Let $b: \mathbb R_+\times\mathbb R_+\times \mathcal P\to\mathbb R$ be Lipschitz, where $\mathcal P$ denotes the set of probability measures $\mu$ on $\mathbb R_+$ of finite first moment and is endowed with the Wasserstein metric of order $1$. Consider the equation

$$(1+\alpha)c(t,\mu, \alpha)=\int_{\mathbb R_+} \big(b(t,x,\mu)+c(t,\mu,\alpha)\big)^+\mu(dx) - (1-\alpha)\big(b(t,0,\mu) + c(t,\mu,\alpha)\big)^+,\quad\quad (\ast)$$

for all $t\in\mathbb R_+$, $\mu\in\mathcal P$ and $\alpha\in [0,1]$. My questions are as follows :

  1. Does $(\ast)$ admit a unique solution $c: \mathbb R_+\times \mathcal P\times [0,1]\to\mathbb R$?

  2. If so, could this solution be Lipschitz? If not, could this solution be Lipschitz w.r.t. $(x,\mu)$ and continuous w.r.t. $\alpha$?

PS : Some special cases have been studied, e.g. $b\ge 0$ or $b\equiv b(t,\mu)$, where we may derive the explicit expression for $c$. So my question is rather for the case where $b$ may change sign and depend on $x$.


1 Answer 1


If $b(\cdot,0,\cdot)\ge 0$, then $(\ast)$ admits a unique solution that is Lipschitz.

For any (fixed) $t\in\mathbb R_+$, $\mu\in\mathcal P$ and $\alpha\in [0,1]$, define the function $F\equiv F_{t,\mu,\alpha}:\mathbb R\to\mathbb R$ by

$$F(z):=(1+\alpha)z + (1-\alpha)\big(b(t,0,\mu)+z\big)^+ - \int_{\mathbb R_+}\big(b(t,x,\mu)+z\big)^+\mu(dx).$$

Then it is clear that $F(\pm\infty)=\pm\infty$. Under the assumption, one has for $z<-b(t,0,\mu)$

$$F(z)\le (1+\alpha) z<0$$

and for $z\ge -b(t,0,\mu)$

$$F(z)=2z + (1-\alpha)b(t,0,\mu) - \int_{\mathbb R_+}\big(b(t,x,\mu)+z\big)^+\mu(dx)$$

is strictly increasing on $[-b(t,0,\mu),+\infty)$. Therefore, $F$ has a unique root, denoted by $c(t,\mu,\alpha)$, i.e. $F\big(c(t,\mu,\alpha)\big)=0$. Namely, $(\ast)$ admits a unique solution $c(t,\mu,\alpha)$.

To show the Lipschitz continuity, it suffices to estimate $|c(t',\mu,\alpha)-c(t,\mu,\alpha)|$, $|c(t,\mu',\alpha)-c(t,\mu,\alpha)|$ and $|c(t,\mu,\alpha')-c(t,\mu,\alpha)|$. The arguments are almost the same, so WLOG we only compute $|c(t',\mu,\alpha)-c(t,\mu,\alpha)|$. Note that the solution $c(t,\mu,\alpha)\ge -b(t,0,\mu)$ and thus $(\ast)$ can be rewritten as

\begin{eqnarray} 2c(t,\mu,\alpha) &=& \int_{\mathbb R_+}\big(b(t,x,\mu)+c(t,\mu,\alpha)\big)^+\mu(dx)-(1-\alpha)b(t,0,\mu) \\ 2c(t',\mu,\alpha) &=& \int_{\mathbb R_+}\big(b(t',x,\mu)+c(t',\mu,\alpha)\big)^+\mu(dx)-(1-\alpha)b(t,0,\mu). \\ \end{eqnarray}

Making the difference of the above two equations, one has

\begin{eqnarray} 2|c(t',\mu,\alpha)-c(t,\mu,\alpha)| &\le & \int_{\mathbb R_+}\Big[C|t'-t|+\big|c(t',\mu,\alpha)-c(t,\mu,\alpha)\big|\Big]\mu(dx) \\ &= & C|t'-t|+\big|c(t',\mu,\alpha)-c(t,\mu,\alpha)\big|, \end{eqnarray} which yields $|c(t',\mu,\alpha)-c(t,\mu,\alpha)|\le C|t'-t|$, where $C$ denotes the Lipschitz constant of $b$.


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