Research Article A Generalized Strong Convergence Algorithm in the Presence of Errors for Variational Inequality Problems in Hilbert Spaces

Research Article

A Generalized Strong Convergence Algorithm in the Presence of

Errors for Variational Inequality Problems in Hilbert Spaces

Mostafa Ghadampour

,

Donal O

’Regan ,

Ebrahim Soori

,

and Ravi P. Agarwal

2_{School of Mathematics, Statistics, National University of Ireland, Galway, Ireland}

3_{Department of Mathematics Texas A&M University-Kingsville, 700 University Blvd., MSC 172 Kingsville, Texas, USA}

Correspondence should be addressed to Ebrahim Soori; [email protected]

Received 3 April 2021; Revised 8 May 2021; Accepted 9 June 2021; Published 21 June 2021 Academic Editor: Ismat Beg

Copyright © 2021 Mostafa Ghadampour et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

In this paper, we study the strong convergence of an algorithm to solve the variational inequality problem which extends a recent paper (Thong et al., Numerical Algorithms. 78, 1045-1060 (2018)). We reduce and refine some of their algorithm conditions and we prove the convergence of the algorithm in the presence of some computational errors. Then, using the MATLAB software, the result will be illustrated with some numerical examples. Also, we compare our algorithm with some other well-known algorithms.

1. Introduction

Let H be a real Hilbert space with the inner product h:, :i and the norm k:k, and C be a nonempty, closed, and convex subset of H. The variational inequality (VI) is tofind a point x∈ C such that

Ax, y − x

h i≥ 0,∀y ∈ C, ð1Þ

where A is a mapping of C into H. The solution set of (1) is denoted by VIðA, CÞ. Variational inequalies arise in the study of network equilibriums, optimization problems, saddle point problem, Nash equilibrium problems in non-cooperative games etc.; see, for example, [1–12] and the references therein.

A new algorithm was proposed by Korpelevich [13] for solving the problem (VI) in the Euclidean space which is known as the extragradient method. Let x₁ be an arbitrary element in H and consider

y_n= PCðxn− λAxnÞ,

x_n+1= PCðxn− λAynÞ:

(

ð2Þ

whereλ is a number in ð0, 1Þ, PC is the Euclidean least

dis-tance projection of H onto C, and A: C ⟶ H is a monotone operator. The next algorithm (3) was introduced by Tseng [14] and applying the modified forward-backward (F-B) method is a good alternative to the extragradient method (TEGM):

y_n= PCðxn− λAxnÞ,

x_n+1= PXðyn− λ Ayð n− AxnÞÞ,

(

ð3Þ

where X = C and X = H if A is Lipschitz continuous. The fol-lowing algorithm (4) was proposed by Shehu and Iyiola [15] which a viscosity type subgradient extragradient method (VSEGM): yn= PCðxn− λnAxnÞ, Tn= z ∈ H : xf h n− λnAxn− yn, z − yni≤ 0g, zn= PTnðxn− λnAynÞ, x_n+1= αnf xð Þ + 1 − αn ð nÞzn, 8 > > > > > < > > > > > : ð4Þ

where the operator A is monotone and Lipschitz continu-ous, f is a strict contraction mapping, l∈ ð0, 1Þ, μ ∈ ð0, 1Þ, and λn= lmn where mn is the smallest nonnegative integer

m such that

λ Axk n− Aynk≤ μ rk lmnð Þxn k, ð5Þ

where rlmnðxÞ = x − P_Cðx − lmnAðxÞÞ for all x∈ C. Recently, the sequence produced by the following algorithm was introduced by Thong and Hieu [3] based on Tseng’s method (THEGM): yn= PCðxn− λnAxnÞ, zn= yn− λnðAyn− AxnÞ, x_n+1= αnf xð Þ + 1 − αn ð nÞzn, 8 > > < > > : ð6Þ

where the operator A is monotone and Lipschitz continu-ous, γ > 0, l ∈ ð0, 1Þ, μ ∈ ð0, 1Þ, and λn is chosen to be the

