Proof. Recall that $\hat\gamma$ is nothing else but the stationary distribution of the location process $X({\cdot})$ of a single particle, corresponding to $\gamma$ . Consider the stationary regime of this process. Then the $1/w$ -Lipschitz property of $\hat\gamma$ in $({-}B_L,B_R]$ is proved in exactly same way as in the proof of Lemma 4.2, with w replacing $v$ . Given that $\hat\gamma$ is right-continuous, it is $1/w$ -Lipschitz in $[{-}B_L,B_R]$ as well. The properties (5.2) are obvious from the structure of the particle movement process $X({\cdot})$ : clearly, in steady state the process spends zero time exactly at the right boundary point $B_R$ and a non-zero fraction of time at the left boundary point $-B_L$ .

Since $\hat\gamma({\cdot})$ is Lipschitz in $({-}B_L,B_R]$ , it is absolutely continuous, and then almost all points $-B_L <x<B_R$ are regular (that is, the proper derivative $\hat\gamma^{\prime}(x)$ exists). Consider any regular point $-B_L <x<B_R$ . To prove (5.3), essentially the same argument as in the proof of Lemma 4 applies. Indeed, in the stationary regime, the average rate at which the particle crosses the point x from right to left (‘leaves’ $(x,B_R]$ , to be precise) is $w \hat\gamma^{\prime}(x)$ (by the same argument as in the proof of Lemma 4.3). The average rate, in the stationary regime, at which the particle crosses from left to right (‘enters’ $(x,B_R]$ , to be precise) is $\hat\zeta(x)$ ; indeed, such crossings may occur only when the particle jumps, and $\hat\zeta(x)$ is the steady-state probability that an urge to jump results in a jump over the point x. These right-to-left and left-to-right crossing rates must be equal, which gives (5.3).

The property (5.4) easily follows from a direct analysis of the expression for $\hat\zeta(x)$ , using the facts that $\,\bar{\!J}({\cdot})$ is non-increasing and RCLL, and that the measure given by $\hat\gamma$ has exactly one atom at $-B_L$ and is absolutely continuous in $({-}B_L,B_R]$ . We omit details. Finally, (5.5) follows from (5.4) and (5.3).

Proof. As $\gamma^{(k)} \stackrel{\Gamma}{\rightarrow} \gamma$ , denote the corresponding single-particle processes by $X_k({\cdot})$ and $X({\cdot})$ . Clearly, the sequence $A^{(w^{(k)};B_L^{(k)},B_R^{(k)})}\gamma^{(k)}$ is tight. Consider any subsequential limit of it, $\hat \gamma$ , along a subsequence of k. Consider each of the processes $X_k({\cdot})$ and $X({\cdot})$ on a finite time interval [0, T]. These processes can be naturally coupled so that the following property holds w.p.1 along the chosen subsequence: if either (a) $X(0)=-B_L$ and $X_k(0) = -B_L^{(k)}$ for all k, or (b) $X_k(0) \to X(0) > -B_L$ , then

\begin{equation*}(X_k(t), 0 \le t \le T) \stackrel{J_1}{\rightarrow} (X(t), 0 \le t \le T).\end{equation*}

Given this kind of continuity, it is straightforward to see that the limit $\hat \gamma({\cdot})$ of the stationary distributions of $X_k({\cdot})$ must be the stationary distribution $A^{(w;B_L,B_R)}\gamma$ of $X({\cdot})$ .

Proof of Lemma 5.3. The proof consists of the following three claims.

Proof of Claim 1. The existence follows from the Brouwer fixed point theorem. (Cf. [Reference Kantorovich and Akilov8, Theorem XVI.5.1] for its more general form—Kakutani’s theorem.) Indeed, by Lemma 5.2, for a fixed set of positive parameters $w, B_L,B_R$ , $A^{(w;B_L,B_R)}\gamma$ continuously (in the uniform convergence topology) maps $\Gamma(w;\,B_L,B_R)$ into itself. The set $\Gamma(w;\,B_L,B_R)$ is convex and compact. Therefore, a fixed point $\gamma = A^{(w;B_L,B_R)}\gamma$ exists.

Proof of Claim 2. The properties (5.6)–(5.10) follow from Lemma 5.1 and the properties (5.1)–(5.5).

Next, observe that $\gamma(x)$ must be strictly increasing in $[{-}B_L,B_R]$ . If not, we could find an interval $[y_1,y_2] \subset [{-}B_L,B_R)$ such that $y_1 < y_2$ , $\gamma(y_2)=\gamma(y_1)$ (and then $(d^+/dx) \gamma(y_1)=0$ ), and $\gamma(x) < \gamma(y_1)$ for $x< y_1$ ; but, by (5.10) and (5.8), $(d^+/dx) \gamma(y_1) = \zeta(y_1)/w > 0$ , a contradiction.

Let us now prove (5.11). First, from (5.8) and the fact that $\gamma(x)$ is strictly increasing in $[{-}B_L,B_R]$ , observe that $\liminf_{x \uparrow B_R} \zeta(x) > 0$ . Therefore, if $\inf_{[{-}B_L, B_R]} \zeta(x) = 0$ were to hold, we would have a point $y\in [{-}B_L,B_R)$ such that $\zeta(y)=0$ . (Here we use (5.9).) But $\gamma(x)$ is strictly increasing in a neighborhood of y, which, again by (5.8), implies $\zeta(y)>0$ . This completes the proof of (5.11) and of Claim 2.

