Proof of Theorem 1.1. We fix $\nu> \nu^*$ in this proof and so $\lambda$ is also fixed. We will prove the theorem in five steps. In the first four steps, we assume the number of offspring is 2, that is, $L=1$ . In the last step, we prove the result for general L.

Step 1. Suppose $L=1$ and thus $m=1$ . Let $\mathbf{w}(t,x) = 1 -\mathbf{u}(t,x)$ ; then $\mathbf{w}(t,x)$ satisfies

(3.1) \begin{equation} \begin{cases} &\frac{\partial \mathbf{w}}{\partial t} = \frac{1}{2} \frac{\partial^2 \mathbf{w}}{\partial x^2} + \mathbf{g}\cdot \big(\mathbf{w}-\mathbf{w}^2\big),\\[7pt] &\mathbf{w}\!\left(t+\frac{1}{\nu},x\right) = \mathbf{w}(t,x-1), \end{cases} \end{equation}

for $t\geq 0$ , $x\in\mathbb{R}$ . Define

\begin{equation*} \mathbf{w}({-}t,x) = \mathbf{w}\!\left({-}t+\frac{\lceil \nu t \rceil}{\nu}, x+\lceil \nu t \rceil\right),\quad \mbox{for } t>0,\, x\in\mathbb{R}. \end{equation*}

By the periodicity of $\mathbf{w}$ , we get that $\mathbf{w}(t,x)$ satisfies (3.1). Put

(3.2) \begin{equation} \widetilde {\mathbf{w}}(t,x)\,:\!=\, \frac{e^{\lambda x-\gamma(\lambda)t}\mathbf{w}(t,x)} {\psi(x,\lambda)}. \end{equation}

Recall that $\big\{Y_t, \Pi_x^{\lambda} \big\}$ is a diffusion with infinitesimal generator (2.6). For $t\ge0$ , let $Y_{[0, t]}$ be the restriction of Y to [0, t]. Define

(3.3) \begin{equation} f\big(Y_{[0, t]}\big)= \widetilde {\mathbf{w}}({-}t,Y_t) e^{-\int_0^t \mathbf{g}(Y_s)\mathbf{w}({-}s,Y_s)\text{d}s},\quad t\geq 0. \end{equation}

In this step we prove that $\left\{\big(f\big(Y_{[0, t]}\big)\big)_{t\geq 0}, \Pi_x^{\lambda} \right\}$ is a positive martingale.

By the Feynman–Kac formula, we have

(3.4) \begin{equation} \mathbf{w}(T,x) = \Pi_x \left[ \mathbf{w}(T-t,B_t) e^{\int_0^t \mathbf{g}(B_s)(1-\mathbf{w}(T-s,B_s))\text{d}s} \right], \quad \mbox{for $T\in\mathbb{R}$, $t>0$.} \end{equation}

Recall that, since $m=1$ ,

\begin{equation*} \frac{\text{d}\Pi_x^{\lambda}}{\text{d}\Pi_x}\bigg{|}_{\mathcal{F}_t^B} = \frac{\Xi_t(\lambda)}{\Xi_0(\lambda)} = \frac{e^{-\gamma(\lambda)t-\lambda B_t +\int_0^t\mathbf{g}(B_s)\text{d}s} \psi(B_t,\lambda) }{e^{-\lambda x} \psi(x,\lambda)}. \end{equation*}

Therefore,

\begin{align*} \mathbf{w}(T,x) &= \Pi_x^{\lambda} \left[\frac{\Xi_0(\lambda)}{\Xi_t(\lambda)} \mathbf{w}(T-t,B_t) e^{\int_0^t \mathbf{g}(B_s)(1-\mathbf{w}(T-s,B_s))\text{d}s} \right]\\ &= \Pi_x^{\lambda} \left[ e^{-\lambda x} \psi(x,\lambda) \frac{e^{\lambda B_t + \gamma(\lambda)t - \int_0^t \mathbf{g}(B_s)\text{d}s} \mathbf{w}(T-t,B_t)}{\psi(B_t,\lambda)} e^{\int_0^t \mathbf{g}(B_s)(1-\mathbf{w}(T-s,B_s))\text{d}s} \right]\\ &= \Pi_x^{\lambda} \left[ e^{-\lambda x +\gamma(\lambda)T } \psi(x,\lambda) \frac{e^{\lambda B_t-\gamma(\lambda)(T-t)} \mathbf{w}(T-t,B_t)}{\psi(B_t,\lambda)} e^{-\int_0^t \mathbf{g}(B_s)\mathbf{w}(T-s,B_s)\text{d}s} \right]. \end{align*}

Thus we have

\begin{equation*} \widetilde {\mathbf{w}}(T,x) = \Pi_x^{\lambda} \left[\widetilde {\mathbf{w}}(T-t,B_t) e^{-\int_0^t \mathbf{g}(B_s)\mathbf{w}(T-s,B_s)\text{d}s} \right]. \end{equation*}

Note that both $\big\{B_t, \Pi_x^{\lambda}\big\}$ and $\big\{Y_t, \Pi_x^{\lambda}\big\}$ are diffusions with infinitesimal generator $\mathcal{A}$ . Thus

(3.5) \begin{equation} \widetilde {\mathbf{w}}(T,x) = \Pi_x^{\lambda} \left[\widetilde {\mathbf{w}}(T-t,Y_t) e^{-\int_0^t \mathbf{g}(Y_s)\mathbf{w}(T-s,Y_s)\text{d}s} \right]. \end{equation}

It follows from $\nu=\frac{\gamma(\lambda)}{\lambda}$ that

(3.6) \begin{equation} \widetilde {\mathbf{w}}\!\left(t+\frac{1}{\nu},x+1\right) = \frac{e^{\lambda (x+1)-\gamma(\lambda)\big(t+\frac{1}{\nu}\big)}\mathbf{w}\big(t+\frac{1}{\nu},x+1\big)} {\psi(x+1,\lambda)} = \frac{e^{\lambda x-\gamma(\lambda)t}\mathbf{w}(t,x)} {\psi(x,\lambda)} = \widetilde {\mathbf{w}}(t,x). \end{equation}

For $0<s<t$ , we have

\begin{align*}& \Pi_{x}^{\lambda} \left[ f\big(Y_{[0, t]}\big) \Big|\mathcal{F}_s \right]= \Pi_{x}^{\lambda} \left[\widetilde {\mathbf{w}}({-}t,Y_t) e^{-\int_0^t \mathbf{g}(Y_r)\mathbf{w}({-}r,Y_r)\text{d}r}\Big|\mathcal{F}_s \right] \\ &=e^{-\int_0^s \mathbf{g}(Y_r)\mathbf{w}({-}r,Y_r)\text{d}r} \Pi_{Y_s}^{\lambda} \left[\widetilde {\mathbf{w}}({-}t,Y_{t-s}) e^{-\int_0^{t-s} \mathbf{g}(Y_r)\mathbf{w}({-}(r+s),Y_r)\text{d}r} \right]\\ &=e^{-\int_0^s \mathbf{g}(Y_r)\mathbf{w}({-}r,Y_r)\text{d}r} \Pi_{Y_s}^{\lambda} \left[\widetilde {\mathbf{w}}\big({-}s-(t-s),Y_{t-s}\big) e^{-\int_0^{t-s} \mathbf{g}(Y_r)\mathbf{w}({-}s-r,Y_r)\text{d}r} \right]\\ &=e^{-\int_0^s \mathbf{g}(Y_r)\mathbf{w}({-}r,Y_r)\text{d}r} \widetilde {\mathbf{w}}({-}s,Y_s)= f(Y_{[0, s]}), \end{align*}

where the penultimate equality follows from (3.5) with $T=-s$ . Hence the process $\big\{\big(f\big(Y_{[0, t]}\big)\big)_{t\geq 0}, \Pi_x^{\lambda} \big\}$ is a positive martingale.

Step 2. Suppose $L=1$ . In this step, we will show that there exists a constant $\beta\geq 0$ such that for any $x\in\mathbb{R}$ ,

(3.7) \begin{equation} \widetilde {\mathbf{w}}({-}t,Y_t) \rightarrow \beta \quad \mbox{ as } t\to\infty,\ \Pi_x^{\lambda}\mbox{-a.s.} \end{equation}

Moreover, we also prove that for any $x,T\in\mathbb{R}$ ,

(3.8) \begin{equation} \widetilde {\mathbf{w}}(T-t,Y_t) \rightarrow \beta \quad \mbox{ as } t\to\infty,\ \Pi_x^{\lambda}\mbox{-a.s.} \end{equation}

It follows from (2.7) and (2.1) that

\begin{equation*}\lim\limits_{s\rightarrow\infty}\dfrac{Y_s+\nu s}{s} = -\gamma^{\prime}(\lambda) + \frac{\gamma(\lambda)}{\lambda}>0.\end{equation*}

Thus $\lim\limits_{s\rightarrow\infty} (Y_s+\nu s) = \infty$ . Since a positive martingale has a non-negative finite limit, taking logarithms in (3.3) and dividing by $Y_t+\nu t$ gives

(3.9) \begin{equation} \limsup_{t\rightarrow\infty} \left\{\frac{\ln \widetilde {\mathbf{w}}({-}t,Y_t)}{Y_t+\nu t} - \frac{1}{Y_t+\nu t} \int_0^t \mathbf{g}(Y_s)\mathbf{w}({-}s,Y_s)\text{d}s \right\} \leq 0 \quad \Pi_x^{\lambda}\mbox{-a.s.} \end{equation}

Put $\|\mathbf{g}\|_\infty=\max_{x\in[0, 1]}\mathbf{g}(x)$ . Taking $T=t$ in (3.4), we get

\begin{align*} \mathbf{w}(t,x) &= \Pi_x \left[ \mathbf{w}(0,B_t) e^{\int_0^t \mathbf{g}(B_s)(1-\mathbf{w}(t-s,B_s))\text{d}s} \right] \leq \Pi_x \left[ \mathbf{w}(0,B_t) e^{\|\mathbf{g}\|_\infty t} \right]\\ &\leq e^{\|\mathbf{g}\|_\infty/\nu} \Pi_0 \left[ \mathbf{w}(0,B_t+x) \right], \quad t\in\left[0,\frac{1}{\nu}\right]. \end{align*}

Since $\mathbf{w}(0,x)\rightarrow 0$ as $x\rightarrow\infty$ , we have, for any $\epsilon>0$ , $\mathbf{w}(0,x/2)\leq \epsilon/2$ when x is large enough. Since $\Pi_0(B_t+x\leq x/2) = \Pi_0(B_t\leq -x/2)$ , we have $\Pi_0(B_t+x\leq x/2)\leq \epsilon/2$ for x large enough. Therefore, for x large enough,

\begin{equation*} \mathbf{w}(t,x) \leq e^{\|\mathbf{g}\|_\infty/\nu} \Pi_0(\mathbf{w}(0,B_t+x)) \leq e^{\|\mathbf{g}\|_\infty/\nu} \epsilon,\quad t\in\left[0,\frac{1}{\nu}\right]. \end{equation*}

This implies that $\mathbf{w}(t,x)\rightarrow 0$ as $x\rightarrow\infty$ uniformly in $t\in\left[0,\frac{1}{\nu}\right]$ . Combining this with the fact that

\begin{equation*}\mathbf{w}({-}t, Y_t) = \mathbf{w}\!\left({-}t+\frac{\lceil \nu t \rceil}{\nu}, Y_t+\lceil \nu t \rceil\right)\end{equation*}

and $\lim\limits_{t\rightarrow\infty} (Y_t+\lceil \nu t \rceil) = \infty$ $\Pi_x^{\lambda}$ -a.s., we get

\begin{equation*} \lim_{t\rightarrow\infty}\frac{1}{t} \int_0^t \mathbf{g}(Y_s)\mathbf{w}({-}s,Y_s)\text{d}s = 0, \quad \Pi_x^{\lambda}\mbox{-a.s.} \end{equation*}

Therefore, by (3.9),

\begin{equation*} \limsup_{t\rightarrow\infty} \left\{\frac{\ln \widetilde {\mathbf{w}}({-}t,Y_t)}{Y_t+\nu t}\right\} \leq 0, \quad \Pi_x^{\lambda}\mbox{-a.s.} \end{equation*}

Hence by (3.2), we have

(3.10) \begin{equation} \limsup_{t\rightarrow\infty} \left\{\frac{\ln \left[ e^{\lambda Y_t + \gamma(\lambda)t} \mathbf{w}({-}t,Y_t) / \psi(Y_t,\lambda) \right] }{Y_t+\nu t} \right\} \leq 0, \quad \Pi_x^{\lambda}\mbox{-a.s.} \end{equation}

Since $\psi(\cdot,\lambda)$ is positive, continuous, and periodic, we have

\begin{equation*}0<\inf_{x\in\mathbb{R}} \psi(x,\lambda) \leq \sup_{x\in\mathbb{R}} \psi(x,\lambda)<\infty.\end{equation*}

Combining this with $\nu = \frac{\gamma(\lambda)}{\lambda}$ and (3.10), we get that

\begin{equation*} \limsup_{t\rightarrow\infty} \left\{\frac{\lambda (Y_t+\nu t) + \ln \mathbf{w}({-}t,Y_t) }{Y_t+\nu t} \right\} \leq 0, \quad \Pi_x^{\lambda}\mbox{-a.s.}; \end{equation*}

that is,

\begin{equation*} \limsup_{t\rightarrow\infty} \left\{\frac{\ln \mathbf{w}({-}t,Y_t)}{Y_t+\nu t}\right\} \leq -\lambda, \quad \Pi_x^{\lambda}\mbox{-a.s.} \end{equation*}

This implies that, for any $\delta>0$ and $\Pi_x^{\lambda}$ -a.s. all $\omega$ , there exists $C(\omega)>0$ such that

(3.11) \begin{equation} \mathbf{w}({-}t,Y_t(\omega)) \leq C(\omega)e^{-(\lambda-\delta)(Y_t(\omega)+\nu t)}, \quad t\geq 0. \end{equation}

Therefore,

\begin{equation*}\int_0^\infty \mathbf{g}(Y_s)\mathbf{w}({-}s,Y_s)\text{d}s<+\infty, \quad \Pi_x^{\lambda}\mbox{-a.s.}\end{equation*}

Consequently, by (3.3), $\widetilde {\mathbf{w}}({-}t,Y_t)$ converges $\Pi_x^{\lambda}$ -a.s. to some limit, say $\xi_x$ .

