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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)

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 $x\neq y,$

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    Hint: cook up something with help of mean value theorem. – 2017-02-08
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    $M=[0,{\pi \over 2}), T = \sin$. – 2017-02-08
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    On $\mathbb R$, $x-\arctan(x)$ works. – 2017-02-08
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    @JonasMeyer: Or just $\arctan$? – 2017-02-08
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    @copper.hat: Oh yeah, but for what it's worth $x-\arctan(x)+2$ would also give a contractive bijection of $\mathbb R$ without fixed points. – 2017-02-08
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    The following link is to a related question, because each example there is also an example here: http://math.stackexchange.com/q/88784/ – 2017-02-09

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$$ \frac{3x + \sqrt{1 + x^2}}{4} $$ 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 $+ \infty$

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    Just a heads up in case you care: I believe we can read most of your email address in one of those tabs. – 2017-02-08
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    @Glitch thanks, it's alright. I make it visible in my profile. I do get some email I don't like from time to time (related to this site) but never by a bot. – 2017-02-08
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    @WillJagy, Thank you, for answering the question with such a great explanation. I get your point, we can take any function whose derivative approaches to $1$ on the given domain. – 2017-02-08