Proof of Theorem 1.Suppose that U satisfies Assumption 1 and let $k<\nu$ . Select a such that $k=a/(1-a)$ , so that $a<1-1/(1+\nu)$ . For any $\tilde{\beta} \in (0,1)$ we see that there exists a $c_0>0$ such that for all $x \notin C$ ,

\begin{align*}\bigl( V_{\tilde{\beta}, \delta}(x,\theta) \bigr)^{1-a}|U^{\prime}(x)|\geq c_0 \exp \bigl\{ \tilde{\beta} (1-a)(1+\nu) \log\! |x| \bigr\}\dfrac{1+\nu}{|x|}=c_0(1+\nu) | x |^{\tilde{\beta} (1-a) (1+\nu)-1} .\end{align*}

Since $(1-a)(1+\nu)-1>0$ , there exists a $\beta$ close to 1 such that $\beta (1-a) (1+\nu)-1>0$ , so

(7) \begin{equation}\lim_{|x| \rightarrow \infty}V^{1-a}_{\beta}(x,\theta)|U^{\prime}(x)|=+\infty.\end{equation}

Now $V_{\beta, \delta} \in C^1$ and $\lim_{|x|\rightarrow \infty}V_{\beta, \delta}(x,\theta)=+\infty$ , so all the level sets are compact. Since the process is positive Harris recurrent and some skeleton is irreducible (see [Reference Bierkens, Roberts and Zitt10]), we get from Proposition 6.1 of [Reference Meyn and Tweedie20] that the level sets are also small. Since $\lim_{|x|\rightarrow \infty}U(x)=+\infty$ , it is clear that $V_{\beta, \delta}$ is bounded below away from 0, so by multiplying with an appropriate constant we can assume that $V_{\beta, \delta}(x,\theta)\geq 1$ for all $(x,\theta)$ . We calculate

\begin{equation*}\mathcal{L}V_{\beta, \delta}(x,\theta)=V_{\beta, \delta}(x,\theta)\bigl( \theta \beta U^{\prime}(x)+([\theta U^{\prime}(x)]^++\gamma(x))({\exp} \{ -2 \theta\ \mathrm{sgn}(x) \delta \}-1 ) \bigr).\end{equation*}

Note that due to Assumption 1 and since

\begin{equation*}U(x)\xrightarrow{|x| \rightarrow \infty} +\infty,\end{equation*}

we get that there exists a compact set $\tilde{C}$ such that for all $x \notin \tilde{C}$ , $\mathrm{sgn}(U^{\prime}(x))=\mathrm{sgn}(x)$ . Therefore, when $x \notin \tilde{C}$ and $\theta\ \mathrm{sgn}(x)>0$ ,

\begin{align*}\dfrac{\mathcal{L}V_{\beta, \delta}(x,\theta)}{V_{\beta, \delta}^a(x,\theta)}\leq V_{\beta, \delta}^{1-a}(x,\theta)|U^{\prime}(x)|\biggl[ \beta+\biggl( \dfrac{\gamma(x)}{|U^{\prime}(x)|}+1 \biggr) ({\exp} \{ -2 \delta \}-1 )\biggr],\end{align*}

and when $\theta \mathrm{sgn}(x) <0$ ,

\begin{equation*}\dfrac{\mathcal{L}V_{\beta, \delta}(x,\theta)}{V_{\beta, \delta}^a(x,\theta)}\leq V_{\beta, \delta}^{1-a}(x,\theta)|U^{\prime}(x)|\biggl[ -\beta +\dfrac{\gamma(x)}{|U^{\prime}(x)|}({\exp} \{ 2 \delta \} -1) \biggr].\end{equation*}

Overall, we have for $x \notin \tilde{C}$

(8) \begin{align} &\dfrac{\mathcal{L}V_{\beta, \delta}(x,\theta)}{V_{\beta, \delta}^a(x,\theta)} \leq V_{\beta, \delta}^{1-a}(x,\theta)|U^{\prime}(x)| \max \biggl\{ \beta+\biggl( \dfrac{\gamma(x)}{|U^{\prime}(x)|}+1 \biggr) ({\exp} \{ -2 \delta \}-1 ) ,\nonumber\\& \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\biggl[\! -\beta +\dfrac{\gamma(x)}{|U^{\prime}(x)|}({\exp} \{ 2 \delta \} -1) \biggr] \biggr\}.\end{align}

Recall that

\begin{equation*}\dfrac{\gamma(x)}{|U^{\prime}(x)|}\xrightarrow{|x| \rightarrow \infty}0.\end{equation*}

Fix $\delta>-1/2\log\!(1-\beta)$ , and by possibly increasing $\tilde{C}$ we have that there exists a constant $c^{\prime}>0$ such that for all $x \notin \tilde{C}$ ,

\begin{equation*}\max \biggl\{ \beta+\biggl(\dfrac{\gamma(x)}{|U^{\prime}(x)|}+1 \biggr) ({\exp} \{ -2 \delta \}-1), -\beta +\dfrac{\gamma(x)}{|U^{\prime}(x)|}({\exp} \{ 2 \delta \} -1 ) \biggr\}<-c^{\prime}<0.\end{equation*}

Combining this with (7) and (8), we find that $V_{\beta, \delta}$ satisfies (6) with $f(u)=cu^a$ for $c=c^{\prime}/2$ and with K being an appropriate constant that bounds the continuous function $\mathcal{L}V_{\beta, \delta}+ f(V_{\beta, \delta})$ inside $\tilde{C}$ .

