Meir keeler contractive mappings and cauchy sequence

[ozkan12]

Denote with ${\varPsi}_{st}$ the family of strictly nondecreasing functions ${\Psi}_{st}:[0,\infty)\to [0,\infty)$ continuous in $t=0$ such that

${\Psi}_{st}=0$ if and only if $t=0$
${\Psi}_{st}(t+s)\le {\Psi}_{st}(t)+{\Psi}_{st}(s)$.

Definition: Let $\left(X,d\right)$ be a metric space and ${\Psi}_{st} \in {\varPsi}_{st}$. Suppose that $f: X\to X$ is a $\alpha-$ admissible mapping satisfying the following condition: for each $\varepsilon >0$ there exists $\delta >0$ such that

$\epsilon\le{\Psi}_{st}\left(M\left(x,y\right)\right)<\varepsilon+\delta$ implies $\alpha\left(x,y\right){\Psi}_{st}\left(d\left(fx,fy\right)\right)<\varepsilon$

for all $x,y\in X$ where $M\left(x,y\right)=max\left\{d\left(x,y\right),d\left(x,fx\right),d\left(y,fy\right),\frac{1}{2}\left[d\left(fx,y\right)+d\left(x,fy\right)\right]\right\}$. Then f is called generalized an $\alpha-{\Psi}_{st}-$ meir-keeler contractive mapping.

Theorem: Let $\left(X,d\right)$ be a complete metric space and $f:X\to X$ a orbitally continuous generalized $\alpha-{\Psi}_{st}-$ meir-keeler contractive mapping, if there exists ${x}_{0}\in X$ such that $\alpha\left({x}_{0},f{x}_{0}\right)\ge 1$. Then, f has a fixed point. ( İn part of proof of this theorem, it is not important definition of $\alpha-$ admissible mapping.)

Proof: Define ${x}_{n+1}={f}^{n+1}{x}_{0}$ for all $n\ge0.$ We want to prove that $\lim_{{m,n}\to{\infty}}d\left({x}_{n},{x}_{m}\right)=0$. İf this not, then there exists $\varepsilon>0$ and a subsequence $\left\{{x}_{n\left(i\right)}\right\}$ of $\left\{{x}_{n}\right\}$ such that $d\left({x}_{n\left(i\right)},{x}_{n\left(i+1\right)}\right)>2\varepsilon$ (2.12). For this $\varepsilon>0$, there exists $\delta>0$ such that $\varepsilon\le{\Psi}_{st}\left(M\left(x,y\right)\right)<\varepsilon+\delta$ implies that $\alpha\left(x,y\right){\Psi}_{st}\left(d\left(fx,fy\right)\right)<\varepsilon$. Put $r=min\left\{\varepsilon,\delta\right\}$ and ${s}_{n}=d\left({x}_{n},{x}_{n+1}\right)$ for all $n\ge1$. From proposition, there exists ${n}_{0}$ such that ${s}_{n}=d\left({x}_{n},{x}_{n+1}\right)<\frac{r}{4}$ for all $n\ge{n}_{0}$ (2.13). ( I didnt write this preposition, İt is enough to know that this is true.) We get $n\left(i\right)\le n\left(i+1\right)-1$. İf $d\left({x}_{n\left(i\right)},{x}_{n\left(i+1\right)-1}\right)\le\varepsilon+\frac{r}{2}$, then
$d\left({x}_{n\left(i\right)},{x}_{n\left(i+1\right)}\right)\le d\left({x}_{n\left(i\right)},{x}_{n\left(i+1\right)-1}\right)+d\left({x}_{n\left(i+1\right)-1},{x}_{n\left(i+1\right)}\right)< \varepsilon +\frac{r}{2}+{s}_{n\left(i+1\right)-1}<\varepsilon+\frac{3r}{4}<2\varepsilon$ which contradicts the assumption (2.12). Therefore, there are values of k such that $n\left(i\right)\le k \le n\left(i+1\right)$ and $d\left({x}_{n\left(i\right)},{x}_{k}\right)>\varepsilon+\frac{r}{2}$. Now if $d\left({x}_{n\left(i\right)},{x}_{n\left(i\right)+1}\right)\ge\varepsilon+\frac{r}{2}$, then ${s}_{n\left(i\right)}=d\left({x}_{n\left(i\right)},{x}_{n\left(i\right)+1}\right)\ge\varepsilon+\frac{r}{2}>r+\frac{r}{2}>\frac{r}{4}$ which is a contradiction with (2.13). Hence, there are values of k with $n\left(i\right)\le k\le n\left(i+1\right)$ such that $d\left({x}_{n\left(i\right)},{x}_{k}\right)<\varepsilon+\frac{r}{2}.$

