Main result

5.4 An inertial extrapolation method for solving

5.4.1 Main result

5.4 An inertial extrapolation method for solving

Step 2. Set

w_n=x_n+θ_n(x_n−xn−1).

Then, compute

un=wn−τnGwn and zn=PC(un−A(un)). (5.96) Step 3. Computey_n:=u_n−η_n(u_n−z_n),whereη_n:=α^mn andm_n is the smallest nonnegative

integer satisfying

⟨A(y_n), u_n−z_n⟩ ≥ σ

2||u_n−z_n||². (5.97) Step 4. Compute

x_n+1 = (1−β_n−α_n)u_n+β_nP_Cn(u_n), (5.98) where C_n :=

x∈H₁ :F_n(x)≤0 and

F_n(x) := ⟨A(y_n), x−y_n⟩. (5.99) Update:

τ_n+1 =

(minn _δ||w

n−un||

||Gw_n−Gu_n||, τ_no

, if Gw_n̸=Gu_n,

τn, otherwise.

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

Remark 5.4.3.

(a) Assumption 5.4.1 (1) (b) requires that the operator A is pseudomonotone and uni- formly continuous which is much more, a weaker assumption than the monotonicity and Lipschitz continuity assumption used in [228,229], as well as the inverse strongly monotonicity assumptions used in other papers for solving split variational inequality problems (see for example [54, 119, 136, 137, 178]). Thus, our method is applicable for a much more general class of Pseudomonotone and uniformly continuous opera- tors.

(b) In addition, our method requires at each iteration, only one projection onto C and another projection onto the half-space C_n, which can be easily computed. In fact,

P_Cn(u_n) :=

(u_n− ^⟨A(y_||A(y^{n),un−yn⟩}

n)||² A(y_n), if ⟨A(y_n), u_n−y_n⟩>0, u_n, if ⟨A(y_n), u_n−y_n⟩ ≤0.

Hence, our method is less computationally expensive than other methods in the lit- erature [54, 228, 229], for solving the split variational inequality problems.

Remark 5.4.4. If {y_n} is bounded, then F_n is Lipschitz continuous.

Indeed, if{y_n}is bounded, then by Lemma 2.5.50, there exists M >¯ 0such that∥A(y_n)∥ ≤ M¯ for all n≥1. Thus, for each x, y ∈C, we obtain

∥F_n(x)−F_n(y)∥=∥⟨A(y_n), x−y_n⟩ − ⟨A(y_n), y −y_n⟩∥

=∥⟨A(y_n), x−y⟩∥

≤ ∥A(yn)∥∥x−y∥

≤M∥x¯ −y∥.

Therefore, F_n is Lipschitz continuous.

We now show that the stopping criterion of Algorithm 5.4.2 is valid.

Lemma 5.4.5. If z_n=w_n =u_n=x_n in Algorithm 5.4.2, then x_n ∈Γ.

Proof. If z_n=w_n=u_n=x_n, then from (5.96), it is clear that x_n =P_C(x_n−A(x_n)).

Also, we obtain that x_n = x_n − τ_nGx_n = x_n − τ_nT^∗(I − S)T x_n, which implies that T^∗(I−S)T x_n = 0. That is,

ST xn =T xn+ ¯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 xn = T xn, which gives that T x_n∈F(S).Therefore, x_n∈Γ.

Next, we show that the limit of the stepsize {τ_n} generated by (5.100) exists.

Lemma 5.4.6. The limit of the stepsize {τ_n} exists and lim

n→∞τ_n>0.

Proof. From (5.100), it is obvious that τ_n+1 ≤τ_n ∀n ∈N.

Also, we know by Lemma 2.5.25 that G is 2∥T∥²-Lipschitz continuous. Thus, we get in the case of Gu_n̸=Gw_n that

τ_n+1 = min

δ∥wn−un∥

∥Gw_n−Gu_n∥, τ_n

≥min δ

2∥T∥², τ_n

. Hence, by induction, we obtain that {τ_n} is bounded below by minn

δ

2∥T∥², τ₁o

. Hence, the limit of {τ_n} exists and lim

n→∞τ_n≥minn

τ₁,_2∥T^δ_∥2

o

>0.

