On the role of the coefficients in the strong convergence of a general type Mann iterative scheme

Abstract

Let H be a Hilbert space. Let $(W_{n})_{n\in\mathbb{N}}$ be a suitable family of mappings. Let S be a nonexpansive mapping and D be a strongly monotone operator. We study the convergence of the general scheme $x_{n+1}=W_{n}(\alpha_{n}Sx_{n}+(1-\alpha_{n})(I-\mu_{n}D)x_{n})$ in dependence on the coefficients $(\alpha_{n})_{n\in\mathbb{N}}$, $(\mu_{n})_{n\in\mathbb{N}}$.

1 Introduction and motivations

The approximation of fixed points of nonlinear mappings is a wide and active research area and its applications occur more and more widely in the calculus of variations and optimization. The starting point of many papers is a modification of Mann’s iterative method [1],

$$x_{n+1}=\alpha_{n} x_{n}+(1-\alpha_{n})Tx_{n}, $$

in order to obtain strong convergence results.

Many of these modified Mann schemes yield approximation sequences by suitable convex combinations like

$$x_{n+1}=\alpha_{n}g(x_{n})+(1-\alpha_{n})Vy_{n}, $$

where g, V, and $(y_{n})_{n\in\mathbb{N}}$ are opportunely chosen (see, for instance, Halpern [2], Ishikawa [3], Moudafi [4], Nakajo and Takahashi [5]).

In this paper, we instead focus on the following iterative method:

$$x_{n+1}=W_{n} \bigl(\alpha_{n} Sx_{n}+(1- \alpha_{n}) (I-\mu_{n}D)x_{n} \bigr). $$

This method is very different from most of existing methods in literature and immediately we discuss on some motivations.

Let H be a Hilbert space and $f:H\to\mathbb{R}$ be a convex and lower semicontinuous. Our interest is focused on the minimization problem

$$ \min_{x\in C} f(x), $$

(1.1)

where C is a constraint closed and convex subset of H.

The following theorem is proved in [6].

Theorem 1.1

Let H be a Hilbert space and $f:H\to\mathbb{R}$ be a convex functional. Then

• (a) $f(x_{0})=\min_{H}f(x)$ if and only if $0\in\partial f(x_{0})$.

• (b) Let $C\subset H$. Then $f(x_{0})=\min_{C}f(x)$ if and only if $(-\partial f(x_{0})\cap\partial\delta_{C}(x_{0}))\neq\emptyset$, where $\delta_{C}$ is the indicator function of C.

Denote by Σ the set of solutions of (1.1). Let us start by the simple case in which $f:H\to\mathbb{R}$ is a convex and continuously Fréchet differentiable functional.

By the definition of an indicator function we recall that (see [6])

$$ \textstyle\partial\delta_{C}(x_{0})= \left \{ \begin{array}{@{}l@{\quad}l} \emptyset, & x_{0}\in H\setminus C,\\ 0, & x_{0}\in\mathring{C}, \\ \{x^{*}\in H:\sup_{C}\langle x^{*},x\rangle=\langle x^{*},x_{0}\rangle\}, & x_{0}\in C\setminus\mathring{C}. \end{array} \right . $$

(1.2)

$f(\cdot)$ being Fréchet differentiable, $\partial f(x_{0})$ is a singleton, $\nabla f(x_{0})$; hence Theorem 1.1(b) of [6] ensures that $x_{0}\in C$ is a solution of (1.1) if and only if $-\nabla f(x_{0})\in\partial\delta_{C}(x_{0})$, i.e.

$$\bigl\langle \nabla f(x_{0}),x_{0} \bigr\rangle \leq \bigl\langle \nabla f(x_{0}),x \bigr\rangle ,\quad \forall x\in C. $$

In other words $x_{0}\in C$ is a solution of (1.1) if and only if

$$ \bigl\langle \nabla f(x_{0}),x-x_{0} \bigr\rangle \geq0,\quad \forall x\in C. $$

(1.3)

From (1.3), for every $\gamma>0$, $x_{0}$ is a solution for (1.1) if and only if

$$ \bigl\langle x_{0}- \bigl(x_{0}-\gamma\nabla f(x_{0}) \bigr),x-x_{0} \bigr\rangle \geq0, \quad\forall x\in C, $$

(1.4)

and, in view of Browder’s characterization of the metric projections $P_{C}$, to solve (1.4) is equivalent to finding $x_{0}$ such that

$$x_{0}=P_{C} (I-\gamma\nabla f )x_{0}. $$

Therefore, to solve problem (1.1) (respectively to approximate solutions of (1.1)) is equivalent to solving (resp. to approximate the solutions of) a fixed point problem which involves the operator ∇f.

It is well known, by the convexity of the functional f, that the operator ∇f is a monotone operator; indeed since

$$\begin{aligned}[b] & f(x)\geq f(y)+ \bigl\langle \nabla f(y),y-x \bigr\rangle , \quad\forall x\in H, \\ & f(y)\geq f(x)+ \bigl\langle \nabla f(x),x-y \bigr\rangle , \quad\forall y\in H, \end{aligned} $$

it easily follows that

$$\bigl\langle \nabla f(y)-\nabla f(x),y-x \bigr\rangle \geq0,\quad \forall x,y\in H. $$

If we assume that ∇f is $L_{f}$-lipschitzian then, by Baillon-Haddad’s results [7], we have

$$\bigl\langle \nabla f(y)-\nabla f(x),y-x \bigr\rangle \geq\frac{1}{L_{f}} \bigl\| \nabla f(x)-\nabla f{y}\bigr\| ^{2}, \quad\forall x,y\in H, $$

i.e. ∇f is $\frac{1}{L_{f}}$-inverse strongly monotone.

Under such a hypothesis on ∇f, Takahashi and Toyoda in [8] proved that $P_{C} (I-\frac{1}{L_{f}}\nabla f)$ is a nonexpansive mapping, hence to solve (resp. to approximate a solution of (1.1)) is equivalent to finding (resp. to approximate) a fixed point of the nonexpansive mapping $P_{C} (I-\frac{1}{L_{f}}\nabla f )$. Xu in 2011 [9] showed that, even if $\Sigma\neq\emptyset$, it is not guaranteed that the natural iteration

$$ x_{n+1}=P_{C} \biggl(I-\frac{1}{L_{f}}\nabla f \biggr)x_{n}= \biggl(P_{C} \biggl(I-\frac{1}{L_{f}} \nabla f \biggr) \biggr)^{n}x_{0}, $$

(1.5)

strongly converges to a solution of Σ. An example is given in the following.

Example 1.2

[9]

Following Hundal [10], there exist in $H=l^{2}$ two closed and convex subset $C_{1}$ and $C_{2}$ such that: (i) $C_{1}\cap C_{2}\neq \emptyset$, and (ii) the sequence generated by $x_{0}\in C_{2}$ and the formula $x_{n}=(P_{C_{2}}P_{C_{1}})^{n}x_{0}$ weakly converges but it does not strongly converge.

Let $f(x)=\frac{1}{2}\|x-P_{C_{1}}x\|^{2}$. We deal with minimized $f(x)$ on $C_{2}$. It follows that $\nabla f(x)=(I-P_{C_{1}})x$. Since $P_{C_{1}}$ is firmly nonexpansive, i.e., 1-inverse strongly monotone, iteration (1.5) becomes

$$x_{n+1}=P_{C_{2}}(I-\nabla f)x_{n}=P_{C_{2}}P_{C_{1}}x_{n}, $$

that is, the sequence generated by (ii).

If we add to the lipschitzianity of ∇f also the (stronger) assumption that ∇f is a $\sigma_{f}$-strongly monotone operator, i.e.

$$\bigl\langle \nabla f(y)-\nabla f(x),y-x \bigr\rangle \geq\sigma_{f} \|x-y\|^{2},\quad \forall x,y\in H, $$

then the mapping $P_{C} (I-\frac{\sigma _{f}}{L_{f}^{2}}\nabla f )$ is a contraction; therefore the contraction principle ensures that problem (1.1) has a unique solution $x^{*}$ and the iterative sequence

$$ x_{n+1}=P_{C} \biggl(I-\frac{\sigma_{f}}{L_{f}^{2}}\nabla f \biggr)x_{n} $$

(1.6)

strongly converges to $x^{*}$.

Notice that, if $C=H$, $P_{C}=I$, then the iteration

$$x_{n+1}= \biggl(I-\frac{\sigma_{f}}{L_{f}^{2}}\nabla f \biggr)x_{n} $$

strongly converges to a zero of ∇f.

Hence a natural question is how to use the good properties of strongly monotone operators to find a solution of ( 1.1 ) if ∇f is only lipschitzian.

A well-known approach is to consider a regularized problem; an example is to appeal to Tikhonov’s regularized problem:

$$\min_{x\in C} \biggl[f(x)+\frac{\varepsilon}{2}\|x\|^{2} \biggr], $$

where $\varepsilon>0$ is given.

This approach arises by the following idea: if ∇f is only lipschitzian (for instance nonexpansive), we can perturb problem (1.1) by a convex and differentiable functional g such that ∇g is a $\sigma_{g}$-strongly monotone and $L_{g}$-lipschizian operator in such a way that

$$ \min_{x\in C}f(x)+\varepsilon g(x). $$

(1.7)

The operator $(\nabla f+\varepsilon\nabla g)$ is a lipschizian and a strongly monotone operator, the minumum problem (1.7) has a unique solution and, for a suitable $\lambda>0$,

$$x_{n+1}=P_{C} \bigl(I-\lambda(\nabla f+\varepsilon\nabla g) \bigr)x_{n} $$

strongly converges to this solution.

Let us observe that

$$\begin{aligned} x_{n+1} =&P_{C} \bigl(I-\lambda(\nabla f+\varepsilon\nabla g) \bigr)x_{n}=P_{C}(I-\lambda\nabla f-\lambda\varepsilon \nabla g)x_{n} \\ =&P_{C} \biggl(\lambda(I-\nabla f)+(1-\lambda) \biggl(I- \frac{\lambda \varepsilon}{(1-\lambda)} \nabla g \biggr) \biggr)x_{n} \\ =&P_{C} \bigl(\lambda(I-\nabla f)+(1-\lambda) (I-\gamma \varepsilon \nabla g ) \bigr)x_{n}, \end{aligned}$$

i.e. $(x_{n})_{n\in\mathbb{N}}$ is generated by the composition of the projection $P_{C}$ and the convex combination of two maps: the first is a nonexpansive mapping; the second is a strongly monotone operator. In fact for an opportune choice of λ (and $\gamma :=\frac{\lambda}{1-\lambda}$), we find the results that

• $(I-\nabla f)$ is a nonexpansive mapping;

• the mapping $(I-\gamma\varepsilon\nabla g )$ is a contraction.

