Krasnoselskii-Mann method for non-self mappings

Abstract

Let H be a Hilbert space and let C be a closed, convex and nonempty subset of H. If $T:C\to H$ is a non-self and non-expansive mapping, we can define a map $h:C\to\mathbb{R}$ by $h(x):=\inf\{\lambda\geq 0:\lambda x+(1-\lambda)Tx\in C\}$. Then, for a fixed $x_{0}\in C$ and for $\alpha_{0}:=\max\{1/2, h(x_{0})\}$, we define the Krasnoselskii-Mann algorithm $x_{n+1}=\alpha _{n}x_{n}+(1-\alpha_{n})Tx_{n}$, where $\alpha_{n+1}=\max\{\alpha_{n},h(x_{n+1})\}$. We will prove both weak and strong convergence results when C is a strictly convex set and T is an inward mapping.

1 Introduction

Let C be a closed, convex and nonempty subset of a Hilbert space H and let $T:C\to H$ be a non-expansive mapping such that the fixed point set $\operatorname{Fix}(T):=\{x\in C:Tx=x\}$ is not empty.

For a real sequence $\{\alpha_{n}\}\subset(0,1)$, we will consider the iterations

$$ \begin{cases} x_{0}\in C, \\ x_{n+1}=\alpha_{n}x_{n}+(1-\alpha_{n})Tx_{n}. \end{cases} $$

(1)

If T is a self-mapping, the iterative scheme above has been studied in an impressive amount of papers (see [1] and the references therein) in the last decades and it is often called ‘segmenting Mann’ [2–4] or ‘Krasnoselskii-Mann’ (e.g., [5, 6]) iteration.

A general result on algorithm (1) is due to Reich [7] and states that the sequence $\{x_{n}\}$ weakly converges to a fixed point of the operator T under the following assumptions:

• (C1) T is a self-mapping, i.e., $T:C\to C$ and

• (C2) $\{\alpha_{n}\}$ is such that $\sum_{n}\alpha _{n}(1-\alpha_{n})=+\infty$.

In this paper, we are interested in lowering condition (C1) by allowing T to be non-self at the price of strengthening the requirements on the sequence $\{\alpha_{n}\}$ and on the set C. Indeed, we will assume that C is a strictly convex set and that the non-expansive map $T:C\to H$ is inward.

Historically, the inward condition and its generalizations were widely used to prove convergence results for both implicit [8–11] and explicit (see, e.g., [1, 12–14]) algorithms. However, we point out that the explicit case was only studied in conjunction with processes involving the calculation of a projection or a retraction $P:H\to C$ at each step.

As an example, in [12], the following algorithm is studied:

$$x_{n+1}=P\bigl(\alpha_{n}f(x_{n})+(1- \alpha_{n})Tx_{n}\bigr), $$

where $T:C\to H$ satisfies the weakly inward condition, f is a contraction and $P:H\to C$ is a non-expansive retraction.

We point out that in many real world applications, the process of calculating P can be a resource consumption task and it may require an approximating algorithm by itself, even in the case when P is the nearest point projection.

To overcome the necessity of using an auxiliary mapping P, for an inward and non-expansive mapping $T:C\to H$, we will introduce a new search strategy for the coefficients $\{\alpha_{n}\}$ and we will prove that the Krasnoselskii-Mann algorithm

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

is well defined for this particular choice of the sequence $\{\alpha _{n}\}$. Also we will prove both weak and strong convergence results for the above algorithm when C is a strictly convex set.

We stress that the main difference between the classical Krasnoselskii-Mann and our algorithm is that the choice of the coefficient $\alpha_{n}$ is not made a priori in the latter, but it is constructed step to step and determined by the values of the map T and the geometry of the set C.

2 Main result

We will make use of the following.

Definition 1

A map $T:C\to H$ is said to be inward (or to satisfy the inward condition) if, for any $x\in C$, it holds

$$ Tx\in I_{C}(x):=\bigl\{ x+c(u-x):c\geq1\mbox{ and }u\in C\bigr\} . $$

(2)

We refer to [15] for a comprehensive survey on the properties of the inward mappings.

Definition 2

A set $C\subset H$ is said to be strictly convex if it is convex and with the property that $x,y\in\partial C$ and $t\in(0,1)$ implies that

$$tx+(1-t)y\in\mathring{C}. $$

In other words, if the boundary ∂C does not contain any segment.