Lemma 5.4.7. Let Assumption 5.4.1 hold and let {x_n} be a sequence generated by Algo- rithm 5.4.2. Assume that the subsequence {x_nk} of {x_n} converges weakly to a point x^∗, and lim

k→∞∥x_nk −u_nk∥= 0, lim

k→∞∥u_nk −w_nk∥= 0 and lim

k→∞∥u_nk −z_nk||= 0, then x^∗ ∈Γ.

Proof. By Lemma 2.5.48 we obtain

⟨u_nk −A(u_nk)−z_nk, x−z_nk⟩ ≤0, ∀ x∈C, which implies that

⟨u_nk −z_nk, x−z_nk⟩ ≤ ⟨A(u_nk), x−z_nk⟩, ∀ x∈C.

Hence

⟨u_nk−z_nk, x−z_nk⟩+⟨A(u_nk), z_nk−u_nk⟩ ≤ ⟨A(u_nk), x−u_nk⟩, ∀x∈C. (5.101) Fix x∈C and let k → ∞in (5.101). Since lim

k→∞∥u_nk−z_nk||= 0, we have 0≤lim inf

k→∞ ⟨A(u_nk), x−u_nk⟩ ∀x∈C. (5.102) Now, choose a sequence{δ_k}of positive numbers such thatδ_k+1 ≤δ_k, ∀ k ≥1 and δ_k → 0 as k → ∞.Then, for each δ_k,we denote by N_k (which exists as a result of (5.102)) the smallest positive integer such that

⟨A(unj), x−unj⟩+δk ≥0 ∀j ≥Nk. (5.103) Since {δ_k} is decreasing, we have that {N_k} is increasing. Furthermore, we set for each k ≥1, m_Nk = _∥A(u^A(uNk)

Nk)∥², providedA(u_Nk)̸= 0.Then it is easy to see that⟨A(u_Nk), m_Nk⟩= 1 for each k≥1. Thus, by (5.103), we have that

⟨A(u_Nk), x+δ_km_Nk−u_Nk⟩ ≥0, which implies by the pseudo-monotonicity of A that

⟨A(x+δ_km_Nk), x+δ_km_Nk−u_Nk⟩ ≥0. (5.104) Since {xn_k} converges weakly to x^∗, we obtain by our hypothesis that {un_k}, {zn_k} and {w_nk} also converge weakly to x^∗. Thus, by the sequentially weakly continuity of A, we have that {A(u_nk)}converges weakly to A(x^∗). IfA(x^∗) = 0, then x^∗ ∈V I(C, A).On the other hand, if we suppose thatA(x^∗)̸= 0,then by the weakly lower semicontinuity of∥ · ∥, we obtain that

0<∥A(x^∗)∥ ≤lim inf

k→∞ ∥A(u_nk)∥.

Since {u_Nk} ⊂ {u_nk}, we obtain that 0≤lim sup

k→∞

∥δ_km_Nk∥= lim sup

k→∞

δ_k

∥A(u_nk)∥

≤

lim sup

k→∞

δ_k lim inf

k→∞ ∥A(u_nk)∥

≤ 0

∥A(x^∗)∥ = 0.

Therefore, lim

k→∞∥δ_km_Nk∥= 0. Thus, lettingk → ∞ in (5.104) yields

⟨A(x), x−x^∗⟩ ≥0 ∀x∈C, (5.105) which implies by Lemma 2.5.9 that x^∗ ∈V I(C, A).

Now, to show that T x^∗ ∈ F(S), recall that Lemma 5.4.6 gives that lim

n→∞τn = τ > 0.

Furthermore, sinceG≡T^∗(I−S)T is Lipschitz continuous,{T^∗(I−S)T w_nk}is bounded.

Hence, we get that

∥(I−τ_nkT^∗(I−S)T)w_nk−(I−τ T^∗(I−S)T)w_nk∥=|τ_nk−τ|∥T^∗(I−S)T w_nk∥ →0, as k → ∞.