largest λ ∈ fγ, γl, γl2, ⋯g satisfying

λ Axk n− Aynk≤ μ xk n− ynk: ð7Þ

In this paper, substituting a sequence fβ_ng ⊂ ð0, 1Þ of coefficients instead of the sequence f1 − αng in the

algo-rithm (6), we extend algoalgo-rithm (6). Moreover, condition (7) will be removed just by a slight change in the coe ffi-cients fλng. Also, a sequence of computational errors in

our algorithm is considered. The strong convergence of the proposed algorithm to a point of the variational inequality VIðC, AÞ will be proved under the presence of computational errors. Finally, some examples will be pre-sented which will examine the convergence of the proposed algorithm in different situations.

2. Preliminaries

In this section, some basic concepts are presented.

Let H be a real Hilbert space with the inner product h:, :i and norm k:k and suppose that C is a nonempty closed con-vex subset of H and A: C ⟶ H is an operator. The operator A is said to be

(i) Monotone if

Ax− Ay, x − y

h i≥ 0,∀x, y ∈ C, ð8Þ

(ii) L-Lipchitz continuous if there exist L > 0 such that Ax− Ay

k k≤ L x − yk k,∀x, y ∈ C: ð9Þ

For the main results of this paper, we need the following useful lemmas.

Lemma 1. Let H be a real Hilbert space. Then, we have the

fol-lowing well-known results: x + y k k2 = xk k2+ 2 x, yh i + yk k2,∀x, y ∈ H: x + y k k2_{≤ xk k}2 + 2 y, x + yh i,∀x, y ∈ H: ð10Þ

Lemma 2 (Xu, see [16]). Let fang be a sequence of

nonnega-tive real numbers satisfying the following relation:

a_n+1≤ 1 − ηð nÞan+ ηnσn+ γn, n ≥ 1, ð11Þ where (a) fηng ⊂ ½0, 1, ∑∞n=1ηn= ∞ (b) lim sup σn≤ 0 (c) γ_n≥ 0ðn ≥ 1Þ, ∑∞_n=1γ_n< ∞ Then, an⟶ 0 as n ⟶ ∞:

Lemma 3 (see [17]). Let C be a closed and convex subset in a

real Hilbert space H. Then, z = PCx if and only if hx− z, y − zi

≤ 0∀y ∈ C.

Lemma 4 (see [18]). Let fang be a sequence of nonnegative

real numbers such that there exists a subsequence fanjg of fang such that anj< anj+1 for all j∈ ℕ. Then, there exists a nondecreasing sequence fmkg of ℕ such that limk⟶∞

mk= ∞ and the following properties are satisfied by all

(sufficiently large) number k ∈ ℕ:

amk≤ amk+1and ak≤ amk+1: ð12Þ In fact, mkis the largest number n in the set f1, 2, ⋯, kg

such that an< an+1.

Lemma 5 (see [3]). Let fxng be a sequence generated by

algo-rithm (3). Then, x_n+1− p k k2_{≤ x} n− p k k2_{− 1 − μ} 2 _x n− yn k k∀p ∈ VI C, Að Þ: ð13Þ

3. Main Results

In this section, we prove a strong convergence theorem for finding a common element of the set of solutions of an equi-librium problem and afixed point problem.

Theorem 6. Let C be a nonempty closed convex subset of a real

Hilbert space H, A be a monotone, and L-Lipschitz continu-ous mapping on C and λ ∈ ð0, 1Þ such that λL < 1. Suppose that f : H ⟶ H is a contraction mapping with a constant ρ ∈ ½0, 1Þ. Let feng ⊆ H be a sequence of computational errors,

x0∈ H be arbitrary and fxng, fyng, and fzng be the

yn= PCðxn− λAxnÞ, zn= yn− λ Ayð n− AxnÞ, x_n+1= αnf xð Þ + βn nzn+ en, 8 > > < > > : ð14Þ

where fαng and fβng are real sequences in ½0, 1 such that αn

+ βn≤ 1 for each n ≥ 1. Also, assume the following conditions:

