viscosity methods for common solutions of equilibrium and variational inequality problems via...

Nonlinear Analysis 75 (2012) 1787–1798

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Nonlinear Analysis

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Viscosity methods for common solutions of equilibrium and variationalinequality problems via multi-step iterative algorithms and commonfixed points

Giuseppe Marino a,∗, Luigi Muglia a, YongHong Yao b

a Dipartimento di Matematica, Universitá della Calabria, 87036 Arcavacata di Rende (CS), Italyb Department of Mathematics, Tianjin Polytechnic University, Tianjin 300160, People’s Republic of China

a r t i c l e i n f o

Article history:Received 9 August 2010Accepted 15 September 2011Communicated by Enzo Mitidieri

MSC:47H0947H1058E35

Keywords:Hierarchical fixed pointsEquilibrium problemNonexpansive mapContractionVariational inequality problemProjection

a b s t r a c t

In this paper, we present a new multi-step iterative method. We prove the strongconvergence of the method to a common fixed point of a finite number of nonexpansivemappings that also solves a suitable equilibrium problem.

© 2011 Elsevier Ltd. All rights reserved.

1. Introduction

Let H be a Hilbert space and C be a closed and convex subset of H . The variational inequality problem (VIP) on C is statedas:

find x∗∈ C such that ⟨Ax∗, x − x∗

⟩ ≥ 0, x ∈ C

where A : C → C is a nonlinear mapping.If we assume that C is the fixed points set of a nonexpansive mapping T and if S is another nonexpansive mapping (not

necessarily with fixed points), VIP becomes

find x∗∈ Fix(T ) such that ⟨(I − S)x∗, x − x∗

⟩ ≥ 0, x ∈ Fix(T ).

This problem, introduced by Mainge and Moudafi in [1,2], is called hierarchical fixed point problem.

Supported by Ministero dell’Universitá e della Ricerca of Italy.∗ Correspondence to: Universitá della Calabria, via P. Bucci, 87036, Arcavacata di Rende (CS), Italy. Tel.: +39 0984 496456; fax: +39 0984 496410.

E-mail addresses: [email protected] (G. Marino), [email protected] (L. Muglia), [email protected] (Y. Yao).

0362-546X/$ – see front matter© 2011 Elsevier Ltd. All rights reserved.doi:10.1016/j.na.2011.09.019

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http://dx.doi.org/10.1016/j.na.2011.09.019

1788 G. Marino et al. / Nonlinear Analysis 75 (2012) 1787–1798

Observe that if S has fixed points, then they are solutions of VIP.If S is a ρ-contraction (i.e. ‖Sx − Sy‖ ≤ ρ‖x − y‖ for some 0 < ρ < 1) the set of solutions of VIP is a singleton and it is

well-known as viscosity problem. This was last introduced by Moudafi in [3] and also developed by Xu in [4].One can check that solving VIP is equivalent to find a fixed point of the nonexpansive map (PFix(T )S), where PFix(T ) is the

metric projection on the closed and convex set Fix(T ).Variational inequalities like VIP covers several topics recently investigated in the literature as monotone inclusions,

convex optimization and quadratic minimization over fixed points sets. We refer to [3–6] for more references.To the best of our knowledge, there are two approaches to VIP. The first, known as a hierarchical fixed point approach, was

introduced by Mainge and Moudafi in [1]. This approach, in the implicit frame, generates a double-index net xs,t : (s, t) ∈

(0, 1) satisfying the fixed point equation

xs,t = tf (xs,t) + (1 − t)(sSxs,t + (1 − s)Txs,t)

where f is a ρ-contraction on C .In [1], the authors proved the following theorem.

Theorem 1.1. The net xs,t strongly converges, as t → 0, to xs, where xs satisfies xs = PFix(sS+(1−s)T )f (xs). Moreover, the net xs, inturn, weakly converges, as s → 0, to a solution x∞ of VIP.

Remark 1.2. In [1], Mainge and Moudafi stated the problem of the strong convergence of the net xt,s when (t, s) → (0, 0)jointly, to a solution of VIP. A negative answer to this question is given in [7].

In [2], Moudafi and Mainge studied the explicit scheme introducing the iterative algorithm

xn+1 = λnf (xn) + (1 − λn)(αnSxn + (1 − αn)Txn), (1.1)

where (αn)n∈N, (λn)n∈N are sequences in (0, 1) and proving the strong convergence to a solution-point of VIP.

Theorem 1.3. Assume that the following hold

(P0) Fix(T ) ∩ int(C) = ∅;(P1) αn = o(λn) and

∑n αn = ∞;

(P2) limn→∞

αn−αn−1αnλn

= limn→∞

λn−λn−1λnλn−1αn

= 0;(P3) there exist two constants θ and k such that

∀x ∈ C, ‖x − Tx‖ ≥ k dist(x, Fix(T ))θ

(P4) λ1+ 1

θn = o(αn).

Suppose that (xn)n is bounded. Then (xn)n strongly converges to a solution of VIP.