For these reasons we are interested in the iteration

$$ x_{n+1}=W_{n} \bigl(\alpha_{n} Sx_{n}+(1-\alpha_{n}) (I-\mu_{n}D)x_{n} \bigr), $$

(1.8)

under the following hypotheses:

Hypotheses (ℋ)

• $(\alpha_{n})_{n\in\mathbb{N}}$ is a sequence in $[0,1)$.

• $S:H\to H$ is a nonexpansive mapping not necessarily with fixed points.

• $D:H \to H$ is a σ-strongly monotone operator and L-lipschitzian.

• $0<\mu_{n}\leq\mu$ with $\mu< \frac{2\sigma}{L^{2}}$, $\rho=\frac{2\sigma-\mu L^{2}}{2}$.

• $(W_{n})_{n\in\mathbb{N}}$ is a sequence of mappings defined on H such that $F:=\bigcap_{n\in\mathbb{N}} \operatorname{Fix}(W_{n})\neq \emptyset$ and

  • (h1) $W_{n}:H\to H$ are nonexpansive mappings, uniformly asymptotically regular on bounded subsets $B\subset H$, i.e.

    $$\lim_{n\to\infty}\sup_{x\in B}\|W_{n+1}x-W_{n}x \|=0, $$

  • (h2) it is possible to define a nonexpansive mapping $W:H\to H$, with $Wx:=\lim_{n\to\infty}W_{n}x$ such that $\operatorname{Fix}(W)=F$.

An interesting example of sequence $(W_{n})_{n\in\mathbb{N}}$ satisfying our hypotheses is the following.

Example 1.3

Let $f(x)$ be functional on H convex and lower semicontinuous. We recall that the proximal operator of f on H is defined as

$$\operatorname{prox}_{\lambda f}(x):=\mathop{\operatorname{argmin}}\limits_{v\in H} \biggl\{ f(x)+ \frac{1}{2\lambda}\|x-v\|^{2} \biggr\} , $$

where $\lambda>0$.

The proximal operator obeys:

• (1) it is a single-value firmly nonexpansive mapping (hence nonexpansive);

• (2) it coincides with $P_{C}$ if $f(x)=\delta_{C}(x)$;

• (3) $\operatorname{prox}_{\lambda f}=(I+\lambda\partial f)^{-1}$ i.e. it is the resolvent of the subdifferential of f;

• (4) $\operatorname{prox}_{\lambda f}x=\operatorname{prox}_{\nu f} (\frac{\nu}{\lambda}x+ (1-\frac{\nu}{\lambda} )\operatorname{prox}_{\lambda f}x )$;

• (5)
  $$x^{*}=\operatorname{prox}_{\lambda f} \bigl(x^{*} \bigr)\quad\Leftrightarrow\quad0\in \partial f \bigl(x^{*} \bigr). $$

If $(\lambda_{n})_{n\in\mathbb{N}}$ converges to $\lambda>0$ then $W_{n}:=\operatorname{prox}_{\lambda_{n} f}(x)$ satisfied (h1) and (h2) where $W:=\operatorname{prox}_{\lambda f}(x)$. In fact, the set of fixed point coincides by (3) and (5). Moreover, by (4),

$$\begin{aligned} \|W_{n+1}-W_{n}x\| =&\bigl\| \operatorname{prox}_{\lambda_{n+1} f}(x)- \operatorname{prox}_{\lambda_{n} f}(x)\bigr\| = \\ =&\biggl\| \operatorname{prox}_{\lambda_{n} f} \biggl(\frac{\lambda _{n}}{\lambda_{n+1}}x+ \biggl(1- \frac{\lambda_{n}}{\lambda_{n+1}} \biggr)\operatorname {prox}_{\lambda_{n+1} f}x \biggr)- \operatorname{prox}_{\lambda_{n} f}(x)\biggr\| \\ \leq&\biggl\| \biggl(\frac{\lambda_{n}}{\lambda_{n+1}}x+ \biggl(1-\frac {\lambda _{n}}{\lambda_{n+1}} \biggr) \operatorname{prox}_{\lambda_{n+1} f}x \biggr)-x\biggr\| \\ =& \biggl\vert 1-\frac{\lambda_{n}}{\lambda_{n+1}} \biggr\vert \| x-\operatorname{prox}_{\lambda _{n+1} f}x \|, \end{aligned}$$

so if x lies in a bounded subset, the uniform asymptotical regularity follows.

In any case we have the following.

Remark 1.4

If $C=\bigcap_{n\in\mathbb{N}}C_{n}$, where $C_{n}\subset H$ are closed and convex for all $n\in\mathbb{N}$, we can always suppose that $C=\bigcap _{n\in\mathbb{N}}\operatorname{Fix}(W_{n})$ where $(W_{n})_{n\in\mathbb {N}}$ is a sequence of nonexpansive mappings satisfying (h1) and (h2). Indeed starting by the sequence of nonexpansive mappings $T_{n}=P_{C_{n}}$ we can always construct a sequence $(W_{n})_{n\in\mathbb{N}}$ such that $C=\bigcap _{n\in \mathbb{N}}C_{n}=\bigcap_{n\in\mathbb{N}}\operatorname{Fix}(T_{n})=\bigcap _{n\in\mathbb {N}}\operatorname{Fix}(W_{n})$ and it satisfies (h1) and (h2) (see for details [11–14]).

Moreover, regarding the strongly monotone operator D we note that the sequence of operators $B_{n}x:=(I-\mu_{n}D)x$ is a sequence of contractions when the sequence $(\mu_{n})_{n\in\mathbb{N}}$ lies in an opportune interval. Such an interval can be detected by the following lemma, proved by Kim and Xu.

Lemma 1.5

[15]

Let $D:H\to H$ be σ-strongly monotone and L-lipschitzian. If $\mu<\frac{2\sigma}{L^{2}}$, $\rho=\frac {2\sigma-\mu L^{2}}{2}$, and $(\mu_{n})_{n\in\mathbb{N}}\subset(0,\mu ]$, then

$$\bigl\| (I-\mu_{n}D)x-(I-\mu_{n}D)y\bigr\| \leq(1-\mu_{n}\rho) \|x-y\|, $$

i.e. $(I-\mu_{n}D)$ is a $(1-\mu_{n}\rho)$-contraction.

In this paper we study some asymptotic behaviors of the sequence generated by iteration (1.8), supposing that there exists (finite or infinite)

$$\tau:=\lim_{n\to\infty}\frac{\alpha_{n}}{\mu_{n}}. $$

We will be able to show that (1.8) strongly converges to a solution of the variational inequality

$$\bigl\langle \tau(I-S)x+Dx,y-x \bigr\rangle \geq0,\quad \forall y\in F, $$

when $\tau\in[0,+\infty)$, and to a special solution of

$$\bigl\langle (I-S)x,y-x \bigr\rangle \geq0, \quad\forall y\in F, $$

if $\tau=+\infty$.

Our research is not far from the research area studied by Moudafi and Maingé and also known as the hierarchical fixed point approach (see [16–19]).

2 Some asymptotic behaviors of the iterative scheme

To study the asymptotic behavior of our method

$$ x_{n+1}=W_{n} \bigl(\alpha_{n} Sx_{n}+(1-\alpha_{n}) (I-\mu_{n}D)x_{n} \bigr) $$

(2.1)

we suppose that there exists

$$\tau:=\lim_{n\to\infty}\frac{\alpha_{n}}{\mu_{n}}. $$

The method can be equivalently written as

$$x_{n+1}=W_{n}y_{n}, $$

where $y_{n}:=\alpha_{n}Sx_{n}+(1-\alpha_{n})B_{n}x_{n}$ and $B_{n}=(I-\mu_{n}D)$. We will use the following convenient notations:

• We say that $\zeta_{n}=o(\eta_{n})$ if $\frac{\zeta_{n}}{\eta _{n}}\to0$ as $n\to\infty$.

• We say that $\zeta_{n}=O(\eta_{n})$ if there exist $K,N>0$ such that $N\leq \vert \frac{\zeta_{n}}{\eta_{n}}\vert \leq K$.

A central role in proving the convergence results is played by the boundedness of the sequence $(x_{n})_{n\in\mathbb{N}}$. We want to put its role in evidence. An expected case occurs when there are common fixed points between S and $(W_{n})_{n\in\mathbb{N}}$.

Proposition 2.1

Suppose that (2.1) satisfies Hypotheses (ℋ).

If $\operatorname{Fix}(S)\cap F\neq\emptyset$ then $(x_{n})_{n\in \mathbb{N}}$ is bounded.

Proof

If $z\in\operatorname{Fix}(S)\cap F$

$$\begin{aligned} \|x_{n+1}-z\| \leq& \bigl\| \alpha_{n}Sx_{n}+(1- \alpha_{n})B_{n}x_{n}-z\bigr\| \\ \leq& \alpha_{n}\|Sx_{n}-z\|+(1- \alpha_{n}) \|B_{n}x_{n}-B_{n}z\| +(1- \alpha_{n}) \|B_{n}z-z\| \\ \leq& \alpha_{n}\|x_{n}-z\|+(1-\alpha_{n}) (1- \mu_{n}\rho)\|x_{n}-z\| +(1-\alpha_{n})\mu_{n}\|Dz\| \\ \leq& \bigl(1-(1-\alpha_{n})\mu_{n}\rho \bigr) \|x_{n}-z\|+(1-\alpha _{n})\mu _{n}\rho \frac{\|Dz\|}{\rho}. \end{aligned}$$

(2.2)

Calling $\beta_{n}:=(1-\alpha_{n})\mu_{n}\rho$ we have

$$\begin{aligned} \|x_{n+1}-z\| \leq (1-\beta_{n})\|x_{n}-z\|+ \beta_{n}\frac{\|Dz\| }{\rho} \leq\max \biggl\{ \|x_{n}-z\|, \frac{\|Dz\|}{\rho} \biggr\} . \end{aligned}$$

Since, by an inductive process, one can see that

$$\|x_{n}-z\|\leq \max \biggl\{ \|x_{0}-z\|,\frac{\|Dz\|}{\rho} \biggr\} , $$

the claim follows. □

Notice that, in this case, boundedness does not depend by any hypotheses on $(\alpha_{n})_{n\in\mathbb{N}}$, $(\mu_{n})_{n\in\mathbb {N}}$, sequences in $[0,1]$.

On the contrary, in the following proposition the boundeness of the sequence is guaranteed by the assumption on the coefficients.