My questions:

1) Firstly, we get $n\left(i\right)\le n\left(i+1\right)-1$. İf $d\left({x}_{n\left(i\right)},{x}_{n\left(i+1\right)-1}\right)\le\varepsilon+\frac{r}{2}$ by using triangle inequality, we get a contradiction...And then we say that there are values of k such that $n\left(i\right)\le k \le n\left(i+1\right)$ and $d\left({x}_{n\left(i\right)},{x}_{k}\right)>\varepsilon+\frac{r}{2}$ and if $d\left({x}_{n\left(i\right)},{x}_{n\left(i\right)+1}\right)\ge\varepsilon+\frac{r}{2}$, then ${s}_{n\left(i\right)}=d\left({x}_{n\left(i\right)},{x}_{n\left(i\right)+1}\right)\ge\varepsilon+\frac{r}{2}>r+\frac{r}{2}>\frac{r}{4}$, again, we get a contradiction...By taking into consideration these cases, how we say that there are values of k with $n\left(i\right)\le k\le n\left(i+1\right)$ such that $d\left({x}_{n\left(i\right)},{x}_{k}\right)<\varepsilon+\frac{r}{2}.$ in last part, if we take $k=n\left(i+1\right)-1$ we can get a contradiction as in first part...

Please can you explain this ? thank you for your attention...best wishes...

 

[jvicens]

2) Additionally, in the proof, we are assuming that $\lim_{{m,n}\to{\infty}}d\left({x}_{n},{x}_{m}\right)=0$, but then we are also assuming that there exists a subsequence $\left\{{x}_{n\left(i\right)}\right\}$ such that $d\left({x}_{n\left(i\right)},{x}_{n\left(i+1\right)}\right)>2\varepsilon$. How can we assume both of these things at the same time? It seems contradictory. Can you please explain this part of the proof as well? Thank you.

3) Lastly, I am not familiar with the proposition that is referenced in the proof. Can you please provide more information on this proposition and why it is important in the proof? Thank you.

1) The contradiction arises when we assume that $\lim_{{m,n}\to{\infty}}d\left({x}_{n},{x}_{m}\right)=0$ and that there exists a subsequence $\left\{{x}_{n\left(i\right)}\right\}$ such that $d\left({x}_{n\left(i\right)},{x}_{n\left(i+1\right)}\right)>2\varepsilon$. This contradiction shows that our assumption must be wrong, and thus, we can conclude that $\lim_{{m,n}\to{\infty}}d\left({x}_{n},{x}_{m}\right)=0$ is true.

The proof then goes on to show that there are values of k such that $n\left(i\right)\le k\le n\left(i+1\right)$ and $d\left({x}_{n\left(i\right)},{x}_{k}\right)<\varepsilon+\frac{r}{2}.$ This is because we have shown that the contradiction cannot occur in the case where $d\left({x}_{n\left(i\right)},{x}_{n\left(i+1\right)-1}\right)\le\varepsilon+\frac{r}{2}$ and $d\left({x}_{n\left(i\right)},{x}_{n\left(i+1\right)}\right)\ge\varepsilon+\frac{r}{2}$. Therefore, there must be values of