Finitely Convergent Deterministic and Stochastic Iterative Methods for Solving Convex Feasibility Problems

TL;DR

This paper introduces new finitely convergent deterministic and stochastic iterative methods for solving convex feasibility problems, ensuring convergence under broad control strategies, including cyclic, random, and remote controls.

Contribution

It presents novel finitely convergent algorithms for convex feasibility problems that work with a wide range of control sequences, including deterministic and stochastic approaches.

Findings

Methods guarantee finite convergence under broad control schemes.

Applicable to infinite constraint sets with nonempty interior.

Convergence holds for deterministic and almost surely for stochastic controls.

Abstract

We propose finitely convergent methods for solving convex feasibility problems defined over a possibly infinite pool of constraints. Following other works in this area, we assume that the interior of the solution set is nonempty and that certain overrelaxation parameters form a divergent series. We combine our methods with a very general class of deterministic control sequences where, roughly speaking, we require that sooner or later we encounter a violated constraint if one exists. This requirement is satisfied, in particular, by the cyclic, repetitive and remotest set controls. Moreover, it is almost surely satisfied for random controls.

Tables

Table 1. Table 1.1: The symbols P f i subscript 𝑃 subscript 𝑓 𝑖 P_{f_{i}} and S ​ ( f i , 0 ) 𝑆 subscript 𝑓 𝑖 0 S(f_{i},0) refer to the subgradient projection and the sublevel set, respectively; see Example 2.4 . A constant overrelaxation parameter r ≤ R 𝑟 𝑅 r\leq R is chosen so that B ​ ( z , 2 ​ R ) ⊂ C 𝐵 𝑧 2 𝑅 𝐶 B(z,2R)\subset C for some z ∈ Q 𝑧 𝑄 z\in Q and applies only to φ i ​ ( x ) = 1 subscript 𝜑 𝑖 𝑥 1 \varphi_{i}(x)=1 . The result of Crombez featuring string averaging is reduced to singleton strings ( [ 17 , n ​ ( t ) = 1 𝑛 𝑡 1 n(t)=1 ] ).

Result	Constraints and Operators	Constraints Qualification	${\phi }_i\left(x\right)$	Overrelaxation and Relaxation	Control Sequence
Iusem, Moledo [24, Theorem 1]	$C_i=S(f_i,0)$ $T_i=P_{f_i}$ $m\in {\mathbb{N}}_{+}$ $Q={ℝ}^n$		${\phi }_i\left(x\right)=\Vert g_i\left(x\right)\Vert$ $g_i\left(x\right)\in \partial f_i\left(x\right)$ for $x\notin C_i$	$r_k\downarrow 0$, $\sum _{k=0}^{\infty }{r}_k=\infty$ ${\alpha }_k\in [\epsilon ,2-\epsilon ]$	$I_k=I$
De Pierro, Iusem [18, Theorem 1]	as in [24]	as in [24]	as in [24]	as in [24]	almost cyclic
Censor, Chen, Pajoohesh [13, Theorem 20]	as in [24]	as in [24]	as in [24]	as in [24]	repetitive (single valued) $+$ [13, Condition 19]
Polyak [29, Theorem 1] $+$ [29, Section 4.2]	$C_i=S(f_i,0)$ $T_i=P_{f_i}$ $m\in {\mathbb{N}}_{+}\cup \left\{\infty \right\}$ $Q\subset {ℝ}^n$	$int\left(C\right)\cap Q\ne \varnothing$	${\phi }_i\left(x\right)=1$	$r_k=r$ or $r_k\to 0$, $\sum _{k=0}^{\infty }{r}_k^2=\infty$ (combined with $\left[k\right]$) ${\alpha }_k=1$	random (single valued)
Crombez [17, Theorem 2.7]	$C_i=Fix{T}_i$ $T_i$ - cutter $m\in {\mathbb{N}}_{+}$ $Q={ℝ}^n$	$int\left(C\right)\ne \varnothing$	${\phi }_i\left(x\right)=1$	$r_k=r$ ${\alpha }_k=1$	$I_k\left(x\right)\subset I_{+}\left(x\right)$
Bauschke, Wang Wang, Xu [7, Theorem 3.1]	$C=FixT$ $T$ - cutter $m=1$ $Q\subset {ℝ}^n$	$int\left(C\cap Q\right)\ne \varnothing$	${\phi }_i\left(x\right)=1$	$r_k\to 0$ $\sum _{k=0}^{\infty }{\alpha }_k{r}_k=\infty$ ${\alpha }_k\in (0,2]$	x
Bauschke, Wang Wang, Xu [7, Theorem 3.2]	as above	$int\left(C\right)\cap Q\ne \varnothing$	as above	$r_k\to 0$ $\sum _{k=0}^{\infty }{\alpha }_k\left(2-{\alpha }_k\right)r_k^2=\infty$ ${\alpha }_k\in (0,2]$	x
Current Paper Theorems 4.7 and 5.4 (see also Ex. 4.9)	$C_i=Fix{T}_i$ $T_i$ - cutter $m\in {\mathbb{N}}_{+}\cup \left\{\infty \right\}$ $Q\subset \mathscr{H}$	$int\left(C\right)\cap Q\ne \varnothing$	$\delta \le {\phi }_i\left(x\right)\le \Delta$ on bounded sets $\delta ,\Delta \in (0,\infty )$ (all above)	$r_k=r$ or $r_k\to 0$, $\sum _{k=0}^{\infty }{\alpha }_k{r}_k=\infty$ (combined with $\left[k\right]$) ${\alpha }_k\in (0,2]$	well matched or random

Equations

$$x_{0}\in Q,\quad x_{k+1}:=P_{Q}\left(x_{k}+\alpha_{k}\sum_{i\in I_{k}(x_{k})}\lambda_{i,k}(x_{k})\Big{(}T_{i}(x_{k})-x_{k}\Big{)}\right),$$

$$\Vert x_{k+1}-z{\Vert }^2\le \Vert x_k-z{\Vert }^2-2MR{r}_{[k]},$$

$$\#(\{k\ge 0:I_k(x)\cap I_{+}(x)\ne \varnothing \})=\infty$$

$$P_f(x):=\{\begin{array}{ll}x-\frac{f(x)}{\Vert g(x){\Vert }^2}g(x),& \text{mbox}iff(x)>0\\ x,& \text{mbox}otherwise\end{array}$$

$$\beta (x):=\{\begin{array}{ll}\frac{\rho (x)+\Vert T(x)-x\Vert }{\Vert T(x)-x\Vert },& \text{mbox}ifT(x)\ne x\\ 0,& \text{mbox}otherwise.\end{array}$$

$$\Vert U(x)-y{\Vert }^2\le \Vert x-y{\Vert }^2-\frac{2-\alpha }{\alpha }\Vert U(x)-x{\Vert }^2.$$

$$w:=y-\rho (x)\frac{T(x)-x}{\Vert T(x)-x\Vert }$$

$$\rho (x)\Vert T(x)-x\Vert +\Vert T(x)-x{\Vert }^2-⟨y-x,T(x)-x⟩=⟨w-T(x),x-T(x)⟩\le 0.$$

$$u:=T(x)-y+\rho (x)\frac{T(x)-x}{\Vert T(x)-x\Vert }\text{and }v:=x-y,$$

$$\Vert U(x)-y{\Vert }^2=\Vert \alpha u+(1-\alpha )v{\Vert }^2=\alpha \Vert u{\Vert }^2+(1-\alpha )\Vert v{\Vert }^2-\alpha (1-\alpha )\Vert u-v{\Vert }^2.$$

$$\Vert u{\Vert }^2$$

$$=\Vert T(x)-y{\Vert }^2+{\rho }^2(x)+2\rho (x)\Vert T(x)-x\Vert -\frac{2\rho (x)}{\Vert T(x)-x\Vert }⟨y-x,T(x)-x⟩$$