Σ∞ n=1αnβn= ∞, Σ∞ n=1ð1− αn− βnÞ < ∞, lim n⟶∞ 1− αn− βn ð Þ αn = lim n⟶∞αn= 0, Σ∞ n=1k k < ∞:en ð15Þ Then,

(i) VIðC, AÞ ≠ ∅ if and only if fxng is bounded and

lim inf

n⟶∞ kxn− ynk = 0

Suppose VIðC, AÞ ≠ ∅. Then, (ii) If lim

n⟶∞∥en∥/αn= 0, then fxng converges strongly to

q = PVIðC,AÞ∘ f ðqÞ, where PVIðC,AÞ∘ f : H ⟶ VIðC,

AÞ is the mapping defined by PVIðC,AÞ∘ f ðxÞ =

PVIðC,AÞðf ðxÞÞ for each x ∈ H

Proof. (i) Assume that fxng is a bounded sequence and

liminfn⟶∞kxn− ynk = 0. Then, there exists a subsequence

fn_ig ⊂ ℕ such that kxni− ynik ⟶ 0 when i ⟶ ∞. Note fx_n

ig is a bounded sequence. Hence, there exists a subse-quence fxn_ikg of fxnig such that fxn_ikg converges weakly to some x∈ C. Now, noting that since y_n

ik= PCðxnik− λAxnikÞ, for all z∈ C, we have

0 ≥ xn_ik− λAxn_ik− yn_ik, z − yn_ik D E = xn_ik− yn_ik, z − yn_ik D E − λ Axn_ik, z − yn_ik D E = xn_ik− yn_ik, z − yn_ik D E − λ Axn_ik, z − xn_ik D E − λ Axn_ik, xn_ik− yn_ik D E = xn_ik− yn_ik, z − yn_ik D E − λ Axn_ik− Az, z − xn_ik D E − λ Az, z − xn_ik D E − λ Axn_ik, xn_ik− yn_ik D E ≥ xn_ik− yn_ik, z − yn_ik D E − λ Az, z − xn_ik D E − λ Axn_ik, xn_ik− yn_ik D E : ð16Þ Now, we have −λ Az, z − xn_ik D E ≤ xn_ik− yn_ik, yn_ik− z D E + λ Axn_ik, xn_ik− yn_ik D E : ð17Þ Therefore, −λ Az, z − xn_ik D E ≤ y n_ik− xn_ik y n_ik− z + λ Ax  n_ik x n_ik− yn_ik: ð18Þ From limk⟶∞kxn_ik− yn_ikk = 0, we have −λhAz, z − xi ≤ 0

for all z∈ C. Now, let y ∈ C and 0 < t < 1, and from the con-vexity C we have y_t= ½ty + ð1 − tÞx ∈ C. Therefore,

0 ≤ Ayh t, yt− xi = Ayh t, ty − txi = t Ayh t, y − xi: ð19Þ

Since 0 < t < 1 then hAyt, y − xi ≥ 0 for all y ∈ C. Because

the mapping A and multiplication are continuous, if t⟶ 0, then we have hAx, y − xi ≥ 0 for all y ∈ C, i.e; x ∈ VIðC, AÞ.

For the converse fix p ∈ VIðC, AÞ, using Lemma 5, we have zn− p k k2_{≤ x} n− p k k2_{− 1 − λL} _{ð Þ}2 _x n− yn k k2_{: ð20Þ} Therefore, zn− p k k≤ xk n− pk: ð21Þ

Using the above inequality, we have 1 xk n+1− pk = αk nf xð Þ + βn nzn+ en− pk = αk nðf xð Þn − pÞ + βnðzn− pÞ− 1 − αð n− βnÞp + enk ≤ αnkf xð Þn − pk + βnkzn− pk + 1 − αð n− βnÞ pk k + ek kn ≤ αnkf xð Þn − f pð Þk + αnkf pð Þ− pk + βnkxn− pk + 1 − αð n− βnÞ pk k + ek kn ≤ αnρ xk n− pk + αnkf pð Þ− pk + βnkxn− pk + 1 − αð n− βnÞ pk k + ek kn = αð nρ + βnÞ xk n− pk + αnð1 − ρÞ ∥f pð Þ− p∥ 1 − ρ + 1 − αð n− βnÞ pk k + ek kn ≤ max kxn− pk, ∥f pð Þ− p∥ 1 − ρ , pk k   + ek k,n ð22Þ so the sequence fxng is bounded.