Next we use a coupling method to prove that $\xi_x$ is a constant $\Pi_x^{\lambda}$ -a.s. Consider $\Big\{\big(Y_t^1,Y_t^2\big), t\geq 0;\, \widetilde{\Pi}_{(x,y)}^{\lambda}\Big\}$ with $\big\{Y_t^1, t\geq 0\big\}$ and $\big\{Y_t^2, t\geq 0\big\}$ being independent, and

\begin{equation*} \big\{Y_t^1, t\geq 0;\, \widetilde{\Pi}_{(x,y)}^{\lambda}\big\} \overset{d}{=} \big\{Y_t, t\geq 0;\,\Pi_{x}^{\lambda}\big\}, \quad \big\{Y_t^2,t\geq 0;\, \widetilde{\Pi}_{(x,y)}^{\lambda}\big\} \overset{d}{=} \big\{Y_t,t\geq 0;\,\Pi_{y}^{\lambda}\big\}. \end{equation*}

Define

\begin{equation*} M_t^i = h\big(Y_t^i\big) + \gamma^{\prime}(\lambda)t - h\big(Y_0^i\big), \quad i=1,2. \end{equation*}

Then $\Big\{M_t^1, t\geq 0;\, \widetilde{\Pi}_{(x,y)}^{\lambda}\Big\}$ and $\Big\{M_t^2, t\geq 0;\, \widetilde{\Pi}_{(x,y)}^{\lambda}\Big\}$ are independent martingales. Hence

\begin{align*} \big\langle M^1-M^2 \big\rangle_t = \langle M^1 \rangle_t + \big\langle M^2 \big\rangle_t \rightarrow \infty, \quad \mbox{as } t\rightarrow\infty. \end{align*}

By the Dambis–Dubins–Schwarz theorem, we get

(3.12) \begin{align} \liminf_{t\rightarrow\infty}\! \big(M_t^1 - M_t^2\big) = -\infty \mbox{ and } \limsup_{t\rightarrow\infty}\! \big(M_t^1 - M_t^2\big) = +\infty, \quad \widetilde{\Pi}_{(x,y)}^{\lambda}\mbox{-a.s.} \end{align}

Since $h(x) = x - \frac{\psi_{\lambda}(x,\lambda)}{\psi(x,\lambda)}$ , we have

\begin{align*} Y_t^1 - Y_t^2 = M_t^1 - M_t^2 + \frac{\psi_{\lambda}\big(Y_t^1,\lambda\big)}{\psi\big(Y_t^1,\lambda\big)} - \frac{\psi_{\lambda}\big(Y_t^2,\lambda\big)}{\psi\big(Y_t^2,\lambda\big)}+ h\big(Y_0^1\big) - h\big(Y_0^2\big). \end{align*}

Combining (3.12) with the boundedness of $\frac{\psi_{\lambda}(x,\lambda)}{\psi(x,\lambda)}$ , we get

\begin{align*} \liminf_{t\rightarrow\infty} \!\big(Y_t^1 - Y_t^2\big) = -\infty \mbox{ and } \limsup_{t\rightarrow\infty} \!\big(Y_t^1 - Y_t^2\big) = +\infty, \quad \widetilde{\Pi}_{(x,y)}^{\lambda}\mbox{-a.s.} \end{align*}

Define $E\,:\!=\,\left\{\omega\,:\, \exists t_n=t_n(\omega)\to\infty \mbox{ with } Y_{t_n}^1 = Y_{t_n}^2 \mbox{ for all } n\right\}$ . Then it follows from the display above that

(3.13) \begin{equation} \widetilde{\Pi}_{(x,y)}^{\lambda} (E)=1. \end{equation}

If we use $\tilde{\xi}_x$ and $\tilde{\xi}_y$ respectively to denote the limits of $\widetilde {\mathbf{w}}\big({-}t,Y_t^1\big)$ and $\widetilde {\mathbf{w}}\big({-}t,Y_t^2\big)$ under $\widetilde{\Pi}_{(x,y)}^{\lambda}$ , then (3.13) implies $\widetilde{\Pi}_{(x,y)}^{\lambda}\big(\tilde{\xi}_x = \tilde{\xi}_y\big) = 1$ . Since $\tilde{\xi}_x$ and $\tilde{\xi}_y$ are independent, there is a constant $\beta\geq 0$ such that $\tilde{\xi}_x = \tilde{\xi}_y=\beta$ . Since $\tilde{\xi}_x \overset{d}{=} \xi_x$ , we have for any $x\in\mathbb{R}$ , $\xi_x = \beta$ , which means (3.7) holds.

Now we consider

\begin{equation*} f\big(Y_{[0, t]},T\big) = \widetilde {\mathbf{w}}(T-t,Y_t) e^{-\int_0^t \mathbf{g}(Y_s)\mathbf{w}(T-s,Y_s)\text{d}s},\quad t\geq 0. \end{equation*}

The proof in Step 1 also works if $f\big(Y_{[0, t]}\big)$ is replaced by $f(Y_{[0, t]},T)$ . Then, using the same argument as above, there exists another constant $\beta_T$ such that

\begin{equation*} \widetilde {\mathbf{w}}(T-t,Y_t) \rightarrow \beta_T,\quad \Pi_x^{\lambda}\mbox{-a.s.} \end{equation*}

We also have

\begin{align*} \liminf_{t\rightarrow\infty} \!\big(Y_{t+T}^1 - Y_t^2\big) = -\infty \mbox{ and } \limsup_{t\rightarrow\infty} \!\big(Y_{t+T}^1 - Y_t^2\big) = +\infty. \end{align*}

Hence, if we put $E_T\,:\!=\,\left\{\omega\,:\, \exists t_n=t_n(\omega)\to\infty \mbox{ with } Y_{t_n+T}^1 = Y_{t_n}^2 \mbox{ for all } n\right\}$ , then

(3.14) \begin{equation} \widetilde{\Pi}_{(x,y)}^{\lambda} (E_T)=1. \end{equation}

Notice that $Y_{t_n+T}^1 = Y_{t_n}^2$ implies $\widetilde {\mathbf{w}}\!\left(T-(t+T),Y_{t+T}^1\right) = \widetilde {\mathbf{w}}\big({-}t,Y_t^2\big)$ . Combining this with (3.14), we have $\beta_T = \beta$ ; that is, for any $x,T\in\mathbb{R}$ , (3.8) holds.

Step 3. In Step 2, we have shown that $\widetilde {\mathbf{w}}({-}t,Y_t) \rightarrow \beta$ $\Pi_x^{\lambda}$ -a.s., that is, $\widetilde {\mathbf{w}}$ converges along each path. In this step, we prove that

(3.15) \begin{equation} \widetilde {\mathbf{w}}(t,x) \rightarrow \beta \quad \mbox{ uniformly in }t\in \left[0,\frac{1}{\nu}\right] \mbox{ as } x\rightarrow\infty, \end{equation}

and $\beta$ is positive.

First, we prove that $\widetilde {\mathbf{w}}$ is bounded. Using the same notation as in Step 2, we also have, for any $k\in \mathbb{Z}$ ,

\begin{equation*} \Big\langle M^1_{\cdot+\frac{k}{\nu}}-M^2_{\cdot} \Big\rangle_t = \Big\langle M^1_{\cdot+\frac{k}{\nu}} \Big\rangle_t + \big\langle M^2_{\cdot} \big\rangle_t \rightarrow \infty, \quad \mbox{as } t\rightarrow\infty. \end{equation*}

By the Dambis–Dubins–Schwarz theorem, it holds that

\begin{equation*} \liminf_{t\rightarrow\infty} \!\Big(M_{t+\frac{k}{\nu}}^1 - M_t^2\Big) = -\infty \mbox{ and } \limsup_{t\rightarrow\infty} \!\Big(M_{t+\frac{k}{\nu}}^1 - M_t^2\Big) = +\infty, \end{equation*}

so the same conclusion holds for $Y_{t+\frac{k}{\nu}}^1 - Y_t^2$ . This implies, with $E_k\,:\!=\,\Big\{\omega\,:\, \exists t_n=t_n(\omega)\to\infty$ with $Y_{t_n+\frac{k}{\nu}}^1+k = Y_{t_n}^2 \mbox{ for all } n\Big\}$ , that

(3.16) \begin{equation} \widetilde{\Pi}_{(x,y)}^{\lambda} (E_k)=1. \end{equation}

Consider the event $\Omega_0$ defined by

\begin{equation*} \bigcap_{k\in\mathbb{Z}} E_k \bigcap\left\{\lim_{t\rightarrow\infty}\widetilde {\mathbf{w}}\big({-}t,Y_t^i\big) = \beta,\, \lim_{t\rightarrow\infty}\frac{Y_t^i}{t} = -\gamma^{\prime}(\lambda), \mbox{ and } Y_t^i {\mbox{ is continuous for } i=1,2} \right\}. \end{equation*}

By (2.7), (3.8), and (3.16), we get $\widetilde{\Pi}_{(x,y)}^{\lambda}(\Omega_0) = 1$ . By (3.6), it suffices to show that $\widetilde {\mathbf{w}}(t,x)$ is bounded in $\big[0,\frac{1}{\nu}\big] \times \mathbb{R}$ . Fix $\omega\in\Omega_0$ , and consider the continuous curves

\begin{equation*} \mathcal{L}_k^i = \left\{\left({-}t+\frac{k}{\nu}, Y_t^i(\omega)+k\right)\,:\, t\in\left(\frac{k-1}{\nu},\frac{k}{\nu}\right) \right\}, \quad k\in\mathbb{N},\, i=1,2. \end{equation*}

In Step 2 we have shown $\lim_{t\rightarrow\infty} (Y_t^i(\omega)+\nu t) = \infty$ for $i=1,2$ ; thus for any $x_0$ large enough, there exist $k=k(\omega)>j=j(\omega)\in N$ such that

\begin{equation*} x_0 \leq Y_t^1(\omega)+k \mbox{ for any } t\in \left[\frac{k-1}{\nu},\frac{k}{\nu}\right], \mbox{ and } x_0 \geq Y_t^2(\omega)+j \mbox{ for any } t\in \left[\frac{j-1}{\nu},\frac{j}{\nu}\right]. \end{equation*}

Define

\begin{equation*} \widetilde{\tau}(\omega) = \inf\bigg\{t\geq j/\nu\,:\, Y_{t+\frac{k-j}{\nu}}^1(\omega)+k = Y_t^2(\omega)+j \bigg\}. \end{equation*}

