Main result

4.2 Generalized split feasibility problems

4.2.1 Main result

Step 1. Given the iterates x_n−1 and x_n (n ≥1), choose α_n such that 0≤α_n ≤α¯_n, where

¯ α_n :=







min _n−1

n+α−1,_∥x ^ϵn

n−xn−1∥ , if x_n ̸=x_n−1

n−1

n+α−1, otherwise.

(4.85)

Step 2. Set

w_n =x_n+α_n(x_n−xn−1).

Then, compute

un =wn−γnGwn and yn =J_λ^B_n(I−λnA)un = (I+λnB)⁻¹(I−λnA)un, (4.86) where

γ_n+1=

(minn _µ∥u

n−wn∥

∥Gun−Gwn∥, γ_no

, if Gu_n̸=Gw_n,

γ_n, otherwise

(4.87)

and

λ_n+1 =

(minn _η∥u

n−y_n∥

∥Aun−Ayn∥, λ_no

, if Au_n ̸=Ay_n,

λ_n, otherwise.

(4.88)

Step 3. Compute

x_n+1 = (1−β_n−θ_n)u_n+β_nz_n, (4.89) where z_n:= (1−ψ_n)u_n+ψ_n

y_n+λ_n A(u_n)−A(y_n)

.

Stopping criterion: If y_n =u_n =w_n =x_n, then stop, otherwise, set n :=n+ 1 and go back to Step 1.

Remark 4.2.4. λ_n+1 ≤λ_n ∀n ∈N and lim

n→∞λ_n ≥min{λ₁,_L^η}.

Indeed, from (4.88), it is obvious that λ_n+1 ≤ λ_n ∀n ∈ N. Furthermore, since A is L-Lipschitz continuous, we get in the case of Aun̸=Ayn, that

λ_n+1 = min

η∥u_n−y_n∥

∥Au_n−Ay_n∥, λ_n

≥minnη L, λ_no

,

which by induction implies that λ_n ≥min{λ₁,_L^η}. This gives that the limit of {λ_n} exists and lim

n→∞λ_n≥min{λ₁,_L^η}.

Since G is 2∥T∥²-Lipschitz continuous, then in a similar manner, we have from (4.87) that γ_n+1 ≤γ_n ∀n∈N and lim

n→∞γ_n≥minn

γ₁,_2∥T^µ_∥2

o .

We now show that the stopping criterion of Algorithm 4.2.3 is valid.

Lemma 4.2.5. If y_n =u_n =w_n=x_n in Algorithm 4.2.3, then x_n∈Γ.

Proof. Letyn =un =wn=xn. Then, from (4.86), it is clear thatxn=J_λ^B_n(xn−λnA(xn)).

This implies thatx_n ∈(A+B)⁻¹(0). From (4.86), we also obtain thatx_n =x_n−γ_nGx_n= x_n−γ_nT^∗(I−S)T x_n,which implies that T^∗(I−S)T x_n = 0. That is,

ST x_n =T x_n+ ¯z, where T^∗z¯= 0. Now, let z ∈Γ, then we obtain that

∥T x_n−T z∥² =∥T x_n−T z∥²+ 2⟨x_n−z, T^∗z⟩¯

=∥T xn−T z∥²+ 2⟨T xn−T z,z⟩¯

=∥T x_n−T z+ ¯z∥²− ∥¯z∥²

=∥ST x_n−T z∥²− ∥¯z∥²

≤ ∥T x_n−T z∥²− ∥¯z∥²,

which implies that ∥¯z∥ = 0. That is, ¯z = 0. Hence ST x_n = T x_n, which gives that T x_n∈F(S).Therefore, x_n∈Γ.

Lemma 4.2.6. Let Assumption 4.2.2 hold and let {x_n} be a sequence generated by Algo- rithm 4.2.3. Then {x_n} is bounded.

