Given $u\in \mathrm{C}(\Omega)$ where $\Omega$ is an open subset of $\mathbb{R}^n$, then the followings are equivalent:
- $u$ is a viscosity solution of $|Du|\geq 1$ in $\Omega$.
- If $V$ is an open bounded subset of $\Omega$ and $\varphi\in \mathrm{C}^\infty(V)$ such that $|D\varphi| \leq 1$ in $V$ and $\varphi \leq u$ on $\partial V$ then $\varphi\leq u$ in $\overline{V}$.
For (i) implies (ii): If $u-\phi$ has a min at $x_0\in \overline{V}$ and $$ \min_{\partial V}(u-\phi) > \min_{V}(u-\phi)$$ then we must have $x_0\in V$, then we can find $r>0$ such that $B(x_0,r)\subset V$ and thus we can actually make it as a strict local minimum by replacing $\phi(x)$ with $\phi(x)-\eta(x-x_0)|x-x_0|^2$ where $\eta\in \mathrm{C}^\infty_c$ and $\eta$ is compactly supported in $B(0,r)$ and $\eta = 1$ near $0$.
The sequence $\phi_\lambda(x) = \lambda \phi(x)$ with $\lambda<1$ satisfies $\phi_\lambda \rightarrow \phi$ uniformly, thus $u-\phi_\lambda$ has a nearby local minimum at $x_\lambda$ and $x_\lambda \rightarrow x_0$ (up to subsequence), therefore the supersolution test of $u$ implies that $$|D\phi_\lambda(x_\lambda)| = \lambda |D\phi(x_\lambda)| \geq 1$$ which is a contradiction since by assumption $\lambda|D\varphi(x_\lambda)|\leq \lambda < 1$. Therefore the asumption is wrong and thus $$ 0\leq \min_{\partial V}(u-\phi) \leq \min_{V}(u-\phi).$$
For (ii) implies (i): Assume that $u$ is not a super-solution, then there exists $x_0\in \Omega$ such that there exists $\varphi\in \mathrm{C}^\infty(\Omega)$ such that $u-\varphi$ has a strict minimum at $x_0\in \Omega$ and $|D\varphi(x_0)| < 1$. By replacing $\varphi(x)$ by $\varphi(x) - |x-x_0|^2 - C$ where $C = (u-\varphi)(x_0)$ we can assume $u(x_0) = \phi(x_0)$ and thus $(u-\phi)(x)>0$ for all $x$ near $x_0$, say $x\in B(x_0,3r)$. By continuity we can also assume that $$ |D\varphi(x)| < 1 \qquad\text{for all}\qquad x\in B(x_0,2r).$$
Let $\varphi^\varepsilon(x) = (1+\varepsilon\eta(x))\varphi(x)$ where $\eta\in \mathrm{C}_c^\infty$ with $\eta = 1$ on $B(x_0, r)$ and $\mathrm{supp}(\eta)\subset B(x_0,2r)$, we have $$ |D\varphi^\varepsilon(x)|\leq (1+\varepsilon)|D\varphi(x)| + C\varepsilon,\qquad x\in B(x_0,2r)$$ where $C = \sup_{\overline{B(x_0,2r)}}|D\eta(x)\varphi(x)|$. Since $|D\varphi(x)|<1$ in $B(x_0,2r)$, we can find $\varepsilon$ small such that $$|D\varphi^\varepsilon(x)|<1, \qquad x\in B(x_0,2r).$$ Applying (ii) with $V = B(x_0,2r)$, we see that on the boundary $\varphi^\varepsilon = \varphi<u$ on $\partial V$, therefore $$\varphi^\varepsilon(x) \leq u(x) \qquad \text{for all}\qquad x\in V$$ which is a contradiction since $$\varphi^\varepsilon(x_0) = (1+\varepsilon)\varphi(x_0) > u(x_0).$$