Approximate Common Solutions for a Family of Inverse Strongly Monotone Mappings

Abstract

In 2003 W. Takahashi and M. Toyoda showed the weak convergence of an iteration process of finding the solution of a variational inequality problem for an inverse strongly monotone mapping. In the present paper, we show that for the same process, most of its iterates are approximate common solutions for a finite family of variational inequalities induced by inverse strongly monotone mappings.

1. Introduction

In ref. [1], W. Takahashi snd M. Toyoda introduced an iteration process for finding a common element of the set of fixed points of a nonexpansive mapping and the set of solutions of a variational inequality problem for an inverse strongly monotone mapping, and they showed its weak convergence. Such convergence results are of interest from a theoretical point of view but in practice only, a finite number of iterates can be carried out. Therefore, it is important to show that the iteration process generates an approximate solution and to obtain an estimation for a number of iterates that is sufficient to obtain such an approximate solution. This is the goal of the current paper for the iteration process of ref. [1]. As a matter of fact, we will show that for the iteration process, most of its iterates are approximate common solutions of a finite family of variational inequalities induced by inverse strongly monotone mappings.

Assume that $(H,\langle ·,·\rangle )$ is a real Hilbert space equipped with an inner product that induces the Euclidean norm

$$\parallel x\parallel ={\langle x,x\rangle }^{1/2},x\in H,$$

where $K\subset H$ is a nonempty, convex, closed set, $\alpha \in (0,1)$, and $A:K\to H$ satisfies, for each $u,v\in K$,

$$\langle A\left(u\right)-A\left(v\right),u-v\rangle \ge {\alpha \parallel A\left(u\right)-A\left(v\right)\parallel }^2.$$

(1)

In other words, A is $\alpha$-inverse strongly monotone [1,2,3].

For each $x\in H$ and each $r>0$, set

$$B(x,r)=\{y\in H:\parallel x-y\parallel \le r\}.$$

Let $I:H\to H$ denote the identity self-mapping of H: $I\left(x\right)=x$, $x\in H$.

Set

$$VI(K,A)=\{z\in K:\langle A(z),u-z\rangle \ge 0,u\in K\}.$$

(2)

The following examples of inverse strongly monotone mappings are given in ref. [1]. If $A=I-T$, where T is a nonexpansive mapping of K into itself, then A is (1/2) inverse strongly monotone, and $VI(K,A)$ is the set of fixed points of T.

A mapping $A:K\to H$ is called strongly monotone if there exists a positive real number $\eta$ such that

$$\langle A\left(u\right)-A\left(v\right),u-v\rangle \ge {\eta \parallel u-v\parallel }^2.$$

Then, we say that A is $\eta$ strongly monotone. If A is $\eta$ strongly monotone and $\kappa$ is Lipschitz continuous, then A is $(\eta /{\kappa }^2)$ inverse strongly monotone. There are many mappings that are inverse strongly monotone but not strongly monotone [1]. In fact, let H be real numbers, let $K=[0,1]$, and define $A:K\to H$ by

$$A\left(x\right)=x^2(1+x),0\le x\le 1.$$

It was shown in [1] that A is inverse strongly monotone but not strongly monotone.

The next result is well known in the literature [4,5,6].

Lemma 1.

Let D be a nonempty closed convex subset of H. Then, for each $x\in H$, there is a unique point $P_D\left(x\right)\in D$ satisfying

$$\parallel x-P_D\left(x\right)\parallel =inf\{\parallel x-y\parallel :y\in D\}.$$

Moreover, the following assertions hold:

1.

$$\parallel P_D\left(x\right)-P_D\left(y\right)\left|\right|\le \parallel x-y\parallel forallx,y\in H$$

and for each $x\in H$ and each $z\in D$,

$$\langle z-P_D\left(x\right),x-P_D\left(x\right)\rangle \le 0,$$

$$\parallel z-P_D{\left(x\right)\parallel }^2+\parallel x-P_D{\left(x\right)\parallel }^2\le {\parallel z-x\parallel }^2.$$

2. If $x\in H$ and $\xi \in D$ and if for each $z\in D$

$$\langle z-\xi ,x-\xi \rangle \le 0,$$

then $\xi =P_D\left(x\right)$.

3. For each $x,y\in H$,

$$\parallel P_D\left(x\right)-P_D{\left(y\right)\parallel }^2+\parallel (I-P_D)\left(x\right)-(I-P_D)\left(y\right){\parallel }^2\le {\parallel x-y\parallel }^2,$$

$$\parallel P_D\left(x\right)-P_D{\left(y\right)\parallel }^2\le \langle P_D\left(x\right)-P_D\left(y\right),x-y\rangle .$$

Proposition 1.