Proof. From (4.88), we have that

∥Au_n−Ay_n∥ ≤ η

λ_n+1∥u_n−y_n∥ ∀n ∈N (4.90) holds for both Au_n=Ay_n and Au_n̸=Ay_n. Now, let x^∗ ∈Γ. Then, by Step 3, we obtain

∥zn−x^∗∥² =∥(1−ψn)un+ψnyn+ψnλn A(un)−A(yn)

−x^∗∥²

=∥(1−ψ_n)(u_n−x^∗) +ψ_n(y_n−x^∗) +ψ_nλ_n A(u_n)−A(y_n)

∥²

= (1−ψn)²∥un−x^∗∥²+ψ_n²∥yn−x^∗∥²+ψ_n²λ²_n∥A(un)−A(yn)∥² + 2ψ_n(1−ψ_n)⟨u_n−x^∗, y_n−x^∗⟩

+ 2ψ_n(1−ψ_n)λ_n⟨u_n−x^∗, A(u_n)−A(y_n)⟩+ 2ψ_n²λ_n⟨y_n−x^∗, A(u_n)−A(y_n)⟩. (4.91) Using Lemma 2.5.18 (i), we obtain

2⟨u_n−x^∗, y_n−x^∗⟩=∥u_n−x^∗∥²+∥y_n−x^∗∥²− ∥u_n−y_n∥². (4.92) Combining (4.91) and (4.92), we get

∥zn−x^∗∥² = (1−ψn)∥un−x^∗∥²+ψn∥yn−x^∗∥²+ψ²_nλ²_n∥A(un)−A(yn)∥²

−ψ_n(1−ψ_n)∥u_n−y_n∥²+ 2ψ_n(1−ψ_n)λ_n⟨u_n−x^∗, A(u_n)−A(y_n)⟩

+ 2ψ²_nλ_n⟨y_n−x^∗, A(u_n)−A(y_n)⟩. (4.93)

Now, from (4.86), we obtain Ay_n+ 1

λ_n(u_n−λ_nAu_n−y_n)∈(A+B)y_n.

Since x^∗ ∈Γ, we have that 0∈(A+B)(x^∗). Hence, we obtain from Lemma 2.5.15that

⟨y_n−u_n−λ_n A(y_n)−A(u_n)

, y_n−x^∗⟩ ≤0, which implies that

⟨y_n−u_n, y_n−x^∗⟩ ≤λ_n⟨A(y_n)−A(u_n), y_n−x^∗⟩. (4.94) Now, observe that

∥y_n−x^∗∥² ≤ ∥y_n−u_n∥²+∥u_n−x^∗∥²+ 2⟨y_n−u_n, u_n−x^∗⟩

≤ ∥y_n−u_n∥²+∥u_n−x^∗∥²−2⟨y_n−u_n, y_n−u_n⟩+ 2⟨y_n−u_n, y_n−x^∗⟩

≤ ∥u_n−x^∗∥²− ∥y_n−u_n∥²−2λ_n⟨A(u_n)−A(y_n), y_n−x^∗⟩, (4.95) where the last inequality follows from (4.94). Substituting (4.95) into (4.93), and using (4.90), we obtain

∥z_n−x^∗∥² ≤(1−ψ_n)∥u_n−x^∗∥²+ψ_n

∥u_n−x^∗∥² − ∥u_n−y_n∥²

−2λn⟨A(un)−A(yn), yn−x^∗⟩

−ψn(1−ψn)∥un−yn∥²+ψ_n²λ²_n∥A(un)−A(yn)∥² + 2ψ_n(1−ψ_n)λ_n⟨u_n−x^∗, A(u_n)−A(y_n)⟩+ 2ψ_n²λ_n⟨y_n−x^∗, A(u_n)−A(y_n)⟩

=∥u_n−x^∗∥²−ψ_n(2−ψ_n)∥u_n−y_n∥²+ψ_n²λ²_n∥A(u_n)−A(y_n)∥²

−2ψ_nλ_n⟨A(u_n)−A(y_n), y_n−x^∗⟩+ 2ψ²_nλ_n⟨y_n−x^∗, A(u_n)−A(y_n)⟩

+ 2ψ_n(1−ψ_n)λ_n⟨u_n−x^∗, A(u_n)−A(y_n)⟩

=∥u_n−x^∗∥²−ψ_n(2−ψ_n)∥u_n−y_n∥²+ψ_n²λ²_n∥A(u_n)−A(y_n)∥² + 2ψ_n(1−ψ_n)λ_n⟨A(u_n)−A(y_n), u_n−y_n⟩

