# Convergence Analysis for a Three-Step Thakur Iteration for Suzuki-Type Nonexpansive Mappings with Visualization

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## Abstract

**:**

## 1. Introduction

## 2. Preliminaries

**Definition**

**1**

**Lemma**

**1**

**Definition**

**2**

**Proposition**

**1**

- (i)
- $\left(\right)open="\parallel "\; close="\parallel ">Tx-{T}^{2}x$ ([12], Lemma 5).
- (ii)
- Either $\frac{1}{2}\left(\right)open="\parallel "\; close="\parallel ">x-Tx\le \left(\right)open="\parallel "\; close="\parallel ">x-y$ or $\frac{1}{2}\left(\right)open="\parallel "\; close="\parallel ">{T}^{2}x-Ty\le \left(\right)open="\parallel "\; close="\parallel ">Tx-y$ holds ([12], Lemma 5).
- (iii)
- $\left(\right)open="\parallel "\; close="\parallel ">x-Ty+\left(\right)open="\parallel "\; close="\parallel ">x-y$ ([12], Lemma 7).

**Definition**

**3**

**Lemma**

**2**

**Definition**

**4**

**Lemma**

**3**

**Theorem**

**1**

## 3. Convergence Theorems

**Lemma**

**4.**

**Proof.**

**Theorem**

**2.**

**Proof.**

**Theorem**

**3.**

**Proof.**

**Theorem**

**4.**

**Proof.**

**Theorem**

**5.**

**Proof.**

## 4. Data Dependence

**Definition**

**5**

**Lemma**

**5**

**Theorem**

**6.**

**Proof.**

## 5. Example and Comparative Study

**Example**

**1.**

**Proof.**

**Case I**: Let $x}_{1}\in \left(\right)open="["\; close=")">0,\frac{1}{7$. If $y}_{1}\in \left(\right)open="["\; close=")">0,\frac{1}{7$, then it can easily be seen that T is nonexpansive and condition $\left(C\right)$ is automatically satisfied. Further, if we take ${y}_{1}\in \left(\right)open="["\; close="]">{\displaystyle \frac{1}{7},2}$, then $\frac{1}{2}{\left(\right)}_{({x}_{1},{x}_{2})}1\le {\left(\right)}_{({x}_{1},{x}_{2})}1$ stands true only for ${y}_{1}\in \left(\right)open="["\; close="]">{\displaystyle \frac{6}{7},2}$. Moreover, evaluating the nonexpansiveness condition ${\left(\right)}_{T}1$ for ${x}_{1}\in \left(\right)open="["\; close="]">{\displaystyle 0,\frac{1}{7}}$ and ${y}_{1}\in \left(\right)open="["\; close="]">{\displaystyle \frac{6}{7},2}$, one finds $|{\displaystyle \frac{2-7{x}_{1}-{y}_{1}}{7}}|\le {y}_{1}-{x}_{1}$ which is obviously true as $|{\displaystyle \frac{2-7{x}_{1}-{y}_{1}}{7}}|\in \left(\right)open="["\; close="]">{\displaystyle 0,\frac{8}{49}}$, while ${y}_{1}-{x}_{1}\in \left(\right)open="("\; close="]">{\displaystyle \frac{5}{7},2}$. Therefore, T satisfies condition C for the case considered.

**Case II**: Let us now consider the rest of the interval i.e., ${x}_{1}\in \left(\right)open="["\; close="]">{\displaystyle \frac{1}{7},2}$. Similarly with

