Recovering a variance from its Gaussian characteristic value

On a bounded interval, $u \mapsto e^{-u/2}$ has a Lipschitz inverse. For $a, b \in [0, B]$,

Why the boundedness matters. The map $u \mapsto e^{-u/2}$ is injective on $[0,\infty)$ but its inverse is not uniformly continuous there: $e^{-a/2}$ and $e^{-b/2}$ can be arbitrarily close while $|a-b|$ is arbitrarily large, provided $a, b$ are both large. Restricting to $[0,B]$ removes this, with the explicit constant $2e^{B/2}$.

Where it is used. The value $e^{-v/2}$ is $\int\cos\,dN(0,v)$. In the truncation argument for the Markov chain central limit theorem, one first shows the numbers $e^{-v_K/2}$ form a Cauchy sequence, by testing the laws against $\cos$; this lemma converts that into the Cauchy property of the variances $v_K$ themselves, using the separately established uniform bound $v_K \le B$.

Proof. By symmetry assume $a \le b$. Then

using $b \le B$ for the first factor and the elementary inequality $e^t \ge 1 + t$ for the second. Multiplying by $2e^{B/2}$ gives $b - a \le 2e^{B/2}(e^{-a/2} - e^{-b/2})$, and both sides of the desired inequality are then read off after removing the absolute values (the right-hand difference is nonnegative in this case).