Show negative exponential of exponential is Lipschitz

Question

Simple analysis question. Let $f(x) = \exp(-\exp(x))$, on $\mathbb R$. Then, how does one show that $f$ is Lipschitz? I tried with the inequality $$ |\exp(a)-\exp(b)|\leq (\exp(a)+\exp(b))|a-b|, $$ which is valid for any real numbers $a$, $b$. When I proceed I obtain \begin{align} |f(x)-f(y)|&\leq(\exp(-\exp(x))+\exp(-\exp(y)))|\exp(x)-\exp(y)|\\ &\leq 2|\exp(x)-\exp(y)|. \end{align} Now the problem is that the exponential is not Lipschitz, so I cannot proceed from here. Another idea would be using $\exp(-a)\leq a^{-1}$ for all $a > 0$, but I do not see how.

Answer 1

In such a case, plotting the curve provides a better acquaintance with the issue:

What do we see on this curve ? That its derivative is negative, bounded and probably at its minimal value (therefore $|f'(x)|$ at its maximal value) for $x=0$.

This has to be confirmed, i.e., made rigorous.

Let us compute :

$$f'(x)=\exp(-\exp(x))(-\exp(x)) \ \ \text{implying}$$ $$f''(x)=\exp(-\exp(x)).\exp(x).(\exp(x)-1)$$

As a consequence:

$$\operatorname{sign}(f''(x))=\operatorname{sign}(\exp(x)-1)$$

which is negative for $x<0$ and positive otherwise ; we can therefore deduce that $f'(x)$ is decreasing until a minimum value $m$ taken in $x=0$, then increasing. This minimum value is

$$m=f'(0)=\exp(-\exp(0))(-\exp(0))=-e^{-1}\approx -0.368$$

Therefore, you can conclude, using the MVT that:

$$\forall a,b \in \mathbb{R}, \ \ \ |f(b)-f(a)| \leq M|b-a|$$

with Lipschitz constant $M=|m|$.