That is, lim

k→∞∥u_nk−(I−τ T^∗(I −S)T)w_nk∥= 0, which implies by our hypothesis that

k→∞lim ||w_nk− I−τ T^∗(I−S)T

w_nk||= 0. (5.106) Thus, we obtain from Lemma2.5.55, thatx^∗ ∈F I−τ T^∗(I−S)T

.Hence, using the same line of argument as in the proof of Lemma 5.4.5, we obtain that T x^∗ ∈ F(S). Therefore x^∗ ∈Γ.

Lemma 5.4.8. Suppose Assumption 5.4.1 holds. Then, the sequence {x_n} generated by Algorithm 5.4.2is bounded.

Proof. From Lemma 2.5.16 (ii), (iii), and (iv), u_n can be written in the form

u_n = (1−β_n)w_n+β_nV_nw_n, (5.107) where β_n=τ_n||T||² and V_n is a nonexpansive mapping for eachn ∈N. That is, (I−τ_nG) is τ_n||T||²-averaged.

Now, let x^∗ ∈Γ, we obtain from (5.107) that

||u_n−x^∗||² =||(1−β_n)w_n+β_nV_nw_n−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 (5.108)

≤ ||w_n−x^∗||².

On the other hand,

||w_n−x^∗||=||x_n+θ_n(x_n−xn−1)−x^∗||

≤ ||x_n−x^∗||+θ_n||x_n−xn−1||

=||x_n−x^∗||+α_nθ_n αn

||x_n−x_n−1||.

Now from (5.95), we observe that θ_n

α_n||x_n−xn−1|| ≤ ϵ_n

α_n → 0. Thus there exists M₁ >0 such that θ_n

α_n||x_n−xn−1|| ≤M₁ ∀n ≥1, and so,

||un−x^∗|| ≤ ||wn−x^∗|| ≤ ||xn−x^∗||+αnM1. (5.109) From (5.98), we obtain

||xn+1−x^∗||=||(1−βn−αn)un+βnPCnun−x^∗||

=||(1−β_n−α_n)(u_n−x^∗) +β_n(P_Cnu_n−x^∗)−α_nx^∗||

≤ ||(1−β_n−α_n)(u_n−x^∗) +β_n(P_Cnu_n−x^∗)||+α_n||x^∗||. (5.110) Note that

||(1−β_n−α_n)(u_n−x^∗) +β_n(P_Cnu_n−x^∗)||²

= (1−β_n−α_n)²||u_n−x^∗||²+ 2(1−β_n−α_n)β_n⟨u_n−x^∗, P_Cnu_n−x^∗⟩ +β_n²||P_Cnu_n−x^∗||²

≤(1−β_n−α_n)²||u_n−x^∗||²+ 2(1−β_n−α_n)β_n||u_n−x^∗||||P_Cnu_n−x^∗||

+β_n²||PCnun−x^∗||²

≤(1−β_n−α_n)²||u_n−x^∗||²+ (1−β_n−α_n)β_n||u_n−x^∗||² + (1−β_n−α_n)β_n||P_Cnu_n−x^∗||²+β_n²||P_Cnu_n−x^∗||²

≤(1−β_n−α_n)(1−α_n)||u_n−x^∗||²+ (1−α_n)β_n||P_Cnu_n−x^∗||² (5.111)

≤(1−β_n−α_n)(1−α_n)||u_n−x^∗||²+ (1−α_n)β_n||u_n−x^∗||²

= (1−α_n)²||u_n−x^∗||², which implies from (5.109) that

||(1−β_n−α_n)(u_n−x^∗) +β_n(P_Cnu_n−x^∗)||= (1−α_n)||x_n−x^∗||+ (1−α_n)α_nM₁

≤(1−α_n)||x_n−x^∗||+α_nM₁. (5.112) Now, combining (5.110) and (5.112) gives

||x_n+1−x^∗|| ≤(1−α_n)||x_n−x^∗||+α_n||x^∗||+α_nM₁

= (1−αn)||xn−x^∗||+αn(||x^∗||+M1).

It follows from Lemma 2.5.24 that {x_n} is bounded. Consequently, {A(u_n)},{u_n},{w_n}, {z_n} and {A(y_n)}are also bounded.

Lemma 5.4.9. Let{x_n}be generated by Algorithm5.4.2such that Assumption5.4.1holds.

