**Step 1: Strict subsolution.** First suppose $u_t + Lu < 0$ strictly in $U_T$. If the maximum of $u$ over $\overline{U_T}$ occurs at an interior point $(x_0, t_0) \in U_T$, then at that point: $\nabla_x u = 0$, $D^2_x u \le 0$ (the spatial Hessian is negative semi-definite), and $u_t \ge 0$ (since we are at a temporal maximum from below). Uniform ellipticity gives $-\sum a_{ij}\partial_{x_ix_j}u \ge 0$ at a spatial maximum, and $c \ge 0$ with $u(x_0, t_0) \ge 0$ (being the maximum) gives $cu \ge 0$. Hence: \begin{align*} u_t + Lu \ge 0 \quad \text{at } (x_0, t_0), \end{align*} contradicting $u_t + Lu < 0$. **Step 2: General case.** For the non-strict inequality $u_t + Lu \le 0$, define $u^\varepsilon(x,t) := u(x,t) - \varepsilon t$ for $\varepsilon > 0$. Then $u^\varepsilon_t + Lu^\varepsilon = (u_t + Lu) - \varepsilon \le -\varepsilon < 0$. By Step 1, $\max_{\overline{U_T}} u^\varepsilon = \max_{\Gamma_T} u^\varepsilon$. Letting $\varepsilon \to 0$ gives the result.