A different approach was introduced by Yao et al. in [8]. This two-step iterative algorithm generates a sequence (xn)n∈N bythe explicit scheme

yn = βnSxn + (1 − βn)xnxn+1 = αnf (xn) + (1 − αn)Tyn, n ≥ 1. (1.2)

Theorem 1.4. Let C be a nonempty closed and convex subset of a real Hilbert space H. Let S and T be two nonexpansivemappingson C into itself. Let f : C → C be a ρ-contraction and (αn)n and (βn)n two real sequences in (0, 1). Assume that the sequence(xn)n generated by scheme (1.2) is bounded and

(i)∑

n∈N αn = ∞

(ii) limn→∞1αn

1βn

−1

βn−1

= 0, limn→∞1βn

1 −αn−1αn

= 0

(iii) limn→∞ βn = 0, limn→∞αnβn

= 0, limn→∞

β2n

αn= 0

(iv) Fix(T ) ∩ int(C) = ∅

(v) there exists a constant k > 0 such that ‖x − Tx‖ ≥ k Dist(x, Fix(T )), for each x ∈ C, where Dist(x, Fix(T )) =

infy∈Fix(T ) ‖x − y‖.

Then the sequence (xn)n strongly converges to x = PΩ f (x) which solves VIP.

On the other hand, if C = Fix(T ) and F(x, y) := ⟨(I − S)x, y − x⟩, the VIP can be reformulated as

find x∗∈ C such that F(x∗, y) ≥ 0, y ∈ C, (1.3)

i.e. as an equilibrium problem. In [9,10], it is shown that formulation (1.3) covers monotone inclusion problems, saddlepoint problems, variational inequality problems, minimization problems, Nash equilibria in noncooperative games, vectorequilibrium problems and certain fixed point problems (see [11]).

G. Marino et al. / Nonlinear Analysis 75 (2012) 1787–1798 1789

It is worth to remark that, in the case of VIP, the induced bifunction F(x, y) := ⟨(I − S)x, y − x⟩ satisfies the followingconditions:

(f1) F(x, x) = 0 for all x ∈ H;(f2) F(x, y) + F(y, x) ≤ 0 for all (x, y) ∈ H × H (i.e. F is monotone);(f3) for each x, y, z ∈ H

lim supt→0

F(tz + (1 − t)x, y) ≤ F(x, y);

i.e. F is hemicontinuous in the first variable.(f4) the function y → F(x, y) is convex and lower semicontinuous for each x ∈ H .

Recently many authors have generalized the classical equilibrium problem introduced by Combettes and Hirstoaga [12] byintroducing ‘‘perturbations’’ to the function F . As an example, Moudafi in [13] studies the equilibrium problem

to find x∗∈ C such that F(x∗, y) + ⟨Ax∗, y − x∗

⟩ ≥ 0, ∀y ∈ C

where A is an α-inverse strongly monotone operator. In [14–16], the authors study the mixed problem

to find x∗∈ C such that F(x∗, y) + ϕ(x∗) − ϕ(y) ≥ 0, ∀y ∈ C

with ϕ being an opportune mapping.Here, we study the equilibrium problem

to find x∗∈ C such that F(x∗, y) + h(x∗, y) ≥ 0, ∀y ∈ C (1.4)

that includes all previous equilibrium problems as particular cases.On the other hand, from a long time,many authorswere interested in the construction of iterative algorithms thatweakly

or strongly converge to a common fixed point of a family of nonexpansive mappings (see e.g. [17–19]). In [20], Xu provesthat the sequence generated by

xn+1 = (I − ϵn+1A)Tn+1xn + ϵn+1u

where Tn = Tn mod N , strongly converges to a solution of a quadratic minimization problem under the assumptionFix(T1T2 · · · TN) = Fix(TNT1 · · · TN−1) = · · · = Fix(T2T3 · · · T1).

In [21], Yao lacks this hypothesis and studies the viscosity approximation of a common fixed point of the family ofmappings.

In [22], Colao et al. use a different approach to obtain the convergence of a more general scheme that involves anequilibrium problem.

In this paper, our aims are as follows:

• to introduce a multi-step iterative method that generalizes the two-stepmethod introduced in [8] for two nonexpansivemappings to a finite family of nonexpansive mappings;

• to prove that this method converges to a common fixed point of the mappings that is also an equilibrium point of (1.4);• to furnish a second reading of our results that tie up the systems of variational inequalities problems and the hierarchical

fixed point problems.

2. Preliminary results

This lemma appears implicitly in the paper of Reineermann [23].

Lemma 2.1 ([23]). Let H be a Hilbert space, x, y, z ∈ H and λ a real number. Then

‖λx + (1 − λ)y − z‖2= λ‖x − z‖2

+ (1 − λ)‖y − z‖2− λ(1 − λ)‖x − y‖2.

In the sequel, we will indicate with EP(F , h) the set of solutions of (1.4).

Lemma 2.2 ([24]). Let C be a convex closed subset of a Hilbert space H.Let F : C × C → R be a bi-function such that

(f1) F(x, x) = 0 for all x ∈ C;(f2) F is monotone and upper hemicontinuous in the first variable;(f3) F is lower semicontinuous and convex in the second variable.

Let h : C × C → R be a bi-function such that

(h1) h(x, x) = 0 for all x ∈ C;(h2) h is monotone and weakly upper semicontinuous in the first variable;(h3) h is convex in the second variable.


Moreover, let us suppose that

(H) for fixed r > 0 and x ∈ C, there exists a bounded K ⊂ C and a ∈ K such that for all z ∈ C \ K , −F(a, z) + h(z, a) +1r ⟨a −

z, z − x⟩ < 0.