Definition 3

A sequence $\{y_{n}\}\subset C$ is Fejér-monotone with respect to a set $D\subset C$ if, for any element $y\in D$,

$$\|y_{n+1}-y\|\leq\|y_{n}-y\| \quad \forall n\in\mathbb{N}. $$

For a closed and convex set C and a map $T:C\to H$, we define a mapping $h:C\to\mathbb{R}$ as

$$ h(x):=\inf\bigl\{ \lambda\geq0:\lambda x+(1-\lambda)Tx\in C\bigr\} . $$

(3)

Note that the above quantity is a minimum since C is closed. In the following lemma, we group the properties of the function defined above.

Lemma 1

Let C be a nonempty, closed and convex set, let $T:C\to H$ be a mapping and define $h:C\to\mathbb{R}$ as in (3). Then the following properties hold:

• (P1) for any $x\in C$, $h(x)\in[0,1]$ and $h(x)=0$ if and only if $Tx\in C$;

• (P2) for any $x\in C$ and any $\alpha\in[h(x),1]$, $\alpha x+(1-\alpha)Tx\in C$;

• (P3) if T is an inward mapping, then $h(x)<1$ for any $x\in C$;

• (P4) whenever $Tx\notin C$, $h(x)x+(1-h(x))Tx\in\partial C$.

Proof

Properties (P1) and (P2) follow directly from the definition of h. To prove (P3), observe that (2) implies

$$\frac{1}{c}Tx+\biggl(1-\frac{1}{c}\biggr)x\in C $$

for some $c\geq1$. As a consequence,

$$h(x)=\inf\bigl\{ \lambda\geq0:\lambda x+(1-\lambda)Tx\in C\bigr\} \leq\biggl(1- \frac{1}{c}\biggr)< 1. $$

In order to verify (P4), we first note that $h(x)>0$ by property (P1) and that $h(x)x+(1-h(x))Tx\in C$. Let $\{\eta_{n}\}\subset(0,h(x))$ be a sequence of real numbers converging to $h(x)$ and note that, by the definition of h, it holds

$$z_{n}:=\eta_{n}x+(1-\eta_{n})Tx\notin C $$

for any $n\in\mathbb{N}$. Since $\eta_{n}\to h(x)$ and

$$\bigl\Vert z_{n}-h(x)x-\bigl(1-h(x)\bigr)Tx\bigr\Vert =\bigl\vert \eta_{n}-h(x)\bigr\vert \|x-Tx\|, $$

it follows that $z_{n}\to h(x)x+(1-h(x))Tx\in C$, so that this last must belong to ∂C. □

Our main result is the following.

Theorem 1

Let C be a convex, closed and nonempty subset of a Hilbert space H and let $T:C\to H$ be a mapping. Then the algorithm

$$ \begin{cases} x_{0}\in C, \\ \alpha_{0}:=\max\{\frac{1}{2},h(x_{0})\}, \\ x_{n+1}:=\alpha_{n}x_{n}+(1-\alpha_{n})Tx_{n}, \\ \alpha_{n+1}:=\max\{\alpha_{n},h(x_{n+1})\} \end{cases} $$

(4)

is well defined.

If we further assume that

• 1. C is strictly convex and

• 2. T is a non-expansive mapping, which satisfies the inward condition (2) and such that $\operatorname{Fix}(T)\neq\emptyset$,

then $\{x_{n}\}$ weakly converges to a point $p\in \operatorname{Fix}(T)$. Moreover, if $\sum_{n=0}^{\infty}(1-\alpha_{n})<\infty$, then the convergence is strong.

Proof

To prove that the algorithm is well defined, it is sufficient to note that $\alpha_{n}\in[h(x_{n}),1]$ for any $n\in\mathbb{N}$; then, by recalling property (P2) from Lemma 1, it immediately follows that

$$x_{n+1}=\alpha_{n}x_{n}+(1-\alpha_{n})Tx_{n} \in C. $$

Assume now that T satisfies the inward condition. In this case, by property (P3) of the previous lemma, we obtain that the non-decreasing sequence $\{\alpha_{n}\}$ is contained in $[\frac{1}{2},1)$. Also, since T is non-expansive and with at least one fixed point, it follows by standard arguments that $\{x_{n}\}$ is Fejér-monotone with respect to $\operatorname{Fix}(T)$ and, as a consequence, both $\{x_{n}\}$ and $\{Tx_{n}\}$ are bounded.

Firstly, assume that $\sum_{n=0}^{\infty}(1-\alpha_{n})=\infty$. Then, since $\alpha_{n}\geq\frac{1}{2}$, we derive that $\sum_{n=0}^{\infty}\alpha_{n}(1-\alpha_{n})=\infty$ and from Lemma 2 of [16] we obtain that

$$\|x_{n}-Tx_{n}\|\to0. $$

This fact, together with the Fejér-monotonicity of $\{x_{n}\}$ proves that the sequence weakly converges in $\operatorname{Fix}(T)$ (see [17], Proposition 2.1).

