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:

C=\{x\in H:c(x)\le 0\}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}Q=\{y\in K:q(y)\le 0\},

(3.1)

where c:H\to \mathbb{R} and q:K\to \mathbb{R} 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

{C}_{n}=\{x\in H:c({x}_{n})\le \u3008{\xi}_{n},{x}_{n}-x\u3009\},

(3.2)

where {\xi}_{n}\in \partial c({x}_{n}), and

{Q}_{n}=\{y\in K:q(A{x}_{n})\le \u3008{\zeta}_{n},A{x}_{n}-y\u3009\},

(3.3)

where {\zeta}_{n}\in \partial q(A{x}_{n}).

Obviously, {C}_{n} and {Q}_{n} are half-spaces and it is easy to verify that {C}_{n}\supset C and {Q}_{n}\supset Q hold for every n\ge 0 from the subdifferentiable inequality. In what follows, we define

{f}_{n}(x)=\frac{1}{2}{\parallel (I-{P}_{{Q}_{n}})Ax\parallel}^{2},\phantom{\rule{1em}{0ex}}n\ge 0,

where {Q}_{n} is given as in (3.3). We then have

\mathrm{\nabla}{f}_{n}(x)={A}^{\ast}(I-{P}_{{Q}_{n}})Ax.

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 {x}_{0}. Assume {x}_{n} has been constructed. If \mathrm{\nabla}{f}_{n}({x}_{n})=0, then stop; otherwise, continue and construct {x}_{n+1} *via* the formula

{x}_{n+1}={P}_{{C}_{n}}({x}_{n}-{\tau}_{n}\mathrm{\nabla}{f}_{n}({x}_{n})),

where {C}_{n} is given as (3.2), and

{\tau}_{n}=\frac{{\rho}_{n}{f}_{n}({x}_{n})}{{\parallel \mathrm{\nabla}{f}_{n}({x}_{n})\parallel}^{2}},\phantom{\rule{1em}{0ex}}0<{\rho}_{n}<4.

(3.4)

López *et al.* proved that under some certain conditions the sequence ({x}_{n}) generated by Algorithm 3.1 converges weakly to a solution of the SFP (1.1). Since the projections onto half-spaces {C}_{n} and {Q}_{n} have closed forms and {\tau}_{n} 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 u\in H, and start an initial guess {x}_{0}\in H arbitrarily. Assume that the *n* th iterate {x}_{n} has been constructed. If \mathrm{\nabla}{f}_{n}({x}_{n})=0, then stop ({x}_{n} is a approximate solution of SFP (1.1)). Otherwise, continue and calculate the (n+1)th iterate {x}_{n+1} *via* the formula:

{x}_{n+1}={P}_{{C}_{n}}[{\alpha}_{n}u+(1-{\alpha}_{n})({x}_{n}-{\tau}_{n}\mathrm{\nabla}{f}_{n}({x}_{n}))],

(3.5)

where the sequence ({\alpha}_{n})\subset (0,1) and ({\tau}_{n}) and ({\rho}_{n}) are given as in (3.4).

The convergence result of Algorithm 3.2 is stated in the next theorem.

**Theorem 3.3** *Assume that* ({\alpha}_{n}) *and* ({\rho}_{n}) *satisfy the assumptions*:

(a1) {lim}_{n\to \mathrm{\infty}}{\alpha}_{n}=0 *and* {\sum}_{n=0}^{\mathrm{\infty}}{\alpha}_{n}=\mathrm{\infty}.

(a2) {inf}_{n}{\rho}_{n}(4-{\rho}_{n})>0.

*Then the sequence* ({x}_{n}) *generated by Algorithm * 3.2 *converges in norm to* {P}_{S}u.

*Proof* We may assume that the sequence ({x}_{n}) is infinite, that is, Algorithm 3.2 does not terminate in a finite number of iterations. Thus \mathrm{\nabla}{f}_{n}({x}_{n})\ne 0 for all n\ge 0. 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 {P}_{S} is well-defined. We set z={P}_{S}u. Since z\in S\subset {C}_{n} and the projection operator {P}_{{C}_{n}} is nonexpansive, we obtain

\begin{array}{rl}{\parallel {x}_{n+1}-z\parallel}^{2}& ={\parallel {P}_{{C}_{n}}[{\alpha}_{n}u+(1-{\alpha}_{n})({x}_{n}-{\tau}_{n}\mathrm{\nabla}{f}_{n}({x}_{n}))]-z\parallel}^{2}\\ \le {\parallel {\alpha}_{n}(u-z)+(1-{\alpha}_{n})({x}_{n}-{\tau}_{n}\mathrm{\nabla}{f}_{n}({x}_{n})-z)\parallel}^{2}\\ \le (1-{\alpha}_{n}){\parallel {x}_{n}-{\tau}_{n}\mathrm{\nabla}{f}_{n}({x}_{n})-z\parallel}^{2}+2{\alpha}_{n}\u3008u-z,{x}_{n+1}-z\u3009.\end{array}