$$\le \Vert T(x)-y{\Vert }^2-{\rho }^2(x)$$

$$=\Vert (x-y)+(T(x)-x){\Vert }^2-{\rho }^2(x)$$

$$=\Vert x-y{\Vert }^2-\Vert T(x)-x{\Vert }^2+2(\Vert T(x)-x{\Vert }^2-⟨y-x,T(x)-x⟩)-{\rho }^2(x)$$

$$\le \Vert x-y{\Vert }^2-\Vert T(x)-x{\Vert }^2-2\rho (x)\Vert T(x)-x\Vert -{\rho }^2(x)$$

$$=\Vert x-y{\Vert }^2-(\rho (x)+\Vert T(x)-x\Vert {)}^2.$$

$$\alpha \Vert u{\Vert }^2\le \alpha \Vert x-y{\Vert }^2-\frac{1}{\alpha }\Vert U(x)-x{\Vert }^2.$$

$$\alpha (1-\alpha )\Vert u-v{\Vert }^2=\frac{1-\alpha }{\alpha }\Vert U(x)-x{\Vert }^2.$$

$$V(x):=x+\alpha \sum _{j\in J(x)}{\lambda }_j(x){\beta }_j(x)(T_j(x)-x),$$

$${\beta }_j(x):=\{\begin{array}{ll}\frac{{\rho }_j(x)+\Vert T_j(x)-x\Vert }{\Vert T_j(x)-x\Vert },& \text{mbox}if{T}_j(x)\ne x\\ 0,& \text{mbox}otherwise.\end{array}$$

$$\operatorname*{Fix}(P_{Q}V)=Q\cap\operatorname*{Fix}V\quad\text{and}\quad\operatorname*{Fix}V=\Big{\{}x\colon x\in\bigcap_{j\in J(x)}\operatorname*{Fix}T_{j}\Big{\}}.$$

$$\Vert P_Q(V(x))-z{\Vert }^2\le \Vert x-z{\Vert }^2-2\alpha \lambda R\rho (x).$$

$$⟨Vx-P_Q(V(x)),z-P_Q(V(x))⟩\le 0$$

$$⟨T_j(x)-x,z-x⟩\ge \Vert T_j(x)-x{\Vert }^2.$$

$$⟨V(x)-P_Q(V(x)),z-P_Q(V(x))⟩$$

$$=\alpha \sum _{j\in J_{+}(x)}{\lambda }_j(x){\beta }_j(x)⟨T_j(x)-x,z-x⟩$$

$$\ge \alpha \sum _{j\in J_{+}(x)}{\lambda }_j(x){\beta }_j(x)\Vert T_j(x)-x{\Vert }^2>0,$$

$$\Vert U_j(x)-y{\Vert }^2\le \Vert x-y{\Vert }^2-\frac{2-\alpha }{\alpha }\Vert U_j(x)-x{\Vert }^2.$$

$$y:=z-R\frac{U_j(x)-x}{\Vert U_j(x)-x\Vert },$$

$$\Vert U_j(x)-y{\Vert }^2$$

$$=\Vert U_j(x)-z{\Vert }^2+\frac{2R}{\Vert U_j(x)-x\Vert }⟨U_j(x)-z,U_j(x)-x⟩+R^2.$$

$$\Vert x-y{\Vert }^2$$

$$=\Vert x-z{\Vert }^2-\frac{2R}{\Vert U_j(x)-x\Vert }⟨z-x,U_j(x)-x⟩+R^2$$

$$\Vert U_j(x)-x\Vert =\alpha ({\rho }_j(x)+\Vert T_j(x)-x\Vert )\ge \alpha {\rho }_j(x).$$

Full text

Keywords: Metric projection, random control, repetitive control, subgradient projection. Mathematics Subject Classification (2010): 47J25, 47N10, 90C25.

1 Introduction

The convex feasibility problem (CFP) is one of the fundamental problems in optimization. It is mainly used for (but not limited to) modeling problems in which finding a solution satisfying a given list of constraints is satisfactory, whereas obtaining the optimal solution is not necessary. In this paper we consider the following variant of the CFP defined in a real Hilbert space $\mathcal{H}$: Find $x\in C\cap Q$ with $C:=\bigcap_{i\in I}C_{i}$. We assume that each one of the sets $C_{i}$, $i\in I$, as well as $Q$, is closed and convex and $I:=\{1,2,\ldots,m\}$ for some $m\in\mathbb{N}_{+}\cup\{\infty\}$.

Oftentimes it is convenient to represent the constraint set $C_{i}$ as the fixed point set of an operator $T_{i}\colon\mathcal{H}\to\mathcal{H}$, which satisfies certain conditions – a variation of firm nonexpansivity. In particular, one can use the metric projection $P_{C_{i}}$, the proximal operator $\operatorname*{Prox}_{f_{i}}$ (when $C_{i}$ is the set of minimizers of a proper, l.s.c. and convex function $f_{i}$) or the subgradient projection $P_{f_{i}}$ (when $C_{i}$ is the sublevel set of a real-valued, l.s.c. and convex function $f_{i}$). Note, however, that in some cases $P_{f_{i}}$ may happen to be discontinuous (see [3, Example 29.47]). For this reason, in this paper we consider a class of operators called cutters, which, by some authors, are also called firmly quasi-nonexpansive; see [3, Definition 4.1]. A comprehensive overview of this class, citing many relevant references, can be found in [10]. Here we only note that in view of [10, Theorem 2.2.5], a cutter can be considered a generalization of a firmly nonexpansive mapping, provided it has a fixed point.

There is a great number of deterministic and stochastic Fejér monotone algorithms designed for solving the CFP, which are governed by cutter operators; see, for example, [1, 2, 6, 8, 10, 11, 12, 15, 16, 23, 27] to name but a few. A typical result ensures the asymptotic convergence of the generated iterates. For example, this can be weak, norm or linear convergence, but also their stochastic versions. A potential drawback, when using these algorithms, is that the produced iterates may never be feasible, unlike the limit point to which they converge. An example of such a situation has recently been given in [26, Theorem 7].

The purpose of this paper is to propose a modification of the iterative method discussed in [1], so that the sequence of the generated iterates reaches the solution set $C\cap Q$ within a finite number of steps in the presence of the following constraint qualification: $\operatorname*{int}(C)\cap Q\neq\emptyset$. We emphasize here that our approach is not the only way of achieving finite convergence. Other results that discuss this topic can be found, for instance, in [4, 5, 14, 20, 22, 25, 28].

To be more precise, we focus on a projected version of the framework considered in [1], which can be formulated as follows:

[TABLE]

where $\alpha_{k}\in(0,2]$ is the relaxation parameter, the weights $\lambda_{i,k}(x_{k})\in[0,1]$ satisfy $\sum_{i\in I_{k}}\lambda_{i,k}(x_{k})=1$ and $I_{k}(x_{k})\subset I$ is a nonempty and finite set of indices. The idea of the above-mentioned modification is to extend each nonzero vector $(T_{i}(x_{k})-x_{k})$ from (1.1) by a scalar $r_{[k]}/\varphi_{i}(x_{k})$, where $\{r_{k}\}_{k=0}^{\infty}\subset(0,\infty)$ is a sequence of small overrelaxation parameters, $[k]\in\{0,\ldots,k\}$ counts all the correction steps up to the $k$-th iterate and $\varphi_{i}(x_{k})\in(0,\infty)$ satisfies a certain boundedness condition; see Theorem 4.7.

We observed, that after such a modification, whenever we do a correction step ($x_{k+1}\neq x_{k}$), we move towards an interior point $z\in\operatorname*{int}(C)\cap Q$ under the assumption that the overrelaxation parameter $r_{[k]}$ is small enough. In particular, we show that

[TABLE]

where $M>0$ is some constant and $B(z,2R)\subseteq C$. It is not difficult to see that by repeatedly applying the above inequality one could arrive at a contradiction knowing that correction steps happen a considerable number of times and $\sum_{k=0}^{\infty}r_{k}=\infty$. This simple argument suggests that eventually we should encounter an iterate $x_{k}\in C\cap Q$ for some $k$.

