In this section, we establish convergence of the sequence
or its weighted average to an element of
. First we recall the following elementary lemma without proof.
Lemma 2.1.
Suppose that
is a nonnegative sequence and
is a positive sequence such that
. If
as
, then
as
.
We start with a weak ergodic theorem which extends a theorem of Lions [11] (see also [12] page 139 Theorem 3.1 as well as [10] Theorem 2.1).
Theorem 2.2.
Assume that
is a solution to (1.5) and
satisfies (1.6). If
and
, then
as
if and only if
is bounded.
Proof.
Suppose that
by (1.5); we get
This implies that
Then
is bounded and this proves necessity. Now, we prove sufficiency. By monotonicity of
, we have
for all
. Multiplying both sides of the above inequality by
and using (1.5), we deduce
Summing both sides of this inequality from
to
, we get
Divide both sides of the above inequality by
and suppose that
and
as
. By assumptions on
,
and Lemma 2.1, we have
This implies that
From (1.6), we get
By (1.6) and boundedness of
, we get
exists. If
, we obtain again
exists. Therefore,
, and hence
exists. This follows that
exists. It implies that
and hence
and
as
. Now we prove
. Suppose that
. By monotonicity of
and Assumption (1.6), we get
Letting
, we get:
. By maximality of
, we get
.
Remark 2.3.
Since range of
is
(the domain of
), as a trivial consequence of Theorem 2.2, we have that If
is bounded then
.
In the following, we prove a weak convergence theorem. Since the necessity is obvious, we omit the proof of necessity in the next theorems.
Theorem 2.4.
Let
be a solution to (1.5) and
. If
satisfies (1.6), then
as
if and only if
is bounded.
Proof.
Since assumption on
implies that
, from (1.5) and (2.7), we get
(The last inequality follows from Assumption (1.6)). Summing both sides of this inequality from
to
and letting
, since
satisfies (1.6), we have
By assumption on
, we have
as
. Assume
as
, by the monotonicity of
, we have
. Letting
, we get
. Similar to the proof of Theorem 2.2,
exists. This implies that
as
.
In two following, theorems we show strong convergence of
under suitable assumptions on operator
and the sequence
.
Theorem 2.5.
Assume that
is compact and
. If
satisfies (1.6), then
as
if and only if
is bounded.
Proof.
By (2.11) and assumption on
, we get
and
as
. Therefore, there exists a subsequence
of
such that
as
and
is bounded. The compacity of
implies that
has a strongly convergent subsequence (we denote again by
) to
. By the monotonicity of
, we have
. Letting
, we obtain
. Now, the proof of Theorem 2.2 shows that
exists. This implies that
as
.
Theorem 2.6.
Assume that
is strongly monotone operator and
. If
satisfies (1.6), then
as
if and only if
is bounded.
Proof.
By the proof of Theorem 2.2,
as
, and
exists. Since
is strongly monotone, we have
Multiplying both sides of (2.12) by
and summing from
to
, we have
(The last inequality follows from Assumption (1.6)). Letting
, we get:
So,
. This implies that
as
.
In the following theorem, we assume that
, where
is a proper, lower semicontinuous and convex function and
.
Theorem 2.7.
Let
, where
is a proper, lower semicontinuous, and convex function. Assume that
is nonempty (i.e.,
has at least one minimum point) and
. If
satisfies (1.6), then
as
.
Proof.
Since
is subdifferential of
and
, by Assumption (1.6), we have
Multiplying both sides of the above inequality by
and summing from
to
and letting
, we get
By assumption on
, we deduce
By convexity of
, we have
Therefore,
From (2.19), by Assumption (1.6), we get
Again by (2.19), we get
for all
. By (2.20) and (2.21), we have that
exists. From Assumptions (1.6), (2.17), and (2.21), we get
If
, then
. This implies that
. On the other hand, for each
by (1.5), we get (2.7). The proof of Theorem 2.2 implies that there exists
. Then the theorem is concluded by Opial's Lemma (see [13]).