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Let ${\varphi }_t(x_0)$ denote the flow map generated by $\dot{x}=f(x)$.

is nonpositive on $E$.

The First Lyapunov theorem says that the existence of such a function makes the origin Lyapunov stable. To prove it, choose a sufficiently small closed ball ${\overline{B}}_{\epsilon }\subset E$. By compactness and positive definiteness,By continuity, some $\delta >0$ satisfies $V(x)<\alpha$ whenever $\Vert x\Vert <\delta$. Along a trajectory beginning in $B_{\delta }$, $V$ cannot increase. Such a trajectory therefore cannot reach the sphere $\Vert x\Vert =\epsilon$, where $V\ge \alpha$. Hence it remains in $B_{\epsilon }$ for every $t\ge 0$, which is Lyapunov stability.

The LaSalle invariance principle says that if a trajectory remains in a compact positively invariant set on which $\dot{V}\le 0$, then it approaches the largest invariant subset of $\{x:\dot{V}(x)=0\}$.

SetThenThe Taylor series at the endpoint isand henceThe Watson lemma, together withtherefore yields

The phase function $\cos t$ attains its global maximum $1$ on $[0,3\pi ]$ at the stationary endpoint $t=0$ and at the interior point $t=2\pi$. At either point,By Laplace's method at a nondegenerate stationary endpoint, the endpoint contributesBy Laplace's method, the interior maximum contributesThe remaining parts of the interval are exponentially smaller. Summing the two contributions according to Laplace's method with multiple global maxima gives

Let $L(t)=f(x,t)$ be the likelihood function. Since the prior distribution is $N(0,I_p)$, the posterior density has the formwhere $Z$ is the normalizing constant.

Choose $0<\delta \le 1/2$ and use the Gaussian autoregressive proposal reversible with respect to a standard normal distributionThuswhose covariance matrix is the required $2\delta I_p$. Reversibility with respect to the standard normal distribution sayssoThe Metropolis–Hastings acceptance probability therefore reduces toThis is the Preconditioned Crank–Nicolson algorithm with proposal scale $\beta =\sqrt{2\delta }$.

Let $K(t,\cdot )$ be the Markov kernel of the Metropolis–Hastings algorithm. Away from the possibility of remaining at $t$, its transition density isMultiplication by the target density givesThis is detailed balance for every pair of distinct states.

Writefor the rejection probability, soFor every measurable $B\subseteq \Theta$, detailed balance and the Tonelli theorem giveHence $\mu$ is an invariant distribution of the chain.

The Kolmogorov zero-one law says that for independent random variables $(Y_n{)}_{n\ge 0}$, every event in the tail sigma-algebrahas probability zero or one.

In the canonical model of an independent and identically distributed sequence, the sample space is a sequence space $E^{\mathbb{N}}$ with the product law, $Y_n$ is the $n$th coordinate, and the left Bernoulli shift $S$ satisfies $Y_n\circ S=Y_{n+1}$. If $A$ is shift-invariant, then for every $N$,so $A\in \mathcal{T}$. The zero-one law therefore says that the canonical one-sided i.i.d. shift is ergodic.

A measure-preserving transformation $T$ on a probability space $(X,\mathcal{A},m)$ is ergodic if every invariant event $A\in \mathcal{A}$, meaningup to an $m$-null set, satisfies $m(A)\in \{0,1\}$. Thus the invariant sigma-algebra is trivial modulo null sets.

Again set $h=\sqrt{G}$ and fix $\theta$. For $0<\alpha <C^{-1/2}$ defineSince $h^{\prime \prime }=-Kh$ and $K\le C$,For $0<r<\pi /\sqrt{C}$, differentiation givesThe expression in parentheses vanishes at zero, and henceIts limit at zero is $(1-\alpha \sqrt{C})/\sqrt{C}>0$, so $f>0$. Letting $\alpha \uparrow C^{-1/2}$ yields the one-dimensional Rauch comparison inequalityIt follows thatNowas $x\to 0$. Given $0<\gamma <1$, choose $x_0\in (0,\pi )$ such that $2(1-\cos x)/x^2\ge \gamma$ for $0<x\le x_0$, and set ${\epsilon }_0=x_0/\sqrt{C}$. Then every geodesic polar coordinate ball of radius $\epsilon \le {\epsilon }_0$ satisfies

Put $h(r,\theta )=\sqrt{G(r,\theta )}$. The Jacobi equation in geodesic polar coordinates and the initial conditions from part (b) areInside a geodesic polar coordinate ball, $h>0$. If $K\le 0$, then $h_{rr}\ge 0$, so $h_r\ge 1$ and therefore $h(r,\theta )\ge r$. The Riemannian area element is $hdrd\theta$, whenceThus nonpositive Gaussian curvature makes such a ball at least as large as the Euclidean disk of the same radius.