≤ ∥u_n−x^∗∥²−ψ_n(2−ψ_n)∥u_n−y_n∥²+ψ²_nλ²_n η²

λ²_n+1∥u_n−y_n∥² + 2ψ_n(1−ψ_n)λ_n η

λ_n+1∥u_n−y_n∥²

=∥un−x^∗∥²−ψn

2−ψn−2(1−ψn)η λ_n

λ_n+1 −ψnη² λ²_n λ²_n+1

∥un−yn∥². (4.96) Since the limit of {λ_n} exists (see Remark 4.2.4 (d)), then lim

n→∞λ_n = lim

n→∞λ_n+1. Also, by Assumption 4.2.2 (2)(d), there exists ψ ∈ (0,1] such that ψ = lim

n→∞ψ_n. Thus, we obtain that

n→∞lim

2−ψ_n−2(1−ψ_n)η λ_n

λ_n+1 −ψ_nη² λ²_n λ²_n+1

= (1−η)(2−ψ+ψη)>0. (4.97) Hence, there exists n₀ ∈ N such that 2−ψ_n−2(1−ψ_n)η_λ^λn

n+1 −ψ_nη²_λ^λ2²ⁿ n+1

>0 ∀n ≥n₀. Thus, we obtain from (4.96) that

∥z_n−x^∗∥ ≤ ∥u_n−x^∗∥ ∀n ≥n₀. (4.98)

Observe from Lemma2.5.25and Lemma2.5.16(ii), (iii), that (I−γ_nG) isγ_n∥T∥²-averaged.

That is, (I−γ_nG) = (1−β_n)I+β_nV_n, ∀ n∈ N, whereβ_n =γ_n∥T∥² ∈(0,1) and V_n is nonexpansive for all n ∈N. Therefore, we can rewriteu_n from (4.86) as:

u_n= (1−β_n)w_n+β_nV_nw_n, n ≥1. (4.99) Thus, we obtain that

∥un−x^∗∥² = (1−βn)∥wn−x^∗∥²+βn∥Vnwn−x^∗∥²−βn(1−βn)∥Vnwn−wn∥²

≤ ∥w_n−x^∗∥²−β_n(1−β_n)∥V_nw_n−w_n∥² (4.100)

≤ ∥w_n−x^∗∥².

Now, from Step 1 and Assumption 4.2.2 (2), we have that α_n∥x_n−x_n−1∥ ≤ϵ_n ∀n ∈ N, which implies that

α_n

θ_n∥x_n−xn−1∥ ≤ ϵ_n

θ_n →0, as n→ ∞. (4.101)

Hence, there exists M₁ >0 such that α_n

θ_n∥xn−xn−1∥ ≤M1 ∀ n ∈N. (4.102) Thus, from Step 2, we obtain

∥w_n−x^∗∥ ≤ ∥x_n−x^∗∥+α_n∥x_n−xn−1∥

=∥x_n−x^∗∥+θ_nα_n

θ_n∥x_n−x_n−1∥

≤ ∥xn−x^∗∥+θnM1, (4.103)

and so,

∥u_n−x^∗∥ ≤ ∥w_n−x^∗∥ ≤ ∥x_n−x^∗∥+θ_nM₁. (4.104) From (4.89), we obtain

∥x_n+1−x^∗∥=∥(1−β_n−θ_n)u_n+β_nz_n−x^∗∥

=∥(1−β_n−θ_n)(u_n−x^∗) +β_n(z_n−x^∗)−θ_nx^∗∥

≤ ∥(1−β_n−θ_n)(u_n−x^∗) +β_n(z_n−x^∗)∥+θ_n∥x^∗∥. (4.105) Note that

∥(1−βn−θn)(un−x^∗) +βn(zn−x^∗)∥²

= (1−β_n−θ_n)²∥u_n−x^∗∥²+ 2(1−β_n−θ_n)β_n⟨u_n−x^∗, z_n−x^∗⟩ +β_n²∥z_n−x^∗∥²

≤(1−β_n−θ_n)²∥u_n−x^∗∥²+ 2(1−β_n−θ_n)β_n∥u_n−x^∗∥.∥z_n−x^∗∥ +β_n²∥z_n−x^∗∥²

≤(1−β_n−θ_n)²∥u_n−x^∗∥²+ (1−β_n−θ_n)β_n∥u_n−x^∗∥² + (1−β_n−θ_n)β_n∥z_n−x^∗∥² +β_n²∥z_n−x^∗∥²

