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Get a function $f:\mathbb{R} \longrightarrow \mathbb{R} \in C^{\infty}$ such that $\forall x \in \mathbb{R}$ :

$$ \vert f'(x) \vert < 1, f(x) \neq x $$

My attempt was to give the function $f : \mathbb{R} \longrightarrow \mathbb{R}$ where $f(x) = \sqrt{1+x^2}$. A quick calculation shows that :

$$f'(x)=\dfrac{x}{\sqrt{1+x^2}}$$

Clearly for $x \in \mathbb{R} \implies x^2+1>x^2$ and :

$$\vert x \vert < \sqrt{1+x^2} \implies -\sqrt{1+x^2}<x<\sqrt{1+x^2} \implies -1<\dfrac{x}{\sqrt{1+x^2}}<1$$

So : $ \vert f'(x) \vert <1$ and it is clear that $f(x) \neq x, \forall x\in \mathbb{R}$.

$f \in C^{\infty}$?

My attempt was to put $f$ as $f=h \circ g$ where $h(x)=\sqrt{x}, x>0$ and $g(x)=1+x^2$. Both are $C^{\infty}$ so $f$ is too.

Is correct?

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  • 1
    $\begingroup$ Yes, to be complete, you may quote the fact that $C^k$ is closed under composition $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 2 '19 at 4:48
  • $\begingroup$ For extra credit, show that if you force $|f'(x)| < c$, for a fixed $c < 1$, then there must be a fixed point. $\endgroup$ – Jair Taylor Mar 2 '19 at 4:54
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    $\begingroup$ Yes, because of the contraction theorem that point exists and it is more, it is unique $\endgroup$ – Juan Daniel Valdivia Fuentes Mar 2 '19 at 4:55
  • $\begingroup$ Alternatively, start with any smooth $h$ such that $h$ is bounded and $h'>0$ (e.g., $h(x)=\arctan x$, $h(x)=\frac{e^x}{1+e^x}$, or - loosely related to your solution - $h(x)=\frac x{\sqrt{1+x^2}}$), and use $f(x)=-a (h(x)-b)+x$ with $a=1/\sup|h'|$ and $b=\sup h$ $\endgroup$ – Hagen von Eitzen Mar 2 '19 at 5:57

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