Proof of Claim 3. First, note that any $\gamma$ satisfying (5.6)–(5.11) is such that the following holds: the inverse $\gamma^{-1}(\nu)$ for $\nu\in (0,1]$ is continuous and Lipschitz, and is strictly increasing in $[\gamma({-}B_L),1]$ . Moreover, the mapping $\nu=\gamma(x)$ gives a one-to-one correspondence between regular points of $\gamma$ in $[{-}B_L,B_R]$ and regular points of $\gamma^{-1}$ in $[\gamma({-}B_R),1]$ , with the derivatives at corresponding regular points satisfying $(d/d\nu) \gamma^{-1}(\nu)= 1/\gamma^{\prime}(x)$ . Furthermore, the latter relation holds for the right derivative for any $x\in [{-}B_L,B_R)$ and corresponding $\nu=\gamma(x)$ , and for the left derivative for any $x\in ({-}B_L,B_R]$ .

The proof of uniqueness is by contradiction. Suppose $\gamma_1$ and $\gamma_2$ are two different distributions satisfying the conditions (5.6)–(5.11). Let us define $\alpha_i(x) \doteq w (d^+/dx) \gamma_i(x)$ , $i=1,2$ ; then, by the first equation in (5.10), $\alpha_i(x)=\zeta_i(x)$ .

Suppose, for example, that $\gamma_1(y) > \gamma_2(y)$ for at least one $y\ge 0$ . Then let us call

\begin{equation*}h = \max_\nu\!\big[\gamma_2^{-1}(\nu) - \gamma_1^{-1}(\nu)\big] >0\end{equation*}

the ‘horizontal distance’ between $\gamma_1$ and $\gamma_2$ . It is well-defined, by the properties of $\gamma_1$ and $\gamma_2$ . Let $\nu_*$ be the minimum $\nu$ at which the horizontal distance is attained. Clearly, $0<\nu_*<1$ . Let $y_1 = \gamma_1^{-1}(\nu_*) < y_2 = \gamma_2^{-1}(\nu_*)$ . Then we must have $(d^+/dx) \gamma_1(y_1) \le (d^+/dx) \gamma_2(y_2)$ or, equivalently, $\alpha_1(y_1)\le \alpha_2(y_2)$ . Given the choice of $\nu_*$ , $y_1 - \gamma_1^{-1}(\nu) \le y_2 - \gamma_2^{-1}(\nu)$ for all $\nu \in [0,1]$ , and then $\zeta_1(y_1) \ge \zeta_2(y_2)$ . Since $\alpha_i(y_i)=\zeta_i(y_i)$ , we must have

\begin{equation*}\alpha_1(y_1)= \alpha_2(y_2) = \zeta_1(y_1) = \zeta_2(y_2).\end{equation*}

Note that if, for example, $\,\bar{\!J} (y) <1$ for all $y>0$ , then we immediately obtain a contradiction, because in this case $\zeta_1(y_1)> \zeta_2(y_2)$ . The case of general jump size distribution J requires some more details, which are as follows.

We must have that $\,\bar{\!J}(y_2{-})=1$ , i.e. the jump size is at least $y_2$ . Indeed, by the choice of $\nu_*$ , $y_1 - \gamma_1^{-1}(\nu) < y_2 - \gamma_2^{-1}(\nu)$ for all $\nu < \nu_*$ . Also, recall that both inverse functions $\gamma_i^{-1}(\nu)$ are continuous. This means that as we increase $\nu$ in the interval $[0,\nu_*)$ , the interval $\big[y_1 - \gamma_1^{-1}(\nu),y_2 - \gamma_2^{-1}(\nu)\big]$ continuously changes (‘moves left’) from $[y_1 + B_L, y_2 + B_L]$ to (in the limit!) [0, 0], while remaining of non-zero length. If at least one of points $y \in [0,y_2)$ were a point of decrease of $\,\bar{\!J}$ , we would have $\,\bar{\!J} \big(y_1 - \gamma_1^{-1}(\nu)\big) > \,\bar{\!J} \big(y_2 - \gamma_2^{-1}(\nu)\big)$ on a subset of $[0,\nu_*)$ having non-zero Lebesgue measure, which would imply $\zeta_1(y_1) > \zeta_2(y_2)$ , contradicting the definition of $\nu_*$ .

Consider now

\begin{equation*}\nu^* = \max\!\big\{\bar\nu \ge \nu_* | \gamma_2^{-1}(\nu) - \gamma_1^{-1}(\nu) = h, \forall \nu \in [\nu_*,\bar\nu]\big\}.\end{equation*}

We must have $\nu^* < 1$ . Let $y_i^* = \gamma_i^{-1}(\nu^*)$ . It is easy to see that $y_2^* = \max\!\big\{y \ge y_2 | \,\bar{\!J}(y{-})=1\big\}$ , i.e. $y_2^*$ must be exactly the smallest point of decrease of $\,\bar{\!J}$ . Indeed, $\,\bar{\!J}$ cannot have a point of decrease $\hat \nu \in [\nu_*,\nu^*)$ . If such a point $\hat \nu$ existed, then using an argument analogous to that we used just above to show that $\,\bar{\!J}(y_2{-})=1$ , we would obtain that $\gamma_1^{-1}(\nu)- \gamma_2^{-1}(\nu)$ would be strictly increasing in a small interval immediately to the right of the point $\hat \nu$ , which would contradict the definition of the horizontal distance h. On the other hand, if $y_2^*$ were not a point of decrease of $\,\bar{\!J}$ , i.e. $\,\bar{\!J}(\hat \nu)=1$ for some $\hat y > y_2^*$ , then, again using essentially the same argument, we would obtain that $\gamma_1^{-1}(\nu)- \gamma_2^{-1}(\nu)$ would have to remain constant in a small interval immediately to the right of the point $\nu^*$ , which would contradict the definition of $\nu^*$ .