Assume that lim

k→∞||P_Cnku_nk−u_nk||= 0 = lim

k→∞η_nk||u_nk−z_nk||², then lim

k→∞∥u_nk−z_nk∥= 0.

Proof. From Step 3, we have that ηn = α^mn with α ∈ (0,1). Hence, {ηn} ⊂ (0,1) is bounded. Thus, there exists a subsequence {η_nk} of {η_n} such that lim inf

k→∞ η_nk ≥0.

Case 1. Suppose that lim inf

k→∞ η_nk >0. Then

0≤ ∥u_nk−z_nk∥² = η_nk∥u_nk−z_nk∥² η_nk , which implies that

lim sup

k→∞

∥u_nk −z_nk∥² ≤lim sup

k→∞

η_nk∥u_nk−z_nk∥²

lim sup

k→∞

1 η_nk

=

lim sup

k→∞

η_nk∥u_nk −z_nk∥² 1 lim inf

k→∞ η_nk

= 0.

Hence, lim sup

k→∞

∥u_nk−z_nk∥= 0. Therefore, lim

k→∞∥u_nk −z_nk∥= 0.

Case 2 . Suppose that lim inf

k→∞ η_nk = 0.Then, we may assume without loss of generality that

k→∞lim η_nk = 0 and lim

k→∞∥u_nk −z_nk∥=a≥0.

Define ¯y_nk−u_nk := 1

αη_nk(z_nk−u_nk).Since {z_nk−u_nk}is bounded and lim

k→∞η_nk = 0, then

k→∞lim ∥¯y_nk −u_nk∥= 0. (5.113) From the step-size rule and definition of ¯y_k, we have

⟨A(¯y_nk), u_nk−z_nk⟩< σ

2∥u_nk −z_nk∥², ∀k ∈N. This implies that,

2⟨A(u_nk), u_nk−z_nk⟩+ 2⟨A(¯y_nk)−A(u_nk), u_nk −z_nk⟩< σ∥u_nk −z_nk∥², ∀k ∈N. Setting µ_nk :=u_nk−A(u_nk), we obtain

2⟨u_nk−µ_nk, u_nk −z_nk⟩+ 2⟨A(¯y_nk)−A(u_nk), u_nk−z_nk⟩< σ∥u_nk−z_nk∥², ∀k ∈N. Using Lemma 2.5.18(ii) we have

2⟨u_nk−µ_nk, u_nk−z_nk⟩=∥u_nk−z_nk∥²+∥u_nk −µ_nk∥² − ∥z_nk −µ_nk∥².

Therefore,

∥u_nk−µ_nk∥²− ∥z_nk−µ_nk∥² <(σ−1)∥u_nk−z_nk∥²−2⟨A(¯y_nk)−A(u_nk), u_nk −z_nk⟩,

∀k∈N. Since A is uniformly continuous on bounded subsets of C and (5.113), if a > 0 then the right hand side of the inequality above converges to (σ −1)a < 0 as k → ∞. From the last inequality, we obtain

lim sup

k→∞

∥un_k−µn_k∥²− ∥zn_k −µn_k∥²

≤(σ−1)a <0.

For ϵ= −(σ−1)a

2 >0,there exists N ∈N such that

∥un_k −µn_k∥²− ∥zn_k −µn_k∥² ≤(σ−1)a+ϵ= (σ−1)a

2 ≤0 ∀k∈N, k ≥N; Thus

∥u_nk −µ_nk∥<∥z_nk −µ_nk∥ ∀k ∈N, k ≥N.

which is a contradiction to the definition of z_nk = P_C u_nk −A(u_nk)

. Hence a = 0, the proof is complete.

Theorem 5.4.10. Suppose that Assumption 5.4.1 holds. Then, the sequence {x_n} gener- ated by Algorithm 5.4.2 converges strongly to the point x^∗ ∈Γ, where

∥x^∗∥= min{∥¯x∥: ¯x∈Γ}.

