Theorem 3.1.
Let
be a nonempty closed convex subset of a Hilbert space
and let
be a sequence of
-strict pseudocontractions mappings on
into itself with
. Assume that
Let
be a bifunction satisfying (A1), (A2), (A3), (A4), and
. Let
and
be sequence generated by
and
Assume that
for all
, where
is a small enough constant, and
is a sequence in
with
Let
for any bounded subset
of
and let
be a mapping of
into itself defined by
and suppose that
Then the sequences
and
converge weakly to an element of
.
Proof.
Pick
. Then from the definition of
in Lemma 2.4, we have
, and therefore
. It follows from (3.1) that
Since
for all
, we get
that is, the sequence
is decreasing. Hence
exists. In particular,
is bounded. Since
is firmly nonexpensive,
is also bounded. Also (3.2) implies that
Taking the limit as
yields that
Since
is bounded, it follows that
We apply Lemma 2.2 to get
Next, we claim that
. Indeed, let
be an arbitrary element of
. Then as above
and hence
Therefore, from (3.2), we have
and hence
So, from the existence of
, we have
Next, we claim that
. since
is bounded and
is reflexive,
is nonempty. Let
be an arbitrary element. Then a subsequence
of
converges weakly to
. Hence, from (3.11) we know that
As
, we obtain that
. Let us show
. Since
, we have
By (A2), we have
and hence
From (A4), we have
Then, for
and
, from (A1), and (A4), we also have
Taking
and using (A3), we get
and hence
. Since
is a strict pseudocontraction mapping, by Lemma 2.1(
) we know that the mapping
is demiclosed at zero. Note that
and
. Thus,
. Consequently, we deduce that
. Since
is an arbitrary element, we conclude that
.
To see that
and
are actually weakly convergent, we take 
Since
exist for every
, by (2.2), we have
Hence
and proof is completed.