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Viscosity methods for common solutions of equilibrium and variational inequality problems via multi-step iterative algorithms and common fixed points
Nonlinear Analysis 75 (2012) 1787–1798
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Viscosity methods for common solutions of equilibrium and variational inequality problems via multi-step iterative algorithms and common fixed points✩ Giuseppe Marino a,∗ , Luigi Muglia a , YongHong Yao b a
Dipartimento di Matematica, Universitá della Calabria, 87036 Arcavacata di Rende (CS), Italy
b
Department of Mathematics, Tianjin Polytechnic University, Tianjin 300160, People’s Republic of China
article
info
Article history: Received 9 August 2010 Accepted 15 September 2011 Communicated by Enzo Mitidieri
abstract In this paper, we present a new multi-step iterative method. We prove the strong convergence of the method to a common fixed point of a finite number of nonexpansive mappings that also solves a suitable equilibrium problem. © 2011 Elsevier Ltd. All rights reserved.
MSC: 47H09 47H10 58E35 Keywords: Hierarchical fixed points Equilibrium problem Nonexpansive map Contraction Variational inequality problem Projection
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 stated as: 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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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 , in turn, 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 = λn f (xn ) + (1 − λn )(αn Sxn + (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 = ∞; α −α
λ −λ
(P2) limn→∞ nα λn−1 = limn→∞ λ nλ n−α1 = 0; n n n n−1 n (P3) there exist two constants θ and k such that
‖x − Tx‖ ≥ k dist(x, Fix(T ))θ
∀x ∈ C , 1+ θ1
(P4) λn
= 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 by the explicit scheme
yn = βn Sxn + (1 − βn )xn xn+1 = αn f (xn ) + (1 − αn )Tyn ,
(1.2)
n ≥ 1.
Theorem 1.4. Let C be a nonempty closed and convex subset of a real Hilbert space H. Let S and T be two nonexpansive mappings on 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 =∞
α
(ii) limn→∞ α1 β1 − β 1 = 0, limn→∞ β1 1 − αn−1 = 0 n n n n n−1 β2 (iii) limn→∞ βn = 0, limn→∞ αβn = 0, limn→∞ αn = 0 n n (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, saddle point problems, variational inequality problems, minimization problems, Nash equilibria in noncooperative games, vector equilibrium problems and certain fixed point problems (see [11]).
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It is worth to remark that, in the case of VIP, the induced bifunction F (x, y) := ⟨(I − S )x, y − x⟩ satisfies the following conditions: (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 sup F (tz + (1 − t )x, y) ≤ F (x, y); t →0
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] by introducing ‘‘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 authors were interested in the construction of iterative algorithms that weakly or strongly converge to a common fixed point of a family of nonexpansive mappings (see e.g. [17–19]). In [20], Xu proves that the sequence generated by xn+1 = (I − ϵn+1 A)Tn+1 xn + ϵn+1 u where Tn = Tn mod N , strongly converges to a solution of a quadratic minimization problem under the assumption Fix(T1 T2 · · · TN ) = Fix(TN T1 · · · TN −1 ) = · · · = Fix(T2 T3 · · · T1 ). In [21], Yao lacks this hypothesis and studies the viscosity approximation of a common fixed point of the family of mappings. In [22], Colao et al. use a different approach to obtain the convergence of a more general scheme that involves an equilibrium problem. In this paper, our aims are as follows:
• to introduce a multi-step iterative method that generalizes the two-step method introduced in [8] for two nonexpansive mappings 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.
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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
Tr x =
z ∈ C : F (z , y) + h(z , y) +
1 r
⟨y − z , z − x⟩ ≥ 0, ∀y ∈ C
(2.1)
called resolvent of F and h. Then (1) (2) (3) (4)
Tr x ̸= ∅; Tr x is a singleton; Tr is firmly nonexpansive; 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
r2 − r1 ‖Tr y − y‖. ‖Tr2 y − Tr1 x‖ ≤ ‖y − x‖ + 2 r 2
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 respect to 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 − Trn xn ‖ → 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 − Trn xn ‖ → 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 = Trn xn 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) +
1 rn
⟨y − un , un − xn ⟩ ≥ F (y, un ).
In particular,
h(unm , y) + y − unm ,
unm − xnm
rnm
≥ F (y, unm ).
(2.2)
By condition (f3), for x ∈ H fixed, the function F (x, ·) is lower semicontinuous and convex, and thus weakly lower semicontinuous. Since ‖xn − un ‖ → 0, as n → ∞ and by the hypothesis on (rn )n , we obtain (unm − xnm )/rnm → 0. Therefore, letting m → ∞ in (2.2) yields F (y, q) ≤ lim F (y, unm ) ≤ lim h(unm , y) ≤ h(q, y), m→∞
m→∞
y ∈ H.
Replacing y with yτ := τ y + (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)).
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Hence we have F (τ y + (1 − τ )q, y) + h(τ y + (1 − τ )q, y) ≥ 0,
τ ∈ (0, 1], y ∈ H .
Letting τ → 0+ and using the hemicontinuity of F and the weak upper semicontinuity of h (in the first variable), we conclude that F (q, y) + h(q, y) ≥ 0, therefore q ∈ EP (f , h).
y ∈ 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 scheme
1 F (un , y) + h(un , y) + rn ⟨y − un , un − xn ⟩ ≥ 0, ∀y ∈ C yn,1 = βn,1 S1 un + (1 − βn,1 )un yn,i = βn,i Si un + (1 − βn,i )yn,i−1 , i = 2 . . . , N xn+1 = αn f (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∈N are 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‖. 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‖ αn ρ‖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‖, we obtain the claim.
