J. Korean Math. Soc. 44 (2007), No. 4, pp. 823–834
A NEW SYSTEM OF GENERALIZED NONLINEAR MIXED QUASIVARIATIONAL INEQUALITIES AND ITERATIVE
ALGORITHMS IN HILBERT SPACES
Jong Kyu Kim and Kyung Soo Kim
Reprinted from the
Journal of the Korean Mathematical Society Vol. 44, No. 4, July 2007
c
°2007 The Korean Mathematical Society
J. Korean Math. Soc. 44 (2007), No. 4, pp. 823–834
A NEW SYSTEM OF GENERALIZED NONLINEAR MIXED QUASIVARIATIONAL INEQUALITIES AND ITERATIVE
ALGORITHMS IN HILBERT SPACES
Jong Kyu Kim and Kyung Soo Kim
Abstract. We introduce a new system of generalized nonlinear mixed quasivariational inequalities and prove the existence and uniqueness of the solution for the system in Hilbert spaces. The main result of this paper is an extension and improvement of the well-known corresponding results in Kim-Kim [16], Noor [21]-[23] and Verma [24]-[26].
1. Introduction and preliminaries
In recent years, many classical variational inequalities and complementarity problems have been extended and generalized to study a large variety of prob- lems arising in mechanics, physics, optimization and control theory, nonlinear programming problems, economics, transportation, equilibrium problems and engineering sciences, etc. (see, [1], [2], [4]-[26])
In 2001, Verma [25] introduced and studied a new system of nonlinear vari- ational inequalities based on a new system of iterative algorithms. And also, Verma [26] investigated the approximation-solvability of a new system of non- linear quasivariational inequalities in Hilbert spaces.
In 2004, Kim-Kim [16] introduced and studied a new system of generalized nonlinear mixed variational inequalities bases on a new system of iterative algorithms in Hilbert spaces.
In this paper, we introduce a new system of generalized nonlinear mixed quasivariational inequalities and prove the existence and uniqueness of solu- tion for the systems in Hilbert spaces. We also construct some new iterative algorithms for the problems and give the convergence analysis of the iterative sequences generated by the algorithms.
Throughout this paper, let H be a real Hilbert space with the inner prod- uct h·, ·i and the norm k · k. Let A, B, S, T, g 1 , g 2 : H → H be single-valued mappings, φ 1 , φ 2 : H → R ∪ {+∞} be proper convex lower semicontinuous
Received January 14, 2006.
2000 Mathematics Subject Classification. 49J40.
Key words and phrases. a system of generalized nonlinear mixed quasivariational inequal- ities, iterative sequence with errors, algorithm.
This work was supported by the Kyungnam University Foundation Grant, 2006.
°2007 The Korean Mathematical Societyc
823
functions and K be a closed convex subset of H. We consider the following problem:
Find x ∗ , y ∗ ∈ H such that g 1 (x ∗ ), g 2 (y ∗ ) ∈ K and
(1.1)
hρ(A(y ∗ ) + S(y ∗ )) + g 1 (x ∗ ) − g 2 (y ∗ ), x − g 1 (x ∗ )i
≥ ρφ 1 (g 1 (x ∗ )) − ρφ 1 (x),
hγ(B(x ∗ ) + T (x ∗ )) + g 2 (y ∗ ) − g 1 (x ∗ ), x − g 2 (y ∗ )i
≥ γφ 2 (g 2 (y ∗ )) − γφ 2 (x)
for all x ∈ H, which is called the new system of generalized nonlinear mixed quasivariational inequalities, where ρ > 0 and γ > 0 are constants.
Special Cases of the problem (1.1):
(I) If A = B = 0, then the problem (1.1) reduces the following:
Find x ∗ , y ∗ ∈ H such that g 1 (x ∗ ), g 2 (y ∗ ) ∈ K and (1.2)
½ hρS(y ∗ ) + g 1 (x ∗ ) − g 2 (y ∗ ), x − g 1 (x ∗ )i ≥ ρφ 1 (g 1 (x ∗ )) − ρφ 1 (x), hγT (x ∗ ) + g 2 (y ∗ ) − g 1 (x ∗ ), x − g 2 (y ∗ )i ≥ γφ 2 (g 2 (y ∗ )) − γφ 2 (x) for all x ∈ H, which is called the system of nonlinear mixed quasivariational inequalities.