Special cases of our framework can be found in the literature; see, for example, [13, 18, 24], where $\varphi_{i}(x_{k}):=\|g_{i}(x_{k})\|$ ($g_{i}(x_{k})$ is a subgradient of a convex function $f_{i}$ describing the sublevel set $C_{i}$) and [7, 17, 29], where $\varphi_{i}(x_{k}):=1$. For a more detailed description of the above-mentioned works, see Table 1.1. Here we only note that the convergence analysis in [13, 18, 24] differs significantly from the one presented in [7, 17, 29]. In this paper we follow the path set in [7, 17, 29], where one can find weaker forms of inequality (1.2). Moreover, in our approach, by introducing general functionals $\varphi_{i}$, we show that this path can also be used for $\varphi_{i}(x)=\|g_{i}(x)\|$, as is the case in [13, 18, 24]. In particular, we establish inequality (1.2) for subgradient projection methods from [13, 18, 24], a property which was not known before.

Various strategies defining the control sequence $\{I_{k}\}_{k=0}^{\infty}$ are known in the literature, ranging from deterministic, where $I_{k}\colon\mathcal{H}\to 2^{I}\setminus\emptyset$, to random, where $I_{k}\colon\Omega\to 2^{I}\setminus\emptyset$ are i.i.d. defined in some probability space. In the former case, the most recognizable are almost cyclic, intermittent and repetitive (chaotic) control sequences; see [1] but also Table 1.1. The common feature of all the above-mentioned deterministic control sequences is that

[TABLE]

for all $x\notin C$, where $I_{+}(x):=\{i\in I\colon x\notin C_{i}\}$. Moreover, one can show that the random controls almost surely satisfy condition (1.3) provided that the probability of performing a correction step is positive whenever the iterate is outside the set $C$. We note that the latter condition, which we formally write in (5.5), was proposed by Polyak in [29, Assumption 2], see also [9, Assumption 1]. What we show in this paper is that condition (1.3) is actually a sufficient one for obtaining the finite convergence of the modified method (1.1) discussed above.

The use of the counter $[k]$ instead of $k$ was originally suggested in [29, Section 4.2] for a particular variant of our general framework (compare with Table 1.1). Note that there are certain situations, where $[k]=k$ or when $[k]$ can simply be omitted; see Remark 4.11. However, it is important to notice that in some cases, not using the counter $[k]$ may lead to lack of finite convergence. To illustrate this, we provide two counterexamples. In the relatively simple Example 5.6, we show that finite convergence does not occur for the alternating projection method with relaxation applied to two halfspaces in the plane. In Example 5.7, which is more important, but also more technical, we assume that the control sequence is repetitive, $r_{k}\downarrow 0$ (monotonically) and $\sum_{k=1}^{\infty}r_{k}=\infty$. In particular, this answers a question raised in [13] related to the use of repetitive controls (called expanding therein), showing that in general [13, Theorem 20] cannot hold without a certain technical assumption [11, Condition 19]. On the other hand, in view of our result, [11, Condition 19] is no longer necessary for repetitive controls, when combined with $[k]$.

The contribution of our paper can be summarized as follows: We develop a unified analysis for a large class of finitely convergent iterative methods, which employ certain overrelaxation parameters. We also introduce a very broad class of control sequences, which covers a wide range of deterministic iterative methods and further allows us to consider stochastic iterative methods. As a byproduct, we improve all the results from [7, 13, 17, 18, 24, 29]; see the last row in Table 1.1.

Our paper is organized as follows. In section 2 we provide basic facts related to cutters. In section 3 we establish two auxiliary lemmata which are the key tools in our analysis. In particular, Lemma 3.2 laid the foundation for establishing inequality (1.2). In section 4 we discuss in detail condition (1.3) together with several examples. We present there our main result, namely Theorem 4.7. In section 5 we formulate a stochastic counterpart of Theorem 4.7 (Theorem 5.4). In the Appendix we provide two counterexamples showing that omitting the counter $[k]$ may cause lack of finite convergence.

2 Preliminaries

Let $\mathcal{H}$ be a real Hilbert space with inner product $\langle\cdot,\cdot\rangle$ and induced norm $\|\cdot\|$.

Definition 2.1**.**

Let $T\colon\mathcal{H}\to\mathcal{H}$ be an operator with $\operatorname*{Fix}T:=\{z\in\mathcal{H}\colon T(z)=z\}\neq\emptyset$. We say that $T$ is a cutter if $\langle x-T(x),z-T(x)\rangle\leq 0$ for all $x\in\mathcal{H}$ and $z\in\operatorname*{Fix}T$.

Example 2.2** (Metric Projection).**

Let $C\subseteq\mathcal{H}$ be nonempty, closed and convex. The metric projection $P_{C}(x):=\operatorname*{argmin}_{z\in C}\|z-x\|$ is a cutter and $\operatorname*{Fix}P_{C}=C$; see, for example, [10, Theorem 1.2.4].

Example 2.3** (Proximal Operator).**

Let $f\colon\mathcal{H}\to\mathbb{R}\cup\{+\infty\}$ be a lower semicontinuous and convex function. The proximal operator $\operatorname*{Prox}\nolimits_{f}(x):=\operatorname*{argmin}_{y\in\mathcal{H}}(f(y)+\frac{1}{2}\|y-x\|^{2})$ is firmly nonexpansive and $\operatorname*{Fix}(\operatorname*{Prox}_{f})=\operatorname*{Argmin}_{x\in\mathcal{H}}f(x)$; see [3, Propositions 12.28 and 12.29]. Thus if $f$ has at least one minimizer, then $\operatorname*{Prox}_{f}$ is a cutter; see [10, Theorem 2.2.5].

Example 2.4** (Subgradient Projection).**

Let $f\colon\mathcal{H}\to\mathbb{R}$ be a lower semicontinuous and convex function with nonempty sublevel set $S(f,0):=\{x\in\mathcal{H}\colon f(x)\leq 0\}\neq\emptyset$. For each $x\in\mathcal{H}$, let $g(x)$ be a chosen subgradient from the subdifferential set $\partial f(x):=\{g\in\mathcal{H}\colon f(y)\geq f(x)+\langle g,y-x\rangle\text{, for all }y\in\mathcal{H}\}$, which, by [3, Proposition 16.27], is nonempty. Note here that we apply [3, Proposition 16.27] to a real-valued functional $f$. The subgradient projection

[TABLE]

is a cutter and $\operatorname*{Fix}P_{f}=S(f,0)$; see, for example, [10, Corollary 4.2.6].

Note that both the proximal operator and the subgradient projection extend the notion of the metric projection; see [3, Examples 12.25 and 29.44]. However, as we mentioned earlier in the introduction, the subgradient projection need not be even continuous; see [3, Example 29.47]. A comprehensive overview of cutters can be found in [10, Chapter 2]. Below we only recall [10, Remark 2.1.31], which will be used in the sequel.

Proposition 2.5**.**

Let $T\colon\mathcal{H}\to\mathcal{H}$ be an operator with $\operatorname*{Fix}T\neq\emptyset$. Then $T$ is a cuttter if and only if $\langle T(x)-x,z-x\rangle\geq\|T(x)-x\|^{2}$ for all $x\in\mathcal{H}$ and $z\in\operatorname*{Fix}T$.

3 Auxiliary Results

Lemma 3.1**.**

Let $T\colon\mathcal{H}\to\mathcal{H}$ be a cutter, let $\alpha\in(0,2]$ and let $\rho\colon\mathcal{H}\to(0,\infty)$. Define the operator $U\colon\mathcal{H}\to\mathcal{H}$ by $U(x):=x+\alpha\beta(x)(T(x)-x)$, where

[TABLE]

Assume that $x\notin\operatorname*{Fix}T$ and $B(y,\rho(x))\subseteq\operatorname*{Fix}T$. Then we have

[TABLE]

• Proof.

