn} is bounded, then, there exists M3 > 0 such that αnM1 + 2∥xn−x∥ ≤¯ M₃, ∀ n≥1. It follows from (4.49) that

∥w_n−x∥¯ ² ≤ ∥x_n−x∥¯ ²+θ_n∥x_n−xn−1∥M₃. (4.50) In the same way, we obtain from Lemma 2.5.18 and (4.30) that

∥l_n−x∥¯ ² =∥(1−α_n) (b_n+ϱ_n(b_n−bn−1))−x∥¯ ²

=∥(1−α_n)(b_n−x) +¯ ϱ_n(1−α_n)(b_n−bn−1)−α_nx∥¯ ²

≤ ∥(1−α_n)(b_n−x) +¯ ϱ_n(1−α_n)(b_n−bn−1)∥²+α²_n∥¯x∥². (4.51) But

∥(1−α_n)(b_n−x) +¯ ϱ_n(1−α_n)(b_n−bn−1)∥²

≤(1−α_n)²∥b_n−x∥¯ ²+ 2(1−α_n)ϱ_n∥b_n−x∥∥b¯ _n−bn−1∥+ϱ²_n∥b_n−bn−1∥²

≤(1−αn)²∥bn−x∥¯ ²+ϱn∥bn−bn−1∥

2(1−αn)∥bn−x∥¯ +αn

ϱ_n

α_n∥bn−bn−1∥

≤(1−α_n)∥b_n−x∥¯ ²+ϱ_n∥b_n−bn−1∥(2∥b_n−x∥¯ +α_nM₂^∗),

for some M₂^∗ >0.Thus, there exists M₄ >0 such that 2∥b_n−x∥¯ +α_nM₂^∗ ≤M₄, ∀n≥1.

Using (4.51), we get

∥l_n−x∥¯ ² ≤(1−α_n)∥b_n−x∥¯ ²+ϱ_n∥b_n−bn−1∥M₄+α²_n∥¯x∥²

= (1−α_n)∥b_n−x∥¯ ²+α_n ϱn

α_n∥b_n−bn−1∥M₄+α_n∥¯x∥²

. (4.52)

Since {l_n} is bounded, then there exists M₅ >0 such that _α^ϱn

n∥b_n−bn−1∥M₄ +α_n∥¯x∥² ≤ M₅, ∀ n≥1. It follows from (4.52) that

∥l_n−x∥¯ ² ≤(1−α_n)∥b_n−x∥¯ ²+α_nM₅. (4.53)

Using Lemma2.5.18, Lemma2.5.13, (4.20), (4.28), (4.50) and (4.53), we obtain fromStep 6 that

∥xn+1−x∥¯ ² =∥(1−ϕn)ln+ϕnSαvn−x∥¯ ²

=∥(1−ϕ_n)(l_n−x) +¯ ϕ_n(S_αv_n−x)∥¯ ²

= (1−ϕ_n)∥l_n−x∥¯ ²+ϕ_n∥S_αv_n−x∥¯ ²−(1−ϕ_n)ϕ_n∥S_αv_n−l_n∥²

≤(1−ϕ_n)∥l_n−x∥¯ ²+ϕ_n∥v_n−x∥¯ ²−(1−ϕ_n)ϕ_n∥S_αv_n−l_n∥²

≤(1−ϕ_n)∥l_n−x∥¯ ²+ϕ_n

∥l_n−x∥¯ ²−

1− λ²_n(σ_n+σ)² λ²_n+1

∥u_n−l_n∥²

−(1−ϕ_n)ϕ_n∥S_αv_n−l_n∥²

=∥l_n−x∥¯ ²−ϕ_n

1− λ²_n(σ_n+σ)² λ²_n+1

∥u_n−l_n∥²

−(1−ϕ_n)ϕ_n∥S_αv_n−l_n∥² (4.54)

≤(1−α_n)∥b_n−x∥¯ ²+α_nM₅−ϕ_n

1− λ²_n(σn+σ)² λ²_n+1

∥u_n−l_n∥²

−(1−ϕ_n)ϕ_n∥S_αv_n−l_n∥² (4.55)