Proposition 2.2

Let us suppose that (2.1) satisfies Hypotheses (ℋ). Let $(\alpha_{n})_{n\in\mathbb{N}}$ be a sequence in $[0,1]$ and let $(\mu_{n})_{n\in\mathbb{N}}$ be a sequence in $(0,\mu)$. Assume that

• (B) either $\alpha_{n}=O (\mu_{n} )$ or $\alpha_{n}=o (\mu_{n} )$ (a sufficient condition is that there exists $\lim_{n\to\infty} \frac{\alpha _{n}}{\mu _{n}}=\tau\in[0,+ \infty)$).

Then $(x_{n})_{n\in\mathbb{N}}$ is bounded.

Proof

Let $z \in F$. Then for every $n\in\mathbb{N}$,

$$\begin{aligned} \|x_{n+1}-z\| \leq& \bigl\| \alpha_{n}Sx_{n}+(1- \alpha_{n})B_{n}x_{n}-z\bigr\| \\ \leq& \alpha_{n}\|Sx_{n}-Sz\|+\alpha_{n} \|Sz-z \|+(1-\alpha_{n})\| B_{n}x_{n}-B_{n}z \|+(1-\alpha_{n})\|B_{n}z-z\| \\ \leq& \alpha_{n}\|x_{n}-z\|+\alpha_{n}\|Sz-z \|+(1-\alpha_{n}) (1-\mu_{n}\rho )\| x_{n}-z\| +(1-\alpha_{n})\mu_{n}\|Dz\| \\ \leq& \bigl(1-(1-\alpha_{n})\mu_{n}\rho \bigr) \|x_{n}-z\|+\alpha_{n}\| Sz-z\| +(1-\alpha_{n}) \mu_{n}\rho\frac{\|Dz\|}{\rho}. \end{aligned}$$

(2.3)

Since (B) holds, there exist $\gamma>0$ and $N_{0}$ such that, for all $n>N_{0}$, $\alpha_{n}\leq\gamma(1-\alpha_{n})\mu_{n}$; hence

$$\begin{aligned} \|x_{n+1}-z\| \leq& \bigl(1-(1-\alpha_{n})\mu_{n} \rho \bigr)\|x_{n}-z\|+ \gamma (1-\alpha_{n})\mu_{n} \|Sz-z\|+(1-\alpha_{n})\mu_{n}\rho\frac{\|Dz\|}{\rho} \\ \leq& \bigl(1-(1-\alpha_{n})\mu_{n}\rho \bigr) \|x_{n}-z\|+ (1-\alpha_{n})\mu_{n}\rho \frac{\gamma\|Sz-z\|+\|Dz\|}{\rho}. \end{aligned}$$

Calling $\beta_{n}:=(1-\alpha_{n})\mu_{n}\rho$ we have

$$\begin{aligned} \|x_{n+1}-z\| \leq& (1-\beta_{n})\|x_{n}-z\|+ \beta_{n}\frac{\gamma\| Sz-z\| +\|Dz\|}{\rho} \\ \leq&\max \biggl\{ \|x_{n}-z\|, \frac{\gamma\|Sz-z\|+\|Dz\|}{\rho } \biggr\} . \end{aligned}$$

Since, by an inductive process, one can see that

$$\|x_{n}-z\|\leq \max \biggl\{ \|x_{i}-z\|, \frac{\|Dz\|+\gamma\|Sz-z\|}{\rho}: i=0,\ldots ,N_{0} \biggr\} , $$

the claim follows. □

It is remarkable that, by boundedness, we can deduce the asymptotical regularity of the iterative sequence, i.e. that

$$\|x_{n+1}-x_{n}\|\to0, \quad\mbox{as } n\to\infty, $$

which is often a key to prove convergent results when the mappings involved are continuous.

To prove it, we use the Xu lemma.

Lemma 2.3

[20]

Assume $(a_{n})_{n \in\mathbb{N}}$ is a sequence of nonnegative numbers such that

$$ a_{n+1}\leq(1-\gamma_{n})a_{n}+ \delta_{n},\quad n\geq0, $$

where $(\gamma_{n})_{n}$ is a sequence in $(0,1)$ and $(\delta_{n})_{n}$ is a sequence in ℝ such that:

• (1) $\sum_{n=1}^{\infty}\gamma_{n}=\infty$;

• (2) $\lim\sup_{n\to\infty}\delta_{n}/\gamma_{n}\leq0$ or $\sum_{n=1}^{\infty}|\delta_{n}|<\infty$.

Then $\lim_{n\to\infty}a_{n}=0$.

Proposition 2.4

Let Hypotheses (ℋ) be satisfied. We suppose that $\lim_{n\to\infty}\frac{\alpha _{n}}{\mu _{n}}=\tau\in[0,+\infty)$ and that:

• (H1) $\sum_{n=1}^{\infty}\mu_{n}=\infty$ and $|\mu _{n}-\mu_{n-1}|=o(\mu_{n})$;

• (H2) $|\alpha_{n}-\alpha_{n-1}|=o(\mu_{n})$;

• (H3) $\sup_{z\in B}\|W_{n}z-W_{n-1}z\|=o(\mu_{n})$, with $B\subset H$ bounded.

Then $(x_{n})_{n\in\mathbb{N}}$ is asymptotically regular.

Remark 2.5

Note that, for $(W_{n})_{n\in\mathbb{N}}$ as in Example 1.3, hypothesis (H3) reduces to an hypothesis on $(\lambda_{n})_{n\in\mathbb {N}}$ since

$$\lim_{n\to\infty}\frac{\|W_{n+1}x-W_{n}x\|}{\mu_{n}}=\lim_{n\to \infty} \frac {|\lambda_{n+1}-\lambda_{n}|}{\mu_{n}}. $$

Proof of Proposition 2.4

First of all, from Proposition 2.2, $(x_{n})_{n\in\mathbb{N}}$ is bounded.

If we denote by $y_{n}=\alpha_{n}Sx_{n}+(1-\alpha_{n})B_{n}x_{n}$ then

$$x_{n+1}-x_{n}=W_{n}y_{n}-W_{n-1}y_{n-1}=W_{n}y_{n}-W_{n}y_{n-1}+W_{n}y_{n-1}-W_{n-1}y_{n-1}, $$

so, passing to the norm and by using the nonexpansivity of $(W_{n})_{n\in \mathbb{N}}$,

$$\begin{aligned} \|x_{n+1}-x_{n}\|\leq\|y_{n}-y_{n-1} \|+\|W_{n}y_{n-1}-W_{n-1}y_{n-1}\|. \end{aligned}$$

(2.4)

Now let us observe that

$$\begin{aligned} y_{n}-y_{n-1} =&\alpha_{n}(Sx_{n}-Sx_{n-1})+( \alpha_{n}-\alpha _{n-1})Sx_{n-1}+(1- \alpha_{n}) (B_{n}x_{n}-B_{n-1}x_{n-1}) \\ &{}+(1-\alpha_{n})B_{n-1}x_{n-1}-(1- \alpha_{n-1})B_{n-1}x_{n-1} \\ =&\alpha_{n}(Sx_{n}-Sx_{n-1})+( \alpha_{n}-\alpha _{n-1}) (Sx_{n-1}-B_{n-1}x_{n-1}) \\ &{} +(1-\alpha_{n}) (B_{n}x_{n}-B_{n-1}x_{n-1}). \end{aligned}$$

Therefore replacing the last equality in (2.4) and by using the boundedness of $(x_{n})_{n\in\mathbb{N}}$, we obtain

$$\begin{aligned} \|x_{n+1}-x_{n}\| \leq& \alpha_{n} \|Sx_{n}-Sx_{n-1}\|+|\alpha _{n}- \alpha_{n-1}|O(1)+(1-\alpha_{n})\|B_{n}x_{n}-B_{n-1}x_{n-1} \| \\ &{}+\|W_{n}y_{n-1}-W_{n-1}y_{n-1}\| \\ \leq&\alpha_{n}\|x_{n}-x_{n-1}\|+| \alpha_{n}-\alpha _{n-1}|O(1)+(1-\alpha_{n}) \|B_{n}x_{n}-B_{n}x_{n-1}\| \\ &{}+(1-\alpha_{n})\|B_{n}x_{n-1}-B_{n-1}x_{n-1} \|+\| W_{n}y_{n-1}-W_{n-1}y_{n-1}\| \\ \leq&\alpha_{n}\|x_{n}-x_{n-1}\|+| \alpha_{n}-\alpha _{n-1}|O(1)+(1-\alpha_{n}) (1- \mu_{n}\rho)\|x_{n}-x_{n-1}\| \\ &{}+(1-\alpha_{n})|\mu_{n-1}-\mu_{n}| \|Dx_{n-1}\|+\| W_{n}y_{n-1}-W_{n-1}y_{n-1} \| \\ \leq& \bigl(1-(1-\alpha_{n})\rho\mu_{n} \bigr) \|x_{n}-x_{n-1}\|+\| W_{n}y_{n-1}-W_{n-1}y_{n-1} \| \\ &{}+ \bigl(|\alpha_{n}-\alpha_{n-1}|+(1- \alpha_{n})| \mu_{n-1}-\mu_{n}| \bigr)O(1). \end{aligned}$$

(2.5)

Denoting

$$\begin{aligned} &a_{n}:=\|x_{n}-x_{n-1}\|,\qquad \gamma_{n}:=(1- \alpha_{n})\rho\mu_{n}, \\ &\delta_{n}:=\|W_{n}y_{n-1}+W_{n-1}y_{n-1} \|+ \bigl(|\alpha_{n}-\alpha _{n-1}|+(1-\alpha_{n})| \mu_{n-1}-\mu_{n}| \bigr)O(1), \end{aligned}$$

(2.4) becomes

$$a_{n+1}\leq(1-\gamma_{n})a_{n}+ \delta_{n}. $$

Thus, our hypotheses (H1), (H2), and (H3), are enough to ensure, by Lemma 2.3, that $(x_{n})_{n\in\mathbb{N}}$ is asymptotically regular. □

Remark 2.6

By the previous proof, it is clear that the hypothesis $\tau\in [0,+\infty)$ is needed only to ensure the boundedness of $(x_{n})_{n\in \mathbb{N}}$. So, more in general, boundedness, (H1), (H2), and (H3) are enough to prove asymptotical regularity.

From now on we will suppose that $\mu_{n}\to0$, as $n\to\infty$; then, since τ is nonnegative, either $\alpha_{n}\to0$, as $n\to\infty$, or $\alpha_{n}=0$.

Since we are searching for solutions of variational inequalities on fixed points sets, we show some sufficient condition for which the set of weak limits of $(x_{n})_{n\in\mathbb{N}}$ lies in F.

Proposition 2.7

Let Hypotheses (ℋ) satisfied. Let us suppose that $\lim_{n\to\infty}\alpha_{n}= \lim_{n\to \infty}{\mu_{n}}=0$. Let us suppose $\lim_{n\to\infty}\frac{\alpha_{n}}{\mu _{n}}=\tau\in[0,+\infty)$ and let $(x_{n})_{n\in\mathbb{N}}$ defined by (2.1) be asymptotically regular. Then $\omega_{w}(x_{n})\subset F$.