For r > 0 and x ∈ H, let Tr : H → 2C be a mapping defined by

Trx =

z ∈ C : F(z, y) + h(z, y) +

1r⟨y − z, z − x⟩ ≥ 0, ∀y ∈ C

(2.1)

called resolvent of F and h.Then

(1) Trx = ∅;(2) Trx is a singleton;(3) Tr is firmly nonexpansive;(4) EP(F , h) = Fix(Tr) and it is closed and convex.

Lemma 2.3 ([24]). Let us suppose that (f1)–(f3), (h1)–(h3) and (H) hold. Let x, y ∈ H, r1, r2 > 0. Then

‖Tr2y − Tr1x‖ ≤ ‖y − x‖ +

r2 − r1r2

‖Tr2y − y‖.

Remark 2.4. In the sequel, given a sequence (zn)n, we will denote with ωw(zn) the set of cluster points of (zn) with respectto the weak topology, i.e.

ωw(zn) = q ∈ H : there exists nk → ∞ for which znk q.

Analogously, we will denote with ωs(zn) the set of cluster points of (zn) with respect the norm-topology, i.e.

ωs(zn) = q ∈ H : there exists nk → ∞ for which znk → q.

Lemma 2.5. Suppose that the hypotheses of Lemma 2.2 are satisfied. Let (rn)n∈N a sequence in (0, +∞) with lim infn rn > 0.Suppose that (xn)n∈N is a bounded sequence. Then the following statements are equivalent and true:

(a) if ‖xn − Trnxn‖ → 0, as n → ∞, the weak cluster points of (xn)n∈N satisfies the problem

F(x, y) + h(x, y) ≥ 0 ∀y ∈ C

i.e. ωw(xn) ⊆ EP(F , h).(b) the demiclosedness principle holds in the sense that, if xn x∗ and ‖xn − Trnxn‖ → 0, as n → ∞, then (I − Trk)x

∗= 0, for

all k ∈ N.

Proof. The equivalence of (a) and (b) immediately follows by (4) of Lemma 2.2. We prove now that (a) is true.Let q be a weak cluster point of (xn)n∈N. Let us call un = Trnxn and let (xnm)m∈N be a subsequence of (xn)n∈N weakly

converging to q. We show that q ∈ EP(F , h).At first, note that by the monotonicity of f and h, we have

h(un, y) +1rn

⟨y − un, un − xn⟩ ≥ F(y, un).

In particular,

h(unm , y) +

y − unm ,

unm − xnmrnm

≥ F(y, unm). (2.2)

By condition (f3), for x ∈ H fixed, the function F(x, ·) is lower semicontinuous and convex, and thus weakly lowersemicontinuous.

Since ‖xn − un‖ → 0, as n → ∞ and by the hypothesis on (rn)n, we obtain (unm − xnm)/rnm → 0. Therefore, lettingm → ∞ in (2.2) yields

F(y, q) ≤ limm→∞

F(y, unm) ≤ limm→∞

h(unm , y) ≤ h(q, y), y ∈ H.

Replacing ywith yτ := τy + (1 − τ)q with τ ∈ [0, 1], we obtain

0 = F(yτ , yτ ) + h(yτ , yτ ) ≤ τ(F(yτ , y) + h(yτ , y)) + (1 − τ)(F(yτ , q) + h(yτ , q))≤ τ(F(yτ , y) + h(yτ , y)) + (1 − τ)(h(q, yτ ) + h(yτ , q)) ≤ (F(yτ , y) + h(yτ , y)).


Hence we have

F(τy + (1 − τ)q, y) + h(τy + (1 − τ)q, y) ≥ 0, τ ∈ (0, 1], y ∈ H.

Letting τ → 0+ and using the hemicontinuity of F and theweak upper semicontinuity of h (in the first variable), we concludethat

F(q, y) + h(q, y) ≥ 0, y ∈ H;

therefore q ∈ EP(f , h).

Lemma 2.6 ([6]). Assume (an)n is a sequence of nonnegative numbers such that

an+1 ≤ (1 − γn)an + δn, n ≥ 0,

where (γn)n is a sequence in (0, 1) and (δn)n is a sequence in R such that,

(1)∑

∞

n=1 γn = ∞;(2) lim supn→∞ δn/γn ≤ 0 or

∑∞

n=1 |δn| < ∞.

Then limn→∞ an = 0.

3. Main results

Let us consider the schemeF(un, y) + h(un, y) +

1rn

⟨y − un, un − xn⟩ ≥ 0, ∀y ∈ C

yn,1 = βn,1S1un + (1 − βn,1)unyn,i = βn,iSiun + (1 − βn,i)yn,i−1, i = 2 . . . ,Nxn+1 = αnf (xn) + (1 − αn)Tyn,N , n ≥ 1

(3.1)

where

• (αn)n∈N, (βn,i)n∈N (i = 1 . . . ,N) are sequences in (0, 1);• (rn)n∈N is a sequence in (0, +∞) with lim infn→∞ rn > 0;• the mapping f is a ρ-contraction on C;• Si, T : C → C are nonexpansive mappings;• F , h : C × C → R be two bi-functions satisfying the hypotheses of Lemma 2.2.

Lemma 3.1. Let us suppose that Ω = Fix(T )∩(∩i Fix(Si))∩EP(F , h) = ∅. Then the sequences (xn)n∈N, (yn,i)n∈N for all i, (un)n∈Nare bounded.

