We first show a strong convergence of an iterative algorithm based on extragradient and hybrid methods which solves the problem of finding a common element of the set of solutions of a generalized equilibrium problem, the set of fixed points of a family of infinitely nonexpansive mappings, and the set of solutions of the variational inequality for a monotone, Lipschitzcontinuous mapping in a Hilbert space.
Theorem 3.1. Let C be a nonempty closed convex subset of a real Hilbert space H. Let F be a bifunction from C × C to R satisfying (A1)(A4). Let A be a monotone and kLipschitzcontinuous mapping of C into H and B be an αinversestrongly monotone mapping of C into H. Let S_{1}, S_{2},... be a family of infinitely nonexpansive mappings of C into itself such that . Assume that for all i ∈ {1, 2,...} and for any bounded subset K of C, thenthere holds
Let {x_{
n
} }, {u_{
n
} }, {y_{
n
} } and {z_{
n
} } be sequences generated by
for every n = 1, 2,... where {λ_{
n
} } ⊂ [a, b] for some , {r_{
n
}} ⊂ [d, e] for some d, e ∈ (0, 2α), and {α_{
n
} }, {β_{
n
} }, {γ_{
n
} } are three sequences in [0, 1] satisfying the conditions:

(i)
α_{
n
} + β_{
n
} ≤ 1 for all n ∈ N;

(ii)
;

(iii)
;

(iv)
and for all n ∈ N;
Then, {x_{
n
} }, {u_{
n
} }, {y_{
n
} } and {z_{
n
} } converge strongly to w = P_{Ω}(x).
Proof. It is obvious that C_{
n
} is closed, and Q_{
n
} is closed and convex for every n = 1, 2,.... Since
we also have that C_{
n
} is convex for every n = 1, 2,.... It is easy to see that 〈x_{
n
}  z, x  x_{
n
} 〉 ≥ 0 for all z ∈ Q_{
n
} and by (2.1), . Let t_{
n
}= P_{
C
}(u_{
n
} λ_{
n
}Ay_{
n
}) for every n = 1, 2,.... Let u ∈ Ω and let >be a sequence of mappings defined as in Lemma 2.2. Then . From and the αinverse strongly monotonicity of B, we have
From (2.2), the monotonicity of A, and u ∈ V I(C, A), we have
Further, Since y_{
n
} = (1  γ_{
n
} )u_{
n
} + γ_{
n
}P_{
C
} (u_{
n
}  λ_{
n
}Au_{
n
} ) and A is kLipschitzcontinuous, we have
In addition, from the definition of P_{
C
} , we have
It follows from , and (3.2) that
In addition, from u ∈ V I(C, A) and (3.2), we have
Therefore, from (3.2) to (3.4) and z_{
n
} = (1  α_{
n
}  β_{
n
} )x_{
n
} + α_{
n
}y_{
n
} + β_{
n
}S_{
n
}t_{
n
} and u = S_{
n
}u, we have
for every n = 1, 2,... and hence u ∈ C_{
n
} . So, Ω ⊂ C_{
n
} for every n = 1, 2,.... Next, let us show by mathematical induction that x_{
n
} is well defined and Ω ⊂ C_{
n
} ∩ Q_{
n
} for every n = 1, 2,.... For n = 1 we have x_{1} = x ∈ C and Q_{1} = C. Hence, we obtain Ω ⊂C_{1} ∩ Q_{1}. Suppose that x_{
k
} is given and Ω ⊂ C_{
k
} ∩ Q_{
k
} for some k ∈ N. Since Ω is nonempty, C_{
k
} ∩ Q_{
k
} is a nonempty closed convex subset of H. Hence, there exists a unique element x_{k+1}∈ C_{
k
}∩ Q_{
k
}such that . It is also obvious that there holds 〈x_{k+1} z, x  x_{k+1}〉 ≥ 0 for every z ∈ C_{
k
}∩ Q_{
k
}. Since Ω ⊂ C_{
k
} ∩ Q_{
k
} , we have 〈x_{k+1} z, x  x_{k+1}〉 ≥ 0 for every z ∈ Ω and hence Ω ⊂ Q_{k+1}. Therefore, we obtain Ω ⊂ C_{k+1}∩ Q_{k+1}.