Proof

The proof is based on Opial’s condition. The condition on τ gives the boundedness of our sequence by Proposition 2.2.

Let thus $z\in\omega_{w}(x_{n})$ and let $(x_{n_{k}})_{k\in\mathbb{N}}$ be a subsequence weak convergent to z. If $z\notin F$ then $z\neq Wz$ and

$$\begin{aligned} \liminf_{k\to\infty} \|x_{n_{k}}-z\| < & \liminf _{k\to\infty} \| x_{n_{k}}-Wz\| \\ \leq& \liminf_{k\to\infty} \bigl[\|x_{n_{k}}-x_{n_{k}+1}\|+ \|x_{n_{k}+1}-Wz\| \bigr] \\ \leq& \bigl(\mbox{by asymptotical regularity of } (x_{n})_{n\in\mathbb{N}} \bigr) \\ \leq& \liminf_{k\to\infty} \bigl[\|W_{n_{k}}y_{n_{k}}-W_{n_{k}}z \|+\| W_{n_{k}}z-Wz\| \bigr] \\ \bigl(\mbox{by condition (h2) on }(W_{n})_{n\in\mathbb{N}} \bigr) \leq& \liminf_{k\to\infty} \|y_{n_{k}}-z\| \\ (\mbox{since }\alpha_{n}\to0) \leq& \liminf_{k\to\infty} (1-\alpha _{n_{k}})\|B_{n_{k}}x_{n_{k}}-z\| \\ =& \liminf_{k\to\infty} (1-\alpha_{n_{k}})\|x_{n_{k}}- \mu _{n_{k}}Dx_{n_{k}}-z\| \\ \leq& \liminf_{k\to\infty} \bigl[\|x_{n_{k}}-z\|+ \mu_{n_{k}}\|Dx_{n_{k}}\|\bigr]. \end{aligned}$$

Therefore, the boundedness of $(x_{n})_{n\in\mathbb{N}}$, along with the hypothesis $\mu_{n}\to0$, produces the contradiction

$$\liminf_{k\to\infty} \|x_{n_{k}}-z\|< \liminf _{k\to\infty} \| x_{n_{k}}-Wz\| \leq\liminf_{k\to\infty} \|x_{n_{k}}-z\|. $$

 □

Now we are able to prove our first convergence result.

Theorem 2.8

Let Hypotheses (ℋ) be satisfied. Let us suppose that $\mu_{n}\to0$ and there exists

$$\lim_{n\to\infty}\frac{\alpha_{n}}{\mu_{n}}=\tau\in[0,+\infty). $$

Moreover, suppose that

• (H1) $\sum_{n=1}^{\infty}\mu_{n}=\infty$ and $|\mu _{n}-\mu_{n-1}|=o(\mu_{n})$;

• (H2) $|\alpha_{n}-\alpha_{n-1}|=o(\mu_{n})$;

• (H3) $\sup_{z\in B}\|W_{n}z-W_{n-1}z\|=o(\mu_{n})$, with $B\subset H$ bounded.

Then $(x_{n})_{n\in\mathbb{N}}$ defined by (2.1) strongly converges in F to $x^{*}$, that is, the unique solution of the variational inequality problem

$$ \bigl\langle \tau(I-S)x+Dx, y-x \bigr\rangle \geq0,\quad \forall y \in F. $$

(2.6)

Proof

Recall that, since S is nonexpansive, $(I-S)$ is $\frac {1}{2}$-inverse strongly monotone, so the operator $(\tau(I-S)+D)$ is a strongly monotone operator. Since F is closed and convex, problem (2.6) has a unique solution in F, which we indicate by $x^{*}$.

The hypotheses on τ furnish, by Proposition 2.2, the boundedness of $(x_{n})_{n\in\mathbb{N}}$. Then, in view of hypotheses (H1), (H2), and (H3), we can apply Proposition 2.4 to obtain asymptotical regularity. This allows one to apply Proposition 2.7 to get $\omega _{w}(x_{n})\subset F$. So, let $x^{*}\in F$, the unique solution of (2.6); by using the convexity of the norm and the subdifferential inequality

$$\|x+y\|^{2}\leq\|x\|^{2}+2\langle y,x+y\rangle,\quad \forall x,y \in H, $$

we have, denoting again $B_{n}=(I-\mu_{n}D)$,

$$\begin{aligned} \bigl\| x_{n+1}-x^{*}\bigr\| ^{2} \leq& \bigl\| \alpha_{n} \bigl(Sx_{n}-x^{*} \bigr)+(1-\alpha _{n}) \bigl(B_{n}x_{n}-x^{*} \bigr)\bigr\| ^{2} \\ =& \bigl\| \alpha_{n} \bigl(Sx_{n}-Sx^{*} \bigr)+ \alpha_{n} \bigl(Sx^{*}-x^{*} \bigr)+(1-\alpha _{n}) \bigl(B_{n}x_{n}-B_{n}x^{*} \bigr) \\ &{}+(1-\alpha_{n}) \bigl(B_{n}x^{*}-x^{*} \bigr) \bigr\| ^{2} \\ =& \bigl\| \alpha_{n} \bigl(Sx_{n}-Sx^{*} \bigr)+(1-\alpha _{n}) \bigl(B_{n}x_{n}-B_{n}x^{*} \bigr) \\ &{}- \bigl(\alpha _{n}(I-S)x^{*}+(1-\alpha_{n})\mu_{n}Dx^{*} \bigr)\bigr\| ^{2} \\ \leq& \alpha_{n}\bigl\| x_{n}-x^{*}\bigr\| ^{2}+(1- \alpha_{n}) (1-\mu_{n}\rho)\bigl\| x_{n}-x^{*} \bigr\| ^{2} \\ &{}-2 \bigl\langle \bigl(\alpha_{n}(I-S)x^{*}+(1- \alpha_{n}) \mu _{n}Dx^{*} \bigr),x_{n+1}-x^{*} \bigr\rangle \\ =& \bigl(1-(1-\alpha_{n})\mu_{n}\rho \bigr) \bigl\| x_{n}-x^{*}\bigr\| ^{2} \\ &{}-2(1-\alpha_{n})\mu_{n} \biggl\langle \frac{\alpha_{n}}{(1-\alpha _{n})\mu _{n}}(I-S)x^{*}+Dx^{*},x_{n+1}-x^{*} \biggr\rangle . \end{aligned}$$

(2.7)

Denoting by

$$\begin{aligned} & a_{n}=\bigl\| x_{n}-x^{*}\bigr\| ^{2},\qquad \gamma_{n}= (1-\alpha_{n})\mu_{n}\rho, \\ &\delta_{n}=-\frac{2}{\rho} \biggl\langle \frac{\alpha_{n}}{(1-\alpha_{n})\mu _{n}}(I-S)x^{*}+Dx^{*},x_{n+1}-x^{*} \biggr\rangle , \end{aligned}$$

(2.7) can be written $a_{n+1}\leq(1-\gamma_{n})a_{n}+\gamma _{n}\delta _{n}$.

To invoke the Xu Lemma 2.3, since $\sum_{n}\gamma _{n}=\infty$ from (H1), we need to prove only that $\limsup_{n\to\infty}\delta_{n}\leq0$.

There exists a subsequence $(x_{n_{k}})_{k\in\mathbb{N}}$ of $(x_{n})_{n\in \mathbb{N}}$ such that

$$\begin{aligned} \limsup_{n\to\infty}\delta_{n}&=\limsup_{n\to\infty} \biggl\langle \frac{\alpha_{n}}{(1-\alpha_{n})\mu _{n}}(I-S)x^{*}+Dx^{*},x^{*}-x_{n+1} \biggr\rangle \\ &=\lim_{k\to\infty} \biggl\langle \frac {\alpha_{n_{k}}}{(1-\alpha_{n_{k}})\mu _{n_{k}}}(I-S)x^{*}+Dx^{*},x^{*}-x_{{n_{k}}+1} \biggr\rangle . \end{aligned}$$

Since $(x_{n_{k}})_{k\in\mathbb{N}}$ is bounded, we can suppose that $(x_{n_{k}})_{k\in\mathbb{N}}$ weakly converges to p. Proposition 2.7 gives $p\in F$. By using the asymptotical regularity of $(x_{n})_{n\in\mathbb{N}}$ we have

$$\begin{aligned} &\limsup_{n\to\infty} \biggl\langle \frac{\alpha_{n}}{(1-\alpha_{n})\mu _{n}}(I-S)x^{*}+Dx^{*},x^{*}-x_{n+1} \biggr\rangle \\ &\quad=\lim_{k\to\infty} \biggl\langle \frac{\alpha_{n_{k}}}{(1-\alpha _{n_{k}})\mu _{n_{k}}}(I-S)x^{*}+Dx^{*},x^{*}-x_{{n_{k}}+1} \biggr\rangle \\ &\quad=\lim_{k\to\infty} \biggl[ \biggl\langle \frac{\alpha_{n_{k}}}{(1-\alpha _{n_{k}})\mu _{n_{k}}}(I-S)x^{*}+Dx^{*},x^{*}-x_{n_{k}} \biggr\rangle \\ &\qquad{}+ \biggl\langle \frac{\alpha _{n_{k}}}{(1-\alpha_{n_{k}})\mu _{n_{k}}}(I-S)x^{*}+Dx^{*},x_{{n_{k}}}-x_{{n_{k}}+1} \biggr\rangle \biggr] \\ &\quad=\lim_{k\to\infty} \biggl\langle \frac{\alpha_{n_{k}}}{(1-\alpha _{n_{k}})\mu _{n_{k}}}(I-S)x^{*}+Dx^{*},x^{*}-x_{{n_{k}}} \biggr\rangle \\ &\quad= \bigl\langle \tau(I-S)x^{*}+Dx^{*},x^{*}-p \bigr\rangle \leq0 \quad\bigl(\mbox{since }x^{*} \mbox{ is the solution of (2.6)} \bigr). \end{aligned}$$

 □

Remark 2.9

Let us remark that, in the study of the behavior of $(x_{n})_{n\in \mathbb {N}}$ for $\tau\in[0,+\infty)$, the set of fixed points of S never appears; all the properties, including the strong convergence, have been proved only by the hypotheses on the control sequences.

Let us now suppose $\lim_{n\to\infty}\frac{\alpha _{n}}{\mu _{n}}=\tau=+\infty$. In this case, necessarily $\mu_{n}\to0$ as $n\to \infty$. Therefore either $\alpha_{n}\to\alpha>0$ or $\alpha_{n}\to0$ too and $\mu_{n}=o(\alpha_{n})$.