Let $\lambda >0$, $u\in K$. Then, $u\in VI(K,A)$ if and only if

$$P_K(u-\lambda A\left(u\right))=u.$$

Proof. 

By definition, $u\in VI(K,A)$ if and only if

$$\langle A\left(u\right),z-u\rangle \ge 0 \mathrm{for} \mathrm{each} z\in K.$$

On the other hand, Lemma 1 implies that

$$P_K(u-\lambda A\left(u\right))=u$$

if and only if for each $z\in K$,

$$\langle z-u,u-\lambda A\left(u\right)-u\rangle \le 0.$$

This completes the proof of Proposition 1. □

Let

$$0<a\le b<2\alpha .$$

The following algorithm was studied in ref. [1].

Let ${\left\{{\lambda }_n\right\}}_{n=0}^{\infty }\subset [a,b].$

1. Initialization. Choose $x_0\in H.$

2. Iterative step. For each integer $t\ge 0$ given a current iterate $x_t\in H$, set

$$x_{t+1}=P_K(x_t-{\lambda }_{t}A\left(x_t\right)).$$

It was shown in [1] that the iterates of the algorithm weakly converge to a point of $VI(K,A)$.

2. The Algorithms

Assume that $U\subset H$ is a nonempty convex set, m is a natural number, $K_i\subset U$, where $i=1,\dots ,m$, are nonempty, closed, convex sets, $\alpha \in (0,1)$, and that for each $i\in \{1,\dots ,m\}$,

$$A_i:U\to H$$

and satisfies, for each $u,v\in H$,

$$\langle A_i\left(u\right)-A_i\left(v\right),u-v\rangle \ge \alpha {\parallel A_i\left(u\right)-A_i\left(v\right)\parallel }^2.$$

(3)

Set

$$S=\{z\in {\cap }_{i=1}^m{K}_i:\langle A_i\left(z\right),u-z\rangle \ge 0,u\in K_i,i=1,\dots ,m\}.$$

(4)

Assume that

$$S\ne Ø.$$

(5)

For each $ϵ>0$, set

$$S_{ϵ}={\cap }_{i=1}^m\{z\in U:B(z,ϵ)\cap K_i\ne Ø \mathrm{and}$$

$$\langle A_i\left(z\right),\xi -z\rangle \ge -ϵ\parallel \xi -z\parallel -ϵ \mathrm{for} \mathrm{all} \xi \in K_i\}.$$

Clearly, S is the set of common solutions of the variational inequalities, while $S_{ϵ}$ is the set of common $ϵ$-approximate solutions of variational inequalities.

Let

$$0<a\le b<2\alpha ,$$

(6)

$$\widehat{\Delta }\in (0,m^{-1}].$$

(7)

In this paper, we study two algorithms. The first of them is associated with the Cimmino algorithm [7], while the second one is a method with the remotest set control [8].

Let us consider our first algorithm:

1. Initialization. Choose $x_0\in U.$

2. Iterative step. For each integer $t\ge 0$ given a current iterate $x_t\in H$, choose

$${\left\{{\lambda }_{t,i}\right\}}_{i=1}^m\subset [a,b],{\left\{{\alpha }_{t,i}\right\}}_{i=1}^m\subset [\widehat{\Delta },1]$$

such that

$$\sum _{i=1}^m{\alpha }_{t,i}=1$$

and set

$$x_{t+1}=\sum _{i=1}^m{\alpha }_{t,i}{P}_{K_i}(x_t-{\lambda }_{t,i}{A}_i\left(x_t\right)).$$

Let us consider our second algorithm.

Assume that $M>1$ and that

$$\{u\in S\cap B(0,M):A_i\in B(0,M),i=1,\dots ,m\}\ne Ø.$$

1. Initialization. Choose $x_0\in U.$

2. Iterative step. For each integer $t\ge 0$ given a current iterate $x_t\in H$, choose

$${\left\{{\lambda }_{t,i}\right\}}_{i=1}^m\subset [a,b],m_t\in \{1,\dots ,m\}$$

such that either

$$\parallel A_{m_t}\left(x_t\right)\parallel >M+1$$

or

$$\parallel A_i\left(x_t\right)\parallel \le M+1,i=1,\dots ,m,$$

$$\parallel x_t-P_{K_{m_t}}(x_t-{\lambda }_{t,m_t}{A}_{m_t}\left(x_t\right))\parallel \le \parallel x_t-P_{K_i}(x_t-{\lambda }_{t,i}{A}_i\left(x_t\right))\parallel ,$$

$i=1,\dots ,m$, and set

$$x_{t+1}=P_{K_{m_t}}(x_t-{\lambda }_{t,m_t}{A}_{m_t}\left(x_t\right)).$$

These two algorithms are known in the literature, where they are used in order to solve a convex feasibility problem, in other words, to find a point belonging to the intersection of a finite family of convex, closed sets. In the classical Cimmino method, all ${\alpha }_{t,i}=1/m$.

