Common Fixed Point Theorems on Weakly Compatible Maps

[ozkan12]

Let $\left(X,d\right)$ be a metric space. Let $A,B,S,T: X\to X$ be mappings satisfying

1) $T\left(X\right)\subset A(X)$ and $S\left(X\right)\subset B(X)$

2) The pairs $(S,A)$ and $(T,B)$ are weakly compatible and

3) $d\left(Sx,Ty\right)\le \alpha.max\left\{d\left(Ax,By\right),d\left(Ax,Sx\right),d\left(By,Ty\right)\right\}$

for all $x,y\in X$, where $0\le\alpha<\frac{1}{2}$ then, $A,B,S,T$ have a unique common fixed point.

İn proof of this theorem, I have several troubles...

Firstly, using condition 1, we define sequences $\left\{{x}_{n}\right\}$ and $\left\{{y}_{n}\right\}$ in X by rule

${y}_{2n}=B{x}_{2n+1}=S{x}_{2n}$ and ${y}_{2n+1}=A{x}_{2n+2}=T{x}_{2n+1}$, n=0,1,2,3...

Question: How we write $\left\{{x}_{n}\right\}$ and $\left\{{y}_{n}\right\}$ in X by rule

${y}_{2n}=B{x}_{2n+1}=S{x}_{2n}$ and ${y}_{2n+1}=A{x}_{2n+2}=T{x}_{2n+1}$, n=0,1,2,3... by using

condition 1,

Question 2: İf ${y}_{n}$ is Cauchy sequence and converges to $x\in X$, we can say that

$S{x}_{2n}\to x, B{x}_{2n+1}\to x, T{x}_{2n+1}\to x ,A{x}_{2n+2}\to x$...Also, using $T(X)\subset A(X)$ can we say

that there exists $u\in X$ such that x=Au..

Common fixed point theorems on weakly compatible maps on dislocated metric spaces - Springer

You can download article from this link..

 

[jvicens]

I am not familiar with the specific article you have provided, so I cannot provide a detailed answer to your questions. However, I can give some general guidance on how to approach this problem.

Firstly, using condition 1, we can define the sequences $\{x_n\}$ and $\{y_n\}$ in $X$ as follows:

$x_0 \in X$ (arbitrary)

$x_{2n+1} = T(x_{2n})$ for $n=0,1,2,...$

$x_{2n+2} = A(x_{2n+1})$ for $n=0,1,2,...$

$y_0 = S(x_0)$

$y_{2n+1} = A(x_{2n+1})$ for $n=0,1,2,...$

$y_{2n+2} = B(x_{2n+1})$ for $n=0,1,2,...$

Note that $x_n$ and $y_n$ are defined in terms of each other, so they are not independent sequences. Also note that we are using the fact that $T(X)\subset A(X)$ and $S(X)\subset B(X)$ to define $y_n$ in terms of $x_n$.

For your second question, if $y_n$ is a Cauchy sequence and converges to $x\in X$, then we can say that $S(x_{2n})\to x$, $B(x_{2n+1})\to x$, $T(x_{2n+1})\to x$, $A(x_{2n+2})\to x$, since these are all just different terms in the sequence $y_n$.

However, we cannot say that there exists $u\in X$ such that $x=Au$, since $A$ is a mapping from $X$ to $X$, not from $X$ to some other set. We can say that $x$ is a fixed point of $A$, but we cannot say that $x$ is equal to $Au$ for some $u\in X$.

I hope this helps in understanding the problem and how to approach it. Please refer to the specific article you have provided for more detailed guidance on the proof.