By the definition of $\Omega_0$ , we know $\widetilde{\tau}(\omega)<+\infty$ . Let $\widetilde{\mathcal{L}}$ denote the line segment

\begin{equation*}\bigg\{\frac{1}{\nu}\bigg\} \times \bigg[Y_{\frac{j-1}{\nu}}^2(\omega)+j, Y_{\frac{k-1}{\nu}}^1(\omega)+k\bigg],\end{equation*}

and define the curves

\begin{align*} &\widetilde{\mathcal{L}}_k^1 = \left\{\left({-}t+\frac{k}{\nu}, Y_t^1(\omega)+k\right)\,:\, t\in \left[\frac{k-1}{\nu}, \widetilde{\tau}(\omega)+\frac{k-j}{\nu}\right] \right\},\\ &\widetilde{\mathcal{L}}_j^2 = \left\{\left({-}t+\frac{j}{\nu}, Y_t^2(\omega)+j\right)\,:\, t\in \left[\frac{j-1}{\nu}, \widetilde{\tau}(\omega)\right] \right\}. \end{align*}

By the definition of $\widetilde{\tau}(\omega)$ , we have

\begin{equation*} \left({-}\left(\widetilde{\tau}(\omega)+\frac{k-j}{\nu}\right) + \frac{k}{\nu}, \, Y_{\widetilde{\tau}(\omega)+\frac{k-j}{\nu}}^1(\omega)+k\right) = \left({-}\widetilde{\tau}(\omega)+\frac{j}{\nu}, \, Y_{\widetilde{\tau}(\omega)}^2(\omega)+j\right). \end{equation*}

Define

\begin{equation*}y(\omega) = Y_{\widetilde{\tau}(\omega)+\frac{k-j}{\nu}}^1(\omega)+k;\end{equation*}

then $y(\omega) = Y_{\widetilde{\tau}(\omega)}^2(\omega)+j$ , and $\widetilde{\mathcal{L}}_k^1$ and $\widetilde{\mathcal{L}}_j^2$ intersect at the point $\big({-}\widetilde{\tau}(\omega)+\frac{j}{\nu}, \, y(\omega)\big)$ . Combining (2.6), (3.5), and the Feynman–Kac formula, we have

\begin{equation*} \frac{\partial \widetilde {\mathbf{w}}}{\partial t} = \frac{1}{2} \frac{\partial^2 \widetilde {\mathbf{w}}}{\partial x^2} + \left(\frac{\psi_x(x,\lambda)}{\psi(x,\lambda)} - \lambda\right) \frac{\partial \widetilde {\mathbf{w}}}{\partial x} - \mathbf{g} \mathbf{w} \widetilde {\mathbf{w}}. \end{equation*}

Let D denote the bounded domain with boundary $\widetilde{\mathcal{L}}\cup\widetilde{\mathcal{L}}_k^1\cup\widetilde{\mathcal{L}}_j^2$ (see Figure 1). By the maximum principle, we have that $\widetilde {\mathbf{w}}$ attains its maximum in $\overline D$ on $\widetilde{\mathcal{L}}_k^1\cup\widetilde{\mathcal{L}}_j^2$ , where $\overline D$ is the closure of D. Hence the maximum of $\widetilde {\mathbf{w}}$ on $\overline D$ is less than or equal to $K\,:\!=\,\max_{t\geq 0} \big\{\widetilde {\mathbf{w}}\big({-}t,Y_t^1(\omega)\big), \widetilde {\mathbf{w}}\big({-}t,Y_t^2(\omega)\big)\big\}$ . By the continuity of $\widetilde {\mathbf{w}}$ and since

\begin{equation*} \lim_{t\rightarrow\infty} \widetilde {\mathbf{w}}({-}t,Y_t^i(\omega)) = \beta, \quad i=1,2, \end{equation*}

we get $K<\infty$ . Notice that, fixing $\omega\in\Omega_0$ , for any $t_0\in \big[0,\frac{1}{\nu}\big]$ and any $x_0$ large enough, there exist $k=k(\omega,x_0)$ , $j=j(\omega, x_0)$ and a bounded domain $D=D(\omega,x_0)$ such that D has boundary $\widetilde{\mathcal{L}}\cup\widetilde{\mathcal{L}}_k^1\cup\widetilde{\mathcal{L}}_j^2$ and $(t_0,x_0) \in \overline D$ . Thus $\widetilde {\mathbf{w}}(t_0,x_0) \leq K$ . Combining this with $\lim\limits_{x\rightarrow-\infty} \widetilde {\mathbf{w}}(t,x) = 0$ and the continuity of $\widetilde {\mathbf{w}}$ , we get that $\widetilde {\mathbf{w}}(t,x)$ is bounded in $\big[0,\frac{1}{\nu}\big]\times \mathbb{R}$ , hence bounded in $\mathbb{R}\times \mathbb{R}$ .

Figure 1.

Bounded domain D with boundary $\widetilde{\mathcal{L}}\cup\widetilde{\mathcal{L}}_k^1\cup\widetilde{\mathcal{L}}_j^2$ .

Since $\widetilde {\mathbf{w}}(t,x)$ is bounded, by (3.5), (3.8), and the dominated convergence theorem,

\begin{equation*} \widetilde {\mathbf{w}}(t,x) = \Pi_x^{\lambda} \left[\beta e^{-\int_0^{\infty} \mathbf{g}(Y_s)\mathbf{w}(t-s,Y_s)\text{d}s} \right]\leq \beta. \end{equation*}

Since $\widetilde {\mathbf{w}}(t,x)>0$ , we have $\beta>0$ .

Next we show that

\begin{equation*} \widetilde {\mathbf{w}}(t,x) \rightarrow \beta \quad \mbox{ uniformly in }t\in \left[0,\frac{1}{\nu}\right] \mbox{ as } x\rightarrow\infty. \end{equation*}

It follows from (3.2) that

(3.17) \begin{equation} \begin{split} \mathbf{w}(t-s, Y_s) &\leq e^{-\lambda Y_s + \gamma(\lambda)(t-s)} \psi(Y_t,\lambda) \widetilde {\mathbf{w}}(t-s,Y_s)\\ &\leq \beta \max_{z\in[0, 1]}\psi(z,\lambda) e^{\gamma(\lambda)\frac{1}{\nu}} e^{-\lambda(Y_s+\nu s)} \\ &\leq C_1 e^{-\lambda(Y_s+\nu s)}, \quad \forall t\in \left[0,\frac{1}{\nu}\right], \end{split} \end{equation}

where $C_1$ is a constant depending only on $\lambda$ . Moreover, we have

\begin{equation*} \mathbf{g}(Y_s) \mathbf{w}(t-s, Y_s) \leq C e^{-\lambda(Y_s+\nu s)}, \quad \forall t\in \left[0,\frac{1}{\nu}\right], \end{equation*}

where $C = C_1 \|\mathbf{g}\|_{\infty}$ . For any $y\in [0, 1]$ and $n\in\mathbb{N}$ , let

\begin{equation*} f_n(y) = \Pi_y^{\lambda} \left[ e^{-\int_0^{\infty} Ce^{-\lambda(Y_s+n+\nu s)}\text{d}s} \right]. \end{equation*}

Recall that $\{x\}$ is the fractional part of x and $\lfloor x \rfloor$ is the integer part of x. Then by the periodicity of $Y_t$ , we have

\begin{equation*} \widetilde {\mathbf{w}}(t,x) \geq \beta \Pi_x^{\lambda} \left[ e^{-\int_0^{\infty} Ce^{-\lambda(Y_s+\nu s)} \text{d}s} \right] = \beta \Pi_{\{x\}}^{\lambda} \left[ e^{-\int_0^{\infty} Ce^{-\lambda(Y_s+\lfloor x \rfloor +\nu s)} \text{d}s} \right] = \beta f_{\lfloor x \rfloor }(\{x\}). \end{equation*}

Since $\int_0^{\infty} e^{-\lambda(Y_s+\nu s)} \text{d}s < +\infty$ $\Pi_x^{\lambda}$ -a.s., by the dominated convergence theorem, we get

\begin{align*} \lim_{n\rightarrow\infty} f_n(y) &= \Pi_y^{\lambda} \left[\lim_{n\rightarrow\infty} e^{-\int_0^{\infty} Ce^{-\lambda(Y_s+n+\nu s)}\text{d}s} \right] = \Pi_y^{\lambda} \left[ e^{- \lim_{n\rightarrow\infty} \int_0^{\infty} Ce^{- \lambda(Y_s+n+\nu s)}\text{d}s} \right]\\ &= \Pi_y^{\lambda} \left[ e^{- \int_0^{\infty} \lim_{n\rightarrow\infty} Ce^{- \lambda(Y_s+n+\nu s)}\text{d}s } \right] = 1. \end{align*}

Notice that $f_n(y) \leq f_{n+1}(y)$ , using Dini’s theorem we get $f_n(y) \rightarrow 1$ uniformly for $y\in[0, 1]$ . Combining this with $\beta f_n(y) \leq \widetilde {\mathbf{w}}(t,y+n) \leq \beta$ , we have (3.15).

By (3.2), we get

(3.18) \begin{equation} \frac{e^{\lambda x-\gamma(\lambda)t}\mathbf{w}(t,x)} {\psi(x,\lambda)} \rightarrow \beta \quad \mbox{ uniformly in } t\in \left[0,\frac{1}{\nu}\right] \mbox{ as } x\rightarrow\infty. \end{equation}

Step 4. Suppose $L=1$ . We will show that

(3.19) \begin{equation} \mathbf{w}\!\left(\frac{y-x}{\nu}, y\right) \sim \beta e^{-\lambda x}\psi(y,\lambda) \quad \mbox{ uniformly in } y\in [0, 1] \mbox{ as } x\rightarrow +\infty. \end{equation}

It is equivalent to show that, for any $\epsilon>0$ , there exists $x_0$ such that for any $x>x_0$ ,

\begin{equation*} \sup_{y\in [0, 1]} \Bigg{|}\frac{e^{\lambda x}\mathbf{w}\!\left(\frac{y-x}{\nu}, y\right)}{\psi(y,\lambda)} - \beta\Bigg{|} < \epsilon. \end{equation*}

By (3.18), for any $\epsilon>0$ , there exists $z_0$ such that for any $z>z_0$ ,

\begin{equation*} \sup_{t\in [0,\frac{1}{v}]} \Bigg{|}\frac{e^{\lambda z-\gamma(\lambda)t}\mathbf{w}\!\left(t, z\right)}{\psi(z,\lambda)} - \beta\Bigg{|} < \epsilon. \end{equation*}

Put

\begin{equation*} t = \frac{\{y-x \}}{\nu}, \quad z=-\lfloor y-x \rfloor + y. \end{equation*}

For any $x>z_0+1$ , we have

\begin{equation*} \sup_{\frac{\{y-x \}}{\nu}\in \big[0,\frac{1}{\nu}\big]} \Bigg{|}\frac{e^{\lambda ({-}\lfloor y-x \rfloor +y)-\gamma(\lambda)\frac{\{y-x \}}{\nu}}\mathbf{w}\!\left(\frac{\{y-x \}}{\nu}, -\lfloor y-x \rfloor+y\right)} {\psi({-}\lfloor y-x \rfloor+y,\lambda)} - \beta\Bigg{|} < \epsilon, \end{equation*}

that is

\begin{equation*} \sup_{y\in [0, 1]} \Bigg{|}\frac{e^{\lambda x}\mathbf{w}\!\left(\frac{y-x}{\nu}, y\right)}{\psi(y,\lambda)} - \beta\Bigg{|} < \epsilon, \end{equation*}

where we used the periodicity of $\psi$ and the fact that

\begin{equation*} \lambda ({-}\lfloor y-x \rfloor +y) - \gamma(\lambda)\frac{\{y-x \}}{\nu} = \lambda ({-}\lfloor y-x \rfloor +y) - \lambda \{y-x \} = \lambda ({-}(y-x)+y) = \lambda x. \end{equation*}

Thus (3.19) holds.

Step 5. In Steps 1–4, we have proven the theorem in the case of binary branching. It suffices to prove Steps 1–4 again for general branching mechanism.

