IS S N 1 3 0 3 –5 9 9 1
WEAK CONVERGENCE THEOREM BY A NEW
EXTRAGRADIENT METHOD FOR FIXED POINT PROBLEMS AND VARIATIONAL INEQUALITY PROBLEMS
·
IBRAHIM KARAHAN AND MURAT ÖZDEMIR
Abstract. We introduce a new extragradient iterative process, motivated and inspired by [S. H. Khan, A Picard-Mann Hybrid Iterative Process, Fixed Point Theory and Applications, doi:10.1186/1687-1812-2013-69], for …nding a common element of the set of …xed points of a nonexpansive mapping and the set of solutions of a variational inequality for an inverse strongly monotone mapping in a Hilbert space. Using this process, we prove a weak convergence theorem for the class of nonexpansive mappings in Hilbert spaces. Finally, as an application, we give some theorems by using resolvent operator and strictly pseudocontractive mapping.
1. Introduction
Let H be a real Hilbert space with the inner product h ; i and the norm k k, respectively. Let C be a nonempty closed convex subset of H; I be the idendity mapping on C; and PC be the metric projection from H onto C:
Recall that a mapping T : C ! C is called nonexpansive if kT x T yk kx yk ; 8x; y 2 C:
We denote by F (T ) the set of …xed points of T , i.e., F (T ) = fx 2 C : T x = xg. For a mapping A : C ! H; it is called monotone if
hAx Ay; x yi 0;
L-Lipschitzian if there exists a constant L > 0 such that kAx Ayk L kx yk ; 8x; y 2 C; and -inverse strongly monotone if
hAx Ay; x yi kAx Ayk2;
for all x; y 2 C:
Received by the editors March 26, 2014, Accepted: May 19, 2014. 2000 Mathematics Subject Classi…cation. 49J30, 47H09, 47J20.
Key words and phrases. Variational inequalities, …xed point problems, weak convergence.
c 2 0 1 4 A n ka ra U n ive rsity
Remark 1.1. It is obvious that any -inverse strongly monotone mapping A is monotone and 1-Lipschitz continuous.
Monotonicity conditions in the context of variational methods for nonlinear op-erator equations were used by Vainberg and Kacurovskii [1] and then many authors have studied on this subject.
In this paper, we consider the following variational inequality problem V I (C; A): …nd a x 2 C such that
hAx; y xi 0; 8y 2 C: The set of solutions of V I (C; A) is denoted by ; i.e.,
= fx 2 C : hAx; y xi 0, 8y 2 Cg :
In the context of the variational inequality problem it is easy to check that x 2 , x 2 F (PC(I A)) ; 8 > 0:
Variational inequalities were initially studied by Stampacchia [2], [3]. Such a prob-lem is connected with convex minimization probprob-lem, the compprob-lementarity probprob-lem, the problem of …nding point x 2 C satisfying 0 2 A and etc. Fixed point problems are also closely related to the variational inequality problems.
For …nding an element of F = F (T ) \ ; many authors have studied widely under suitable assumptions (see [4, 5, 6, 7, 8, 9]). For example, in 2006, Takahashi and Toyoda [10] introduced following iterative process:
x02 C
xn+1= nxn+ (1 n) T PC(I nA) xn; 8n 0, (1.1)
where C is a nonempty closed convex subset of a real Hilbert space H; A : C ! H is an -inverse strongly monotone mapping, PC : H ! C is a metric projection,
T : C ! C is a nonexpansive mapping, f ng [a; b] for some a; b 2 (0; 1) ; and
f ng [c; d] for some c; d 2 (0; 2 ) : They proved that if F = F (T )\ is nonempty,
then the sequence fxng generated by (1.1) converges weakly to some z 2 F where
z = limn!1PFxn: In the same year, Nadezkhina and Takahashi [11] generalized
the iterative process (1.1) and motivated by this process they introduced following iterative scheme for nonexpansive mapping S and monotone k-Lipschitzian mapping
A 8 < : x02 C xn+1= nxn+ (1 n) SPC(xn nyn) yn= PC(I nA) xn; 8n 0: (1.2) They proved the weak convergence of fxng under the suitable conditions.
