 Solution

For fixed $L$, expansion of the matrix in part (b) givesHence its largest eigenvalue is $1+O({\epsilon }^3)$. With $z=(x(0),p(0))$ and $z^{\prime }=M_{\epsilon }^{L}z$,For each fixed starting state, or uniformly on any bounded set of starting states, this impliesUsing $e^{-u}\ge 1-u$ for $u\ge 0$ in the acceptance formula givesfor sufficiently small $\epsilon$. The constant necessarily depends on a bound for $\Vert z\Vert$: a uniform pointwise constant over all of ${\mathbb{R}}^{2d}$ would be impossible because the energy error is quadratic in the starting state.