Proof. Let us observe, first of all that, if v ∈ Ω , then

‖yn,1 − v‖ ≤ ‖un − v‖ ≤ ‖xn − v‖.

For all from i = 2 to i = N , by induction, one proves that

‖yn,i − v‖ ≤ βn,i‖un − v‖ + (1 − βn,i)‖yn,i−1 − v‖ ≤ ‖un − v‖ ≤ ‖xn − v‖.

Thus we obtain that for every i = 1, . . . ,N

‖yn,i − v‖ ≤ ‖un − v‖ ≤ ‖xn − v‖. (3.2)

Moreover,

‖xn+1 − v‖ ≤ αn‖f (xn) − v‖ + (1 − αn)‖Tyn,N − v‖

≤ αn‖f (xn) − f (v)‖ + αn‖f (v) − v‖ + (1 − αn)‖yn,N − v‖

≤ αnρ‖xn − v‖ + αn‖f (v) − v‖ + (1 − αn)‖xn − v‖

≤ (1 − (1 − ρ)αn)‖xn − v‖ + αn‖f (v) − v‖.

So, calling

M = max‖x0 − v‖,

‖f (v) − v‖

1 − ρ

,

we obtain the claim.


Lemma 3.2. Let us suppose that Ω = ∅. Moreover, let us suppose that

(H1) αn → 0, as n → ∞ and∑

n∈N αn = ∞;(H2)

∑n∈N |αn − αn−1| < ∞ or limn→∞

|αn−αn−1|αn

= 0;

(H3)∑

n∈N |βn,i − βn−1,i| < ∞ or limn→∞

|βn,i−βn−1,i|

αn= 0, for all i ∈ 1, . . . ,N;

(H4)∑

n∈N |rn − rn−1| < ∞ or limn→∞

|rn−rn−1|αn

= 0;

hold. Then limn→∞ ‖xn+1 − xn‖ = 0, i.e. (xn)n∈N is asymptotically regular.

Proof. Observing that

xn+1 − xn = αnf (xn) + (1 − αn)Tyn,N − αn−1f (xn−1) − (1 − αn−1)Tyn−1,N

= αn(f (xn) − f (xn−1)) + (f (xn−1) − Tyn−1,N)(αn − αn−1) + (1 − αn)(Tyn,N − Tyn−1,N)

then, passing to the norm we have

‖xn+1 − xn‖ ≤ αn‖f (xn) − f (xn−1)‖ + ‖f (xn−1) − Tyn−1,N‖ |αn − αn−1| + (1 − αn)‖Tyn,N − Tyn−1,N‖

≤ αnρ‖xn − xn−1‖ + ‖f (xn−1) − Tyn−1,N‖ |αn − αn−1| + (1 − αn)‖yn,N − yn−1,N‖. (3.3)

By definition of yn,i one obtains that, for all i = N, . . . , 2

‖yn,i − yn−1,i‖ ≤ βn,i‖un − un−1‖ + ‖Siun−1 − yn−1,i−1‖ |βn,i − βn−1,i| + (1 − βn,i)‖yn,i−1 − yn−1,i−1‖. (3.4)

In the case i = 1, we have

‖yn,1 − yn−1,1‖ ≤ βn,1‖un − un−1‖ + ‖S1un−1 − un−1‖ |βn,1 − βn−1,1| + (1 − βn,1)‖un − un−1‖

= ‖un − un−1‖ + ‖S1un−1 − un−1‖ |βn,1 − βn−1,1|. (3.5)

Substituting (3.5) in all (3.4)-type one obtains for i = 2, . . . ,N

‖yn,i − yn−1,i‖ ≤ ‖un − un−1‖ +

i−k=2

‖Skun−1 − yn−1,k−1‖ |βn,k − βn−1,k| + ‖S1un−1 − un−1‖ |βn,1 − βn−1,1|.

So we conclude that

‖xn+1 − xn‖ ≤ αnρ‖xn − xn−1‖ + ‖f (xn−1) − Tyn−1,N‖ |αn − αn−1| + (1 − αn)‖un − un−1‖

+

N−k=2


By Lemma 2.3, we know that

‖un − un−1‖ ≤ ‖xn − xn−1‖ + L1 −

rn−1

rn

(3.6)

where L = supn ‖un − xn‖ so, substituting (3.6) in (3.3) we obtain

‖xn+1 − xn‖ ≤ αnρ‖xn − xn−1‖ + |αn − αn−1| ‖f (xn−1) − Tyn−1,N‖ + (1 − αn)‖xn − xn−1‖ + L rn − rn−1

rn

+

N−k=2


If we call

M := maxsupn∈N

‖f (xn−1) − Tyn−1,N‖, L, supn∈N,i=2,...,N

‖Siun−1 − yn−1,i−1‖, supn∈N

‖S1un−1 − un−1‖

and b > 0 a minorant for (rn)n∈N, we have

‖xn+1 − xn‖ ≤ [1 − αn(1 − ρ)]‖xn − xn−1‖ + M

|αn − αn−1| +

|rn − rn−1|

b+

N−k=1

|βn,k − βn−1,k|

. (3.7)

By hypotheses (H1)–(H4) and Lemma 2.6, we obtain the claim.

Lemma 3.3. Let us suppose that Ω = ∅. Let us suppose that (xn)n∈N is asymptotically regular. Then ‖xn−un‖ = ‖xn−Trnxn‖ →

0, as n → ∞.


