Show that $f:[0,\infty)\to [0,\infty)$, $f(x)=e^{-x}$ is a contraction mapping.
I want to show that $|f(x)-f(y)|\le c|x-y|$ where $0\lt c\lt 1$.
I tried to write $f(x)=e^{-x}$ into Taylor polynomials but It's not helpful. Can anyone give me any idea of how to prove this?
I thought about suggesting using the mean value theorem. But $$\left|\frac{f(0)-f(x)}{0-x}\right|=\frac{1-e^{-x}}x\to1$$ as $x\to0$ so there is no $c<1$ with $|f(x)-f(y)|\le c|x-y|$ for all $x$, $y\in[0,\infty)$.
If you consider $f$ on $[a,\infty)$ for $a>0$ it does become a contraction.
This is in addition to Lord Shark the Unknown's answer. In general, if your function $f(x)$ is nice enough, you can show that it is a contraction mapping if there's a $c<1$ with $|f'(x)|<c$ for all $x$.
