Schröder's method and the infinity point

1.   Introduction

A self-mapping $\mathcal{F}$ on a convex, closed, and bounded subset $K$ of a Banach space $U$ is known as nonexpansive if $\lVert{\mathcal{F}u -\mathcal{F}v}\rVert$ $\leq$ $\lVert{u-v}\rVert$, $u, v\in U$ and need not essentially possess a fixed point. It is widely known that a point $u \in U$ is a fixed point or an invariant point if $\mathcal{F}u = u$. However, some researchers ensured the survival of a fixed point of nonexpansive mapping in Banach spaces utilizing suitable geometric postulates. Numerous mathematicians have extended and generalized these conclusions to consider several nonlinear mappings. One such special class of mapping is Suzuki generalized nonexpansive mapping (SGNM). Many extensions, improvements and generalizations of nonexpansive mappings are given by eminent researchers (see ^{[8,9,10,13,15,17,19,21,22,25]}, and so on). On the other hand, Krasnosel'skii ^[16] investigated a novel iteration of approximating fixed points of nonexpansive mapping. A sequence $\{u_i\}$ utilizing the Krasnosel'skii iteration is defined as: $u_1 = u, u_{i+1} = (1-\alpha)u_i + \alpha \mathcal{F}u_i$, where $\alpha \in(0, 1)$ is a real constant. This iteration is one of the iterative methods which is the extension of the celebrated Picard iteration ^[24], $u_{i+1} = \mathcal{F}u_i$. The convergence rate of the Picard iteration ^[24] is better than the Krasnosel'skii iteration although the Picard iterative scheme is not essentially convergent for nonexpansive self-mappings. It is interesting to see that the fixed point of a self-mapping $\mathcal{F}$ is also a fixed point of the iteration $\mathcal{F}^n$ $(n\in \mathbb{N})$, of the self-mapping $\mathcal{F}$ but the reverse implication is not feasible. Recently several authors presented extended and generalized results for better approximation of fixed points (see ^{[1,3,11,23,26,27]}).

We present convergence and common fixed point conclusions for the associated $\alpha$-Krasnosel'skii mappings satisfying condition (E) in the current work. Also, we support these with nontrivial illustrative examples to demonstrate that our conclusions improve, generalize and extend comparable conclusions of the literature.

2.   Preliminaries

We symbolize $F(\mathcal{F})$, to be the collection of fixed points of a self-mapping $\mathcal{F}$, that is, $F(\mathcal{F})$ = $\{u\in U : \mathcal{F}u = u\}$. We begin with the discussion of convex Banach spaces, $\alpha$-Krasnosel'skii mappings and the condition (E) (see ^{[12,18,20,23]}).

Definition 2.1. ^[14] A Banach space $U$ is uniformly convex if, for $\epsilon \in (0, 2]\ $ $\exists$ $\delta > 0$ satisfying, $\lVert\frac{u + v}{2}\rVert$ $\leq 1-\delta$ so that $\lVert u-v\rVert > \epsilon$ and $\lVert u \rVert = \lVert v\rVert = 1$, $u, v \in U$.

Definition 2.2. ^[14] A Banach space $U$ is strictly convex if, $\lVert \frac{u+v}{2}\rVert < 1$ so that $u \neq v, \lVert u\rVert = \lVert v\rVert = 1$, $u, v \in U$.

Theorem 2.1. ^[5] Suppose $U$ is a uniformly convex Banach space. Then $\exists$ a $\gamma > {0}$, satisfying $\lVert{\frac{1}{2}(\, u+v)\, }\rVert$$\leq[\, 1-\gamma\frac{\epsilon}{\delta}]\, \delta$ for every $\epsilon, \; {\delta > 0}$ so that $\lVert{u-v}\rVert\geq{\epsilon}$, $\; \lVert{u}\rVert\leq{\delta}$ and $\lVert{v}\rVert\leq\delta$, for $u, v \in U$.

Theorem 2.2. ^[14] The subsequent postulates are equivalent in a Banach space $U$:

(i) $U$ is strictly convex.