Finally, if $y_2^*$ is a point of decrease of $\,\bar{\!J}$ , then, using essentially the same argument once again, we obtain that $\gamma_1^{-1}(\nu)- \gamma_2^{-1}(\nu)$ must be strictly increasing in a small interval immediately to the right of the point $\nu^*$ , which contradicts the definition of the horizontal distance h.

Proof. (i) The continuity follows immediately from the continuity of the operator $A^{(w;B_L,B_R)}$ on $\Gamma$ (Lemma 5.2) and the uniqueness of $\gamma=\mathcal T(w;\,B_L,B_R)$ (Lemma 5.3(i)).

(ii) The monotonicity is proved by contradiction, using the horizontal distance, in almost exactly the same way as in the proof of uniqueness (Claim 3) in Lemma 5.3.

(iii) The case $w\uparrow\infty$ is obvious, because $\gamma$ is $1/w$ -Lipschitz in $[B_L,B_R]$ and $\gamma(B_R)=1$ . Suppose $w\downarrow 0$ . Then $\gamma=\mathcal T(w;\,B_L,B_R)$ is monotone non-increasing in w. It suffices to show that, for each $y< B_R$ , $\lim \gamma(y) =0$ . Indeed, fix $y<B_R$ and any $y_1\in (y,B_R)$ . Clearly, $\lim \gamma(y_1) < \nu_1 < 1$ . Fix small $\epsilon>0$ , so that $y < y_1-\epsilon < y_1$ . Then, uniformly in all sufficiently small $w>0$ , the corresponding particle process $X({\cdot})$ is such that when it is to the left of the point $y_1$ , it will jump with rate at least $\eta(\nu_1)$ , while it moves left at the small speed w; this in turn easily implies that the steady-state probability of the particle being to the left of $y_1-\epsilon$ vanishes.

(iv) This is obvious from the definition of a finite-frame traveling wave shape.

Proof. The proof is by contradiction. The basic intuition is as follows. Suppose, for example, that $\limsup_{B\to\infty} w_B > v$ . Then there exists a constant $w_\infty > v$ such that $w_B > w_\infty > v$ along some subsequence of B. Then we show that the distributions $\gamma_B$ would have to stay tight around the left end $-B$ of the frame. This is, informally speaking, due to the fact that, if in steady state $X_B({\cdot})$ were to stay away from $-B$ most of the time, then its average drift would be bounded above by $v - w_\infty < 0$ , while it must be 0 for any B. But if $\gamma_B$ stays tight around $-B$ , then as $B\to\infty$ , the median of $\gamma_B$ cannot stay at 0, which contradicts the $\gamma_B$ definition. The contradiction in the case $\liminf_{B\to\infty} w_B < v$ is obtained similarly—we would have the distributions $\gamma_B$ staying tight around the right end B of the frame.

We proceed with the formal proof. Suppose $w_B$ does not converge to $v$ . Consider, first, the case $\limsup_{B\to\infty} w_B > v$ . Then we can choose a constant $w_\infty > v$ and a subsequence of B along which $w_B > w_\infty > v$ . For all large B, consider $\gamma_B^{up}=\mathcal T(w_\infty;\,B,B)$ corresponding to the speed $w_\infty$ ; also consider the corresponding single-particle process $X_B^{up}({\cdot})$ . By monotonicity in w (Lemma 5.4(ii)), $\gamma_B \preceq \gamma_B^{up}$ for all large B. By monotonicity in the right end of the frame (Lemma 5.4(ii)) and the shift property (Lemma 5.4(iv)), we observe that the distribution $\theta_B \gamma_B^{up}$ (i.e., the distribution $\gamma_B^{up}$ shifted right by B, and therefore concentrated on [0, 2B]) will be monotone non-decreasing (in the sense of $\preceq$ ) in B. Then $\theta_B \gamma_B^{up}$ must uniformly converge, $\theta_B \gamma_B^{up} \stackrel{u}{\rightarrow} \gamma_{\infty}$ , to some distribution $\gamma_{\infty}$ concentrated on $[0,\infty)$ , and the function $\gamma_{\infty}$ is $1/w_\infty$ -Lipschitz on $[0,\infty)$ . (This follows from the fact that all functions $\theta_B \gamma_B^{up}$ are uniformly $1/w_\infty$ -Lipschitz in $[0,\infty)$ .) Moreover,

(6.1) \begin{equation}\gamma_{\infty}(0)>0.\end{equation}

Indeed the particle process $X_B^{up}({\cdot})$ is such that the steady-state average drift of the particle is bounded above by $-\big(1-\gamma_B^{up}({-}B)\big)w_\infty + v$ ; if (6.1) did not hold, then, as $B\to\infty$ , this upper bound would converge to $-\big(1-\gamma_{\infty}(0)\big)w_\infty + v =-w_\infty + v <0$ , which is impossible.

To obtain a contradiction, it remains to show that the distribution $\gamma_{\infty}$ is proper, i.e. $\lim_{y\to\infty} \gamma_{\infty}(y) = 1$ . Indeed, this will imply that the median of $\gamma_B^{up}$ , and then of $\gamma_B$ as well, will eventually become less than 0, which contradicts the definition of $\gamma_B$ .