Note that I-{P}_{{Q}_{n}} is firmly nonexpansive and \mathrm{\nabla}{f}_{n}(z)=0, it is deduced from Lemma 2.2 that

\begin{array}{rl}\u3008\mathrm{\nabla}{f}_{n}({x}_{n}),{x}_{n}-z\u3009& =\u3008(I-{P}_{{Q}_{n}})A{x}_{n},A{x}_{n}-Az\u3009\\ \ge {\parallel (I-{P}_{{Q}_{n}})A{x}_{n}\parallel}^{2}\\ =2{f}_{n}({x}_{n}),\end{array}

which implies that

\begin{array}{rl}{\parallel {x}_{n}-{\tau}_{n}\mathrm{\nabla}{f}_{n}({x}_{n})-z\parallel}^{2}& ={\parallel {x}_{n}-z\parallel}^{2}+{\parallel {\tau}_{n}\mathrm{\nabla}{f}_{n}({x}_{n})\parallel}^{2}-2{\tau}_{n}\u3008\mathrm{\nabla}{f}_{n}({x}_{n}),{x}_{n}-z\u3009\\ \le {\parallel {x}_{n}-z\parallel}^{2}+{\tau}_{n}^{2}{\parallel \mathrm{\nabla}{f}_{n}({x}_{n})\parallel}^{2}-4{\tau}_{n}{f}_{n}({x}_{n})\\ \le {\parallel {x}_{n}-z\parallel}^{2}-{\rho}_{n}(4-{\rho}_{n})\frac{{f}_{n}^{2}({x}_{n})}{{\parallel \mathrm{\nabla}{f}_{n}({x}_{n})\parallel}^{2}}.\end{array}

Thus, we have

\begin{array}{rl}{\parallel {x}_{n+1}-z\parallel}^{2}\le & (1-{\alpha}_{n}){\parallel {x}_{n}-{\tau}_{n}\mathrm{\nabla}{f}_{n}({x}_{n})-z\parallel}^{2}+2{\alpha}_{n}\u3008u-z,{x}_{n+1}-z\u3009\\ \le & (1-{\alpha}_{n}){\parallel {x}_{n}-z\parallel}^{2}+2{\alpha}_{n}\u3008u-z,{x}_{n+1}-z\u3009\\ -(1-{\alpha}_{n}){\rho}_{n}(4-{\rho}_{n})\frac{{f}_{n}^{2}({x}_{n})}{{\parallel \mathrm{\nabla}{f}_{n}({x}_{n})\parallel}^{2}}.\end{array}

(3.6)

Now we prove ({x}_{n}) is bounded. Indeed, we have from (3.6) that

\begin{array}{rl}{\parallel {x}_{n+1}-z\parallel}^{2}& \le (1-{\alpha}_{n}){\parallel {x}_{n}-z\parallel}^{2}+2{\alpha}_{n}\u3008u-z,{x}_{n+1}-z\u3009\\ \le (1-{\alpha}_{n}){\parallel {x}_{n}-z\parallel}^{2}+\frac{1}{4}{\alpha}_{n}{\parallel {x}_{n+1}-z\parallel}^{2}+4{\alpha}_{n}{\parallel u-z\parallel}^{2},\end{array}

and consequently

{\parallel {x}_{n+1}-z\parallel}^{2}\le \frac{1-{\alpha}_{n}}{1-\frac{1}{4}{\alpha}_{n}}{\parallel {x}_{n}-z\parallel}^{2}+\frac{\frac{3}{4}{\alpha}_{n}}{1-\frac{1}{4}{\alpha}_{n}}\frac{16}{3}{\parallel u-z\parallel}^{2}.