Re-cently, independetly from the above processes, Khan [12] and Sahu [13], individually, introduced the following iterative process which Khan referred to as Picard-Mann hybrid iterative process:
8 < : x02 C xn+1= T yn yn = nxn+ (1 n) T xn; 8n 0; (1.3)
where f ng is a sequence in (0; 1) : Khan proved a strong and a weak convergence
theorems in a Banach space for iterative process (1.3) under the suitable conditions where T is a nonexpansive mapping. Also, he proved that the iterative process given by (1.3) converges faster than the Picard, Mann and Ishikawa processes for the contraction mappings.
In this paper, motivated and inspired by the idea of extragradient method and the above processes, we introduce the following process:
8 < : x02 C xn+1= T PC(I nA) yn yn= nxn+ (1 n) T PC(I nA) xn; 8n 0; (1.4) where T is a nonexpansive mapping and PC is a metric projection from H onto C:
Our iterative process is independent from all of the above processes. Also, under the suitable conditions, we establish a weak convergence theorem.
2. Preliminaries
In this section, we collect some useful lemmas that will be used for our main result in the next section. We write xn * x to indicate that the sequence fxng
converges weakly to x; and xn ! x for the strong convergence. It is well known
that for any x 2 H; there exists a unique point y02 C such that
kx y0k = inf fkx yk : y 2 Cg :
We denote y0by PCx; where PC is called the metric projection of H onto C: It is
known that PC has the following properties:
(i) kPCx PCyk kx yk ; for all x; y 2 H;
(ii) kx yk2 kx PCxk2+ ky PCxk2; for all x 2 H; y 2 C;
(iii) hx PCx; y PCxi 0; for all x 2 H; y 2 C;
On the other hand, it is known that a Hilbert space H satis…es the Opial condi-tion that, for any sequence fxng with xn * x; the inequality
lim inf
n!1 kxn xk < lim infn!1 kxn yk
holds for every y 2 H with y 6= x:
Lemma 2.1. [10] Let C be a nonempty closed convex subset of a real Hilbert space H and fxng be a sequence in H: Suppose that, for all z 2 C;
kxn+1 zk kxn zk
for every n = 0; 1; 2; : : : : Then, fPCxng converges strongly to some u 2 C:
Lemma 2.2. [10] Let C be a nonempty closed convex subset of a real Hilbert space H and let A be an -inverse strongly monotone mapping of C into H: Then the set of solutions of V I (C; A), ; is nonempty.
For a set-valued mapping S : H ! 2H, if the inequality
hf g; u vi 0
holds for all u; v 2 C; f 2 Su; g 2 Sv; then S is called monotone mapping. A monotone mapping S : H ! 2H is maximal if the graph G (S) of S is not
properly contained in the graph of any other monotone mappings. It is known that a monotone mapping S is maximal if and only if, for (u; f ) 2 H H; hu v; f wi 0 for every (v; w) 2 G (S) implies f 2 Su: Let A be an inverse strongly monotone mapping of C into H; let NCv be the normal cone to C at v 2 C;
i.e.,
NCv = fw 2 H : hv u; wi 0; 8u 2 Cg ;
and de…ne
Sv = Av + NCv v 2 C
; v =2 C:
Then S is maximal monotone and 0 2 Sv if and only if v 2 :
Lemma 2.3. [14] Let C be a nonempty closed convex subset of a real Hilbert space H; and T be a nonexpansive self-mapping of C: If F (T ) 6= ;; then I T is demiclosed; that is whenever fxng is a sequence in C weakly converging to some
x 2 C and the sequence f(I T ) xng strongly converges to some y, it follows that
(I T ) x = y: Here I is the identity operator of H:
Lemma 2.4. [15] Let H be a real Hilbert space, let f ng be a sequence of real
numbers such that 0 < a n b < 1 for all n = 0; 1; 2; : : : ; and let fxng and
fyng be sequences of H such that
lim sup n!1 kxnk c; lim sup n!1 kynk c and lim n!1k nxn+ (1 n) ynk = c;
for some c > 0: Then,
lim
n!1kxn ynk = 0:
3. Main result
In this section, we introduced a new extragradient method and proved that the sequence generated by this iteration method converges weakly to a …xed point of nonexpansive mapping and to a solution of variational inequality V I(C; A). Theorem 3.1. Let C be a nonempty closed convex subset of a real Hilbert space H: Let A : C ! H be an -inverse strongly monotone mapping and T : C ! C be a nonexpansive mapping such that F = F (T ) \ 6= ;: For arbitrary initial value x02 H; let fxng be a sequence de…ned by (1.4) where f ng [a; b] for some
a; b 2 (0; 2 ) and f ng [c; d] for some c; d 2 (0; 1). Then, the sequence fxng
Proof. We devide our proof into four steps.