(ii) Let p∈ VIðC, AÞ. From part (i), we have that fxng is

bounded. Then ff ðxnÞg, fyng and fzng are bounded. Now,

using Lemma 5, we have zn− p k k2_{≤ x} n− p k k2_{− 1 − λL} _{ð Þ}2 _x n− yn k k2_{∀p ∈ VI C, Að Þ,} ð23Þ (note that our sequence fzng in the algorithm (14)

replaces fx_n+1g in Lemma 5). ☐

Sinceαn+ βn+ ð1 − αn− βnÞ = 1, by the convexity of k:k 2

x_n+1− p k k2_{= α} nf xð Þ + βn nzn+ en− p k k2 = αk nðf xð Þ + en n− pÞ + βnðzn+ en− pÞ + 1 − αð n− βnÞ eðn− pÞk2 ≤ αnkf xð Þ + en n− pk2+ βnkzn+ en− pk2+ 1 − αð n− βnÞ ekn− pk2 ≤ αnkf xð Þn − pk2+ 1 − αð nÞ zk n− pk2+ 1 − αð n− βnÞÞ pk k2 + αð n+ βn+ 1 − αð n− βnÞÞ ek kn 2 + 2 αh nf xð Þn − αnp + βnzn− βnp− 1 − αð n− βnÞp, eni ≤ αnkf xð Þn − pk2+ 1 − αð nÞ xk n− pk2 − 1 − αð nÞ 1 − λLð Þ2   xn− yn k k2_{+ 1 − α} n− βn ð Þ pk k2 + ek kn 2+ 2 αh nf xð Þ + βn nzn− p, eni by 3:3ð ð ÞÞ ≤ αnkf xð Þn − pk2+ xk n− pk2− 1 − αð nÞ 1 − λLð Þ2   xn− yn k k2 + 1 − αð n− βnÞ pk k2+ ek kn 2+ 2 αk nf xð Þ + βn nzn− pk ek kn : ð24Þ Therefore, 1 − αn ð Þ 1 − λL ð Þ2 _x n− yn k k2_{≤ x} n− p k k2_{− x} n+1− p k k2 + αnkf xð Þn − pk2+ 1 − αð n− βnÞ pk k2+ ek kn 2 + 2 αk nf xð Þ + βn nzn− pk ek kn : ð25Þ Also, we have, x_n+1− x_n k k = αk nf xð Þ + βn nzn+ en− xnk = αk nðf xð Þn − xnÞ + βnðzn− xnÞ− 1 − αð n− βnÞxn+ enk ≤ αnkf xð Þn − xnk + βnkzn− xnk + 1 − αð n− βnÞ xk k + en k kn = αnkf xð Þn − xnk + βnkyn− λ Ayð n− AxnÞ− xnk + 1 − αð n− βnÞ xk k + en k kn ≤ αnkf xð Þn − xnk + βnkyn− xnk + βnλL yk n− xnk + 1 − αð n− βnÞ xk k + en k kn : ð26Þ Note that P_VIðC,AÞ∘ f is a contraction mapping. Then, by the Banach contraction principle, there exists a unique element q∈ H such that q = P_VIðC,AÞ∘ f ðqÞ. Now, we claim that fxng converges strongly to q = PVIðC,AÞ∘ f ðqÞ. It is

enough to consider two cases:

Case 1. Suppose there exists some n₀∈ ℕ such that ∥x_n+1− q∥2≤ ∥xn− q∥2 for all n≥ n0. Then, limn⟶∞∥xn− q∥ exists.