‖f (v) − v‖ , 1−ρ
(3.2)
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Lemma 3.2. Let us suppose that Ω ̸= ∅. Moreover, let us suppose that (H1) αn → 0, as n → ∞ and
n∈N αn = ∞; |α −α | |α − α | < ∞ or limn→∞ n α n−1 = 0; n n−1 n∈N n ∑ |β −β | (H3) |βn,i − βn−1,i | < ∞ or limn→∞ n,i αnn−1,i = 0, for all i ∈ {1, . . . , N }; ∑n∈N |rn −rn−1 | (H4) = 0; n∈N |rn − rn−1 | < ∞ or limn→∞ αn
(H2)
∑
∑
hold. Then limn→∞ ‖xn+1 − xn ‖ = 0, i.e. (xn )n∈N is asymptotically regular. Proof. Observing that xn+1 − xn = αn f (xn ) + (1 − αn )Tyn,N − αn−1 f (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 ‖ ≤ αn ρ‖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 ‖ + ‖Si un−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 ‖ + ‖S1 un−1 − un−1 ‖ |βn,1 − βn−1,1 | + (1 − βn,1 )‖un − un−1 ‖ = ‖un − un−1 ‖ + ‖S1 un−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 −
‖Sk un−1 − yn−1,k−1 ‖ |βn,k − βn−1,k | + ‖S1 un−1 − un−1 ‖ |βn,1 − βn−1,1 |.
k=2
So we conclude that
‖xn+1 − xn ‖ ≤ αn ρ‖xn − xn−1 ‖ + ‖f (xn−1 ) − Tyn−1,N ‖ |αn − αn−1 | + (1 − αn )‖un − un−1 ‖ N − + ‖Sk un−1 − yn−1,k−1 ‖ |βn,k − βn−1,k | + ‖S1 un−1 − un−1 ‖ |βn,1 − βn−1,1 |. k =2
By Lemma 2.3, we know that
r n −1 ‖un − un−1 ‖ ≤ ‖xn − xn−1 ‖ + L 1 −
(3.6)
rn
where L = supn ‖un − xn ‖ so, substituting (3.6) in (3.3) we obtain
rn − rn−1 ‖xn+1 − xn ‖ ≤ αn ρ‖xn − xn−1 ‖ + |αn − αn−1 | ‖f (xn−1 ) − Tyn−1,N ‖ + (1 − αn )‖xn − xn−1 ‖ + L r n
+
N −
‖Sk un−1 − yn−1,k−1 ‖ |βn,k − βn−1,k | + ‖S1 un−1 − un−1 ‖ |βn,1 − βn−1,1 |.
k =2
If we call M := max{sup ‖f (xn−1 ) − Tyn−1,N ‖, L, n∈N
sup
n∈N,i=2,...,N
‖Si un−1 − yn−1,i−1 ‖, sup ‖S1 un−1 − un−1 ‖} n∈N
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 | + By hypotheses (H1)–(H4) and Lemma 2.6, we obtain the claim.
|rn − rn−1 | b
+
N −
|βn,k − βn−1,k | .
(3.7)
k=1
Lemma 3.3. Let us suppose that Ω ̸= ∅. Let us suppose that (xn )n∈N is asymptotically regular. Then ‖xn − un ‖ = ‖xn − Trn xn ‖ → 0, as n → ∞.
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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 points of (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 lim
n→∞
‖ x n +1 − x n ‖ = 0 ∀i ∈ {1, . . . , N }. βn,i
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 | αn βn,k0
= limn→∞
|αn −αn−1 | αn βn,k0
(H6) there exists a constant K > 0 such that α1 n
| r −r
|
= limn→∞ αnn βnn,−k 1 = 0; 0 1 βn,k − βn−11,k < K , for all n > 1 0
0
hold. Then lim
n→∞
‖ x n +1 − x n ‖ = 0. βn,k0
(3.8)
Proof. We start by (3.7). Dividing both the terms by βn,k0 we have
|α − α | |r − r | ‖xn+1 − xn ‖ ‖ x n − x n −1 ‖ n −1 n n−1 n ≤ [1 − αn (1 − ρ)] +M + + βn,k0 βn,k0 βn,k0 bβn,k0
N ∑ k=1
|βn,k − βn−1,k | . βn,k0
So
1 ‖xn+1 − xn ‖ ‖ x n − x n −1 ‖ 1 ≤ [1 − αn (1 − ρ)] + [1 − αn (1 − ρ)]‖xn − xn−1 ‖ − βn,k0 βn−1,k0 βn,k0 βn−1,k0 N ∑ |βn,k − βn−1,k | |α − α | |r − r | n n −1 n−1 n k=1 +M + + βn,k0 bβn,k0 βn,k0 1 1 ‖ x n − x n −1 ‖ ≤ [1 − αn (1 − ρ)] + ‖ x n +1 − x n ‖ − β β β n−1,k0 n,k0 n−1,k0 N ∑ |βn,k − βn−1,k | |α − α | |r − r | n−1 n n −1 n k=1 +M + + βn,k0 bβn,k0 βn,k0
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‖xn − xn−1 ‖ + αn K ‖xn+1 − xn ‖ βn−1,k0 N ∑ |β − β | n , k n − 1 , k |α − α | |r − r | n −1 n n −1 n k=1 +M + + βn,k0 bβn,k0 βn,k0
by (H6) ≤ [1 − αn (1 − ρ)]
‖ x n − x n −1 ‖ + αn KM βn−1,k0 N ∑ |βn,k − βn−1,k | |α − α | |r − r | n−1 n n −1 n k=1 +M + + . βn,k0 bβn,k0 βn,k0
by Lemma 3.1 ≤ [1 − αn (1 − ρ)]
By the boundedness of (xn )n∈N , by (H1), (H5) and Lemma 2.6, we conclude that lim
n→∞
‖xn+1 − xn ‖ = 0. βn,k0
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, ‖Si un − 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 ‖Si un − 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 ‖SN un − v‖2 + (1 − αn )(1 − βn,N )‖yn,N −1 − v‖2 − (1 − αn )(1 − βn,N )βn,N ‖SN un − yn,N −1 ‖2 ≤ αn ‖f (xn ) − v‖2 + (1 − αn )‖un − v‖2 − (1 − αn )βn,N (1 − βn,N )‖SN un − yn,N −1 ‖2 ≤ αn ‖f (xn ) − v‖2 + ‖xn − v‖2 − (1 − αn )βn,N (1 − βn,N )‖SN un − yn,N −1 ‖2 . So we have the inequality