Proof. We recall that, by the firm nonexpansivity of Trn , a standard calculation (see [22]) shows that if p ∈ EP(F , h)

‖un − p‖2≤ ‖xn − p‖2

− ‖xn − un‖2.

So, let v ∈ Ω; then by (3.2)

‖xn+1 − v‖2

≤ αn‖f (xn) − v‖2+ (1 − αn)‖Tyn,N − v‖

2

≤ αn‖f (xn) − v‖2+ (1 − αn)‖yn,N − v‖

2

≤ αn‖f (xn) − v‖2+ (1 − αn)‖un − v‖

2

≤ αn‖f (xn) − v‖2+ (1 − αn)‖xn − v‖

2− (1 − αn)‖xn − un‖

2.

So we observe that

(1 − αn)‖xn − un‖2

≤ αn‖f (xn) − v‖2+ ‖xn − v‖

2− ‖xn+1 − v‖

2

≤ αn‖f (xn) − v‖2+ ‖xn+1 − xn‖(‖xn − v‖ + ‖xn+1 − v‖).

By Lemma 3.1, we have that (xn)n∈N is bounded and by the its asymptotically regularity we have that limn→∞ ‖xn − un‖

= 0.

Remark 3.4. By previous lemma we have ωw(xn) = ωw(un) and ωs(xn) = ωs(un) i.e. the sets of strong/weak cluster pointsof (xn)n∈N and (un)n∈N coincides.

Of course, if βn,i → βi = 0, as n → ∞, for all index i, the assumptions of Lemma 3.2 are enough to assure that

limn→∞

‖xn+1 − xn‖βn,i

= 0 ∀i ∈ 1, . . . ,N.

In the next lemma, we examine the case in which at least one sequence (βn,k0)n∈N is a null sequence.

Lemma 3.5. Let us suppose that Ω = ∅. Let us suppose that (H1) holds. Moreover, for a index k0 ∈ 1, . . . ,N, limn→∞ βn,k0 =

0 and

(H5) for all i, limn→∞

|βn,i−βn−1,i|

αnβn,k0= limn→∞

|αn−αn−1|αnβn,k0

= limn→∞

|rn−rn−1|αnβn,k0

= 0;

(H6) there exists a constant K > 0 such that 1αn

1βn,k0

−1

βn−1,k0

< K, for all n > 1

hold. Then

limn→∞

‖xn+1 − xn‖βn,k0

= 0. (3.8)

Proof. We start by (3.7). Dividing both the terms by βn,k0 we have


≤ [1 − αn(1 − ρ)]‖xn − xn−1‖

βn,k0+ M

|αn − αn−1|

βn,k0+

|rn − rn−1|

bβn,k0+

N∑k=1


βn,k0

.

So


≤ [1 − αn(1 − ρ)]‖xn − xn−1‖

βn−1,k0+ [1 − αn(1 − ρ)]‖xn − xn−1‖

1βn,k0

−1

βn−1,k0

+M

|αn − αn−1|

βn,k0+

|rn − rn−1|

bβn,k0+

N∑k=1


βn,k0

≤ [1 − αn(1 − ρ)]

‖xn − xn−1‖

βn−1,k0+ ‖xn+1 − xn‖

1βn,k0

−1

βn−1,k0

+M

|αn − αn−1|

βn,k0+

|rn − rn−1|

bβn,k0+

N∑k=1


βn,k0


by (H6) ≤ [1 − αn(1 − ρ)]‖xn − xn−1‖

βn−1,k0+ αnK‖xn+1 − xn‖

+M

|αn − αn−1|

βn,k0+

|rn − rn−1|

bβn,k0+

N∑k=1


βn,k0

by Lemma 3.1 ≤ [1 − αn(1 − ρ)]

‖xn − xn−1‖

βn−1,k0+ αnKM

+M

|αn − αn−1|

βn,k0+

|rn − rn−1|

bβn,k0+

N∑k=1


βn,k0

.

By the boundedness of (xn)n∈N, by (H1), (H5) and Lemma 2.6, we conclude that

limn→∞


= 0.

Lemma 3.6. Let us suppose that Ω = ∅. Let us suppose that βn,i → βi ∈ (0, 1) as n → ∞ for all i = 1, . . . ,N. Moreover,suppose that (H1)–(H4) are satisfied. Then, for all i, ‖Siun − un‖ → 0, as n → ∞.

Proof. First of all, we note that (xn)n∈N is asymptotically regular.First we prove that for every i ∈ 1, . . . ,N one has ‖Siun − yn,i−1‖ → 0 as n → ∞. Let v ∈ Ω . When i = N , by

Lemma 2.1, we have

‖xn+1 − v‖2

≤ αn‖f (xn) − v‖2+ (1 − αn)‖yn,N − v‖

2

= αn‖f (xn) − v‖2+ (1 − αn)βn,N‖SNun − v‖

2

+ (1 − αn)(1 − βn,N)‖yn,N−1 − v‖2− (1 − αn)(1 − βn,N)βn,N‖SNun − yn,N−1‖

2

≤ αn‖f (xn) − v‖2+ (1 − αn)‖un − v‖

2− (1 − αn)βn,N(1 − βn,N)‖SNun − yn,N−1‖

2

≤ αn‖f (xn) − v‖2+ ‖xn − v‖

2− (1 − αn)βn,N(1 − βn,N)‖SNun − yn,N−1‖

2.