The argument follows the proof of [7, Corollary 2.1(v)], which is only presented for a constant overrelaxation $\rho$. Define

[TABLE]

and observe that $w\in B(y,\rho(x))\subseteq\operatorname*{Fix}T$. Since $T$ is a cutter, we have

[TABLE]

On the other hand, by [3, Corollary 2.14] applied to

[TABLE]

we obtain

[TABLE]

By (3.4), we have

[TABLE]

Consequently,

[TABLE]

Moreover,

[TABLE]

Combining this with (3.6), we arrive at (3.2), which completes the proof. $\blacksquare$

The following lemma is a key tool in our analysis. We use it, in particular, to derive estimate (1.2). Before proceeding, recall that $I_{+}(x):=\{i\in I\colon x\notin C_{i}\}$, $x\in\mathcal{H}$.

Lemma 3.2**.**

Assume that $C_{i}=\operatorname*{Fix}T_{i}$ for given cutter operators $T_{i}\colon\mathcal{H}\to\mathcal{H}$, $i\in I$. Moreover, let $\alpha\in(0,2]$, let $\rho_{i}\colon\mathcal{H}\to(0,\infty)$, $i\in I$, and let $J\colon\mathcal{H}\to 2^{I}\setminus\{\emptyset\}$ satisfy $\sup_{x\in\mathcal{H}}\#(J(x))<\infty$. Furthermore, let $\lambda_{j}\colon\mathcal{H}\to[0,1]$ be such that $\sum_{j\in J(x)}\lambda_{j}(x)=1$. Define the operator $V\colon\mathcal{H}\to\mathcal{H}$ by

[TABLE]

where

[TABLE]

Assume that $C\cap Q\neq\emptyset$ and that the weights $\lambda_{j}$ satisfy the inequality $\lambda_{j}(x)\geq\lambda>0$ for all $x\in\mathcal{H}$ and $j\in J_{+}(x):=J(x)\cap I_{+}(x)$. Then

[TABLE]

Moreover, assume that there are $z\in Q$ and $R>0$ such that $B(z,2R)\subseteq C$. Then for all $x\notin C$ with $\rho(x):=\max_{j\in J_{+}(x)}\rho_{j}(x)\leq R$, we have

[TABLE]

• Proof.

We first show (3.12). To this end, assume that $Q\cap C\neq\emptyset$ and define $F:=\{x\colon x\in\bigcap_{j\in J(x)}\operatorname*{Fix}T_{j}\}$. Observe that $C\subseteq F$ and thus $F\neq\emptyset$. Moreover, it is not difficult to see that the inclusion $Q\cap F\subseteq\operatorname*{Fix}P_{Q}V$ follows from the definition of $V$. It suffices to show that $\operatorname*{Fix}P_{Q}V\subseteq Q\cap F$. Clearly, by the definition of the metric projection, $P_{Q}V\subseteq Q$ and consequently, $\operatorname*{Fix}P_{Q}V\subseteq Q$. Let $x\in\operatorname*{Fix}P_{Q}V$ and suppose to the contrary that $x\notin F$, that is, $J_{+}(x)\neq\emptyset$. Since $P_{Q}$ is a cutter, we have

[TABLE]

for all $z\in Q$. On the other hand, since each $T_{j}$ is a cutter, for all $x\in\mathcal{H}$ and $z\in\operatorname*{Fix}T_{j}$, we have, by Proposition 2.5,

[TABLE]

Since $x=P_{Q}(V(x))$, for any $z\in Q\cap C$, we arrive at

[TABLE]

which is in contradiction with (3.14). Consequently $J_{+}(x)=\emptyset$ and $Q\cap F=\operatorname*{Fix}P_{Q}V$, as claimed.

Next we show that (3.13) holds for all $x\notin C$ with $\rho(x)\leq R$. To this end, for each $i\in I$, we define an auxiliary operator $U_{i}$ by $U_{i}(x):=x+\alpha\beta_{i}(x)(T_{i}(x)-x)$. Thus $V(x)=\sum_{j\in J(x)}\lambda_{j}(x)U_{j}(x)$. Let $x\notin C$ be such that $\rho(x)\leq R$ and let $j\in J_{+}(x)$. Observe that for any $y\in B(z,R)$, we have $B(y,\rho(x))\subseteq B(z,2R)\subseteq\operatorname*{Fix}T_{j}$. Consequently, by Lemma 3.1 applied to $U_{j}$, we have

[TABLE]

In particular, the above inequality holds for

[TABLE]

which by the choice of $j\in J_{+}(x)$ is well defined. By expanding the left-hand side of the inequality (3.17) with $y$ defined as above, we obtain

[TABLE]

On the other hand,

[TABLE]

and

[TABLE]

By combining (3.17) with (3.18), (Proof.) and (Proof.), we arrive at

[TABLE]

Observe that (Proof.) holds for all $j\in J_{+}(x)$. Moreover, for all $j\in J(x)\setminus J_{+}(x)$, we have

[TABLE]

Hence, by the nonexpansivity of the metric projection $P_{Q}$ and the convexity of the squared norm $\|\cdot\|^{2}$, we have

[TABLE]

The above inequality completes the proof. $\blacksquare$

The following lemma extends [7, Lemma 2.2].

Lemma 3.3** (Slater Condition).**

For each $i\in I$, let $f_{i}\colon\mathcal{H}\rightarrow\mathbb{R}$ be a convex and lower semicontinuous function, and assume that $f(z):=\sup_{i\in I}f_{i}(z)<0$ for some $z\in\mathcal{H}$. Then for all $r>0$, we have

[TABLE]

• Proof.

Let $x\in B(z,r)$ and $i\in I_{+}(x)$. By the subgradient inequality, we have $f_{i}(z)\geq f_{i}(x)+\langle g_{i}(x),z-x\rangle,$ where $g_{i}(x)\in\partial f_{i}(x)$. Since $i\in I_{+}(x)$, we have $f_{i}(x)>0$ and consequently,

[TABLE]

from which (3.25) follows. $\blacksquare$

4 Deterministic Methods

In this section we return to the CFP defined in the introduction.

Definition 4.1**.**

We call the sequence $\{I_{k}\}_{k=0}^{\infty}$ a control sequence in $I$ if each $I_{k}\colon\mathcal{H}\to 2^{I}\setminus\{\emptyset\}$ is a set-valued mapping with $M:=\sup_{x,k}\#(I_{k}(x))<\infty$. If each $I_{k}$ is single-valued, say $I_{k}(x)=\{i_{k}(x)\}$, where $i_{k}\colon\mathcal{H}\to I$, then we also call the sequence $\{i_{k}\}_{k=0}^{\infty}$ a control sequence in $I$. We say that the control sequence in $I$ is nonadaptive if each set-valued mapping $I_{k}$ (single-valued mapping $i_{k}$) is constant, that is, when $I_{k}(x)=I_{k}(y)$ ($i_{k}(x)=i_{k}(y)$) for all $x,y\in\mathcal{H}$. In this case we omit the argument.

Definition 4.2**.**

We say that the control sequence $\{I_{k}\}_{k=0}^{\infty}$ in $I$ is well matched with the set $C$ if

[TABLE]

for all $x\notin C$. In particular, a single-valued control sequence $\{i_{k}\}_{k=0}^{\infty}$ in $I$ is well matched with $C$ if

[TABLE]

for all $x\notin C$.

Example 4.3**.**

In the literature (see [1, 10]) one can find the following examples of a nonadaptive control sequences $\{I_{k}\}_{k=0}^{\infty}$ in $I$ which are well matched with $C$: (a) cyclic control defined in a finite $I$ which satisfies $I_{k}=I_{(k\operatorname*{mod}s)}$ for all $k=0,1,2,\ldots$, where $s\geq 2$; (b) intermittent control defined in a finite $I$ which satisfies $I=\bigcup_{k=n}^{n+s-1}I_{k}$ for all $n=0,1,2,\ldots$ and some integer $s\geq 2$; and the more general (c) repetitive control defined in both finite and infinite $I$, which satisfies $I=\bigcup_{k=n}^{\infty}I_{k}$ for all $n=0,1,2,\ldots$. All three definitions simplify when reduced to a single-valued control sequences $\{i_{k}\}_{k=0}^{\infty}$.