≤(1−αn)∥wn−x∥¯ ²+αnM5−ϕn

1− λ²_n(σn+σ)² λ²_n+1

∥un−ln∥²

−(1−ϕ_n)ϕ_n∥S_αv_n−l_n∥²

≤(1−α_n)∥x_n−x∥¯ ²+θ_n(1−α_n)∥x_n−xn−1∥M₃+α_nM₅

−ϕ_n

1− λ²_n(σn+σ)² λ²_n+1

∥u_n−l_n∥²−(1−ϕ_n)ϕ_n∥S_αv_n−l_n∥². (4.56) Furthermore, we obtain from (4.50), (4.54), and Lemma 2.5.18(ii) that

∥x_n+1−x∥¯ ² ≤ ∥l_n−x∥¯ ²

=∥(1−α_n)(b_n−x) + (1¯ −α_n)ϱ_n(b_n−bn−1)−α_nx∥¯ ²

≤ ∥(1−α_n)(b_n−x) + (1¯ −α_n)ϱ_n(b_n−b_n−1)∥² + 2α_n⟨−¯x, l_n−x⟩¯

≤(1−αn)²∥bn−x∥¯ ²+ 2(1−αn)ϱn∥bn−x∥∥b¯ n−bn−1∥+ϱ²_n∥bn−bn−1∥² + 2αn⟨−¯x, ln−xn+1⟩+ 2αn⟨−¯x, xn+1−x⟩¯

≤(1−α_n)∥w_n−x∥¯ ²+ 2(1−α_n)ϱ_n∥b_n−x∥∥b¯ _n−bn−1∥+ϱ²_n∥b_n−bn−1∥² + 2α_n∥x∥∥l¯ _n−x_n+1∥+ 2α_n⟨−¯x, x_n+1−x⟩¯

≤(1−α_n)[∥x_n−x∥¯ ² +θ_n∥x_n−xn−1∥M₃] + 2(1−α_n)ϱ_n∥b_n−x∥∥b¯ _n−bn−1∥ +ϱ²_n∥b_n−bn−1∥² + 2α_n∥¯x∥∥l_n−x_n+1∥+ 2α_n⟨−¯x, x_n+1−x⟩¯

≤(1−α_n)∥x_n−x∥¯ ²+α_nh

(1−α_n)θ_n αn

∥x_n−xn−1∥M₃ + 2(1−α_n)ϱ_n

α_n∥b_n−bn−1∥∥b_n−x∥¯ +ϱ_n∥b_n−bn−1∥ϱn

α_n∥b_n−bn−1∥+ 2∥¯x∥∥l_n−x_n+1∥+ 2⟨−¯x, x_n+1−x⟩¯ i

≤(1−α_n)∥x_n−x∥¯ ²+α_nΦ_n, (4.57)

where

Φ_n= (1−α_n)_α^θn

n∥x_n−xn−1∥M₃+2(1−α_n)_α^ϱn

n∥b_n−bn−1∥∥b_n−¯x∥+ϱ_n∥b_n−bn−1∥_α^ϱn

n∥b_n−bn−1∥ + 2∥¯x∥∥l_n−x_n+1∥+ 2⟨−¯x, x_n+1−x⟩.¯

Next, we claim that the sequencen

∥x_n−x∥¯ o

converges to zero. Indeed, in view of Lemma 2.5.55 and by Remark 4.1.3, it is enough to establish that lim sup

k→∞

⟨−¯x, x_nk+1−x⟩ ≤¯ 0 and

k→∞lim ∥ln_k −xn_k+1∥= 0, for every subsequence n

∥xn_k−x∥¯ o of

n

∥xn−x∥¯ o

satisfying lim inf

k→∞

∥x_nk+1−x∥ − ∥x¯ _nk −x∥¯

≥0. (4.58)

Now, letn

∥x_nk−x∥¯ o

be a subsequence of n

∥x_n−x∥¯ o

such that (4.58) holds. From (4.56), we get

lim sup

k→∞

"

ϕn_k 1−λ²_n

k(σ_nk +σ)² λ²_n

k+1

!