For general branching mechanism, $\mathbf{w}=1-\mathbf{u}$ satisfies

\begin{equation*} \frac{\partial \mathbf{w}}{\partial t} = \frac{1}{2} \frac{\partial^2 \mathbf{w}}{\partial x^2} + \mathbf{g}\cdot(1-\mathbf{w}-\mathbf{f}(1-\mathbf{w})), \end{equation*}

where $\mathbf{f}(s) = \mathbf{E} s^{1+L} = \sum\limits_{k=0}^{\infty} p_k s^{1+k}$ and $m = \sum\limits_{k=0}^{\infty} k p_k$ . By the Feynman–Kac formula,

\begin{equation*} \mathbf{w}(T,x) = \Pi_x \left[ \mathbf{w}(T-t,B_t) e^{\int_0^t \mathbf{g}(B_s) \frac{1-\mathbf{w}-\mathbf{f}(1-\mathbf{w})}{\mathbf{w}}(T-s,B_s) \text{d}s} \right]. \end{equation*}

By the definition in (2.5),

\begin{align*} &\mathbf{w}(T,x) = \Pi_x^{\lambda} \left[\frac{\Xi_0(\lambda)}{\Xi_t(\lambda)} \mathbf{w}(T-t,B_t) e^{\int_0^t \mathbf{g}(B_s)\frac{1-\mathbf{w}-\mathbf{f}(1-\mathbf{w})}{\mathbf{w}}(T-s,B_s)\text{d}s} \right]\\ &= \Pi_x^{\lambda} \left[ e^{-\lambda x +\gamma(\lambda)T } \psi(x,\lambda) \frac{e^{\lambda B_t-\gamma(\lambda)(T-t)} \mathbf{w}(T-t,B_t)}{\psi(B_t,\lambda)} e^{\int_0^t \mathbf{g}(B_s)\big(\frac{1-\mathbf{w}-\mathbf{f}(1-\mathbf{w})}{\mathbf{w}}(T-s,B_s)-m\big)\text{d}s} \right]. \end{align*}

Put

\begin{equation*} A(w) =\left\{\begin{array}{ll} m - \frac{1-w-\mathbf{f}(1- w)}{w}, &\quad w\in(0, 1],\\[4pt] 0, &\quad w=0.\end{array}\right. \end{equation*}

Since $\big(B_t, \Pi^\lambda_x\big)$ and $\big(Y_t, \Pi^\lambda_x\big)$ have the same law, by (3.2), we have

\begin{align*} \widetilde {\mathbf{w}}(T,x) = \Pi_x^{\lambda} \left[\widetilde {\mathbf{w}}(T-t,Y_t) e^{-\int_0^t \mathbf{g}(Y_s)A(\mathbf{w})(T-s,Y_s)\text{d}s} \right]. \end{align*}

It follows from [Reference Athreya and Ney1, Corollary 2, p. 26] that $A({\cdot})$ is non-negative and non-decreasing. Moreover, for any $r,c\in (0, 1)$ ,

(3.20) \begin{equation} \sum_{n=0}^{\infty} A\big(cr^n\big) <+\infty \quad \mbox{ iff } \quad \mathbf{E}\big(L\log^+L\big) < +\infty. \end{equation}

Using the argument of Step 1, we get that

\begin{equation*}\Big\{\widetilde {\mathbf{w}}({-}t,Y_t) e^{-\int_0^t \mathbf{g}(Y_s)A(\mathbf{w})({-}s,Y_s)\text{d}s}, \Pi_x^{\lambda}\Big\}_{t\geq 0}\end{equation*}

is a non-negative martingale. Using the argument at the beginning of Step 2, we get that $\mathbf{w}(t,x) \rightarrow 0$ uniformly in $t\in \big[0,\frac{1}{\nu}\big]$ as $x\rightarrow\infty$ and $\mathbf{w}({-}t,Y_t)$ decays exponentially with rate at least $-\lambda$ . We will get that $\widetilde {\mathbf{w}}({-}t,Y_t)$ converges $\Pi_x^{\lambda}$ -a.s., if we can show that

\begin{align*} \int_0^{\infty} \mathbf{g}(Y_s)A(\mathbf{w})({-}s,Y_s)\text{d}s < +\infty, \quad \mbox{ $\Pi_x^{\lambda}$-a.s.} \end{align*}

By (3.11) and (3.20), if $\mathbf{E}(L\log^+L) < \infty$ , we have

\begin{align*} \int_0^{\infty} \mathbf{g}(Y_s) A(\mathbf{w})({-}s,Y_s)\text{d}s &\leq \|\mathbf{g}\|_\infty \int_0^{\infty} A\!\left(C e^{-(\lambda-\delta)(Y_s+\nu s)} \right) \text{d}s \\ &\leq \|\mathbf{g}\|_\infty \int_0^{\infty} A\!\left(C \big(e^{-(\lambda-\delta)(\nu-\gamma^{\prime}(\lambda))s}\big) \right)\text{d}s\\ &\leq \|\mathbf{g}\|_\infty \sum_{n=0}^{\infty} A\!\left(C \big(e^{-(\lambda-\delta) (\nu-\gamma^{\prime}(\lambda))n}\big) \right) < +\infty \quad \mbox{ $\Pi_x^{\lambda}$-a.s., } \end{align*}

where C is a constant depending on $\omega$ and may change in value from line to line.

Also by the arguments in Steps 2 and 3, $\widetilde {\mathbf{w}}(t,x)$ satisfies

\begin{equation*} \frac{\partial \widetilde {\mathbf{w}}}{\partial t} = \frac{1}{2} \frac{\partial^2 \widetilde {\mathbf{w}}}{\partial x^2} + \left(\frac{\psi_x(x,\lambda)}{\psi(x,\lambda)} - \lambda\right) \frac{\partial \widetilde {\mathbf{w}}}{\partial x} - \mathbf{g} \cdot A({\mathbf{w}}) \widetilde {\mathbf{w}}, \end{equation*}

and the maximum principle holds. Hence, $\widetilde {\mathbf{w}}(t,x)$ is bounded in $\big[0,\frac{1}{\nu}\big]\times \mathbb{R}$ . Since $A({\cdot})$ is non-decreasing, (3.17) implies

\begin{equation*} A(\mathbf{w})(t-s,Y_s) \leq A\!\left(C_1e^{-\lambda(Y_s+\nu s)}\right). \end{equation*}

By the periodicity of $\{Y_t\}$ , we have

\begin{align*} \widetilde {\mathbf{w}}(t,y+n)& \geq \beta \Pi_{y+n}^{\lambda} \left[ e^{-\int_0^{\infty} \|\mathbf{g}\|_\infty A\!\left(C_1e^{-\lambda(Y_s+\nu s)}\right) \text{d}s } \right]\\ &= \beta \Pi_y^{\lambda} \left[ e^{-\int_0^{\infty} \|\mathbf{g}\|_\infty A\!\left(C_1e^{-\lambda(Y_s+n+\nu s)}\right) \text{d}s } \right]. \end{align*}

To prove the theorem, it suffices to show

(3.21) \begin{equation} \Pi_y^{\lambda} \left[ e^{-\int_0^{\infty} \|\mathbf{g}\|_\infty A\!\left(C_1e^{-\lambda(Y_s+n+\nu s)}\right) \text{d}s } \right] \rightarrow 1, \quad \mbox{ as $n\rightarrow\infty$ uniformly for $y\in [0, 1]$.} \end{equation}

We know

\begin{equation*} \int_0^{\infty} \|\mathbf{g}\|_\infty A\!\left(C_1e^{-\lambda(Y_s+n+\nu s)}\right) \text{d}s < +\infty, \quad \mbox{ $\Pi_x^{\lambda}$-a.s.} \end{equation*}

By an argument similar to that in Step 3, (3.21) follows from $\lim_{u\downarrow 0} A(u) = 0$ , the dominated convergence theorem, and Dini’s theorem. This completes the proof.

Proof of Theorem 1.2. We prove the theorem in seven steps. In the first six steps, we consider the case that $L=1$ . In the last step we consider general L.

Step 1. Suppose $L=1$ . Put $\mathbf{w}(t,x) = 1 -\mathbf{u}(t,x)$ . We first prove that $\mathbf{w}(t,x)$ decays exponentially with rate at least $-\lambda$ uniformly in $t\in \big[0,\frac{1}{\nu}\big]$ . Using an argument similar to that of Theorem 1.1, we have that, for $t, x\in\mathbb{R}$ , $\mathbf{w}(t,x)$ satisfies

\begin{equation*} \begin{cases}\frac{\partial \mathbf{w}}{\partial t} = \frac{1}{2} \frac{\partial^2 \mathbf{w}}{\partial x^2} + \mathbf{g}\cdot\big(\mathbf{w}-\mathbf{w}^2\big),\\[6pt]\mathbf{w}\!\left(t+\frac{1}{\nu^*},x\right) = \mathbf{w}(t,x-1). \end{cases} \end{equation*}

By the Feynman–Kac formula, we get

\begin{equation*} \mathbf{w}(T,x) = \Pi_x \left[\mathbf{w}(T-t,B_t) e^{\int_0^t \mathbf{g}(B_s)(1-\mathbf{w}(T-s,B_s))\text{d}s} \right], \quad \mbox{for $T\in\mathbb{R}$, $t>0$.} \end{equation*}

For any $\lambda<\lambda^*$ , define

(4.1) \begin{equation} \widetilde{\mathbf{w}}(t,x) = \frac{e^{\lambda x-\gamma(\lambda)t}\mathbf{w}(t,x)} {\psi(x,\lambda)}. \end{equation}

Changing measure with $\Xi_t(\lambda)$ and following the same ideas as in Step 1 of the proof of Theorem 1.1, we get that

\begin{equation*} \widetilde{\mathbf{w}}(T,x) = \Pi_x^{\lambda} \left[\widetilde{\mathbf{w}}(T-t,Y_t) e^{-\int_0^t \mathbf{g}(Y_s)\mathbf{w}(T-s,Y_s)\text{d}s} \right], \end{equation*}

and

\begin{equation*}\left\{\widetilde{\mathbf{w}}({-}t,Y_t) e^{-\int_0^t \mathbf{g}(Y_s) \mathbf{w}({-}s,Y_s)\text{d}s}, \Pi_x^{\lambda} \right\}_{t\geq 0}\end{equation*}

is a positive martingale. Therefore, we have

\begin{equation*} \limsup_{t\rightarrow\infty} \left\{\frac{\ln \widetilde{\mathbf{w}}({-}t,Y_t)}{Y_t+\nu^*t} - \frac{1}{Y_t+\nu^*t} \int_0^t \mathbf{g}(Y_s)\mathbf{w}({-}s,Y_s)\text{d}s \right\} \leq 0, \quad \Pi_x^{\lambda}\mbox{-a.s.} \end{equation*}

Notice that for $\lambda<\lambda^*$ ,

\begin{align*} \lim_{t\rightarrow\infty} \frac{Y_t+\nu^*t}{t} \rightarrow -\gamma^{\prime}(\lambda) + \nu^* > 0. \end{align*}

Combining this with (4.1), we get

\begin{align*} \limsup_{t\rightarrow\infty} \frac{\ln \widetilde{\mathbf{w}}({-}t,Y_t)}{Y_t+\nu^*t} &= \limsup_{t\rightarrow\infty} \frac{\ln\!\left(e^{\lambda(Y_t+\nu^*t) + (\gamma(\lambda)-\lambda \nu^*)t} \mathbf{w}({-}t,Y_t)\right)}{Y_t+\nu^*t}\\ &= \limsup_{t\rightarrow\infty} \frac{\ln \mathbf{w}({-}t,Y_t)}{Y_t+\nu^*t} + \lambda + \frac{\gamma(\lambda)-\lambda \nu^*}{\nu^*-\gamma^{\prime}(\lambda)} \leq 0, \quad \Pi_x^{\lambda}\mbox{-a.s.,} \end{align*}

where $\gamma(\lambda) - \lambda \nu^*>0$ . Hence $\mathbf{w}({-}t,Y_t)$ decays exponentially with rate at least $-\lambda$ $\Pi_x^{\lambda}$ -a.s. This implies

\begin{equation*} \int_0^\infty \mathbf{g}(Y_s)\mathbf{w}({-}s,Y_s)\text{d}s<+\infty, \quad \Pi_x^{\lambda}\mbox{-a.s.} \end{equation*}

Then $\widetilde {\mathbf{w}}({-}t,Y_t)$ converges $\Pi_x^{\lambda}$ -a.s. Using a coupling method similar to that of Step 2 in the proof of Theorem 1.1, we get that the limit of $\widetilde {\mathbf{w}}({-}t,Y_t)$ is a constant. Notice that

\begin{equation*} \widetilde {\mathbf{w}}\!\left(t+\frac{1}{\nu^*},x+1\right) = \frac{e^{\lambda (x+1)-\gamma(\lambda)\big(t+\frac{1}{\nu^*}\big)}\mathbf{w}\!\left(t+\frac{1}{\nu^*},x+1\right)} {\psi(x+1,\lambda)} = e^{\lambda-\frac{\gamma(\lambda)}{\nu^*}} \widetilde {\mathbf{w}}(t,x) < \widetilde {\mathbf{w}}(t,x), \end{equation*}

where we used $\mathbf{w}\!\left(t+\frac{1}{\nu^*},x+1\right) = \mathbf{w}(t,x)$ and $\nu^* < \frac{\gamma(\lambda)}{\lambda}$ for $\lambda<\lambda^*$ . Therefore, for any $k\in\mathbb{N}$ ,

\begin{equation*} \widetilde {\mathbf{w}}\!\left({-}t+\frac{k}{\nu},Y_t+k\right) \leq \widetilde {\mathbf{w}}({-}t,Y_t). \end{equation*}

Using an argument similar to that of Step 3 in the proof of Theorem 1.1, we get that $\widetilde {\mathbf{w}}(t,x)$ is bounded in $\Big[0,\frac{1}{\nu^*}\Big]\times \mathbb{R}$ . Then, by (4.1), $\mathbf{w}(t,x)$ decays exponentially with rate at least $-\lambda$ uniformly in $t\in \Big[0,\frac{1}{\nu^*}\Big]$ .