3. Main Results

In this paper, we prove the following two results.

Theorem 1.

Assume that $ϵ\in (0,1),$ $M>1,$

$$\{u\in S\cap B(0,M):A_i\left(u\right)\in B(0,M),i=1,\dots ,m\}\ne Ø,$$

(8)

$${\left\{{\lambda }_{t,i}\right\}}_{i=1}^m\subset [a,b],t=0,1,\dots ,$$

(9)

${\left\{x_t\right\}}_{t=0}^{\infty }\subset U,$

$$\parallel x_0\parallel \le M$$

(10)

and for each integer $t\ge 0$,

$${\left\{{\alpha }_{t,i}\right\}}_{i=1}^m\subset [\widehat{\Delta },1],\sum _{i=1}^m{\alpha }_{t,i}=1,$$

(11)

$$x_{t+1}=\sum _{i=1}^m{\alpha }_{t,i}{P}_{K_i}(x_t-{\lambda }_{t,i}{A}_i\left(x_t\right)).$$

(12)

Let a positive number ϵ satisfy

$${ϵ}_0\le min\{ϵ,aϵ,2^{-1}{(M+1)}^{-1}ϵ,2^{-1}aϵ\}.$$

(13)

Then,

$$Card\left(\right\{t\in \{0,1,\dots \}:max\{\parallel A_i\left(x_t\right)\parallel :i=1,\dots ,m\}>M+1\})$$

$$\le 4M^2{\widehat{\Delta }}^{-1}{a}^{-1}{(2\alpha -b)}^{-1},$$

$$Card\left(\right\{t\in \{0,1,\dots \}:max\{\parallel x_t-P_{K_i}(x_t-{\lambda }_{t,i}{A}_i\left(x_t\right))\parallel ,i=1,\dots ,m\}>{ϵ}_0\left\}\right)$$

$$\le 32M^2{\widehat{\Delta }}^{-1}\alpha {(2\alpha -b)}^{-1}{ϵ}_0^{-2}{a}^{-1}b.$$

Moreover, if $t\ge 0$ is an integer and for each integer $i\in \{1,\dots ,m\}$,

$$\parallel A_i\left(x_t\right)\parallel \le M+1,\parallel x_t-P_{K_i}(x_t-{\lambda }_{t,i}{A}_i\left(x_t\right))\parallel <{ϵ}_0,$$

then

$$x_t\in S_{ϵ}.$$

Theorem 2.

Assume that $ϵ\in (0,1),$ $M>1,$ (8) holds,

$${\left\{{\lambda }_{t,i}\right\}}_{i=1}^m\subset [a,b],t=0,1,\dots ,$$

(14)

${\left\{x_t\right\}}_{t=0}^{\infty }\subset U,$

$$\parallel x_0\parallel \le M,{\left\{m_t\right\}}_{t=0}^{\infty }\subset \{1,\dots ,m\}$$

(15)

and for each integer $t\ge 0$ either

$$\parallel A_{m_t}\left(x_t\right)\parallel >M+1$$

or

$$\parallel A_i\left(x_t\right)\parallel \le M+1,i=1,\dots ,m,$$

$$\parallel x_t-P_{K_{m_t}}(x_t-{\lambda }_{t,m_t}{A}_{m_t}\left(x_t\right))\parallel \ge \parallel x_t-P_{K_i}(x_t-{\lambda }_{t,i}{A}_i\left(x_t\right))\parallel ,$$

(16)

$i=1,\dots ,m$ and

$$x_{t+1}=P_{K_{m_t}}(x_t-{\lambda }_{t,m_t}{A}_{m_t}\left(x_t\right)).$$

(17)

Let a positive number ${ϵ}_0$ satisfy

$${ϵ}_0\le min\{ϵ,aϵ,2^{-1}{(M+1)}^{-1}ϵ,2^{-1}aϵ\}.$$

Then,

$$Card\left(\right\{t\in \{0,1,\dots \}:max\{\parallel A_i\left(x_t\right)\parallel :i=1,\dots ,m\}>M+1\})$$

$$\le 4M^2{a}^{-1}{(2\alpha -b)}^{-1},$$

$$Card\left(\right\{t\in \{0,1,\dots \}:max\{\parallel A_i\left(x_t\right)\parallel :i=1,\dots ,m\}\le M+1,$$

