Let
be an equilibrium bifunction from
to
, that is
for every
In addition, assume that
is convex and lower semicontinuous in the variable
for each fixed
.
The equilibrium problem for
is to find
such that
First, we recall several well-known facts in [12, 13] which are necessary in the proof of our results.
The equilibrium bifunction
is said to be
(i)monotone, if for all
, we have
(ii)strongly monotone with constant
, if, for all
, we have
(iii)hemicontinuous in the variable
for each fixed
, if
We can get the following proposition from the above definitions.
Proposition 2.1.
-
(i)
If
is hemicontinuous in the first variable for each fixed
and
is monotone, then
, where
is the solution set of (2.1),
is the solution set of
for all
, and they are closed and convex.
-
(ii)
If
is hemicontinuous in the first variable for each
and
is strongly monotone, then
is a nonempty singleton.
Lemma 2.2 (see [14]).
Let
be the sequences of positive numbers satisfying the conditions:
(i)
,
(ii)
, 
Then, 
Lemma 2.3 (see [15]).
Let
be a nonempty closed convex subset of a Hilbert space
and
a strictly pseudocontractive mapping. Then
is demiclosed at zero.
We construct a regularization solution
for (1.9) by solving the following variational inequality problem: find
such that
where
, is the regularization parameter.
We have the following result.
Theorem 2.4.
Let
be a nonempty closed convex subset of
. Let
be a
-inverse strongly monotone mapping of
into
, and let
be a monotone hemicontinuous mapping of
into
such that
. Then, we have
(i)For each
, the problem (2.5) has a unique solution
;
(ii)If
then
and
for all 
-
(iii)
where
is a positive constant.
Proof.
-
(i)
Let
Then, problem (2.5) has the following form: find
such that
where
It is not difficult to verify that
are the monotone bifunctions, and for each fixed
, they are hemicontinuous in the variable
. Therefore,
also is monotone hemicontinuous in the variable
for each fixed
. Moreover, it is strongly monotone with constant
. Hence, (2.8) (consequently (2.5)) has a unique solution
for each
.
-
(ii)
Now we prove that
Since 
and 
,
and
By adding the last inequality to (2.8) in which
is replaced by
and using the properties of
, we obtain
that implies (2.10). It means that
is bounded. Let
, as
. Since
is closed and convex,
is weakly closed. Hence
. We prove that
. From the monotone property of
and (2.8), it follows
Letting
, we obtain
for any
By virtue of Proposition 2.1, we have 
Now we show that
for all
From (2.8),
for any
, and the monotone property of
, it implies that
On the base of
-inverse strongly monotone property of
, the monotone property of
,
,
, for all
,
. From the last inequality, we have
Tending
in the last inequality, we obtain
Since
is
-inverse strongly monotone, the mapping
satisfies (1.6), where
Because
, we have
. When
, this inequality will not be changed if
is replaced by
. Thus,
is strictly pseudocontractive. Applying Lemma 2.2, we can conclude that
It means that
It is well known that the sets
are closed and convex. Therefore,
is also closed and convex. Then, from (2.10) it implies that
is the unique element in
having a minimal norm. Consequently, we have
(iii) From (2.8) and the properties of
, for each
, it follows
or
For each 
is bounded since the operator
is Lipschitzian with Lipschitz constants
. Using (2.10), the boundedness of
and the Lagrange's mean-value theorem for the function
,
,
, on
if
or
if
, we have conclusion (iii). This completes the proof.
Remark 2.5.
Obviously, if
, where
is the solution of (2.8) with
, as
, then 
Now, we consider the regularization inertial proximal point algorithm
Clearly,
is a bifunction. Moreover, it is strongly monotone with
By Proposition 2.1, there exists a unique element
satisfying (2.20).
Theorem 2.6.
Let
be a nonempty closed convex subset of a Hilbert space
. Let
be a
-inverse strongly monotone mapping of
into
, and let
be a monotone hemicontinuous mapping of
into
such that
. Assume that the parameters
, and
are chosen such that
(i)
,
,
(ii)
,
(iii)
-
(iv)
Then the sequence
defined by (2.20) converges strongly to the element
as
.
Proof.
From (2.20) it follows
By the similar argument, from (2.5) it implies
where
is the solution of (2.5) when
is replaced by
. By setting
in (2.23) and
in (2.24) and adding one obtained result to the other, we have,
Therefore, from the monotone property of the mappings
,
, it follows
From (2.23), (2.5) with
and
, we have
where
Since the seris in (iii) is convergent,
Consequently,
From Lemma 2.2, it follows that
as
.
On the other hand,
as
. Therefore, we have
as
This completes the proof.
Remark 2.7.
The sequences
and
which are defined by
with
satisfy all conditions in Theorem 2.6.