Step 2. We will show that $\Big\{f\big(B_{[0, t]}\big), \Pi_x^{(y,\lambda^*)}\Big\}$ is a local martingale, where $f\big(B_{[0, t]}\big)$ is defined in (4.7) below. Recall the definitions of h, $\tau_{\lambda}^x$ , $\Lambda_t^{(x,\lambda)}$ , and $\Pi^{(x,\lambda)}_y$ given in (2.4), (2.9), (2.10), and (2.11) respectively, with $\lambda = \lambda^*$ in (2.4). Fix $y\in \mathbb{R}$ . For any (t, x) such that $y-\gamma^{\prime}(\lambda^*)t+$ $h(x)>0$ , define

(4.2) \begin{equation} \widehat {\mathbf{w}}(t,x,y) \,:\!=\, \frac{e^{\lambda^*x-\gamma(\lambda^*)t} \mathbf{w}(t,x)} {\psi(x,\lambda^*)(y-\gamma^{\prime}(\lambda^*)t+h(x))}, \end{equation}

and for any $z>0$ , define

(4.3) \begin{equation} \tau_z \,:\!=\, \inf\big\{t\geq 0\,:\, y+\gamma^{\prime}(\lambda^*)t+h(B_t) \leq z \big\}. \end{equation}

We mention here that $\tau_z$ actually depends on y. For any $x\in\mathbb{R}$ , we may define

\begin{equation*} \tau_z(x) \,:\!=\, \inf\big\{t\geq 0\,:\, x+\gamma^{\prime}(\lambda^*)t+h(B_t) \leq z \big\}. \end{equation*}

Then $\tau_z$ is shorthand for $\tau_z(y)$ , and for any $x\in\mathbb{R}$ , $\tau_z(y-x) = \tau_{z+x}$ . Using (4.2), it is easy to show

(4.4) \begin{equation} \widehat{\mathbf{w}}\!\left(t+\frac{1}{\nu^*}, x+1, y\right) = \widehat{\mathbf{w}}(t,x,y). \end{equation}

We first prove that for any $T\in\mathbb{R}$ and $t>0$ ,

(4.5) \begin{equation} \widehat {\mathbf{w}}\big(T,x,y+\nu^*T\big) = \Pi_x^{(y,\lambda^*)} \Big( \widehat {\mathbf{w}}\big(T-t\wedge\tau_z, B_{t\wedge\tau_z}, y+\nu^*T\big) e^{-\int_0^{t\wedge\tau_z} \mathbf{g}(B_s) \mathbf{w}(T-s,B_s) \text{d}s } \Big). \end{equation}

First note that, by the Feynman–Kac formula and the optional stopping theorem,

(4.6) \begin{equation} \mathbf{w}(T,x)= \Pi_x \left(\mathbf{w}(T-t\wedge\tau_z, B_{t\wedge\tau_z}) e^{\int_0^{t\wedge\tau_z} \mathbf{g}(B_s) (1-\mathbf{w}(T-s,B_s)) \text{d}s} \right),\quad T\in \mathbb{R}, \ t>0. \end{equation}

Noticing that $\Lambda_{t\wedge\tau_z}^{(y,\lambda^*)} > 0$ and $\nu^* = \gamma^{\prime}(\lambda^*)$ , a direct calculation shows that for $x>h^{-1}(y)$ ,

\begin{align*} &\Pi_x^{(y,\lambda^*)} \!\left( \psi(x,\lambda^*)(y+h(x)) e^{-\lambda^*x+\gamma(\lambda^*)T}\right.\\&\quad\times\left.\widehat {\mathbf{w}}\big(T-t\wedge\tau_z, B_{t\wedge\tau_z}, y+\nu^*T\big) e^{-\int_0^{t\wedge\tau_z} \mathbf{g}(B_s) \mathbf{w}(T-s,B_s) \text{d}s} \right)\\ = &\Pi_x^{(y,\lambda^*)} \left( \frac{\Lambda_0^{(y,\lambda^*)}}{\Lambda_{t\wedge\tau_z}^{(y,\lambda^*)}} \mathbf{w}\big(T-t\wedge\tau_z, B_{t\wedge\tau_z}\big) e^{\int_0^{t\wedge\tau_z} \mathbf{g}(B_s) (1-\mathbf{w}(T-s,B_s)) \text{d}s} \right)\\ = &\Pi_x \left( \frac{\Lambda_0^{(y,\lambda^*)}}{\Lambda_{t\wedge\tau_z}^{(y,\lambda^*)}} \mathbf{w}\big(T-t\wedge\tau_z, B_{t\wedge\tau_z}\big) e^{\int_0^{t\wedge\tau_z} \mathbf{g}(B_s) (1-\mathbf{w}(T-s,B_s)) \text{d}s} \frac{\Lambda_{t\wedge\tau_z}^{(y,\lambda^*)}}{\Lambda_0^{(y,\lambda^*)}} \mathbf{1}_{\Big\{\Lambda_{t\wedge\tau_z}^{(y,\lambda^*)}>0 \Big\} } \right)\\ = &\Pi_x \left( \mathbf{w}\big(T-t\wedge\tau_z, B_{t\wedge\tau_z}\big) e^{\int_0^{t\wedge\tau_z} \mathbf{g}(B_s) (1-\mathbf{w}(T-s,B_s)) \text{d}s} \right) = \mathbf{w}(T,x), \end{align*}

where in the last equality we used (4.6). Then using the definition of $\widehat {\mathbf{w}}$ given in (4.2), we get (4.5).

Notice that $\Big\{B_t, t\geq 0;\,\Pi_x^{(y,\lambda^*)}\Big\}$ is not a Brownian motion. By the argument in the last paragraph of Subsection 2.2, we have that $\left\{y+\gamma^{\prime}(\lambda^*)T(t)+h(B_{T(t)}), \, \Pi_x^{(y,\lambda^*)}\right\}$ is a Bessel-3 process starting from $y+h(x)$ . Define

(4.7) \begin{equation} f\big(B_{[0, t]}\big) = \widehat {\mathbf{w}}({-}t,B_t,y) e^{-\int_0^t \mathbf{g}(B_s)\mathbf{w}({-}s,B_s)\text{d}s}. \end{equation}

By the Markov property, for any $0<s<t$ ,

\begin{align*} \Pi_x^{(y,\lambda^*)}\Big[f\big(B_{[0, t\wedge\tau_z]}\big)| \mathcal{F}_{s}\Big]= f\big(B_{[0, s\wedge\tau_z]}\big). \end{align*}

The proof of the display above is given in the appendix; see Lemma A.1. This implies that $\Big\{f\big(B_{[0,t\wedge\tau_z]}\big), \Pi_x^{(y,\lambda^*)}\Big\}$ is a martingale. Since $\tau_z\rightarrow\infty$ as $z\downarrow 0$ , $\Big\{f\big(B_{[0, t]}\big), \Pi_x^{(y,\lambda^*)}\Big\}$ is a local martingale.

Step 3. In this step, we will show that $\widehat {\mathbf{w}}(T-t,B_t,y)$ converges $\Pi_x^{(y-\nu^*T,\lambda^*)}$ -a.s. to a constant $\beta\geq 0$ as $t\rightarrow\infty$ . Non-negative local martingales are non-negative super-martingales and hence must converge. Therefore $f(B_{[0, t]})$ , defined by (4.7), converges $\Pi_x^{(y,\lambda^*)}$ -a.s. as $t\to\infty$ . Notice that $\left\{y+\gamma^{\prime}(\lambda^*)T(t)+h(B_{T(t)}), \, \Pi_x^{(y,\lambda^*)}\right\}$ is a Bessel-3 process starting from $y+h(x)$ . It is known (see, for example, [Reference Shiga and Watanabe17, Theorem 3.2]) that the Bessel-3 process grows no slower than $t^{1/2-\epsilon}$ for any $\epsilon>0$ . Recall that $T(t) = \inf\{s>0\,:\, \langle M \rangle_s > t\}$ . Equation (2.8) yields $T(t) \in \Big[\frac{t}{c_2}, \frac{t}{c_1}\Big]$ . Therefore, $y+\gamma^{\prime}(\lambda^*)t+h(B_{t})$ grows no slower than $t^{1/2-\epsilon}$ for any $\epsilon>0$ . By (2.4) and the boundedness of $\psi_{\lambda}(\cdot,\lambda)/\psi(\cdot,\lambda)$ , we get that $B_t+\nu^*t = B_t + \gamma^{\prime}(\lambda^*)t$ grows no slower than $t^{1/2-\epsilon}$ . By Step 1, $\mathbf{w}(t,x)$ decays exponentially fast uniformly in $t\in \Big[0,\frac{1}{\nu^*}\Big]$ . Hence we have

\begin{equation*}\int_0^{\infty} \mathbf{g}(B_t) \mathbf{w}({-}t,B_t) \text{d}t < +\infty,\quad \Pi_x^{(y,\lambda^*)}\mbox{-a.s.}\end{equation*}

Therefore, by (4.7), the convergence of $f(B_{[0,t]})$ implies that $\widehat {\mathbf{w}}({-}t,B_t,y)$ converges $\Pi_x^{(y,\lambda^*)}$ -a.s. to some limit, say $\xi_x$ .

By (4.5), we have

\begin{align*} &\widehat {\mathbf{w}}(T,x,y) \nonumber\\ & = \Pi_x^{(y-\nu^*T,\lambda^*)} \!\left( \widehat {\mathbf{w}}\big(T-t\wedge\tau_z(y-\nu^*T), B_{t\wedge\tau_z(y-\nu^*T)}, y\big) e^{-\int_0^{t\wedge\tau_z(y-\nu^*T)} \mathbf{g}(B_s) \mathbf{w}(T-s,B_s) \text{d}s } \right). \end{align*}

By the same method, we can get

\begin{equation*} \left\{\widehat {\mathbf{w}}\big(T-t\wedge\tau_z(y-\nu^*T), B_{t\wedge\tau_z(y-\nu^*T)}, y\big) e^{-\int_0^{t\wedge\tau_z(y-\nu^*T)} \mathbf{g}(B_s) \mathbf{w}(T-s,B_s) \text{d}s}, \, \Pi_x^{(y-\nu^*T,\lambda^*)} \right\} \end{equation*}

is a local martingale and $\widehat {\mathbf{w}}(T-t,B_t,y)$ converges $\Pi_x^{(y-\nu^*T,\lambda^*)}$ -a.s. to some limit, say $\xi_x^T$ .

