How Does a Quasi-Nonexpansive Operator Function in Metric Spaces?

[ozkan12]

Let (X,d) be a metric space. An operator $T:X\to X$ is said to be quasi nonexpansive if T has at least one fixed point in X and, for each fixed point p, we have

$d\left(Tx,p\right)\le d\left(x,p\right)$ (1)

And also we give a mapping such that

$d\left(Tx,Ty\right)\le2\delta d\left(x,Tx\right)+\delta d\left(x,y\right)$ (2) for all x,y in X. Also $\delta \in [0,1)$.

İn (2) if we take x:= p and y:=x then we get,

$d\left(Tx,p\right)\le\delta d\left(x,p\right)<d\left(x,p\right)$. İn there, p is fixed point of T. (3)

İn (2), İf we take x:= p and y:=x we obtain d(Tx,p)=0, d(x,p)=0...So, How we write (3) ?...

 

[Jhoneine]

From (2) we can see that $d\left(Tx,p\right)\le2\delta d\left(x,Tx\right)+\delta d\left(x,p\right)$ for all x in X. Since $d\left(x,p\right)=0$ and $d\left(Tx,p\right)\ge 0$, we obtain $d\left(Tx,p\right)\le 2\delta d\left(x,Tx\right)$. Taking $\delta \in [0,1)$ implies $d\left(Tx,p\right)\le d\left(x,Tx\right)$. Hence, we can conclude that an operator T is quasi nonexpansive if it has at least one fixed point p, and for all x in X, we have $d\left(Tx,p\right)\le d\left(x,Tx\right)$.

 

[math771]

Based on (2), we can rewrite (3) as:

$d(Tp,p)\le\delta d(p,p)$

Since $p$ is a fixed point, $d(p,p)=0$, so we have:

$d(Tp,p)\le0$

This implies that $d(Tp,p)=0$, which makes sense since $p$ is a fixed point. Therefore, (3) is true.