So we have the inequality

(1 − αn)βn,N(1 − βn,N)‖SNun − yn,N−1‖2

≤ αn‖f (xn) − v‖2+ ‖xn − v‖

2− ‖xn+1 − v‖

2

≤ αn‖f (xn) − v‖2+ L‖xn+1 − xn‖

where L := supn(‖xn − v‖ + ‖xn+1 − v‖).Since βn,N → βN ∈ (0, 1) with n → ∞ and (xn)n∈N is asymptotically regular, then (‖SNun − yn,N−1‖)n∈N is a null

sequence.Let i ∈ 1, . . . ,N − 1. Then one has

‖xn+1 − v‖2

≤ αn‖f (xn) − v‖2+ (1 − αn)‖yn,N − v‖

2

≤ αn‖f (xn) − v‖2+ (1 − αn)[βn,N‖SNun − v‖

2+ (1 − βn,N)‖yn,N−1 − v‖

2]

≤ αn‖f (xn) − v‖2+ (1 − αn)βn,N‖xn − v‖

2+ (1 − αn)(1 − βn,N)‖yn,N−1 − v‖

2

≤ αn‖f (xn) − v‖2+ (1 − αn)βn,N‖xn − v‖

2

+ (1 − αn)(1 − βn,N)[βn,N−1‖SN−1un − v‖2+ (1 − βn,N−1)‖yn,N−2 − v‖

2]

= αn‖f (xn) − v‖2+ (1 − αn)(βn,N + (1 − βn,N)βn,N−1)‖xn − v‖

2

+ (1 − αn)

N∏k=N−1

(1 − βn,k)‖yn,N−2 − v‖2

and so, after (N − i + 1)-iterations

‖xn+1 − v‖2

≤ αn‖f (xn) − v‖2+ (1 − αn)

βn,N +

N−j=i+2

N∏p=j

(1 − βn,p)

βn,j−1

‖xn − v‖

2

+ (1 − αn)

N∏k=i+1

(1 − βn,k)‖yn,i − v‖2


≤ αn‖f (xn) − v‖2+ (1 − αn)

βn,N +

N−j=i+2

N∏p=j

(1 − βn,p)

βn,j−1

× ‖xn − v‖2+ (1 − αn)

N∏k=i+1

(1 − βn,k)

× [βn,i‖Siun − v‖2+ (1 − βn,i)‖yn,i−1 − v‖

2− βn,i(1 − βn,i)‖Siun − yn,i−1‖

2]

≤ αn‖f (xn) − v‖2+ (1 − αn)‖xn − v‖

2− βn,i(1 − αn)

N∏k=i

(1 − βn,k)‖Siun − yn,i−1‖2.

Again we obtain that

(1 − αn)βn,i

N∏k=i

(1 − βn,k)‖Siun − yn,i−1‖2

≤ αn‖f (xn) − v‖2+ ‖xn − v‖

2− ‖xn+1 − v‖

2

≤ αn‖f (xn) − v‖2+ L‖xn+1 − xn‖

so ‖Siun − yn,i−1‖ → 0 as n → ∞.Obviously for i = 1, we have ‖S1un − un‖ → 0.To conclude, we have that

‖S2un − un‖ ≤ ‖S2un − yn,1‖ + ‖yn,1 − un‖ = ‖S2un − yn,1‖ + βn,1‖S1un − un‖

from which ‖S2un − un‖ → 0. Thus by induction ‖Siun − un‖ → 0 for all i = 2, . . . ,N since it is enough to observe that

‖Siun − un‖ ≤ ‖Siun − yn,i−1‖ + ‖yn,i−1 − Si−1un‖ + ‖Si−1un − un‖

≤ ‖Siun − yn,i−1‖ + (1 − βn,i−1)‖Si−1un − yn,i−2‖ + ‖Si−1un − un‖.

Example 3.7. As an example, if we consider N = 2 and the sequences

αn =1

√n, rn = 2 −

1n

βn,1 =12

−1n, βn,2 =

12

−1n2

, n > 2

then they satisfy the hypotheses of Lemma 3.6.

Lemma 3.8. Let us suppose that Ω = ∅ and (βn,i)n∈N → βi for all i as n → ∞. Suppose there exists k ∈ 1, . . . ,N such thatβn,k → 0, as n → ∞.

Let k0 ∈ 1, . . . ,N the largest index such that (βn,k0)n∈N → 0, as n → ∞. Suppose that(i) αn

βn,k0→ 0, as n → ∞;

(ii) if i ≤ k0 and (βn,i)n∈N → 0 thenβn,k0βn,i

→ 0, as n → ∞.

Moreover, suppose that (H1), (H5) and (H6) hold. Then, for all i, ‖Siun − un‖ → 0, as n → ∞.

Proof. First of all we note that if (H5) holds then also (H2)–(H4) are satisfied. So (xn)n∈N is asymptotically regular.Let k0 be as in the hypotheses. As in Lemma 3.6, for every index i ∈ 1, . . . ,N such that βn,i → βi = 0, one has

‖Siun − yn,i−1‖ → 0 as n → ∞.For all the other indexes i ≤ k0, we can prove that ‖Siun − yn,i−1‖ → 0 as n → ∞ in a similar manner. By

‖xn+1 − v‖2

≤ αn‖f (xn) − v‖2+ ‖xn − v‖

2− (1 − αn)βn,i

N∏k=i

(1 − βn,k)‖Siun − yn,i−1‖2

thus

(1 − αn)

N∏k=i

(1 − βn,k)‖Siun − yn,i−1‖2

≤αn

βn,i‖f (xn) − v‖

2+

‖xn − xn+1‖Lβn,i

.