Let l_{0} = P_{Ω}x. From and l_{0} v Ω ⊂ C_{
n
} ∩ Q_{
n
} , we have
for every n = 1, 2,.... Therefore, {x_{
n
} } is bounded. From (3.2) to (3.5) and the lipschitz continuity of A, we also obtain that {u_{
n
} }, {y_{
n
} }, {Au_{
n
} }, {t_{
n
} } and {z_{
n
} } are bounded. Since x_{n+1}∈ C_{
n
} ∩ Q_{
n
} ⊂ C_{
n
} and , we have
for every n = 1, 2,.... It follows from (3.6) that lim_{n→∞}x_{
n
} x exists.
Since and x_{n+1}∈ Q_{
n
} , using (2.2), we have
for every n = 1, 2,.... This implies that
Since x_{n+1}∈ C_{
n
}, we have z_{
n
} x_{n+1}^{2} ≤ x_{
n
} x_{n+1}^{2} + (3  3γ_{
n
}+ α_{
n
})b^{2}Au_{
n
}^{2} and hence it follows from lim_{n→∞}γ_{
n
}= 1 and lim_{n→∞}α_{
n
}= 0 that lim_{n→∞}z_{
n
} x_{n+1} = 0. Since
for every n = 1, 2,..., we have x_{
n
}  z_{
n
}  → 0.
For u ∈ Ω, from (3.5), we obtain
Since lim_{n→∞}γ_{
n
}= 1 and lim_{n→∞}α_{
n
}= 0, {x_{
n
} }, {y_{
n
} }, {Au_{
n
} }, and {z_{
n
} } are bounded, we have
By lim inf_{n→∞}β_{
n
}> 0, we get
From (3.3) and u = S_{
n
}u, we have
Thus, lim_{n→∞}t_{
n
}  u^{2} x_{
n
}  u^{2} = 0.
From (3.3) and (3.2), we have
It follows that
The assumptions on γ_{
n
} and λ_{
n
} imply that and . Consequently, lim_{n→∞}u_{
n
}  y_{
n
}  = lim_{n→∞}t_{
n
}  y_{
n
}  = 0. Since A is Lipschitzcontinuous, we have lim_{n→∞}At_{
n
}  Ay_{
n
}  = 0. It follows from u_{
n
}  t_{
n
}  ≤ u_{
n
}  y_{
n
}  + t_{
n
}  y_{
n
}  that lim_{n→∞}u_{
n
}  t_{
n
}  = 0.
We rewrite the definition of z_{
n
} as
From lim_{n→∞}z_{
n
}  x_{
n
}  = 0, lim_{n→∞}α_{
n
}= 0, the boundedness of {x_{
n
} }, {y_{
n
}} and lim inf_{n→∞}β_{
n
}> 0 we infer that lim_{n→∞}S_{
n
}t_{
n
} x_{
n
} = 0.
By (3.2)(3.5), we have
Hence, we have
Since lim_{n→∞}α_{
n
}= 1, lim inf_{n→∞}β_{
n
}> 0, lim_{n→∞}γ_{
n
}= 1, x_{
n
}  z_{
n
}  → 0 and the sequences {x_{
n
} } and {z_{
n
} } are bounded, we obtain Bx_{
n
}  B_{
u
}  → 0.
For u ∈ Ω, we have, from Lemma 2.2,
Hence,
Then, by (3.5), we have
Hence,
Since lim_{n→∞}α_{
n
}= 0, lim inf_{n→∞}β_{
n
}> 0, lim_{n→∞}γ_{
n
}= 1, x_{
n
} z_{
n
} → 0, Bx_{
n
} Bu → 0 and the sequences {x_{
n
}}, {u_{
n
}} and {z_{
n
}} are bounded, we obtain x_{
n
} u_{
n
} → 0. From z_{
n
}  t_{
n
}  ≤ z_{
n
}  x_{
n
} +x_{
n
}  u_{
n
} +u_{
n
}  t_{
n
} , we have z_{
n
}  t_{
n
}  → 0.
From t_{
n
}  x_{
n
}  ≤ t_{
n
}  u_{
n
}  + x_{
n
}  u_{
n
} , we also have t_{
n
}  x_{
n
}  → 0.
