Let $z=F(x,y)$ be a function from $\mathbb R^d\times \mathbb R$ to $\mathbb R$ satisfying the following conditions:
- $z=F(x,y)$ is Lipschitz continuous w.r.t. $(x,y)$;
- Given $x$, $F(x,y)$ is non-decreasing w.r.t $y$ (not necessarily strictly increasing);
- For any $x\in\mathbb R^d$, there is a unique $y$ such that $F(x,y)=0$;
Based on the above conditions, $F(x,y)=0$ is an implicit function, denoted as $y=y(x)$. Can we prove that $y=y(x)$ is also Lipschitz continuous? Thank you!
A highly related problem: this problem. It is shown that the answer is negative without the condition that $F(x,y)$ is non-decreasing w.r.t $y$.
