Convergence of Inexact Iterates of an Algorithm Based on Unions of Nonexpansive Mappings in Metric Spaces

Abstract

In our recent paper we obtained an extension of a theorem of Tam (2018), which was established for iterates of a set-valued paracontracting operators in a finite-dimensional Euclidean space. This extension was obtained for exact iterates of maps in a metric space. In the present paper we prove two analogs of this result for inexact iterates under the presence of computational errors. In the first result, the errors are summable, while for the second one they converge to zero. In a particular case of a Banach space, if all the operators are symmetric, then the set of all limit points of iterates is symmetric too.

1. Introduction

The study of the fixed point theory of nonlinear mappings has been a rapidly growing area of research. See, for example, [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23], and the references mentioned therein. The origin of this research activity is Banach’s classical work [24]. Since that seminal paper, many important and significant results have been obtained in this area, which have applications in engineering, medical and the natural sciences [22,23,25,26,27,28,29]. In particular, in [30], a new framework for the study of an iterative algorithm was introduced. This algorithm is presented in terms of a structured set-valued operator. More precisely, we assume that at each point in the ambient space the value of the mapping is a set of values of a finite collection of single-valued paracontracting operators. For this algorithm, a convergence of iterates was shown, which is a generalization of a convergent result proved in [31].

In our recent paper [32], we obtained an extension of a theorem of Tam [30], which was established for iterates of set-valued paracontracting operators in a finite-dimensional Euclidean space. This extension was obtained for exact iterates of maps in a metric space. In the present paper, we prove two analogs of this result for inexact iterates under the presence of computational errors. In the first result, the errors are summable, while for the second one they just converge to zero. In a particular case of a Banach space, if all the operators are symmetric, then the set of all limit points of iterates is also symmetric. It should be mentioned that a convex feasibility problem, which has applications in the computer tomography [22,23,25,26,27,28,29], is a particular case of the problem considered in the paper.

2. Preliminaries

Assume that $(\mathcal{X},\rho )$ is a metric space and that a set $\mathcal{C}\subset \mathcal{X}$ is nonempty and closed. For each $\xi \in \mathcal{X}$ and each $\Delta >0$ put

$$B(\xi ,\Delta )=\{\eta \in \mathcal{X}:\rho (\xi ,\eta )\le \Delta \}$$

and for each $\xi \in \mathcal{C}$ and each nonempty subset $D\subset \mathcal{C}$ set

$$\rho (\xi ,D)=sup\left\{\rho \right(\xi ,\eta ):\eta \in D\}.$$

For each map $A:\mathcal{C}\to \mathcal{C}$ set

$$\mathrm{Fix}\left(A\right)=\{z\in \mathcal{C}:A(z)=z\}.$$

Assume that $m\ge 1$ is an integer, $T_j:\mathcal{C}\to \mathcal{C}$, $j=1,\cdots ,m$, $\overline{c}\in (0,1]$ and that for each $j\in \{1,\cdots ,m\}$, each $\xi \in Fix\left(T_j\right)$ and each $\eta \in \mathcal{C}$, we have

$$\rho {(\xi ,\eta )}^2-\rho {(\xi ,T_j\left(\eta \right))}^2\ge \overline{c}\rho {(\eta ,T_j\left(\eta \right))}^2.$$

(1)

It should be mentioned that Equation (1) is true for many nonlinear mappings [22,23].

Assume that for each $\xi \in \mathcal{X}$,

$$\Phi \left(\xi \right)\subset \{1,\cdots ,m\}$$

(2)

is nonempty. Set

$$T\left(\xi \right)=\{T_j\left(\xi \right):j\in \Phi \left(\xi \right)\}$$

(3)

for every $\xi \in \mathcal{C}$,

$$\overline{F}\left(T\right)=\{\xi \in \mathcal{C}:T_j\left(\xi \right)=\xi ,j=1,\cdots ,m\}$$

(4)

and

$$F\left(T\right)=\{\xi \in \mathcal{C}:\xi \in T(\xi \left)\right\}.$$

(5)

We assume that

$$\overline{F}\left(T\right)\ne \varnothing$$

(6)

and fix

$$\theta \in C.$$

Denote by Card $\left(B\right)$ the cardinality of a set B. For every $\xi \in R^1$ put

$$\lfloor \xi \rfloor =inf\{j:j \mathrm{i}\mathrm{s} \mathrm{a}\mathrm{n} \mathrm{i}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{g}\mathrm{e}\mathrm{r} \mathrm{a}\mathrm{n}\mathrm{d} j\le \xi \}.$$

3. The First Result

In this section, we obtain the following theorem showing that almost all inexact iterates of our set-valued mappings with summable computational errors are approximated solutions of the corresponding fixed point problem. For other results of this type see [22,23].