(1 − αn )βn,N (1 − βn,N )‖SN un − 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 (‖SN un − 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 ‖SN un − 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 −1 un − 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 N ∏ + (1 − αn ) (1 − βn,k )‖yn,N −2 − v‖2 k=N −1
and so, after (N − i + 1)-iterations
‖xn+1 − v‖ ≤ αn ‖f (xn ) − v‖ + (1 − αn ) βn,N + 2
2
N − j=i+2
+ (1 − αn )
N ∏ k=i+1
(1 − βn,k )‖yn,i − v‖2
N ∏ p=j
(1 − βn,p ) βn,j−1 ‖xn − v‖2
G. Marino et al. / Nonlinear Analysis 75 (2012) 1787–1798
≤ αn ‖f (xn ) − v‖2 + (1 − αn ) βn,N +
N −
j=i+2 N ∏
× ‖xn − v‖2 + (1 − αn )
N ∏
1795
(1 − βn,p ) βn,j−1
p=j
(1 − βn,k )
k=i+1
× [βn,i ‖Si un − v‖2 + (1 − βn,i )‖yn,i−1 − v‖2 − βn,i (1 − βn,i )‖Si un − yn,i−1 ‖2 ] N ∏ ≤ αn ‖f (xn ) − v‖2 + (1 − αn )‖xn − v‖2 − βn,i (1 − αn ) (1 − βn,k )‖Si un − yn,i−1 ‖2 . k=i
Again we obtain that
(1 − αn )βn,i
N ∏
(1 − βn,k )‖Si un − yn,i−1 ‖2 ≤ αn ‖f (xn ) − v‖2 + ‖xn − v‖2 − ‖xn+1 − v‖2
k=i
≤ αn ‖f (xn ) − v‖2 + L‖xn+1 − xn ‖ so ‖Si un − yn,i−1 ‖ → 0 as n → ∞. Obviously for i = 1, we have ‖S1 un − un ‖ → 0. To conclude, we have that
‖S2 un − un ‖ ≤ ‖S2 un − yn,1 ‖ + ‖yn,1 − un ‖ = ‖S2 un − yn,1 ‖ + βn,1 ‖S1 un − un ‖ from which ‖S2 un − un ‖ → 0. Thus by induction ‖Si un − un ‖ → 0 for all i = 2, . . . , N since it is enough to observe that
‖Si un − un ‖ ≤ ‖Si un − yn,i−1 ‖ + ‖yn,i−1 − Si−1 un ‖ + ‖Si−1 un − un ‖ ≤ ‖Si un − yn,i−1 ‖ + (1 − βn,i−1 )‖Si−1 un − yn,i−2 ‖ + ‖Si−1 un − un ‖. Example 3.7. As an example, if we consider N = 2 and the sequences 1
αn = √ , βn,1 =
n 1 2
rn = 2 − 1
− ,
1 n
βn,2 =
n
1 2
−
1 n2
,
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 that βn,k → 0, as n → ∞. Let k0 ∈ {1, . . . , N } the largest index such that (βn,k0 )n∈N → 0, as n → ∞. Suppose that (i) βαn → 0, as n → ∞; n,k0
βn,k
(ii) if i ≤ k0 and (βn,i )n∈N → 0 then β 0 → 0, as n → ∞. n,i Moreover, suppose that (H1), (H5) and (H6) hold. Then, for all i, ‖Si un − 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 ‖Si un − yn,i−1 ‖ → 0 as n → ∞. For all the other indexes i ≤ k0 , we can prove that ‖Si un − 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 ∏
(1 − βn,k )‖Si un − yn,i−1 ‖2
k=i
thus
(1 − αn )
N ∏
(1 − βn,k )‖Si un − yn,i−1 ‖2 ≤
k=i
αn ‖ x n − x n +1 ‖ L ‖f (xn ) − v‖2 + . βn,i βn,i
By Lemma 3.5 or by hypothesis (ii) on the sequences
‖xn − xn+1 ‖L ‖xn − xn+1 ‖L βn,k0 = →0 βn,i βn,k0 βn,i so the thesis follows.
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Example 3.9. Let us consider N = 3 and the following sequences 1
αn = √ , n 1
1
rn = 2 −
βn,1 = √ , 4
βn,2 =
n 1
n2 1 2
−
1 n2
,
n>1
βn,3 = √ 3
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 lim ‖Si un − yn,i−1 ‖ = 0,
n→∞
∀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 ‖S1 un − un ‖. By Lemma 3.6, ‖S1 un − 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 ‖ N − ≤ ‖xn − xn+1 ‖ + αn ‖f (xn ) − Tyn,N ‖ + ‖yn,k − yn,k−1 ‖ + ‖yn,1 − xn ‖ k=2
≤ ‖xn − xn+1 ‖ + αn ‖f (xn ) − Tyn,N ‖ +
N −
βn,k ‖Sk un − yn,k−1 ‖ + ‖yn,1 − xn ‖
k=2
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 , explicitly defined 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 the map 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∗ ⟩ ≤ αn ρ‖xn − x∗ ‖2 + (1 − αn )‖yn,N − x∗ ‖2 + 2αn ⟨f (x∗ ) − x∗ , xn+1 − x∗ ⟩ ≤ αn ρ‖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 sup⟨f (x∗ ) − x∗ , xn+1 − x∗ ⟩ = lim ⟨f (x∗ ) − x∗ , xnk − x∗ ⟩ = ⟨f (x∗ ) − x∗ , z − x∗ ⟩ ≤ 0. k→∞
n→∞
∗
By Lemma 2.6 xn → x , as n → ∞.