Proposition 4.4**.**

Let $\{I_{k}\}_{k=0}^{\infty}$ be a nonadaptive control sequence in $I$ and consider the following statements:

1. (i)

$\{I_{k}\}_{k=0}^{\infty}$* is well matched with $C$.* 2. (ii)

$F_{n}:=\bigcap_{k=n}^{\infty}\bigcap_{i\in I_{k}}C_{i}=C$* for all $n=0,1,2,\ldots$.* 3. (iii)

$\{I_{k}\}_{k=0}^{\infty}$* is repetitive in $I^{\prime}$ for some $\emptyset\neq I^{\prime}\subseteq I$ (that is, $I^{\prime}\subseteq\bigcup_{k=n}^{\infty}I_{k}$ for all $n=0,1,2,\ldots$) and $C=\bigcap_{i\in I^{\prime}}C_{i}$ .*

Then (i)$\Leftrightarrow$(ii)$\Leftarrow$ (iii). Moreover, if $I$ is finite, then (ii)$\Rightarrow$ (iii).

• Proof.

We first show that (i) implies (ii). Suppose to the contrary that the control sequence $\{I_{k}\}_{k=0}^{\infty}$ is well matched with $C$ and that for some $n\geq 0$, there exists a point $x\in F_{n}\setminus C$. Then $x\in C_{i}$ for all $i\in I_{k}$ and $k\geq n$, that is, $\#(x,\{I_{k}\}_{k=0}^{\infty})\leq n$. On the other hand, since $x\notin C$, by condition (4.1), $\#(x,\{I_{k}\}_{k=0}^{\infty})=\infty$, which is a contradiction.

To show that (ii) implies (i), assume that the equality $F_{n}=C$ holds for all $n=0,1,2,\ldots$ and let $x\notin C$. Then for each $n=0,1,2,\ldots$, consider the smallest $k_{n}\geq n$ such that $x\notin F_{k_{n}}$. By the definition of the set $F_{n}$ and by eventually passing to a subsequence, we may assume that $x\notin\bigcap_{i\in I_{k_{n}}}C_{i}$ in view of which, $\#(x,\{I_{k}\}_{k=0}^{\infty})=\infty$.

It is not difficult to see that (iii) implies (i). Indeed, if $x\notin C$, then, by (iii), $x$ violates at least one constraint $C_{i}$ for some $i\in I^{\prime}$. Since the control is repetitive in $I^{\prime}$, we see that $i\in I_{k}$ for infinitely many $k$’s and thus $\#(x,\{I_{k}\}_{k=0}^{\infty})=\infty$.

Assume now that $I$ is finite. We show that (iii) follows from (ii). To this end, define $I^{\prime}:=\limsup_{k\to\infty}I_{k}=\bigcap_{n=0}^{\infty}\bigcup_{k=n}^{\infty}I_{k}$ and observe that $i\in I^{\prime}$ if and only if $i\in I_{k}$ for infinitely many $k$’s. Since $I$ is finite, we see that $I^{\prime}\neq\emptyset$. Assume that $I^{\prime}$ is a proper subset of $I$. For each $i\in I\setminus I^{\prime}$, there is $n_{i}\geq 0$ such that $i\notin\bigcup_{k=n_{i}}^{\infty}I_{k}$ and since $I$ is finite, we have $n:=\max_{i\in I\setminus I^{\prime}}n_{i}<\infty$. Consequently, $I^{\prime}=\bigcup_{k=n}^{\infty}I_{k}$ and, by (ii), we arrive at $C=F_{n}=\bigcap_{i\in I^{\prime}}C_{i}$. This completes the proof. $\blacksquare$

Remark 4.5**.**

The implication (i)$\Rightarrow$ (iii) may not be true when $I$ is infinite. To see this, consider a decreasing sequence of sets $C_{k+1}\subseteq C_{k}$ with nonempty intersection $C$ and define $i_{k}:=k$. The control $\{i_{k}\}_{k=0}^{\infty}$ is well matched with $C$, but clearly it is not repetitive in any subset $I^{\prime}\subseteq I$.

Example 4.6**.**

Observe that in view of Definition 4.2, if for all $x\in\mathcal{H}$ and all $k=0,1,2,\ldots$, the set $I_{k}(x)$ contains at least one index from the set of violated constraints $I_{+}(x)$, then the control sequence $\{I_{k}\}_{k=0}^{\infty}$ is well matched with $C$. In particular, when $I$ is finite, one could use maximal control sequences such as: (a) the remotest set control $i_{k}(x):=\operatorname*{argmax}_{i\in I}d(x,C_{i})$; (b) the maximal displacement control $i_{k}(x):=\operatorname*{argmax}_{i\in I}\|T_{i}(x)-x\|$, when $C_{i}=\operatorname*{Fix}T_{i}$; or (c) the maximal violation control $i_{k}(x):=\operatorname*{argmax}_{i\in I}f_{i}^{+}(x)$, when $C_{i}=S(f_{i},0)$.

Theorem 4.7**.**

Assume that $C_{i}=\operatorname*{Fix}T_{i}$ for given cutter operators $T_{i}\colon\mathcal{H}\to\mathcal{H}$, $i\in I$. Let $\{I_{k}\}_{k=0}^{\infty}$ be a given control sequence in $I$ and let the weights $\lambda_{i,k}\colon\mathcal{H}\to[0,1]$ be such that $\sum_{i\in I_{k}(x)}\lambda_{i,k}(x)=1$. Moreover, let $\{\alpha_{k}\}_{k=0}^{\infty}\subset(0,2]$ be a sequence of relaxations and let $\{r_{k}\}_{k=0}^{\infty}\subset(0,\infty)$ be a sequence of overrelaxations. Finally, let $\varphi_{i}\colon\mathcal{H}\to(0,\infty)$. Define the sequence $\{x_{k}\}_{k=0}^{\infty}$ by

[TABLE]

where

[TABLE]

and

[TABLE]

Assume that

1. (i)

$\operatorname*{int}(C)\cap Q\neq\emptyset$. 2. (ii)

$r_{k}\to 0$* and $\sum_{k=0}^{\infty}\alpha_{k}r_{k}=\infty$.* 3. (iii)

For each bounded subset $S\subset\mathcal{H}$, there are $\delta,\Delta\in(0,\infty)$ s.t. $\delta\leq\varphi_{i}(x)\leq\Delta$ for all $x\in S$. 4. (iv)

There is $\lambda>0$ such that $\lambda_{i,k}(x)\geq\lambda>0$ for all $x$, $k$ and $i\in I_{k}(x)\cap I_{+}(x)$. 5. (v)

$\{I_{k}\}_{k=0}^{\infty}$* is well matched with the set $C$.*

If the sequence $\{x_{k}\}_{k=0}^{\infty}$ is bounded (see Examples 4.8 and 4.9), then $x_{k}\in C\cap Q$ for some $k$.

• Proof.

Since $x_{k}\in Q$, it suffices to show that $x_{k}\in C$ for some $k$. For each $k=0,1,2,\ldots$, define the operator $V_{k}\colon\mathcal{H}\to\mathcal{H}$ by

[TABLE]

where $I_{k}^{+}(x):=I_{k}(x)\cap I_{+}(x)$. Clearly, we can write $x_{k+1}=P_{Q}(V_{k}(x_{k}))$. We divide the rest of the proof into two cases.

Case 1. Assume that $n:=\sup_{k\geq 0}[k]<\infty$, in which case $x_{n}=x_{k}$ for all $k\geq n$. We show that $x_{n}\in C$. By Lemma 3.2, the equality $x_{k+1}=x_{k}$ implies that $x_{k}\in\operatorname*{Fix}V_{k}$, that is, $x_{k}\in\bigcap_{i\in I_{k}(x_{k})}C_{i}$. Consequently $\#(x_{n},\{I_{k}\}_{k=0}^{\infty})\leq n$ and thus $x_{n}\in C$. Otherwise, since the control $\{I_{k}\}_{k=0}^{\infty}$ is well matched with $C$, we would get $\#(x_{n},\{I_{k}\}_{k=0}^{\infty})=\infty$, a contradiction.