Theorem 1.

Let M be a positive number, $ϵ\in (0,1]$,

$$\overline{F}\left(T\right)\cap B(\theta ,M)\ne \varnothing$$

(7)

and let a sequence ${\left\{r_i\right\}}_{i=0}^{\infty }\subset (0,1]$ satisfy

$$\sum _{i=0}^{\infty }{r}_i<\infty .$$

(8)

Assume that ${\left\{x_i\right\}}_{i=0}^{\infty }\subset \mathcal{C}$ satisfies

$$\rho (\theta ,x_0)\le M$$

(9)

and

$$inf\{\rho (x_{p+1},y):y\in T\left(x_p\right)\}\le r_p,p=0,1,\cdots ,$$

(10)

$Q_1\ge 1$ is an integer such that

$$r_j\le ϵ/2forallnaturalnumbersj\ge Q_1$$

(11)

and that

$$Q_0=\lfloor 4{ϵ}^{-2}(4M^2+6\sum _{j=0}^{\infty }{r}_j(2M+\sum _{j=0}^{\infty }{r}_j))\rfloor +1.$$

(12)

Then,

$$Card\left(\{j\in \{0,1,\cdots ,\}:\rho (x_j,x_{j+1})\ge ϵ/2\}\right)\le Q_0,$$

if an integer $p\ge Q_1$ and $\rho (x_p,x_{p+1})\le ϵ/2$, then

$$min\{\rho (x_p,T_s\left(x_p\right)):s\in \Phi \left(x_p\right)\}\le ϵ$$

and

$$Card\left(\{p\in \{0,1,\cdots ,\}:min\{\rho (x_p,T_s\left(x_p\right)):s\in \Phi \left(x_p\right)\}>ϵ\}\right)\le Q_0+Q_1.$$

Proof. 

By (7), there exists

$$z\in \overline{F}\left(T\right)\cap B(\theta ,M).$$

(13)

By (13),

$$z=T_j\left(z\right),j=1,\cdots ,m.$$

(14)

It follows from (9) and (13) that

$$\rho (x_0,z)\le 2M.$$

(15)

Assume that $i\ge 0$ is an integer. It follows from (3) and (10) that there exists

$$j\left(i\right)\in \Phi \left(x_i\right)$$

(16)

such that

$$\rho (x_{i+1},T_{j\left(i\right)}\left(x_i\right))\le r_i.$$

(17)

Equations (1) and (14) implies that

$$\rho (z,T_{j\left(i\right)}\left(x_i\right))\le \rho (z,x_i).$$

(18)

In view of (17) and (18),

$$\rho (x_{i+1},z)\le \rho (x_{i+1},T_{j\left(i\right)}\left(x_i\right))+\rho (T_{j\left(i\right)}\left(x_i\right),z)\le r_i+\rho (z,x_i).$$

(19)

By (15) and (19), for every natural number j,

$$\rho (x_j,z)\le \rho (z,x_0)+\sum _{k=0}^{j-1}{r}_k\le 2M+\sum _{k=0}^{\infty }{r}_k.$$

(20)

By (1) and (14),

$$\rho {(z,x_i)}^2\ge \rho {(T_{j\left(i\right)}\left(x_i\right),z)}^2+\overline{c}\rho {(T_{j\left(i\right)}\left(x_i\right),x_i)}^2.$$

(21)

By (17), (18) and (20),

$$|\rho {(x_{i+1},z)}^2-\rho {(T_{j\left(i\right)}\left(x_i\right),z)}^2|$$

$$\le (\rho (x_{i+1},z)+\rho (T_{j\left(i\right)}\left(x_i\right),z))|\rho (x_{i+1},z)-\rho (T_{j\left(i\right)}\left(x_i\right),z)|$$

$$\le (\rho (x_{i+1},z)+\rho (x_i,z))\rho (x_{i+1},T_{j\left(i\right)}\left(x_i\right))\le 2r_i(2M+\sum _{k=0}^{\infty }{r}_k).$$

(22)

By (17), (18) and (20),

$$|\rho {(x_{i+1},x_i)}^2-\rho {(T_{j\left(i\right)}\left(x_i\right),x_i)}^2|$$

$$\le (\rho (x_{i+1},x_i)+\rho (T_{j\left(i\right)}\left(x_i\right),x_i))|\rho (x_{i+1},x_i)-\rho (T_{j\left(i\right)}\left(x_i\right),x_i)|$$

$$\le (\rho (x_{i+1},z)+2\rho (x_i,z)+\rho (z,T_{j\left(i\right)}\left(x_i\right)))\rho (x_{i+1},T_{j\left(i\right)}\left(x_i\right))$$

$$\le r_i(3\rho (x_i,z)+\rho (x_{i+1},z))\le 4r_i(2M+\sum _{k=0}^{\infty }{r}_k).$$