(II) If φ 1 = φ 2 = δ K (the indicator function of a nonempty closed convex subset K), then the problem (1.1) reduces the following:
Find x ∗ , y ∗ ∈ K such that g 1 (x ∗ ), g 2 (y ∗ ) ∈ K and (1.3)
½ hρ(A(y ∗ ) + S(y ∗ )) + g 1 (x ∗ ) − g 2 (y ∗ ), x − g 1 (x ∗ )i ≥ 0, hγ(B(x ∗ ) + T (x ∗ )) + g 2 (y ∗ ) − g 1 (x ∗ ), x − g 2 (y ∗ )i ≥ 0
for all x ∈ K, which is called the system of generalized nonlinear quasivaria- tional inequalities.
(III) If φ 1 = φ 2 = δ K (the indicator function of a nonempty closed convex subset K) and A = B = 0, then the problem (1.1) reduces the following:
Find x ∗ , y ∗ ∈ K such that g 1 (x ∗ ), g 2 (y ∗ ) ∈ K and (1.4)
½ hρS(y ∗ ) + g 1 (x ∗ ) − g 2 (y ∗ ), x − g 1 (x ∗ )i ≥ 0, hγT (x ∗ ) + g 2 (y ∗ ) − g 1 (x ∗ ), x − g 2 (y ∗ )i ≥ 0
for all x ∈ K, which is called the system of nonlinear quasivariational inequal- ities.
Now, we give some definitions and lemmas for the main theorems.
Definition 1.1. Let T, g : H → H be mappings.
(1) T : H → H is said to be k-strongly monotone if there exists a constant k > 0 such that
hT (x) − T (y), x − yi ≥ kkx − yk 2 for all x, y ∈ H. This implies that
kT (x) − T (y)k ≥ kkx − yk,
that is, T is k-expanding and when k = 1, it is expanding.
(2) T : H → H is called g-k-strongly monotone if there exists a constant k > 0 such that
hT (x) − T (y), g(x) − g(y)i ≥ kkg(x) − g(y)k 2 for all x, y ∈ H. This implies that
kT (x) − T (y)k ≥ kkg(x) − g(y)k, that is, T is g-k-expanding and when k = 1, it is g-expanding.
(3) T : H → H is said to be s-Lipschitz continuous if there exists a constant s > 0 such that
kT (x) − T (y)k ≤ skx − yk for all x, y ∈ H.
(4) T : H → H is called g-s-Lipschitz continuous if there exists a constant s > 0 such that
kT (x) − T (y)k ≤ skg(x) − g(y)k for all x, y ∈ H.
Lemma 1.1 ([16]). Let {a n }, {b n } and {c n } be sequences of nonnegative num- bers satisfying the following condition: there exists n 0 such that
(1.5) a n+1 ≤ (1 − t n )a n + b n t n + c n
for all n ≥ n 0 , where t n ∈ [0, 1], P ∞
n=0
t n = ∞, lim
n→∞ b n = 0, P ∞
n=0
c n < ∞. Then
n→∞ lim a n = 0.
Lemma 1.2 ([3]). For any given u ∈ H, a point z ∈ H satisfies hu − z, v − ui ≥ ρφ(u) − ρφ(v)
for all v ∈ H if and only if
u = J φ ρ (z),
where J φ ρ = (I + ρ∂φ) −1 and ∂φ denotes the subdifferential of a proper convex lower semicontinuous function φ : H → R ∪ {+∞}.
Remark 1.1. It is well known that J φ ρ is nonexpansive (see [3]).
Lemma 1.3. For any given x ∗ , y ∗ ∈ H, (x ∗ , y ∗ ) is a solution of the problem (1.1) if and only if
( g 1 (x ∗ ) = J φ ρ
1
(g 2 (y ∗ ) − ρ(A(y ∗ ) + S(y ∗ ))), g 2 (y ∗ ) = J φ γ
2
(g 1 (x ∗ ) − γ(B(x ∗ ) + T (x ∗ ))).
Proof. It is easy to prove that Lemma 1.3 holds from Lemma 1.2. ¤
2. Existence and uniqueness
Now, we shall show the existence and uniqueness of solution for the problems (1.1).