Case 2. Assume now that $\sup_{k\geq 0}[k]=\infty$, that is, the set $\mathcal{N}:=\{n\geq 0\colon x_{n}\neq x_{n+1}\}$ is infinite. There is a point $z\in\operatorname*{int}(C)\cap Q$ and a radius $R>0$ such that the ball $B(z,2R)\subseteq C$. Since the sequence $\{x_{k}\}_{k=0}^{\infty}$ is assumed to be bounded, there are $0<\delta\leq\Delta<\infty$ such that $\delta\leq\varphi_{i}(x_{k})\leq\Delta$ for all $k=0,1,2,\ldots$. Consequently, the fraction $r_{[k]}/\varphi_{i}(x_{k})$ can be made arbitrarily small because of the estimate $r_{[k]}/\varphi_{i}(x_{k})\leq r_{[k]}/\delta$. In particular, we may assume that $r_{[k]}/\varphi_{i}(x_{k})\leq R$ for all large enough $k$, say $k\geq K$ and $K\in\mathcal{N}$. Thus, by Lemma 3.2 applied to $V:=V_{k}$ and $\rho_{i}(x):=r_{[k]}/\varphi_{i}(x)$, we get

[TABLE]

for all $k\in\mathcal{N}$, $k\geq K$ (compare with (1.2)). On the other hand,

[TABLE]

for all $k\notin\mathcal{N}$. Consequently, by inductively applying (4.7) and (4.8) to large enough $k$ ($k\geq K$), we obtain

[TABLE]

and therefore, since $K\in\mathcal{N}$, we arrive at

[TABLE]

Since, by assumption, $\sup_{k\geq 0}[k]=\infty$, we see that the left-hand side in (4.10) tends to infinity as $k\to\infty$, which is a contradiction. Consequently, we must have $n=\sup_{k\geq 0}[k]<\infty$, in which case we have already shown that $x_{n}\in C\cap Q$. $\blacksquare$

There are at least two known situations, where we can ensure that the sequence $\{x_{k}\}_{k=0}^{\infty}$ is indeed bounded.

Example 4.8**.**

In the setting of Theorem 4.7, for each $x\in\mathcal{H}$, define

[TABLE]

Obviously, $\varphi_{i}(x)$ satisfies (iii). Furthermore, by applying Lemma 3.2, it is not difficult to see that the sequence $\{x_{k}\}_{k=0}^{\infty}$ is bounded. Consequently, $x_{k}\in C\cap Q$ for some $k$.

Example 4.9**.**

In the setting of Theorem 4.7, assume that $C_{i}=\{x\colon f_{i}(x)\leq 0\}$ for some convex and lower semicontinuous functions $f_{i}\colon\mathcal{H}\to\mathbb{R}$, $i\in I$. Let $T_{i}:=P_{f_{i}}$ and let $g_{i}\colon\mathcal{H}\to\mathcal{H}$ be the associated subgradient mapping (see Example 2.4). For each $x\in\mathcal{H}$, define

[TABLE]

Using (2.1) and the convention that the summation over the empty set is zero, method (4.3) becomes

[TABLE]

Assume that (i’) $f(z):=\sup_{i\in I}f_{i}(z)<0$ for some $z\in Q$; (ii’) = (ii); and (iii’) $\bigcup_{i\in I}\partial f_{i}(S)$ is bounded for bounded subsets $S\subset\mathcal{H}$. Then the sequence $\{x_{k}\}_{k=0}^{\infty}$ is bounded and assumptions (i’)–(iii’) imply conditions (i)–(iii). Consequently, $x_{k}\in C\cap Q$ for some $k$.

• Proof.

We first demonstrate that the sequence $\{x_{k}\}_{k=0}^{\infty}$ is bounded. Obviously, the statement is trivial when $n:=\sup_{k\geq 0}[k]<\infty$. Assume now that $\sup_{k\geq 0}[k]=\infty$. It suffices to show that $\|x_{k+1}-z\|\leq\|x_{k}-z\|$ for all $k$ large enough.

If $I_{k}^{+}(x_{k})=\emptyset$, then $x_{k+1}=x_{k}$ and thus $\|x_{k+1}-z\|=\|x_{k}-z\|$. On the other hand, if $I_{k}^{+}(x_{k})\neq\emptyset$, then, by using the nonexpansivity of the metric projection $P_{Q}$ and the convexity of $\|\cdot\|^{2}$, we get

[TABLE]

where

[TABLE]

Moreover, by combining the subgradient inequality with the inequality $r_{[k]}\leq-f(z)$, which holds for all $k$ large enough (since $r_{[k]}\to 0$ as $k\to\infty$), we get

[TABLE]

Consequently, $\theta_{i,k}(x_{k})\leq-(2-\alpha_{[k]})(f_{i}(x_{k})+r_{[k]})\leq 0$, which leads to $\|x_{k+1}-x\|\leq\|x_{k}-z\|$, as asserted.

Let $S$ be nonempty and bounded subset of $\mathcal{H}$ and choose $r>0$ so that $S\subset B(z,r)$. The assumed boundedness of the subdifferential and Lemma 3.3 imply that there are $\Delta\geq\delta>0$ such that $\delta\leq\|g_{i}(x)\|\leq\Delta$, where the first inequality holds for all $x\in S$ and all $i\in I_{+}(x)$, whereas the second one holds for all $x\in S$ and all $i\in I$. This, when combined with (4.12), implies condition (iii). $\blacksquare$

Remark 4.10** (Constant overrelaxations).**

A careful analysis of the proof of Theorem 4.7 shows that when $\varphi_{i}(x)=1$, $i\in I$, we can use constant overrelaxations $r_{k}:=r\leq R$. This, however, requires the explicit knowledge of the radius $R$, which is not always possible.

Remark 4.11** (Using $k$ instead of $[k]$).**

Observe that if $I_{k}^{+}(x)\neq\emptyset$ for all $x\notin C$, then $[k]=k$ as long as $x_{k}\notin C$. This condition is trivially satisfied if, for example, the number of constraints $m=1$ or when $I_{k}=I$ for all $k=0,1,2,\ldots$. Furthermore, the square brackets can be partially omitted if $\alpha_{k}\geq\alpha>0$ and $\sum_{k=0}^{\infty}r_{k}=\infty$. In this case we may replace “$\alpha_{[k]}$” by “$\alpha_{k}$” in (4.3) without losing the finite convergence property. If, in addition, we assume that the sequence $\{r_{k}\}_{k=0}^{\infty}$ is decreasing and the control is $s$-intermittent for some $s\geq 1$, then we may fully drop the square brackets notation and write “$r_{k}$” instead of “$r_{[k]}$” in (4.5). This, however, requires a more detailed discussion, which we present below. We note that if $r_{k}$ is not monotone, then by using $r_{k}$ instead of $r_{[k]}$, we may indeed lose the finite convergence property; see Example 5.6.

• Proof.

For each $k=0,1,2,\ldots,$ define $V_{k}$ in the same way as in (4.6) with “$[k]$” replaced by “$k$” and assume that $\mathcal{N}$ is infinite. Since the control is $s$-intermittent, we have $\{k,k+1,\ldots,k+s-1\}\cap\mathcal{N}\neq\emptyset$ for all $k=0,1,2,\ldots$. Since $r_{k+l}\geq r_{k+s}$ for all $l\in\{0,\ldots,s-1\}$, inequality (4.10) becomes

[TABLE]

for all $k\geq K$. Observe that due to monotonicity of the $r_{k}$’s, the left-hand side tends to infinity as $k\to\infty$, which leads to a contradiction. Thus $\mathcal{N}$ has to be finite and by using the same argument as in the proof of Theorem 4.7, we see that $x_{k}\in C\cap Q$ for some $k$. $\blacksquare$

Remark 4.12**.**

The uniform boundedness of the subdifferential on bounded sets, which corresponds to condition (iii’) of Example 4.9, is a rather standard assumption; see [1, Proposition 7.8]. Note that condition (iii’) is not mentioned explicitly in [18, 24], but only in [13, Remark 16]. Nevertheless, it is satisfied therein because the set $I$ is finite and $\mathcal{H}=\mathbb{R}^{n}$. Thus Example 4.9 (in view of Remark 4.11) improves upon the results established in [13, 18, 24].