(23)

In view of (21)–(23),

$$\rho {(z,x_i)}^2\ge \rho {(T_{j\left(i\right)}\left(x_i\right),z)}^2+\overline{c}\rho {(T_{j\left(i\right)}\left(x_i\right),x_i)}^2$$

$$\ge \rho {(x_{i+1},z)}^2-2r_i(2M+\sum _{k=0}^{\infty }{r}_k)+\overline{c}\rho {(x_{i+1},x_i)}^2-4r_i(2M+\sum _{k=0}^{\infty }{r}_k).$$

(24)

Let q be a natural number. Equations (9), (13) and (24) imply that that

$$4M^2\ge {(M+\rho (z,\theta ))}^2\ge {(\rho (z,\theta )+\rho (\theta ,x_0))}^2$$

$$\ge \rho {(z,z_0)}^2\ge \rho {(z,x_0)}^2-\rho {(z,x_q)}^2$$

$$=\sum _{i=0}^{q-1}(\rho {(z,x_i)}^2-\rho {(z,x_{i+1})}^2)$$

$$\ge \overline{c}\sum _{i=0}^{q-1}\rho {(x_i,x_{i+1})}^2-6(2M+\sum _{j=0}^{\infty }{r}_j)\sum _{j=0}^{\infty }{r}_j,$$

$$4M^2+6(2M+\sum _{j=0}^{\infty }{r}_j)\sum _{j=0}^{\infty }{r}_j$$

$$\ge \overline{c}\sum _{i=0}^{q-1}\rho {(x_i,x_{i+1})}^2$$

$$\ge \overline{c}Card\left(\{i\in \{0,\cdots ,q-1\}:\rho (x_i,x_{i+1})\ge ϵ/2\}\right){ϵ}^2/4$$

and

$$Card\left(\{i\in \{0,\cdots ,q-1\}:\rho (x_i,x_{i+1})\ge ϵ/2\}\right)\le 4{ϵ}^{-2}(4M^2+6\sum _{j=0}^{\infty }{r}_j(2M+\sum _{j=0}^{\infty }{r}_j)).$$

By the equation above and (12),

$$Card\left(\{i\in \{0,1,\cdots \}:\rho (x_i,x_{i+1})\ge ϵ/2\}\right)\le Q_0.$$

(25)

Assume that $i\ge Q_1$ is an integer and

$$\rho (x_i,x_{i+1})\le ϵ/2.$$

(26)

By (11), (16), (17) and (26) that

$$\rho (T_{j\left(i\right)}\left(x_i\right),x_i)\le \rho (x_{i+1},x_i)+\rho (T_{j\left(i\right)}\left(x_i\right),x_{i+1})\le r_i+ϵ/2\le ϵ$$

and

$$min\{\rho (x_i,T_j\left(x_i\right)):j\in \varphi \left(x_i\right)\}\le ϵ.$$

Together with (25), this completes the proof of Theorem 1. □

4. The Second Result

In our second main result, we obtain an extension of Theorem 1 under the presence of computational errors tends to zero but not necessarily summable.

Theorem 2.

Assume that ${\left\{x_i\right\}}_{i=0}^{\infty }\subset C$ satisfies

$$\underset{t\to \infty }{lim\; inf}\rho (\theta ,x_t)<\infty$$

(27)

and

$$\underset{t\to \infty }{lim}inf\{\rho (x_{t+1},y):y\in T\left(x_t\right)\}=0.$$

(28)

Then there exists a strictly increasing sequence of natural numbers ${\left\{t_k\right\}}_{k=1}^{\infty }$ such that

$$\underset{k\to \infty }{lim}\rho (x_{t_k},x_{t_k+1})=0$$

(29)

and

$$\underset{k\to \infty }{lim}inf\{\rho (x_{t_k},\xi ):\xi \in T\left(x_{t_k}\right)\}=0.$$

Proof. 

By (27), there exists

$$M>\underset{t\to \infty }{lim\; inf}\rho (\theta ,x_t).$$

(30)

In view of (28), in order to prove the theorem it is sufficient to show the existence of a strictly increasing sequence of natural numbers ${\left\{t_k\right\}}_{k=1}^{\infty }$ such that (29) holds.