Theorem 2.1. Let S : H → H be a g 2 -k 2 -strongly monotone and g 2 -s 2 - Lipschitz continuous mapping, T : H → H be a g 1 -k 1 -strongly monotone and g 1 -s 1 -Lipschitz continuous mapping, A : H → H be a g 2 -l 2 -Lipschitz continu- ous mapping and B : H → H be a g 1 -l 1 -Lipschitz continuous mapping. Suppose that g 1 , g 2 : H → H are invertible. If
(2.1)
0 < ρ < 2(k 2 − l 2 )
s 2 2 − l 2 2 , l 2 < k 2 , 0 < γ < 2(k 1 − l 1 )
s 2 1 − l 1 2 , l 1 < k 1 , then the problem (1.1) has a unique solution (x ∗ , y ∗ ).
Proof. First, in order to prove the existence of the solution. Define a mapping F : H → H as follows :
(2.2) F (x) = J φ ρ1[J φ γ2(x − γ(B(g 1 −1 (x)) + T (g −1 1 (x))))
(x − γ(B(g 1 −1 (x)) + T (g −1 1 (x))))
− ρ(A + S)(g 2 −1 (J φ γ
2
(x − γ(B(g −1 1 (x)) + T (g 1 −1 (x))))))]
for each x ∈ H. Since J φ ρ
1
is nonexpansive, for any x, y ∈ H, we have
(2.3)
kF (x) − F (y)k
= kJ φ ρ1[J φ γ2(x − γ(B(g 1 −1 (x)) + T (g −1 1 (x))))
(x − γ(B(g 1 −1 (x)) + T (g −1 1 (x))))
− ρ(A + S)(g 2 −1 (J φ γ2(x − γ(B(g −1 1 (x)) + T (g 1 −1 (x))))))]
− J φ ρ
1
[J φ γ
2
(y − γ(B(g 1 −1 (y)) + T (g 1 −1 (y))))
− ρ(A + S)(g 2 −1 (J φ γ2(y − γ(B(g 1 −1 (y)) + T (g 1 −1 (y))))))]k
≤ kJ φ γ2(x − γ(B(g −1 1 (x)) + T (g −1 1 (x))))
− J φ γ
2
(y − γ(B(g 1 −1 (y)) + T (g −1 1 (y))))
− ρ{S(g −1 2 (J φ γ2(x − γ(B(g 1 −1 (x)) + T (g −1 1 (x))))))
− S(g −1 2 (J φ γ2(y − γ(B(g 1 −1 (y)) + T (g 1 −1 (y))))))}k + ρkA(g 2 −1 (J φ γ
2
(x − γ(B(g 1 −1 (x)) + T (g 1 −1 (x))))))
− A(g −1 2 (J φ γ2(y − γ(B(g −1 1 (y)) + T (g 1 −1 (y))))))k.
It follows from the conditions of S, T, A, B, g 1 , g 2 , that kJ φ γ2(x − γ(B(g 1 −1 (x)) + T (g −1 1 (x))))
− J φ γ
2
(y − γ(B(g 1 −1 (y)) + T (g 1 −1 (y))))
(2.4)
− ρ{S(g 2 −1 (J φ γ
2
(x − γ(B(g −1 1 (x)) + T (g 1 −1 (x))))))
− S(g 2 −1 (J φ γ2(y − γ(B(g 1 −1 (y)) + T (g −1 1 (y))))))}k 2
= kJ φ γ2(x − γ(B(g −1 1 (x)) + T (g 1 −1 (x))))
− J φ γ
2
(y − γ(B(g −1 1 (y)) + T (g 1 −1 (y))))k 2
− 2ρhJ φ γ2(x − γ(B(g 1 −1 (x)) + T (g −1 1 (x))))
− J φ γ2(y − γ(B(g 1 −1 (y)) + T (g 1 −1 (y)))), S(g −1 2 (J φ γ
2
(x − γ(B(g 1 −1 (x)) + T (g −1 1 (x))))))