Therefore all the conditions of Theorem 2 are satisfied. Note that

\begin{equation*} H_{f}(s)=\int_1^s1/f(u){\mathrm{d}} u=c^{-1}\int_1^su^{-a}{\mathrm{d}} u=\dfrac{1}{c(1-a)}\bigl(s^{1-a}-1\bigr), \end{equation*}

so

\begin{equation*}H^{-1}_{f}(t)= ( 1+c(1-a)t )^{1/(1-a)}\end{equation*}

and therefore

\begin{equation*}f \circ H^{-1}_{f}(t)= c ( 1+c(1-a)t )^{a/(1-a)}.\end{equation*}

Since we picked a such that $k=a/(1-a)$ , meaning that $k+1=1/(1-a)$ , (3) follows.

Proof of Remark 2.Fix $k<\nu$ and set a such that $k=a/(1-a)$ , therefore $1-a=1/(1+k)$ . Our goal is to find an appropriate upper bound on ${\gamma(x)}/{|U^{\prime}(x)|}$ that will guarantee that the right-hand side of (8) is negative for appropriately chosen $\beta$ and $\delta$ . For this it suffices to prove that for some $\beta,\delta >0$ , the maximum appearing in the right-hand side of (8) is negative, bounded away from zero and that (7) holds. These two conditions will guarantee that the drift condition (6) holds for $V_{\beta , \delta}$ and we can conclude similarly to the proof of Theorem 1. Since from the proof of that theorem

\begin{align*}\bigl( V_{\tilde{\beta}, \delta}(x,\theta) \bigr)^{1-a}|U^{\prime}(x)|\geq c_0(1+\nu) | x |^{\tilde{\beta} (1-a) (1+\nu)-1},\end{align*}

in order to guarantee (7) it suffices to pick $\beta=(1+\eta)(1+k)/(1+\nu)$ for some $\eta >0$ . From the discussion after (8), we can pick

\begin{equation*}\delta=-\frac{1}{2}(1+\eta)\log\!(1-\beta)=-\frac{1}{2}(1+\eta)\log\!(1-(1+\eta)(1+k)/(1+\nu)).\end{equation*}

With this choice of $\beta, \delta$ , the first part of the maximum of the right-hand side of (8) will be negative, bounded away from zero. The second part of the maximum is given by

\begin{align*}-\beta +\dfrac{\gamma(x)}{|U^{\prime}(x)|}({\exp} \{ 2 \delta \} -1) & =-(1+\eta)\dfrac{1+k}{1+\nu}\\& \quad +\dfrac{\gamma(x)}{|U^{\prime}(x)|}\biggl( \biggl(\dfrac{1}{(1-(1+\eta)(1+k)/(1+\nu))}\biggr)^{(1+\eta)} -1\biggr).\end{align*}

Therefore, solving the inequality

\begin{equation*}-\beta +\frac{\gamma(x)}{|U^{\prime}(x)|}({\exp} \{ 2 \delta \} -1)<-\eta\end{equation*}

(which would guarantee that the maximum on the right-hand side of (8) is negative, bounded away from zero), we get

\begin{equation*} \dfrac{\gamma(x)}{|U^{\prime}(x)|} \leq M(k){,}\end{equation*}

where M(k) as in (4).

Proof of Corollary 1.Let $\pi$ as in (5). Then for all $\delta^{\prime}>0$ there exists a compact set C with $0 \in C$ , such that for all $x \notin C$ ,

\begin{align*}|U^{\prime}(x)|=\dfrac{(\nu+1)|x|}{\nu+x^2}\geq \dfrac{1+\nu-\delta^{\prime}}{|x|}.\end{align*}

Therefore, for every $\delta^{\prime}>0$ , the distribution satisfies Assumption 1, where the $\nu$ in that assumption is equal to $\nu-\delta^{\prime}$ . From Proposition 1, for all $k<\nu$ , there exists a $\beta \in (0,1), \delta >0$ and $B>0$ such that for all $(x,\theta) \in \mathbb{R} \times \{ -1,+1 \}$ ,

\begin{equation*}\| \mathbb{P}_{x,\theta}(Z_t \in \cdot)-\mu({\cdot}) \|_{TV} \leq \dfrac{B V_{\beta, \delta}(x)}{t^{1+k}}+\dfrac{B}{t^{k}}.\end{equation*}

Now suppose that the Zig-Zag starts from $x=0$ , $\theta=+1$ . There exists a $C_0$ and $K>0$ such that for all $|x| \geq K$ , $\pi(x)\geq C_0 |x|^{-\nu-1}$ . Fix a time $t>K$ . Let $A_t= \{ x\,:\, x>t \}$ . After time less than or equal to t has passed, the Zig-Zag will not have hit $A_t$ . We therefore get for all $t > K$

\begin{align*}\|\mathbb{P}_{0,+1}(Z_t \in \cdot)-\mu({\cdot})\|_{TV} &\geq| \mathbb{P}_{0,+1}(X_t \in A_t)-\pi(A_t) |\\ &=\pi(A_t)\\&= \int_{|x|>t} \pi(x){\mathrm{d}} x \\& \geq 2C_0\int_t^{+\infty}x^{-\nu-1}{\mathrm{d}} x\\ &=2\dfrac{C_0}{\nu}\dfrac{1}{t^{\nu}}.\end{align*}