Throughout this section, unless otherwise stated, we assume that
is a maximal monotone operator with domain
,
is a relatively nonexpansive mapping,
is an α-inverse-strongly monotone mapping and
is a bifunction satisfying (A1)–(A4), where
is a nonempty closed convex subset of a reflexive, strictly convex, and smooth Banach space
. In this section, we study the following algorithm.
Algorithm 3.1.
where
is a sequence in
and
,
are sequences in
.
First we investigate the condition under which the Algorithm 3.1 is well defined. Rockafellar [40] proved the following result.
Lemma 3.2 (Rockafellar [40]).
Let
be a reflexive, strictly convex, and smooth Banach space and let
be a multivalued operator, then there hold the following:
(i)
is closed and convex if
is maximal monotone such that
;
(ii)
is maximal monotone if and only if
is monotone with
for all
.
Utilizing this result, we can show the following lemma.
Lemma 3.3.
Let
be a reflexive, strictly convex, and smooth Banach space. If
, then the sequence
generated by Algorithm 3.1 is well defined.
Proof.
For each
, define two sets
and
as follows:
It is obvious that
is closed and
are closed convex sets for each
. Let us show that
is convex. For
and
, put
. It is sufficient to show that
. Indeed, observe that
is equivalent to
Note that there hold the following:
Thus we have
This implies that
. Therefore,
is convex and hence
is closed and convex.
On the other hand, let
be arbitrarily chosen, then
and
. From Algorithm 3.1, it follows that
So
for all
. Now, from Lemma 3.2 it follows that there exists
such that
and
. Since
is monotone, it follows that
, which implies that
and hence
. Furthermore, it is clear that
, then
, and therefore
is well defined. Suppose that
and
is well defined for some
. Again by Lemma 3.2, we deduce that
such that
and
, then from the monotonicity of
we conclude that
, which implies that
and hence
. It follows from Lemma 2.4 that
which implies that
. Consequently,
and so
. Therefore
is well defined, then, by induction, the sequence
generated by Algorithm 3.1, is well defined for each integer
.
Remark 3.4.
From the above proof, we obtain that
for each integer
.
We are now in a position to prove the main theorem.
Theorem 3.5.
Let
be a uniformly smooth and uniformly convex Banach space. Let
be a sequence in
and
be sequences in
such that
Let
. If
is uniformly continuous, then the sequence
generated by Algorithm 3.1 converges strongly to
.
Proof.
First of all, if follows from the definition of
that
. Since
, we have
Thus
is nondecreasing. Also from
and Lemma 2.3, we have that
for each
and for each
. Consequently,
is bounded. Moreover, according to the inequality
we conclude that
is bounded. Thus, we have that
exists. From Lemma 2.3, we derive the following:
for all
. This implies that
. So it follows from Lemma 2.1 that
. Since
, from the definition of
, we also have
Observe that
At the same time,
Since
and
, it follows that
and hence that
. Further, from
, we have
, which yields
Then it follows from
that
. Hence it follows from Lemma 2.1 that
. Since from (3.15) we derive
we have
Thus, from
,
, and
, we know that
. Consequently from (3.16),
, and
it follows that
So it follows from (3.15),
, and
that
. Utilizing Lemma 2.1 we deduce that
Furthermore, for
arbitrarily fixed, it follows from Proposition 2.8 that
Since
is uniformly norm-to-norm continuous on bounded subsets of
, it follows from (3.23) that
and
, which hence yield
. Utilizing Lemma 2.1, we get
. Observe that
due to (3.23). Since
is uniformly norm-to-norm continuous on bounded subsets of
, we have that
On the other hand, we have
Noticing that
we have
From (3.26) and
, we obtain
Since
is also uniformly norm-to-norm continuous on bounded subsets of
, we obtain
Observe that
Since
is uniformly continuous, it follows from (3.27), (3.31) and
that
.
Now let us show that
, where
Indeed, since
is bounded and
is reflexive, we know that
. Take
arbitrarily, then there exists a subsequence
of
such that
. Hence
. Let us show that
. Since
, we have that
. Moreover, since
is uniformly norm-to-norm continuous on bounded subsets of
and
, we obtain
It follows from
and the monotonicity of
that
for all
and
. This implies that
for all
and
. Thus from the maximality of
, we infer that
. Therefore,
. Further, let us show that
. Since
and
, from
we obtain that
and
.
