# Existence, uniqueness and regularity of the solution to some integral equation

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 $$x$$.

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 $$b$$.