From (25) and our assumptions, lim

n⟶∞∥xn− yn∥ = 0. From

the conditions (b), (c), and (d), lim

n⟶∞ð1 − αn− βnÞ = limn⟶∞

αn= lim_n⟶∞kenk = 0. Then, from (26) and the boundedness of

the sequences f∥xn∥g, f∥f ðxnÞ − xn∥g and fβng, we obtain that

lim n⟶∞kxn+1− xnk = 0: ð27Þ Now, we have x_n+p− xn   ≤ x n+p− xn+p−1+ x n+p−1− xn+p−2+⋯+ xk n+1− xnk, ð28Þ for all p∈ ℕ. Note that from (27), it is concluded that lim

n⟶∞k

x_n+p−i− x_{n+p−ði+1Þ}k = 0, for all 0 ≤ i ≤ p − 1 and p ∈ ℕ. Then, from (28), lim

n⟶∞∥xn+p− xn∥ = 0, for all p ∈ ℕ, i.e., fxng is a

Cauchy sequence in the Hilbert space H; therefore, fxng is

convergent. Now we show that fxng converges strongly to q.

Note that x_n+1− q k k2_{= α} nf xð Þ + βn nzn+ en− q k k2 = αk nðf xð Þ + en n− qÞ + βnðzn+ en− qÞ + 1 − αð n− βnÞ eð n− qÞk2 ≤ β2 nkzn+ en− qk2+ 2 αh nðf xð Þ + en n− qÞ + 1 − αð n− βnÞ eð n− qÞ, xn+1− qi by Lemma 2:1ð Þ = β2nkzn+ en− qk2+ 2αnhðf xð Þ + en n− qÞ, xn+1− qi + 2 1 − αð n− βnÞ ehn− q, xn+1− qi = β2nkzn+ en− qk2+ 2αnhðf xð Þn − qÞ, xn+1− qi + 2αnhen, xn+1− qi + 2 1 − αð n− βnÞ eh n− q, xn+1− qi ≤ β2 nkzn+ en− qk2+ 2αnρ xk n− qk xk n+1− qk + 2αnhf qð Þ− q, xn+1− qi + 2αnhen, xn+1− qi + 2 1 − αð n− βnÞ ehn− q, xn+1− qi≤ β2nkzn− qk2 + 2β2nhen, zn+ en− qi + 2αnρ xk n− qk2ðby Lemma 2:1Þ + 2αnhf qð Þ− q, xn+1− qi + 2αnhen, xn+1− qi + 2 1 − αð n− βnÞ ehn− q, xn+1− qi ≤ 1 − αð nÞ2kxn− qk2+ 2αnρ xk n− qk2 + 2αnhf qð Þ− q, xn+1− qi + 2βn2hen, zn+ en− qi + 2αnhen, xn+1− qi + 2 1 − αð n− βnÞ eh n− q, xn+1− qi = 1 − 2αð nð1 − ρÞÞ xk n− qk2+ 2αnð1 − ρÞ  αnkxn− qk2 2 1 − ρð Þ + f qð Þ− q, xn+1− q h i 1 − ρ   + 2β2nhen, zn+ en− qi + 2αnhen, xn+1− qi + 2 1 − αð n− βnÞ ehn− q, xn+1− qi ≤ 1 − 2αð nð1 − ρÞÞ xk n− qk2+ 2αnð1 − ρÞ  αnkxn− qk2 2 1 − ρð Þ + f qð Þ− q, x_n+1− q h i 1 − ρ   + 2β2nk k zen kn+ en− qk + 2αnk k xen k n+1− qk + 2 1 − αð n− βnÞ ekn− qk xk n+1− qk: ð29Þ Since fxng is strongly convergent, so xn⟶ z for some