− S(g 2 −1 (J φ γ2(y − γ(B(g −1 1 (y)) + T (g −1 1 (y))))))i + ρ 2 kS(g 2 −1 (J φ γ2(x − γ(B(g −1 1 (x)) + T (g 1 −1 (x))))))
(x − γ(B(g −1 1 (x)) + T (g 1 −1 (x))))))
− S(g 2 −1 (J φ γ
2
(y − γ(B(g −1 1 (y)) + T (g −1 1 (y))))))k 2
≤ kJ φ γ2(x − γ(B(g −1 1 (x)) + T (g 1 −1 (x))))
− J φ γ2(y − γ(B(g −1 1 (y)) + T (g 1 −1 (y))))k 2
− 2ρk 2 kJ φ γ
2
(x − γ(B(g 1 −1 (x)) + T (g 1 −1 (x))))
− J φ γ2(y − γ(B(g 1 −1 (y)) + T (g 1 −1 (y))))k 2 + ρ 2 s 2 2 kJ φ γ
2
(x − γ(B(g −1 1 (x)) + T (g 1 −1 (x))))
− J φ γ2(y − γ(B(g 1 −1 (y)) + T (g 1 −1 (y))))k 2
= (1 − 2ρk 2 + ρ 2 s 2 2 )kJ φ γ2(x − γ(B(g 1 −1 (x)) + T (g −1 1 (x))))
− J φ γ
2
(y − γ(B(g −1 1 (y)) + T (g 1 −1 (y))))k 2
≤ (1 − 2ρk 2 + ρ 2 s 2 2 )kx − y − γ(B(g 1 −1 (x)) + T (g −1 1 (x))
− (B(g 1 −1 (y)) + T (g −1 1 (y))))k 2
≤ (1 − 2ρk 2 + ρ 2 s 2 2 )
· [(kx − yk 2 − 2γhx − y, T (g 1 −1 (x)) − T (g −1 1 (y))i + γ 2 kT (g −1 1 (x)) − T (g 1 −1 (y))k 2 )
12+ γl 1 kx − yk] 2
≤ (1 − 2ρk 2 + ρ 2 s 2 2 )
³q
1 − 2γk 1 + γ 2 s 2 1 + γl 1
´ 2
kx − yk 2
and
(2.5)
ρkA(g −1 2 (J φ γ2(x − γ(B(g 1 −1 (x)) + T (g −1 1 (x))))))
− A(g −1 2 (J φ γ2(y − γ(B(g 1 −1 (y)) + T (g 1 −1 (y))))))k
≤ ρl 2 kx − y − γ(B(g 1 −1 (x)) + T (g −1 1 (x))
− (B(g 1 −1 (y)) + T (g 1 −1 (y))))k
≤ ρl 2 {kx − y − γ(T (g 1 −1 (x)) − T (g −1 1 (y)))k + γl 1 kx − yk}
≤ ρl 2
³q
1 − 2γk 1 + γ 2 s 2 1 + γl 1
´
kx − yk.
From (2.3)-(2.5), we have
(2.6) kF (x) − F (y)k ≤ θ 1 θ 2 kx − yk for all x, y ∈ H, where
(2.7) θ 1 = q
1 − 2ρk 2 + ρ 2 s 2 2 + ρl 2 , θ 2 = q
1 − 2γk 1 + γ 2 s 2 1 + γl 1 . It follows from (2.1) that θ 1 < 1 and θ 2 < 1. Thus (2.6) implies that F is a contractive mapping and so there exists a point g 1 (x ∗ ) ∈ H such that g 1 (x ∗ ) = F (g 1 (x ∗ )). Therefore, from the definition of F, we have
g 1 (x ∗ ) = J φ ρ1(g 2 (y ∗ ) − ρ(A + S)(y ∗ )) and
g 2 (y ∗ ) = J φ γ
2
(g 1 (x ∗ ) − γ(B + T )(x ∗ )).
By Lemma 1.3, we know that (x ∗ , y ∗ ) is a solution of the problem (1.1).
Next, we show the uniqueness of the solution. Let (u ∗ , v ∗ ) be an another solution of the problem (1.1). Then we have
g 1 (u ∗ ) = J φ ρ1(g 2 (v ∗ ) − ρ(A + S)(v ∗ )) and
g 2 (v ∗ ) = J φ γ2(g 1 (u ∗ ) − γ(B + T )(u ∗ )).