By Lemma 3.5 or by hypothesis (ii) on the sequences

‖xn − xn+1‖Lβn,i

=‖xn − xn+1‖L

βn,k0

βn,k0

βn,i→ 0

so the thesis follows.


Example 3.9. Let us consider N = 3 and the following sequences

αn =1

√n, rn = 2 −

1n2

βn,1 =14√n, βn,2 =

12

−1n2

, n > 1

βn,3 =13√n

satisfy all hypotheses (i)–(iii), (H1), (H5) and (H6) of Lemma 3.8.

Remark 3.10. Under the hypotheses of Lemma 3.8, analogously to Lemma 3.6, one can see that

limn→∞

‖Siun − yn,i−1‖ = 0, ∀i ∈ 2, . . . ,N. (3.9)

Corollary 3.11. Let us suppose that the hypotheses of either Lemma 3.6 or Lemma 3.8 are satisfied. Then ωw(xn) = ωw(un) =

ωw(yn), ωs(xn) = ωs(un) = ωs(yn,1) and ωw(xn) ⊂ Ω .

Proof. By Remark 3.4, we have ωw(xn) = ωw(un) and ωs(xn) = ωs(un).Now we observe that

‖xn − yn,1‖ ≤ ‖xn − un‖ + ‖yn,1 − un‖ = ‖xn − un‖ + βn,1‖S1un − un‖.

By Lemma 3.6, ‖S1un − un‖ → 0, as n → ∞, hence

‖xn − yn,1‖ → 0, if n → ∞ (3.10)

and ωw(xn) = ωw(yn,1) and ωs(xn) = ωs(yn,1).Let p ∈ ωw(xn). Since p ∈ ωw(un), by Lemma 3.6 and demiclosedness principle, we have p ∈ Fix(Si) for all index i, i.e.

p ∈ ∩i Fix(Si). Moreover,

‖xn − Txn‖ ≤ ‖xn − xn+1‖ + ‖xn+1 − Tyn,N‖ + ‖Tyn,N − Txn‖≤ ‖xn − xn+1‖ + αn‖f (xn) − Tyn,N‖ + ‖yn,N − xn‖

≤ ‖xn − xn+1‖ + αn‖f (xn) − Tyn,N‖ +

N−k=2

‖yn,k − yn,k−1‖ + ‖yn,1 − xn‖

≤ ‖xn − xn+1‖ + αn‖f (xn) − Tyn,N‖ +

N−k=2

βn,k‖Skun − yn,k−1‖ + ‖yn,1 − xn‖

so, ‖xn − Txn‖ → 0, as n → ∞, by using (3.10), Lemma 3.2 and Remark 3.10.By demiclosedness principle p ∈ Fix(T ).By Lemma 3.3 and Lemma 2.5, we note that p ∈ EP(F , h).

Theorem 3.12. Let us suppose that Ω = ∅. Let (αn)n∈N, (βn,i)n∈N, i = 1, . . . ,N sequences in (0, 1) such that βn,i → βi ∈

(0, 1) as n → ∞, for all index i. Moreover, let us suppose that (H1)–(H4) hold. Then the sequences (xn)n∈N and (un)n∈N, explicitlydefined by scheme (3.1), both strongly converge to the unique solution x∗

∈ Ω of the variational inequality

⟨f (x∗) − x∗, z − x∗⟩ ≤ 0, ∀z ∈ Ω. (3.11)

Proof. Since themap PΩ f is a ρ-contraction, it has a unique fixed point x∗; it is the unique solution of (3.11). Since (H1)–(H4)hold, the sequence (xn)n∈N is asymptotically regular (Lemma 3.2). By Lemma 3.3, ‖xn − un‖ → 0, as n → ∞. Moreover,

‖xn+1 − x∗‖2

≤ ‖αn(f (xn) − f (x∗)) + (1 − αn)(Tyn,N − x∗)‖2+ 2αn⟨f (x∗) − x∗, xn+1 − x∗

⟩

≤ αnρ‖xn − x∗‖2+ (1 − αn)‖yn,N − x∗

‖2+ 2αn⟨f (x∗) − x∗, xn+1 − x∗

⟩

≤ αnρ‖xn − x∗‖2+ (1 − αn)‖xn − x∗

‖2+ 2αn⟨f (x∗) − x∗, xn+1 − x∗

⟩

≤ [1 − (1 − ρ)αn]‖xn − x∗‖2+ 2αn⟨f (x∗) − x∗, xn+1 − x∗

⟩.

Recalling that (by Lemma 2.5) every weak cluster point z of (xn)n∈N is in Ω , then for an opportune subsequence (xnk) z

lim supn→∞

⟨f (x∗) − x∗, xn+1 − x∗⟩ = lim

k→∞

⟨f (x∗) − x∗, xnk − x∗⟩ = ⟨f (x∗) − x∗, z − x∗

⟩ ≤ 0.

By Lemma 2.6 xn → x∗, as n → ∞.


In a similar way, we can conclude as follows.