Since
is uniformly norm-to-norm continuous on bounded subsets of
, from
we derive
From
, it follows that
By the definition of
, we have
where
Replacing
by
, we have from (A2) that
Since
is convex and lower semicontinuous, it is also weakly lower semicontinuous. Letting
in the last inequality, from (3.38) and (A4) we have
For
, with
, and
, let
. Since
and
, then
and hence
. So, from (A1) we have
Dividing by
, we have
Letting
, from (A3) it follows that
So,
. Therefore, we obtain that
by the arbitrariness of
.
Next, let us show that
and
.
Indeed, put
. From
and
, we have
. Now from weakly lower semicontinuity of the norm, we derive for each 
It follows from the definition of
that
and hence
So we have
. Utilizing the Kadec-Klee property of
, we conclude that
converges strongly to
. Since
is an arbitrary weakly convergent subsequence of
, we know that
converges strongly to
. This completes the proof.
Theorem 3.5 covers [25, Theorem 3.1] by taking
and
. Also Theorem 3.5 covers [24, Theorem 2.1] by taking
,
and
.
Theorem 3.6.
Let
be a nonempty closed convex subset of a uniformly smooth and uniformly convex Banach space
. Let
be a maximal monotone operator with domain
,
be a relatively nonexpansive mapping,
be an α-inverse-strongly monotone mapping and
be a bifunction satisfying (A1)–(A4). Assume that
is a sequence in
satisfying
and that
is a sequences in
satisfying
.
Define a sequence
.
Algorithm 3.7.
where
is the single valued duality mapping on
. Let
. If
is uniformly continuous, then
converges strongly to
.
Proof.
For each
, define two sets
and
as follows:
It is obvious that
is closed and
are closed convex sets for each
. Let us show that
is convex and so
is closed and convex. Similarly to the proof of Lemma 3.3, since
is equivalent to
we know that
is convex and so is
. Next, let us show that
for each
. Indeed, utilizing Proposition 2.8, we have, for each
,
So
for all
and
. As in the proof of Lemma 3.3, we can obtain
and hence
. It follows from Lemma 2.4 that
which implies that
. Consequently,
and so
for all
. Therefore, the sequence
generated by Algorithm 3.7 is well defined. As in the proof of Theorem 3.5, we can obtain
. Since
, from the definition of
we also have
As in the proof of Theorem 3.5, we can deduce not only from
that
but also from
,
and
that
Since 
, from the definition of
, we also have
It follows from (3.55) and
that
Utilizing Lemma 2.1 we have
Furthermore, for
arbitrarily fixed, it follows from Proposition 2.8 that
Since
is uniformly norm-to-norm continuous on bounded subsets of
, it follows from (3.58) that
and
, which together with
, yield
. Utilizing Lemma 2.1, we get
. Observe that
due to (3.58). Since
is uniformly norm-to-norm continuous on bounded subsets of
, we have
Note that
Therefore, from
we get
Since
is also uniformly norm-to-norm continuous on bounded subsets of
, we obtain
It follows that
Since
is uniformly continuous, it follows from (3.58) and (3.64) that
.
Finally, we prove that
. Indeed, for
arbitrarily fixed, there exists a subsequence
of
such that
, then
. Now let us show that
. Since
, we have that
. Moreover, since
is uniformly norm-to-norm continuous on bounded subsets of
, and
, we obtain that
. It follows from
and the monotonicity of
that
for all
and
. This implies that
for all
and
. Thus from the maximality of
, we infer that
. Further, let us show that
. Since
and
, from
we obtain that
and
.
Since
is uniformly norm-to-norm continuous on bounded subsets of
, from
we derive
. From
it follows that
By the definition of
, we have
where
. Replacing
by
, we have from (A2) that
Since
is convex and lower semicontinuous, it is also weakly lower semicontinuous. Letting
in the last inequality, from (3.66) and (A4) we have
, for all
. For
, with
, and
, let
. Since
and
, then
and hence
. So, from (A1) we have
Dividing by
, we have
, for all
. Letting
, from (A3) it follows that
, for all
. So,
. Therefore, we obtain that
by the arbitrariness of
.
Next, let us show that
and
.
Indeed, put
. From
and
, we have
. Now from weakly lower semicontinuity of the norm, we derive for each 
It follows from the definition of
that
and hence
. So we have
. Utilizing the Kadec-Klee property of
, we know that
. Since
is an arbitrary weakly convergent subsequence of
, we know that
. This completes the proof.
Theorem 3.6 covers [25, Theorem 3.2] by taking
and
. Also Theorem 3.6 covers [24, Theorem 2.2] by taking
and
.