Next we use a coupling method to prove that there is a constant $\beta\geq 0$ such that

(4.8) \begin{equation} \xi_x^T = \beta, \quad \Pi_x^{(y-\nu^*T,\lambda^*)}\mbox{-a.s.}\quad \forall T\geq 0. \end{equation}

Similarly to Step 2 in the proof of Theorem 1.1, consider a process $\Big\{\big(B_t^1,B_t^2\big), t\geq 0;\, \widetilde{\Pi}_x^{(y,T)} \Big\}$ with $\big\{B_t^1, t\geq 0\big\}$ and $\big\{B_t^2, t\geq 0\big\}$ being independent, and

\begin{align*}&\left\{B_t^1, t\geq 0;\, \widetilde{\Pi}_x^{(y,T)}\right\} \overset{d}{=} \left\{B_t, t\geq 0;\, \Pi_x^{(y,\lambda^*)}\right\}, \\& \left\{B_t^2, t\geq 0;\, \widetilde{\Pi}_x^{(y,T)}\right\} \overset{d}{=} \left\{B_t, t\geq 0;\, \Pi_x^{(y-v^*T,\lambda^*)}\right\}.\end{align*}

Define random curves

\begin{equation*} \mathcal{L}_k^1 = \left\{\left({-}t+\frac{k}{\nu^*}, B_t^1+k\right)\,:\, t\geq\frac{k-1}{\nu^*} \right\}, \quad k\in\mathbb{N}, \end{equation*}

and

\begin{equation*} \mathcal{L}_k^2 = \left\{\left(T-t+\frac{k}{\nu^*}, B_t^2+k\right)\,:\, t\geq\frac{k-1}{\nu^*}+T \right\}, \quad k\in\mathbb{N}. \end{equation*}

Notice that all the curves start from the line $\big\{\frac{1}{\nu^*}\big\} \times \mathbb{R}$ , and for each $i=1, 2$ , if $\mathcal{L}_1^i$ is given, we can get all the curves $\mathcal{L}_k^i$ by translation. Using the fact that $y + h(B_{T(t)}) + \gamma^{\prime}(\lambda^*)T(t)$ is a Bessel-3 process, $\gamma^{\prime}(\lambda^*) = \nu^*$ , and $|h(x) -x|$ is bounded, we have

\begin{equation*} \lim_{t\rightarrow\infty} B_t + \nu^*t = \infty \quad \mbox{$\Pi_x^{(y,\lambda^*)}$-a.s.} \end{equation*}

Now we show that for $\widetilde{\Pi}_x^{(y,T)}$ -a.s. all $\omega$ , it holds that for any $k\in\mathbb{N}$ , $\mathcal{L}_k^1$ and $\mathcal{L}_{k+1}^1$ intersect each other. It is also equivalent to show that for any $k\in\mathbb{N}$ , $\mathcal{L}_k^1$ and $\mathcal{L}_{k+1}^1$ intersect each other $\widetilde{\Pi}_x^{(y,T)}$ -a.s. If there exists $t\geq \frac{k-1}{v^*}$ such that $B_{t+\frac{1}{\nu^*}}^1 + 1 = B_t^1 $ , then

\begin{equation*} \left({-}t+\frac{k}{\nu^*}, B_t^1+k\right) = \left({-}\left(t+\frac{1}{\nu^*}\right)+\frac{k+1}{\nu^*}, B_{t+\frac{1}{\nu^*}}^1+k+1\right), \end{equation*}

which implies $\mathcal{L}_k^1$ and $\mathcal{L}_{k+1}^1$ intersect each other. Notice that

\begin{align*} B_{t+\frac{1}{\nu^*}}^1 + 1 = B_t^1 &\Longleftrightarrow B_{t+\frac{1}{\nu^*}}^1 + \nu^*t + 1 = B_t^1 + \nu^*t\\ &\Longleftrightarrow h\Big(B_{t+\frac{1}{\nu^*}}^1\Big) + \nu^*\Big(t +\frac{1}{\nu^*}\Big) = h\big(B_t^1\big) + \nu^*t\\ &\Longleftrightarrow \widehat R_{\langle M^1 \rangle_{t+\frac{1}{\nu^*}}} = \widehat R_{\langle M^1 \rangle_{t}}, \end{align*}

where $\widehat R_t\,:\!=\,y+h\big(B_{T(t)}^1\big) + \nu^*T(t)$ is a standard Bessel-3 process starting at $y+h(x)$ . By (2.8), we have

\begin{equation*} \langle M^1 \rangle_{t+\frac{1}{\nu^*}} - \langle M^1 \rangle_t \in \left[ \frac{c_1}{\nu^*}, \frac{c_2}{\nu^*} \right], \end{equation*}

where $\langle M^1 \rangle_t = \int_0^t \big(h^{\prime}\big(B_s^1\big)\big)^2 \text{d}s$ . Put

\begin{equation*} l(t) = \widehat R_{\langle M^1 \rangle_{t+\frac{1}{\nu^*}}} -\widehat R_{\langle M^1 \rangle_{t}};\, \end{equation*}

then l(t) is continuous $\widetilde{\Pi}_x^{(y,T)}$ -a.s. Since $\widetilde{\Pi}_x^{(y,T)}\!\left(\lim_{t\to\infty}\widehat R_t =\infty\right)=1$ , we have

\begin{equation*} \widetilde{\Pi}_x^{(y,T)} \left(\! \mbox{ for any }T>0, \exists\ t>T \mbox{ s.t. } l(t)>0 \right) = 1. \end{equation*}

To prove $\mathcal{L}_k^1$ and $\mathcal{L}_{k+1}^1$ intersect $\widetilde{\Pi}_x^{(y,T)}$ -a.s., it suffices to show that

\begin{equation*} \widetilde{\Pi}_x^{(y,T)} \left( \!\mbox{ for any } T>0, \exists\ t>T \mbox{ s.t. } l(t)<0\right) = 1. \end{equation*}

Notice that

\begin{equation*} l(t) \leq \max_{s\in \big[\frac{c_1}{\nu^*},\frac{c_2}{\nu^*}\big]}\bigg( \widehat R_{s+\langle M^1 \rangle_{t}} - \widehat R_{\langle M^1 \rangle_{t}}\bigg). \end{equation*}

For simplicity, put $b_1 = \frac{c_1}{\nu^*}$ and $b_2 = \frac{c_2}{\nu^*}$ . It suffices to show that

(4.9) \begin{equation} \widetilde{\Pi}_x^{(y,T)} \left( \!\mbox{ for any }T>0, \exists\ t>T \mbox{ s.t. } \max_{s\in [b_1,b_2]}( \widehat R_{t+s} - \widehat R_t) < 0 \right) = 1. \end{equation}

A classical result shows $\widehat R_t$ satisfies

(4.10) \begin{equation} \text{d} \widehat R_t = \text{d} \widehat{B}_t + \frac{1}{\widehat R_t} \text{d}t, \end{equation}

where $\Big(\widehat{B}_t;\, \widetilde{\Pi}_x^{(y,T)}\Big)$ is a standard Brownian motion. By (4.10), we have

\begin{align*} & \widetilde{\Pi}_x^{(y,T)} \left(\max_{s\in [b_1,b_2]}\big( \widehat{B}_{t+s} - \widehat{B}_t \big)< -1 \right)\\ \geq &\widetilde{\Pi}_x^{(y,T)} \left(\widehat{B}_{t+b_1} - \widehat{B}_t < -2,\, \max_{s\in [b_1,b_2]} \widehat{B}_{t+s} - \widehat{B}_{t+b_1} < 1 \right)\\ =& \Pi_0 (B_{b_1}<-2) \cdot \Pi_0 \left(\max_{s\in[0,b_2-b_1]} B_s < 1 \right) \geq C > 0, \end{align*}

where the constant C does not depend on t. Hence

\begin{equation*} \sum_{j=0}^{\infty} \widetilde{\Pi}_x^{(y,T)} \left(\max_{s\in [b_1,b_2]} (\widehat{B}_{jb_2+s} - \widehat{B}_{jb_2}) < -1\right) = +\infty. \end{equation*}

By the second Borel–Cantelli lemma and the independent increments property of the Brownian motion, we have

(4.11) \begin{equation} \widetilde{\Pi}_x^{(y,T)} \left(\max_{s\in [b_1,b_2]}( \widehat{B}_{jb_2+s} - \widehat{B}_{jb_2} )< -1, \mbox{ i.o. }\right) = 1. \end{equation}

By (4.10), it holds that

\begin{equation*} \widehat R_{t+s} - \widehat R_t = \widehat{B}_{t+s} - \widehat{B}_t + \int_t^{t+s} \frac{1}{\widehat R_r} \text{d}r. \end{equation*}

If $\widehat R_{t+r}>b_2$ for $r\in [0,b_2]$ and $\widehat{B}_{t+s} - \widehat{B}_t < -1$ for $s\leq b_2$ , then

(4.12) \begin{equation} \widehat R_{t+s} -\widehat R_t < -1 + \frac{s}{b_2} \leq 0. \end{equation}

Since the Bessel-3 process is transient, we have

(4.13) \begin{equation} \widetilde{\Pi}_x^{(y,T)}\!\left( \exists\ T>0, \mbox{ s.t. for any } t>T, \widehat R_t>b_2 \right) = 1.\end{equation}

Combining (4.11), (4.12), and (4.13), we get (4.9). So $\widetilde{\Pi}_x^{(y,T)}$ -a.s., it holds that for any $k\in\mathbb{N}$ , $\mathcal{L}_k^1$ and $\mathcal{L}_{k+1}^1$ intersect each other. Using the same method, we can also prove that $\widetilde{\Pi}_x^{(y,T)}$ -a.s., for any $k,j\in\mathbb{N}$ , $\mathcal{L}_k^1$ and $\mathcal{L}_{k+j}^1$ intersect each other.

Consider

\begin{align*} \Omega_0 = &\bigcap_{k,j\in\mathbb{N}} \left\{\mbox{$\mathcal{L}_k^1$, $\mathcal{L}_{k+j}^1$ intersect}\right\} \bigcap\left\{\lim_{t\rightarrow\infty} \widehat {\mathbf{w}}\big({-}t, B_t^1,y\big),\, \lim_{t\rightarrow\infty} \widehat {\mathbf{w}}\big(T-t,B_t^2,y\big) \mbox{ exists} \right\} \\ &\;\;\bigcap \left\{ \lim_{t\rightarrow\infty}B_t^i + \nu^*t = +\infty, \mbox{ and $B_t^i$ is continuous for $i=1,2$} \right\}; \end{align*}

then $\widetilde{\Pi}_x^{(y,T)}(\Omega_0) = 1$ . For any $\omega\in \Omega_0$ and $j\in\mathbb{N}$ , we know $\mathcal{L}_j^2$ starts from the point $\left(\frac{1}{\nu^*}, B^2_{\frac{j-1}{\nu^*}+T}+j\right)$ and there exists $k\in\mathbb{N}$ such that

\begin{equation*} B_{\frac{k-1}{\nu^*}}^1(\omega) + k \leq B^2_{\frac{j-1}{\nu^*}+T}(\omega)+j \leq B_{\frac{k}{\nu^*}}^1(\omega) + k + 1. \end{equation*}

Hence the starting point of $\mathcal{L}_j^2(\omega)$ is between the starting points of $\mathcal{L}_k^1(\omega)$ and $\mathcal{L}_{k+1}^1(\omega)$ . Since $\mathcal{L}_k^1(\omega)$ and $\mathcal{L}_{k+1}^1(\omega)$ intersect, we have that $\mathcal{L}_j^2(\omega)$ must intersect either $\mathcal{L}_k^1(\omega)$ or $\mathcal{L}_{k+1}^1(\omega)$ . We use $(s_1(\omega),x_1(\omega))$ to denote the intersection point. By (4.4), there exist $t_1^1(\omega)$ , $t_1^2(\omega)$ satisfying

\begin{equation*} \widehat {\mathbf{w}}\!\left({-}t_1^1(\omega), B_{t_1^1}^1(\omega), y\right) = \widehat {\mathbf{w}}\!\left(s_1(\omega),x_1(\omega),y\right) = \widehat {\mathbf{w}}\!\left(T-t_1^2(\omega), B_{t_1^2}^2(\omega),y\right). \end{equation*}

Since j is arbitrary, we can find $\{t_n^i(\omega)\,:\, n\in\mathbb{N}, \, i=1,2\}$ by induction such that

\begin{equation*} \widehat {\mathbf{w}}\!\left({-}t_n^1(\omega), B_{t_n^1}^1(\omega), y\right) = \widehat {\mathbf{w}}\!\left(T-t_n^2(\omega), B_{t_n^2}^2(\omega),y\right) \end{equation*}

and satisfying

\begin{equation*} \lim_{n\rightarrow\infty} t_n^i(\omega) = \infty \quad \mbox{ for $i=1,2$.} \end{equation*}

Therefore, we have

\begin{equation*} \widetilde{\Pi}_x^{(y,T)} \left(\lim_{t\rightarrow\infty} \widehat {\mathbf{w}}\big({-}t,B_t^1,y\big) = \lim_{t\rightarrow\infty} \widehat {\mathbf{w}}\big(T-t,B_t^2,y\big)\right) = 1. \end{equation*}

By the independence of $\big\{B_t^1, t\geq 0\big\}$ and $\big\{B_t^2, t\geq 0\big\}$ , we get that the limits must be the same. So there is a constant $\beta\geq 0$ such that

\begin{equation*} \xi_x = \beta, \quad \Pi_x^{(y,\lambda^*)}\mbox{-a.s.}, \quad \mbox{ and } \quad \xi_x^T = \beta, \quad \Pi_x^{(y-v^*T,\lambda^*)}\mbox{-a.s.} \end{equation*}

Thus (4.8) is valid.