Step 1. Let tn = PC(I nA) xn: First, we show that fxng and ftng are
bounded sequences. Let z 2 F (T ) \ ; then, we have ktn zk2 = kPC(I nA) xn zk2
k(I nA) xn (I nA) zk2
= kxn z n(Axn Az)k2
kxn zk2 2 nhxn z; Axn Azi + 2nkAxn Azk2
kxn zk2+ n( n 2 ) kAxn Azk2
kxn zk2 (3.1)
and from (3.1) we get
kxn+1 zk2 = kT PC(I nA) yn zk2 = kT PC(I nA) yn T PC(I nA) zk2 kyn zk2 = k n(xn z) + (1 n) (T tn z)k2 nkxn zk2+ (1 n) kT tn zk2 nkxn zk2+ (1 n) ktn zk2 nkxn zk2 + (1 n) h kxn zk2+ n( n 2 ) kAxn Azk2 i = kxn zk2+ (1 n) n( n 2 ) kAxn Azk2 kxn zk2+ (1 d) a (b 2 ) kAxn Azk2 kxn zk2:
Therefore, there exists limn!1kxn zk and Axn Az ! 0: Hence fxng and ftng
are bounded.
Step 2. We will show that limn!1kxn ynk = 0: Before that, we shall show
limn!1kT tn xnk = 0. From Step 1, we know that limn!1kxn zk exists for
all z 2 F (T ) \ . Let limn!1kxn zk = c: Since
kxn+1 zk kyn zk kxn zk ;
we get
lim
n!1kyn zk = c: (3.2)
On the other hand, since
kT tn zk ktn zk kxn zk ;
we have
lim sup
n!1 kT tn zk
Also, we know that lim sup n!1 kxn zk c (3.4) and lim n!1kyn zk = limn!1k n(xn z) + (1 n) (T tn z)k = c: (3.5)
Hence, from (3.3), (3.4), (3.5), and Lemma 2.4 , we get that lim
n!1kxn T tnk = 0: (3.6)
We have also
kxn ynk = kxn nxn (1 n) T tnk
= (1 n) kxn T tnk :
So, from (3.6) we obtain that lim
n!1kxn ynk = 0: (3.7)
Since A is Lipschitz continuous, we have Axn Ayn! 0:
Step 3. Next, we show that limn!1kT xn xnk = 0: Using the properties of
metric projections, since
ktn zk2 = kPC(I nA) xn PC(I nA) zk2 htn z; (I nA) xn (I nA) zi = 1 2 h ktn zk2+ k(I nA) xn (I nA) zk2 ktn z [(I nA) xn (I nA) z]k2 i 1 2 h ktn zk2+ kxn zk2 k(tn xn) + n(Axn Az)k2 i = 1 2 h ktn zk2+ kxn zk2 ktn xnk2
2 nhtn xn; Axn Azi 2nkAxn Azk2
i ; it follows that
ktn zk2 kxn zk2 ktn xnk2
So, using the inequality (3.8) we get kxn+1 zk2 = kT PC(I nA) yn zk2 = kT PC(I nA) yn T PC(I nA) zk2 kyn zk2 = k n(xn z) + (1 n) (T tn z)k2 nkxn zk2+ (1 n) kT tn zk2 nkxn zk2+ (1 n) ktn zk2 kxn zk2 (1 n) ktn xnk2 2 n(1 n) htn xn; Axn Azi 2 n(1 n) kAxn Azk2 kxn zk2 (1 d) ktn xnk2 2 n(1 n) htn xn; Axn Azi 2 n(1 n) kAxn Azk2:
Since limn!1kxn+1 zk = limn!1kxn zk and Axn Az ! 0; we obtain
lim
n!1kxn tnk = 0: (3.9)
On the other hand, we have
kT xn xnk kT xn T tnk + kT tn xnk
kxn tnk + kT tn xnk :
So, it follows from (3.6) and (3.9) that lim
n!1kT xn xnk = 0: (3.10)