≤(1−βn−θn)(1−θn)∥un−x^∗∥²+ (1−θn)βn∥zn−x^∗∥² (4.106)

≤(1−β_n−θ_n)(1−θ_n)∥u_n−x^∗∥²+ (1−θ_n)β_n∥u_n−x^∗∥²

= (1−θ_n)²∥u_n−x^∗∥²,

which implies from (4.104) and (4.105) that

∥x_n+1−x^∗∥ ≤(1−θ_n)∥x_n−x^∗∥+θ_n(1−θ_n)M₁+θ_n∥x^∗∥

≤(1−θn)∥xn−x^∗∥+θn(∥x^∗∥+M1).

It follows from Lemma 2.5.24 that {x_n} is bounded.

Lemma 4.2.7. Let{x_n}be a sequence generated by Algorithm4.2.3such that Assumption 4.2.2 holds. If there exists a subsequence {x_nk} of {x_n} which converges weakly to a point z ∈H1 and lim

n→∞∥un_k−yn_k∥= 0 = lim

n→∞∥un_k −xn_k∥. Then, z ∈Γ.

Proof. From Step 2 and (4.101), we have

∥w_n−x_n∥=θ_nα_n

θ_n∥x_n−xn−1∥ →0, as n→ ∞. (4.107) Let {x_nk} be a subsequence of{x_n} which is weakly convergent to a point z ∈H₁. Then, the subsequences {u_nk},{y_nk} and {w_nk} are weakly convergent toz ∈H₁.

We first show that T z ∈ F(S). To do this, recall from Remark 4.2.4 that lim

n→∞γ_n ≥ minn

γ₁,_2∥T^µ_∥2

o

>0. Let lim

n→∞γ_n=γ >0,since {T^∗(I−S)T w_nk}is bounded, we get that

∥(I−γ_nkT^∗(I−S)T)w_nk−(I−γT^∗(I−S)T)w_nk∥=|γ_nk−γ|∥T^∗(I−S)T w_nk∥ →0, as k → ∞.

That is, lim

k→∞∥u_nk − (I −γT^∗(I −S)T)w_nk∥ = 0, which implies from (4.107) and our hypothesis that

k→∞lim ∥w_nk− I −γT^∗(I−S)T

w_nk∥= 0. (4.108)

Thus, we obtain from Lemma2.5.55, thatz ∈F I−γT^∗(I−S)T

.Hence, using the same line of argument as in the proof of Lemma 4.2.5, we obtain that T z ∈F(S).

Now, let (v, w)∈G(A+B),then w−Av ∈B(v).Also, we obtain from (4.86) that 1

λ_nk u_nk −λ_nkAu_nk −y_nk

∈B(y_nk).

Thus, by the monotonicity of B, we obtain that

⟨v −y_nk, w−Av− 1

λ_nk u_nk −λ_nkAu_nk −y_nk

⟩ ≥0. (4.109)

Using (4.109) and the monotonicity ofA, we obtain

⟨v−y_nk, w⟩ ≥ ⟨v−y_nk, Av+ 1

λ_nk(u_nk−y_nk)−Au_nk⟩

=⟨v−y_nk, Av−A(y_nk)⟩+⟨v−y_nk, A(y_nk)−A(u_nk)⟩

+⟨v−yn_k, 1

λ_nk(un_k −yn_k)⟩

≥ ⟨v−y_nk, A(y_nk)−A(u_nk)⟩+⟨v−y_nk, 1 λn_k

(u_nk−y_nk)⟩. (4.110)

Recall that lim

k→∞λ_nk ≥ min{λ₁,_L^η} > 0, lim

k→∞∥u_nk −y_nk∥ = 0 and by the Lipschitz continuity of A, lim

k→∞∥A(u_nk)−A(y_nk)∥ = 0. Hence, passing limit as k → ∞in (4.110), we obtain

⟨v−z, w⟩ ≥0.

Also, by Lemma 2.5.15, A+B is maximal monotone, thus, 0 ∈ (A+B)z. This together with the fact that T z ∈F(S) gives that z ∈Γ.