$$max\{\parallel x_t-P_{K_i}(x_t-{\lambda }_{t,i}{A}_i\left(x_t\right))\parallel ,i=1,\dots ,m\}>{ϵ}_0\left\}\right)$$

$$\le 32M^2{a}^{-1}{(2\alpha -b)}^{-1}\alpha b{ϵ}_0^{-2}.$$

Moreover, if $t\ge 0$ is an integer and

$$\parallel A_{m_t}\left(x_t\right)\parallel \le M+1,\parallel x_t-P_{K_{m_t}}(x_t-{\lambda }_{t,m_t}{A}_{m_t}\left(x_t\right))\parallel \le {ϵ}_0,$$

then

$$x_t\in S_{ϵ}.$$

Theorems 1 and 2 show that the number of iterates that are not $ϵ$-approximate solutions of our problem is bounded by a constant that depends on $a,b,\alpha$, and $ϵ$.

4. Auxiliary Results

Let $(K,A)\in \{(K_i,A_i):i=1,\dots ,m\}.$

Proposition 2.

Let $u,v\in U$ and $\lambda >0$. Then,

$${\parallel (I-\lambda A)\left(u\right)-(I-\lambda A)\left(v\right)\parallel }^2\le {\parallel u-v\parallel }^2+\lambda (\lambda -2\alpha ){\parallel A\left(u\right)-A\left(v\right)\parallel }^2.$$

If $\lambda \le 2\alpha$, then $I-\lambda A$ is nonexpansive.

Proof. 

By (3),

$${\parallel (I-\lambda A)\left(u\right)-(I-\lambda A)\left(v\right)\parallel }^2={\parallel u-v-\lambda (A\left(u\right)-A\left(v\right))\parallel }^2$$

$$={\parallel u-v\parallel }^2-2\lambda \langle u-v,A\left(u\right)-A\left(v\right)\rangle +{\lambda }^2{\parallel A\left(u\right)-A\left(v\right)\parallel }^2$$

$$\le {\parallel u-v\parallel }^2+\lambda (\lambda -2\alpha ){\parallel A\left(u\right)-A\left(v\right)\parallel }^2.$$

Proposition 2 is proved. □

Lemma 2.

Assume that

$$x\in U,\lambda \in [a,b],u\in K,$$

(18)

$$u=P_K(u-\lambda A\left(u\right)),y=P_K(x-\lambda A\left(x\right)).$$

(19)

Then,

$${\parallel y-u\parallel }^2\le {\parallel x-u\parallel }^2+\lambda (\lambda -2\alpha ){\parallel A\left(u\right)-A\left(x\right)\parallel }^2,$$

$${\parallel y-u\parallel }^2\le {\parallel x-u\parallel }^2-{\parallel y-x+\lambda (A\left(x\right)-A\left(u\right))\parallel }^2$$

and

$${\parallel y-u\parallel }^2\le {\parallel x-u\parallel }^2-4^{-1}{\parallel y-x\parallel }^{2}min\{1,b^{-2}a(2\alpha -b)\}.$$

Proof. 

In view of (6) and (18),

$$\lambda \le b<2\alpha .$$

(20)

Proposition 2 implies that for each $v_1,v_2\in U$,

$$\parallel (I-\lambda A)\left(v_1\right)-(I-\lambda A)\left(v_2\right){\parallel }^2 \le \parallel v_1-v_2{\parallel }^2+\lambda (\lambda -2\alpha ){\parallel A\left(v_1\right)-A\left(v_2\right)\parallel }^2.$$

(21)

Lemma 1 and (18) and (19) imply that

$$\parallel y-u\parallel = \parallel P_K(x-\lambda A\left(x\right))-P_K(u-\lambda A\left(u\right))\parallel = \parallel x-\lambda A\left(x\right)-(u-\lambda A\left(u\right))\parallel .$$

(22)

It follows from (18), (20), and (22) that

$${\parallel y-u\parallel }^2\le {\parallel x-u\parallel }^2+\lambda (\lambda -2\alpha ){\parallel A\left(u\right)-A\left(x\right)\parallel }^2.$$

(23)

Lemma 1, Proposition 2, the inequality $b<2\alpha$, and relations (18) and (19) imply that

$${\parallel y-u\parallel }^2={\parallel P_K(x-\lambda A\left(x\right))-P_K(u-\lambda A\left(u\right))\parallel }^2$$

$$\le \langle y-u,(x-\lambda A(x\left)\right)-(u-\lambda A(u\left)\right)\rangle$$

$$=2^{-1}{[\parallel y-u\parallel }^2+{\parallel x-\lambda A\left(x\right)-(u-\lambda A\left(u\right))\parallel }^2$$

$${-\parallel y-u-((x-\lambda A\left(x\right))-(u-\lambda A\left(u\right)))\parallel }^2]$$

$$\le 2^{-1}{[\parallel y-u\parallel }^2{+\parallel x-u\parallel }^2{-\parallel y-x+\lambda (A\left(x\right)-A\left(u\right))\parallel }^2].$$