Remark 4.13** (Comparison with [7]).**

Assume that $C=\operatorname*{Fix}T$, where $T\colon\mathcal{H}\to\mathcal{H}$ is a cutter ($m=1$) and $\varphi(x):=1$. Then (4.3) becomes the iterative method proposed in [7, equation (3)]. Furthermore, Theorem 4.7 guarantees finite convergence if one only assumes that $r_{k}\to 0$, $\sum_{k=0}^{\infty}\alpha_{k}r_{k}=\infty$ and $C\cap\operatorname*{int}(Q)\neq\emptyset$. This extends both [7, Theorems 3.1 and Theorem 3.2]; compare with Table 1.1.

5 Stochastic Methods

In this section we consider a stochastic version of Theorem 4.7. Let $(\Omega,\mathcal{F},\operatorname*{Pr})$ be a given probability space.

Definition 5.1**.**

We call the sequence $\{I_{k}\}_{k=0}^{\infty}$ a random control sequence in $I$ if $I_{k}\colon\Omega\to 2^{I}\setminus\{\emptyset\}$ are independent and identically distributed (set-valued) random variables on $(\Omega,\mathcal{F},\operatorname*{Pr})$ with $M:=\sup_{\omega,k}\#(I_{k}(\omega))<\infty$. If each $I_{k}(\omega)$ is single-valued, say $I_{k}(\omega)=\{i_{k}(\omega)\}$ for $i_{k}\colon\Omega\to I$, then we also call the sequence $\{i_{k}\}_{k=0}^{\infty}$ a random control in $I$.

Remark 5.2**.**

The phrase “identically distributed” means that $\operatorname*{Pr}(\{\omega\in\Omega\colon I_{k}(\omega)=J\})=\operatorname*{Pr}(\{\omega\in\Omega\colon I_{n}(\omega)=J\})$ for all $k,n$ and all nonempty $J\subseteq I$ with $\#(J)\leq M$. The phrase “independent” means that $\operatorname*{Pr}(\bigcap_{k\in K}\{\omega\in\Omega\colon I_{k}(\omega)=J_{k}\})=\prod_{k\in K}\operatorname*{Pr}(\{\omega\in\Omega\colon I_{k}(\omega)=J_{k}\})$ for all finite $K$ and all nonempty $J_{k}\subseteq I$ with $\#(J_{k})\leq M$.

Before formulating our next result, we establish a very intuitive lemma in view of which a random control is repetitive almost surely. We recall that $\{I_{k}(\omega)\}_{k=0}^{\infty}$ is repetitive in $I^{\prime}$ if $I^{\prime}\subseteq\bigcup_{k=n}^{\infty}I_{k}(\omega)$ for all $n=0,1,2,\ldots$.

Lemma 5.3**.**

Let $\{I_{k}\}_{k=0}^{\infty}$ be a random control in $I$ and assume that $\operatorname*{Pr}(\{\omega\in\Omega\colon i\in I_{k}(\omega)\})>0$ for all $i\in I^{\prime}\subseteq I$. Then $\operatorname*{Pr}\left(\{\omega\in\Omega\colon\{I_{k}(\omega)\}_{k=0}^{\infty}\text{ is repetitive in }I^{\prime}\}\right)=1.$

• Proof.

Define the events $A_{i}^{k}:=\{\omega\in\Omega\colon i\in I_{k}(\omega)\}$ and the family $\mathcal{J}:=\{J\subseteq I\colon i\in J\text{ and }\#(J)\leq M\}$, where $M:=\sup_{\omega,k}\#(I_{k}(\omega))$. Since the events $A_{J}^{k}:=\{\omega\in\Omega\colon I_{k}(\omega)=J\}$ are disjoint for different values of $J\in\mathcal{J}$, we have

[TABLE]

By assumption, the variables $I_{k}$ are identically distributed. Consequently, we see that $\operatorname*{Pr}(A_{i}^{k})=\operatorname*{Pr}(A_{i}^{n})$ for all $i\in I$ and all $k,n=1,2,\ldots$. Hence the probability $p_{i}:=\operatorname*{Pr}(A_{i}^{k})$ does not depend on $k$ and, by assumption, $p_{i}>0$ for all $i\in I^{\prime}$.

Moreover, the events $A_{i}^{k}$ are independent over $k$. For simplicity, we only show this for a pair $K=\{k,n\}$, $k\neq n$, although the argument holds for any finite set of indices $K$. Indeed, by disjointness (over $J$) and independence (over $k$) of the events $A_{J}^{k}$, we have

[TABLE]

Consequently, for all $i\in I^{\prime}$, we obtain $\sum_{k=0}^{\infty}\operatorname*{Pr}(A_{i}^{k})=\sum_{k=0}^{\infty}p_{i}=\infty$ and, by applying the Borel-Cantelli lemma (see [19, Theorem 8.3.4]) to $A_{i}:=\limsup_{k\to\infty}A_{i}^{k}$, we have $\operatorname*{Pr}(A_{i})=1$.

Consider a decreasing sequence of sets $E_{k}:=\bigcap_{t=1}^{k}A_{i_{t}}$ where $k=1,2,\ldots,n$ for finite $I^{\prime}=\{i_{1},\ldots,i_{n}\}$, whereas $k=1,2,\ldots$ for infinite $I^{\prime}=\{i_{1},i_{2},\ldots\}$. Clearly, the set $A_{i}$ consists of all $\omega\in\Omega$ for which the membership $i\in I_{k}(\omega)$ happens infinitely many times in the sequence $\{I_{k}(\omega)\}_{k=0}^{\infty}$. Bearing this in mind, we get

[TABLE]

We now show, by induction, that $\operatorname*{Pr}(E_{k})=1$ for all $k$. Indeed, by definition, $\operatorname*{Pr}(E_{1})=\operatorname*{Pr}(A_{i_{1}})=1$, as we have already observed above. Moreover,

[TABLE]

By induction, $\operatorname*{Pr}(E_{k})=1$ and, since $\operatorname*{Pr}(A_{i_{k+1}})=1$, we conclude that $\operatorname*{Pr}(E_{k}\cup A_{i_{k+1}})=1$. Otherwise $\operatorname*{Pr}(E_{k+1})$ would be greater than one. Hence $\operatorname*{Pr}(E_{k+1})=1$, as asserted.

Observe that if $I^{\prime}$ is finite, then $E=E_{n}$ and consequently, $\operatorname*{Pr}(E)=\operatorname*{Pr}(E_{n})=1$. On the other hand, if $I^{\prime}$ is infinite, then $\{E_{k}\}_{k=0}^{\infty}$ is a decreasing sequence of events, where $E_{k+1}\subseteq E_{k}$ and, by the continuity of $\operatorname*{Pr}$ (see [21, Chapter II.9, Theorem E]), we have $\operatorname*{Pr}(E)=\lim_{k\to\infty}\operatorname*{Pr}(E_{k})=1.$ This completes the proof. $\blacksquare$

Theorem 5.4**.**

Let the sequence $\{x_{k}\}_{k=0}^{\infty}$ be defined as in Theorem 4.7 using a random control $\{I_{k}\}_{k=0}^{\infty}$ in $I$ (that is, for each $\omega\in\Omega$ we define a sequence $\{x_{k}(\omega)\}_{k=0}^{\infty}$, where at each step $k$ the iterate $x_{k+1}(\omega)$ is obtained by using the realization $I_{k}(\omega)$ instead of $I_{k}(x_{k})$, and where $x_{0}(\omega)=x_{0}\in Q$). Assume that conditions (i)–(iv) of Theorem 4.7 hold and that for all $x\notin C$, we have

[TABLE]

Then, by (5.5), almost surely, the control sequence $\{I_{k}\}_{k=0}^{\infty}$ is well matched with $C$ (condition (v)). Furthermore, almost surely, $x_{k}\in C\cap Q$ for some $k$ given that $\{x_{k}\}_{k=0}^{\infty}$ is bounded (with some positive probability).