Let us show that the distribution $\gamma_{\infty}$ is proper. Suppose not; that is, $\lim_{y\to\infty} \gamma_{\infty}(y) = $ $\nu^*_\infty < 1$ . Let $\nu^* = \eta^{-1}(w_\infty)$ .

The case $\nu^*_\infty > \nu^*$ is impossible. Indeed, in this case, pick a small $\epsilon>0$ and any fixed $H>0$ such that $\gamma_{\infty}(H) > \nu^* + \epsilon$ . Then, uniformly in all large B, when the particle process $X_B^{up}({\cdot})+B$ (‘living’ in [0, 2B]) is to the right of H, the particle jumps right at most at the rate $\eta(\nu^* + \epsilon) < w_\infty$ . We also have the following: when the process $X_B^{up}({\cdot})+B$ ‘jumps over the point H ^′, we have that uniformly in all B and on the jump starting point in [0, H], the distribution of the ‘overshoot-over-H ^′ distance V is uniformly stochastically bounded above by some proper distribution with finite mean. (This is obvious if the jump size distribution $J({\cdot})$ has finite support. Otherwise, $\mathbb{P}\{V > y\} \le \,\bar{\!J}(y) / \,\bar{\!J}(H)$ .) Using these properties, we can construct a proper stochastic upper bound, uniform in B, on the stationary distribution of $X_B^{up}({\cdot})+B$ , which is $\theta_B \gamma_B^{up}$ . This contradicts the fact that the distribution $\gamma_{\infty}$ is not proper.

Consider the case $\nu^*_\infty \le \nu^*$ . Consider the particle process $X_\infty({\cdot})$ , taking values in $\mathbb{R}_+=[0,\infty)$ , that is defined in exactly the same way as the process defining the (finite-frame) operator value $A^{(w;B_L,B_R)} \gamma$ , except that $X_\infty({\cdot})$ corresponds to the speed $w_\infty$ and distribution $\gamma_\infty$ on $\mathbb{R}_+=[0,\infty)$ (which is not necessarily proper); the lower boundary is 0 (at which regulation does occur); and there is no upper boundary (and no regulation from above—no forward jump is ever ‘truncated’). The process $X_\infty({\cdot})$ is regenerative, with regenerations occurring when the particle hits 0. Given that for any finite $y\ge 0$ , $\eta(\gamma_\infty(y)) > w_\infty$ , the process cannot possibly be positive recurrent. (Indeed, even if the rate of jumps is constant at $w_\infty$ , i.e. lower than it actually is, the process $X_\infty({\cdot})$ models the workload of a queue where the workload is depleted at the constant rate $w_\infty$ and new workload arrives as a random process with average rate exactly $w_\infty$ . Therefore, $X_\infty({\cdot})$ is stochastically bounded below by a non-positive recurrent process.) Consider two sub-cases: (a) $X_\infty({\cdot})$ is transient (which is certainly the case if $\nu^*_\infty < \nu^*$ , but logically also possible if $\nu^*_\infty = \nu^*$ ), and (b) $X_\infty({\cdot})$ is null-recurrent (which is possible only when $\nu^*_\infty = \nu^*$ ).

Observe that the processes $X_B^{up}({\cdot})+B$ for all B and the process $X_\infty({\cdot})$ , all starting at 0, can be naturally coupled so that, as $B\to\infty$ , the trajectory of $X_B^{up}({\cdot})+B$ converges (in the Skorokhod topology) to that of $X_\infty({\cdot})$ w.p.1.

Sub-case (a): If $X_\infty({\cdot})$ is transient, then for some fixed $\delta>0$ and any $H>0$ , for all sufficiently large B, the probability that $X_B^{up}({\cdot})+B$ will up-cross level H in a regeneration cycle is at least $\delta$ . This implies that the mean duration of the regeneration cycle of $X_B^{up}({\cdot})+B$ has to go to $\infty$ as $B\to\infty$ (because the time to return to 0 from a point $y\ge H$ is at least $H/(w_\infty/2)$ ). This implies that the steady-state probability $\theta_B \gamma_B^{up}(0)$ of $X_B^{up}({\cdot})+B$ being at 0 will vanish. Indeed, the mean time the process spends at 0 within one regeneration cycle is uniformly bounded above by an exponential random variable with mean $1/v$ (because the rate at which the particle jumps forward when it sits at the left boundary point is at least $\int_0^1 \eta(\nu) d\nu = v$ ). But vanishing $\theta_B \gamma_B^{up}(0)$ would mean $\gamma_\infty(0)=0$ , which contradicts (6.1).

Sub-case (b): If $X_\infty({\cdot})$ is null-recurrent, then the regeneration cycle of $X_\infty({\cdot})$ is finite w.p.1, and has infinite mean. Recall the natural coupling of $X_B^{up}({\cdot})+B$ and $X_\infty({\cdot})$ . Given this coupling, for any fixed $H>0$ , on the event that $X_\infty({\cdot})$ does not up-cross H in a regeneration cycle, the regeneration cycle length of $X_B^{up}({\cdot})+B$ converges to that of $X_\infty({\cdot})$ almost surely. Using Fatou’s lemma, we see that, as $B\to\infty$ , $\liminf$ of the mean regeneration cycle duration of $X_B^{up}({\cdot})+B$ is at least that of $X_\infty({\cdot})$ , which is infinity. This, again, leads to vanishing $\theta_B \gamma_B^{up}(0)$ and a contradiction with (6.1).

This completes the proof that $\limsup_{B\to\infty} w_B > v$ is impossible.