Proof. Using Lemma 2.5.8 and Lemma 2.5.49, we obtain

||P_Cn(u_n)−x^∗|| ≤ ||u_n−x^∗||²− ||P_Cn(u_n)−u_n||²

=||u_n−x^∗||² −dist²(u_n, C_n)

≤ ||u_n−x^∗||²−1

mF_n(u_n)2

=||u_n−x^∗||² −1

m⟨A(y_n), u_n−y_n⟩2

=||u_n−x^∗||² −η_n

m⟨A(y_n), u_n−z_n⟩2

≤ ||u_n−x^∗||²−σηn

2m||u_n−z_n||²2

. (5.114)

Furthermore, we have

xn+1 = (1−βn−αn)un+βnPCn(un) = (1−βn)un+βnPCn(un)−αnun.

Let ϕ_n = (1−β_n)u_n+β_nP_Cn(u_n), then we obtain

||ϕ_n−x^∗||² =||(1−β_n)u_n+β_nP_Cn(u_n)−x^∗||²

=||(1−β_n)(u_n−x^∗) +β_n(P_Cn(u_n)−x^∗)||²

= (1−β_n)²||u_n−x^∗||² +β_n²||P_Cn(u_n)−x^∗||²+ 2(1−β_n)β_n⟨u_n−x^∗, P_Cn(u_n)−x^∗⟩

≤(1−β_n)²||u_n−x^∗||²+β_n²||P_Cn(u_n)−x^∗||²+ 2(1−β_n)β_n||u_n−x^∗||||P_Cn(u_n)−x^∗||

≤(1−βn)²||un−x^∗||²+β_n²||PCn(un)−x^∗||²+ (1−βn)βn||un−x^∗||² (5.115) + (1−β_n)β_n||P_Cn(u_n)−x^∗||²

= (1−β_n)||u_n−x^∗||²+β_n||P_Cn(u_n)−x^∗||²

≤ ||w_n−x^∗||². (5.116)

On the other hand, we have

||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−xn−1⟩+θ_n²||x_n−xn−1||²

≤ ||x_n−x^∗||²+ 2θ_n||x_n−x^∗||||x_n−xn−1||+θ_n²||x_n−xn−1||²

=||x_n−x^∗||²+θ_n||x_n−xn−1||(2||x_n−x^∗||+θ_n||x_n−xn−1||)

≤ ||x_n−x^∗||²+θ_n||x_n−x_n−1||M₂, (5.117) for some M₂ >0. Combining (5.116) and (5.117) we obtain

||ϕ_n−x^∗||² ≤ ||x_n−x^∗||²+θ_n||x_n−xn−1||M₂. (5.118) Since ϕ_n = (1−β_n)u_n+β_nP_Cn(u_n), we have u_n−ϕ_n = β_n(u_n−P_Cn(u_n)). Therefore it follows that

x_n+1 =ϕ_n−α_nu_n = (1−α_n)ϕ_n−α_n(u_n−ϕ_n) = (1−α_n)ϕ_n−α_nβ_n(u_n−P_Cn(u_n)).

Thus, from Lemma 2.5.18(ii), we obtain

||x_n+1−x^∗||² =||(1−α_n)ϕ_n−α_nβ_n(u_n−P_Cn(u_n))−x^∗||²

=||(1−α_n)(ϕ_n−x^∗)− α_nβ_n(u_n−P_Cn(u_n)) +α_nx^∗

||²

≤(1−α_n)²||ϕ_n−x^∗||²−2⟨α_nβ_n(u_n−P_Cn(u_n)) +α_nx^∗, x_n+1−x^∗⟩

≤(1−αn)||ϕn−x^∗||²+ 2⟨αnβn(un−PCn(un)), x^∗−xn+1⟩ + 2α_n⟨x^∗, x^∗−x_n+1⟩

≤(1−α_n)||ϕ_n−x^∗||²+ 2α_nβ_n||u_n−P_Cn(u_n)||||x^∗−x_n+1||

+ 2α_n⟨x^∗, x^∗−x_n+1⟩. (5.119)

Therefore, using (5.118) and (5.119), we have

||x_n+1−x^∗||² ≤(1−α_n)||x_n−x^∗||²+ (1−α_n)θ_n||x_n−xn−1||M₂ + 2α_nβ_n||u_n−P_Cn(u_n)||||x^∗−x_n+1||+ 2α_n⟨x^∗, x^∗−x_n+1⟩

= (1−α_n)||x_n−x^∗||²+α_nhθ_n

α_n(1−α_n)||x_n−xn−1||M₂+ 2β_n||u_n−P_Cn(u_n)||||x^∗−x_n+1||

+ 2⟨x^∗, x^∗−xn+1⟩i

= (1−α_n)||x_n−x^∗||²+α_nd_n, (5.120)

whered_n :=hθ_n

α_n(1−α_n)||x_n−x_n−1||M₂+2β_n||u_n−P_Cn(u_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.

