MHB Generalization of Banach contraction principle

[ozkan12]

Let $\left(X,d\right)$ be a complete metric space and suppose that $f:X\to X$ satisfies

$d\left(fx,fy\right)\le\beta\left(d\left(x,y\right)\right)d\left(x,y\right)$ for all x,y $\in$ X where $\beta$ is a decreasing function on ${R}^{+}$ to $[0,1)$. Then $f$ has a unique fixed point.

The mapping $d\left(fx,fy\right)\le\beta\left(d\left(x,y\right)\right)d\left(x,y\right)$ more general than banach contraction...How this happens ? In my opinion, If we take $\beta\left(t\right)=c$, $c\in [0,1)$ we get banach principle...İs this true ? Can you help me ? Thank you for your attention...Best wishes...

 

[Ackbach]

Let $\left(X,d\right)$ be a complete metric space and suppose that $f:X\to X$ satisfies

$d\left(fx,fy\right)\le\beta\left(d\left(x,y\right)\right)d\left(x,y\right)$ for all x,y $\in$ X where $\beta$ is a decreasing function on ${R}^{+}$ to $[0,1)$. Then $f$ has a unique fixed point.

The mapping $d\left(fx,fy\right)\le\beta\left(d\left(x,y\right)\right)d\left(x,y\right)$ more general than banach contraction...How this happens?

A mathematician somewhere wanted to know if he could generalize the Banach Fixed Point Theorem - still get guaranteed fixed points while relaxing the hypotheses of the theorem.

In my opinion, If we take $\beta\left(t\right)=c$, $c\in [0,1)$ we get banach principle...İs this true?

Yes - so $d\left(fx,fy\right)\le\beta\left(d\left(x,y\right)\right)d\left(x,y\right)$ is more general than a contraction, because $\beta(t)$ wouldn't need to be constant, necessarily.