Suppose that

$$ \sum_{n=0}^{\infty}(1-\alpha_{n})< \infty. $$

(5)

Since

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

and by the boundedness of $\{x_{n}\}$ and $\{Tx_{n}\}$, it is promptly obtained that

$$\sum_{n=0}^{\infty}\|x_{n+1}-x_{n} \|< \infty, $$

i.e., $\{x_{n}\}$ is a strongly Cauchy sequence and hence $x_{n}\to x^{*}\in C$.

Note that T satisfies the inward condition. Then, by applying properties (P2) and (P3) from Lemma 1, we obtain that $h(x^{*})<1$ and that for any $\mu\in(h(x^{*}),1)$ it holds

$$ \mu x^{*}+(1-\mu)Tx^{*}\in C. $$

(6)

On the other hand, we observe that since $\lim_{n\to\infty}\alpha_{n}=1$ by (5) and since $\alpha_{n}=\max\{\alpha _{n-1}, h(x_{n})\}$ holds, it follows that we can choose a sub-sequence $\{x_{n_{k}}\}$ with the property that $\{h(x_{n_{k}})\}$ is non-decreasing and $h(x_{n_{k}})\to1$. In particular, for any $\mu<1$,

$$ \mu x_{n_{k}}+(1-\mu)Tx_{n_{k}}\notin C $$

(7)

eventually holds.

Choose $\mu_{1},\mu_{2}\in(h(x^{*}),1)$ with $\mu_{1}\neq\mu_{2}$ and set $v_{1}:=\mu_{1}x^{*}+(1-\mu_{1})Tx^{*}$ and $v_{2}:=\mu _{2}x^{*}+(1-\mu_{2})Tx^{*}$. Then, whenever $\mu\in[\mu_{1},\mu_{2}]$, by (6) we have that $v:=\mu x^{*}+(1-\mu)Tx^{*}\in C$. Moreover,

$$\mu x_{n_{k}}+(1-\mu)Tx_{n_{k}}\to v $$

since $x_{n}\to x^{*}$. This last, together with (7), implies that $v\in\partial C$ and $[v_{1},v_{2}]\subset\partial C$, since μ is arbitrary.

By the strict convexity of C, we derive that

$$\mu_{1}x^{*}+(1-\mu_{1})Tx^{*}= \mu_{2}x^{*}+(1-\mu_{2})Tx^{*} $$

and $x^{*}=Tx^{*}$ must necessarily hold, i.e., $\{x_{n}\}$ strongly converges to a fixed point of T. □

Remark 1

Following the same line of proof, it can be easily seen that the same results hold true if the starting coefficient $\alpha_{0}=\max\{ \frac{1}{2},h(x_{0})\}$ is substituted by $\alpha_{0}=\max\{b,h(x_{0})\}$, where $b\in(0,1)$ is a fixed and arbitrary value. In the statement of Theorem 1, the value $b=\frac{1}{2}$ was taken to ease the notation.

We also note that the value $h(x_{n})$ can be replaced, in practice, by $h_{n}=1-\frac{1}{2^{j_{n}}}$, where $j_{n}:=\min\{j\in\mathbb {N}:(1-\frac{1}{2^{j}})x_{n}+\frac{1}{2^{j}}Tx_{n}\in C\}$.

Remark 2

As it follows from the proof, the condition $\sum_{n}(1-\alpha _{n})<\infty$ provides a localization result for the fixed point $x^{*}$ as a side result. Indeed, in this case, it holds that $x^{*}=v_{1}=v_{2}$ belongs to the boundary ∂C of the set C.

Remark 3

In [18], for a closed and convex set C, the map

$$f(x):=\inf\bigl\{ \lambda\in[0,1]:x\in\lambda C\bigr\} $$

was introduced and used in conjunction with an iterative scheme to approximate a fixed point of minimum norm (see also [19]). Indeed, in the above mentioned paper, it is proved that the iterative scheme

$$\begin{cases} \lambda_{n}=\max\{f(x_{n}),\lambda_{n-1}\}, \\ y_{n}=\alpha_{n}x_{n}+(1-\alpha_{n})Tx_{n}, \\ x_{n+1}=\alpha_{n}\lambda_{n}x_{n}+(1-\alpha_{n})y_{n} \end{cases} $$

strongly converges under the assumptions that $\{\alpha_{n}\}$ is a sequence in $(0,1)$ such that $\lim_{n}\frac{\alpha_{n}}{(1-\lambda_{n})}=0$ and that $\sum_{n}(1-\lambda_{n})\alpha_{n}=\infty$. We point out that the mentioned conditions appear to be difficult to be checked as they involve the geometry of the set C.