z∈ C, then we conclude that xn⇀ z. Also, as in the proof of

part (i), we conclude z∈ VIðC, AÞ. Therefore, from Lemma 3 lim sup

n⟶∞ hf qð Þ− q, xn+1− qi = f qh ð Þ− q, z − qi≤ 0: ð30Þ

Next, we consider the sequences fη_ng, fang, fσng and

From the above, we have

a_n+1≤ 1 − ηð nÞan+ ηnσn+ γn: ð32Þ

From ðcÞ, there exists an integer n₁∈ ℕ such that fηng ⊂ ½0, 1, for each n ≥ n1. Without loss of generality, we

may assume that fηng ⊂ ½0, 1, for each n ≥ 1. From condition

ðaÞ, note ∞ = Σ∞

n=1αnβn≤ Σ∞n=1αn, soΣ∞n=1αn= ∞, and hence,

Σ∞

n=1ηn= 2ð1 − ρÞΣ∞n=1αn= ∞. Therefore, condition ðaÞ of

Lemma 2 holds. From ðcÞ, the boundedness of kxn− qk2and

(3.9), we have lim sup n⟶∞ σn= lim supn⟶∞ αnkxn− qk2 2 1 − ρð Þ + f qð Þ− q, xn+1− q h i 1 − ρ ≤ lim sup n⟶∞ αnkxn− qk2 2 1 − ρð Þ + lim supn⟶∞ f qð Þ− q, xn+1− q h i 1 − ρ ≤ 0: ð33Þ 0 5 10 15 20 25 30 35 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 f(n) g(n) k(n)

Figure 1: Example 1. Convergence behavior ff_ng in Example 1.

1 2 3 4 5 6 7 8 9 10 –0.5 0 0.5 1 1.5 2 Algo 3.1 THEGM TEGM VSEGM

Therefore, condition ðbÞ of Lemma 2 holds. Next, note γn≥ 0 and kzn+ en− qk, kxn+1− qk, and ken− qk are

bounded. Hence, there exist some positive constants M₁, M₂, and M₃ such that, kzn+ en− qk ≤ M1, kxn+1− qk ≤ M2

and ken− qk ≤ M3, for each n≥ 1. Then, from ðbÞ and ðdÞ,

we have Σ∞ n=1γn= Σ∞n=1 2β2nk k zen k n+ en− qk + 2αnk k xen k n+1− qk  + 2 1 − αð n− βnÞ ekn− qk xk n+1− qk ≤ Σ∞ n=1 2β2nk kMen 1+ 2αnk kMen 2+ 2 1 − αð n− βnÞM3M2   ≤ 2M1Σ∞n=1k k + 2Men 2Σ∞n=0k k + 2Men 3M2Σ∞n=1ð1 − αn− βnÞ < ∞: ð34Þ Therefore, condition ðcÞ of Lemma 2 holds. Then, from Lemma 2, it follows that an⟶ 0 as n ⟶ ∞, i.e., limn⟶∞

kx_n− qk2

= 0.

Case 2. Suppose there exists a subsequence fkxnj− qk

2_{g of} fkx_n− qk2_{g such that kx} nj− qk 2 < kxnj+1− qk 2_{for all j∈ ℕ.}

Now from Lemma 4, there exists a nondecreasing sequence fm_kg of ℕ such that limk⟶∞mk= ∞ and the following

inequalities hold for all k∈ ℕ:

xm_k− q

 2_{≤ x} m_k+1− q

 2

and xk k− qk2≤ x m_k+1− q2: ð35Þ Now, from (25), we have

1 − αmk   1 − λLð Þ2   xmk− ymk   2 ≤ x mk− p 2 − x mk+1− p 2 + αmk f xmk   − p  2 + 1 − αmk− βmk  p k k2_{+ e} mk  2 + 2 αmkf xmk   + βmkzmk− p    e :mk ð36Þ Hence, lim k⟶∞kxmk− ymkk = 0, therefore from (26) (adjusted) lim k⟶∞kxmk+1− xmkk = 0.