(ii) $u = 0$ or $v = 0$ or $v = cu$ for $c > {0}$, whenever $\lVert { u + v }\rVert$ = $\lVert{u}\rVert + \lVert{v}\rVert, u, v \in U$.

Definition 2.3. Suppose $\mathcal{F}$ is a self-mapping on a non-void subset $V$ of a Banach space $U$.

(i) Suppose for $u \in U$, $\exists$ $v \in V$ so that for all $w \in V$, $\lVert{v-u}\rVert$ $\leq$$\lVert{w-u}\rVert$. Then $v$ is a metric projection ^[6] of $U$ onto $V$, and is symbolized by $P_V(.)$. The mapping $P_V(u) : U \rightarrow V$ is the metric projection if $P_V(x)$ exists and is determined uniquely for each $x\in U$.

(ii) $\mathcal{F}$ satisfies condition $(E_{\mu})$ ^[23] on $V$ if $\exists\;$ $\; \mu\geq{1}$, satisfying $\lVert{u-\mathcal{F}v}\rVert \leq \mu\lVert{u-\mathcal{F}u}\rVert+\lVert{u-v}\rVert, \; \; u, v\in V$. Moreover, $\mathcal{F}$ satisfies condition $(E)$ on $V$, if $\mathcal{F}$ satisfies $(E_{\mu})$.

(iii) $\mathcal{F}$ satisfies condition (E) ^[23] and $F(\mathcal{F}) \neq 0$, then $\mathcal{F}$ is quasi-nonexpansive.

(iv) $\mathcal{F}$ is a generalized $\alpha$-Reich-Suzuki nonexpansive ^[21] if for an $\alpha \in [\, 0, 1)\,$, $\frac{1}{2}\lVert{u-\mathcal{F}u}\rVert \leq \lVert{u-v}\rVert \implies \lVert{\mathcal{F}u-\mathcal{F}v}\rVert \leq$ max $\{ \alpha\lVert{\mathcal{F}u-u}\rVert + \alpha \lVert{\mathcal{F}v-v}\rVert + (\, 1-2\alpha)\, \lVert{u-v}\rVert, \; \alpha\lVert{\mathcal{F}u-v}\rVert + \alpha\lVert{\mathcal{F}v-u}\rVert + (\, 1-2\alpha)\, \lVert{u-v}\rVert\}$, $\forall \; u, v \in V$.

(v) A self-mapping $\mathcal{F}_{\alpha}:V\to V$ is an $\alpha$-Krasnosel'skii associated with $\mathcal{F}$ ^[2] if, $\mathcal{F}_{\alpha} u = (1-\alpha) u + \alpha \mathcal{F}u$, for $\alpha \in(0, 1)$, $u \in V$.

(vi) $\mathcal{F}$ is asymptotically regular ^[4] if $\lim\limits_{n\to\infty} \lVert{\mathcal{F}^nu-\mathcal{F}^{n+1}u}\rVert = 0$.

(vii) $\mathcal{F}$ is a generalized contraction of Suzuki type ^[2], if $\exists$ $\beta\in(0, 1)$ and $\alpha_1, \alpha_2, \alpha_3 \in [\, 0, 1]\,$, where $\alpha_1 + 2\alpha_2 + 2\alpha_3 = 1$, satisfying $\beta\lVert{u-\mathcal{F}u}\rVert \leq \lVert{u-v}\rVert$ implies

(viii) $\mathcal{F}$ is $\alpha$-nonexpansive ^[7] if $\exists$ an $\alpha < 1$ satisfying

Theorem 2.3. ^[5] A continuous mapping on a non-void, convex and compact subset $V$ of a Banach space $U$ has a fixed point in $V$.

Pant et al.^[23] derived a proposition that if $\beta = \frac{1}{2}$, then a generalized contraction of Suzuki type is a generalized $\alpha$-Reich-Suzuki nonexpansive. Moreover, the reverse implication may not necessarily hold.

