# Proving $e^{-|x|}$ is Lipschitz

As the title suggests, I'm trying to prove that $e^{-|x|}$ is Lipschitz in $\mathbb{R}$. I tried applying the mean value theorem, however $e^{-|x|}$ isn't differentiable at $x=0$. I know it looks simple, but how can I prove that?

## 2 Answers

Recall that $e^{a}\geq 1+a$ for all $a\in\Bbb R$. Thus, if $|x|<|y|$, then \begin{align} e^{-|x|}-e^{-|y|} &= e^{-|x|}(1-e^{|x|-|y|})\\ &\leq e^{-|x|}(1-(1+(|x|-|y|)) \\ &=e^{-|x|}(|y|-|x|)\\ &\leq|y|-|x| \\ &\leq |x-y|. \end{align} Interchanging the roles of $x$ and $y$ gives that $$|e^{-|x|}-e^{-|y|}|\leq |x-y|,\quad\text{for all }x,y\in\mathbb{R}.$$

Note that if $x,y>0,x+y=d$, then

$$\left\vert\frac{e^{-x}-e^{-y}}{x+y}\right\vert=\frac{1}{d}\vert e^{-x}-e^{x-d}\vert\le\frac{1-e^{-d}}{d}\le1$$

This gives that if $-y<0<x$, $\vert f(x)-f(-y) \vert \le \vert x - (-y) \vert$, i.e. that $f$ is still Lipschitz when the arguments are on opposite sides of $0$.

The cases where $x,y>0$, resp. $x,y<0$ are handed easily by the Mean Value Theorem.