Now, suppose that n

||xn_k−x^∗||o

is a subsequence of n

||xn−x^∗||o

such that lim inf

k→∞

||x_nk+1−x^∗|| − ||x_nk −x^∗||

≥0, 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. (5.121) Now,

||x_n+1−x^∗||² =||(1−β_n−α_n)u_n+β_nP_Cn(u_n)−x^∗||²

=||(1−β_n−α_n)(u_n−x^∗) +β_n(P_Cn(u_n)−x^∗)−α_nx^∗||²

=||(1−β_n−α_n)(u_n−x^∗) +β_n(P_Cn(u_n)−x^∗)||²

−2α_n⟨(1−β_n−α_n)(u_n−x^∗) +β_n(P_Cn(u_n)−x^∗), x^∗⟩+α²_n||x^∗||²

≤ ||(1−β_n−α_n)(u_n−x^∗) +β_n(P_Cn(u_n)−x^∗)||²+α_nM₃, (5.122) for some M₃ >0. Substituting (5.111) into (5.122), we have

||x_n+1−x^∗||² ≤(1−β_n−α_n)(1−α_n)||u_n−x^∗||²+ (1−α_n)β_n||P_Cn(u_n)−x^∗||²+α_nM₃. (5.123) Using (5.114), (5.117), and (5.123), we have

||xn+1−x^∗||² ≤(1−βn−αn)(1−αn)||un−x^∗||² + (1−αn)βn||un−x^∗||²

−(1−α_n)β_nσηn

2m||u_n−z_n||²2

+α_nM₃

= (1−αn)²||un−x^∗||²−(1−αn)βn

ση_n

2m||un−zn||²2

+αnM3

≤ ||w_n−x^∗||²−(1−α_n)β_n ση_n

2m||u_n−z_n||² 2

+α_nM₃

≤ ||x_n−x^∗||²+θ_n||x_n−xn−1||M₂ −(1−α_n)β_n ση_n

2m||u_n−z_n||² 2

+α_nM₃

=||x_n−x^∗||²+α_n θ_n

α_n∥x_n−xn−1∥M₂+M₃

−(1−α_n)β_n ση_n

2m||u_n−z_n||² 2

.

Hence

(1−αn)βn

ση_n

2m||un−zn||² 2

≤ ||xn−x^∗||²− ||xn+1−x^∗||² +αn

θ_n

α_n∥xn−xn−1∥M2+M3

. (5.124)

By (5.121) and (5.124), we obtain lim sup

k→∞

(1−α_nk)β_nk σηn_k

2m ||u_nk −z_nk||² 2

≤lim sup

k→∞

n

||xn_k −x^∗||²− ||xn_k+1−x^∗||²o

+ lim sup

k→∞

αn_k

M1M2+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 that

k→∞lim η_nk||u_nk −z_nk||² = 0. (5.125) Now, from Step 3, we have

y_nk =u_nk −α^mn

u_nk−z_nk

=u_nk−η_nk

u_nk−z_nk . Thus, by (5.125), we obtain that

k→∞lim ∥u_nk−y_nk∥= 0. (5.126) Also, observe that equation (5.111) and (5.122), gives

∥x_n+1−x^∗∥² ≤(1−α_n)∥u_n−x^∗∥²+α_nM₃. Substituting (5.108) into the last inequality above, we obtain

∥x_n+1−x^∗∥² ≤(1−α_n)h

∥w_n−x^∗∥² −β_n(1−β_n)∥V_nw_n−w_n∥²i

+α_nM₃

≤(1−α_n)h

∥x_n−x^∗∥²+θ_n∥x_n−x_n−1∥M₂i

−β_n(1−β_n)(1−α_n)∥V_nw_n−w_n∥² +α_nM₃.

