Let $l:\Omega\to\mathbb{R}$ a sufficiently smooth function on an open set $\Omega$. Let the equations $$(I):\ \|\nabla u(x)\|=l(x)$$ $$(II):\ \|\nabla v(x)\|+l(x)v(x)=0$$ Prove that $u(x)$ is a viscosity solution of $(I)$ iff $v(x):=-e^{-u(x)}$ is a viscosity solution of $(II)$.
I don't even know how to start. Any help would be appreciated.
Here is one part of the argument; the others are similar. Let $u$ be a viscosity solution of (I) and set $v=-e^{-u}$. Let $\phi$ be a smooth function such that $v-\phi$ has a local maximum at some $x\in \Omega$. We need to show that $$\|\nabla \phi(x)\| + l(x)v(x) \leq 0.$$ We can assume that $v(x)=\phi(x)$ and $v\leq \phi$ near $x$ (by adding a constant to $\phi$). Set $\psi(y)=-\log(-\phi(y))$. Since $t \mapsto -\log(-t)$ is increasing, $u(x)=\psi(x)$ and $u\leq \psi$ near $x$. This means $u-\psi$ has a local maximum at $x$ and hence $$ \|\nabla \psi(x)\|\leq l(x)$$ by the viscosity subsolution property for (I). By the chain rule $\nabla \psi(x) = -\nabla \phi(x)/\phi(x)$. Since $\phi(x)=v(x) < 0$ we have $|\phi(x)|=-\phi(x)=-v(x)$ and so $$\|\nabla \phi(x)\| \leq -l(x)v(x).$$ This is exactly the subsolution property for (II).