By Proposition 2.1, if $\operatorname{Fix}(S)\cap F$ is nonempty, the boundedness of $(x_{n})_{n\in\mathbb{N}}$ follows. On the contrary, if there are no common fixed points, the boundedness is not guaranteed as shown by the following counterexample.

Example 2.10

Let us consider $H=\mathbb{R}$, $x_{0}=1$, $W_{n}x=Dx=x$, $Sx=x+1$, $\alpha_{n}=\frac{1}{\sqrt{n}}$, and $\mu _{n}=\frac{1}{n}$.

Our method gives the positive number sequence:

$$x_{n+1}=\frac{1}{\sqrt{n}}(x_{n}+1)+ \biggl(1- \frac{1}{\sqrt{n}} \biggr) \biggl(1-\frac{1}{n} \biggr)x_{n}. $$

If there exists $M>0$ such that $x_{n}< M$ then we note that, for every k,

$$\begin{aligned} x_{k+1}-x_{k} =&\frac{x_{k}}{\sqrt{k}}+\frac{1}{\sqrt{k}}+ \biggl(1-\frac {1}{\sqrt{k}} \biggr) \biggl(1-\frac{1}{k} \biggr)x_{k}-x_{k} \\ =&\frac{1}{\sqrt{k}}-\frac{x_{k}}{k} \biggl(1-\frac{1}{\sqrt {k}} \biggr) \simeq\frac{1}{\sqrt{k}}-\frac{M}{k} \\ >&\frac{1}{\sqrt{k}} \biggl(1-\frac{M}{\sqrt{k}} \biggr)= \frac {1}{\sqrt{k}}, \end{aligned}$$

and this is in contradiction with the boundedness of $(x_{n})_{n\in \mathbb{N}}$.

Nevertheless, we explicitly note that if $W_{n}=P_{C}$ and there exist solutions of the variational inequality problem

$$\bigl\langle (I-S)x,y-x \bigr\rangle \geq0,\quad \forall y\in C, $$

then the boundedness is ensured even if $F\cap\operatorname {Fix}(S)=\emptyset$. This is shown in the following proposition.

Proposition 2.11

Let C be a closed and convex subset of H. Let us suppose that the variational inequality problem

$$\bigl\langle (I-S)x,y-x \bigr\rangle \geq0, \quad\forall y\in C, $$

has at least a solution $x^{*}$. Then the sequence defined by

$$x_{n+1}=P_{C} \bigl(\alpha_{n}Sx_{n}+(1- \alpha_{n})B_{n}x_{n} \bigr) $$

is bounded.

Proof

We know that, for all $\eta\in(0,1]$, we have

$$ x^{*}=P_{C} \bigl(\eta Sx^{*}+(1-\eta)x^{*} \bigr). $$

(2.8)

Taking $W_{n}=P_{C}$, we have

$$\begin{aligned} \bigl\| x_{n+1}-x^{*}\bigr\| \leq& \bigl\| P_{C} \bigl(\alpha_{n}Sx_{n}+(1- \alpha _{n})B_{n}x_{n} \bigr)-P_{C} \bigl( \alpha _{n}Sx^{*}+(1-\alpha_{n})B_{n}x^{*} \bigr) \bigr\| \\ &{}+\bigl\| P_{C} \bigl(\alpha_{n}Sx^{*}+(1-\alpha_{n})B_{n}x^{*} \bigr)-x^{*}\bigr\| \quad \bigl(\mbox{as in Proposition 2.1 in (2.8)} \bigr) \\ \leq& \bigl(1-(1-\alpha_{n})\mu_{n}\rho \bigr) \bigl\| x_{n}-x^{*}\bigr\| \\ &{}+\bigl\| P_{C} \bigl(\alpha _{n}Sx^{*}+(1-\alpha_{n})B_{n}x^{*} \bigr)-x^{*}\bigr\| \quad \bigl(\mbox{taking }\eta=\alpha_{n}\mbox{ in (2.8)} \bigr) \\ \leq& \bigl(1-(1-\alpha_{n})\mu_{n}\rho \bigr) \bigl\| x_{n}-x^{*}\bigr\| \\ &{}+\bigl\| P_{C} \bigl(\alpha _{n}Sx^{*}+(1-\alpha_{n})B_{n}x^{*} \bigr)-P_{C} \bigl(\alpha_{n} Sx^{*}+(1-\alpha_{n})x^{*} \bigr)\bigr\| \\ \leq& \bigl(1-(1-\alpha_{n})\mu_{n}\rho \bigr) \bigl\| x_{n}-x^{*}\bigr\| +(1-\alpha_{n})\mu_{n}\rho \frac {\|Dx^{*}\|}{\rho}. \end{aligned}$$

So the boundedness follows as in Proposition 2.1. □

Therefore it is meaningful to prove convergence results if $\operatorname{Fix}(S)\cap F\neq\emptyset$.

Theorem 2.12

Let Hypotheses (ℋ) satisfied. Let us suppose that

$$\lim_{n\to\infty} \mu_{n}= 0,\qquad \lim _{n\to\infty} \alpha_{n}=\alpha \in[0,1),\qquad \lim _{n\to\infty} \frac{\alpha_{n}}{\mu_{n}}=\tau=+\infty, $$

and $\operatorname{Fix}(S)\cap F\neq\emptyset$. Moreover, suppose that:

• (H1s) $\sum_{n=1}^{\infty}\mu_{n}=\infty$ and $|\mu _{n}-\mu_{n-1}|=o(\alpha_{n}\mu_{n})$;

• (H2s) $|\alpha_{n}-\alpha_{n-1}|=o(\alpha_{n}\mu_{n})$;

• (H3s) $\sup_{z\in B}\|W_{n}z-W_{n-1}z\|=o(\alpha_{n}\mu_{n})$, with $B\subset H$ bounded.

• (H4) $\vert \frac{1}{\alpha_{n}}-\frac {1}{\alpha _{n-1}}\vert =O(\mu_{n})$.

(Note that (H1s), (H2s), (H3s) are stronger than (H1), (H2), (H3) of Theorem  2.8.)

Then $(x_{n})_{n\in\mathbb{N}}$ defined by (2.1) strongly converges to $\bar{x}\in F\cap\operatorname{Fix}(S)$, that is, the unique solution of the variational inequality problem

$$ \langle Dx, y-x\rangle\geq0, \quad\forall y\in F\cap \operatorname{Fix}(S). $$

(2.9)

Remark 2.13

Note that, if $\alpha_{n}\to\alpha>0$, the requirements (H1s), (H2s), (H3s) reduce to (H1), (H2), (H3).

Proof

If $\operatorname{Fix}(S)\cap F\neq\emptyset$, $(x_{n})_{n\in\mathbb {N}}$ is bounded by Proposition 2.1. Since (H1s)-(H2s)-(H3s) imply (H1)-(H2)-(H3), by using Proposition 2.4, we see that $(x_{n})_{n\in\mathbb{N}}$ is asymptotically regular. Let us divide the proof in steps.

Step 1. $\|x_{n+1}-x_{n}\|=o(\alpha_{n})$.

Proof of Step 1

We need to prove that

$$\lim_{n\to\infty}\frac{\|x_{n+1}-x_{n}\|}{\alpha_{n}}=0. $$

If $\alpha_{n}\to\alpha>0$ we do not need to prove anything; so let $\alpha=0$. Dividing by $\alpha_{n}$ in (2.5) of Proposition 2.4 we have

$$\begin{aligned} \frac{\|x_{n+1}-x_{n}\|}{\alpha_{n}} \leq& \bigl(1-(1-\alpha_{n})\rho\mu _{n} \bigr)\frac{\| x_{n}-x_{n-1}\|}{\alpha_{n}}+\frac{\|W_{n}y_{n-1}+W_{n-1}y_{n-1}\|}{\alpha _{n}} \\ &{}+\frac{(|\alpha_{n}-\alpha_{n-1}|+(1-\alpha_{n})|\mu_{n-1}-\mu _{n}|)}{\alpha_{n}}O(1) \\ =& \bigl(1-(1-\alpha_{n})\rho\mu_{n} \bigr) \frac{\|x_{n}-x_{n-1}\|}{\alpha_{n}}\pm \bigl(1-(1-\alpha_{n})\rho\mu_{n} \bigr) \frac{\|x_{n}-x_{n-1}\|}{\alpha_{n-1}} \\ &{}+\frac{\|W_{n}y_{n-1}+W_{n-1}y_{n-1}\|}{\alpha_{n}}+\frac{(|\alpha _{n}-\alpha_{n-1}|+(1-\alpha_{n})|\mu_{n-1}-\mu_{n}|)}{\alpha_{n}}O(1) \\ \leq& \bigl(1-(1-\alpha_{n})\rho\mu_{n} \bigr) \frac{\|x_{n}-x_{n-1}\|}{\alpha _{n-1}}+ \biggl\vert \frac{1}{\alpha_{n}}-\frac{1}{\alpha_{n-1}} \biggr\vert \| x_{n}-x_{n-1}\| \\ &{}+\frac{\|W_{n}y_{n-1}+W_{n-1}y_{n-1}\|}{\alpha_{n}}+\frac{(|\alpha _{n}-\alpha_{n-1}|+(1-\alpha_{n})|\mu_{n-1}-\mu_{n}|)}{\alpha_{n}}O(1). \end{aligned}$$

The boundedness of $(x_{n})_{n\in\mathbb{N}}$ and (H4) give

$$\begin{aligned} \frac{\|x_{n}-x_{n+1}\|}{\alpha_{n}} \leq& \bigl(1-(1-\alpha_{n})\rho\mu _{n} \bigr)\frac {\|x_{n}-x_{n-1}\|}{\alpha_{n-1}}+O(\mu_{n})\|x_{n-1}-x_{n}\| \\ &{}+\frac{\|W_{n}y_{n-1}+W_{n-1}y_{n-1}\|}{\alpha_{n}}+\frac{(|\alpha _{n}-\alpha_{n-1}|+|\mu_{n-1}-\mu_{n}|)}{\alpha_{n}}O(1), \end{aligned}$$

so denoting

$$\begin{aligned}[b] &a_{n}=\frac{\|x_{n}-x_{n-1}\|}{\alpha_{n-1}}, \qquad\gamma_{n}= (1-\alpha _{n})\mu _{n}\rho, \\ &\delta_{n}= \biggl[O(\mu_{n})\|x_{n-1}-x_{n} \|+\frac{\| W_{n}y_{n-1}+W_{n-1}y_{n-1}\|}{\alpha_{n}}+\frac{(|\alpha_{n}-\alpha _{n-1}|+|\mu_{n-1}-\mu_{n}|)}{\alpha_{n}} \biggr]O(1), \end{aligned} $$

our inequality can be written as $a_{n+1}\leq(1-\gamma_{n})a_{n}+\delta _{n}$. In view of (H1s), (H2s), and (H3s), we can apply the Xu Lemma 2.3 to conclude that $\|x_{n+1}-x_{n}\|=o(\alpha_{n})$. □

Step 2. $\omega_{w}(x_{n})\subset F\cap\operatorname{Fix}(S)$.

