in the proof of Proposition \[prp:steady-stat-equat\], and recall further that the inverse distribution function $X^\star$ of $\mu^\star$ satisfies $U(X^\star(z))=2z-1$ for all $z\in[0,1)$.

Fix some $z\in[0,1)$. By Definition \[dfn:transient\] of transient solutions, $X$ is a continuously differentiable curve in ${\mathrm{D}}([0,1))$. Hence $y:[0,\infty)\to{\mathbb{R}}$ with $y(t)=X(t,z)$ is continuously differentiable, and satisfies — in view of — the ordinary differential equation $$\dot y(t) = U(y^\star) - U(y(t)),
    $$ where $y^\star=X^\star(z)$. Since $U:{\mathbb{R}}\to{\mathbb{R}}$ is a Lipschitz continuous function (see Lemma \[lem:18\]), different solution curves to this differential equations cannot cross, so either $y(t)\le y^\star$ or $y(t)\ge y^\star$ consistently for all $t\ge0$. Since further $U$ is monotonically increasing, it follows that $y(t)$ is bounded and monotone in time — increasing if $y(0)\le y^\star$ and decreasing if $y(0)\ge y^\star$. Thus $y(t)$ converges (monotonically) to some limit $y_\infty$ as $t\to\infty$. By continuity of $U$, it follows that $0=U(y_\infty)-U(y^\star)$.

If $q_a>1$, then *strict* monotonicity of $U$ directly implies $y_\infty=y^\star$. If $q_a=1$, then $$\begin{aligned}
    U(x) = - \int_{-\infty}^0 \omega(x-y)dy +\int_{0}^{+\infty} \omega(x-y)dy =1 -2 G(x) 
  \end{aligned}$$ is strictly increasing *except* possibly for an at most countable number of intervals on which it is constant. By definition of the pseudo-inverse, $X^\star$ never takes values *inside* these intervals. Instead, there is an at 