G. Marino et al. / Nonlinear Analysis 75 (2012) 1787–1798
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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 → βi for all i as n → ∞. Suppose that there exists k ∈ {1, . . . , N } for which βn,k → 0 as n → ∞. Let k0 ∈ {1, . . . , N } the largest index for which βn,k → 0. Moreover, let us suppose that (H1), (H5)and (H6) hold and (i) βαn → 0, as n → ∞; n,k0
βn,k
(ii) if i ≤ k0 and (βn,i )n∈N → 0 then β 0 → 0, as n → ∞; n,i (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 the variational inequality
⟨f (x∗ ) − x∗ , z − x∗ ⟩ ≤ 0,
∀z ∈ Ω .
(3.12)
If A : C → H is a nonlinear mapping, let us consider the VIP to 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 constant α > 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 ≥ 1 and that every α -inverse strongly monotone operator is also α1 Lipschitz 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 > 0 since they are compositions of nonexpansive mappings (see page 419 in [25]). Let us consider S˜1 , . . . , S˜M a finite number of nonexpansive self-mappings on C and A1 , . . . , AN be a finite number of α inverse strongly monotone operators. Let T be a nonexpansive mapping on C with fixed points. Let us consider the following mixed problem. To find x∗ ∈ Fix(T ) ∩ EP (F , h) such that
⟨(I − S˜1 )x∗ , y − x∗ ⟩ ≥ 0, ⟨(I − S˜2 )x∗ , y − x∗ ⟩ ≥ 0, · · · ⟨(I − S˜M )x∗ , y − x∗ ⟩ ≥ 0, ∗ ∗ ⟨A1 x∗ , y − x∗ ⟩ ≥ 0, ⟨A2 x , y − x ⟩ ≥ 0, · · · ⟨AN x∗ , y − x∗ ⟩ ≥ 0,
y ∈ Fix(T ) ∩ EP (F , h) y ∈ Fix(T ) ∩ EP (F , h) y ∈ Fix(T ) ∩ EP (F , h) y∈C y∈C
(3.14)
y ∈ C.
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) S˜i )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. Then the sequences (xn )n∈N and (un )n∈N explicitly defined by scheme
1 F (un , y) + h(un , y) + ⟨y − un , un − xn ⟩ ≥ 0, rn yn,1 = βn,1 PFixT ∩EP (G) S˜1 un + (1 − βn,1 )un yn,i = βn,i PFixT ∩EP (G) S˜i un + (1 − βn,i )yn,i−1 , yn,j = βn,i PC (I − λAj )un + (1 − βn,i )yn,i−1 , xn+1 = αn f (xn ) + (1 − αn )Tyn,N ,
∀y ∈ C i = 2...,M j = 1...,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 → ∞.
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Let k0 ∈ {1, . . . , N + M } be the largest index for which βn,k → 0. Moreover, let us suppose that (H1), (H5)and (H6) hold and (i) βαn → 0, as n → ∞; n,k0
βn,k
(ii) if i ≤ k0 and (βn,i )n∈N → 0 then β 0 → 0, as n → ∞. n,i 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 problems introduced by Mainge and Moudafi in [1,2]. References [1] A. Moudafi, P.-E. Mainge, Towards viscosity approximations of hierarchical fixed points problems, Fixed Point Theory Appl. (2006) Art. ID 95453, pp. 10. [2] A. Moudafi, P.-E. Mainge, Strong convergence of an iterative method for hierarchical fixed point problems, Pac. J. Optim. 3 (3) (2007) 529–538. [3] A. Moudafi, Viscosity approximation methods for fixed-points problems, J. Math. Anal. Appl. 241 (1) (2000) 46–55. [4] H.K. Xu, Viscosity approximation methods for nonexpansive mappings, J. Math. Anal. Appl. 298 (2004) 279–291. [5] G. Marino, H.K. Xu, A general iterative method for nonexpansive mappings in Hilbert spaces, J. Math. Anal. Appl. 318 (1) (2006) 43–52. [6] H.K. Xu, Iterative algorithms for nonlinear operators, J. Lond. Math. Soc. 2 (2002) 1–17. [7] F. Cianciaruso, V. Colao, L. Muglia, H.K. Xu, On implicit methods for variational inequalities via hierarchical fixed point approach, Bull. Aust. Math. Soc. 80 (1) (2009) 117–124. [8] Y. Yao, Y.-C. Liou, G. Marino, Two-step iterative algorithms for hierarchical fixed point problems and variational inequality problems, J. Appl. Math. Comput. 31 (1–2) (2009) 433–445. [9] E. Blum, W. Oettli, From optimization and variational inequalities to equilibrium problems, Math. Student 63 (1–4) (1994) 123–145. [10] W. Oettli, A remark on vector-valued equilibria and generalized monotonicity, Acta Math. Vietnam. 22 (1997) 215–221. [11] A. Göpfert, H. Riahi, C. Tammer, C. Zălinescu, Variational Methods in Partially Ordered Spaces, Springer-Verlag, New York, 2003. [12] Patrick L. Combettes, Sever A. Hirstoaga, Equilibrium programming in Hilbert spaces, J. Nonlinear Convex Anal. 6 (1) (2005) 117–136. [13] A. Moudafi, Weak convergence theorems for nonexpansive mappings and equilibrium problems, J. Nonlinear Convex Anal. 9 (1) (2008) 37–43. [14] J.W. Peng, Iterative algorithms for mixed equilibrium problems, strictly pseudo-contractions and monotone mappings, J. Optim. Theory Appl. 144 (1) (2010) 107–119. [15] L.C. Ceng, J.C. Yao, A hybrid iterative scheme for mixed equilibrium problems and fixed point problems, J. Comput. Appl. Math. 214 (1) (2008) 186–201 (English summary). [16] J.W. Peng, J.C. Yao, A new hybrid-extragradient method for generalized mixed equilibrium problems, fixed point problems and variational inequality problems, Taiwanese J. Math. 12 (6) (2008) 1401–1432. [17] S. Atsushiba, W. Takahashi, Strong convergence theorems for a finite family of