[Math] The exponential function is locally Lipschitz continuous with the Lipschitz constant $K=1$

Question

The exponential function $x→e^{x}$ becomes arbitrarily steep as $x → ∞$, and therefore is not globally Lipschitz continuous, despite being an analytic function. I am asking if it is possible to find certain regions for the variable $x$ in which the exponential function $x→e^{x}$ is locally Lipschitz continuous with the Lipschitz constant $K=1$.

Answer 1

$|e^{x}-e^{y}| \leq |x-y|$ in an interval $[a,b]$ if and only if $e^{x} \leq 1$ or $x \leq 0$ on $[a,b]$ if and only if $b \leq 0$. [ In one direction divide by $x-y$ and take limit as $y \to x$].