Kannan mapping and quasi-nonexpansive mapping

[ozkan12]

Definition of Kannan Mapping

Let (X,d) be a complete metric space...İf for each $x,y\in X$ following condition holds, then $T:X\to X$ is Kannan mapping

$d(Tx,Ty)\le\alpha\left[d\left(x,Tx\right),d(\left(y,Ty\right)\right)]$ $\alpha\in[0,\frac{1}{2})$

Definition of Quasi-Nonexpansive Mapping

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

$d(Tx,p)\le d(x,p)$ for each x $\in$ X.

İf T is a Kannan mapping, then T is a quasi nonexpansive operator.Proof:

İndeed, if T is a Kannan operator, then from definition of Kannan mapping with y=p in set of fixed point of T we get

$d(Tx,p)\le\alpha d\left(x,Tx\right)\le\alpha\left[d(x,p)+d(p,Tx)\right]$ and hence

$d\left(Tx,p\right)\le\frac{\alpha}{1-\alpha}d\left(x,p\right)<d(x,p)$İn there we get d(Tx,p)<d(x,p)...But in definition of quasi nonexpansive operator we use "$\le$"...But in last step we use "<" İn this case how we say that every kannan mapping is a quasi non expansive operator

 

[Ackbach]

$< \; \implies \; \le$. That is, $<$ is a stronger condition than $\le$, and implies it. You could just as well have used the weaker $\le$ in the very last line.

 

[ozkan12]

Dear Ackbach,

First of all, Thank you for your attention...

How "$<$" implies $\le$ ? İn nigt, I thougt that how we write this inequeality...Then I found some articles...İf we take p=x in

$$d(Tx,p)<d(x,p)$$ then we get $d(Tx,p)=d(x,p)=0$...So, we get $d(Tx,p)\le d(x,p)$...

But, how "$<\implies \le$" ? I didnt understand...as you say I can use " $\le$" in last inequeality...That is, I can write $d\left(Tx,p\right)\le d\left(x,p\right)$...İs this true ? Thank you for your attention..$<\implies \le$

 

[Ackbach]

This is a property of logic. In Natural Deduction, we call it the "Disjunction Introduction Rule". Copi calls it something else. It goes like this:

A implies A or B. If A is true, then A or B is true.

If $x<y$ is true, then it is also true that either $x<y$ or $x=y$. That is, if $x<y$ is true, then $x\le y$. Therefore, $x<y \; \implies \; x\le y$. So, my shorthand notation there, $< \; \implies \; \le$ is what I meant.

 

[ozkan12]

Dear Ackbach,

That is, we can use $\le$ instead of $<$ in last inequality...

Also, can we say that if we take x=p, we will obtain last inequality ? Thank you for your attention...

 

[ozkan12]

Dear Ackbach

İf $x\le y$ then we can say $x<y$ or $x=y$..İn my research it is not as you say...

 

[Ackbach]

If $x\le y$, then you can say $x<y$ or $x=y$, but you don't know which. It could be either. You definitely CANNOT narrow it down to only one of those two possibilities.

 

[ozkan12]

yes, this is true...but we don't say if $x<y$ then $x<y$ or $x=y$ ...And still in first post is strange for me ...