nonexpansive mappings and applications, in: B.N. Prasad Birth Centenary Commemoration Volume, Indian J. Math. 41 (3) (1999) 435–453. [18] R.E. Bruck, Properties of fixed point sets of nonexpansive mapping in Banach spaces, Trans. Amer. Math. Soc. 179 (1973) 251–262. [19] P.K.F. Kuhfittig, Common fixed points of nonexpansive mappings by iteration, Pacific J. Math. 97 (1) (1981) 137–139. [20] H.K. Xu, An iterative approach to quadratic optimization, J. Optim. Theory Appl. 116 (3) (2003) 659–678. [21] Y. Yao, A general iterative method for a finite family of nonexpansive mappings, Nonlinear Anal. 66 (2007) 2676–2687. [22] V. Colao, G. Marino, H.K. Xu, An iterative method for finding common solutions of equilibrium and fixed point problems, J. Math. Anal. Appl. 344 (1) (2008) 340–352 (English summary). [23] J. Reineermann, Uber fixpunkte kontrahierender abbildungen und schwach konvergente Toeplitz-verfahren, Arch. Math. (Basel) 20 (1969) 59–64. [24] F. Cianciaruso, G. Marino, L. Muglia, Y. Yao, A hybrid projection algorithm for finding solutions of mixed equilibrium problem and variational inequality problem, Fixed Point Theory Appl. (2010) Art. ID 383740, 19 pp. [25] W. Takahashi, M. Toyoda, Weak convergence theorems for nonexpansive mappings and monotone operators, J. Optim. Theory Appl. 118 (2) (2003) 417–428.
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Viscosity methods for common solutions of equilibrium and variational inequality problems via multi-step iterative algorithms and common fixed points✩ Giuseppe Marino a,∗ , Luigi Muglia a , YongHong Yao b a
Dipartimento di Matematica, Universitá della Calabria, 87036 Arcavacata di Rende (CS), Italy
b
Department of Mathematics, Tianjin Polytechnic University, Tianjin 300160, People’s Republic of China
article
info
Article history: Received 9 August 2010 Accepted 15 September 2011 Communicated by Enzo Mitidieri
abstract In this paper, we present a new multi-step iterative method. We prove the strong convergence of the method to a common fixed point of a finite number of nonexpansive mappings that also solves a suitable equilibrium problem. © 2011 Elsevier Ltd. All rights reserved.
MSC: 47H09 47H10 58E35 Keywords: Hierarchical fixed points Equilibrium problem Nonexpansive map Contraction Variational inequality problem Projection
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 stated as: 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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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 , in turn, 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 = λn f (xn ) + (1 − λn )(αn Sxn + (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 = ∞; α −α
λ −λ
(P2) limn→∞ nα λn−1 = limn→∞ λ nλ n−α1 = 0; n n n n−1 n (P3) there exist two constants θ and k such that
‖x − Tx‖ ≥ k dist(x, Fix(T ))θ
∀x ∈ C , 1+ θ1
(P4) λn
= 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 by the explicit scheme
yn = βn Sxn + (1 − βn )xn xn+1 = αn f (xn ) + (1 − αn )Tyn ,
(1.2)
n ≥ 1.
Theorem 1.4. Let C be a nonempty closed and convex subset of a real Hilbert space H. Let S and T be two nonexpansive mappings on 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 =∞
α
(ii) limn→∞ α1 β1 − β 1 = 0, limn→∞ β1 1 − αn−1 = 0 n n n n n−1 β2 (iii) limn→∞ βn = 0, limn→∞ αβn = 0, limn→∞ αn = 0 n n (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, saddle point problems, variational inequality problems, minimization problems, Nash equilibria in noncooperative games, vector equilibrium problems and certain fixed point problems (see [11]).
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It is worth to remark that, in the case of VIP, the induced bifunction F (x, y) := ⟨(I − S )x, y − x⟩ satisfies the following conditions: (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 sup F (tz + (1 − t )x, y) ≤ F (x, y); t →0
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] by introducing ‘‘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 authors were interested in the construction of iterative algorithms that weakly or strongly converge to a common fixed point of a family of nonexpansive mappings (see e.g. [17–19]). In [20], Xu proves that the sequence generated by xn+1 = (I − ϵn+1 A)Tn+1 xn + ϵn+1 u where Tn = Tn mod N , strongly converges to a solution of a quadratic minimization problem under the assumption Fix(T1 T2 · · · TN ) = Fix(TN T1 · · · TN −1 ) = · · · = Fix(T2 T3 · · · T1 ). In [21], Yao lacks this hypothesis and studies the viscosity approximation of a common fixed point of the family of mappings. In [22], Colao et al. use a different approach to obtain the convergence of a more general scheme that involves an equilibrium problem. In this paper, our aims are as follows:
• to introduce a multi-step iterative method that generalizes the two-step method introduced in [8] for two nonexpansive mappings 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.