Lemma 2.1. ^[2] Let $\mathcal{F}$ be a generalized contraction of the Suzuki type on a non-void subset $V$ of a Banach space $U$. Let $\beta\in[\, \frac{1}{2}, 1)\,$, then

Proposition 2.1. ^[23] Let $\mathcal{F}$ be a generalized contraction of the Suzuki type on a non-void subset $V$ of a Banach space $U$, then $\mathcal{F}$ satisfies condition (E).

The converse of this proposition is not true, which can be verified by the following example.

Example 2.1. Suppose $U = (\mathbb{R}^2, \left\|.\right\|)$ with the Euclidean norm and $V = [-1, 1]\times[-1, 1]$ be a subset of $U$. Let $\mathcal{F}:V\to V$ be defined as

Case I. Let $x = (u_1, u_2), y = (v_1, v_2)$ with $|u_1|\leq\frac{1}{2}$, $|v_1|\leq\frac{1}{2}$. Then,

which implies

Case II. If $|u_1|\leq\frac{1}{2}$, $|v_1| > \frac{1}{2}$

Consider

Hence,

Here $\mu = 8$ satisfies the inequality.

Case III. If $|u_1| > \frac{1}{2}$, $|v_1|\leq\frac{1}{2}$

Consider

So,

Case IV. If $|u_1| > \frac{1}{2}$ and $|v_1| > \frac{1}{2}$, then

Since $|u_1| > \frac{1}{2}$ and $|v_1| > \frac{1}{2}$, by simple calculation as above, we attain

Thus, $\mathcal{F}$ satisfies condition (E) for $\mu = 4$.

Now, suppose $x = (\frac{1}{2}, 1)$ and $y = (1, 1)$, so

Clearly, $\left\|\mathcal{F}x-\mathcal{F}y\right\| = \sqrt{(\frac{5}{4})^2+(1-1)^2} = \frac{5}{4}$.

Consider

Since $\alpha_1, \alpha_2, \alpha_3\geq 0$, therefore

which is a contradiction.

Thus, $\mathcal{F}$ is not a generalized contraction of the Suzuki type.

3.   Results

Now, we establish results for a pair of $\alpha$-Krasnosel'skii mappings using condition (E).

Theorem 3.1. Let $\mathcal{F}_i$, for $i\in\{1, 2\}$, be self-mappings on a non-void convex subset $V$ of a uniformly convex Banach space $U$ and satisfy condition (E) so that $F(\mathcal{F}_1\cap \mathcal{F}_2)\neq \phi$. Then the $\alpha$-Krasnosel'skii mappings $\mathcal{F}_{i_{\alpha}}$, $\alpha \in (\, 0, 1)\,$ and $i\in\{1, 2\}$ are asymptotically regular.

Proof. Let $v_0 \in V$. Define $v_{n+1} = \mathcal{F}_{i_{\alpha}} v_n$ for $i\in\{1, 2\}$ and $n \in N\cup\{0\}$. Thus,

and

It is sufficient to show that $\lim\limits_{n\to\infty} \lVert{\mathcal{F}_i v_n-v_n}\rVert = 0$ to prove $\mathcal{F}_{i_{\alpha}}$ is asymptotically regular.

By definition, for $u_0 \in F(\mathcal{F}_1\cap \mathcal{F}_2)$, we have

and for $i\in\{1, 2\}$,

Thus, the sequence $\{\lVert u_0-v_n\rVert\}$ is bounded by $s_0 = \lVert u_0-v_0\rVert$. From inequality (3.2), $v_n \to u_0$ as $n \to \infty$, if $v_{n_0} = u_0$, for some $n_0\in \mathbb{N}$. So, assume $v_n\neq u_0$, for $n \in \mathbb{N}$, and

If $\alpha\leq\frac{1}{2}$ and using Eq (3.3), we obtain

As the space $U$ is uniformly convex with $\lVert w_n\rVert \leq 1$, $\lVert e_n \rVert \leq 1$ and $\lVert w_n-e_n\rVert = \frac{\lVert v_n-\mathcal{F}_iv_n\rVert}{ \lVert u_0-v_n\rVert} \geq \frac{\lVert v_n-\mathcal{F}_iv_n\rVert}{s_0} = \epsilon$ (say) for $i\in\{1, 2\}$, we obtain