Theorem 4.2.8. Suppose that Assumption4.2.2holds. Then, the sequence{x_n}generated by Algorithm 4.2.3 converges strongly to x^∗ ∈Γ, where

∥x^∗∥= min{∥z∥:z ∈Γ}.

Proof. Let x^∗ ∈Γ. Then, we obtain from (4.98), (4.100) and Step 2 that

∥(1−βn)un+βnzn−x^∗∥² =∥(1−βn)(un−x^∗) +βn(zn−x^∗)∥²

= (1−βn)²∥un−x^∗∥²+β_n²∥zn−x^∗∥²+ 2(1−βn)βn⟨un−x^∗, zn−x^∗⟩

≤(1−β_n)²∥u_n−x^∗∥² +β_n²∥z_n−x^∗∥²+ 2(1−β_n)β_n∥u_n−x^∗∥.∥z_n−x^∗∥

≤(1−β_n)²∥u_n−x^∗∥² +β_n²∥z_n−x^∗∥² + (1−β_n)β_n ∥u_n−x^∗∥²+∥z_n−x^∗∥²

= (1−β_n)∥u_n−x^∗∥²+β_n∥z_n−x^∗∥²

≤ ∥w_n−x^∗∥². (4.111)

On the other hand, we have

∥w_n−x^∗∥² =∥x_n−x^∗∥²+ 2α_n⟨x_n−x^∗, x_n−xn−1⟩+α²_n∥x_n−xn−1∥²

≤ ∥x_n−x^∗∥²+ 2α_n∥x_n−x^∗∥.∥x_n−x_n−1∥+α²_n∥x_n−x_n−1∥²

=∥xn−x^∗∥²+αn∥xn−xn−1∥(2∥xn−x^∗∥+αn∥xn−xn−1∥)

≤ ∥x_n−x^∗∥²+ 3α_n∥x_n−xn−1∥M₂, (4.112) for some M₂ >0. Combining (4.111) and (4.112), we obtain

∥(1−β_n)u_n+β_nz_n−x^∗∥² ≤ ∥x_n−x^∗∥²+ 3α_n∥x_n−xn−1∥M₂. (4.113) Thus, from Step 3 and Lemma 2.5.18(ii), we obtain

∥x_n+1−x^∗∥² =∥(1−θ_n)

(1−β_n)u_n+β_nz_n−x^∗

−

θ_nβ_n(u_n−z_n) +θ_nx^∗

≤(1−θ_n)²∥(1−β_n)u_n+β_nz_n−x^∗∥²−2⟨θ_nβ_n(u_n−z_n) +θ_nx^∗, x_n+1−x^∗⟩

≤(1−θ_n)∥(1−β_n)u_n+β_nz_n−x^∗∥²+ 2⟨θ_nβ_n(u_n−z_n), x^∗ −x_n+1⟩ + 2θn⟨x^∗, x^∗−xn+1⟩

≤(1−θ_n)∥(1−β_n)u_n+β_nz_n−x^∗∥²+ 2θ_nβ_n∥u_n−z_n∥.∥x^∗−x_n+1∥

+ 2θ_n⟨x^∗, x^∗−x_n+1⟩. (4.114)

Therefore, using (4.113) and (4.114), we have

∥x_n+1−x^∗∥² ≤(1−θ_n)∥x_n−x^∗∥²+ 3(1−θ_n)α_n∥x_n−xn−1∥M₂ + 2θ_nβ_n∥u_n−z_n∥.∥x^∗−x_n+1∥+ 2θ_n⟨x^∗, x^∗−x_n+1⟩

= (1−θ_n)∥x_n−x^∗∥² +θ_nh

3α_n

θ_n(1−θ_n)∥x_n−x_n−1∥M₂+ 2β_n∥u_n−z_n∥.∥x^∗−x_n+1∥ + 2⟨x^∗, x^∗ −x_n+1⟩i

= (1−θ_n)∥x_n−x^∗∥²+θ_nd_n, ∀n≥n₀, (4.115) whered_n:=h

3α_n

θ_n(1−θ_n)∥x_n−x_n−1∥M₂+ 2β_n∥u_n−z_n∥.∥x^∗−x_n+1∥+ 2⟨x^∗, x^∗−x_n+1⟩i . To show that the sequence n

