Unique fixed point

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $|f(x)-f(y)|\leq k |x-y|$ for some $k \in (0, 1)$ and all $x, y \in \mathbb{R}$. Show that $f$ has a unique fixed point.

Solution



For $f:\mathbb{R}\to\mathbb{R}$ such that for some $c\in(0,1)$ and for all $x_1,x_2\in\mathbb{R}$, $|f(x_1)-f(x_2)|\leq c|x_1-x_2|$, there is a unique $\bar{x}\in\mathbb{R}$ with $f(\bar{x})=\bar{x}$.  First it should be clear that $f$ is continuous.  Now let $x_0\in\mathbb{R}$ and let $x_1=f(x_0), x_2=f(x_1), \dots$.  Then for $n>0$,
\begin{align*}
    |x_n-x_{n-1}|=|f(x_{n-1})-f(x_{n-2})|
        &\leq c|x_{n-1}-x_{n-2}|=c|f(x_{n-2})-f(x_{n-3})|\\
        &\leq c^2|x_{n-2}-x_{n-3}|=c^2|f(x_{n-3})-f(x_{n-4})|\\
        &\leq c^3|x_{n-3}-x_{n-4}|\\
        &\vdots\\
        &\leq c^{n-1}|x_1-x_0|
    \end{align*}
so that for $n>m$,
\begin{align*}
    |x_n-x_m|
        &\leq\sum_{k=0}^{n-m-1}|x_{m+k+1}-x_{m+k}|\leq\sum_{k=0}^{n-m-1}c^{m+k}|x_1-x_0|\\
        &=c^m|x_1-x_0|\sum_{k=0}^{n-m-1}c^k=c^m|x_1-x_0|\big(\frac{1-c^{n-m}}{1-c}\big)\\
        &=\frac{|x_1-x_0|}{1-c}(c^m-c^n)\xrightarrow[n,m]{}0
    \end{align*}
thus $\{x_n\}_n$ is cauchy in $\mathbb{R}$; let $\bar{x}=\lim\limits_{n\to\infty}x_n$.  Since $f$ is continuous,
\begin{align*}
    |f(\bar{x})-\bar{x}|
        &=\lim\limits_{n\to\infty}|f(x_n)-\bar{x}|\\
        &=\lim\limits_{n\to\infty}|x_{n-1}-\bar{x}|\\
        &=0
    \end{align*}
thus $f(\bar{x})=\bar{x}$.  If $x$ is any fixed point of $f$ distinct from $\bar{x}$ then we would have
$0<|x-\bar{x}|=|f(x)-f(\bar{x})|\leq c|x-\bar{x}|<|x-\bar{x}|$ which is absurd hence it follows that $\bar{x}$ is unique.