# Function without fixed points

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}

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?

• Yes, to be complete, you may quote the fact that $C^k$ is closed under composition – GNUSupporter 8964民主女神 地下教會 Mar 2 '19 at 4:48
• For extra credit, show that if you force $|f'(x)| < c$, for a fixed $c < 1$, then there must be a fixed point. – Jair Taylor Mar 2 '19 at 4:54
• Yes, because of the contraction theorem that point exists and it is more, it is unique – Juan Daniel Valdivia Fuentes Mar 2 '19 at 4:55
• 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 $b=\sup h$ – Hagen von Eitzen Mar 2 '19 at 5:57