**Case I**, if ${x}_{1}$ and ${y}_{1}$ belongs to the same interval, then T is a contraction and satisfies condition C since all contractions are included in the class of Suzuki mappings. On the other side, if ${y}_{1}\in \left(\right)open="["\; close=")">{\displaystyle 0,\frac{1}{7}}$ then $\frac{1}{2}{\left(\right)}_{({x}_{1},{x}_{2})}1\le {\left(\right)}_{({x}_{1},{x}_{2})}1$ becomes $\frac{1}{2}|{\displaystyle {x}_{1}-\frac{{x}_{1}+12}{7}}|+\frac{1}{2}|{x}_{2}-{x}_{2}|\le |{x}_{1}-{y}_{1}|+|{x}_{2}-{y}_{2}|$, or, even more, $\frac{6-3{x}_{1}}{7}\le {x}_{1}-{y}_{1}+\frac{1}{7}$, as $|{x}_{2}-{y}_{2}|\in \left(\right)open="["\; close="]">{\displaystyle 0,\frac{1}{7}}$. Further, this implies $\frac{10{x}_{1}-5}{7}\ge {y}_{1}$ i.e., ${x}_{1}\in \left(\right)open="["\; close="]">{\displaystyle \frac{1}{2},2}$. For ${x}_{1}\in \left(\right)open="["\; close="]">{\displaystyle \frac{1}{2},2}$ and ${y}_{1}\in \left(\right)open="["\; close="]">{\displaystyle 0,\frac{1}{7}}$, the nonexpansive condition is $|{\displaystyle \frac{{x}_{1}+7{y}_{1}-2}{7}}|\le |{x}_{1}-{y}_{1}|$ which is true as $|{\displaystyle \frac{{x}_{1}+7{y}_{1}-2}{7}}|\in \left(\right)open="["\; close="]">{\displaystyle 0,\frac{3}{14}}$ and $|{x}_{1}-{y}_{1}|\in \left(\right)open="["\; close="]">{\displaystyle \frac{5}{14},2}$, so T satisfies condition C for this case also.

Algorithm 1: Convergence visualization |

**Example**

**2.**

**Proof.**

**Case 1**: Let us presume that $max\left(\right)open="\{"\; close="\}">\left|f\left(0\right)-g\left(0\right)\right|,{\mathrm{ess}\phantom{\rule{0.166667em}{0ex}}\mathrm{sup}\phantom{\rule{0.166667em}{0ex}}}_{(0,\infty )}\left|f\left(x\right)-g\left(x\right)\right|$. For T to satisfy condition $\left(C\right)$, this must imply $max\left(\right)open="\{"\; close="\}">\left|Tf\left(0\right)-Tg\left(0\right)\right|,{\mathrm{ess}\phantom{\rule{0.166667em}{0ex}}\mathrm{sup}\phantom{\rule{0.166667em}{0ex}}}_{(0,\infty )}\left|Tf\left(x\right)-Tg\left(x\right)\right|$. Because of this last inequality, it is expected the problem to be divided again into two sub-cases. We will analyze just the nontrivial one i.e., $\frac{1}{2}\left|Tf\left(0\right)-f\left(0\right)\right|\le \left|f\left(0\right)-g\left(0\right)\right|$ implies $\left|Tf\left(0\right)-Tg\left(0\right)\right|\le \left|f\left(0\right)-g\left(0\right)\right|$, as the desired result follows easly from the other one. If $f\left(0\right)\ne 7$ and $g\left(0\right)\ne 7$, or $f\left(0\right)=7$ and $g\left(0\right)=7$, it can be easly noticed that T is nonexpansive, and therefore condition $\left(C\right)$ is automatically fulfilled. For $f\left(0\right)\ne 7$ and $g\left(0\right)=7$, T is nonexpansive just for $f\left(0\right)\in \left(\right)open="["\; close="]">{\displaystyle 0,\frac{28}{5}}$, and again, condition $\left(C\right)$ is satisfied. For $f\left(0\right)\in \left(\right)open="("\; close=")">{\displaystyle \frac{28}{5},7}$ and $g\left(0\right)=7$, $\frac{1}{2}\left|Tf\left(0\right)-f\left(0\right)\right|\le \left|f\left(0\right)-g\left(0\right)\right|$ becomes $\frac{5f\left(0\right)}{14}\le 7-f\left(0\right)$ which is not true as $\frac{5f\left(0\right)}{14}\in \left(\right)open="("\; close=")">{\displaystyle 2,\frac{5}{2}}$ and $7-f\left(0\right)\in \left(\right)open="("\; close=")">{\displaystyle 0,\frac{7}{5}}$. The same result is obtained if we take $f\left(0\right)=7$ and $g\left(0\right)\ne 7$. Considering all the situations analyzed, we conclude that T is a Suzuki-mapping for the current case.

