Lipschitz constant of $\log(1 + \exp(\lvert x \rvert ))$

Question

Following the Wikipedia article we have that "An everywhere differentiable function $g : \mathbb{R} \mapsto \mathbb{R}$ is Lipschitz continuous (with $K = \sup \lvert g'(x)\rvert$) if and only if it has bounded first derivative."

Hence, the Lipschitz constant of $\log(1+ \exp(-x))$ equals $1$ since its derivative is bounded in $(-1,0)$.

I was wondering whether the same result can be applied for $\log(1+ \exp(\lvert x\rvert))$. The analysis would be mostly the same, but the condition "everywhere differentiable" does not hold anymore.

Answer 1

Let $g(x) = \log\left(1 + \exp(|x|\right)$,

• if $x > 0$ and $y < 0$:

$$\left|g(x) - g(y)\right| \le |g(x)| + |g(y)| \le |x| + |y| = |x - y|$$

This is the only problematic case. Other cases can be handled easily.