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]).