∥un_k−ln_k∥²

#

≤lim sup

k→∞

(1−α_nk)∥x_nk −x∥¯ ²− ∥x_nk+1 −x∥¯ ² +α_nk

θ_nk

α_nk(1−α_nk)∥x_nk−x_nk−1∥M₃+M₅

=−lim inf

k→∞ ∥x_nk+1−x∥¯ ²− ∥x_nk−x∥¯ ²

≤0.

Since lim

k→∞α_nk = 0, it follows from Assumption 4.1.1 (2) that

k→∞lim ∥u_nk −l_nk∥= 0. (4.59) In like manner, we obtain from (4.56) that

lim sup

k→∞

(1−ϕ_nk)ϕ_nk∥S_αv_nk−l_nk∥²

≤lim sup

k→∞

(1−α_nk)∥x_nk −x∥¯ ²− ∥x_nk+1−x∥¯ ² +α_nk

θ_nk

α_nk(1−α_nk)∥x_nk −x_nk−1∥M₃+M₅

=−lim inf

k→∞ ∥x_nk+1 −x∥¯ ² − ∥x_nk −x∥¯ ²

≤0.

Consequently, by the conditions of the control parameters, we obtain

k→∞lim ∥Sαvn_k −ln_k∥= 0. (4.60)

In addition, we obtain from (4.55) and (4.37) that

∥x_nk+1−x∥¯ ² ≤(1−α_nk)∥b_nk −x∥¯ ²+α_nkM₅

≤(1−αn_k)∥wn_k −x∥¯ ²−ηn_k[1−(ϑn_k +ϑ)]∥T wn_k−zn_k∥² +αn_kM5

≤(1−α_nk)∥x_nk−x∥¯ ²−η_nk[1−(ϑ_nk+ϑ)]∥T w_nk −z_nk∥² +θ_nk(1−α_nk)∥x_nk −x_nk−1∥M₃+α_nkM₅.

By applying (4.58), Assumption4.1.1 (2), Lemma4.1.5 and the fact that lim

k→∞α_nk = 0 we have

lim sup

k→∞

η_nk[1−(ϑ_nk+ϑ)]∥T w_nk−z_nk∥²

≤lim sup

k→∞

(1−α_nk)∥x_nk −x∥¯ ²− ∥x_nk+1−x∥¯ ² +α_nk

θ_nk

α_nk(1−α_nk)∥x_nk −x_nk−1∥M₃+M₅

=−lim inf

k→∞ ∥x_nk+1 −x∥¯ ² − ∥x_nk −x∥¯ ²

≤0.

Therefore,

k→∞lim ∥T w_nk−z_nk∥= 0.

It is easy to see from Step 3 that

∥b_nk−w_nk∥=η_nk∥T^∗(z_nk −T w_nk)∥= (ϑ_nk+ϑ)∥T w_nk −z_nk∥ →0 as k → ∞. (4.61) Observe that when ∥T^∗(z_nk −T w_nk)∥= 0, clearly, ∥b_nk−w_nk∥= 0.

Also, from Step 5, we obtain

∥l_nk −b_nk∥ ≤(1−α_nk)∥b_nk −b_nk∥+ (1−α_nk)ϱ_nk∥b_nk−b_nk−1∥+α_nk∥b_nk∥

=α_nk

(1−α_nk)ϱ_nk αn_k

∥b_nk−b_nk−1∥+∥b_nk∥

. Since lim

k→∞αn_k = 0, we get

k→∞lim ∥l_nk−b_nk∥= 0. (4.62) Using (4.29) and (4.59), we obtain

k→∞lim ∥v_nk −u_nk∥= 0. (4.63) Combining (4.59) and (4.63), we obtain

k→∞lim ∥v_nk −l_nk∥ ≤ lim

k→∞(∥v_nk −u_nk∥+∥u_nk −l_nk∥) = 0. (4.64)