It turns out that

\parallel {x}_{n+1}-z\parallel \le max\{\parallel {x}_{n}-z\parallel ,\frac{16}{3}\parallel u-z\parallel \},

and inductively

\parallel {x}_{n}-z\parallel \le max\{\parallel {x}_{0}-z\parallel ,\frac{16}{3}\parallel u-z\parallel \},

and this means that ({x}_{n}) is bounded. Since {\alpha}_{n}\to 0, with no loss of generality, we may assume that there is \sigma >0 so that {\rho}_{n}(4-{\rho}_{n})(1-{\alpha}_{n})\ge \sigma for all *n*. Setting {s}_{n}={\parallel {x}_{n}-z\parallel}^{2}, from the last inequality of (3.6), we get the following inequality:

{s}_{n+1}-{s}_{n}+{\alpha}_{n}{s}_{n}+\frac{\sigma {f}_{n}^{2}({x}_{n})}{{\parallel \mathrm{\nabla}{f}_{n}({x}_{n})\parallel}^{2}}\le 2{\alpha}_{n}\u3008u-z,{x}_{n+1}-z\u3009.

(3.7)

Now, following an idea in [28], we prove {s}_{n}\to 0 by distinguishing two cases.

Case 1: ({s}_{n}) is eventually decreasing (*i.e.* there exists k\ge 0 such that {s}_{n}>{s}_{n+1} holds for all n\ge k). In this case, ({s}_{n}) must be convergent, and from (3.7) it follows that

\frac{\sigma {f}_{n}^{2}({x}_{n})}{{\parallel \mathrm{\nabla}{f}_{n}({x}_{n})\parallel}^{2}}\le M{\alpha}_{n}+({s}_{n}-{s}_{n+1}),

(3.8)

where M>0 is a constant such that 2\parallel {x}_{n+1}-z\parallel \parallel u-z\parallel \le M for all n\in \mathbb{N}. Using the condition (a1), we have from (3.8) that \frac{{f}_{n}^{2}({x}_{n})}{{\parallel \mathrm{\nabla}{f}_{n}({x}_{n})\parallel}^{2}}\to 0. Thus, to verify that {f}_{n}({x}_{n})\to 0, it suffices to show that (\mathrm{\nabla}f({x}_{n})) is bounded. In fact, it follows from Lemma 2.1 that (noting that \mathrm{\nabla}{f}_{n}(z)=0 due to z\in S)

\parallel \mathrm{\nabla}{f}_{n}({x}_{n})\parallel =\parallel \mathrm{\nabla}{f}_{n}({x}_{n})-\mathrm{\nabla}{f}_{n}(z)\parallel \le {\parallel A\parallel}^{2}\parallel {x}_{n}-z\parallel .

This implies that \parallel \mathrm{\nabla}{f}_{n}({x}_{n})\parallel is bounded and it yields {f}_{n}({x}_{n})\to 0, namely \parallel (I-{P}_{{Q}_{n}})A{x}_{n}\parallel \to 0.

Since *∂q* is bounded on bounded sets, there exists a constant \eta >0 such that \parallel {\zeta}_{n}\parallel \le \eta for all n\ge 0. From (3.3) and the trivial fact that {P}_{{Q}_{n}}(A{x}_{n})\in {Q}_{n}, it follows that

q(A{x}_{n})\le \u3008{\zeta}_{n},A{x}_{n}-{P}_{{Q}_{n}}(A{x}_{n})\u3009\le \eta \parallel (I-{P}_{{Q}_{n}})A{x}_{n}\parallel \to 0.

(3.9)

If {x}^{\ast}\in {\omega}_{w}({x}_{n}), and ({x}_{{n}_{k}}) is a subsequence of ({x}_{n}) such that {x}_{{n}_{k}}\rightharpoonup {x}^{\ast}, then the *w-lsc* of *q* and (3.9) imply that

q\left(A{x}^{\ast}\right)\le \underset{k\to \mathrm{\infty}}{lim\hspace{0.17em}inf}q(A{x}_{{n}_{k}})\le 0.

It turns out that A{x}^{\ast}\in Q. Next, we turn to prove {x}^{\ast}\in C. For convenience, we set {y}_{n}:={\alpha}_{n}u+(1-{\alpha}_{n})({x}_{n}-{\tau}_{n}\mathrm{\nabla}{f}_{n}({x}_{n})). In fact, since the {P}_{{C}_{n}} is firmly nonexpansive, it concludes that

{\parallel {P}_{{C}_{n}}{x}_{n}-{P}_{{C}_{n}}z\parallel}^{2}\le {\parallel {x}_{n}-z\parallel}^{2}-{\parallel (I-{P}_{{C}_{n}}){x}_{n}\parallel}^{2}.