Proof of Step 2

Let $z\in F\cap\operatorname{Fix}(S)$; then by the boundedness and the subdifferential inequality

$$\begin{aligned} \|x_{n+1}-z\|^{2} \leq& \bigl\| \alpha_{n}(Sx_{n}-z)+(1- \alpha_{n}) (B_{n}x_{n}-z)\bigr\| ^{2} \\ \leq&\bigl\| \alpha_{n}(Sx_{n}-z)+(1-\alpha_{n}) (x_{n}-z)\bigr\| ^{2}-2\mu_{n}\langle Dx_{n},x_{n+1}-z\rangle \\ \leq& \alpha_{n}\|Sx_{n}-z\|^{2}+(1- \alpha_{n})\|x_{n}-z\|^{2}-\alpha _{n}(1- \alpha _{n})\|Sx_{n}-x_{n}\|^{2} \\ &{} +2\mu_{n}\langle Dx_{n},z-x_{n+1}\rangle \\ \leq& \|x_{n}-z\|^{2}-\alpha_{n}(1- \alpha_{n})\|Sx_{n}-x_{n}\|^{2}+2 \mu_{n}O(1), \end{aligned}$$

we have

$$\begin{aligned} \alpha_{n}(1-\alpha_{n})\|Sx_{n}-x_{n} \|^{2} \leq&\|x_{n}-z\|^{2}-\|x_{n+1}-z\| ^{2}+2\mu_{n}O(1) \\ \leq&\|x_{n}-x_{n+1}\|O(1)+2\mu_{n}O(1). \end{aligned}$$

Dividing by $\alpha_{n}$ we obtain

$$\begin{aligned} (1-\alpha_{n})\|Sx_{n}-x_{n}\|^{2} \leq& \frac{\|x_{n}-x_{n+1}\|}{\alpha _{n}}O(1)+2\frac{\mu_{n}}{\alpha_{n}}O(1). \end{aligned}$$

Since $\tau=+\infty$ and by using Step 1, $\|x_{n}-Sx_{n}\|\to0$, as $n\to \infty$, the demiclosedness principle for nonexpansive mappings guarantees that $\omega_{w}(x_{n})\subset\operatorname{Fix}(S)$. By Opial’s condition, if $z\in\omega_{w}(x_{n})\subset\operatorname{Fix}(S)$, $(x_{n_{k}})_{k\in\mathbb{N}}$ weakly converges to z and $z\notin F$ then

$$\begin{aligned} \liminf_{k\to\infty} \|x_{n_{k}}-z\| < & \liminf _{k\to\infty} \| x_{n_{k}}-Wz\| \\ \leq& \liminf_{k\to\infty} \bigl[\|x_{n_{k}}-x_{n_{k}+1}\|+ \|x_{n_{k}+1}-Wz\| \bigr] \\ \leq& \liminf_{k\to\infty} \bigl[\|x_{n_{k}}-x_{n_{k}+1}\|+\| W_{n_{k}}y_{n_{k}}-W_{n}z\|+\|W_{n_{k}}z-Wz\|\bigr] \\ \leq& \liminf_{k\to\infty} \bigl[\|x_{n_{k}}-x_{n_{k}+1}\|+ \|y_{n_{k}}-z\|+\| W_{n_{k}}z-Wz\|\bigr] \\ \leq& \liminf_{k\to\infty} \bigl[\|x_{n_{k}}-x_{n_{k}+1} \|+\alpha_{n_{k}}\| x_{n_{k}}-z\| \\ &{}+(1-\alpha_{n_{k}})\|B_{n_{k}}x_{n_{k}}-z\|+ \|W_{n_{k}}z-Wz\| \bigr] \\ \leq& \liminf_{k\to\infty} \bigl[\|x_{n_{k}}-x_{n_{k}+1} \|+\|x_{n_{k}}-z\| \\ &{}+(1-\alpha_{n_{k}})\mu_{n_{k}}\|Dx_{n_{k}}\|+ \|W_{n_{k}}z-Wz\| \bigr] \\ \leq& \liminf_{k\to\infty} \|x_{n_{k}}-z\|, \end{aligned}$$

which is absurd. So we have $\omega_{w}(x_{n})\subset F\cap\operatorname{Fix}(S)$. □

Finally we conclude our proof, showing the convergence of the sequence.

Step 3. $(x_{n})_{n\in\mathbb{N}}$ strongly converges to $\bar{x}$ satisfying (2.9).

Proof of Step 3

Let $\bar{x}$ the unique solution of the variational inequality problem (2.9). Since $\bar{x}\in F\cap\operatorname{Fix}(S)$, we have

$$\begin{aligned} \|x_{n+1}-\bar{x}\|^{2} \leq& \bigl\| \alpha_{n}(Sx_{n}- \bar{x})+(1-\alpha _{n}) (B_{n}x_{n}-\bar{x}) \bigr\| ^{2} \\ =& \bigl\| \alpha_{n}(Sx_{n}-\bar{x})+(1-\alpha_{n}) (B_{n}x_{n}-B_{n}\bar {x})+(1-\alpha _{n}) (B_{n}\bar{x}-\bar{x})\bigr\| ^{2} \\ =& \bigl\| \alpha_{n}(Sx_{n}-\bar{x})+(1-\alpha_{n}) (B_{n}x_{n}-B_{n}\bar {x})-(1-\alpha _{n})\mu_{n}D\bar{x}\bigr\| ^{2} \\ \leq& \alpha_{n}\|x_{n}-\bar{x}\|^{2}+(1- \alpha_{n}) (1-\mu_{n}\rho)\| x_{n}-\bar {x} \|^{2}-2 \bigl\langle (1-\alpha_{n})\mu_{n}D \bar{x},x_{n+1}-\bar{x} \bigr\rangle \\ =& \bigl(1-(1-\alpha_{n})\mu_{n}\rho \bigr) \|x_{n}-\bar{x}\|^{2}-2(1-\alpha_{n})\mu _{n}\langle D\bar{x},x_{n+1}-\bar{x}\rangle. \end{aligned}$$

Denoting

$$\begin{aligned} a_{n}=\|x_{n}-\bar{x}\|^{2},\qquad \gamma_{n}= (1-\alpha_{n})\mu_{n}\rho, \qquad\delta _{n}=\langle D\bar{x},\bar{x}-x_{n+1}\rangle, \end{aligned}$$

our inequality can be written as

$$a_{n+1}\leq(1-\gamma_{n})a_{n}+\frac{2}{\rho} \gamma_{n}\delta_{n}. $$

To invoke the Xu Lemma 2.3 we need to prove that $\limsup_{n\to\infty}\delta_{n}\leq0$.

There exists a subsequence $(x_{n_{k}})_{k\in\mathbb{N}}$ of $(x_{n})_{n\in \mathbb{N}}$ such that

$$\limsup_{n\to\infty}\langle D\bar{x},\bar{x}-x_{n+1}\rangle= \lim_{k\to \infty}\langle D\bar{x},\bar{x}-x_{{n_{k}}+1}\rangle. $$

Since $(x_{n_{k}})_{k\in\mathbb{N}}$ is bounded, we suppose that $(x_{n_{k}})_{k\in\mathbb{N}}$ weakly converges to p. Step 3 guarantees that $p\in F\cap\operatorname{Fix}(S)$. By using the asymptotical regularity of $(x_{n_{k}})_{k\in\mathbb{N}}$ we have

$$\begin{aligned} \limsup_{n\to\infty}\langle D\bar{x},\bar{x}-x_{n+1} \rangle &=\lim_{k\to\infty}\langle D\bar{x},\bar{x}-x_{{n_{k}}+1} \rangle \\ &=\lim_{k\to\infty} \bigl[\langle D\bar{x},\bar{x}-x_{n_{k}} \rangle +\langle D\bar{x},x_{{n_{k}}}-x_{{n_{k}}+1}\rangle \bigr]=\lim _{k\to\infty}\langle D\bar {x},\bar{x}-x_{{n_{k}}}\rangle \\ &=\langle D\bar{x},\bar{x}-p\rangle\leq0. \end{aligned}$$

 □

 □

Theorem 2.14

Let Hypotheses (ℋ). Let us suppose that

$$\lim_{n\to\infty}\mu_{n}=\lim_{n\to\infty} \alpha_{n}=0 \quad\textit{and}\quad \tau =\lim_{n\to\infty} \frac{\alpha_{n}}{\mu_{n}}=+\infty. $$

Let us suppose that $(x_{n})_{n\in\mathbb{N}}$ is bounded. Moreover, suppose that

• (H1s) $\sum_{n=1}^{\infty}\mu_{n}=\infty$ and $|\mu _{n}-\mu_{n-1}|=o(\alpha_{n}\mu_{n})$;

• (H2s) $|\alpha_{n}-\alpha_{n-1}|=o(\alpha_{n}\mu_{n})$;

• (H3s) $\sup_{z\in B}\|W_{n}z-W_{n-1}z\|=o(\alpha_{n}\mu_{n})$, with $B\subset H$ bounded;

• (H4) $\vert \frac{1}{\alpha_{n}}-\frac {1}{\alpha _{n-1}}\vert =O(\mu_{n})$.

Let $\bar{\Sigma}$ be the set of solutions of the variational inequality problem

$$ \bigl\langle (I-S)x, y-x \bigr\rangle \geq0,\quad \forall y\in F, $$

(2.10)

and let us suppose that $\bar{\Sigma}\neq\emptyset$.

Then $(x_{n})_{n\in\mathbb{N}}$ defined by (2.1) strongly converges to $\tilde{x,}$ that is, the unique solution of the variational inequality problem

$$ \langle Dx, y-x\rangle\geq0,\quad \forall y\in\bar{\Sigma}. $$

(2.11)

Proof

Since $\bar{\Sigma}$ coincides with the set of fixed point of the nonexpansive mapping $P_{F}S$, it is closed and convex. So (2.11) has a unique solution.

Let us note that (H1s)-(H2s)-(H3s) imply (H1)-(H2)-(H3); hence, by using Proposition 2.4, $(x_{n})_{n\in\mathbb{N}}$ is asymptotically regular. We divide the proof in steps.

Step 1. $\|x_{n+1}-x_{n}\|=o(\alpha_{n})$.

Proof

As for Step 1 of Theorem 2.12. □

Step 2. $\omega_{w}(x_{n})\subset\bar{\Sigma}$.

