Can anyone please provide me an example of a Contractive mapping which is not a Contraction mapping.
Definitions:
A mapping $T: M\to M$ is said to be contractive if $d(Tx, Ty)<d(x,y)$ for each $x,y\in M$ with $x\neq y,$
A mapping $T: M\to M$ is said to be contraction if there exist a constant $0\leq k<1$ such that $d(Tx, Ty)\leq k d(x,y)$ for each $x,y\in M$ with $x\neq y,$
$$ \frac{3x + \sqrt{1 + x^2}}{4} $$ is a bijection of the real line with no fixpoint. The bit about contraction is the Mean Value Theorem together with the observation that the derivative is always between $0$ and $1,$ while getting arbitrarily close to $1$ as $x$ goes to $+ \infty$
How about this one :
Let $f(x)=\sqrt x$ be defined on $D= (\frac{1}{4},\frac{1}{4}+\epsilon)$ for any $\epsilon>0$.
Now $|f(x)-f(y)|=|\sqrt x-\sqrt y|=\frac{|x-y|}{\sqrt x+\sqrt y}<|x-y|$ so that $f$ is contractive. If possible let $f$ be contraction i.e. $\exists\alpha$ with $0<\alpha<1$ such that $\forall x,y\in D$ we have $|\sqrt x-\sqrt y|\leq\alpha|x-y|$
or $1\leq\alpha|\sqrt x+\sqrt y|$ or $\alpha\geq\frac{1}{\sqrt x+\sqrt y}$ for all $x,y\in D$.
Now given any $\beta<1$ we can find $x_{\beta},y_{\beta}\in D$ such that $\frac{1}{\sqrt x_{\beta}+\sqrt y_{\beta}}>\beta$ so that $\alpha>\beta$. Which means we must have $\alpha\geq 1$; a contradiction.
