In this section, we propose and analyze the following relaxed viscosity iterative algorithms for finding a common element of the solution set of GSVI (1.3), the solution set of SFP (1.6) and the fixed point set of a strictly pseudo-contractive mapping .
Algorithm 3.1 Let for , , and such that for all . For given arbitrarily, let , , be the sequences generated by the Mann-type viscosity iterative scheme with regularization
Algorithm 3.2 Let for , , and such that for all . For given arbitrarily, let , , be the sequences generated by the Mann-type viscosity iterative scheme with regularization
Next, we first give the strong convergence criteria of the sequences generated by Algorithm 3.1.
Theorem 3.1 Let C be a nonempty closed convex subset of a real Hilbert space . Let and be -inverse strongly monotone for . Let be a k-strictly pseudocontractive mapping such that . Let be a ρ-contraction with . For given arbitrarily, let , , be the sequences generated by Algorithm 3.1, where for , , and such that
-
(i)
;
-
(ii)
and for all ;
-
(iii)
and ;
-
(iv)
and ;
-
(v)
;
-
(vi)
and .
Then the sequences , , converge strongly to the same point if and only if . Furthermore, is a solution of GSVI (1.3), where .
Proof First, taking into account , without loss of generality, we may assume that for some .
Now, let us show that is ζ-averaged for each , where
Indeed, it is easy to see that is -ism, that is,
Observe that
Hence, it follows that is -ism. Thus, is -ism according to Proposition 2.2(ii). By Proposition 2.2(iii), the complement is -averaged. Therefore, noting that is -averaged and utilizing Proposition 2.3(iv), we know that for each , is ζ-averaged with
This shows that is nonexpansive. Furthermore, for with , we have
Without loss of generality, we may assume that
Consequently, it follows that for each integer , is -averaged with
This immediately implies that is nonexpansive for all .
Next, we divide the remainder of the proof into several steps.
Step 1. is bounded.
Indeed, take arbitrarily. Then , for , and
For simplicity, we write
for each . Then for each . From Algorithm 3.1 it follows that
(3.1)
Utilizing Lemma 2.5, we also have
(3.2)
Since is -inverse strongly monotone for and for , we know that for all ,
(3.3)
Hence it follows from (3.1) and (3.3) that
(3.4)
Since for all , utilizing Lemma 2.4, we obtain from (3.4)
(3.5)
Now, we claim that
(3.6)
As a matter of fact, if , then it is clear that (3.6) is valid, that is,
Assume that (3.6) holds for , that is,
(3.7)
Then we conclude from (3.5) and (3.7) that
By induction, we conclude that (3.6) is valid. Hence, is bounded. Since , , and are Lipschitz continuous, it is easy to see that , , and are bounded, where for all .
Step 2. .
Indeed, define for all . It follows that
(3.8)
Since for all , utilizing Lemma 2.4, we have
(3.9)
Next, we estimate . Observe that
(3.10)
and
(3.11)
Combining (3.10) with (3.11), we get
(3.12)
which hence implies that
(3.13)
Hence it follows from (3.8), (3.9) and (3.13) that
Since , , and are bounded, it follows from conditions (i), (iii), (v) and (vi) that
Hence by Lemma 2.1, we get . Thus,
(3.14)
Step 3. and , where .
Indeed, utilizing Lemma 2.4 and the convexity of , we obtain from Algorithm 3.1 and (3.2)-(3.3) that
Therefore,
Since , , , and for some , it follows that
Step 4. .
Indeed, by firm nonexpansiveness of , we have
that is,
(3.15)
Moreover, using the argument technique similar to the above one, we derive
that is,
(3.16)
Utilizing (3.2), (3.15) and (3.16), we have
(3.17)
So, from Algorithm 3.1 and (3.17), it follows that
which hence implies that
Since , , , , , and , it follows from the boundedness of , , and that
Consequently, it immediately follows that . Also, since and , we have
Thus, we have
(3.18)
Note that
It hence follows that
Step 5. , where .
Indeed, since is bounded, there exists a subsequence of such that
(3.19)
Also, since H is reflexive and is bounded, without loss of generality, we may assume that weakly for some . First, it is clear from Lemma 2.2 that . Now, let us show that . We note that
where is defined as that in Lemma 1.1. According to Lemma 2.2, we obtain . Further, let us show that . As a matter of fact, since , and , we deduce that weakly and weakly. Let
where . Then T is maximal monotone and if and only if ; see [41] for more details. Let . Then we have
and hence
So, we have
On the other hand, from
we have
and hence,
Therefore, from
we have