This implies that

$${\parallel y-u\parallel }^2\le {\parallel x-u\parallel }^2-{\parallel y-x+\lambda (A\left(x\right)-A\left(u\right))\parallel }^2.$$

Together with (23), this implies that

$${\parallel y-u\parallel }^2\le {\parallel x-u\parallel }^2{-max\{a(2\alpha -b)\parallel A\left(u\right)-A\left(x\right)\parallel }^2{,\parallel y-x+\lambda (A\left(x\right)-A\left(u\right))\parallel }^2\}.$$

In view of the Cauchy–Schwartz inequality,

$${\parallel y-x\parallel }^2\le {\parallel y-x+\lambda (A\left(x\right)-A\left(u\right))+\lambda (A\left(u\right)-A\left(x\right))\parallel }^2$$

$$\le {2\parallel y-x+\lambda (A\left(x\right)-A\left(u\right))\parallel }^2+2{\lambda }^2{\parallel A\left(u\right)-A\left(x\right)\parallel }^2$$

$$\le {2\parallel y-x+\lambda (A\left(x\right)-A\left(u\right))\parallel }^2+2b^2{\parallel A\left(u\right)-A\left(x\right)\parallel }^2.$$

Together with the relation above, this implies that

$${\parallel y-u\parallel }^2\le {\parallel x-u\parallel }^2-4^{-1}{\parallel y-x\parallel }^{2}min\{1,b^{-2}a(2\alpha -b)\}.$$

Lemma 2 is proved. □

Lemma 3.

Assume that ${ϵ}_0>0$, $x\in U$, $\lambda \in [a,b]$, and

$$\parallel x-P_K(x-\lambda A\left(x\right))\parallel \le {ϵ}_0.$$

(24)

Then, for each $\xi \in K$,

$$\langle Ax,\xi -x\rangle \ge -{\lambda }^{-1}{ϵ}_0\parallel \xi -x\parallel -{\lambda }^{-1}{ϵ}_0^2-\parallel A\left(x\right)\parallel {ϵ}_0.$$

Proof. 

Lemma 1 implies that for each $\xi \in K$,

$$0\ge \langle x-\lambda A\left(x\right)-P_K(x-\lambda A\left(x\right)),\xi -P_K(x-\lambda A\left(x\right))\rangle .$$

By (24) and the relation above, for each $\xi \in K$,

$$0\ge \langle x-P_K(x-\lambda A\left(x\right)),\xi -P_K(x-\lambda A\left(x\right))\rangle .$$

$$-\lambda \langle A\left(x\right),\xi -P_K(x-\lambda A\left(x\right))\rangle$$

$$\ge -\parallel x-P_K(x-A\left(x\right))\parallel (\parallel \xi -x\parallel +\parallel x-P_K(x-\lambda A\left(x\right))\parallel$$

$$-\lambda \langle A\left(x\right),\xi -x\rangle -\lambda \langle A\left(x\right),x-P_K(x-\lambda A\left(x\right))\rangle$$

$$\ge -{ϵ}_0\parallel \xi -x\parallel -{ϵ}_0^2-\lambda \langle A\left(x\right),\xi -x\rangle -\lambda \parallel A\left(x\right)\parallel {ϵ}_0$$

and

$$\langle A\left(x\right),\xi -x\rangle \ge -{\lambda }^{-1}{ϵ}_0\parallel \xi -x\parallel -{\lambda }^{-1}{ϵ}_0^2-\parallel A\left(x\right)\parallel {ϵ}_0.$$

Lemma 3 is proved. □

5. Proof of Theorem 1

In view of (8), there exists

$$u_{*}\in S\cap B(0,M)$$

(25)

such that

$$\parallel A_i\left(u_{*}\right)\parallel \le M,i=1,\dots ,m.$$

(26)

Let $t\ge 0$ be an integer and $i\in \{1,\dots ,m\}$. Lemma 2 and (9) and (25) imply that

$$\parallel u_{*}-P_{K_i}(x_t-{\lambda }_{t,i}{A}_i\left(x_t\right)){\parallel }^2$$

$$\le \parallel x_t-u_{*}{\parallel }^2+{\lambda }_{t,i}({\lambda }_{t,i}-2\alpha ){\parallel A_i\left(u_{*}\right)-A_i\left(x_t\right)\parallel }^2$$

$$\le \parallel x_t-u_{*}{\parallel }^2-(2\alpha -b)a\parallel A_i\left(u_{*}\right)-A_i\left(x_t\right){)\parallel }^2,$$