Step 4. Finally, we show that fxng converges weakly to a p 2 F: Since fxng is a
bounded sequence, there is a subsequence fxnig of fxng such that fxnig converges
weakly to p: We need to show that p belongs to F: First, we show that p 2 : From (3.9), we have tni * p: Let
Sv = Av + NCv ; v 2 C;
; ; v =2 C:
Then S is maximal monotone mapping. Let (v; w) 2 G (S) : Since w Av 2 NCv
and tn 2 C; we get
hv tn; w Avi 0: (3.11)
On the other hand, from the de…niton of tn; we have that
hxn nAxn tn; tn vi 0 and hence, v tn; tn xn n + Axn 0:
Therefore, using (3.11), we get hv tni; wi hv tni; Avi hv tni; Avi v tni; tni xni ni + Axni = v tni; Av Axni tni xni ni = hv tni; Av Atnii + hv tni; Atni Axnii v tni; tni xni ni hv tni; Atni Axnii v tni; tni xni ni : Hence, for i ! 1 we have
hv p; wi 0:
Since S is maximal monotone, we have p 2 S 10 and hence p 2 : Next, we show that p 2 F (T ) : From (3.10), Lemma 2.3 and by using xni * p, we have that
p 2 F (T ) : So desired conclusion (p 2 F ) is obtained.
Now it remains to show that fxng converges weakly to p 2 F and p = limn!1PFxn:
Let assume that there exists an another subsequence xnj of fxng and xnj * p02
F: We shall show that p = p0: Conversely, let suppose that p 6= p0. By using Opial
condition, we obtain that lim n!1kxn pk = lim infi!1 kxni pk < lim inf i!1 kxni p0k = lim n!1kxn p0k = lim inf j!1 xnj p0 < lim inf j!1 xnj p = lim n!1kxn pk :
This is a contradiction, so we get p = p0: This implies that xn * p 2 F:
Finally, we need to show p = limn!1PFxn: Since p 2 F; we have
hp PFxn; PFxn xni 0:
By Lemma 2.1, fPFxng converges strongly to u02 F: Then, we get
hp u0; u0 pi 0;
Corollary 1. Let C be a nonempty closed convex subset of a real Hilbert space H: Let A : C ! H be an -inverse strongly monotone mapping such that 6= ;: For arbitrary initial value x02 H; let fxng be a sequence de…ned by
xn+1= PC(I nA) yn
yn= nxn+ (1 n) PC(I nA) xn; 8n 0;
where f ng [a; b] for some a; b 2 (0; 2 ) and f ng [c; d] for some c; d 2
(0; 1). Then, the sequence fxng converges weakly to a point p 2 where p =
limn!1P xn:
4. Applications
Let B : H ! 2H be a maximal monotone mapping. The resolvent of B of order r > 0 is the single valued mapping JrB: H ! H de…ned by
JrBx = (I + rB) 1x for any x 2 H. It is easy to check that F JB
r = B 10. Moreover, the resolvent
JB
r is a nonexpansive mapping. So, we can give the following theorem.