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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
Tr x =
z ∈ C : F (z , y) + h(z , y) +
1 r
⟨y − z , z − x⟩ ≥ 0, ∀y ∈ C
(2.1)
called resolvent of F and h. Then (1) (2) (3) (4)
Tr x ̸= ∅; Tr x is a singleton; Tr is firmly nonexpansive; 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
r2 − r1 ‖Tr y − y‖. ‖Tr2 y − Tr1 x‖ ≤ ‖y − x‖ + 2 r 2
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 respect to 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 − Trn xn ‖ → 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 − Trn xn ‖ → 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 = Trn xn 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) +
1 rn
⟨y − un , un − xn ⟩ ≥ F (y, un ).
In particular,
h(unm , y) + y − unm ,
unm − xnm
rnm
≥ F (y, unm ).
(2.2)
By condition (f3), for x ∈ H fixed, the function F (x, ·) is lower semicontinuous and convex, and thus weakly lower semicontinuous. Since ‖xn − un ‖ → 0, as n → ∞ and by the hypothesis on (rn )n , we obtain (unm − xnm )/rnm → 0. Therefore, letting m → ∞ in (2.2) yields F (y, q) ≤ lim F (y, unm ) ≤ lim h(unm , y) ≤ h(q, y), m→∞
m→∞
y ∈ H.
Replacing y with yτ := τ y + (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)).
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Hence we have F (τ y + (1 − τ )q, y) + h(τ y + (1 − τ )q, y) ≥ 0,
τ ∈ (0, 1], y ∈ H .
Letting τ → 0+ and using the hemicontinuity of F and the weak upper semicontinuity of h (in the first variable), we conclude that F (q, y) + h(q, y) ≥ 0, therefore q ∈ EP (f , h).
y ∈ 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 scheme
1 F (un , y) + h(un , y) + rn ⟨y − un , un − xn ⟩ ≥ 0, ∀y ∈ C yn,1 = βn,1 S1 un + (1 − βn,1 )un yn,i = βn,i Si un + (1 − βn,i )yn,i−1 , i = 2 . . . , N xn+1 = αn f (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∈N are 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‖. 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‖ αn ρ‖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‖, we obtain the claim.
‖f (v) − v‖ , 1−ρ
(3.2)
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Lemma 3.2. Let us suppose that Ω ̸= ∅. Moreover, let us suppose that (H1) αn → 0, as n → ∞ and
n∈N αn = ∞; |α −α | |α − α | < ∞ or limn→∞ n α n−1 = 0; n n−1 n∈N n ∑ |β −β | (H3) |βn,i − βn−1,i | < ∞ or limn→∞ n,i αnn−1,i = 0, for all i ∈ {1, . . . , N }; ∑n∈N |rn −rn−1 | (H4) = 0; n∈N |rn − rn−1 | < ∞ or limn→∞ αn
(H2)
∑
∑
hold. Then limn→∞ ‖xn+1 − xn ‖ = 0, i.e. (xn )n∈N is asymptotically regular. Proof. Observing that xn+1 − xn = αn f (xn ) + (1 − αn )Tyn,N − αn−1 f (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 ‖ ≤ αn ρ‖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 ‖ + ‖Si un−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 ‖ + ‖S1 un−1 − un−1 ‖ |βn,1 − βn−1,1 | + (1 − βn,1 )‖un − un−1 ‖ = ‖un − un−1 ‖ + ‖S1 un−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 −
‖Sk un−1 − yn−1,k−1 ‖ |βn,k − βn−1,k | + ‖S1 un−1 − un−1 ‖ |βn,1 − βn−1,1 |.
k=2
So we conclude that
‖xn+1 − xn ‖ ≤ αn ρ‖xn − xn−1 ‖ + ‖f (xn−1 ) − Tyn−1,N ‖ |αn − αn−1 | + (1 − αn )‖un − un−1 ‖ N − + ‖Sk un−1 − yn−1,k−1 ‖ |βn,k − βn−1,k | + ‖S1 un−1 − un−1 ‖ |βn,1 − βn−1,1 |. k =2
By Lemma 2.3, we know that
r n −1 ‖un − un−1 ‖ ≤ ‖xn − xn−1 ‖ + L 1 −
(3.6)
rn
where L = supn ‖un − xn ‖ so, substituting (3.6) in (3.3) we obtain
rn − rn−1 ‖xn+1 − xn ‖ ≤ αn ρ‖xn − xn−1 ‖ + |αn − αn−1 | ‖f (xn−1 ) − Tyn−1,N ‖ + (1 − αn )‖xn − xn−1 ‖ + L r n
+
N −
‖Sk un−1 − yn−1,k−1 ‖ |βn,k − βn−1,k | + ‖S1 un−1 − un−1 ‖ |βn,1 − βn−1,1 |.
k =2
If we call M := max{sup ‖f (xn−1 ) − Tyn−1,N ‖, L, n∈N
sup
n∈N,i=2,...,N
‖Si un−1 − yn−1,i−1 ‖, sup ‖S1 un−1 − un−1 ‖} n∈N
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 | + By hypotheses (H1)–(H4) and Lemma 2.6, we obtain the claim.
|rn − rn−1 | b
+
N −
|βn,k − βn−1,k | .
(3.7)
k=1
Lemma 3.3. Let us suppose that Ω ̸= ∅. Let us suppose that (xn )n∈N is asymptotically regular. Then ‖xn − un ‖ = ‖xn − Trn xn ‖ → 0, as n → ∞.