• Proof.

Define the following events: $E_{0}:=\{\omega\in\Omega\colon\{x_{k}(\omega)\}_{k=0}^{\infty}\text{ is bounded}\}$, $E_{1}:=\{\omega\in\Omega\colon\{I_{k}(\omega)\}_{k=0}^{\infty}$ $\text{ is repetitive in }I^{\prime}\}$, $E_{2}:=\{\omega\in\Omega\colon\{I_{k}(\omega)\}_{k=0}^{\infty}$ is well matched with $C\}$ and $E_{3}:=\{\omega\in\Omega\colon x_{k}(\omega)\in C\cap Q\text{ for some }k\}$, where, as in Lemma 5.3, $I^{\prime}:=\{i\in I\colon\operatorname*{Pr}(A_{i}^{k})>0\}$ and $A_{i}^{k}:=\{\omega\in\Omega\colon i\in I_{k}(\omega)\}$. It suffices to show that $\operatorname*{Pr}(E_{3}\mid E_{0})=1$.

We first demonstrate that $C=\bigcap_{i\in I^{\prime}}C_{i}$. Suppose to the contrary that $x\in\bigcap_{i\in I^{\prime}}C_{i}\setminus C$. Clearly, $I_{+}(x)\subseteq I\setminus I^{\prime}$ and thus $\operatorname*{Pr}(A_{i}^{k})=0$ for all $i\in I_{+}(x)$. By (5.5), we obtain

[TABLE]

a contradiction.

Consequently, by Proposition 4.4, we get $E_{1}\subseteq E_{2}$. Moreover, by Lemma 5.3, we get $\operatorname*{Pr}(E_{1})=1$. Thus $\operatorname*{Pr}(E_{2})=1$, but also $\operatorname*{Pr}(E_{0}\cup E_{2})=1$. On the other hand, by Theorem 4.7, we have $E_{0}\cap E_{2}\subseteq E_{3}=E_{0}\cap E_{3}$, where, by assumption, $\operatorname*{Pr}(E_{0})>0$. Hence,

[TABLE]

and we arrive at $\operatorname*{Pr}(E_{3}\mid E_{0})=\operatorname*{Pr}(E_{0}\cap E_{3})/\operatorname*{Pr}(E_{0})\geq\operatorname*{Pr}(E_{0})/\operatorname*{Pr}(E_{0})=1$. $\blacksquare$

Remark 5.5** (Comparison with [29]).**

If $\{i_{k}\}_{k=0}^{\infty}$ is a single-valued random control in $I$. Then condition (5.5) recovers [29, Assumption 2], that is, $\operatorname*{Pr}(\{\omega\in\Omega\colon i_{k}(\omega)\in I_{+}(x)\})>0$ for all $x\notin C$. In particular, Theorem 5.4 recovers [29, Theorem 1] reduced to at most countably infinite number of constraints.

Acknowledgements. We thank two anonymous referees for their useful comments and helpful suggestions. This work was partially supported by the Israel Science Foundation (Grants 389/12 and 820/17), the Fund for the Promotion of Research at the Technion and by the Technion General Research Fund.

Appendix

Example 5.6**.**

We show that the relaxed alternating projection method, where $\alpha=1/2$, may fail to converge in finitely many steps if the sequence of overrelaxations $\{r_{k}\}_{k=0}^{\infty}$ is not monotone and when we use $r_{k}$ instead of $r_{[k]}$; compare with Remark 4.11 (e). To this end, consider the CFP with $Q=\mathcal{H}=\mathbb{R}^{2}$, $C_{1}:=\{(x,y)\colon x\leq 0\}$ and $C_{2}:=\{(x,y)\colon y\leq 0\}$. Clearly, $C_{1}\cap C_{2}=(-\infty,0]\times(-\infty,0]$. Define

[TABLE]

Set $(x_{0},y_{0}):=(1,1)$ and

[TABLE]

Then $r_{k}\to 0$ and $\sum_{k=0}^{\infty}r_{k}=\infty$, but $(x_{k},y_{k})\notin C_{1}\cap C_{2}$ for all $k=0,1,2,\ldots$. Indeed, observe that $x_{k}=0$ for all $k=1,2,\ldots$. Moreover, $y_{0}=y_{1}=1$ and, by induction,

[TABLE]

Example 5.7**.**

We show that the subgradient projection method (4.13), when combined with a repetitive control (as in [13]), may fail to converge in finitely many steps if we choose to use “$r_{k}$” instead of “$r_{[k]}$”, even though the sequence of overrelaxations $\{r_{k}\}_{k=0}^{\infty}$ is decreasing. Indeed, consider the CFP with $Q=\mathcal{H}=\mathbb{R}^{2}$, $C_{1}:=\{(x,y)\colon f_{1}(x,y)\leq 0\}$ and $C_{2}:=\{(x,y)\colon f_{2}(x,y)\leq 0\}$, where $f_{1}(x,y):=|y|-1$ and $f_{2}(x,y):=x^{2}-1$. Thus $C_{1}\cap C_{2}=[-1,1]\times[-1,1]$ and the Slater condition is satisfied since $f_{1}(0,0)=f_{2}(0,0)=-1<0$. Define two auxiliary sequences $\{a_{k}\}_{k=0}^{\infty}$ and $\{b_{k}\}_{k=0}^{\infty}$ by

[TABLE]

and let the sequence of overrelaxations $\{r_{k}\}_{k=0}^{\infty}$ consist of all the elements of $\{a_{k}\}_{k=0}^{\infty}$ and $\{b_{k}\}_{k=0}^{\infty}$ sorted in a decreasing order, that is,

[TABLE]

Observe that $r_{k}\to 0$ monotonically and $\sum_{k=0}^{\infty}r_{k}=\infty$, as required in [13]. Indeed, the former condition follows from the inequality $b_{k+1}\leq\frac{b_{k}}{16}$ and the latter one from the definition of $a_{k}$. Let $m_{k}$ and $n_{k}$ denote the position (we start counting from [math]) of $a_{k}$ and $b_{k}$ in the sequence $\{r_{k}\}_{k=0}^{\infty}$, respectively, and define the control sequence $\{i_{k}\}_{k=0}^{\infty}$ by $i_{m_{k}}:=1$ and $i_{n_{k}}:=2$. It is not difficult to see that $\{i_{k}\}_{k=0}^{\infty}$ is repetitive. Following (4.13), we define

[TABLE]

whenever $f_{i_{k}}(x_{k},y_{k})>0$ and $(x_{k+1},y_{k+1}):=(x_{k},y_{k})$ otherwise, where $g_{i_{k}}(x_{k},y_{k})\in\partial f_{i_{k}}(x_{k},y_{k})$. Then $(x_{k},y_{k})\notin C_{1}\cap C_{2}$ for all $k=0,1,2,\ldots$.

• Proof.

Observe that by (5.13), we get

[TABLE]

Moreover,

[TABLE]

On the other hand, by the definition of $m_{k}$ and $n_{k}$, we have $r_{m_{k}}=a_{k}$ and $r_{n_{k}}=b_{k}$. Consequently, by the choice of the starting point, we obtain $y_{k}=0$ for all $k=1,2,\ldots$.

We claim that $x_{n_{k}}=1+\sqrt{2b_{k}}$. Indeed, by the equality $n_{0}=2$ and by (5.14), we have

[TABLE]

Observe that, by (5.14), we also obtain $x_{n_{k+1}}=x_{n_{k}+1}$. Consequently, by (5.15) and by induction,

[TABLE]

Using the positivity of $b_{k}$, we see that $x_{n_{k}}>1$ which, when combined with (5.14), yields that $x_{k}>1$ for all $k=0,1,2,\ldots$. This implies that $(x_{k},y_{k})\notin C_{1}\cap C_{2}$ for all $k=0,1,2,\ldots,$ as claimed. $\blacksquare$