Now we are in a position to prove the main theorems of this paper.
Theorem 3.1.
Let
be a uniformly convex and uniformly smooth Banach space, let
be a nonempty closed convex subset of
, let
be a continuous mapping and, let
be a relatively nonexpansive mapping such that
. Assume that
are sequences in
and
is a sequence in
such that
and
. Define a sequence
in
by the following algorithm:
where
is assumed to exist for each
,
If
is uniformly continuous and
, then
converges strongly to
, which is a solution of the
(1.1).
Proof.
First of all, let us show that
and
are closed and convex for each
. Indeed, from the definition of
and
, it is obvious that
is closed and
is closed and convex for each
. We claim that
is convex. For any
and any
, put
. It is sufficient to show that
. Note that the inequality
is equivalent to the one
Observe that there hold the following:
and
. Thus, we have
This implies that
. So,
is convex. Next let us show that
for all
. Indeed, we have for all 
So
for all
. Next let us show that
We prove this by induction. For
, we have
. Assume that
. Since
is the projection of
onto
, by Lemma 2.3, we have
As
by the induction assumption, the last inequality holds, in particular, for all
. This together with the definition of
implies that
. Hence (3.7) holds for all
. This implies that
is well defined.
On the other hand, it follows from the definition of
that
. Since
, we have
Thus
is nondecreasing. Also from
and Lemma 2.4, it follows that
for each
for each
. Consequently,
is bounded. Moreover, according to the inequality
we conclude that
is bounded and so is
. Indeed, since
is relatively nonexpansive, we derive for each 
and hence
is bounded. Again from
we know that
is also bounded.
On account of the boundedness and nondecreasing property of
we deduce that
exists. From Lemma 2.4, we derive
for all
. This implies that
. So it follows from Lemma 2.2 that
. Since
, from the definition of
, we also have
Observe that
On the other hand, since from (3.1) we have for each 
utilizing Lemma 2.3 we obtain
. Thus, in terms of Lemmas 2.4 and 2.6 we conclude that
Since
and
, we obtain
. Thus by Lemma 2.2 we have
. From
it follows that
is bounded. At the same time, observe that
and hence
From
and the boundedness of
and
, we derive
. Note that
is uniformly continuous. Hence
by virtue of
. Since
it is known that
is bounded. Consequently, from (3.15),
and
it follows that
Further, it follows from (3.14),
and
that
Utilizing Lemma 2.2, we obtain
Since
is uniformly norm-to-norm continuous on bounded subsets of
, we have
Furthermore, we have
It follows from
and
that
Noticing that
we have
From (3.24) 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 (3.26), (3.30) and
that
.
Finally, let us show that
converges strongly to
, which is a solution of the
(1.1). Indeed, assume that
is a subsequence of
such that
. Then
. Next let us show that
and convergence is strong. Put
. From
and
, we have
. Now from weakly lower semicontinuity of the norm, we derive
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 arbitrarily weakly convergent subsequence of
, we know that
converges strongly to
. Now observe that from (3.1) we have for each 
Since
is uniformly norm-to-norm continuous on bounded subsets of
, from
we infer that
. Noticing that
and
is a continuous mapping, we obtain that
and
. Therefore, from
it follows that
that is,
Letting
we conclude from (3.34) that
and hence
This shows that
is a solution of the
(1.1). This completes the proof.
Corollary 3.2 ([11, Theorem?2.1]).
Let
be a uniformly convex and uniformly smooth Banach space, let
be a nonempty closed convex subset of
, and let
be a relatively nonexpansive mapping such that
. Assume that
and
are sequences in
such that
and
. Define a sequence
in
by the following algorithm:
where
is the single-valued duality mapping on
. If
is uniformly continuous, then
converges strongly to
.
Proof.
In Theorem 3.1, we know from (3.1) and Lemma 2.3 that
is equivalent to
. Now, put
for all
. Then we have
for all
. Thus algorithm (3.1) reduces to algorithm (3.39). By Theorem 3.1 we obtain the desired result.
Theorem 3.3.
Let
be a uniformly convex and uniformly smooth Banach space, let
be a nonempty closed convex subset of
, let
be a continuous mapping, and let
be a relatively nonexpansive mapping such that
. Assume that
satisfies
and
satisfies
. Define a sequence
in
by the following algorithm:
where
is assumed to exist for each
,
If
is uniformly continuous and
, then
converges strongly to
, which is a solution of the
(1.1).
Proof.
We only derive the difference. First, let us show that
is closed and convex for each
. From the definition of
, it is obvious that
is closed for each
. We prove that
is convex. Similarly to the proof of Theorem 3.1, since
is equivalent to
we know that
is convex. Next, let us show that
for each
. Indeed, we have for each 
So
for all
and
. Similarly to the proof of Theorem 3.1, we also obtain
for all
. Consequently,
for all
. Therefore, the sequence
generated by (3.42) is well defined. As in the proof of Theorem 3.1, we can obtain
. Since
, from the definition of
, we also have
As in the proof of Theorem 3.1, we can deduce from
and
that
and hence
by Lemma 2.2. Further, it follows from
and the boundedness of
and
that
Since
, from the definition of
, we also have
It follows from (3.47) and
that
Utilizing Lemma 2.2, we have
Since
is uniformly norm-to-norm continuous on bounded subsets of
we have
Note that
Therefore, from
we have
Since
is also uniformly norm-to-norm continuous on bounded subsets of
, we obtain
It follows that
Since
is uniformly continuous, it follows from (3.50) and (3.54) that
.
Finally, let us show that
converges strongly to
, which is a solution of the
(1.1). Indeed, assume that
is a subsequence of
such that
. Then
. Next let us show that
and convergence is strong. Put
. From
and
, we have
. Now from weakly lower semicontinuity of the norm, we derive
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 arbitrarily weakly convergent subsequence of
, we know that
converges strongly to
. Now observe that from (3.1), we have for each 
Since
is uniformly norm-to-norm continuous on bounded subsets of
, from
we infer that
. Noticing that
and
is a continuous mapping, we obtain that
and
. Observe that
It follows from
that
Letting
we conclude from (3.34) that
This shows that
is a solution of the
(1.1). This completes the proof.
Corollary 3.4 ([11, Theorem?2.2]).
Let
be a uniformly convex and uniformly smooth Banach space, let
be a nonempty closed convex subset of
, and let
be a relatively nonexpansive mapping. Assume that
is a sequence in
such that
. Define a sequence
in
by the following algorithm:
where
is the single-valued duality mapping on
. If
is nonempty, then
converges strongly to
.
Proof.
In Theorem 3.3, we know from (3.42) and Lemma 2.3 that
is equivalent to
. Now, put
for all
. Then we have
for all
. Thus algorithm (3.42) reduces to algorithm (3.61). Thus under the lack of the uniform continuity of
it follows from (3.55) that
. By the careful analysis of the proof of Theorem 3.3, we can obtain the desired result.