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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 points of (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 lim
n→∞
‖ x n +1 − x n ‖ = 0 ∀i ∈ {1, . . . , N }. βn,i
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 | αn βn,k0
= limn→∞
|αn −αn−1 | αn βn,k0
(H6) there exists a constant K > 0 such that α1 n
| r −r
|
= limn→∞ αnn βnn,−k 1 = 0; 0 1 βn,k − βn−11,k < K , for all n > 1 0
0
hold. Then lim
n→∞
‖ x n +1 − x n ‖ = 0. βn,k0
(3.8)
Proof. We start by (3.7). Dividing both the terms by βn,k0 we have
|α − α | |r − r | ‖xn+1 − xn ‖ ‖ x n − x n −1 ‖ n −1 n n−1 n ≤ [1 − αn (1 − ρ)] +M + + βn,k0 βn,k0 βn,k0 bβn,k0
N ∑ k=1
|βn,k − βn−1,k | . βn,k0
So
1 ‖xn+1 − xn ‖ ‖ x n − x n −1 ‖ 1 ≤ [1 − αn (1 − ρ)] + [1 − αn (1 − ρ)]‖xn − xn−1 ‖ − βn,k0 βn−1,k0 βn,k0 βn−1,k0 N ∑ |βn,k − βn−1,k | |α − α | |r − r | n n −1 n−1 n k=1 +M + + βn,k0 bβn,k0 βn,k0 1 1 ‖ x n − x n −1 ‖ ≤ [1 − αn (1 − ρ)] + ‖ x n +1 − x n ‖ − β β β n−1,k0 n,k0 n−1,k0 N ∑ |βn,k − βn−1,k | |α − α | |r − r | n−1 n n −1 n k=1 +M + + βn,k0 bβn,k0 βn,k0
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‖xn − xn−1 ‖ + αn K ‖xn+1 − xn ‖ βn−1,k0 N ∑ |β − β | n , k n − 1 , k |α − α | |r − r | n −1 n n −1 n k=1 +M + + βn,k0 bβn,k0 βn,k0
by (H6) ≤ [1 − αn (1 − ρ)]
‖ x n − x n −1 ‖ + αn KM βn−1,k0 N ∑ |βn,k − βn−1,k | |α − α | |r − r | n−1 n n −1 n k=1 +M + + . βn,k0 bβn,k0 βn,k0
by Lemma 3.1 ≤ [1 − αn (1 − ρ)]
By the boundedness of (xn )n∈N , by (H1), (H5) and Lemma 2.6, we conclude that lim
n→∞
‖xn+1 − xn ‖ = 0. βn,k0
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, ‖Si un − 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 ‖Si un − 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 ‖SN un − v‖2 + (1 − αn )(1 − βn,N )‖yn,N −1 − v‖2 − (1 − αn )(1 − βn,N )βn,N ‖SN un − yn,N −1 ‖2 ≤ αn ‖f (xn ) − v‖2 + (1 − αn )‖un − v‖2 − (1 − αn )βn,N (1 − βn,N )‖SN un − yn,N −1 ‖2 ≤ αn ‖f (xn ) − v‖2 + ‖xn − v‖2 − (1 − αn )βn,N (1 − βn,N )‖SN un − yn,N −1 ‖2 . So we have the inequality
(1 − αn )βn,N (1 − βn,N )‖SN un − 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 (‖SN un − 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 ‖SN un − 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 −1 un − 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 N ∏ + (1 − αn ) (1 − βn,k )‖yn,N −2 − v‖2 k=N −1
and so, after (N − i + 1)-iterations
‖xn+1 − v‖ ≤ αn ‖f (xn ) − v‖ + (1 − αn ) βn,N + 2
2
N − j=i+2
+ (1 − αn )
N ∏ k=i+1
(1 − βn,k )‖yn,i − v‖2
N ∏ p=j
(1 − βn,p ) βn,j−1 ‖xn − v‖2
G. Marino et al. / Nonlinear Analysis 75 (2012) 1787–1798
≤ αn ‖f (xn ) − v‖2 + (1 − αn ) βn,N +
N −
j=i+2 N ∏
× ‖xn − v‖2 + (1 − αn )
N ∏
1795
(1 − βn,p ) βn,j−1
p=j
(1 − βn,k )
k=i+1
× [βn,i ‖Si un − v‖2 + (1 − βn,i )‖yn,i−1 − v‖2 − βn,i (1 − βn,i )‖Si un − yn,i−1 ‖2 ] N ∏ ≤ αn ‖f (xn ) − v‖2 + (1 − αn )‖xn − v‖2 − βn,i (1 − αn ) (1 − βn,k )‖Si un − yn,i−1 ‖2 . k=i
Again we obtain that
(1 − αn )βn,i
N ∏
(1 − βn,k )‖Si un − yn,i−1 ‖2 ≤ αn ‖f (xn ) − v‖2 + ‖xn − v‖2 − ‖xn+1 − v‖2
k=i
≤ αn ‖f (xn ) − v‖2 + L‖xn+1 − xn ‖ so ‖Si un − yn,i−1 ‖ → 0 as n → ∞. Obviously for i = 1, we have ‖S1 un − un ‖ → 0. To conclude, we have that
‖S2 un − un ‖ ≤ ‖S2 un − yn,1 ‖ + ‖yn,1 − un ‖ = ‖S2 un − yn,1 ‖ + βn,1 ‖S1 un − un ‖ from which ‖S2 un − un ‖ → 0. Thus by induction ‖Si un − un ‖ → 0 for all i = 2, . . . , N since it is enough to observe that
‖Si un − un ‖ ≤ ‖Si un − yn,i−1 ‖ + ‖yn,i−1 − Si−1 un ‖ + ‖Si−1 un − un ‖ ≤ ‖Si un − yn,i−1 ‖ + (1 − βn,i−1 )‖Si−1 un − yn,i−2 ‖ + ‖Si−1 un − un ‖. Example 3.7. As an example, if we consider N = 2 and the sequences 1
αn = √ , βn,1 =
n 1 2
rn = 2 − 1
− ,
1 n
βn,2 =
n
1 2
−
1 n2
,
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 that βn,k → 0, as n → ∞. Let k0 ∈ {1, . . . , N } the largest index such that (βn,k0 )n∈N → 0, as n → ∞. Suppose that (i) βαn → 0, as n → ∞; n,k0
βn,k
(ii) if i ≤ k0 and (βn,i )n∈N → 0 then β 0 → 0, as n → ∞. n,i Moreover, suppose that (H1), (H5) and (H6) hold. Then, for all i, ‖Si un − 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 ‖Si un − yn,i−1 ‖ → 0 as n → ∞. For all the other indexes i ≤ k0 , we can prove that ‖Si un − 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 ∏
(1 − βn,k )‖Si un − yn,i−1 ‖2
k=i
thus
(1 − αn )
N ∏
(1 − βn,k )‖Si un − yn,i−1 ‖2 ≤
k=i
αn ‖ x n − x n +1 ‖ L ‖f (xn ) − v‖2 + . βn,i βn,i
By Lemma 3.5 or by hypothesis (ii) on the sequences
‖xn − xn+1 ‖L ‖xn − xn+1 ‖L βn,k0 = →0 βn,i βn,k0 βn,i so the thesis follows.