**Case 2**: Suppose $max\left(\right)open="\{"\; close="\}">\left|f\left(0\right)-g\left(0\right)\right|,{\mathrm{ess}\phantom{\rule{0.166667em}{0ex}}\mathrm{sup}\phantom{\rule{0.166667em}{0ex}}}_{(0,\infty )}\left|f\left(x\right)-g\left(x\right)\right|$. The inequality $\frac{1}{2}\left|Tf\left(0\right)-f\left(0\right)\right|\le {\mathrm{ess}\phantom{\rule{0.166667em}{0ex}}\mathrm{sup}\phantom{\rule{0.166667em}{0ex}}}_{(0,\infty )}\left|f\left(x\right)-g\left(x\right)\right|$ must imply $max\left(\right)open="\{"\; close="\}">\left|Tf\left(0\right)-Tg\left(0\right)\right|,{\mathrm{ess}\phantom{\rule{0.166667em}{0ex}}\mathrm{sup}\phantom{\rule{0.166667em}{0ex}}}_{(0,\infty )}\left|Tf\left(x\right)-Tg\left(x\right)\right|$. If we consider that $\left|Tf\left(0\right)-Tg\left(0\right)\right|\le {\mathrm{ess}\phantom{\rule{0.166667em}{0ex}}\mathrm{sup}\phantom{\rule{0.166667em}{0ex}}}_{(0,\infty )}\left|f\left(x\right)-g\left(x\right)\right|$, it follows ${\left(\right)}_{T}\infty $; but as $\left|f\left(0\right)-g\left(0\right)\right|\le {\mathrm{ess}\phantom{\rule{0.166667em}{0ex}}\mathrm{sup}\phantom{\rule{0.166667em}{0ex}}}_{(0,\infty )}\left|f\left(x\right)-g\left(x\right)\right|$, it follows that ${\left(\right)}_{f}\infty $ too, so T is nonexpansive. If we suppose $\left|Tf\left(0\right)-Tg\left(0\right)\right|>{\mathrm{ess}\phantom{\rule{0.166667em}{0ex}}\mathrm{sup}\phantom{\rule{0.166667em}{0ex}}}_{(0,\infty )}\left|f\left(x\right)-g\left(x\right)\right|$, kipping in mind that ${\mathrm{ess}\phantom{\rule{0.166667em}{0ex}}\mathrm{sup}\phantom{\rule{0.166667em}{0ex}}}_{(0,\infty )}\left|f\left(x\right)-g\left(x\right)\right|\ge \left|f\left(0\right)-g\left(0\right)\right|$ on one side, ${\mathrm{ess}\phantom{\rule{0.166667em}{0ex}}\mathrm{sup}\phantom{\rule{0.166667em}{0ex}}}_{(0,\infty )}\left|f\left(x\right)-g\left(x\right)\right|\ge \frac{1}{2}\left|Tf\left(0\right)-f\left(0\right)\right|$ on the other side and considering all combinations that T could involve, we find that the assumption is absurd and $\left|Tf\left(0\right)-Tg\left(0\right)\right|$ could not be grater than ${\mathrm{ess}\phantom{\rule{0.166667em}{0ex}}\mathrm{sup}\phantom{\rule{0.166667em}{0ex}}}_{(0,\infty )}\left|f\left(x\right)-g\left(x\right)\right|$. So, overall, T is a Suzuki mapping in Case 2 also. □

## 6. Conclusions

## Author Contributions

## Funding

## Conflicts of Interest

## References

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**Table 1.**Number of iterative steps required for approximating a fixed point, with error $\epsilon ={10}^{-15}$, starting from the initial point $({x}_{1}^{*},{x}_{2}^{*})=(1,0.1)$.

TTP | Picard | Mann | Ishikawa | Agarwal | Noor | Abbas-Nazir | |
---|---|---|---|---|---|---|---|

(1, 0.1) | 15 | 19 | 25 | - | 19 | - | 11 |

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**MDPI and ACS Style**

Usurelu, G.I.; Postolache, M.
Convergence Analysis for a Three-Step Thakur Iteration for Suzuki-Type Nonexpansive Mappings with Visualization. *Symmetry* **2019**, *11*, 1441.
https://doi.org/10.3390/sym11121441

**AMA Style**

Usurelu GI, Postolache M.
Convergence Analysis for a Three-Step Thakur Iteration for Suzuki-Type Nonexpansive Mappings with Visualization. *Symmetry*. 2019; 11(12):1441.
https://doi.org/10.3390/sym11121441

**Chicago/Turabian Style**

Usurelu, Gabriela Ioana, and Mihai Postolache.
2019. "Convergence Analysis for a Three-Step Thakur Iteration for Suzuki-Type Nonexpansive Mappings with Visualization" *Symmetry* 11, no. 12: 1441.
https://doi.org/10.3390/sym11121441