(3.10)

On the other hand, we have

\begin{array}{rl}{\parallel {x}_{n+1}-z\parallel}^{2}& ={\parallel {P}_{{C}_{n}}{y}_{n}-{P}_{{C}_{n}}z\parallel}^{2}\\ ={\parallel {P}_{{C}_{n}}{y}_{n}-{P}_{{C}_{n}}{x}_{n}+{P}_{{C}_{n}}{x}_{n}-{P}_{{C}_{n}}z\parallel}^{2}\\ \le {\parallel {P}_{{C}_{n}}{x}_{n}-{P}_{{C}_{n}}z\parallel}^{2}+2\u3008{P}_{{C}_{n}}{y}_{n}-{P}_{{C}_{n}}{x}_{n},{x}_{n+1}-z\u3009,\end{array}

(3.11)

and

\begin{array}{rl}\parallel {P}_{{C}_{n}}{y}_{n}-{P}_{{C}_{n}}{x}_{n}\parallel & \le \parallel {y}_{n}-{x}_{n}\parallel \\ =\parallel {\alpha}_{n}u+(1-{\alpha}_{n})({x}_{n}-{\tau}_{n}\mathrm{\nabla}{f}_{n}({x}_{n}))-{x}_{n}\parallel \\ \le {\alpha}_{n}\parallel u-{x}_{n}\parallel +{\tau}_{n}\parallel \mathrm{\nabla}{f}_{n}({x}_{n})\parallel .\end{array}

(3.12)

Noting that ({x}_{n}) is bounded, we have from (3.10)-(3.12) that

\begin{array}{rcl}{\parallel {x}_{n+1}-z\parallel}^{2}& \le & {\parallel {x}_{n}-z\parallel}^{2}-{\parallel (I-{P}_{{C}_{n}}){x}_{n}\parallel}^{2}+2({\alpha}_{n}\parallel u-{x}_{n}\parallel +{\tau}_{n}\parallel \mathrm{\nabla}{f}_{n}({x}_{n})\parallel )\parallel {x}_{n+1}-z\parallel \\ \le & {\parallel {x}_{n}-z\parallel}^{2}-{\parallel (I-{P}_{{C}_{n}}){x}_{n}\parallel}^{2}+({\alpha}_{n}+\frac{{f}_{n}({x}_{n})}{\parallel \mathrm{\nabla}{f}_{n}({x}_{n})\parallel})M,\end{array}

(3.13)

where *M* is some positive constant. Clearly, from (3.13), it turns out that

{\parallel (I-{P}_{{C}_{n}}){x}_{n}\parallel}^{2}\le {s}_{n}-{s}_{n+1}+({\alpha}_{n}+\frac{{f}_{n}({x}_{n})}{\parallel \mathrm{\nabla}{f}_{n}({x}_{n})\parallel})M.

(3.14)

Thus, we assert that \parallel (I-{P}_{{C}_{n}}){x}_{n}\parallel \to 0 due to the fact that {s}_{n}-{s}_{n+1}+({\alpha}_{n}+\frac{{f}_{n}({x}_{n})}{\parallel \mathrm{\nabla}{f}_{n}({x}_{n})\parallel})M\to 0. Moreover, by the definition of {C}_{n}, we obtain

c({x}_{n})\le \u3008{\xi}_{n},{x}_{n}-{P}_{{C}_{n}}({x}_{n})\u3009\le \delta \parallel {x}_{n}-{P}_{{C}_{n}}{x}_{n}\parallel \to 0\phantom{\rule{1em}{0ex}}(n\to \mathrm{\infty}),

where *δ* is a constant such that \parallel {\xi}_{n}\parallel \le \delta for all n\ge 0. The w-lsc of *c* then implies that

c\left({x}^{\ast}\right)\le \underset{k\to \mathrm{\infty}}{lim\hspace{0.17em}inf}c({x}_{{n}_{k}})=0.

Consequently, {x}^{\ast}\in C, and hence {\omega}_{w}({x}_{n})\subset S. Furthermore, due to (2.2), we get

\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\u3008u-z,{x}_{n}-z\u3009=\underset{w\in {\omega}_{w}({x}_{n})}{max}\u3008u-{P}_{S}u,w-{P}_{S}u\u3009\le 0.

(3.15)

Taking into account of (3.7), we have

{s}_{n+1}\le (1-{\alpha}_{n}){s}_{n}+2{\alpha}_{n}\u3008u-z,{x}_{n+1}-z\u3009.

(3.16)

Applying Lemma 2.3 to (3.16), we obtain {s}_{n}\to 0.

Case 2: ({s}_{n}) is not eventually decreasing, that is, we can find an integer {n}_{0} such that {s}_{{n}_{0}}\le {s}_{{n}_{0}+1}. Now we define

{V}_{n}:=\{{n}_{0}\le k\le n:{s}_{k}\le {s}_{k+1}\},\phantom{\rule{1em}{0ex}}n>{n}_{0}.