From (29) (adjusted), we conclude that xmk+1− q  2_{≤ 1 − 2α} mkð1 − ρÞ   xmk− q  2 + 2αmkð1 − ρÞ  αmkxmk− q 2 2 1 − ρð Þ + f qð Þ− q, xmk+1− q   1 − ρ " # + 2β2mk emk zmk+ emk− q+ 2αmk emk xmk+1− q + 2 1 − αmk− βmk  emk− q   _xm k+1− q   ≤ 1 − 2αmkð1 − ρÞ   xmk+1− q  2 + 2αmkð1 − ρÞ  αmkxmk− q 2 2 1 − ρð Þ + f qð Þ− q, xmk+1− q   1 − ρ " # + 2βm2k emk zmk+ emk− q + 2αmk emk xmk+1− q + 2 1 − αmk− βmk  emk− q   _xm k+1− q  _: ð37Þ Therefore, we have xk− q k k2_{≤ x} mk+1− q  2_≤ αmkxmk− q 2 2 1 − ρð Þ + f qð Þ− q, xmk+1− q   1 − ρ + emk   αmk M₁ + 1 − αmk− βmk  αmk M₂, ð38Þ where M₁= β 2 mk 1 − ρ zmk+ emk− q+ 2αmkxmk+1− q   , M₂= 1 1 − ρemk− qxmk+1− q: ð39Þ

From our assumptions limk⟶∞αmk= 0, _k⟶∞lim ð1 − αmk− βmkÞ/αmk= 0, limk⟶∞∥emk∥/αmk= 0, and (30) (adjusted), we conclude lim supk⟶∞∥xk− q∥2= 0. Hence,

xk⟶ q which completes the proof of part (ii).

Open problem 1. Can we remove the condition liminfn⟶∞∥xn− yn∥ = 0 in (i) in Theorem 6?

4. Numerical Example

In this section, the algorithm (14) is illustrated with some examples.

Example 1. Put αn= 1/n + 1, βn= 1 − 1/n + 1, en= 0, H = L2

½0, 1, A ≡ I, FðxÞ ≡ 1, λ = 1/2, C = Bð0, 1Þ = ff ∈ L2½0, 1: kf k

≤ 1g.

Table 1: Numerical results of convergence for x₁= 2 in Example 3.

Then, from the algorithm (14), we have the following sequences: g_n= PCðfn− λAfnÞ = PC f_n 2 , kn= gn− λ Agð n− AfnÞ = 1 2ðfn+ gnÞ, f_n+1= αnF fð Þ + βn nkn+ en= 1 n + 1 + n n + 1kn, ð40Þ where PCis as follows PB zð Þ,ρð Þ =x x kx− zk≤ ρ, z + ρ x− z k kðx− zÞ kx− zk > ρ, 8 < : ð41Þ

whereρ > 0 (see [19]). We have,

VI C, Að Þ = f ∈ C : Af , g − ff h i≥ 0,∀g ∈ Cg

= f ∈ C : f , g − ff h i≥ 0,∀g ∈ Cg: ð42Þ Obviously, 0 ∈ VIðC, AÞ; hence, VIðC, AÞ ≠ ∅.

Now, with f₁= 2 and using the MATLAB software, we see that f f_ng converges to 0 (Figure 1).

In the following example, using the MATLAB software, we compare some similar algorithms and their convergence speed and behavior. In particular, the TEGM algorithm (3), VSEGM algorithm (4), THEGM algorithm (6), and the algo-rithm (14) are compared. We see that the algoalgo-rithm (14) has a higher convergence speed than the other algorithms (Figure 2). Example 2. Let αn= 1/ ffiffiffi n p , βn= 1 − 1/ ffiffiffi n p − 1/n2_{, λ = 1/2, e} n = 0, H = ℝ, A ≡ I, f ðxÞ = x/2, and C = ½0,∞Þ, x1= 2 (Figure 2).

Now, we examine the convergence of the sequences fxng, fyng, and fzng in Theorem 6 in the following example.

