In this section, we turn to consider a new relaxed CQ algorithm in the setting of infinite-dimensional Hilbert spaces for solving SFP (1.1) where the closed convex subsets C and Q are of the particular structure, i.e. level sets of convex functions given as follows:
(3.1)
where and are convex functions. We assume that both c and q are subdifferentiable on H and K, respectively, and that ∂c and ∂q are bounded operators (i.e., bounded on bounded sets). By the way, we mention that every convex function defined on a finite-dimensional Hilbert space is subdifferentiable and its subdifferential operator is a bounded operator (see [24]).
Set
(3.2)
where , and
(3.3)
where .
Obviously, and are half-spaces and it is easy to verify that and hold for every from the subdifferentiable inequality. In what follows, we define
where is given as in (3.3). We then have
Firstly, we recall the relaxed CQ algorithm of López et al. [19] for solving the SFP (1.1) where C and Q are given in (3.1) as follows.
Algorithm 3.1 Choose an arbitrary initial guess . Assume has been constructed. If , then stop; otherwise, continue and construct via the formula
where is given as (3.2), and
(3.4)
López et al. proved that under some certain conditions the sequence generated by Algorithm 3.1 converges weakly to a solution of the SFP (1.1). Since the projections onto half-spaces and have closed forms and is obtained adaptively via the formula (3.4) (no need to know a priori the norm of operator A), the above relaxed CQ algorithm 3.1 is implementable. But the weak convergence is its a weakness. To overcome this weakness, inspired by Algorithm 3.1, we will introduce a new relaxed CQ algorithm for solving the SFP (1.1) where C and Q are given in (3.1) so that the strong convergence is guaranteed.
It is well known that Halpern’s algorithm has a strong convergence for finding a fixed point of a nonexpansive mapping [25, 26]. Then we are in a position to give our algorithm. The algorithm given below is referred to as a Halpern-type algorithm [27].
Algorithm 3.2 Let , and start an initial guess arbitrarily. Assume that the n th iterate has been constructed. If , then stop ( is a approximate solution of SFP (1.1)). Otherwise, continue and calculate the th iterate via the formula:
(3.5)
where the sequence and and are given as in (3.4).
The convergence result of Algorithm 3.2 is stated in the next theorem.
Theorem 3.3 Assume that and satisfy the assumptions:
(a1) and .
(a2) .
Then the sequence generated by Algorithm 3.2 converges in norm to .
Proof We may assume that the sequence is infinite, that is, Algorithm 3.2 does not terminate in a finite number of iterations. Thus for all . Recall that S is the solution set of the SFP (1.1),
In the consistent case of the SFP (1.1), S is nonempty, closed and convex. Thus, the metric projection is well-defined. We set . Since and the projection operator is nonexpansive, we obtain
Note that is firmly nonexpansive and , it is deduced from Lemma 2.2 that
which implies that
Thus, we have
(3.6)
Now we prove is bounded. Indeed, we have from (3.6) that
and consequently
It turns out that
and inductively
and this means that is bounded. Since , with no loss of generality, we may assume that there is so that for all n. Setting , from the last inequality of (3.6), we get the following inequality:
(3.7)
Now, following an idea in [28], we prove by distinguishing two cases.
Case 1: is eventually decreasing (i.e. there exists such that holds for all ). In this case, must be convergent, and from (3.7) it follows that
(3.8)
where is a constant such that for all . Using the condition (a1), we have from (3.8) that . Thus, to verify that , it suffices to show that is bounded. In fact, it follows from Lemma 2.1 that (noting that due to )
This implies that is bounded and it yields , namely .
Since ∂q is bounded on bounded sets, there exists a constant such that for all . From (3.3) and the trivial fact that , it follows that
(3.9)
If , and is a subsequence of such that , then the w-lsc of q and (3.9) imply that
It turns out that . Next, we turn to prove . For convenience, we set . In fact, since the is firmly nonexpansive, it concludes that
(3.10)
On the other hand, we have
(3.11)
and
(3.12)
Noting that is bounded, we have from (3.10)-(3.12) that
(3.13)
where M is some positive constant. Clearly, from (3.13), it turns out that
(3.14)
Thus, we assert that due to the fact that . Moreover, by the definition of , we obtain
where δ is a constant such that for all . The w-lsc of c then implies that
Consequently, , and hence . Furthermore, due to (2.2), we get
(3.15)
Taking into account of (3.7), we have
(3.16)
Applying Lemma 2.3 to (3.16), we obtain .
Case 2: is not eventually decreasing, that is, we can find an integer such that . Now we define
It is easy to see that is nonempty and satisfies . Let
It is clear that as (otherwise, is eventually decreasing). It is also clear that for all . Moreover,
(3.17)
In fact, if , then the inequity (3.17) is trivial; if , from the definition of , there exists some such that , we deduce that
and the inequity (3.17) holds again. Since for all , it follows from (3.8) that
so that as (noting that is bounded). By the same argument to the proof in case 1, we have . On the other hand, noting again, we have from (3.12) and (3.14) that
where M is a positive constant. Letting yields that
(3.18)
from which one can deduce that
(3.19)
Since , it follows from (3.7) that
(3.20)
Combining (3.19) and (3.20) yields
(3.21)
and hence , which together with (3.18) implies that
(3.22)
which, together with (3.17), in turn implies that , that is, . □
Remark 3.4 Since u can be chosen in H arbitrarily, one can compute the minimum-norm solution of SFP (1.1) where C and Q are given in (3.1) by taking in Algorithm 3.2 whether or .