It is easy to see that {V}_{n} is nonempty and satisfies {V}_{n}\subseteq {V}_{n+1}. Let

\psi (n):=max{V}_{n},\phantom{\rule{1em}{0ex}}n>{n}_{0}.

It is clear that \psi (n)\to \mathrm{\infty} as n\to \mathrm{\infty} (otherwise, ({s}_{n}) is eventually decreasing). It is also clear that {s}_{\psi (n)}\le {s}_{\psi (n)+1} for all n>{n}_{0}. Moreover,

{s}_{n}\le {s}_{\psi (n)+1},\phantom{\rule{1em}{0ex}}n>{n}_{0}.

(3.17)

In fact, if \psi (n)=n, then the inequity (3.17) is trivial; if \psi (n)<n, from the definition of \psi (n), there exists some i\in \mathbb{N} such that \psi (n)+i=n, we deduce that

{s}_{\psi (n)+1}>{s}_{\psi (n)+2}>\cdots >{s}_{\psi (n)+i}={s}_{n},

and the inequity (3.17) holds again. Since {s}_{\psi (n)}\le {s}_{\psi (n)+1} for all n>{n}_{0}, it follows from (3.8) that

\frac{\sigma {f}_{\psi (n)}^{2}({x}_{\psi (n)})}{{\parallel \mathrm{\nabla}{f}_{\psi (n)}({x}_{\psi (n)})\parallel}^{2}}\le M{\alpha}_{\psi (n)}\to 0,

so that {f}_{\psi (n)}({x}_{\psi (n)})\to 0 as n\to \mathrm{\infty} (noting that \parallel \mathrm{\nabla}{f}_{\psi (n)}({x}_{\psi (n)})\parallel is bounded). By the same argument to the proof in case 1, we have {\omega}_{w}({x}_{\psi (n)})\subset S. On the other hand, noting {s}_{\psi (n)}\le {s}_{\psi (n)+1} again, we have from (3.12) and (3.14) that

\begin{array}{rl}\parallel {x}_{\psi (n)}-{x}_{\psi (n)+1}\parallel \le & \parallel {x}_{\psi (n)}-{P}_{{C}_{\psi (n)}}{x}_{\psi (n)}\parallel +\parallel {P}_{{C}_{\psi (n)}}{x}_{\psi (n)}-{P}_{{C}_{\psi (n)}}{y}_{\psi (n)}\parallel \\ \le & \sqrt{{\alpha}_{\psi (n)}+\frac{{f}_{\psi (n)}({x}_{\psi (n)})}{\parallel \mathrm{\nabla}{f}_{\psi (n)}({x}_{\psi (n)})\parallel}}\\ \times (\sqrt{{\alpha}_{\psi (n)}+\frac{{f}_{\psi (n)}({x}_{\psi (n)})}{\parallel \mathrm{\nabla}{f}_{\psi (n)}({x}_{\psi (n)})\parallel}}+1)M,\end{array}

where *M* is a positive constant. Letting n\to \mathrm{\infty} yields that

\parallel {x}_{\psi (n)}-{x}_{\psi (n)+1}\parallel \to 0,

(3.18)

from which one can deduce that

\begin{array}{rl}\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\u3008u-z,{x}_{\psi (n)+1}-z\u3009& =\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\u3008u-z,{x}_{\psi (n)}-z\u3009\\ =\underset{w\in {\omega}_{w}({x}_{\psi}(n))}{max}\u3008u-{P}_{S}u,w-{P}_{S}u\u3009\le 0.\end{array}

(3.19)

Since {s}_{\psi (n)}\le {s}_{\psi (n)+1}, it follows from (3.7) that

{s}_{\psi}(n)\le 2\u3008u-z,{x}_{\psi (n)+1}-z\u3009,\phantom{\rule{1em}{0ex}}n>{n}_{0}.

(3.20)

Combining (3.19) and (3.20) yields

\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}sup}{s}_{\psi (n)}\le 0,

(3.21)

and hence {s}_{\psi (n)}\to 0, which together with (3.18) implies that

\begin{array}{rl}\sqrt{{s}_{\psi (n)+1}}& \le \parallel ({x}_{\psi (n)}-z)+({x}_{\psi (n)+1}-{x}_{\psi (n)})\parallel \\ \le \sqrt{{s}_{\psi (n)}}+\parallel {x}_{\psi (n)+1}-{x}_{\psi (n)}\parallel \to 0,\end{array}

(3.22)

which, together with (3.17), in turn implies that {s}_{n}\to 0, that is, {x}_{n}\to z. □

**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 u=0 in Algorithm 3.2 whether 0\in C or 0\notin C.