Step 4. In this step we prove $\beta>0$ by contradiction. If $\beta=0$ , then $\widehat {\mathbf{w}}({-}t,B_t,y) \rightarrow 0$ as $t\rightarrow\infty$ . Hence the positive local martingale

(4.14) \begin{equation} \widehat {\mathbf{w}}({-}t,B_t,y)e^{-\int_0^t \mathbf{g}(B_s)\mathbf{w}({-}s,B_s) \text{d}s} \rightarrow 0, \quad \mbox{ as } t\rightarrow\infty,\, \Pi_x^{(y,\lambda^*)}\mbox{-a.s.} \end{equation}

In Step 2 we proved that $\Big\{f\big(B_{[0, t\wedge\tau_z]}\big), \Pi_x^{(y,\lambda^*)}\Big\}$ is a martingale. Thus if $y+h(x)>z>0$ , we have

\begin{align*} \widehat {\mathbf{w}}(0,x,y) &= \Pi_x^{(y,\lambda^*)} \left( \widehat {\mathbf{w}}({-}t\wedge\tau_z, B_{t\wedge\tau_z}, y) e^{-\int_0^{t\wedge\tau_z} \mathbf{g}(B_s) \mathbf{w}({-}s,B_s) \text{d}s } \right)\\ &= \Pi_x^{(y,\lambda^*)} \left( \widehat {\mathbf{w}}({-}\tau_z, B_{\tau_z}, y) e^{-\int_0^{\tau_z} \mathbf{g}(B_s) \mathbf{w}({-}s,B_s) \text{d}s } \mathbf{1}_{\{\tau_z<t\}} \right)\\ &\quad\; + \Pi_x^{(y,\lambda^*)} \left( \widehat {\mathbf{w}}({-}t, B_{t}, y) e^{-\int_0^{t} \mathbf{g}(B_s) \mathbf{w}({-}s,B_s) \text{d}s } \mathbf{1}_{\{\tau_z\geq t\}} \right). \end{align*}

Letting $t\rightarrow\infty$ and using (4.14), we have

\begin{align*} \widehat {\mathbf{w}}(0,x,y) = \Pi_x^{(y,\lambda^*)} \left( \widehat {\mathbf{w}}\big({-}\tau_z, B_{\tau_z}, y\big) e^{-\int_0^{\tau_z} \mathbf{g}(B_s) \mathbf{w}({-}s,B_s) \text{d}s } \mathbf{1}_{\{\tau_z<\infty\}} \right). \end{align*}

It follows from (4.2) and (4.3) that

\begin{align*} \widehat {\mathbf{w}}(0,x,y) &= \Pi_x^{(y,\lambda^*)} \left( \frac{e^{\lambda^*B_{\tau_z}+\gamma(\lambda^*)\tau_z} \mathbf{w}\big({-}\tau_z, B_{\tau_z}\big)} {\psi\big( B_{\tau_z},\lambda^*\big)\big(y+\gamma^{\prime}(\lambda^*)\tau_z + h\big(B_{\tau_z}\big)\big)} e^{-\int_0^{\tau_z} \mathbf{g}(B_s) \mathbf{w}({-}s,B_s) \text{d}s } \mathbf{1}_{\{\tau_z<\infty\}} \right)\\ &= \Pi_x^{(y,\lambda^*)} \left( \frac{e^{\lambda^*B_{\tau_z}+\gamma(\lambda^*)\tau_z} \mathbf{w}\big({-}\tau_z, B_{\tau_z}\big)} {\psi\big( B_{\tau_z},\lambda^*\big)z} e^{-\int_0^{\tau_z} \mathbf{g}(B_s) \mathbf{w}({-}s, B_s) \text{d}s } \mathbf{1}_{\{\tau_z<\infty\}} \right). \end{align*}

Since $\Big\{y+\gamma^{\prime}(\lambda^*)T(t)+h(B_{T(t)}), \, \Pi_x^{(y,\lambda^*)}\Big\}$ is a Bessel-3 process starting from $y+h(x)$ , we have (see, for example, Karatzas and Shreve [Reference Karatzas and Shreve11, p. 162, Problem 3.23])

(4.15) \begin{equation} \Pi_x^{(y,\lambda^*)}(\tau_z<\infty) = \Pi_x^{(y,\lambda^*)}\!\left(\inf_{t\geq 0}\big\{ y+\gamma^{\prime}(\lambda^*)T(t)+h\big(B_{T(t)}\big)\big\} \leq z\right) = \frac{z}{y+h(x)}. \end{equation}

By (4.2), we have

\begin{align*} &\frac{e^{\lambda^*x}\mathbf{w}(0,x)}{\psi(x,\lambda^*)(y+h(x))} = \widehat {\mathbf{w}}(0,x,y) \leq \Pi_x^{(y,\lambda^*)} \left( \frac{e^{\lambda^*B_{\tau_z}+\gamma(\lambda^*)\tau_z} \mathbf{w}\big({-}\tau_z, B_{\tau_z}\big)} {\psi\big( B_{\tau_z},\lambda^*\big)z} \mathbf{1}_{\{\tau_z<\infty\}} \right)\\ &= \Pi_x^{(y,\lambda^*)} \left( \frac{e^{\lambda^*\big(z-y+\psi_{\lambda}\big(B_{\tau_z},\lambda^*\big)/ \psi\big(B_{\tau_z},\lambda^*\big)\big)} \mathbf{w}\big({-}\{\tau_z\}, B_{\tau_z}+\nu^*\big[\tau_z\big]\big)} {\psi\big( B_{\tau_z},\lambda^*\big)z} \mathbf{1}_{\{\tau_z<\infty\}} \right)\\ &\leq C_1 \Pi_x^{(y,\lambda^*)}(\tau_z<\infty) \leq C_1 \frac{z}{y+h(x)}, \end{align*}

where we used the facts that $\psi_{\lambda}/\psi$ is bounded and that $\mathbf{w}(t,x)$ is bounded in $(t,x)\in \big[{-}\frac{1}{\nu^*},0\big]\times [{-}C_2,C_2]$ . Here $C_1,C_2$ are constants depending only on $y,z,\lambda^*$ . Hence we get that for $x>h^{-1}(z-y)$ , $e^{\lambda^*x}\mathbf{w}(0,x)/\psi(x,\lambda^*)$ is bounded. Combining this with the fact that $e^{\lambda^*x}\mathbf{w}(0,x)/\psi(x,\lambda^*) \rightarrow 0$ as $x\rightarrow-\infty$ , we have that $e^{\lambda^*x}\mathbf{w}(0,x)/\psi(x,\lambda^*)$ is bounded on $\mathbb{R}$ . Similarly, we can prove that $e^{\lambda^*x-\gamma(\lambda^*)t} \mathbf{w}(t,x)/\psi(x,\lambda^*)$ is bounded on $\Big[0,\frac{1}{\nu^*}\Big]\times\mathbb{R}$ .

By Step 1 of the proof of Theorem 1.1, we know that

\begin{equation*} \left\{\frac{e^{\lambda^*Y_t+\gamma(\lambda^*)t}\mathbf{w}({-}t,Y_t)}{\psi(Y_t,\lambda^*)} e^{-\int_0^t \mathbf{g}(Y_s)\mathbf{w}({-}s,Y_s) \text{d}s }, t\geq 0;\, \Pi_x^{\lambda^*}\right\} \end{equation*}

is a martingale and satisfies

(4.16) \begin{equation} \frac{e^{\lambda x} \mathbf{w}(0,x)}{\psi(x,\lambda^*)} = \Pi_x^{\lambda^*} \left[ \frac{e^{\lambda^*Y_t+\gamma(\lambda^*)t}\mathbf{w}({-}t,Y_t)}{\psi(Y_t,\lambda^*)} e^{-\int_0^t \mathbf{g}(Y_s)\mathbf{w}({-}s,Y_s) \text{d}s } \right], \end{equation}

where $\big\{Y_t, \Pi_x^{\lambda^*}\big\}$ is a diffusion with infinitesimal generator

\begin{equation*} (\mathcal{A}f)(x) = \frac{1}{2} \frac{\partial^2 f(x)}{\partial x^2} + \left( \frac{\psi_x(x,\lambda^*)}{\psi(x,\lambda^*)} - \lambda^* \right) \frac{\partial f(x)}{\partial x}. \end{equation*}

So

\begin{equation*}\left\{\frac{e^{\lambda^*Y_t+\gamma(\lambda^*)t} \mathbf{w}({-}t,Y_t)}{\psi(Y_t,\lambda^*)} , \Pi_x^{\lambda^*}\right\}\end{equation*}

is a positive submartingale that is bounded and hence must converge. Using an argument similar to that of Step 2 in the proof of Theorem 1.1, we have that the limit of

\begin{equation*}\frac{e^{\lambda^*Y_t+\gamma(\lambda^*)t} \mathbf{w}({-}t,Y_t)}{\psi(Y_t,\lambda^*)}\end{equation*}

is a constant. Since

\begin{equation*}\mathbf{w}({-}t,Y_t) \leq 1, \; \liminf_{t\rightarrow\infty} (Y_t+\nu^*t) = -\infty \mbox{ and } \inf_{x\in\mathbb{R}}\psi(x,\lambda^*) > 0,\end{equation*}

we have

\begin{align*} \liminf_{t\rightarrow\infty} \frac{e^{\lambda^*Y_t+\gamma(\lambda^*)t} \mathbf{w}({-}t,Y_t)}{\psi(Y_t,\lambda^*)} \leq \frac{\liminf_{t\rightarrow\infty} e^{\lambda^*Y_t+\gamma(\lambda^*)t} \mathbf{w}({-}t,Y_t)}{\inf_{x\in\mathbb{R}}\psi(x,\lambda^*)} = 0. \end{align*}

So the constant must be 0. By (4.16), we have

\begin{align*} \frac{e^{\lambda x} \mathbf{w}(0,x)}{\psi(x,\lambda^*)} \leq \Pi_x^{\lambda^*} \left[ \frac{e^{\lambda^*Y_t+\gamma(\lambda^*)t}\mathbf{w}({-}t,Y_t)}{\psi(Y_t,\lambda^*)} \right]. \end{align*}

Letting $t\rightarrow\infty$ and noticing that

\begin{equation*}\frac{e^{\lambda^*Y_t+\gamma(\lambda^*)t} \mathbf{w}({-}t,Y_t)}{\psi(Y_t,\lambda^*)} \rightarrow 0, \qquad \Pi_x^{\lambda^*}\text{-a.s.},\end{equation*}

we have

\begin{align*} \frac{e^{\lambda x} \mathbf{w}(0,x)}{\psi(x,\lambda^*)} & \leq \lim_{t\rightarrow\infty} \Pi_x^{\lambda^*} \left[ \frac{e^{\lambda^*Y_t+\gamma(\lambda^*)t}\mathbf{w}({-}t,Y_t)}{\psi(Y_t,\lambda^*)} \right]\\ &= \Pi_x^{\lambda^*} \left[ \lim_{t\rightarrow\infty} \frac{e^{\lambda^*Y_t+\gamma(\lambda^*)t}\mathbf{w}({-}t,Y_t)}{\psi(Y_t,\lambda^*)} \right] = 0, \end{align*}

where we used the bounded dominated convergence theorem. This implies $\mathbf{w}(0,x) \equiv 0$ , which contradicts the definition of pulsating traveling waves. Therefore, we have $\beta>0$ .

Step 5. In this step, we show that

(4.17) \begin{equation} \widehat {\mathbf{w}}(t,x,y) \rightarrow \beta \quad \mbox{ uniformly in } t\in \left[0,\frac{1}{\nu^*}\right]\mbox{ as }x\rightarrow\infty. \end{equation}

Fix $y\in\mathbb{R}$ . First we will show that $\widehat {\mathbf{w}}(t,x,y)$ is bounded for $(t,x)\in \Big[0,\frac{1}{\nu^*}\Big]\times [h^{-1}(z-y+\nu^*t),\infty)$ , which implies that (t, x) satisfies $y-\nu^*t+h(x)\geq z>0$ . Recall that $\phi \,:\!=\, \phi(x,\lambda^*) = e^{-\lambda^*x} \psi(x,\lambda^*)$ . By (4.2), we can rewrite $\widehat {\mathbf{w}}$ as

\begin{equation*} \widehat {\mathbf{w}}(t,x,y) = \frac{e^{-\gamma(\lambda^*)t}\mathbf{w}(t,x)} {\phi(x,\lambda^*)\left(y-\gamma^{\prime}(\lambda^*)t\right) - \phi_{\lambda}(x,\lambda^*)}. \end{equation*}

By (2.2) and (2.3), a direct calculation yields

\begin{equation*} \frac{\partial \widehat {\mathbf{w}}}{\partial t} = \frac{1}{2} \frac{\partial^2 \widehat {\mathbf{w}}}{\partial x^2} + \frac{(y-\gamma^{\prime}(\lambda^*))\phi_x - \phi_{\lambda x}} {(y-\gamma^{\prime}(\lambda^*))\phi - \phi_{\lambda}} \frac{\partial \widehat {\mathbf{w}}}{\partial x} -\mathbf{g} \mathbf{w} \widehat {\mathbf{w}}. \end{equation*}

Similarly to Step 3 in the proof of Theorem 1.1, fix $\omega\in\Omega_0$ ; then for any $x_0$ large enough, there exist $k,j\in\mathbb{N}$ such that $(t,x_0)$ is located between the curves $\mathcal{L}_j^1(\omega)$ and $\mathcal{L}_k^1(\omega)$ for any $t\in \Big[0,\frac{1}{\nu^*}\Big]$ . Since $\mathcal{L}_j^1(\omega)$ and $\mathcal{L}_k^1(\omega)$ must intersect each other, it follows from the maximum principle that $\widehat {\mathbf{w}}(t,x_0,y)$ is bounded by the maximum on boundary $\mathcal{L}_j^1(\omega)$ and $\mathcal{L}_k^1(\omega)$ . We know that $\widehat {\mathbf{w}}(t,x,y)$ along $(t,x) \in\mathcal{L}_k^1(\omega)$ converges as $t\to\infty$ , and then $\widehat {\mathbf{w}}(t,x,y)$ is bounded on $\bigcup_{j\geq 1}\mathcal{L}_j^1(\omega)$ . Since $\widehat {\mathbf{w}}(t,x,y)$ is continuous and $\widehat {\mathbf{w}}(t,h^{-1}(z-y+\nu^*t),y)$ is bounded, we have that $\widehat {\mathbf{w}}(t,x_0,y)$ is bounded when $x_0$ is small. Therefore, $\widehat {\mathbf{w}}(t,x,y)$ is bounded in $\Big[0,\frac{1}{\nu^*}\Big]\times [h^{-1}(z-y+\nu^*t),\infty)$ . We denote the bound by $K_z$ .