It is easy to see from (4.60) and (4.64) that

k→∞lim ∥S_αv_nk−v_nk∥ ≤ lim

k→∞(∥S_αv_nk−l_nk∥+∥l_nk −v_nk∥) = 0. (4.65) Using (4.62) and (4.64), we obtain

k→∞lim ∥b_nk−v_nk∥ ≤ lim

k→∞(∥l_nk−b_nk∥+∥v_nk−l_nk∥) = 0. (4.66) Now, by Remark 4.1.3, we obtain

∥w_nk −x_nk∥=θ_nk∥x_nk−x_nk−1∥ →0, k → ∞. (4.67) Since {b_n} is bounded, it follows from (4.62), (4.65) and (4.66) and Step 6 that

∥x_nk+1 −b_nk∥ ≤(1−ϕ_nk)∥l_nk −b_nk∥+ϕ_nk∥S_αv_nk−b_nk∥ →0 as k → ∞. (4.68) Also, using (4.61) and (4.68), we obtain

k→∞lim ∥x_nk+1−w_nk∥ ≤ lim

k→∞(∥x_nk+1−b_nk∥+∥b_nk−w_nk∥) = 0. (4.69) Moreover, we obtain using (4.62) and (4.68) that

k→∞lim ∥l_nk −x_nk+1∥ ≤ lim

k→∞(∥l_nk −b_nk∥+∥x_nk+1−b_nk∥) = 0. (4.70) It follows from (4.67) and (4.69) that

k→∞lim ∥x_nk+1 −x_nk∥= 0. (4.71) Since

x_nk is bounded, then w_ω(x_n) is nonempty. Now, suppose x^∗ ∈ w_ω(x_n) be arbi- trarily selected. Then, there exists a subsequence {xn_k} of {xn} such that xn_k ⇀ x^∗ as k → ∞. Using (4.67), we getw_nk ⇀ x^∗ ask → ∞. Now, employing Lemma4.1.9together with (4.61) and (4.64), we have

x^∗ ∈Ω. (4.72)

From (4.64), (4.70) and (4.71), we have that v_nk ⇀ x^∗ as k → ∞. Furthermore, since I−S_α is demiclosed at zero, then by Lemma2.5.13and the fact that ∥S_αv_nk−v_nk∥ →0, we have

x^∗ ∈Fix(S_α) = Fix(S). (4.73)

Since x^∗ ∈w_ω(x_n) is arbitrary, we obtain using (4.72) and (4.73) that w_ω(x_n)⊂Υ.

Additionally, since {x_nk}is bounded, then there exists a subsequence{x_nkj}of{x_nk}that weakly converges to x^† ∈H₁ and

j→∞lim⟨−¯x, x_nkj −x⟩¯ = lim sup

k→∞

⟨−¯x, x_nk−x⟩.¯

Since ¯x= P_Υ(0),we have lim sup

k→∞

⟨−¯x, x_nk −x⟩¯ = lim

j→∞⟨−¯x, x_nkj −x⟩¯ =⟨−¯x, x^†−x⟩ ≤¯ 0. (4.74) From (4.71) and (4.74), we obtain

lim sup

k→∞

⟨−¯x, x_nk+1 −x⟩¯ = lim sup

k→∞

⟨−¯x, x_nk −x⟩¯ =⟨−¯x, x^†−x⟩ ≤¯ 0. (4.75) Applying Lemma2.5.55to (4.57) and using (4.75) combined with (4.70), and the fact that

k→∞lim

θ_nk

α_nk∥x_nk −x_nk−1∥= 0, lim

k→∞

ϱ_nk

α_nk∥b_nk −b_nk−1∥= 0 and lim

k→∞α_nk = 0, we obtain that

n→∞lim ∥x_n−x^∗∥= 0.

Consequently, we conclude that {xn} converges strongly to the point ¯x=PΥ(0).