We illustrate the statement of our results with a brief example.

Example 1

Let $H=l^{2}(\mathbb{R})$ and let $C:=B_{1}\cap B_{2}$, where $B_{1}:=\{ (t_{i})_{i\in\mathbb{N}}:(t_{1}-49.995)^{2}+\sum_{i=2}^{\infty }t_{i}^{2}\leq(50.005)^{2}\}$ and $B_{2}:=\{(t_{i})_{i\in\mathbb{N}}:\sum_{i=1}^{\infty}t_{i}^{2}\leq 1\}$. Then C is a nonempty, closed and strictly convex subset of H. Let $T:C\to H$ be the map defined by $T(t_{1},t_{2},\ldots,t_{i},\ldots ):=(-t_{1},t_{2},\ldots,t_{i},\ldots)$, then T is a non-expansive inward map with $\operatorname{Fix}(T)=\{(0,t_{2},\ldots ,t_{i},\ldots):\sum_{i=2}^{\infty}t_{i}^{2}\leq1\}$. If we use the algorithm

$$\begin{cases} x_{0}=(t_{i})_{i\in\mathbb{N}}\in C, \\ \alpha_{0}:=\max\{\frac{1}{2},h(x_{0})\}, \\ x_{n+1}:=\alpha_{n}x_{n}+(1-\alpha_{n})Tx_{n}, \\ \alpha_{n+1}:=\max\{\alpha_{n},h(x_{n+1})\}, \end{cases} $$

then, by the natural symmetry of the problem, we obtain the constant sequence

$$x_{1}=\cdots=x_{n}=(0,t_{2}, \ldots,t_{i},\ldots)\in \operatorname{Fix}(T). $$

If we use the algorithm

$$\begin{cases} x_{0}=(t_{i})_{i\in\mathbb{N}}\in C, \\ \alpha_{0}:=\max\{0.01,h(x_{0})\}, \\ x_{n+1}:=\alpha_{n}x_{n}+(1-\alpha_{n})Tx_{n}, \\ \alpha_{n+1}:=\max\{\alpha_{n},h(x_{n+1})\}, \end{cases} $$

then $\{x_{n}\}$ still converges in $\operatorname{Fix}(T)$, but $\{x_{n}\}\cap \operatorname{Fix}(T)=\emptyset$ whenever $t_{i}\neq0$.

We conclude the paper by including few question that appear to be still open to the best of our knowledge.

Question 1

It has been proved that the Krasnoselskii-Mann algorithm converges for general classes of mappings (see, e.g., [20] and [21]). By maintaining the same assumption on the set C and the inward condition of the involved map, it appears to be natural to ask for which classes of mappings the same result of Theorem 1 still holds.

Question 2

Under which assumptions can algorithm (4) be adapted to produce a converging sequence to a common fixed point for a family of mappings? In other words, does the algorithm

$$\begin{cases} x_{0}\in C, \\ \alpha_{0}:=\max\{\frac{1}{2},h_{n}(x_{0})\}, \\ x_{n+1}:=\alpha_{n}x_{n}+(1-\alpha_{n})T_{n}x_{n}, \\ \alpha_{n+1}:=\max\{\alpha_{n},h_{n+1}(x_{n+1})\} \end{cases} $$

converge to a common fixed point of the family $\{T_{n}\}$, where

$$h_{n}(x):=\inf\bigl\{ \lambda\geq0:\lambda x+(1-\lambda)T_{n}x \in C\bigr\} $$

and under suitable hypotheses?

We refer to [22] and [23] for two examples regarding the classical Krasnoselskii-Mann algorithm.

Question 3

In the classical literature, it has been proved that the inward condition can be often dropped in favor of a weaker condition. For example, a mapping $T:C\to X$ is said to be weakly inward (or to satisfy the weakly inward condition) if

$$Tx\in\overline{I_{C}(x)}\quad \forall x\in C. $$

Does Theorem 1 hold even for weakly inward mappings?

On the other hand, we observe that the strict convexity of the set C does appear to be unusual for results regarding the convergence of Krasnoselskii-Mann iterations. We do not know if our result can hold for a convex and closed set C, even at the price of strengthening the requirements on the map T.