z<1.$$ The function $U$ is invertible, and so the unique solution $X^\star:[0,1]\to{\mathbb{R}}$ is given by $$X^\star(z) = U^{-1}(2z - 1).$$ By the properties of $U$, the function $X^\star$ so defined is strictly increasing and bounded, hence it is the inverse distribution function of some absolutely continuous measure $\mu^\star\in{\mathcal{P}}({\mathbb{R}})$ with density in $L^\infty_c$. The associated CDF $ F^\star=F_{\mu^\star} $ is given by $$F^\star(x) = 
    \begin{cases}
      0 & \text{if $U(x) < -1$},\\
      \frac{1}{2} \left( U(x) + 1 \right) & \text{if $U(x) \in [-1,1]$},\\
      1 & \text{if $U(x) > 1$}.
    \end{cases}$$ In conlusion, the steady state $\mu^\star$ has its median at the unique zero of $U$, which is $x_0$. And its density coincides with $ \frac12U' = \frac12 \psi''_a\ast\omega $, extending from $x_0$ in both directions until mass $1/2$ is reached on each side. This proves the claim for $q_a>1$.

*Now assume $q_a=1$.* The right-hand side of varies in $[-m,m]$. If $m<1$, then clearly there are values $z\in[0,1)$ for which cannot be true. Hence, in that case there is no steady state in the sense of Definition \[dfn:steady\].

Let $m\ge1$ from now on. Since $\omega\in L^\infty_c$, its CDF $G$ is continuous and monotonically increasing, and $G(x)=m$, $G(-x)=0$ for all sufficiently large $x>0$. Hence 