Proof of Step 2

Denoting by $y_{n}=\alpha_{n}Sx_{n}+(1-\alpha_{n})B_{n}x_{n}$, we have

$$\begin{aligned} x_{n}-y_{n} =&x_{n}- \alpha_{n}Sx_{n}-(1-\alpha_{n}) (x_{n}- \mu_{n}Dx_{n}) \\ =& x_{n}-\alpha_{n}Sx_{n}-(1- \alpha_{n})x_{n}+(1-\alpha_{n})\mu _{n}Dx_{n}) \\ =& \alpha_{n}(I-S)x_{n}+(1-\alpha_{n}) \mu_{n}Dx_{n}. \end{aligned}$$

(2.12)

Hypotheses $\alpha_{n}\to0$ and $\mu_{n}\to0$ allow one to conclude that $\|x_{n}-y_{n}\|\to0$. As a rule

$$\|y_{n}-W_{n}y_{n}\|\leq\|y_{n}-x_{n} \|+\|x_{n}-W_{n}y_{n}\|=\|y_{n}-x_{n} \|+\| x_{n}-x_{n+1}\| \to0, $$

as $n\to\infty$. Moreover,

$$\begin{aligned} x_{n}-x_{n+1} =&x_{n}-W_{n}y_{n}=(x_{n}-y_{n})+(y_{n}-W_{n}y_{n}) \\ =&\alpha_{n}(I-S)x_{n}+(1-\alpha_{n}) (x_{n}-B_{n}x_{n})+(I-W_{n})y_{n} \\ =&\alpha_{n}(I-S)x_{n}+(1-\alpha_{n}) \mu_{n}Dx_{n}+(I-W_{n})y_{n}. \end{aligned}$$

Dividing by $\alpha_{n}$ we have

$$w_{n}:=\frac{x_{n}-x_{n+1}}{\alpha_{n}}=(I-S)x_{n}+\frac{(1-\alpha_{n})\mu _{n}}{\alpha_{n}}Dx_{n}+ \frac{1}{\alpha_{n}}(I-W_{n})y_{n}. $$

For all $z\in F$,

$$\begin{aligned} \langle w_{n},x_{n}-z\rangle =& \bigl\langle (I-S)x_{n},x_{n}-z \bigr\rangle +\frac {(1-\alpha _{n})\mu_{n}}{\alpha_{n}}\langle Dx_{n},x_{n}-z\rangle \\ &{}+\frac{1}{\alpha_{n}} \bigl\langle (I-W_{n})y_{n},x_{n}-z \bigr\rangle \quad \bigl(\mbox{by monotonicity of }(I-S) \bigr) \\ \geq& \bigl\langle (I-S)z,x_{n}-z \bigr\rangle +\frac{(1-\alpha_{n})\mu_{n}}{\alpha _{n}} \langle Dx_{n},x_{n}-z\rangle \\ &{}+\frac{1}{\alpha_{n}} \bigl\langle (I-W_{n})y_{n},x_{n}-y_{n} \bigr\rangle +\frac {1}{\alpha _{n}} \bigl\langle (I-W_{n})y_{n},y_{n}-z \bigr\rangle . \end{aligned}$$

Since $z\in F$, $z=W_{n}z$ for all $n\in\mathbb{N,}$ and $(I-W_{n})$ is monotone:

$$\begin{aligned} \langle w_{n},x_{n}-z\rangle \geq& \bigl\langle (I-S)z,x_{n}-z \bigr\rangle +\frac {(1-\alpha _{n})\mu_{n}}{\alpha_{n}}\langle Dx_{n},x_{n}-z\rangle \\ &{}+\frac{1}{\alpha_{n}} \bigl\langle (I-W_{n})y_{n},x_{n}-y_{n} \bigr\rangle +\frac {1}{\alpha _{n}} \bigl\langle (I-W_{n})y_{n}+(I-W_{n})z,y_{n}-z \bigr\rangle \\ \geq& \bigl\langle (I-S)z,x_{n}-z \bigr\rangle +\frac{(1-\alpha_{n})\mu_{n}}{\alpha _{n}} \langle Dx_{n},x_{n}-z\rangle +\frac{1}{\alpha_{n}} \bigl\langle (I-W_{n})y_{n},x_{n}-y_{n} \bigr\rangle . \end{aligned}$$

By using (2.12)

$$\begin{aligned} \langle w_{n},x_{n}-z\rangle \geq& \bigl\langle (I-S)z,x_{n}-z \bigr\rangle +\frac {(1-\alpha_{n})\mu_{n}}{\alpha_{n}}\langle Dx_{n},x_{n}-z\rangle \\ &{}+ \bigl\langle (I-W_{n})y_{n},(I-S)x_{n} \bigr\rangle +\frac{(1-\alpha_{n})\mu _{n}}{\alpha _{n}} \bigl\langle (I-W_{n})y_{n},Dx_{n} \bigr\rangle . \end{aligned}$$

Let us denote by $(x_{n_{k}})_{k\in\mathbb{N}}$ a subsequence weakly converging to p; by the same proof as Proposition 2.7 one can see that the boundedness of $(x_{n})$, combined with the assumptions $\mu_{n}\to0$ and $\alpha_{n}\to0$, is enough to guarantee that $p\in F$. We have

$$\begin{aligned} \langle w_{n_{k}},x_{n}-z\rangle \geq& \bigl\langle (I-S)z,x_{n_{k}}-z \bigr\rangle +\frac{(1-\alpha_{n_{k}})\mu _{n_{k}}}{\alpha_{n_{k}}}\langle Dx_{n_{k}},x_{n_{k}}-z\rangle \\ &{}+ \bigl\langle (I-W_{n_{k}})y_{n_{k}},(I-S)x_{n_{k}} \bigr\rangle +\frac{(1-\alpha _{n_{k}})\mu_{n_{k}}}{\alpha_{n_{k}}} \bigl\langle (I-W_{n_{k}})y_{n_{k}},Dx_{n_{k}} \bigr\rangle . \end{aligned}$$

Passing $k\to\infty$, since $w_{n}\to0$ by Step 1, $\|(I-W_{n})y_{n}\|\to0$ and $\tau=+\infty$, we have

$$0\geq \bigl\langle (I-S)z,p-z \bigr\rangle , \quad\forall z\in F. $$

If we replace z by $p+\eta(z-p)$, $\eta\in(0,1)$, we have

$$\bigl\langle (I-S) \bigl(p+\eta(z-p) \bigr),p-z \bigr\rangle \leq0. $$

Letting $\eta\to0$, finally,

$$\bigl\langle (I-S)p,p-z \bigr\rangle \leq0,\quad \forall z\in F, $$

i.e. the claim follows. □

Step 3. Convergence of the sequence.

Proof of Step 3

Let $\tilde{x}$ be the unique solution of the variational inequality problem (2.11). As in Theorem 2.8 we have

$$\begin{aligned} \|x_{n+1}-\tilde{x}\|^{2} \leq& \bigl(1-(1- \alpha_{n})\mu_{n}\rho \bigr)\|x_{n}-\tilde {x}\| ^{2} \\ &{}-2(1-\alpha_{n})\mu_{n} \biggl\langle \frac{\alpha_{n}}{(1-\alpha_{n})\mu _{n}}(I-S) \tilde{x}+D\tilde{x},x_{n+1}-\tilde{x} \biggr\rangle . \end{aligned}$$

Denoting

$$\begin{aligned} & a_{n}=\|x_{n}-\tilde{x}\|^{2},\qquad \gamma_{n}= (1-\alpha_{n})\mu_{n}\rho, \\ &\delta_{n}=\frac{2}{\rho} \biggl\langle \frac{\alpha_{n}}{(1-\alpha_{n})\mu _{n}}(I-S) \tilde{x}+D\tilde{x},\tilde{x}-x_{n+1} \biggr\rangle , \end{aligned}$$

our inequality can be written as

$$a_{n+1}\leq(1-\gamma_{n})a_{n}+\frac{2}{\rho} \gamma_{n}\delta_{n}. $$

To invoke the Xu Lemma 2.3 we need to prove that $\limsup_{n\to\infty}\delta_{n}\leq0$.

There exists a subsequence $(x_{n_{k}})_{k\in\mathbb{N}}$ of $(x_{n})_{n\in \mathbb{N}}$ such that

$$\limsup_{n\to\infty} \biggl\langle \frac{\alpha_{n}}{(1-\alpha_{n})\mu _{n}}(I-S)\tilde{x}+D \tilde{x},\tilde{x}-x_{n+1} \biggr\rangle =\lim_{k\to \infty} \biggl\langle \frac{\alpha_{n_{k}}}{(1-\alpha_{n_{k}})\mu_{n_{k}}}(I-S) \tilde {x}+D\tilde{x},\tilde{x}-x_{{n_{k}}+1} \biggr\rangle . $$

Since $(x_{n_{k}})_{k\in\mathbb{N}}$ is bounded, we can suppose that $(x_{n_{k}})_{k\in\mathbb{N}}$ weakly converges to p. We know, by Step 2, that $p\in\Sigma\subset F$. By using the asymptotical regularity of $(x_{n})_{n\in\mathbb{N}}$ we have

$$\begin{aligned} &\limsup_{n\to\infty} \biggl\langle \frac{\alpha_{n}}{(1-\alpha_{n})\mu _{n}}(I-S) \tilde{x}+D\tilde{x},\tilde{x}-x_{n+1} \biggr\rangle \\ &\quad=\lim_{k\to \infty} \biggl\langle \frac{\alpha_{n_{k}}}{(1-\alpha_{n_{k}})\mu_{n_{k}}}(I-S) \tilde {x}+D\tilde{x}, \tilde{x}-x_{{n_{k}}+1} \biggr\rangle \\ &\quad=\lim_{k\to\infty} \biggl[ \biggl\langle \frac{\alpha_{n_{k}}}{(1-\alpha _{n_{k}})\mu _{n_{k}}}(I-S) \tilde{x}+D\tilde{x},\tilde{x}-x_{n_{k}} \biggr\rangle \\ &\qquad{}+ \biggl\langle \frac{\alpha_{n_{k}}}{(1-\alpha_{n_{k}})\mu_{n_{k}}}(I-S) \tilde {x}+D\tilde{x},x_{n_{k}}-x_{{n_{k}}+1} \biggr\rangle \biggr] \\ &\quad=\lim_{k\to\infty} \biggl\langle \frac{\alpha_{n_{k}}}{(1-\alpha _{n_{k}})\mu _{n_{k}}}(I-S) \tilde{x}+D \tilde{x},\tilde{x}-x_{{n_{k}}} \biggr\rangle . \end{aligned}$$

Since $\tau=\infty$, $p\in F$, and $\tilde{x}\in\Sigma$,

$$\bigl\langle (I-S)\tilde{x},\tilde{x}-x_{n_{k}} \bigr\rangle \to \bigl\langle (I-S)\tilde {x},\tilde{x}-p \bigr\rangle \leq0. $$

Moreover, since $p\in\Sigma$ and $\tilde{x}$ is the solution of (2.11)

$$\langle D\tilde{x},\tilde{x}-x_{n_{k}}\rangle\to\langle D\tilde {x}, \tilde {x}-p\rangle\leq0, $$

so we have

$$\lim_{k\to\infty} \biggl\langle \frac{\alpha_{n_{k}}}{(1-\alpha_{n_{k}})\mu _{n_{k}}}(I-S)\tilde{x}+D \tilde{x},\tilde{x}-x_{{n_{k}}} \biggr\rangle \leq0, $$

and the claim is proved. □

 □

Before we show some applications, we would like to focus on some open questions.