From inequalities (3.4) and (3.5),

By induction, it follows that

We shall prove that $\lim\limits_{n\to\infty}\lVert \mathcal{F}_iv_n-v_n\rVert = 0$ for $i\in\{1, 2\}$. On the contrary, consider that $\{ \lVert \mathcal{F}_iv_n-v_n\rVert\}$ for $i\in\{1, 2\}$ is not converging to zero, and we have a subsequence $\{v_{n_k}\},$ of $\{v_n\},$ satisfying $\lVert \mathcal{F}_iv_{n_k}-v_{n_k} \rVert$ converges to $\zeta > 1$. As $\delta\in [\, 0, 1]\,$ is increasing and $\alpha\leq\frac{1}{2}$, $1-2\alpha\delta\frac{\lVert v_k-\mathcal{F}_iv_k\rVert}{s_0}\in [\, 0, 1]\,$, $i\in\{1, 2\}$, for all $k \in \mathbb{N}$. Since $\lVert \mathcal{F}_iv_{n_k}-v_{n_k}\rVert \to \zeta$ so, for sufficiently large $k, \; \; \lVert \mathcal{F}_iv_{n_k}-v_{n_k}\rVert\geq\frac{\zeta}{2}$, from inequality (3.7), we have

Making $k\to\infty$, it follows that $v_{n_{k}}\to u_0$. By inequality (3.1), we get $\mathcal{F}_{i} v_{n_{k}} \to u_0$ and $\lVert v_{n_{k}} -\mathcal{F}_{i} v_{n_{k}} \rVert \to 0$ as $k\to \infty$, which is a contradiction. If $\alpha > \frac{1}{2}$, then $1-\alpha < \frac{1}{2}$, because $\alpha \in (\, 0, 1)\,$. Now, for $i\in\{1, 2\}$

By the uniform convexity of $U$, we attain, for $i\in\{1, 2\}$,

By induction, we get

Similarly, it can be easily proved that $\lVert \mathcal{F}_{i}v_n-v_n\rVert \to 0$ as $n \to \infty$, which implies that $\mathcal{F}_{i_{\alpha}}$ for $i\in\{1, 2\}$, is asymptotically regular.

Next, we demonstrate by a numerical experiment that a pair of $\alpha$-Krasnosel'skii mappings are asymptotically regular for fix $\alpha \in (0, 1)$.

Example 3.1. Assume $U = (R^2, ||.||)$ with Euclidean norm and $V = \{ u\in R^2 : \lVert u\rVert \leq1 \}$, to be a convex subset of $U$. $\mathcal{F}_i$ for $i\in \{1, 2\}$ be self-mappings on $V$, satisfying

Then, clearly both $\mathcal{F}_1$ and $\mathcal{F}_2$ satisfy the condition $(E)$ and $F(\mathcal{F}_1\cap \mathcal{F}_2) = (0, 0)$. Now, we will show that the $\alpha$-Krasnosel'skii mappings $\mathcal{F}_{i_{\alpha}}$ for $\alpha \in (\, 0, 1)\,$ and $i\in\{1, 2\}$ are asymptotically regular.

Since $\mathcal{F}_1$ is the identity map, $\alpha$- Krasnosel'skii mapping $\mathcal{F}_{1\alpha}$ is also identity and hence asymptotically regular.

Now, we show $\mathcal{F}_{2\alpha}$ is asymptotically regular, let $u = (u_1, u_2) \in V$

Continuing in this manner, we get

Since $(u_1, u_2) \in V$ and $\alpha \in(0, 1)$, we get that $\lim\limits_{n\to \infty}(u_1-\frac{\alpha}{2})^n = 0$ and $\lim\limits_{n\to \infty}(1-\alpha)^n = 0$. Now, consider

Hence, $\mathcal{F}_{2\alpha}$ is also asymptotically regular.