As in the proof of (2.6), we have
kg 1 (x ∗ ) − g 1 (u ∗ )k 2 ≤ θ 1 θ 2 kg 1 (x ∗ ) − g 1 (u ∗ )k 2 .
Since θ 1 < 1 and θ 2 < 1, it follows that g 1 (x ∗ ) = g 1 (u ∗ ) and so g 2 (y ∗ ) = g 2 (v ∗ ).
This completes the proof. ¤
If A = B = 0 in Theorem 2.1, then we have the following theorem as a special case:
Theorem 2.2. Let S and T be the same as in Theorem 2.1. If 0 < ρ < 2k s22 2
and 0 < γ < 2k s21
1
, then the problem (1.2) has a unique solution (x ∗ , y ∗ ).
If φ 1 = φ 2 = δ K in Theorem 2.1, then we have the following theorem as a special case:
Theorem 2.3. Let K be a nonempty closed convex subset of a Hilbert space H. Suppose that A, B, S and T are the same in Theorem 2.1. If the condition (2.1) holds, then the problem (1.3) has a unique solution (x ∗ , y ∗ ).
If φ 1 = φ 2 = δ K and A = B = 0 in Theorem 2.1, then we have the following theorem as a special case:
Theorem 2.4. Let K be a nonempty closed convex subset of a Hilbert space H. Suppose that S and T are the same in Theorem 2.1. If 0 < ρ < 2k s22
2
and
0 < γ < 2k s21
1
, then the problem (1.4) has a unique solution (x ∗ , y ∗ ).
Remark 2.1. (1) If g 1 = g 2 = I (the identity mapping) and φ 1 = φ 2 = φ (a proper convex lower semicontinuous function) in (1.1), then the problem (1.1) reduces to find x ∗ , y ∗ ∈ H such that
½ hρ(A(y ∗ ) + S(y ∗ )) + x ∗ − y ∗ , x − x ∗ i ≥ ρφ(x ∗ ) − ρφ(x), hγ(B(x ∗ ) + T (x ∗ )) + y ∗ − x ∗ , x − y ∗ i ≥ γφ(y ∗ ) − γφ(x)
for all x ∈ H, which is defined by Kim-Kim [16] and is called a new system of generalized nonlinear mixed variational inequalities. Hence Theorem 2.1 is the extension of the result of Kim-Kim [16].
(2) Theorems 2.2, 2.3 and 2.4 are extensions and improvements of the cor- responding results in Noor [21]-[23] and Verma [24]-[26].
3. Algorithms and convergence
In this section, we construct some new iterative algorithms for the problems (1.1)-(1.4). We also give the convergence analysis of the iterative sequences generated by the algorithms.
Now we give the algorithm for solving the problem (1.1) as follows:
Algorithm 3.1. For given x 0 ∈ H, define the iterative sequences {g 1 (x n )} and {g 2 (y n )} with mixed errors as follows:
(3.1)
g 1 (x n+1 ) = (1 − α n )g 1 (x n )
+ α n J φ ρ1g(g 2 (y n ) − ρ(A(y n ) + S(y n ))) + α n u n + w n ,
g 2 (y n ) = J φ γ2(g 1 (x n ) − γ(B(x n ) + T (x n ))) + v n
for all n ≥ 0, where {α n } is a sequence in [0, 1], {u n }, {w n }, {v n } are sequences in H and g 1 , g 2 : H → H are mappings.
If A = B = 0, then Algorithm 3.1 reduces to the following algorithm for solving the problem (1.2).
Algorithm 3.2. For given x 0 ∈ H, define the iterative sequences {g 1 (x n )} and {g 2 (y n )} with mixed errors as follows:
g 1 (x n+1 ) = (1 − α n )g 1 (x n ) + α n J φ ρ1(g 2 (y n ) − ρS(y n )) + α n u n + w n ,
g 2 (y n ) = J φ γ
2
(g 1 (x n ) − γT (x n )) + v n
for all n ≥ 0, where {α n }, {u n }, {w n }, {v n }, g 1 and g 2 are the same as in Algorithm 3.1.
If φ 1 = φ 2 = δ K in Algorithms 3.1 and 3.2, then the following algorithms
for solving the problems (1.3) and (1.4), respectively.