Open Question 1

Since $F\cap\operatorname{Fix}(S)\subset\bar {\Sigma}$, we conjecture that the solution of (2.9) is a solution of (2.11) too, i.e. if $F\cap\operatorname {Fix}(S)\neq\emptyset$, $\bar{x}$ of Theorem 2.8 coincides with $\tilde{x}$ of Theorem 2.14.

Open Question 2

As we have seen in the above, Proposition 2.11, the existence of solutions of the variational inequality problem

$$\bigl\langle (I-S)x,y-x \bigr\rangle \geq0, \quad\forall y\in C, $$

implies the boundedness of the sequence generated by

$$x_{n+1}=P_{C} \biggl(I-\alpha_{n} \biggl((I-S)+ \frac{(1-\alpha_{n})\mu _{n}}{\alpha _{n}}D \biggr) \biggr)x_{n}. $$

By Proposition 2.1, if $\operatorname{Fix}(S)\cap F\neq \emptyset$, our method

$$x_{n+1}=W_{n} \bigl(\alpha_{n}Sx_{n}+(1- \alpha_{n}) (I-\mu_{n}D)x_{n} \bigr) $$

is bounded. We do not know if the existence of solutions of

$$\bigl\langle (I-S)x,y-x \bigr\rangle \geq0, \quad\forall y\in F, $$

implies the boundedness of the sequence generated by

$$x_{n+1}=W_{n} \bigl(\alpha_{n}Sx_{n}+(1- \alpha_{n}) (I-\mu_{n}D)x_{n} \bigr) $$

(i.e., in general, when $W_{n}$ replaces $P_{C}$).

3 Applications

Let $f(x)$ and $g(x)$ be functionals convex and Fréchet differentiable. Let ∇f be $L_{f}$-lipschitzian and let ∇g be $\sigma_{g}$-strongly monotone and $L_{g}$-lipschitzian. Let us consider

$$\min_{C} \bigl(f(x)+\varepsilon g(x) \bigr), $$

where $\varepsilon>0$ is given and C is a closed and convex subset of H. Without loss of generality we can suppose that $C=\bigcap_{n\in \mathbb {N}}\operatorname{Fix}(W_{n})$ with $(W_{n})_{n\in\mathbb{N}}$ is an opportune nonexpansive mapping, We have the following.

Theorem 3.1

Pick two sequences such that $(\mu_{n})_{n\in\mathbb{N}}\subset (0,\frac {2\sigma_{g}}{L_{g}^{2}})$ and

$$\lim_{n\to\infty}\frac{\alpha_{n}}{\mu_{n}}=\frac{1}{\varepsilon}, $$

where $\mu_{n}\to0$, as $n\to\infty$, and

• (H1) $\sum_{n=1}^{\infty}\mu_{n}=\infty$ and $|\mu _{n}-\mu_{n-1}|=o(\mu_{n})$;

• (H2) $|\alpha_{n}-\alpha_{n-1}|=o(\mu_{n})$;

• (H3) $\sup_{z\in B}\|W_{n}z-W_{n-1}z\|=o(\mu_{n})$, with $B\subset H$ bounded.

Then $(x_{n})_{n\in\mathbb{N}}$ generated by

$$x_{n+1}=W_{n} \biggl(\alpha_{n} \biggl(I- \frac{1}{L_{f}}\nabla f \biggr) (x_{n})+(1-\alpha_{n}) \biggl(I-\frac{\mu_{n}}{L_{f}}\nabla g \biggr) (x_{n}) \biggr) $$

strongly converges to $x^{*}$, that is, the unique solution of the variational inequality problem

$$ \bigl\langle \nabla f(x)+\varepsilon\nabla g(x), y-x \bigr\rangle \geq0, \quad\forall y\in C. $$

(3.1)

Proof

The proof follows by Theorem 2.8 since $(I-\frac{1}{L_{f}}\nabla f)$ is nonexpansive and $(\frac{1}{L_{f}}\nabla g)$ is a strongly monotone and lipschitzian operator. □

Choosing $\mu_{n}=\frac{1}{n}$ we immediately obtain the following.

Corollary 3.2

The sequence generated by

$$x_{n+1}=W_{n} \biggl(I-\frac{1}{nL_{f}} \biggl(\nabla f(x_{n})+ \biggl(1-\frac {1}{n} \biggr)\frac{\nabla g(x_{n})}{\varepsilon} \biggr) \biggr) $$

strongly converges to $x^{*}$, that is, the unique solution of the variational inequality problem

$$ \bigl\langle \nabla f(x)+\varepsilon\nabla g(x), y-x \bigr\rangle \geq0, \quad\forall y\in C. $$

(3.2)

Following [21], let $f(x)=\frac{1}{2}\|Ax-b\|^{2}$ where A is a linear and bounded operator and $b\in H$. Let $g(x)=\frac{1}{2}\|x\| ^{2}$. The next corollary easily follows.

Corollary 3.3

The $(x_{n})_{n\in\mathbb{N}}$ generated by

$$x_{n+1}=W_{n} \biggl(I-\frac{1}{n\|A\|^{2}} \biggl(A^{*}Ax_{n}+A^{*}b+ \biggl(1-\frac {1}{n} \biggr) \frac{x_{n}}{\varepsilon} \biggr) \biggr), $$

strongly converges to $x^{*}$, that is, the unique solution of the variational inequality problem

$$ \bigl\langle A^{*}Ax+A^{*}b+\varepsilon x, y-x \bigr\rangle \geq0,\quad \forall y\in C, $$

(3.3)

i.e. $x^{*}$ is the unique solution of

$$\min_{C} \frac{1}{2}\|Ax-b\|^{2}+ \frac{1}{2}\varepsilon\|x\|^{2}. $$

Let us consider a least absolute shrinkage and selection operator, called briefly the lasso problem. Let $H=\mathbb{R}^{n}$; the lasso problem is the minimization problem defined as

$$\min_{C}\frac{1}{2}\|Ax-b\|_{2}^{2}+ \frac{1}{2}\|x\|_{1}, $$

where A is a $m\times n$ matrix, $x\in\mathbb{R}^{n}$, $b\in\mathbb {R}^{m}$ [22]. We consider a lasso problem with solutions. This ill-posed problem can be regularized as

$$\min_{\mathbb{R}^{n}}\frac{1}{2}\|Ax-b\|_{2}^{2}+ \gamma\|x\|_{1}+\frac {1}{2}\varepsilon\|x\|_{2}^{2}+ \delta_{C}(x). $$

This regularization, called an elastic net, is studied in [23].

Taking in account Example 1.3 the proximal operator of $\| \cdot\|_{1}$ on $\mathbb{R}^{n}$ is defined as

$$\operatorname{prox}_{\gamma\|\cdot\|_{1}}(x):=\mathop{\operatorname{argmin}}\limits_{v\in\mathbb {R}^{n}} \biggl\{ \gamma\|x\| _{1}+\frac{1}{2}\|x-v\|^{2} \biggr\} . $$

In [22] the author proved the following.

Proposition 3.4

[22]

If g is a convex and Fréchet differentiable functional on H, a point $x^{*}$ is a solution of the lasso problem if and only if

$$x^{*}=\operatorname{prox}_{\lambda f}(I-\lambda\nabla g)x^{*}. $$

Thus, by Theorem 2.8, we have the following.

Theorem 3.5

Pick two sequences such that

$$\lim_{n\to\infty}\frac{\alpha_{n}}{\mu_{n}}=0 $$

and $\mu_{n}\to0$, as $n\to\infty$. Moreover, suppose that

• (H1) $\sum_{n=1}^{\infty}\mu_{n}=\infty$ and $|\mu _{n}-\mu_{n-1}|=o(\mu_{n})$;

• (H2) $|\alpha_{n}-\alpha_{n-1}|=o(\mu_{n})$.

Then $(x_{n})_{n\in\mathbb{N}}$ generated by

$$x_{n+1}=P_{C} \bigl(\alpha_{n} \operatorname{prox}_{\gamma\|\cdot\| _{1}} \bigl(I-A^{*}A+A^{*}b \bigr)x_{n}+(1- \alpha _{n}) (1- \mu_{n})x_{n} \bigr) $$

strongly converges to $x^{*}\in C$, that is, the unique solution of

$$\langle x,y-x\rangle\geq0,\quad \forall y\in\operatorname {Fix} \bigl( \operatorname{prox}_{\gamma\|\cdot \|_{1}} \bigl(I-A^{*}A+A^{*}b \bigr) \bigr)\cap C, $$

i.e. the solution of the lasso problem with minimum $\|\cdot \| _{2}$-norm solution.

Proof

It is enough to choose $S=\operatorname{prox}_{\gamma\|\cdot\| _{1}}(I-A^{*}A+A^{*}b)$, $P_{C}$. □

By Theorem 2.12, one can prove the following.

Theorem 3.6

Pick $u\in H$. Let $\mu_{n}=\frac{1}{n}$ and $\alpha_{n}=\alpha>0$. Let $(W_{n})_{n\in \mathbb{N}}$ such that $\sup_{z\in B}\|W_{n}z-W_{n-1}z\|=o(\frac{1}{n})$, with $B\subset H$ bounded. Then $(x_{n})_{n\in\mathbb{N}}$ generated by

$$x_{n+1}=W_{n}\bigl(\alpha\operatorname{prox}_{\gamma\|\cdot\| _{1}} \bigl(I-A^{*}A+A^{*}b \bigr)x_{n}+(1-\alpha ) \bigl(\mu_{n}u+(1- \mu_{n})x_{n} \bigr)\bigr) $$

strongly converges to $x^{*}$, that is, the unique solution of the variational inequality problem

$$ \langle x-u, y-x\rangle\geq0, \quad\forall y\in F\cap \operatorname{Fix} \bigl(\operatorname{prox}_{\gamma\|\cdot\|_{1}} \bigl(I-A^{*}A+A^{*}b \bigr) \bigr), $$

(3.4)

i.e. the solution of the lasso problem nearest to u.