Theorem 3.2. Let $\mathcal{F}_i$ be quasi-nonexpansive self-mappings on a non-void and closed subset $V$ of a Banach space $U$ for $i\in \{1, 2\}$, and satisfy condition (E) so that $F(\mathcal{F}_1\cap \mathcal{F}_2) \neq 0$. Then, $F(\mathcal{F}_1\cap \mathcal{F}_2)$ is closed in $V$. Also, if $U$ is strictly convex, then $F(\mathcal{F}_1 \cap \mathcal{F}_2)$ is convex. Furthermore, if $U$ is strictly convex, $V$ is compact, and $\mathcal{F}$ is continuous, then for any $s_0 \in V, \alpha \in (0, 1)$, the $\alpha$-Krasnosel'skii sequence $\{\mathcal{F}^{n}_{i_{\alpha}} (s_0)\},$ converges to $s\in F(\ \mathcal{F}_1\cap \mathcal{F}_2)\ $.

Proof. (i) We assume $\{s_n\} \in F(\ \mathcal{F}_1\cap \mathcal{F}_2)\ $ so that $s_n \to s\in F(\mathcal{F}_1\cap \mathcal{F}_2)$ as $n \to \infty$. Hence, $\mathcal{F}_{i}s_n = s_n$ for $i\in\{1, 2\}$. Next, we show that $\mathcal{F}_is = s$ for $i\in \{1, 2\}$. Since $\mathcal{F}_i$ are quasi-nonexpansive, we get

that is, $\mathcal{F}_is = s$ for $i = 1, 2$, hence $F(\mathcal{F}_2\cap \mathcal{F}_2)$ is closed.

(ii) $V$ is convex since $U$ is strictly convex. Also fix $\gamma \in (\ 0, 1)\ $ and $u, v \in F(\mathcal{F}_1\cap \mathcal{F}_2)\ $ so that $u\neq v$. Take $s = \gamma u + (1-\gamma)v \in V$. Since mapping $\mathcal{F}_i$ satisfy condition (E),

Similarly,

Using strict convexity of $U$, there is a $\theta \in [\ 0, 1]\ $ so that $\mathcal{F}_is = \theta u + (1-\theta) v$ for $i = 1, 2$

and

From inequalities (3.10) and (3.11), we obtain

Hence, $\mathcal{F}_{i}s = s$ for i = 1, 2, implies $s\in F(\mathcal{F}_1\cap \mathcal{F}_2)\ $.

(iii) Let us define $\{ s_n\}$ by $s_n = \mathcal{F}^{n}_{i_{\alpha}}s_0, s_0 \in V$, where $\mathcal{F}_{i_{\alpha}}s_0 = (1-\alpha) s_0 + \alpha \mathcal{F}_is_0, \alpha\in (\ 0, 1)\ $. We have a subsequence $\{s_{n_k}\}$ of $\{s_n\}$ converging to some $s\in V$, since $V$ is compact. Using the Schauder theorem and the continuity of $\mathcal{F}_i$, we have $F(\mathcal{F}_1\cap \mathcal{F}_2)\ \neq\phi$. We shall demonstrate that $s \in F(\mathcal{F}_1\cap \mathcal{F}_2)$. Let $w_0 \in F(\mathcal{F}_1\cap \mathcal{F}_2)$, consider

Therefore, $\{ \lVert s_n-w_0\rVert \}$ converges as it is a decreasing sequence that is bounded below by $0$. Moreover, since $\mathcal{F}_{i_{\alpha}}$ for $i = 1, 2$ is continuous, we have

Since $\alpha > 0$, we get

Since $\mathcal{F}_i$ are quasi-nonexpansive maps, we get

and from inequalities (3.13) and (3.14), we get

Now, from inequality (3.12), we have

which implies that

Since $U$ is strictly convex, either $\mathcal{F}_is-s_0 = a(s-s_0)$ for some $a \gneq 0$ or $s = s_0$. From Eq (15), it follows that $a = 1$, then, $\mathcal{F}_{i}s = s$ for $i = 1, 2$ and $s \in F(\mathcal{F}_1 \cap \mathcal{F}_2)$. Since $\lim\limits_{n \to \infty} \lVert s_n-s_0\rVert$ exists and $\{s_{nk}\}$ converges strongly to $s$. Hence, $\{s_n\}$ converges strongly to $s \in F(\mathcal{F}_1 \cap \mathcal{F}_2)$.