The contradiction to $\liminf_{B\to\infty} w_B < v$ is obtained similarly. In this case, we can construct lower bounds on $\gamma_B$ . The condition (6.1) is replaced by the condition that the steady-state rate of the particle hitting the right boundary B converges down to a positive number. Regeneration time points are when the particle hits the right boundary B. We omit details.

Proof. The proof is by contradiction. Its basic outline is as follows. If $\{\gamma_B\}$ is not tight, we can find a subsequence of B, and a corresponding sequence of scaling factors $C=C(B)\uparrow\infty$ , such that the rescaled distributions $\psi_{(C)}(y)= \gamma_B(Cy)$ are such that, say, some $\nu$ -quantile ( $\nu > 1/2)$ of $\psi_{(C)}$ remains at the point 1. Each $\psi_{(C)}$ is a finite-frame traveling wave shape; its frame is $[{-}B/C,B/C]$ , the corresponding single-particle process is $X_{(C)}(t)=X_B(Ct)/C$ , and the speed is still $w_B$ . We show that the family of distributions $\{\psi_{(C)}\}$ is tight. This is done by considering a quadratic Lyapunov function for $X_{(C)}(t)$ and its steady-state drift, and showing that $\mathbb{E} |X_{(C)}(\infty)|$ remains uniformly bounded. Given the tightness of the sequence $\{\psi_{(C)}\}$ , we consider a subsequence of it that converges to some proper distribution $\psi_{(\infty)}$ . Using Lemma 6.4 we show that $\psi_{(\infty)}$ must be concentrated on at most two atoms, $z_1\le 0$ and $z_2 \ge 0$ . The case of one atom, $z_1=z_2$ , is impossible because it leads to a contradiction with the $\nu$ -quantile ( $\nu > 1/2)$ of $\psi_{(C)}$ remaining at the point 1. Then we must have either $z_1<0$ or $z_2>0$ . In this case we again use a quadratic Lyapunov function for $X_{(C)}(t)$ to show that, for large C, its steady-state drift would be negative; roughly, this is because the probability of $X_{(C)}(\infty)\ge z_2$ is about $1-\nu$ and the probability of $X_{(C)}(\infty)\le z_1$ , and the drift of the quadratic Lyapunov function is strictly negative when $X_{(C)}(\infty)\ge z_2 > 0$ and non-positive when $X_{(C)}(\infty)\le z_1\le 0$ .

We proceed with the formal proof by contradiction. Suppose $\{\gamma_B\}$ is not tight. Then there exists fixed $\nu>1/2$ and a subsequence $B\uparrow \infty$ along which $C=C(B)=\max\big\{\gamma_B^{-1}(\nu), - \gamma_B^{-1}(1-\nu)\big\} \uparrow \infty$ . Without loss of generality, we can choose $\nu$ sufficiently close to 1 so that

(6.2) \begin{equation}\eta(1-\nu) > v > \eta(\nu).\end{equation}

Note that, for each B, either $\gamma_B^{-1}(\nu)=C$ or $\gamma_B^{-1}(1-\nu)=-C$ ; also, $\gamma_B(C)-\gamma_B({-}C) \ge \nu-(1-\nu)=2\nu-1$ .

For each B (with the corresponding C), consider the particle process with space compressed by C and time sped up by C, i.e. the process $X_{(C)}(t)= X_B(Ct)/C$ . For each B, the process lives within the frame $[{-}C_0,C_0]$ , where $C_0=B/C$ . It is possible that $C_0\to\infty$ as $B\to\infty$ . Denote by $\psi_{(C)}(y)=\gamma_B(Cy)$ the corresponding scaled distribution. From now on in this proof, when we say ‘for a given B ^′ or ‘for a given C ^′ we mean ‘for a given pair (B, C) of B and C, corresponding to each other’.

Note that, for each C, either $\psi_{(C)}^{-1}(\nu)=1$ or $\psi_{(C)}^{-1}(1-\nu)=-1$ ; also, $\psi_{(C)}(1)-\psi_{(C)}({-}1) \ge 2\nu-1$ .

Let us prove that

(6.3) \begin{equation}\mbox{the family of distributions $\{\gamma_C\}$ is tight.}\end{equation}

This is trivially true if $C_0$ remains bounded as $B\to\infty$ . Therefore, it suffices to prove (6.3) for a subsequence of B such that $C_0=B/C \to \infty$ . Let us consider such a subsequence.

Let $g(y)=y^2/2$ . Denote by $P_{(C)}^t(y,H)$ the transition function of the process $X_{(C)}(t)$ . According to our definitions, for any fixed C, $\psi_{(C)} P_{(C)}^t g = \psi_{(C)} g$ for all $t\ge 0$ ; here we view the transition function $P_{(C)}^t$ and the distribution $\psi_{(C)}$ as operators (see Section 2).