Next, we will show (4.17). Recall that

\begin{align*} &\widehat {\mathbf{w}}(T,x,y)\\ & = \Pi_x^{\left(y-\nu^*T,\lambda^*\right)} \left( \widehat {\mathbf{w}}\big(T-t\wedge\tau_z(y-\nu^*T), B_{t\wedge\tau_z(y-\nu^*T)}, y\big) e^{-\int_0^{t\wedge\tau_z(y-\nu^*T)} \mathbf{g}(B_s) \mathbf{w}(T-s,B_s) \text{d}s } \right).\end{align*}

Letting $t\rightarrow\infty$ , we have for $T\in \Big[0,\frac{1}{\nu^*}\Big]$

\begin{align*} \widehat {\mathbf{w}}(T,x,y) &\leq \Pi_x^{\big(y-\nu^*T, \lambda^*\big)}\!\left(\beta \mathbf{1}_{\{\tau_z(y-\nu^*T)=\infty\}} + K_z \mathbf{1}_{\{\tau_z(y-\nu^*T)<\infty\}}\right)\\ &= \beta \left(1-\frac{z}{h(x)+y-\nu^*T}\right) + K_z \frac{z}{h(x)+y-\nu^*T}, \end{align*}

where we used (4.15). Therefore, we have

(4.18) \begin{equation} \limsup_{x\rightarrow\infty} \widehat {\mathbf{w}}(T,x,y) \leq \beta \quad \mbox{ uniformly for } T\in \left[0,\frac{1}{\nu^*}\right]. \end{equation}

On the other hand, we have the following estimate:

\begin{equation*} \widehat {\mathbf{w}}(T,x,y) \geq \Pi_x^{\big(y-\nu^*T, \lambda^*\big)} \left(\beta \mathbf{1}_{\{\tau_z(y-\nu^*T)=\infty\}} e^{-\int_0^{\infty} \mathbf{g}(B_t) \mathbf{w}(T-t,B_t) \text{d}t}\right). \end{equation*}

By (4.2), we get

\begin{align*} \mathbf{g}(B_t)\mathbf{w}(T-t,B_t) &\leq \|\mathbf{g}\|_\infty z K_z \max_{x\in[0, 1]} \psi(x,\lambda^*) e^{-\lambda^*B_t+\gamma(\lambda^*)(T-t)}\\ &\leq C e^{-\lambda^*(y-\nu^*T + h(B_t)+\nu^*t)} = C e^{-\lambda^* \widehat R^T_{\langle M \rangle_t}}, \end{align*}

where $\widehat R^T_t = y-\nu^*T + h(B_{T(t)})+\nu^*T(t)$ is a Bessel-3 process starting at $y-\nu^*T+h(x)$ , $T(t) = \inf\{s>0\,:\,\langle M \rangle_s>t \}$ , and the constant C does not depend on T. Thus we have

\begin{align*} \widehat {\mathbf{w}}(T,x,y) &\geq \Pi_x^{\big(y-\nu^*T, \lambda^*\big)}\!\left(\beta \mathbf{1}_{\{\tau_z(y-\nu^*T)=\infty\}} e^{-\int_0^{\infty} C e^{-\lambda^* \widehat R^T_{\langle M \rangle_t}} \text{d}t}\right)\\ &\geq \Pi_x^{\big(y-\nu^*T, \lambda^*\big)}\!\left(\beta \mathbf{1}_{\{\tau_{z+\nu^*T}=\infty\}} e^{-\int_0^{\infty} C e^{-\lambda^* \widehat R^T_t} \text{d}t}\right), \end{align*}

where $\tau_z(y-\nu^*T) = \tau_{z+\nu^*T}$ . Define the stopping time

\begin{equation*} \sigma_x(\widehat R^T) \,:\!=\, \inf\!\left\{t\geq 0\,:\, \widehat R^T_t = y-\nu^*T+h(x) \right\} \end{equation*}

and the function

\begin{equation*} f(x) \,:\!=\, \Pi_{x}^{\big(y-\nu^*T, \lambda^*\big)}\!\left(\beta \mathbf{1}_{\{\tau_{z+\nu^*T}=\infty\}} e^{-\int_0^{\infty} C e^{-\lambda^* \widehat R^T_t} \text{d}t}\right). \end{equation*}

By the Markov property, for $x_1<x_2$ we have

\begin{align*} f(x_1) = \Pi_{x_1}^{\big(y-\nu^*T, \lambda^*\big)}\left[ \mathbf{1}_{\{\sigma_{x_2}(\widehat R^T) < \tau_{z+\nu^*T}\}} e^{-\int_0^{\sigma_{x_2}(\widehat R^T)} C e^{-\lambda^* \widehat R^T_t} \text{d}t} f(x_2)\right] \leq f(x_2); \end{align*}

that is, f(x) is increasing. Put

\begin{equation*} f(\infty) \,:\!=\, \lim_{x\rightarrow\infty} f(x). \end{equation*}

Since $\beta>0$ and $\int_0^{\infty} e^{-\lambda^*\widehat R^T_t} \text{d}t < \infty$ , $\Pi_{x}^{\big(y-\nu^*T, \lambda^*\big)}$ -a.s., we have $f(x) > 0$ . Moreover,

\begin{align*} \beta f(x) &= \lim_{n\rightarrow\infty}\Pi_{x}^{\big(y-\nu^*T, \lambda^*\big)}\left[\beta \mathbf{1}_{\{\sigma_{x+n}(\widehat R^T) < \tau_{z+\nu^*T}\}} e^{-\int_0^{\sigma_{x+n}(\widehat R^T)} C e^{-\lambda^* \widehat R^T_t} \text{d}t} f(x+n)\right] \\ &= \Pi_{x}^{\big(y-\nu^*T, \lambda^*\big)}\left[ \lim_{n\rightarrow\infty} \beta \mathbf{1}_{\{\sigma_{x+n}(\widehat R^T) < \tau_{z+\nu^*T}\}} e^{-\int_0^{\sigma_{x+n}(\widehat R^T)} C e^{-\lambda^* \widehat R^T_t} \text{d}t} \right] \lim_{n\rightarrow\infty} f(x+n)\\ &= \Pi_{x}^{\big(y-\nu^*T, \lambda^*\big)}\left[ \beta \mathbf{1}_{\{\tau_{z+\nu^*T}=\infty\}} e^{-\int_0^{\infty} C e^{-\lambda^* \widehat R^T_t} \text{d}t} \right] f(\infty) = f(x) f(\infty). \end{align*}

Therefore, $f(\infty) = \beta$ . Combining this with $\widehat {\mathbf{w}}(T,x,y) \geq f(x)$ , we have

(4.19) \begin{equation} \liminf_{x\rightarrow\infty} \widehat {\mathbf{w}}(T,x,y) \geq \beta \quad \mbox{ uniformly for }T\in \left[0,\frac{1}{\nu^*}\right]. \end{equation}

By (4.18) and (4.19), we get (4.17).

Step 6. Since $h(x) = x - \psi_{\lambda}(x,\lambda^*)/\psi(x,\lambda^*)$ and $\psi_{\lambda}/\psi$ is bounded, we have

\begin{align*} \lim_{x\rightarrow\infty} \widehat {\mathbf{w}} (t,x,y) = \lim_{x\rightarrow\infty} \frac{e^{\lambda^*x-\gamma(\lambda^*)t}\mathbf{w}(t,x)} {\psi(x,\lambda^*)x}. \end{align*}

Thus

\begin{equation*} \frac{e^{\lambda^*z-\gamma(\lambda^*)t}\mathbf{w}(t,z)} {\psi(z,\lambda^*)z} \rightarrow \beta \quad \mbox{ uniformly in }t\in \left[0,\frac{1}{\nu^*}\right] \mbox{as }z\rightarrow\infty. \end{equation*}

Using an argument similar to that in Step 3 of the proof of Theorem 1.1, we have

\begin{equation*} \frac{e^{\lambda^* x}\mathbf{w}\!\left(\frac{y-x}{\nu^*}, y\right)}{ \psi(y,\lambda^*)({-}\lfloor y-x \rfloor + y)} \rightarrow \beta \quad \mbox{ uniformly in }y\in [0, 1] \mbox{ as }x\rightarrow\infty. \end{equation*}

By $|- \lfloor y-x \rfloor +y-x|\leq 1$ , we have

\begin{equation*} \frac{e^{\lambda^* x}\mathbf{w}\!\left(\frac{y-x}{\nu^*}, y\right)}{ x\psi(y,\lambda^*)} \rightarrow \beta \quad \mbox{ uniformly in }y\in [0, 1] \mbox{ as }x\rightarrow\infty. \end{equation*}

Step 7. For general branching mechanism, $\mathbf{w}=1-\mathbf{u}$ satisfies

\begin{equation*} \frac{\partial \mathbf{w}}{\partial t} = \frac{1}{2} \frac{\partial^2 \mathbf{w}}{\partial x^2} + \mathbf{g}\cdot(1-\mathbf{w}-\mathbf{f}(1-\mathbf{w})). \end{equation*}

Recall that $A(w) = m - \frac{1-w-\mathbf{f}(1-w)}{w}$ . As in Step 3, it suffices to show that

\begin{equation*} \int_0^{\infty} \mathbf{g}(B_t) \mathbf{w}({-}t,B_t) \text{d}t < \infty,\quad \Pi_x^{(y,\lambda^*)}\mbox{-a.s.} \end{equation*}

Since $B_t+\nu^*t$ behaves like $\sqrt{t}$ and $\mathbf{w}(t,x)$ decays exponentially fast, we have

\begin{align*} \int_0^{\infty} A\Big(e^{-c\sqrt{t}}\Big) \text{d}t < \infty \, \mbox{ for some $c>0$} \Longrightarrow\int_0^{\infty} \mathbf{g}(B_t) A(\mathbf{w})({-}t,B_t) \text{d}t < \infty. \end{align*}

Set $s = e^{-c\sqrt{t}}$ ; then

\begin{equation*} \int_0^{\infty} A\Big(e^{-c\sqrt{t}}\Big) \text{d}t < \infty \Longleftrightarrow \int_0^1 A(s) \frac{|\log s|}{s} \text{d}s < \infty. \end{equation*}

By [Reference Chauvin and Rouault5, Theorem 2], it holds that for $a>1$ ,

(4.20) \begin{equation} \int_0^1 A(s) \frac{|\log s|^a}{s} \text{d}s < \infty \Longleftrightarrow \mathbf{E}\big(L(\!\log^+L)^{1+a}\big)<\infty. \end{equation}

Actually the proof of [Reference Chauvin and Rouault5, Theorem 2] also works for $a=1$ . So if $\mathbf{E}\big(L(\!\log^+L)^2\big)<\infty$ , there exists $\beta>0$ such that

\begin{equation*} 1-\mathbf{u}\!\left(\frac{y-x}{\nu^*}, y\right) \sim \beta xe^{-\lambda^* x}\psi(y,\lambda^*)\quad \mbox{ uniformly in } y\in [0, 1] \mbox{ as } x\rightarrow +\infty. \end{equation*}

This completes the proof.