Since z_{
n
} = (1  α_{
n
}  β_{
n
} )x_{
n
} + α_{
n
}y_{
n
} + β_{
n
}S_{
n
}t_{
n
} , we have β_{
n
} (S_{
n
}t_{
n
}  t_{
n
} ) = (1  α_{
n
}  β_{
n
} )(t_{
n
}  x_{
n
} ) + α_{
n
} (t_{
n
}  y_{
n
} ) + (z_{
n
}  t_{
n
} ). Then
and hence S_{
n
}t_{
n
}  t_{
n
}  → 0. At the same time, observe that for all i ∈ {1, 2,...},
It follows from (3.8) and the condition (*) that for all i ∈ {1, 2,...},
As {x_{
n
} } is bounded, there exists a subsequence of {x_{
n
} } such that x_{
ni
} ⇀ w. From x_{
n
}  u_{
n
}  → 0, we obtain that u_{
ni
} ⇀ w. From u_{
n
}  t_{
n
}  → 0, we also obtain that t_{
ni
} ⇀ w. Since {u_{
ni
} } ⊂ C and C is closed and convex, we obtain w ∈ C.
First, we show w ∈ GEP(F, B). By , we know that
It follows from (A2) that
Hence,
For t with 0 < t ≤ 1 and y ∈ C, let y_{
t
} = t_{
y
} + (1  t)w. Since y ∈ C and w ∈ C, we obtain y_{
t
} ∈ C. So, from (3.10) we have
Since , we have . Further, from the inversestrongly monotonicity of B, we have . Hence, from (A4), and , we have
as i → ∞. From (A1), (A4) and (3.11), we also have
and hence
Letting t → 0, we have, for each y ∈ C,
This implies that w ∈ GEP(F, B).
We next show that . Assume . Since and for some i_{0} ∈ {1, 2,...} from the Opial condition, we have
This is a contradiction. Hence, we get .
Finally we show w ∈ V I(C, A). Let
where N_{
C
}v is the normal cone to C at v ∈ C. We have already mentioned that in this case the mapping T is maximal monotone, and 0 ∈ Tv if and only if v ∈ V I(C, A). Let (v, g) ∈ G(T). Then Tv = Av + N_{
C
}v and hence g  Av ∈ N_{
C
}v.
Hence, we have 〈v  t, g  Av〉 ≥ 0 for all t ∈ C. On the other hand, from t_{
n
} = P_{
C
} (u_{
n
}  λ_{
n
}Ay_{
n
} ) and v ∈ C, we have
and hence
Therefore, we have
Hence, we obtain 〈v  w, g〉 ≥ 0 as i → ∞. Since T is maximal monotone, we have w ∈ T^{1}0 and hence w ∈ V I(C, A). This implies that w ∈ Ω.
From l_{0} = P_{Ω}x, w ∈ Ω and (3.6), we have
Hence, we obtain
From , we have , and hence . Since and l_{0} ∈ Ω ⊂ C_{
n
} ∩ Q_{
n
} ⊂ Q_{
n
} , we have
As i → ∞, we obtain  l_{0} w^{2} ≥ 〈l_{0} w, x  l_{0}〉 ≥ 0 by l_{0} = P_{Ω}x and w ∈ Ω. Hence, we have w = l_{0}. This implies that x_{
n
} → l_{0}. It is easy to see u_{
n
} → l_{0}, y_{
n
} → l_{0} and z_{
n
} → l_{0}. The proof is now complete.
By combining the arguments in the proof of Theorem 3.1 and those in the proof of Theorem 3.1 in [3], we can easily obtain the following weak convergence theorem for an iterative algorithm based on the extragradient method which solves the problem of finding a common element of the set of solutions of a generalized equilibrium problem, the set of fixed points of a family of infinitely nonexpansive mappings and the set of solutions of the variational inequality for a monotone, Lipschitzcontinuous mapping in a Hilbert space.
Theorem 3.2. Let C be a nonempty closed convex subset of a real Hilbert space H. Let F be a bifunction from C × C to R satisfying (A1)(A4). Let A be a monotone, and kLipschitzcontinuous mapping of C into H and B be an αinversestrongly monotone mapping of C into H. Let S_{1}, S_{2},... be a family of infinitely nonexpansive mappings of C into itself such that . Assume that for all i ∈ {1, 2,...} and for any bounded subset K of C, thenthere holds
Let {x_{
n
} }, {u_{
n
} } and {y_{
n
} } be the sequences generated by
for every n = 1, 2,.... If {λ_{
n
} } ⊂ [a, b] for some , {β_{
n
} } ⊂ [δ, ε] for some δ, ε ∈ (0, 1) and {r_{
n
} } ⊂ [d, e] for some d, e ∈ (0, 2α). Then, {x_{
n
} }, {u_{
n
} } and {y_{
n
} } converge weakly to w ∈ Ω, where w = lim_{n→∞}P_{Ω}x_{
n
}.