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Example 3.9. Let us consider N = 3 and the following sequences 1
αn = √ , n 1
1
rn = 2 −
βn,1 = √ , 4
βn,2 =
n 1
n2 1 2
−
1 n2
,
n>1
βn,3 = √ 3
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 lim ‖Si un − yn,i−1 ‖ = 0,
n→∞
∀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 ‖S1 un − un ‖. By Lemma 3.6, ‖S1 un − 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 ‖ N − ≤ ‖xn − xn+1 ‖ + αn ‖f (xn ) − Tyn,N ‖ + ‖yn,k − yn,k−1 ‖ + ‖yn,1 − xn ‖ k=2
≤ ‖xn − xn+1 ‖ + αn ‖f (xn ) − Tyn,N ‖ +
N −
βn,k ‖Sk un − yn,k−1 ‖ + ‖yn,1 − xn ‖
k=2
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 , explicitly defined 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 the map 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∗ ⟩ ≤ αn ρ‖xn − x∗ ‖2 + (1 − αn )‖yn,N − x∗ ‖2 + 2αn ⟨f (x∗ ) − x∗ , xn+1 − x∗ ⟩ ≤ αn ρ‖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 sup⟨f (x∗ ) − x∗ , xn+1 − x∗ ⟩ = lim ⟨f (x∗ ) − x∗ , xnk − x∗ ⟩ = ⟨f (x∗ ) − x∗ , z − x∗ ⟩ ≤ 0. k→∞
n→∞
∗
By Lemma 2.6 xn → x , as n → ∞.
G. Marino et al. / Nonlinear Analysis 75 (2012) 1787–1798
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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 → βi for all i as n → ∞. Suppose that there exists k ∈ {1, . . . , N } for which βn,k → 0 as n → ∞. Let k0 ∈ {1, . . . , N } the largest index for which βn,k → 0. Moreover, let us suppose that (H1), (H5)and (H6) hold and (i) βαn → 0, as n → ∞; n,k0
βn,k
(ii) if i ≤ k0 and (βn,i )n∈N → 0 then β 0 → 0, as n → ∞; n,i (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 the variational inequality
⟨f (x∗ ) − x∗ , z − x∗ ⟩ ≤ 0,
∀z ∈ Ω .
(3.12)
If A : C → H is a nonlinear mapping, let us consider the VIP to 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 constant α > 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 ≥ 1 and that every α -inverse strongly monotone operator is also α1 Lipschitz 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 > 0 since they are compositions of nonexpansive mappings (see page 419 in [25]). Let us consider S˜1 , . . . , S˜M a finite number of nonexpansive self-mappings on C and A1 , . . . , AN be a finite number of α inverse strongly monotone operators. Let T be a nonexpansive mapping on C with fixed points. Let us consider the following mixed problem. To find x∗ ∈ Fix(T ) ∩ EP (F , h) such that
⟨(I − S˜1 )x∗ , y − x∗ ⟩ ≥ 0, ⟨(I − S˜2 )x∗ , y − x∗ ⟩ ≥ 0, · · · ⟨(I − S˜M )x∗ , y − x∗ ⟩ ≥ 0, ∗ ∗ ⟨A1 x∗ , y − x∗ ⟩ ≥ 0, ⟨A2 x , y − x ⟩ ≥ 0, · · · ⟨AN x∗ , y − x∗ ⟩ ≥ 0,
y ∈ Fix(T ) ∩ EP (F , h) y ∈ Fix(T ) ∩ EP (F , h) y ∈ Fix(T ) ∩ EP (F , h) y∈C y∈C
(3.14)
y ∈ C.
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) S˜i )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. Then the sequences (xn )n∈N and (un )n∈N explicitly defined by scheme
1 F (un , y) + h(un , y) + ⟨y − un , un − xn ⟩ ≥ 0, rn yn,1 = βn,1 PFixT ∩EP (G) S˜1 un + (1 − βn,1 )un yn,i = βn,i PFixT ∩EP (G) S˜i un + (1 − βn,i )yn,i−1 , yn,j = βn,i PC (I − λAj )un + (1 − βn,i )yn,i−1 , xn+1 = αn f (xn ) + (1 − αn )Tyn,N ,
∀y ∈ C i = 2...,M j = 1...,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 → ∞.
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Let k0 ∈ {1, . . . , N + M } be the largest index for which βn,k → 0. Moreover, let us suppose that (H1), (H5)and (H6) hold and (i) βαn → 0, as n → ∞; n,k0
βn,k
(ii) if i ≤ k0 and (βn,i )n∈N → 0 then β 0 → 0, as n → ∞. n,i 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 .
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