Throughout the rest of this paper, we always assume that
is a
-lipschitzian continuous and
-strongly monotone operator with
and assume that
.
. Let
be mappings defined as Lemma 2.1. Define a mapping
by
, for all
, where
, then, by Lemma 2.3,
is nonexpansive. We consider the mapping
on
defined by
where
. By Lemmas 2.1 and 2.3, we have
It is easy to see that
is a contraction. Therefore, by the Banach contraction principle,
has a unique fixed-point
such that
For simplicity, we will write
for
provided no confusion occurs. Next, we prove that the sequence
converges strongly to a
which solves the variational inequality
Equivalently,
.
Theorem 3.1.
Let
be a nonempty closed convex subset of a real Hilbert space
and
a bifunction from
into
satisfying (A1), (A2), (A3), and (A4). Let
be a
-strictly pseudocontractive nonself mapping such that
. Let
be an
-Lipschitzian continuous and
-
monotone operator on
with
and
,
. Let
be asequence generated by
where
,
, and
satisfy
if
and
satisfy the following conditions:
(i)
,
,
(ii)
and
,
then
converges strongly to a point
which solves the variational inequality (3.4).
Proof.
First, take
. Since
and
, from Lemma 2.1, for any
, we have
Then, since
, we obtain that
Further, we have
It follows that
.
Hence,
is bounded, and we also obtain that
and
are bounded. Notice that
By Lemma 2.1, we have
It follows that
Thus, from Lemma 2.4, (3.7), and (3.11), we obtain that
It follows that
Since
, therefore
From (3.9), we derive that
Define
by
, then
is nonexpansive with
by Lemma 2.3. We note that
So by (3.15) and
, we obtain that
Since
is bounded, so there exists a subsequence
which converges weakly to
. Next, we show that
. Since
is closed and convex,
is weakly closed. So we have
. Let us show that
. Assume that
, Since
and
, it follows from the Opial's condition that
This is a contradiction. So, we get
and
.
Next, we show that
. Since
, for any
, we obtain
From (A2), we have
Replacing
by
, we have
Since
and
, it follows from (A4) that
, for all
. Let
for all
and
, then we have
and hence
. Thus, from (A1) and (A4), we have
and hence
. From (A3), we have
for all
and hence
. Therefore,
. On the other hand, we note that
Hence, we obtain
It follows that
This implies that
In particular,
Since
, it follows from (3.27) that
as
. Next, we show that
solves the variational inequality (3.4).
As a matter of fact, we have
and we have
Hence, for
,
Since
is monotone (i.e.,
, for all
. This is due to the nonexpansivity of
).
Now replacing
in (3.30) with
and letting
, we obtain
That is,
is a solution of (3.4). To show that the sequence
converges strongly to
, we assume that
. Similiary to the proof above, we derive
. Moreover, it follows from the inequality (3.31) that
Interchange
and
to obtain
Adding up (3.32) and (3.33) yields
Hence,
, and therefore
as
,
This is equivalent to the fixed-point equation
Theorem 3.2.
Let
be a nonempty closed convex subset of a real Hilbert space
and
a bifunction from
into
satisfying (A1), (A2), (A3) and (A4). Let
be a
-strictly pseudocontractive nonself mapping such that
. Let
be an
-Lipschitzian continuous and
-strongly monotone operator on
with
. Suppose that
,
. Let
and
be sequences generated by
and
where
,
if
,
, and
satisfy the following conditions:
(i)
,
,
,
,
(ii)
and
,
,
(iii)
,
and
,
then
and
converge strongly to a point
which solves the variational inequality(3.4).
Proof.
We first show that
is bounded. Indeed, pick any
to derive that
By induction, we have
and hence
is bounded. From (3.6) and (3.7), we also derive that
and
are bounded. Next, we show that
. We have
where
On the other hand, we have
From
and
, we note that
Putting
in (3.43) and
in (3.44), we have
So, from (A2), we have
and hence
Since
, without loss of generality, let us assume that there exists a real number a such that
for all
. Thus, we have
where
. Next, we estimate
. Notice that
From (3.48), (3.49), and (3.42), we obtain that
where
is an appropriate constant such that
From (3.41) and (3.50), we obtain
where
. Hence, few by Lemma 2.5, we have
From (3.48) and (3.50),
and
, we have
Since
it follows that
From
and (3.53), we have
For
, we have
This implies that
Then, from (3.7) and (3.59), we derive that
Since
,
, we have
From (3.57) and (3.61), we obtain that
Define
by
, then
is nonexpansive with
by Lemma 2.3. Notice that
By (3.62) and
, we obtain that
Next, we show that
, where
is a unique solution of the variational inequality (3.4). Indeed, take a subsequence
of
such that
Since
is bounded, there exists a subsequence
of
which converges weakly to
.
Without loss of generality, we can assume that
. From (3.61) and (3.64), we obtain
and
. By the same argument as in the proof of Theorem 3.1, we have
. Since
, it follows that
From
, we have
This implies that
where
,
, and
.
It is easy to see that
,
, and
by (3.66). Hence by Lemma 2.5, the sequence
converges strongly to
.