(27)

$$\parallel u_{*}-P_{K_i}(x_t-{\lambda }_{t,i}{A}_i\left(x_t\right)){\parallel }^2$$

$$\le \parallel x_t-u_{*}{\parallel }^2-8^{-1}{b}^{-1}a{\alpha }^{-1}(2\alpha -b){\parallel P_{K_i}(x_t-{\lambda }_{t,i}{A}_i\left(x_t\right))-x_t\parallel }^2.$$

(28)

It follows from (11) and (12) and the convexity of the function ${\parallel ·\parallel }^2$ that

$$\parallel u_{*}-x_{t+1}{\parallel }^2=\parallel u_{*}-\sum _{i=1}^m{\alpha }_{t,i}{P}_{K_i}(x_t-{\lambda }_{t,i}{A}_i\left(x_t\right)){\parallel }^2$$

$$\le \sum _{i=1}^m{\alpha }_{t,i}{\parallel u_{*}-P_{K_i}(x_t-{\lambda }_{t,i}{A}_i\left(x_t\right))\parallel }^2.$$

(29)

By (29) and Lemma 2,

$$\parallel u_{*}-x_{t+1}{\parallel }^2\le \sum _{i=1}^m{\alpha }_{t,i}(\parallel x_t-u_{*}{\parallel }^2$$

$$-max\{(2\alpha -b)a\parallel A_i\left(u_{*}\right)-A\left(x_t\right){\parallel }^2,{\left(8\alpha b\right)}^{-1}a(2\alpha -b)\parallel P_{K_i}(x_t-{\lambda }_{t,i}{A}_i\left(x_t\right))-x_t{\parallel }^2\})$$

$$\le \parallel u_{*}-x_t{\parallel }^2-\widehat{\Delta }\underset{i=1,\dots ,m}{max}max\{(2\alpha -b)a\parallel A_i\left(u_{*}\right)-A\left(x_t\right){\parallel }^2,$$

$${\left(8\alpha b\right)}^{-1}a(2\alpha -b)\parallel P_{K_i}(x_t-{\lambda }_{t,i}{A}_i\left(x_t\right))-x_t{\parallel }^2\},$$

$$\parallel u_{*}-x_{t+1}{\parallel }^2 \le \parallel u_{*}-x_t{\parallel }^2-\widehat{\Delta }\underset{i=1,\dots ,m}{max}{\{\left(8\alpha b\right)}^{-1}a(2\alpha -b)\parallel P_{K_i}(x_t-{\lambda }_{t,i}{A}_i\left(x_t\right))-x_t{\parallel }^2\},$$

(30)

$$\parallel u_{*}-x_{t+1}\parallel \le \parallel u_{*}-x_t{\parallel }^2-\widehat{\Delta }\underset{i=1,\dots ,m}{max}\{(2\alpha -b)a\parallel A_i\left(u_{*}\right)-A\left(x_t\right){\parallel }^2\}.$$

(31)

Set

$$E_1=\{t\in \{0,1,\dots \}:max\{\parallel P_{K_i}(x_t-{\lambda }_{t,i}{A}_i\left(x_t\right))-x_t\parallel :i=1,\dots ,m\}>{ϵ}_0\},$$

(32)

$$E_2=\{t\in \{0,1,\dots \}:max\{\parallel A_i\left(x_t\right)\parallel :i=1,\dots ,m\}>M+1\}.$$

(33)

Let Q be a natural number. Set

$$E_{1,Q}=E_1\cap \{0,\dots ,Q-1\},$$

(34)

$$E_{2,Q}=E_2\cap \{0,\dots ,Q-1\}.$$

(35)

By (8), (10), (31), (33), and (35),

$$4M^2\ge \parallel x_0-u_{*}{\parallel }^2 \ge \parallel x_0-u_{*}{\parallel }^2-{\parallel x_Q-u_{*}\parallel }^2$$

$$=\sum _{t=0}^{Q-1}(\parallel x_t-u_{*}{\parallel }^2-\parallel x_{t+1}-u_{*}{\parallel }^2)$$

$$\ge \sum \{\parallel x_t-u_{*}{\parallel }^2-\parallel x_{t+1}-u_{*}{\parallel }^2:t\in E_{2,Q}\}$$

$$\ge a(2\alpha -b)\widehat{\Delta }\mathrm{Card}\left(E_{2,Q}\right),$$

$$\mathrm{Card}\left(E_{2,Q}\right)\le 4M^2{\left(a(2\alpha -b)\right)}^{-1}{\widehat{\Delta }}^{-1}.$$