Assume the contrary. Then by (30), there is a real number $ϵ\in (0,1)$ and an integer ${\tau }_0\ge 1$ such that

$$\rho (x_{{\tau }_0},\theta )\le M$$

(31)

and that for every integer $t\ge {\tau }_0$,

$$\rho (x_t,x_{t+1})\ge ϵ.$$

(32)

Fix

$$z\in \overline{F}\left(T\right)$$

(33)

and $\delta >0$ such that

$$\delta (1+\rho (\theta ,z)+M)\le \overline{c}{ϵ}^2/16.$$

(34)

By (28) and (30), there exists an integer ${\tau }_1\ge {\tau }_0$ for which

$$\rho (x_{{\tau }_1},\theta )\le M$$

(35)

and that for every integer $t\ge {\tau }_1$,

$$inf\{\rho (x_{t+1},\xi ):\xi \in T\left(x_t\right)\}\le \delta .$$

(36)

In view of (35),

$$\rho (x_{{\tau }_1},z)\le \rho (x_{{\tau }_1},\theta )+\rho (\theta ,z)\le \rho (\theta ,z)+M.$$

(37)

Let $t\ge {\tau }_1$ be an integer. By (3), there exists $j\left(t\right)\in \varphi \left(x_t\right)$ such that

$$\rho (x_{t+1},T_{j\left(t\right)}\left(x_t\right))=inf\{\rho (x_{t+1},\xi ):\xi \in T\left(x_t\right)\}.$$

(38)

It follows from (1), (4), (32), (33), (36) and (38) that

$$\rho {(z,x_t)}^2\ge \rho {(T_{j\left(t\right)}\left(x_t\right),z)}^2+\overline{c}\rho {(T_{j\left(t\right)}\left(x_t\right),x_t)}^2$$

$$\ge \rho {(T_{j\left(t\right)}\left(x_t\right),z)}^2+\overline{c}{(\rho (x_t,x_{t+1})-\rho (T_{j\left(t\right)}\left(x_t\right),x_{t+1}))}^2$$

$$\ge \rho {(T_{j\left(t\right)}\left(x_t\right),z)}^2+\overline{c}{ϵ}^2/4.$$

(39)

Equations (1), (4), (33), (35) and (38) imply that

$$|\rho {(x_{t+1},z)}^2-\rho {(T_{j\left(t\right)}\left(x_t\right),z)}^2|$$

$$\le (\rho (x_{t+1},z)+\rho (T_{j\left(t\right)}\left(x_t\right),z))|\rho (x_{t+1},z)-\rho (T_{j\left(t\right)}\left(x_t\right),z)|$$

$$(2\rho (z,T_{j\left(t\right)}\left(x_t\right))+\rho (T_{j\left(t\right)}\left(x_t\right),x_{t+1}))\rho (x_{t+1},T_{j\left(t\right)}\left(x_t\right))$$

$$\le \delta (2\rho (x_t,z)+\delta ).$$

(40)

By (39) and (40),

$$\rho {(z,x_t)}^2\ge \rho {(x_{t+1},z)}^2-\delta (2\rho (x_t,z)+\delta )+\overline{c}{ϵ}^2/4.$$

(41)

Assume that $t\ge {\tau }_1$ is an integer and that

$$\rho (x_t,z)\le \rho (\theta ,z)+M.$$

(42)

(Note that in view of (37), Equation (42) holds for $t={\tau }_1$.) By (34), (41) and (42),

$$\rho {(x_{t+1},z)}^2\le \rho {(x_t,z)}^2+2\delta (\rho (\theta ,z)+M+\delta )-\overline{c}{ϵ}^2/4$$

$$\le \rho {(x_t,z)}^2+2\delta (\rho (\theta ,z)+M+1)-\overline{c}{ϵ}^2/4$$

$$\le \rho {(x_t,z)}^2-\overline{c}{ϵ}^2/8.$$

(43)

In view of (42) and (43),

$$\rho (x_{t+1},z)\le \rho (x_t,z)\le \rho (\theta ,z)+M.$$

Thus, by induction we showed that for every integer $t\ge {\tau }_1$,

$$\rho (x_t,z)\le \rho (\theta ,z)+M$$

and

$$\rho {(x_{t+1},z)}^2\le \rho {(x_t,z)}^2-\overline{c}{ϵ}^2/8$$

and $\rho {(x_{t+1},z)}^2\to -\infty$ as $t\to \infty$. The contradiction we have reached completes the proof of Theorem 1. □

5. Examples

An interesting example of the problem studied in [30] and here was considered in Section 3 of [30]. This finite-dimensional example is related to a sparsity constrained minimization. Another important example of the problem considered here is obtained when C is a Hilbert space X, for each $i\in \{1,\cdots ,m\}$, $T_i$ is a projector of X on a nonempty, closed, convex set $C_i$. This is a well-known convex feasibility problem [22,23,25,26,27].