Note that $\psi_{(C)} = \mathcal T(w_B;\, C_0,C_0)$ , and that $X_{(C)}({\cdot})$ is the corresponding (finite-frame) particle process. Denote by $\hat X_{(C)}({\cdot})$ a single-particle process, corresponding (like $X_{(C)}({\cdot})$ ) to the distribution function $\psi_{(C)}$ and speed $w_B$ , but different from $X_{(C)}({\cdot})$ in that it evolves in $({-}\infty, \infty)$ , i.e. no regulation keeping the process within the finite frame $[{-}C_0,C_0]$ is applied. The particle continues to move left at speed $w_B$ even when it is at or to the left of $-C_0$ , in which case its jump rate remains the same as it would be at the point $-C_0$ , namely $\bar \eta({-}C_0, \psi_{(C)})$ . The particle also can ‘jump over’ the point $C_0$ ; when it is at or to the right of $C_0$ , it moves left at speed $w_B$ and its jump rate is as it would be at the point $C_0$ , namely $\bar \eta(C_0, \psi_{(C)}) = \eta(\psi_{(C)}(C_0)) = \eta(1) =0$ . Denote by $\hat P_{(C)}^t(y,H)$ the transition function of the process $\hat X_{(C)}(t)$ . We claim that, for any sufficiently large fixed C, uniformly in $y\in [{-}C_0,C_0]$ ,

(6.4) \begin{equation}\big[P_{(C)}^t g\big] (y) - \big[\hat P_{(C)}^t g\big] (y) \le o(t),\end{equation}

where o(t) is a positive function (which may depend on C) such that $o(t)/t\to 0$ as $t\to 0$ . The proof of (6.4) is given in Section 6.2.

It is easy to check directly that, uniformly in $y\in [{-}C_0,C_0]$ ,

\begin{equation*} \lim_{t\downarrow 0} \frac{1}{t} \big[\hat P_{(C)}^t - I\big] g (y) = \hat G_{(C)} g (y) \doteq - w_B g^{\prime}(y) + C \eta(\psi_{(C)}(y))\int_0^\infty dJ(C\xi) [g(y+\xi)-g(y)],\end{equation*}

and we have

\begin{align*} \hat G_{(C)} g (y) & = - w_B y + C \eta(\psi_{(C)}(y))\int_0^\infty dJ(\zeta) \big[y \zeta/C +(\zeta/C)^2/2\big]\\& = - w_B y + \eta(\psi_{(C)}(y))\big[y + (1/2)m^{(2)}/C\big],\end{align*}

and then

(6.5) \begin{equation}\hat G_{(C)} g (y) \le y ({-}w_B+\eta(\psi_{(C)}(y))) + m^{(2)}/(2C).\end{equation}

Note that, by our construction and (6.2), for all sufficiently large C (recall that $w_B \to v$ ), we have, for some fixed $\epsilon>0$ ,

(6.6) \begin{equation}-w_B+\eta(\psi_C(y)) \ge \epsilon, y\le -1, \quad \mbox{and}\quad -w_B+\eta(\psi_C(y)) \le -\epsilon, y\ge 1.\end{equation}

From (6.4) we can see that, informally speaking, the operator $\hat G_{(C)}$ is an upper bound on the generator $G_{(C)}$ of the process $X_{(C)}({\cdot})$ , when these operators are applied to the function g. This observation is only informal, because the function g is not even within the domain of $G_{(C)}$ . Formally, using (6.4) and the fact that the distribution $\psi_{(C)}$ is concentrated on $[{-}C_0,C_0]$ , we can write

(6.7) \begin{equation}0 = \lim_{t\downarrow 0} \frac{1}{t} \psi_{(C)} \big[P_{(C)}^t - I\big] g \le \lim_{t\downarrow 0} \frac{1}{t} \psi_{(C)} \big[\hat P_{(C)}^t - I\big] g= \psi_{(C)} \hat G_{(C)} g\end{equation}

and observe that

\begin{equation*}\psi_{(C)} \hat G_{(C)} g \le -\epsilon \psi_{(C)} h_1 + C_2,\end{equation*}

where $h_1(y)\doteq |y| I(|y|\ge 1)$ , and $C_2$ does not depend on (sufficiently large) C. We obtain that, uniformly in large C,

\begin{equation*}\psi_{(C)} h_1 \le C_2/\epsilon.\end{equation*}

This implies the tightness of $\{\psi_{(C)}\}$ , thus completing the proof of the claim (6.3).

Given the tightness of $\{\psi_{(C)}\}$ , consider a subsequence of C along which $\psi_{(C)} \stackrel{w}{\rightarrow} \psi_{(\infty)}$ , where $\psi_{(\infty)}$ is a proper distribution, not necessarily continuous. Recall that $w_B \to v$ . Then we can apply Lemma 6.4. We obtain that the entire distribution $\psi_{(\infty)}$ must be concentrated either (a) at a single point $z\in[{-}1,1]$ , or (b) on a segment $[z_1,z_2]$ of non-zero length, where $z_1 = \min\{y\ |\ \psi_{(\infty)}(y) = \nu_v\}$ , $z_2 = \sup\{y\ |\ \psi_{(\infty)}(y) = \nu_v\}$ , $\nu_v \doteq \eta^{-1}(v)$ .

The case (a) is impossible, because the property that either $\psi_{(C)}^{-1}(\nu)=1$ or $\psi_{(C)}^{-1}(1-\nu)=-1$ cannot hold for all large C.

Let us show that the case (b) is also impossible. Suppose it holds. Then $z_1 \le 0 \le z_2$ , because otherwise the median of $\psi_{(C)}$ could not stay at 0 for all C. We also know that (since $\psi_{(\infty)}(y)=\nu_v \in [z_1,z_2)$ ) the distribution $\psi_{(\infty)}$ consists of two atoms at $z_1$ and $z_2$ , with positive weights $\nu_v$ and $1-\nu_v$ , respectively. So, either $z_1<0$ or $z_2>0$ . Consider, for concreteness, the case $z_2>0$ . (The treatment of the case $z_1<0$ is analogous.) Fix small $\epsilon>0$ and, for each C, consider the points

\begin{equation*}z_{1,\epsilon} = z_{1,\epsilon}(C) \doteq \psi_{(C)}^{-1}(\nu_v -\epsilon), \qquad z_{2,\epsilon} = z_{2,\epsilon}(C) \doteq \psi_{(C)}^{-1}(\nu_v +\epsilon).\end{equation*}

Given the form of the limiting distribution $\psi_{(\infty)}$ , as $C\to\infty$ , we have

\begin{equation*}z_{1,\epsilon} \to z_1, \qquad z_{2,\epsilon} \to z_2.\end{equation*}

Let $\epsilon_1= \eta(\nu_v-\epsilon)- \eta(\nu_v+\epsilon)$ . Obviously, $\epsilon_1 \downarrow 0$ as $\epsilon\downarrow 0$ .

