We begin with some observations which are related to the problem (1).
Remark 1
If (x∗,y∗,z∗)∈ SNSVVID, by (2) we have that
(3)
provided K⊂g1(H).
Consequently if S is a Lipschitz mapping such that x∗∈F(S), then it follows from (3) that
(4)
By virtue of (4) and Nadler’s Theorem (Nadler 1969), we suggest the following iterative algorithm.
Algorithm 1 Let ε
n
be a sequence of nonnegative real number with ε
n
→0 as n→∞. Let r1,r2,r3 be three given positive real numbers in (0,1). For arbitrary chosen initial x0∈H, compute the sequences {x
n
},{y
n
} and {z
n
} in H, such that
(5)
where
(6)
and {α
n
} is a sequence in (0,1) and S:H→H is a mapping.
Theorem 1
Let K be a nonempty closed and convex subset of a real Hilbert space H and ϕ:H→(−∞,+∞) be a proper convex lower semicontinuous function. Let A
i
:H→2H be a μ
i
-Lipschitz continuous mapping with μ
i
<1 and B
i
:H→2H be a σ
i
-Lipschitz continuous mapping with σ
i
<1, i=1,2,3. Let N
i
:H×H→H be a ρ
i
-Lipschitz continuous with respect to the first variable and η
i
-Lipschitz continuous with respect to the second variable and N
i
be A
i
-strongly monotone with constant υ
i
>0 and B
i
-relaxed monotone with constant ξ
i
>0, i=1,2,3. Let g
i
:H→H be a λ
i
-strongly monotone and γ
i
-Lipschitz continuous mapping, i=1,2,3. Let S:H→H be a τ-Lipschitz continuous mapping with 0<τ≤1. If , and the following conditions are satisfied:
-
(i)
where ;
-
(ii)
-
(iii)
for each i=1,2,3
where
(7)
where M= supn≥1ε
n
.
-
(iv)
{α
n
}⊂(0,1) such that .
Then the sequences {x
n
},{y
n
},{z
n
},{un,i},{vn,i} suggested by Algorithm 1 converge strongly to respectively, and SNSVVID, x∗∈F(S).
Proof. Let and x∗∈F(S). By (2) and (4) we have
(8)
Consequently, by (5) and (6), we have
(9)
Since N1(·,·) is ρ1-Lipschitz continuous with respect to the first variable and η1-Lipschitz continuous with respect to the second variable, and A1 is μ1-Lipschitz continuous, and B1 is σ1-Lipschitz continuous, we have
(10)
Since N1 is A1-strongly monotone with constant υ1>0 and B1-relaxed monotone with constant ξ
i
>0, it follows from (10) that
i.e.,
(11)
where
Note that
(12)
Since g2 is λ2-strongly monotone and γ2-Lipschitz continuous mapping, we have
(13)
where .
On the other hand, by (2) and (5), we have
(14)
In view of the assumptions of N2,A2,B2, g2 and by using the same method as given in the proofs in (11) and (13), we can obtain that
(15)
where
and
(16)
From (15), (16) and (14), we have
(17)
Combining (12), (13) and (17) we obtained
(18)
Observe that
(19)
and in view of (2) and (5), we have
(20)
By using the assumptions on N3,A3,B3 and g3, we have
(21)
where
(22)
(23)
Substituting (21) and (22) into (20), we have
(24)
Combining (19), (23) and (24), it yields that
(25)
This imply that
(26)
Substituting (26) into (18) we have
(27)
that is
(28)
From (11) and (28), we get
(29)
On the other hand, since g1 is λ1-strongly monotone and γ1-Lipschitz continuous mapping, we have
i.e.,
(30)
Similarly, we have
(31)
Substituting (28) into (31), we have
(32)
Set
(33)
Substituting (30), (31), (32) and (33) into (9), we get
(34)
Since
letting , then we have ℓ
n
≤ℓ. Therefore from (34) we have that
(35)
By condition (iii)
(36)
this imply that
(37)
that is
(38)
Put
(39)
By the assumption that 0<τ≤1, it follows that
This imply that t
n
∈(0,1). From assumption (iv) we have
These show that all conditions in Lemma 2 are satisfied. Hence x
n
→x∗ as n→∞. Consequently from (26) and (28), we have z
n
→z∗ and y
n
→y∗ as n→∞, respectively. Moreover since A
i
is μ
i
-Lipschitz continuous and B
i
is σ
i
-Lipschitz continuous with μ
i
<1, σ
i
<1, we can also prove that {un,i} and {vn,i}, i=1,2,3 are Cauchy sequences. Thus there exists such that as n→∞. Moreover by using the continuity of mappings , i=1,2,3, it follows from (5) that
Hence from Lemma 2 it follows that SNSVVID Finally we prove that and Indeed we have
That is . Since A1(y∗)∈C B(H), we must have . Similarly we can show that and This complete the proof. â–