I'm having some troubles solving a question in an exercise. The set-up is the following:
Let $\sigma:\mathbb R^d \rightarrow \mathbb R^{d\times d}$ be a $\mathcal C^2$ map which is bounded and has bounded derivatives. We assume that $\sigma$ is invertible with bounded inverse. We let $a(x) = \sigma(x)\sigma^*(x)$ where $*$ is the transpose operator.
Let $\phi:\mathbb R^d \rightarrow \mathbb R$ be a $\mathcal C^2$ map. We assume that the open set $\mathcal O = \left\{x \in \mathbb R^d | \phi(x)<0 \right\}$ is nonempty and bounded and that $D\phi(x) \neq 0$ if $\phi(x)=0$. Then it is known that $\overline{\mathcal O} = \left\{x \in \mathbb R^d | \phi(x)\le0 \right\}$ and $\partial{\mathcal O} = \left\{x \in \mathbb R^d | \phi(x)=0 \right\}$.
For $\epsilon>0$ we consider the following PDE: $$ \begin{cases} -\sqrt{\epsilon}\text{Tr}\left(a(x) D^2U_{\epsilon}(x) \right) + \left|\sigma^*(x) DU_{\epsilon}(x)\right|^2 =1 \qquad \text{on }\mathcal O\\ U_{\epsilon}(x)=0 \qquad \text{on }\partial{\mathcal O}. \end{cases} $$
The question I have troubles with is the following one:
Assume in this question only that $\phi$ is strictly convex in $\mathbb R^d$, i.e. $D^2\phi >0$ in $\mathbb R^d$. Let $m=\min \phi$. Show that there exists a $\lambda>0$ (independent of $\epsilon$) such that the map $$\psi(x):= -\lambda \left(\sqrt{\phi(x)-m} - \sqrt{-m} \right)$$ is a viscosity supersolution of the above equation for all $\epsilon \in (0,1]$.
Of course dealing with the boundary of $\mathcal O$ is straightforward. On $\mathcal O$ I have tried different things, mainly using the fact that if $D^2\phi >0$ in $\mathbb R^d$, then there exists $\delta>0$ such that $D^2\phi \ge \delta$ in $\mathcal O$, hence $\phi$ is a subsolution of $$D^2v = \delta \qquad \text{on } \mathcal O$$ and trying to build a proof from there, but I always end up stuck...
If someone has an idea that would be great !
Thank you very much