In this section, we discuss the parallel algorithm and the cyclic algorithm, respectively, for solving the variational inequality over the set of the common fixed points of finite strict pseudo-contractions.
Let
be a real Hilbert space and
a
-strongly monotone and boundedly Lipschitzian operator. Let
be a positive integer and
a
-strict pseudo-contraction for some
such that
We consider the problem of finding
such that
Since
is a nonempty closed convex subset of
,
(4.1) has a unique solution. Throughout this section,
is an arbitrary fixed point,
,
is the Lipschitz constant of
on
, the fixed constant
satisfies
, and the sequence
belongs to
.
Firstly we consider the parallel algorithm. Take a positive sequence
such that
and let
By using Lemma 2.8, we assert that
is a
-strict pseudo-contraction with
and
holds. Thus VI(4.1) is equivalent to VI
and we can use scheme (3.2) to solve VI(4.1). In fact, taking
in the scheme (3.2), we get the so-called parallel algorithm
Using Lemma 2.8 and Thorem 3.2, the following conclusion can be deduced directly.
Theorem 4.1.
Suppose that
and
satisfy the same conditions as in Theorem 3.2. Then the sequence
generated by the parallel algorithm (4.3) converges strongly to the unique solution
of
(4.1).
For each
let
where the constant
such that
. Then we turn to defining the cyclic algorithm as follows:
Indeed, the algorithm above can be rewritten as
where 
, namely,
is one of
circularly. For convenience, we denote (4.6) as
We get the following result
Theorem 4.2.
If
satisfies the following conditions:
(i)
;
(ii)
;
(iii)
, or 
then the sequence
generated by (4.6) converges strongly to the unique solution
of
.
Proof.
We break the proof process into six steps.
-
(1)
. We prove it by induction. Definitely
. Suppose
, that is,
We have from
, (4.8), and Lemma 2.7 that
where
Observing
, we get
This together with (4.9) implies that
It suggests that
. Therefore,
for all
. We can also prove that the sequences
,
,
are all bounded.
-
(2)
By (4.6) and Lemma 2.7, we have
where
Since
satisfies (i)–(iii), using Lemma 2.2, we get
-
(3)
By (4.3) and
, we have
Recursively,
By Lemma 2.6,
is nonexpansive, we obtain
Adding all the expressions above, we get
Using this together with the conclusion of step (2), we obtain
-
(4)
. Assume that
such that
, we prove
. By the conclusion of step (3), we get
Observe that, for each
is some permutation of the mappings
, since
are finite, all the full permutation are
, there must be some permutation that appears infinite times. Without loss of generality, suppose that this permutation is
, we can take a subsequence
such that
It is easy to prove that
is nonexpansive. By Lemma 2.3, we get
Using Lemmas 2.6 and 2.9, we obtain
-
(5)
In fact, there exists a subsequence
such that
Without loss of generality, we may further assume that
Since
is the solution of
, we obtain
-
(6)
By (4.6), Lemmas 2.1(ii), and 2.7, we obtain
From the conclusion of step (5) and Lemma 2.2, we get