A globally Lipschitz Hamiltonian force

Given: AC, $U(q)=\sqrt{1+q^2}$, unit mass, and the one-dimensional Euclidean metric.

The HMC target is $\pi (q)=Z_U^{-1}{e}^{-U(q)}$ and its Hamiltonian vector field is $F(q,p)=(p,-U^{\mathrm{\prime }}(q))$. Hamiltonian Monte Carlo

A globally state-Lipschitz ODE vector field has a unique global solution. A globally state-Lipschitz vector field on $\mathbb{R}\times {\mathbb{R}}^n$ has global solutions

Solutions of a state-Lipschitz ODE obey the Gronwall stability estimate. The Grönwall estimate for two solutions of a Lipschitz ODE

Under Countable Choice, a complete $C^2$ Hamiltonian flow preserves $H$ and phase volume; a globally Lipschitz gradient gives the stated sufficient completeness and stability bound. A complete Hamiltonian flow for a C2 potential preserves energy and phase volume

If a function is continuous on a closed interval and differentiable on its interior, its derivative at some interior point equals the average rate of change. The mean value theorem, as the case $g(x)=x$ of Cauchy's: for $f$ continuous on $[a,b]$ with $a<b$ and differentiable on $(a,b)$ there is $c\in (a,b)$ with $f(b)-f(a)=f^{\mathrm{\prime }}(c)(b-a)$

AC implies Countable Choice, the axiom required for the exact-flow volume result. AC supplies countable selections and prescribed serial paths

Under Countable Choice, the function $x\mapsto e^{-\mathrm{\mid }x\mathrm{\mid }}$ is nonnegative, measurable, and Lebesgue integrable. The real-line exponential tail is Lebesgue integrable

The exponential is positive and strictly increasing; under Countable Choice, the measure of $[-1,1]$ is its length $2$. The exponential is positive and satisfies $\exp (-x)=1\mathrm{/}\exp (x)$, The exponential function is strictly increasing, A box in ${\mathbb{R}}^n$ with parameters $a_i\le b_i$ is Lebesgue measurable of measure ${\prod }_{i<n}(b_i-a_i)$, whichever of its faces are included

For nonnegative measurable functions, the nonnegative integral is monotone under pointwise domination. Monotonicity and nonnegative homogeneity of the nonnegative integral

On a measure space, for nonnegative measurable $f$ and measurable $E$, ${\int }_{E}fd\lambda =\int f{\mathbf{1}}_{E}d\lambda$. If $f\ge c\ge 0$ on $E$, monotonicity and the simple-function definition give ${\int }_{E}fd\lambda \ge c\lambda (E)$. Integral over a measurable subset, Monotonicity and nonnegative homogeneity of the nonnegative integral, The integral of a nonnegative simple function