∥x_n−x^∗∥o

converges to zero, by Lemma2.5.55, it is enough to show that lim sup

k→∞

d_nk ≤0 (where{d_nk}is a subsequence of{d_n}), for every subsequence n∥x_nk−x^∗∥o

of n

∥x_n−x^∗∥o

satisfying lim inf

k→∞

∥x_nk+1−x^∗∥ − ∥x_nk −x^∗∥

≥0. (4.116)

Now, suppose thatn

∥x_nk−x^∗∥o

is a subsequence of n

∥x_n−x^∗∥o

such that (4.116) holds.

Then, lim inf

k→∞

∥x_nk+1−x^∗∥²− ∥x_nk−x^∗∥²

= lim

k→∞inf

(∥x_nk+1−x^∗∥ − ∥x_nk−x^∗∥)(∥x_nk+1−x^∗∥+∥x_nk−x^∗∥)

≥0. (4.117) Now, by Step 3, we obtain

∥x_n+1−x^∗∥² =∥(1−β_n−θ_n)(u_n−x^∗) +β_n(z_n−x^∗)−θ_nx^∗∥²

=∥(1−β_n−θ_n)(u_n−x^∗) +β_n(z_n−x^∗)∥²−2θ_n⟨(1−β_n−θ_n)(u_n−x^∗) +β_n(z_n−x^∗), x^∗⟩+θ_n²∥x^∗∥²

≤ ∥(1−β_n−θ_n)(u_n−x^∗) +β_n(z_n−x^∗)∥²+θ_nM₃, (4.118) for some M₃ >0. Substituting (4.106) into (4.118), we have

∥x_n+1−x^∗∥² ≤(1−β_n−θ_n)(1−θ_n)∥u_n−x^∗∥²+ (1−α_n)β_n∥z_n−x^∗∥²+θ_nM₃. (4.119)

Using (4.96), (4.112) and (4.119), we obtain

∥x_n+1−x^∗∥²

≤(1−β_n−θ_n)(1−θ_n)∥u_n−x^∗∥²+ (1−θ_n)β_n ∥u_n−x^∗∥²

−ψ_n

2−ψ_n−2(1−ψ_n)η λn

λ_n+1 −ψ_nη² λ²_n λ²_n+1

∥u_n−y_n∥²

!

+θ_nM₃

= (1−θ_n)²∥u_n−x^∗∥²−(1−θ_n)β_nψ_n

2−ψ_n−2(1−ψ_n)η λ_n

λ_n+1 −ψ_nη² λ²_n λ²_n+1

∥u_n−y_n∥² +θ_nM₃

≤(1−θ_n)²

∥x_n−x^∗∥²+ 3α_n∥x_n−xn−1∥M₂

−(1−θ_n)β_nψ_n

2−ψ_n−2(1−ψ_n)η λ_n λ_n+1

−ψ_nη² λ²_n λ²_n+1

∥u_n−y_n∥²+θ_nM₃

≤ ∥x_n−x^∗∥²+θ_n

3α_n θn

∥x_n−xn−1∥M₂+M₃

−(1−θ_n)β_nψ_n

2−ψ_n−2(1−ψ_n)η λ_n λn+1

−ψnη² λ²_n λ²_n+1

∥un−yn∥².

Hence, by (4.102) and (4.117), we obtain lim sup

k→∞

(1−θ_nk)β_nkψ_nk

2−ψ_nk −2(1−ψ_nk)η λ_nk

λ_nk+1 −ψ_nkη² λ²_nk λ²_n

k+1

∥u_nk −y_nk∥²

≤lim sup

k→∞

(

∥x_nk −x^∗∥²− ∥x_nk+1−x^∗∥²+θ_nk

3M₁M₂+M₃ )

= lim sup

k→∞

n∥x_nk−x^∗∥²− ∥x_nk+1−x^∗∥²o

=−lim inf

k→∞

n∥x_nk+1−x^∗∥²− ∥x_nk−x^∗∥²o

≤0, which implies from (4.97) that

k→∞lim ∥u_nk−y_nk∥= 0. (4.120) Using (4.119), (4.98) and (4.100), we obtain

∥xn+1−x^∗∥² ≤(1−βn−θn)(1−θn)∥un−x^∗∥²+ (1−θn)βn∥zn−x^∗∥²+θnM3

≤(1−θ_n)²∥u_n−x^∗∥²+θ_nM₃

≤ ∥w_n−x^∗∥²−β_n(1−β_n)∥V_nw_n−w_n∥²+θ_nM₃

≤ ∥x_n−x^∗∥² + 3α_n∥x_n−xn−1∥M₂−β_n(1−β_n)∥V_nw_n−w_n∥²+θ_nM₃.

