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 wellknown 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 meanvalue 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.