Since Q is an arbitrary natural number, we conclude that

$$\mathrm{Card}\left(E_2\right)\le 4M^2{\left(a(2\alpha -b)\right)}^{-1}{\widehat{\Delta }}^{-1}.$$

By (8), (10), (30), (32), and (34),

$$4M^2\ge \parallel x_0-u_{*}{\parallel }^2 \ge \parallel x_0-u_{*}{\parallel }^2- {\parallel x_Q-u_{*}\parallel }^2$$

$$=\sum _{t=0}^{Q-1}(\parallel x_t-u_{*}{\parallel }^2-\parallel x_{t+1}-u_{*}{\parallel }^2)$$

$$\ge \sum \{\parallel x_t-u_{*}{\parallel }^2-\parallel x_{t+1}-u_{*}{\parallel }^2:t\in E_{1,Q}\}$$

$$\ge {\left(8\alpha b\right)}^{-1}a(2\alpha -b)\widehat{\Delta }\mathrm{Card}\left(E_{1,Q}\right),$$

$$\mathrm{Card}\left(E_{1,Q}\right)\le 32M^2\alpha b{\left(a(2\alpha -b)\right)}^{-1}{\widehat{\Delta }}^{-1}{ϵ}_0^{-2}.$$

Since Q is an arbitrary natural number, we conclude that

$$\mathrm{Card}\left(E_1\right)\le 32M^2\alpha b{\left(a(2\alpha -b)\right)}^{-1}{\widehat{\Delta }}^{-1}{ϵ}_0^{-2}.$$

Let $t\ge 0$ be an integer and for every $i\in \{1,\dots ,m\}$,

$$\parallel A_i\left(x_t\right)\parallel \le M+1,$$

$$\parallel P_{K_i}(x_i-{\lambda }_{t,i}{A}_i\left(x_t\right))-x_t\parallel \le {ϵ}_0.$$

By Lemma 3, (13), and the relations above, for each $i\in \{1,\dots ,m\}$ and each $\xi \in K_i$,

$$\langle A_i{x}_t,\xi -x_t\rangle \ge -{\lambda }_{t,i}^{-1}{ϵ}_0\parallel \xi -x_t\parallel -{\lambda }_{t,i}^{-1}{ϵ}_0^2-\parallel A_i\left(x_t\right)\parallel {ϵ}_0$$

$$\ge -a^{-1}{ϵ}_0\parallel \xi -x_t\parallel -a^{-1}{ϵ}_0^2-(M+1){ϵ}_0$$

$$\ge -ϵ\parallel \xi -x_t\parallel -ϵ.$$

Theorem 1 is proved.

6. Proof of Theorem 2

In view of (8), there exists $u_{*}\in U$ such that (25) and (26) hold. Let $t\ge 0$ be an integer. Lemma 2, the inequality $b<2\alpha$, and (14) and (17) imply that

$$\parallel u_{*}-x_{t+1}{\parallel }^2={\parallel u_{*}-P_{K_{m_t}}(x_t-{\lambda }_{t,m_t}{A}_{m_t}\left(x_t\right))\parallel }^2$$

$$\le \parallel x_t-u_{*}{\parallel }^2-(2\alpha -b)a\parallel A_{m_t}\left(u_{*}\right)-A_{m_t}\left(x_t\right){)\parallel }^2,$$

(36)

$$\parallel u_{*}-x_{t+1}{\parallel }^2={\parallel u_{*}-P_{K_{m_t}}(x_t-{\lambda }_{t,m_t}{A}_{m_t}\left(x_t\right))\parallel }^2$$

$$\le \parallel x_t-u_{*}{\parallel }^2-{\left(8\alpha b\right)}^{-1}a(2\alpha -b){\parallel P_{K_{m_t}}(x_t-{\lambda }_{t,m_t}{A}_{m_t}\left(x_t\right))-x_t\parallel }^2.$$

(37)

Set

$$E_0=\{t\in \{0,1,\dots \}:max\{\parallel A_i\left(x_t\right)\parallel :i=1,\dots ,m\}>M+1\},$$

(38)

$$E_1=\{t\in \{0,1,\dots \}\setminus E_0:\parallel P_{K_{m_t}}(x_t-{\lambda }_{t,m_t}{A}_{m_t}\left(x_t\right))-x_t\parallel >{ϵ}_0\}.$$

(39)

Let Q be a natural number. Set

$$E_{0,Q}=E_0\cap \{0,\dots ,Q-1\},$$

(40)

$$E_{1,Q}=E_1\cap \{0,\dots ,Q-1\}.$$

(41)