Example 3. Put αn= 1/ ffiffiffi n p , βn= 1 − 1/ ffiffiffi n p − 1/n2_{, λ = 1/2, e} n = 1/n2_{, H = ℝ, A ≡ I, f ≡ 1, and C = ½0,∞Þ, x} 1= 2. Then, we have y_n= PCðxn− λAxnÞ = PC 1 2xn , z_n= yn− λ Ayð n− AxnÞ = 1 2ðyn+ xnÞ, x_n+1= αnf xð Þ + βn nzn+ en= 1ffiffiffi n p + 1 − 1ffiffiffi n p − 1 n2 z_n+ 1 n2: ð43Þ Hence, VI C, Að Þ = x ∈ C : Ax, y − xf h i≥ 0,∀y ∈ Cg = x ∈ 0,∞f ½ Þ: x, y − xh i≥ 0,∀y ∈ Cg = 0f g: ð44Þ Therefore, q = PVIðA,CÞ∘ f ðqÞ = Pf0g∘ ðq/2Þ = 0, and now,

by Theorem 6, the sequence fxng converges strongly to 0.

In the following example, we examine the case that VIðC, AÞ = ∅, and the sequence generated by the algo-rithm (14) is divergent (Table 1 and Figure 3).

Example 4. Putαn= 1/n, βn= 1 − 1/n, λ = 1/2, en= 0, H = ℝ, A≡ −1, f ≡ 1, and C = ½0,∞Þ, x₁= 2. 10 8 6 4 2 0 12 14 16 18 20 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 x_n yn z_n

Then, we have y_n= PCðxn− λAxnÞ = PC xn+ 1 2 , zn= yn− λ Ayð n− AxnÞ = yn, x_n+1= αnf xð Þ + βn nzn+ en= 1 n + 1 − 1 n PC xn+ 1 2 : ð45Þ Hence, VI C, Að Þ = x ∈ C : Ax, y − xf h i≥ 0,∀y ∈ Cg = x ∈ 0,∞f ½ Þ: −1, y − xh i≥ 0,∀y ∈ Cg = ∅: ð46Þ

Now, we will prove by induction that, for all n∈ ℕ

y_n= PC xn+ 1 2 = xn+ 1 2: ð47Þ

When n = 1, then we have x1= 2 ≥ −1/2.

Induction step: suppose for n = k the inequality xk≥ −1/2 holds. Then, x_k+1=1 k+ 1 − 1 k PC xk+ 1 2 = 1 k+ 1 − 1 k xk+ 1 2 = 1 k+ 1 − 1 k xk+ 1 − 1 k 1 2 ≥1 k+ 1 − 1 k −1 2 + 1 −1 k 1 2 ≥ −1 2: ð48Þ

Hence, yn= xn+ 1/2. Consequently lim_n⟶∞kxn− ynk = 1/

2 ≠ 0.

Next, we show that fxng is an unbounded sequence. Note

x₃= 5/4 > 3/4.

Induction step: suppose for n = k the inequality xk> k/4

holds. Then, x_k+1= 1 k+ 1 − 1 k xk+ 1 2 >1 k + 1 − 1 k k 4 + 1 2 = 1 k+ k 4− 1 4+ 1 2 − 1 2k = 1 2k + k 4 + 1 4 > k + 1 4 , ð49Þ thus fxng is an unbounded sequence (Figure 4).

8 6 4 2 0 10 12 14 16 18 20 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 xn yn zn

5. Conclusions

In this paper, we proposed a Tseng-type viscosity algorithm based on the viscosity method, which is an extension of the Thong et al.’s algorithm [3]. We showed that the sequence generated by the proposed algorithm strongly converges to an element of VI ðC, AÞ.

The following are the results in this paper:

(i) We extended the results of Thong et al.’s. method [3] and provided necessary and sufficient conditions for the VI ðC, AÞ to be nonempty

(ii) In our generated sequence fxng, in the algorithm (14),

we put a sequence of computational errors feng and

proved the convergence of the sequence in the presence of computational errors

(iii) We provided some numerical examples to compare our algorithm with the algorithms TEGM, VSEGM, and THEGM

Data Availability

No data were used to support the study.

Conflicts of Interest

This work does not have any conflicts of interest.

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