Theorem 4.1. Let H be a real Hilbert space: Let > 0, A : H ! H be an -inverse strongly monotone mapping and B : H ! 2H be a maximal monotone mapping such that A 10 \ B 10 6= ;: For arbitrary initial value x02 H; let fxng
be a sequence de…ned by
xn+1= JrB(yn nAyn)
yn = nxn+ (1 n) JrB(xn nAxn) ; 8n 0;
where f ng [a; b] for some a; b 2 (0; 2 ) and f ng [c; d] for some c; d 2 (0; 1).
Then the sequence fxng converges weakly to a point p 2 A 10 \ B 10 where p =
limn!1PA 10\B 10xn:
Proof. We have A 10 = V I (H; A), F JB
r = B 10 and PH = I. Since the
resolvent JB
r is a nonexpansive mapping, we obtain the desired conclusion.
Now, we give a theorem for a pair of nonexpansive mapping and strictly pseudo-contractive mapping. A mapping S : C ! C is called k- strictly pseudopseudo-contractive mapping if there exists k with 0 k < 1 such that
kSx Syk2 kx yk2+ k k(I S) x (I S) yk2
for all x; y 2 C: Let A = I S: Then, it is known that the mapping A is inverse strongly monotone mapping with (1 k) =2, i.e.,
hAx Ay; x yi 1 k
2 kAx Ayk
2
Theorem 4.2. Let C be a nonempty closed convex subset of a real Hilbert space H: Let T : C ! C be a nonexpansive mapping and S : C ! C be a k- strictly pseudocontractive mapping such that F (T ) \ F (S) 6= ;: For arbitrary initial value x02 H; let fxng be a sequence de…ned by
xn+1= T ((I n) yn+ nSyn)
yn= nxn+ (1 n) T ((I n) xn+ nSxn) ; 8n 0;
where f ng [a; b] for some a; b 2 (0; 1 k) and f ng [c; d] for some c; d 2
(0; 1). Then the sequence fxng converges weakly to a point p 2 F (T ) \ F (S) where
p = limn!1PF (T )\F (S)xn:
Proof. Let A = I S: Then, we know that A is inverse strongly monotone mapping. Also, It is clear that F (S) = V I (C; A) : Since, A is a mapping from C into itself, we get
(I n) xn+ nSxn= xn n(I S) xn= PC(xn nAxn) :
So, from Theorem 3.1, we obtain the desired conclusion.
Theorem 4.3. Let H be a real Hilbert space: Let > 0, A : H ! H be an -inverse strongly monotone mapping and T : H ! H be a nonexpansive mapping such that F (T ) \ A 10 6= ;: For arbitrary initial value x
02 H; let fxng be a sequence de…ned
by
xn+1= T (yn nAyn)
yn= nxn+ (1 n) T (xn nAxn) ; 8n 0;
where f ng [a; b] for some a; b 2 (0; 2 ) and f ng [c; d] for some c; d 2 (0; 1).
Then the sequence fxng converges weakly to a point p 2 V I (F (T ) ; A) where p =
limn!1PF (T )\A 10xn:
Proof. We have A 10 = V I (H; A) and P
H = I: Also, it is clear that F (S)\A 10
V I (F (S) ; A). So, by Theorem 3.1, we get the desired conclusion. References
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Current address : ·I. Karahan: Department of Mathematics, Erzurum Technical University, Erzurum 25240, TURKEY, M. Özdemir Department of Mathematics, Ataturk University, Erzu-rum 25240, TURKEY,
E-mail address : ibrahimkarahan@erzurum.edu.tr, mozdemir@atauni.edu.tr URL: http://communications.science.ankara.edu.tr/index.php?series=A1