Thus, by (4.117) we obtain lim sup

k→∞

h

β_nk(1−β_nk)∥V_nkw_nk −w_nk∥²i

≤lim sup

k→∞

"

∥x_nk −x^∗∥²− ∥x_nk+1−x^∗∥²

+θ_nk

3M₁M₂+M₃

#

=−lim inf

k→∞

h∥x_nk+1−x^∗∥²− ∥x_nk −x^∗∥²i

≤0, which implies that

k→∞lim ∥Vn_kwn_k−wn_k∥= 0. (4.121) Thus, we obtain from (4.99) that

k→∞lim ∥u_nk −w_nk∥= lim

k→∞β_nk∥V_nkw_nk −w_nk∥= 0. (4.122) Using (4.122) and (4.107), we obtain

k→∞lim ∥u_nk −x_nk∥= 0. (4.123) Observe from Step 3 and (4.120) that

∥z_nk −y_nk∥= (1−ψ_nk)∥u_nk −y_nk∥+ψ_nk∥λ_nk(Au_nk −Ay_nk)∥

≤(1−ψn_k)∥un_k −yn_k∥+ψn_kλn_kL∥un_k −yn_k∥ →0 as k → ∞. (4.124) Thus, from (4.120), we obtain

∥z_nk −u_nk∥= 0. (4.125)

Also, from Step 3, we obtain

∥x_nk+1−u_nk∥ ≤β_nk∥z_nk−u_nk∥+θ_nk∥u_nk∥ →0 as k→ ∞. (4.126) Thus, we obtain from (4.123) and (4.126) that

k→∞lim ∥x_nk+1 −x_nk∥= 0. (4.127) Since

xn_k is bounded, there exists a subsequence

xn_kj of

xn_k which converges weakly to some z ∈H₁, such that

lim sup

k→∞

⟨x^∗, x^∗−x_nk⟩= lim

j→∞⟨x^∗, x^∗−x_nkj⟩=⟨x^∗, x^∗ −z⟩. (4.128) Also, we obtain from (4.120), (4.123) and Lemma4.2.7 that z ∈Γ.

Thus, since x^∗ =P_Γ0, we obtain from (4.128) that lim sup

k→∞

⟨x^∗, x^∗−x_nk⟩=⟨x^∗, x^∗−z⟩ ≤0,

which implies from (4.127) that lim sup

k→∞

⟨x^∗, x^∗−x_nk+1⟩ ≤0. (4.129) Now, recall that dn_k :=

h 3αn_k

θ_nk(1−θn_k)∥xn_k−xn_k−1∥M2+ 2βn_k∥un_k−zn_k∥.∥x^∗−xn_k+1∥+ 2⟨x^∗, x^∗−x_nk+1⟩i

.

Thus, by (4.129), (4.101) and (4.125), we obtain lim sup

k→∞

d_nk ≤ 0. Hence, we get that

n→∞lim ∥x_n−x^∗∥= 0. Therefore, we conclude that{x_n} converges strongly to x^∗ =P_Γ0.

Some Corollaries

If we set B =NC in Algorithm 4.2.3, then we obtain from the definition of normal cone (see (2.5.44)) thatJ_λ^NC(I−λA) = P_C(I−λA).In this case, we know that (A+N_C)⁻¹(0) = V IP(C, A). Thus, we obtain the following consequent result.

Algorithm 4.2.9. Initialization: Choose sequences{θ_n}^∞_n=1, {β_n}^∞_n=1,{ϵ_n}^∞_n=1 and{ψ_n}^∞_n=1 such that the conditions from Assumption 4.2.2 (2) hold and let γ₁, λ₁ > 0, µ, η ∈ (0,1), α≥3 and x0, x1 ∈H1 be given arbitrarily.