Algorithm 3.3. For given x 0 ∈ H, define the iterative sequences {g 1 (x n )} and {g 2 (y n )} with mixed errors as follows:
g 1 (x n+1 ) = (1 − α n )g 1 (x n ) + α n P K (g 2 (y n ) − ρ(A(y n ) + S(y n ))) + α n u n + w n ,
g 2 (y n ) = P K (g 1 (x n ) − γ(B(x n ) + T (x n ))) + v n
for all n ≥ 0, where {α n }, {u n }, {w n }, {v n }, g 1 and g 2 are the same as in Algorithm 3.1.
Algorithm 3.4. For given x 0 ∈ H, define the iterative sequences {g 1 (x n )} and {g 2 (y n )} with mixed errors as follows:
g 1 (x n+1 ) = (1 − α n )g 1 (x n ) + α n P K (g 2 (y n ) − ρS(y n )) + α n u n + w n ,
g 2 (y n ) = P K (g 1 (x n ) − γT (x n )) + v n
for all n ≥ 0, where {α n }, {u n }, {w n }, {v n }, g 1 and g 2 are the same as in Algorithm 3.1.
Now, by using Algorithm 3.1, we prove the following theorem.
Theorem 3.1. Let A, B, S, T , g 1 , g 2 be the same as in Theorem 2.1 and {g 1 (x n )} be the iterative sequence generated by Algorithm 3.1 satisfying the conditions
(3.2)
X ∞ n=0
α n = ∞, X ∞ n=0
kw n k < ∞, lim
n→∞ ku n k = lim
n→∞ kv n k = 0.
If the condition (2.1) holds, then (x n , y n ) converges to the unique solution (x ∗ , y ∗ ) of the problem (1.1).
Proof. By Theorem 2.1, we know that the problem (1.1) has a unique solution (x ∗ , y ∗ ). It follows from Lemma 1.3 that
cg 1 (x ∗ ) = J φ ρ
1
(g 2 (y ∗ ) − ρ(A(y ∗ ) + S(y ∗ ))) and
(3.4) g 2 (y ∗ ) = J φ γ2(g 1 (x ∗ ) − γ(B(x ∗ ) + T (x ∗ ))).
From (3.1) and (3.3), we have kg 1 (x n+1 ) − g 1 (x ∗ )k
= k(1 − α n )g 1 (x n ) + α n J φ ρ1(g 2 (y n ) − ρ(A(y n ) + S(y n )))
+ α n u n + w n − g 1 (x ∗ )k
(3.5)
≤ (1 − α n )kg 1 (x n ) − g 1 (x ∗ )k + α n kJ φ ρ
1
(g 2 (y n ) − ρ(A(y n ) + S(y n )))
− J φ ρ
1
(g 2 (y ∗ ) − ρ(A(y ∗ ) + S(y ∗ )))k + α n ku n k + kw n k
≤ (1 − α n )kg 1 (x n ) − g 1 (x ∗ )k + α n kg 2 (y n ) − g 2 (y ∗ )
− ρ(A(y n ) + S(y n ) − (A(y ∗ ) + S(y ∗ )))k + α n ku n k + kw n k
≤ (1 − α n )kg 1 (x n ) − g 1 (x ∗ )k
+ α n kg 2 (y n ) − g 2 (y ∗ ) − ρ(S(y n ) − S(y ∗ ))k + α n ρl 2 kg 2 (y n ) − g 2 (y ∗ )k + α n ku n k + kw n k.
Since S is g 2 -k 2 -strongly monotone and g 2 -s 2 -Lipschitz continuous, we get
(3.6)
kg 2 (y n ) − g 2 (y ∗ ) − ρ(S(y n ) − S(y ∗ ))k
≤ q
1 − 2ρk 2 + ρ 2 s 2 2 kg 2 (y n ) − g 2 (y ∗ )k.