For any fixed $\epsilon$ (and corresponding $\epsilon_1$ ), and arbitrarily small $\delta>0$ , the following holds for all sufficiently large C:

(6.8) \begin{equation}\hat G_{(C)} g (y) \le y ({-}w_B+\eta(\psi_{(C)}(y))) + \delta \quad \mbox{(from (6.5))}.\end{equation}

We will now estimate $\psi_{(C)} \hat G_{(C)} g$ by considering intervals $({-}\infty, z_{1,\epsilon}]$ , $(z_{1,\epsilon}, z_{2,\epsilon}]$ , $(z_{2,\epsilon}, \infty)$ , and using the following: the $\psi_{(C)}$ -measures of those intervals, estimates of values of $-w_B+\eta(\psi_{(C)}(y))$ over those intervals, and estimates of $|z_{1,\epsilon} - z_1|$ and $|z_{2,\epsilon} - z_2|$ . The values and estimates listed just above are

\begin{equation*}|z_{1,\epsilon} - z_1| < \delta, \qquad |z_{2,\epsilon} - z_2| < \delta,\end{equation*}

\begin{equation*}-w_B+\eta(\psi_{(C)}(y)) > \epsilon_1, \quad y \le z_{1,\epsilon},\end{equation*}

\begin{equation*}-w_B+\eta(\psi_{(C)}(y)) < \epsilon_1, \quad y \ge z_{2,\epsilon},\end{equation*}

\begin{equation*}|-w_B+\eta(\psi_{(C)}(y))| \le \epsilon_1, \quad z_{1,\epsilon}\le y \le z_{2,\epsilon},\end{equation*}

\begin{equation*}\psi_{(C)}(z_{1,\epsilon}) = \nu_v-\epsilon , \qquad \psi_{(C)}(z_{2,\epsilon}) - \psi_{(C)}(z_{1,\epsilon}) = 2\epsilon, \qquad 1- \psi_{(C)}(z_{2,\epsilon}) = 1-\nu_v - \epsilon.\end{equation*}

Then, from (6.8),

\begin{align*}\psi_{(C)} \hat G_{(C)} g & \le (\nu_v-\epsilon) (z_1+\delta) \epsilon_1 + 2\epsilon [\!\max\{|z_1|,z_2\} + 2\delta] \epsilon_1 + (1-\nu_v + \epsilon) (z_2-\delta) ({-}\epsilon_1) +\delta\\& \le (\nu_v-\epsilon) \delta \epsilon_1 + 2\epsilon [\!\max\{|z_1|,z_2\} + 2\delta] \epsilon_1 + (1-\nu_v + \epsilon) (z_2-\delta) ({-}\epsilon_1) +\delta,\end{align*}

where the last inequality is because $z_1 \le 0$ . If we pick sufficiently small $\epsilon>0$ (with the corresponding, also small, $\epsilon_1>0$ ), and then choose sufficiently small $\delta$ , then the right-hand side in the last display is negative. This implies $\psi_{(C)} \hat G_{(C)} g < 0$ , which is a contradiction with (6.7). Therefore, the case (b) is impossible.

Thus, the assumption that the family of distributions $\{\gamma_B\}$ is not tight leads to a contradiction. This completes the proof of the lemma.

Proof. Using the tightness of $\{\gamma_B\}$ , we can find a further subsequence of B along which $\gamma_B \stackrel{w}{\rightarrow} \phi$ ; but, since the functions $\gamma_B$ are uniformly Lipschitz in $({-}B,\infty)$ , we conclude that the function $\phi$ is $1/v$ -Lipschitz, and then $\gamma_B \stackrel{u}{\rightarrow} \phi$ . Also, clearly $\phi(0)=1/2$ .

If $P_{B}^t(y,H)$ is the transition function of the process $X_B({\cdot})$ , then it is easy to see that, for each y and $t\ge 0$ , $P_{B}^t(y,{\cdot}) \stackrel{w}{\rightarrow} P_\infty^t(y,{\cdot})$ , where $P_\infty^t(y,{\cdot})$ is the transition function of the particle process $X_\infty({\cdot})$ , corresponding to the distribution $\phi$ and speed $v$ . Moreover, $P_\infty^t(y,\,{\cdot})$ is continuous in y, i.e. the process $X_\infty({\cdot})$ is Feller continuous. Then (see, e.g., [Reference Liptser and Shiryaev9]) the limit $\phi$ of the stationary distributions $\gamma_B$ of $X_B({\cdot})$ is the stationary distribution of the limiting process $X_\infty({\cdot})$ . This means that $\phi$ is in fact the fixed point $\phi=A \phi$ , and therefore, by Lemma 4, $\phi$ is a traveling wave shape.