Iterative Steps: Set n := 1.

Step 1. Given the iterates x_n−1 and x_n (n ≥1), choose α_n such that 0≤α_n ≤α¯_n, where

¯ α_n :=







min _n−1

n+α−1,_∥x ^ϵn

n−xn−1∥ , if x_n ̸=x_n−1

n−1

n+α−1, otherwise.

(4.130)

Step 2. Set

w_n =x_n+α_n(x_n−xn−1).

Then, compute

u_n =w_n−γ_nGw_n and y_n=P_C(I−λ_nA)u_n, (4.131) where

γ_n+1=

(minn _µ∥u

n−wn∥

∥Gu_n−Gw_n∥, γ_no

, if Gu_n̸=Gw_n,

γn, otherwise

(4.132)

and

λ_n+1 =

(minn _η∥u

n−y_n∥

∥Aun−Ayn∥, λ_no

, if Au_n ̸=Ay_n,

λ_n, otherwise.

(4.133)

Step 3. Compute

x_n+1 = (1−β_n−θ_n)u_n+β_nz_n, (4.134) where z_n:= (1−ψ_n)u_n+ψ_n

y_n+λ_n A(u_n)−A(y_n)

.

Stopping criterion: If y_n =u_n =w_n =x_n, then stop, otherwise, set n :=n+ 1 and go back to Step 1.

Corollary 4.2.10. Suppose that Assumption 4.2.2 holds with B = N_C and Γ := {z ∈ V IP(C, A) : T z ∈ F(S)}. Then, the sequence {x_n} generated by Algorithm 4.2.9 con- verges strongly to x^∗ ∈Γ, where ∥x^∗∥= min{∥z∥:z ∈Γ}.

If A≡0, then we obtain the following consequent result.

Algorithm 4.2.11. Initialization: Choose sequences {θ_n}^∞_n=1, {β_n}^∞_n=1, {ϵ_n}^∞_n=1, {ψ_n}^∞_n=1 such that the conditions from Assumption 4.2.2 (2) hold and letγ1 >0, µ∈(0,1), α≥3 and x₀, x₁ ∈H₁ be given arbitrarily.

Iterative Steps: Set n := 1.

Step 1. Given the iterates xn−1 and x_n (n ≥1), choose α_n such that 0≤α_n ≤α¯_n, where

¯ α_n :=







min _n−1

n+α−1,_∥x ^ϵn

n−xn−1∥ , if x_n ̸=xn−1 n−1

n+α−1, otherwise.

(4.135)

Step 2. Set

w_n =x_n+α_n(x_n−xn−1).

Then, compute

u_n=w_n−γ_nGw_n and y_n=J_λ^B

nu_n, (4.136)

where inf

n≥1λ_n≥λ >0 and γ_n+1=

(minn _µ∥u

n−wn∥

∥Gu_n−Gw_n∥, γ_no

, if Gu_n̸=Gw_n.

γn, otherwise

(4.137)

Step 3. Compute

x_n+1 = (1−β_n−θ_n)u_n+β_nz_n, (4.138) where zn:= (1−ψn)un+ψnyn.

Stopping criterion: If y_n =u_n =w_n =x_n, then stop, otherwise, set n :=n+ 1 and go back to Step 1.

Corollary 4.2.12. Suppose that Assumption 4.2.2 holds with A ≡ 0 and Γ := {z ∈ B⁻¹(0) : T z ∈ F(S)}. Then, the sequence {x_n} generated by Algorithm 4.2.11 converges strongly to x^∗ ∈Γ, where ∥x^∗∥= min{∥z∥:z ∈Γ}.

Remark 4.2.13. If we set H₁ =H₂ =H and T =I =S in Algorithms 4.2.3, 4.2.9 and 4.2.11, we recover relaxed inertial methods for solving the classical MVIP (see [17]), VIP (see [14]) and NPP (see [126]), respectively. Furthermore, under appropriate settings, Al- gorithms 4.2.3, 4.2.9and 4.2.11reduce to relaxed inertial methods for solving the classical SMVIP of Moudafi [171], SVIP of Censor et al. [54] and SCNPP, respectively.