Combining (3.5) and (3.6), we have
(3.7)
kg 1 (x n+1 ) − g 1 (x ∗ )k ≤ (1 − α n )kg 1 (x n ) − g 1 (x ∗ )k + α n θ 1 kg 2 (y n ) − g 2 (y ∗ )k + α n ku n k + kw n k, where θ 1 = p
1 − 2ρk 2 + ρ 2 s 2 2 + ρl 2 . Again from (3.1) and (3.4), we have
(3.8)
kg 2 (y n ) − g 2 (y ∗ )k
= kJ φ γ
2
(g 1 (x n ) − γ(B(x n ) + T (x n ))) + v n
− J φ γ
2
(g 1 (x ∗ ) − γ(B(x ∗ ) + T (x ∗ )))k
≤ kg 1 (x n ) − g 1 (x ∗ ) − γ(B(x n ) + T (x n ) − (B(x ∗ ) + T (x ∗ )))k + kv n k
≤ kg 1 (x n ) − g 1 (x ∗ ) − γ(T (x n ) − T (x ∗ ))k + γl 1 kg 1 (x n ) − g 1 (x ∗ )k + kv n k.
Since T is g 1 -k 1 -strongly monotone and g 1 -s 1 -Lipschitz continuous, we have
(3.9)
kg 1 (x n ) − g 1 (x ∗ ) − γ(T (x n ) − T (x ∗ ))k
≤ q
1 − 2γk 1 + γ 2 s 2 1 kg 1 (x n ) − g 1 (x ∗ )k.
Combining (3.8) and (3.9), we obtain
(3.10) kg 2 (y n ) − g 2 (y ∗ )k ≤ θ 2 kg 1 (x n ) − g 1 (x ∗ )k + kv n k,
where θ 2 = p
1 − 2γk 1 + γ 2 s 2 1 + γl 1 . It follows from (3.7) and (3.10) that
(3.11)
kg 1 (x n+1 ) − g 1 (x ∗ )k
≤ (1 − α n )kg 1 (x n ) − g 1 (x ∗ )k + α n θ 1 θ 2 kg 1 (x n ) − g 1 (x ∗ )k + α n θ 1 kv n k + α n ku n k + kw n k
=
³ 1 − α n
³
1 − θ 1 θ 2
´´
kg 1 (x n ) − g 1 (x ∗ )k + kw n k + α n (1 − θ 1 θ 2 ) · 1
1 − θ 1 θ 2
³
θ 1 kv n k + ku n k
´ . Let a n = kg 1 (x n ) − g 1 (x ∗ )k, b n = 1−θ 1
1
θ
2(θ 1 kv n k + ku n k), c n = kw n k, t n = α n (1 − θ 1 θ 2 ). Then (3.11) can be rewritten as follows:
a n+1 ≤ (1 − t n )a n + b n t n + c n .
From the assumption, we know that {a n }, {b n }, {c n } and {t n } satisfying the conditions of Lemma 1.1. Thus lim
n→∞ a n = 0 and so g 1 (x n ) → g 1 (x ∗ ) as n → ∞.
Since lim
n→∞ g 1 (x n ) = g 1 (x ∗ ), it follows from (3.2), (3.8) and (3.9) that g 2 (y n ) → g 2 (y ∗ ) as n → ∞. Since g 1 , g 2 are invertible,
n→∞ lim x n = x ∗ , lim
n→∞ y n = y ∗ .
This completes the proof. ¤
If A = B = 0 in Theorem 3.1, then we have the following:
Theorem 3.2. Let S, T be the same as in Theorem 2.2 and let {g 1 (x n )}, {g 2 (y n )} be the iterative sequences generated by Algorithm 3.2. If 0 < ρ < 2k s22 2
and 0 < γ < 2k s21
1
, then (x n , y n ) converges to the unique solution (x ∗ , y ∗ ) of the problem (1.2).
If φ 1 = φ 2 = δ K in Theorem 3.1, then we have the following theorem.
Theorem 3.3. Let K, A, B, S, T be the same as in Theorem 2.3 and let {g 1 (x n )}, {g 2 (y n )} be the iterative sequences generated by Algorithm 3.3. If the condition (2.1) holds, then (x n , y n ) converges to the unique solution (x ∗ , y ∗ ) of the problem (1.3).
If φ 1 = φ 2 = δ K and A = B = 0 in Theorem 3.1, then we have the following theorem.
Theorem 3.4. Let K, S, T be the same as in Theorem 2.4 and let {g 1 (x n )}, {g 2 (y n )} be the iterative sequences generated by Algorithm 3.4. If 0 < ρ < 2k s22 2
and 0 < γ < 2k s21
1