Proof of Lemma 6.4. The proof uses a fairly standard fluid-limit-type argument. When C is large, the process $X_{(C)}({\cdot})$ becomes ‘almost deterministic’. Indeed, when C is large, and then the distribution $\psi_{(C)}$ is close to a proper distribution $\psi_{(\infty)}$ , the process makes right jumps at the rate O(C), with the jump sizes being $O(1/C)$ ; as a result, in the vicinity of time t the process trajectory is almost deterministic, with the derivative being close to $\eta(\psi_{(\infty)}\big(X_{(C)}(t)\big)) - w_\infty$ . All properties stated in the lemma stem from this basic behavior.

We proceed with the formal proof. Consider the following natural common probability space construction for the particle processes $X_B$ , for all B. A unit-rate Poisson process $\Pi(t)$ , $t\ge 0,$ drives the particle urges to jump. An i.i.d. sequence $Z_1, Z_2, \ldots$ of random variables with distribution $J({\cdot})$ determines the sequence of particle jump sizes, when it does jump. An i.i.d. sequence $\Xi_1, \Xi_2, \ldots$ of random variables uniformly distributed in [0, 1] determines whether or not the particle actually jumps when it gets the ith urge to jump; specifically, if a particle gets the ith urge to jump at time t, when its location is $y=X_B(t)$ , it actually jumps if $\Xi_i \le \bar \eta(y,\gamma_B)$ . This construction ensures that, for each B, the process $X_B$ is (up to stochastic equivalence) as defined.

The driving sequences satisfy the functional strong law of large numbers (FSLLN) properties: as $B\to\infty$ (and the corresponding $C=C(B) \to\infty$ ), w.p.1,

(6.17) \begin{equation}\left( \frac{1}{C} \Pi(Ct), t\ge 0 \right) \stackrel{u.o.c.}{\rightarrow} (t, t\ge 0),\end{equation}

(6.18) \begin{equation}\left( \frac{1}{C} \sum_{i=1}^{\lfloor C s \rfloor} Z_i, s\ge 0 \right) \stackrel{u.o.c.}{\rightarrow} (s, s\ge 0),\end{equation}

(6.19) \begin{equation}\left( \frac{1}{C} \sum_{i=1}^{\lfloor C s \rfloor} I\{\Xi_i \le \xi\}, 0 \le \xi \le 1, s\ge 0 \right)\stackrel{u.o.c.}{\rightarrow} (\xi s, 0 \le \xi \le 1, s\ge 0).\end{equation}

(i) We have $X_{(C)}(t)=X_{(C)}(0) + X^{\uparrow}_{(C)}(t) - X^{\downarrow}_{(C)}(t)$ , where $X^{\uparrow}_{(C)}(t)$ accounts for the forward (right) jumps in [0, t] and $X^{\downarrow}_{(C)}(t)$ accounts for the total distance traveled by the particle backwards (left) in [0, t] (at a speed which may be either $w_B$ or 0). Both $X^{\uparrow}_{(C)}$ and $X^{\downarrow}_{(C)}$ are non-decreasing; $X^{\downarrow}_{(C)}$ is $w_B$ -Lipschitz; given the FSLLN properties (6.17)–(6.19), we observe that the sequence of processes $X^{\uparrow}_{(C)}$ is asymptotically Lipschitz, meaning, w.p.1, for any $0\le t_1 \le t_2 < \infty$ ,

\begin{equation*}\limsup_{C\to\infty} \Big[X^{\uparrow}_{(C)}(t_2) - X^{\uparrow}_{(C)}(t_1)\Big] \le t_2-t_1.\end{equation*}

This implies that, w.p.1., any subsequence of trajectories $X_{(C)}({\cdot})$ has a further subsequence along which

\begin{equation*}X_{(C)}({\cdot}) \stackrel{u.o.c.}{\rightarrow} X_{(\infty)}({\cdot}),\end{equation*}

where $X_{(\infty)}(t)$ is $(1+w_\infty)$ -Lipschitz, and $X_{(\infty)}(0)=y_0$ . It remains to show that a limit trajectory $X_{(\infty)}({\cdot})$ satisfies (6.15)–(6.16). This easily follows from (6.17)–(6.19). Consider (6.15), for example. If $X_{(\infty)}(t)=y$ , then for an arbitrarily small $\epsilon>0$ there exists $\delta>0$ such that for all $t^{\prime} \in (t-\delta,t+\delta)$ and all sufficiently large C we have

\begin{equation*}\psi_{(C)}\big(X_{(C)}(t^{\prime})\big) < \psi_{(\infty)}(y)+\epsilon\end{equation*}

and then

\begin{equation*}\bar \eta \big(X_{(C)}(t^{\prime}), \psi_{(C)}\big) > \eta\big(\psi_{(\infty)}(y)+\epsilon\big).\end{equation*}

In other words, when t ^′ is close to t and C is sufficiently large, the probability of the particle of $X_{(C)}$ jumping when it gets an urge is bounded below by $\eta(\psi_{(\infty)}(X_{(\infty)}(t))+\epsilon)$ . Application of (6.17)–(6.19) easily gives (6.15). We omit further details. The property (6.16) is shown similarly.

(ii) This follows from (i).

(iii) This follows easily from (ii). Indeed, (ii) implies that for any compact set $K \in \mathbb{R}$ there exists $T>0$ such that, for all sufficiently large C, with uniformly high probability, if $X_{(C)}(0) \in K$ then $X_{(C)}(T)$ is close to $\big[q^{low}, q^{up}\big]$ . Therefore, the stationary distribution of $X_{(C)}({\cdot})$ , i.e. $\psi_{(\infty)}$ , must in the limit concentrate on $\big[q^{low}, q^{up}\big]$ .