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Definition of contraction: A function $f:\mathbb{R}\to\mathbb{R}$ is called a contraction of $\mathbb{R}$ if and only if there exists a constant $r\in[0,1)$ such that for all $x$ and $x'$ in $\mathbb{R}$ we have

$$|f(x)-f(x')| \leq r|x-x'|.$$

Proof: The function $f$ is differentiable on $\mathbb{R}$. Hence, for any $x\in\mathbb{R}$, $f'(x)$ exists and we will denote it as $r$. We are given that $|r|<1$ since $||f'||_{sup} < 1$. Then,

$$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=r$$ $$f(x+h)-f(x)=rh$$

Letting $x' = x+h$, we get

$$f(x')-f(x)=r(x-x')$$ $$\implies |f(x')-f(x)|=r|(x-x')|$$

and $f$ must be a contraction of $\mathbb{R}$.

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    What is the reasoning from the first to the second line? – 2017-02-03
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    Did you mean to type $\|f'\|_{sup}$ in the second line? – 2017-02-03
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    What about using the mean value theorem and a simple estimate? – 2017-02-03
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    @rubik Yes, I edited it now. – 2017-02-03
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    @ detnvvp do you mean from first equation to second equation or first sentence to the second sentence? – 2017-02-03
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    @ThePuix I haven't learned mean value theorem and a simple estimate yet. Should I use it to prove this instead of only using the definition of differential? – 2017-02-03
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    Well, if the question doesn't specifically disallows its use, I suggest you look it up. The wording 'simple estimate' didn't mean anything more specific than what is given in the inequality for the derivative. – 2017-02-03

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