Theorem 3.13. Let us suppose that Ω = ∅. Let (αn)n∈N, (βn,i)n∈N, i = 1, . . . ,N, sequences in (0, 1) such that (βn,i)n∈N → βifor all i as n → ∞. Suppose that there exists k ∈ 1, . . . ,N for which βn,k → 0 as n → ∞. Let k0 ∈ 1, . . . ,N the largestindex for which βn,k → 0. Moreover, let us suppose that (H1), (H5)and (H6) hold and(i) αn

βn,k0→ 0, as n → ∞;


→ 0, as n → ∞;(iii) if (βn,i)n∈N → βi = 0 thus βi lies in (0, 1).Then the sequences (xn)n∈N and (un)n∈N explicitly defined by scheme (3.1) strongly converge to the unique solution x∗

∈ Ω of thevariational inequality

⟨f (x∗) − x∗, z − x∗⟩ ≤ 0, ∀z ∈ Ω. (3.12)

If A : C → H is a nonlinear mapping, let us consider the VIPto find x ∈ C s.t. ⟨Ax, y − x⟩ ≥ 0 ∀y ∈ C .

We will indicate with VI(C, A) the set of solutions of VIP.Recall that a point u ∈ C is a solution to a problem VI(C, A) if and only if

u = PC (I − λA)u ∀λ > 0. (3.13)

Definition 3.14. An operator A : C → H is said to be an α-inverse strongly monotone operator if there exists a constantα > 0 such that

⟨Ax − Ay, x − y⟩ ≥ α‖Ax − Ay‖2, ∀x, y ∈ C .

As an example, we recall that the α-inverse strongly monotone operators are firmly nonexpansive mappings if α ≥ 1and that every α-inverse strongly monotone operator is also 1

αLipschitz continuous (see [25]).

Let us observe also that, if A is α-inverse strongly monotone, the mapping PC (I − λA) are nonexpansive for all λ > 0since they are compositions of nonexpansive mappings (see page 419 in [25]).

Let us consider S1, . . . , SM a finite number of nonexpansive self-mappings on C and A1, . . . , AN be a finite number of α-inverse stronglymonotone operators. Let T be a nonexpansivemapping on C with fixed points. Let us consider the followingmixed problem.

To find x∗∈ Fix(T ) ∩ EP(F , h) such that

⟨(I − S1)x∗, y − x∗⟩ ≥ 0, y ∈ Fix(T ) ∩ EP(F , h)

⟨(I − S2)x∗, y − x∗⟩ ≥ 0, y ∈ Fix(T ) ∩ EP(F , h)

· · ·

⟨(I − SM)x∗, y − x∗⟩ ≥ 0, y ∈ Fix(T ) ∩ EP(F , h)

⟨A1x∗, y − x∗⟩ ≥ 0, y ∈ C

⟨A2x∗, y − x∗⟩ ≥ 0, y ∈ C

· · ·

⟨ANx∗, y − x∗⟩ ≥ 0, y ∈ C .

(3.14)

Let us call (SVIP) the set of solution of the (N + M)-system. This problem is equivalent to find a common fixed point of T ,(PFixT∩EP(F ,h)Si)i=1,...,N , (PC (I − λAi))i=1...,M . So we claim that

Theorem 3.15. Let us suppose that Ω = Fix(T )∩(SVIP)∩EP(f , h) = ∅. Fix λ > 0. Let (αn)n∈N, (βn,i)n∈N, i = 1, . . . , (N+M)sequences in (0, 1) such that βn,i → βi ∈ (0, 1) as n → ∞, for all index i. Moreover, let us suppose that (H1)–(H4) hold. Thenthe sequences (xn)n∈N and (un)n∈N explicitly defined by scheme

F(un, y) + h(un, y) +1rn

⟨y − un, un − xn⟩ ≥ 0, ∀y ∈ C

yn,1 = βn,1PFixT∩EP(G)S1un + (1 − βn,1)un

yn,i = βn,iPFixT∩EP(G)Siun + (1 − βn,i)yn,i−1, i = 2 . . . ,Myn,j = βn,iPC (I − λAj)un + (1 − βn,i)yn,i−1, j = 1 . . . ,Nxn+1 = αnf (xn) + (1 − αn)Tyn,N , n ≥ 1

(3.15)

both strongly converge to the unique solution x∗∈ Ω of the variational inequality

⟨f (x∗) − x∗, z − x∗⟩ ≤ 0, ∀z ∈ Ω.

Theorem 3.16. Let us suppose that Ω = ∅. Fix λ > 0. Let (αn)n∈N, (βn,i)n∈N, i = 1, . . . , (N + M), sequences in (0, 1) and(βn,i)n∈N → βi for all i as n → ∞. Suppose there exists k ∈ 1, . . . ,N + M such that βn,k → 0, as n → ∞.


Let k0 ∈ 1, . . . ,N + M be the largest index for which βn,k → 0. Moreover, let us suppose that (H1), (H5)and (H6) holdand

(i) αnβn,k0

→ 0, as n → ∞;


→ 0, as n → ∞.

Then the sequences (xn)n∈N and (un)n∈N explicitly defined by scheme (3.15) strongly converge to the unique solution x∗∈ Ω of

the ariational inequality

⟨f (x∗) − x∗, z − x∗⟩ ≤ 0, ∀z ∈ Ω.

Remark 3.17. If we choose A1 = · · · = AN = 0 in system (3.14), we obtain a system of hierarchical fixed point problemsintroduced by Mainge and Moudafi in [1,2].

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