By (8), (15), (36), (37), (40), and the definition of $m_t$,

$$4M^2\ge \parallel x_0-u_{*}{\parallel }^2\ge \parallel x_0-u_{*}{\parallel }^2-{\parallel x_Q-u_{*}\parallel }^2$$

$$=\sum _{t=0}^{Q-1}(\parallel x_t-u_{*}{\parallel }^2-\parallel x_{t+1}-u_{*}{\parallel }^2)$$

$$\ge \sum \{\parallel x_t-u_{*}{\parallel }^2-\parallel x_{t+1}-u_{*}{\parallel }^2:t\in E_{0,Q}\}$$

$$\ge a(2\alpha -b)\mathrm{Card}\left(E_{0,Q}\right),$$

$$\mathrm{Card}\left(E_{0,Q}\right)\le 4M^2{\left(a(2\alpha -b)\right)}^{-1}.$$

Since Q is an arbitrary natural number, we conclude that

$$\mathrm{Card}\left(E_0\right)\le 4M^2{\left(a(2\alpha -b)\right)}^{-1}.$$

By (8), (15), (36), (37), (39), and (41),

$$4M^2\ge \parallel x_0-u_{*}{\parallel }^2 \ge \parallel x_0-u_{*}{\parallel }^2-{\parallel x_Q-u_{*}\parallel }^2$$

$$=\sum _{t=0}^{Q-1}(\parallel x_t-u_{*}{\parallel }^2-\parallel x_{t+1}-u_{*}{\parallel }^2)$$

$$\ge \sum \{\parallel x_t-u_{*}{\parallel }^2-\parallel x_{t+1}-u_{*}{\parallel }^2:t\in (\{0,1,\dots \}\setminus E_0)\cap E_{1,Q}\}$$

$$\ge {\left(8\alpha b\right)}^{-1}a(2\alpha -b){ϵ}_0^2\mathrm{Card}\left(E_{1,Q}\right\},$$

$$\mathrm{Card}\left(E_{1,Q}\right)\le 8M^2\alpha b{\left(a(2\alpha -b)\right)}^{-1}{ϵ}_0^{-2}.$$

Since Q is an arbitrary natural number, we conclude that

$$\mathrm{Card}\left(E_1\right)\le 8M^2\alpha b{\left(a(2\alpha -b)\right)}^{-1}{ϵ}_0^{-2}.$$

By (38), (39), the relations above, and the definition of $m_t$,

$$\mathrm{Card}\left(\right\{t\in \{0,1,\dots \}:max\{\parallel A_i\left(x_t\right)\parallel :i=1,\dots ,m\}\le M+1\})=\mathrm{Card}\left(E_0\right)$$

$$\le 4M^2{\left(a(2\alpha -b)\right)}^{-1},$$

$$\mathrm{Card}\left(\right\{t\in \{0,1,\dots \}:max\{\parallel A_i\left(x_t\right)\parallel :i=1,\dots ,m\}\le M+1\},$$

$$max\{\parallel x_t-P_{K_i}(x_t-{\lambda }_{t,i}{A}_i\left(x_t\right))\parallel ,i=1,\dots ,m\}>{ϵ}_0\left\}\right\})$$

$$=\mathrm{Card}\left(E_1\right)\le 32M^2{a}^{-1}{(2\alpha -b)}^{-1}\alpha b{ϵ}_0^{-2}.$$

Let $t\ge 0$ be an integer, and for every $i\in \{1,\dots ,m\}$,

$$\parallel A_i\left(x_t\right)\parallel \le M+1,$$

$$\parallel P_{K_{m_t}}(x_t-{\lambda }_{t,m_t}{A}_{m_t}\left(x_t\right))-x_t\parallel \le {ϵ}_0.$$

Together with (16), this implies that for each $i\in \{1,\dots ,m\}$,

$$\parallel P_{K_i}(x_t-{\lambda }_{t,it}{A}_i\left(x_t\right))-x_t\parallel \le {ϵ}_0.$$

By Lemma 3, the choice of ${ϵ}_0$, and the relations above, for each $i\in \{1,\dots ,m\}$ and each $\xi \in K_i$,

$$\langle A_i{x}_t,\xi -x_t\rangle \ge -{\lambda }_{t,i}^{-1}{ϵ}_0\parallel \xi -x_t\parallel$$

$$-{\lambda }_{t,i}^{-1}{ϵ}_0^2-\parallel A_i\left(x_t\right)\parallel {ϵ}_0$$

$$\ge -a^{-1}{ϵ}_0\parallel \xi -x_t\parallel -a^{-1}{ϵ}_0^2-(M+1){ϵ}_0$$

$$\ge -ϵ\parallel \xi -x_t\parallel -ϵ.$$

Theorem 2 is proved.