Show $\phi$ has a unique fixed point

Question

Let $\phi : \mathbb{R} \rightarrow \mathbb{R}$ a function of classe $\mathscr{C}^{1}$ such that $$ \underset{x \in \mathbb{R}}{\text{sup}}\left|\phi'\left(x\right)\right|<1 $$

I need to show it exists a unique fixed point $\phi\left(x^{\ast}\right) = x^{\ast}$. The original exercise suggests to assess the sign of $x - \phi\left(x\right)$ for a sufficiently large $\left|x\right|$.

My try is to work on $\left[a;+\infty\right[$ with $a>0$ use mean value inequality : $$ \left|\phi\left(x\right) - \phi\left(a\right) \right|<\left|x-a\right| $$ using the hypothesis on $\phi'$. However I can't see how I can use this to solve the question. Any hint ?

Answer 1

Let $c:=\max_{\mathbb R}|\varphi '|$.

For the unicity

Let $x,y\in \mathbb R$ being two fixed point of $\varphi $. As you said, using mean value theorem, $$|x-y|=|\varphi (x)-\varphi (y)|\leq c|x-y|<|x-y|,$$ which is a contradiction.

For existence (Hint)

Define $x_0=0$ and $x_{n+1}=\varphi (x_n)$. Show that $(x_n)$ is a Cauchy sequence. Let $x$ be its limit. Using continuity of $\varphi $, it will be a fix point of $\varphi $.

Answer 2

Do you know Banach fixed-point theorem ?

Here, by hypothesis and mean value inequality, for all $x,y\in \mathbb{R}$, $$|\phi(x)-\phi(y)|<\sup_{t\in\mathbb{R}}|\phi'(t)||x-y|,$$ so that $\phi$ is a contraction mapping since $\sup_{t\in\mathbb{R}}|\phi'(t)|<1$.