Let be a monotone operator on a Hilbert space . Then is a single-valued nonexpansive mapping from to . When is a nonempty closed convex subset of such that for all (here is closure of ), then we have for and all , and hence the following iteration is well defined

Next we will show strong convergence of defined by (3.1) to find a zero of . For reaching this objective, we always assume in the sequel.

Theorem 3.1.

Let be a monotone operator on a Hilbert space with . Assume that is a nonempty closed convex subset of such that for all and for an anchor point and an initial value , is iteratively defined by (3.1). If and satisfy

(i)

(ii)

(iii)

then the sequence converges strongly to , where is the metric projection from onto .

Proof.

The proof consists of the following steps:

Step 1.

The sequence is bounded. Let , then and for some , we have

So, the sequences , , and are bounded.

Step 2.

for each . Since

we have

Step 3.

Indeed, we can take a subsequence of such that

We may assume that by the reflexivity of and the boundedness of . Then . In fact, since

then, for some constant , we have

Thus,

Take on two sides of the above equation by means of (3.4), we must have . So, . Hence, noting the projection inequality (2.3), we obtain

Step 4.

. Indeed,

Therefore,

where So, an application of Lemma 2.1 onto (3.11) yields the desired result.

Theorem 3.2.

Let be as Theorem 3.1, the condition (iii) is replaced by the following condition:

Then the sequence converges strongly to , where is the metric projection from onto .

Proof.

From the proof of Theorem 3.1, we can observe that Steps 1, 3 and 4 still hold. So we only need show to Step 2: for each .

We first estimate From the resolvent identity (2.2), we have

Therefore, for a constant with ,

It follows from Lemma 2.1 that

As then

Since , then there exists and a positive integer such that for all , . Thus for each , we also have

we have

Corollary 3.3.

Let be as Theorem 3.1 or 3.2. Suppose that is a maximal monotone operator on and for , is defined by (3.1). Then the sequence converges strongly to , where is the metric projection from onto .

Proof.

Since is a maximal monotone, then is monotone and satisfies the condition for all . Putting , the desired result is reached.

Corollary 3.4.

Let be as Theorem 3.1 or 3.2. Suppose that is a monotone operator on satisfying the condition for all and for , is defined by (3.1). If is convex, then the sequence converges strongly to , where is the metric projection from onto .

Proof.

Taking , following Theorem 3.1 or 3.2, we easily obtain the desired result.