The next conclusion for metric projection is slightly more fascinating.

Theorem 3.3. Let $\mathcal{F}_i$ be quasi-nonexpansive self-mappings on a non-void, closed, and convex subset $V$ of a uniformly convex Banach space $U$ for $i\in\{1, 2\}$, and satisfies condition (E) so that $F(\mathcal{F}_1 \cap \mathcal{F}_2)\neq \phi$. Let $P: U \to F(\mathcal{F}_1 \cap \mathcal{F}_2)$ be the metric projection. Then, for every $u \in U$, the sequence $\{P\mathcal{F}^{n}_iu\}$ for $i = \{1, 2\}$, converges to $s \in F(\mathcal{F}_1 \cap \mathcal{F}_2)$.

Proof. Let $u \in V$. For $n, m \in N$

Since $u\in F(\mathcal{F}_1\cap \mathcal{F}_2)\ $, $n \in N$ and $\mathcal{F}_i$ are quasi-nonexpansive maps, $\mbox{for}\; i\in \{1, 2\}$ we have

Therefore, for $n \geq m$, it follows that

From inequalities (3.16) and (3.17), we have

which implies that $\lim\limits_{n \to \infty} \lVert P\mathcal{F}^{n}_iu-\mathcal{F}^{n}_iu\rVert$ exists. Taking $\lim\limits_{n \to \infty} \lVert P\mathcal{F}^{n}_iu-\mathcal{F}^{n}_iu\rVert = l$.

If $l = 0$, then we have an integer $n_0(\ \epsilon)\ $ for $\epsilon > 0$, satisfying

for $n \geq n_0$. Therefore, if $n \geq m \geq n_0$ and using inequalities (3.17) and (3.18), we have, for $i \in \{1, 2\}$,

That is, $\{P\mathcal{F}^{n}_iu\}$ for $i = \{1, 2\}$ is a Cauchy sequence in $F(\mathcal{F}_1\cap \mathcal{F}_2)$. Using the completeness of $U$ and the closedness of $F(\mathcal{F}_1\cap \mathcal{F}_2)$ from the above theorem, $\{P\mathcal{F}^{n}_ix\}$ for $i = 1, 2,$ converges in $F(\mathcal{F}_1 \cap \mathcal{F}_2)$. Taking $l > 0$, we claim that the sequence $\{P\mathcal{F}^{n}_iu\}$ for $i = 1, 2,$ is a Cauchy sequence in $U$. Also we have, an $\epsilon _{0} > 0$ so that, for each $n_0 \in N$, we have some $r_0, s_0 \geq n_0$ satisfying

Now, we choose a $\theta > 0$

Let $m_0$ be as large as possible such that for $q\geq m_0$

For this $m_0$, there exist $q_1, q_2$ such that $q_1, q_2 > m_0$ and

Thus, for $q_0 \geq max \{q_1, q_2\}$, we attain

and

Now, using the uniform convexity of $U$, we attain

a contradiction. Hence for every $u \in V$, the sequence $\{P\mathcal{F}^{n}_iu\}$ for $i = 1, 2,$ converges to some $s \in F(\mathcal{F}_1 \cap \mathcal{F}_2)$.

4.   Conclusions

We have proved some properties of common fixed points and also showed that if two mappings have common fixed points, then their $\alpha$-Krasnosel'skii mappings are asymptotically regular. To show the superiority of our results, we have provided an example. Further, we have proved that the $\alpha$-Krasnosel'skii sequence and its projection converge to a common fixed whose collection is closed.

Acknowledgments

Researchers would like to thank the Deanship of Scientific Research, Qassim University for funding publication of this project.

Conflict of interest

The authors declare no conflict of interest.