Let $\varepsilon > 0$, consider the following viscous Hamilton-Jacobi equation $$ \begin{cases} u ^\varepsilon_t + H(x, Du ^\varepsilon) = \varepsilon\Delta u^\varepsilon& \text{in }\mathbb{R}^n \times (0,\infty) \\ u ^{\varepsilon} (x,0) = u_{0}(x) \text{on }\mathbb{R}^n \end{cases} $$
We assume that $u _{0}(x) \in C^2 (\mathbb{R}^n) \quad and \quad \| u _{0} \| _{C^2(\mathbb{R}^n)} < \infty$
$$ \begin{cases} H \in C^2 (\mathbb{R}^n\times\mathbb{R}^n)\\ H,D_{p}H \in BUC(\mathbb{R}^n\times B(0,R))\quad \text{for each } R > 0\\ \lim_{|p| \to \infty}\inf_{x \in \mathbb{R}^n} ((1/2) H(x,p)^2 + D_xH(x,p)\cdot p) = +\infty \end{cases} $$
For sufficiently large C>$0$, let $f^{\pm}(x,t)=u_0(x)\pm Ct\quad \text{for } (x,t)\in \mathbb{R}^n \times (0,\infty)$ From the assumption, we have $| H(x, Du_0) - \varepsilon\Delta u_0|<C$. Then, we have $f ^\pm_t + H(x, Df ^\pm) - \varepsilon\Delta f^\pm =\pm C + H(x,Du_0)-\varepsilon\Delta u_0 > \text{or } < 0\quad \text{on } \mathbb{R}^n \times [0,\infty)$
We conclude that $f^\pm$ are a supersolution/subsolution to above equation, respectively.
Question
Then, in my textbook, it says, therefore,$f^- \leq u^\varepsilon \leq f^+$. I do not understand why this is true. Is it true if we have supersolution, then classical solution is smaller than it? Otherwise, why is this true?