Hence

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

θ_n αn

∥x_n−x_n−1∥M₂ +M₃

. (5.127)

By (5.121) and (5.127), we obtain lim sup

k→∞

h

β_nk(1−β_nk)∥V_nkw_nk −w_nk∥²i

≤lim sup

k→∞

h∥x_nk −x^∗∥²− ∥x_nk+1−x^∗∥²i + lim sup

k→∞

α_nk

M₁M₂+M₃

=−lim inf

k→∞

h

∥xn_k+1−x^∗∥²− ∥xn_k −x^∗∥²i

≤0, which implies that

k→∞lim ∥V_nkw_nk−w_nk∥= 0. (5.128) Thus, we obtain from (5.107) that

k→∞lim ∥u_nk −w_nk∥= lim

k→∞β_nk∥V_nkw_nk −w_nk∥= 0. (5.129) Also, combining (5.117) and (5.123) we have

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

−(1−α_n)β_n∥P_Cnu_n−u_n∥²+α_nM₃

= (1−α_n)²∥u_n−x^∗∥²−(1−α_n)β_n∥P_Cnu_n−u_n∥²+α_nM₃

≤ ∥x_n−x^∗∥²+θ_n∥x_n−xn−1∥M₂−(1−α_n)β_n∥P_Cnu_n−u_n∥²+α_nM₃

≤ ∥x_n−x^∗∥²+α_n θn

α_n∥x_n−xn−1∥M₂+M₃

−(1−α_n)β_n∥P_Cnu_n−u_n∥², which implies that

(1−α_n)β_n∥P_Cnu_n−u_n∥² ≤ ∥x_n−x^∗∥²− ∥x_n+1−x^∗∥²+α_n

M₁M₂+M₃

. (5.130) Thus, we obtain by (5.121) that

k→∞lim ∥P_Cnku_nk−u_nk∥= 0. (5.131) Combining this with (5.125), we obtain from Lemma 5.4.9 that

k→∞lim ∥u_nk −z_nk∥= 0. (5.132) Furthermore, from Step 2 we obtain

∥w_nk −x_nk∥=α_nk θ_nk

α_nk∥x_nk −x_nk−1∥

→0 as k→ ∞. (5.133) Using (5.129) and (5.133), we obtain

k→∞lim ∥u_nk −x_nk∥= 0. (5.134)

From (5.98) and (5.131), we have

∥x_nk+1−u_nk∥ ≤β_nk∥P_Cnku_nk −u_nk∥+α_nk||u_nk|| →0.

Thus, we obtain from (5.134) that

k→∞lim ∥x_nk+1 −x_nk∥= 0.

Since

x_nk is bounded, there exists a subsequence

x_nkj of

x_nk which converges weakly to some ¯x∈H₁, such that

lim sup

k→∞

⟨x^∗, x^∗−x_nk⟩= lim

j→∞⟨x^∗, x^∗−x_nj⟩=⟨x^∗, x^∗ −x⟩.¯ Also, we obtain from (5.129), (5.132), (5.134) and Lemma 5.4.7 that ¯x∈Γ.

From x^∗ =P_Γ0, we obtain lim sup

k→∞

⟨x^∗, x^∗−x_nk⟩=⟨x^∗, x^∗−x⟩ ≤¯ 0.

Since limk→∞∥x_nk+1 −x_nk∥ →0,we obtain lim sup

k→∞

⟨x^∗, x^∗−x_nk+1⟩ ≤0. (5.135) Now, recall that d_nk :=hθ_nk

α_nk(1−α_nk)∥x_nk −x_nk−1∥M₂+ 2β_nk∥u_nk −P_Cnk(u_nk)∥

∥x^∗−x_nk+1∥+ 2⟨x^∗, x^∗−x_nk+1⟩i . Hence, by (5.135), limn→∞

θ_n

α_n∥x_n−xn−1∥= 0, (5.131) and Lemma2.5.55, we obtain

n→∞lim ∥x_n−x^∗∥= 0.

Therefore, {x_n} converges strongly to x^∗ =P_Γ0.