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# Krasnoselskii-Mann method for non-self mappings

Fixed Point Theory and Applications20152015:39

https://doi.org/10.1186/s13663-015-0287-4

• Accepted: 3 March 2015
• Published:

## 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.

## Keywords

• Hilbert Space
• Convergence Result
• Resource Consumption
• Nonempty Subset
• Real World Application

## 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’ [24] 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:
1. (C1)

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

2. (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 [811] and explicit (see, e.g., [1, 1214]) 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:
1. (P1)

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

2. (P2)

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

3. (P3)

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

4. (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. 1.

C is strictly convex and

2. 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.

## Declarations

### Acknowledgements

This project was funded by Ministero dell’Istruzione, dell’Universitá e della Ricerca (MIUR).

Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

## Authors’ Affiliations

(1)
Department of Mathematics and Computer Science, Universitá della Calabria, Rende, CS, Italy
(2)
Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah, 21589, Saudi Arabia

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