[proofplan] Subtract the two functions and prove that the difference cannot become positive inside the cylinder. To rule out a positive interior maximum, perturb the difference by the strictly decreasing time function $-\varepsilon t$, which converts the differential inequality into a strict one. If the perturbed function had a positive maximum away from the parabolic boundary, the first- and second-derivative conditions at that maximum would contradict the strict inequality. Finally let $\varepsilon \downarrow 0$ to remove the perturbation. [/proofplan] [step:Subtract the two functions and obtain a subsolution with nonpositive parabolic boundary values] Define the difference function \begin{align*} w: \overline{U} \times [0,T] \to \mathbb{R}, \qquad w(x,t) := u(x,t) - v(x,t). \end{align*} Since $u,v \in C^{1,2}(Q_T) \cap C(\overline{U} \times [0,T])$, we have $w \in C^{1,2}(Q_T) \cap C(\overline{U} \times [0,T])$. For every $(x,t) \in Q_T$, \begin{align*} \partial_t w(x,t) - \Delta w(x,t) = \partial_t u(x,t) - \Delta u(x,t) - \partial_t v(x,t) + \Delta v(x,t) \leq 0. \end{align*} The boundary hypothesis gives \begin{align*} w(x,t) \leq 0 \end{align*} for every $(x,t) \in \partial_p Q_T$. [/step] [step:Perturb the subsolution to make the parabolic inequality strict] Fix $\varepsilon > 0$. Define the perturbed function \begin{align*} w_\varepsilon: \overline{U} \times [0,T] \to \mathbb{R}, \qquad w_\varepsilon(x,t) := w(x,t) - \varepsilon t. \end{align*} Then $w_\varepsilon \in C^{1,2}(Q_T) \cap C(\overline{U} \times [0,T])$. For every $(x,t) \in Q_T$, \begin{align*} \partial_t w_\varepsilon(x,t) - \Delta w_\varepsilon(x,t) = \partial_t w(x,t) - \Delta w(x,t) - \varepsilon \leq -\varepsilon < 0. \end{align*} On the parabolic boundary, $t \geq 0$ and $w \leq 0$, hence \begin{align*} w_\varepsilon(x,t) = w(x,t) - \varepsilon t \leq 0 \end{align*} for every $(x,t) \in \partial_p Q_T$. [/step] [step:Exclude a positive maximum of the perturbed function on each truncated cylinder] Fix $\tau \in (0,T)$. We first prove that \begin{align*} w_\varepsilon(x,t) \leq 0 \end{align*} for every $(x,t) \in \overline{U} \times [0,\tau]$. Suppose instead that there exists $(x_1,t_1) \in \overline{U} \times [0,\tau]$ such that $w_\varepsilon(x_1,t_1) > 0$. Since $U$ is bounded, $\overline{U}$ is compact in $\mathbb{R}^n$, and therefore $\overline{U} \times [0,\tau]$ is compact. Since $w_\varepsilon$ is continuous, it attains its maximum on $\overline{U} \times [0,\tau]$ at some point $(x_0,t_0) \in \overline{U} \times [0,\tau]$. The maximum value is positive. Since $w_\varepsilon \leq 0$ on $\partial_p Q_T$, the point $(x_0,t_0)$ cannot lie in the initial slice $\overline{U} \times \{0\}$ or on the lateral boundary $\partial U \times [0,\tau]$. Hence \begin{align*} x_0 \in U, \qquad 0 < t_0 \leq \tau < T. \end{align*} If $t_0 < \tau$, then $(x_0,t_0)$ is an interior maximum in both space and time. Because $w_\varepsilon \in C^{1,2}(Q_T)$ and $(x_0,t_0) \in Q_T$, the first time derivative and the spatial second derivatives exist at $(x_0,t_0)$. The one-variable maximum condition in time and the second-derivative test in the spatial variables give \begin{align*} \partial_t w_\varepsilon(x_0,t_0) = 0, \qquad \Delta w_\varepsilon(x_0,t_0) \leq 0. \end{align*} Consequently \begin{align*} \partial_t w_\varepsilon(x_0,t_0) - \Delta w_\varepsilon(x_0,t_0) \geq 0, \end{align*} contradicting $\partial_t w_\varepsilon - \Delta w_\varepsilon < 0$ in $Q_T$. It remains to consider $t_0 = \tau$. This is still a time strictly below $T$, so $(x_0,\tau) \in Q_T$ and all derivatives appearing below are classical derivatives supplied by $w_\varepsilon \in C^{1,2}(Q_T)$. Since $x_0 \in U$ and $w_\varepsilon$ has a maximum on $\overline{U} \times [0,\tau]$ at $(x_0,\tau)$, the spatial function $x \mapsto w_\varepsilon(x,\tau)$ has a local maximum at $x_0$, so \begin{align*} \Delta w_\varepsilon(x_0,\tau) \leq 0. \end{align*} Also, because the map $t \mapsto w_\varepsilon(x_0,t)$ has a maximum on $[0,\tau]$ at the right endpoint $\tau$ and is differentiable at $\tau$, its derivative satisfies \begin{align*} \partial_t w_\varepsilon(x_0,\tau) \geq 0. \end{align*} Thus \begin{align*} \partial_t w_\varepsilon(x_0,\tau) - \Delta w_\varepsilon(x_0,\tau) \geq 0, \end{align*} again contradicting $\partial_t w_\varepsilon - \Delta w_\varepsilon < 0$ in $Q_T$. Therefore $w_\varepsilon \leq 0$ on $\overline{U} \times [0,\tau]$ for every $\tau \in (0,T)$. Now let $(x,t) \in \overline{U} \times [0,T)$. Choose $\tau \in (t,T)$. The truncated-cylinder estimate gives $w_\varepsilon(x,t) \leq 0$. Since $w_\varepsilon$ is continuous on $\overline{U} \times [0,T]$, taking $t \uparrow T$ with $x$ fixed gives $w_\varepsilon(x,T) \leq 0$ for every $x \in \overline{U}$. Hence $w_\varepsilon \leq 0$ on $\overline{U} \times [0,T]$. [guided] The point of introducing a truncated cylinder is to avoid differentiating at the terminal time $T$, where the hypotheses only give continuity. Fix a number $\tau \in (0,T)$. We will prove the maximum estimate on $\overline{U} \times [0,\tau]$, where the top time slice lies strictly inside the original cylinder $Q_T$. Assume for contradiction that $w_\varepsilon(x_1,t_1) > 0$ for some $(x_1,t_1) \in \overline{U} \times [0,\tau]$. Since $U$ is bounded, the [Heine-Borel theorem](/theorems/309) gives compactness of $\overline{U}$ in $\mathbb{R}^n$, and therefore $\overline{U} \times [0,\tau]$ is compact. The function $w_\varepsilon$ is continuous on this compact set, so it attains a maximum at some point $(x_0,t_0) \in \overline{U} \times [0,\tau]$. This maximum is positive because it is at least $w_\varepsilon(x_1,t_1)$. Where can this positive maximum occur? It cannot occur on the initial slice $\overline{U} \times \{0\}$ or on the lateral boundary $\partial U \times [0,\tau]$, because those points are contained in the parabolic boundary $\partial_p Q_T$, and the previous step proved $w_\varepsilon \leq 0$ there. Hence the maximum point satisfies \begin{align*} x_0 \in U, \qquad 0 < t_0 \leq \tau < T. \end{align*} The inequality $\tau < T$ is the crucial repair: it ensures that even if $t_0 = \tau$, the point $(x_0,t_0)$ is still in $Q_T$, where $w_\varepsilon$ has the classical regularity $C^{1,2}$. First suppose $t_0 < \tau$. Then $(x_0,t_0)$ is an interior maximum in all variables. Since $w_\varepsilon \in C^{1,2}(Q_T)$ and $(x_0,t_0) \in Q_T$, the time derivative and the spatial second derivatives exist. The one-variable first-derivative test in time gives $\partial_t w_\varepsilon(x_0,t_0) = 0$. The spatial second-derivative test says that the spatial Hessian at $(x_0,t_0)$ is negative semidefinite, so its trace satisfies \begin{align*} \Delta w_\varepsilon(x_0,t_0) \leq 0. \end{align*} Therefore \begin{align*} \partial_t w_\varepsilon(x_0,t_0) - \Delta w_\varepsilon(x_0,t_0) \geq 0. \end{align*} This contradicts the strict inequality \begin{align*} \partial_t w_\varepsilon(x,t) - \Delta w_\varepsilon(x,t) < 0 \end{align*} for every $(x,t) \in Q_T$. Now suppose $t_0 = \tau$. This is the endpoint of the truncated time interval, but it is not the terminal time $T$ of the original problem. Thus $(x_0,\tau) \in Q_T$, so the derivatives $\partial_t w_\varepsilon(x_0,\tau)$ and $\Delta w_\varepsilon(x_0,\tau)$ exist classically. The spatial variables are interior because $x_0 \in U$, and the spatial function $x \mapsto w_\varepsilon(x,\tau)$ has a local maximum at $x_0$. Hence \begin{align*} \Delta w_\varepsilon(x_0,\tau) \leq 0. \end{align*} For the time derivative, the function $t \mapsto w_\varepsilon(x_0,t)$ has a maximum on $[0,\tau]$ at the right endpoint $\tau$. Therefore for every sufficiently small $h > 0$, \begin{align*} w_\varepsilon(x_0,\tau) - w_\varepsilon(x_0,\tau-h) \geq 0. \end{align*} Dividing by $h$ and letting $h \downarrow 0$ gives \begin{align*} \partial_t w_\varepsilon(x_0,\tau) \geq 0, \end{align*} because the two-sided derivative exists at $\tau$ as a point of the open time interval $(0,T)$. Combining the endpoint time inequality with the spatial Laplacian inequality gives \begin{align*} \partial_t w_\varepsilon(x_0,\tau) - \Delta w_\varepsilon(x_0,\tau) \geq 0, \end{align*} again contradicting the strict differential inequality in $Q_T$. Thus no positive value of $w_\varepsilon$ can occur on $\overline{U} \times [0,\tau]$. Since $\tau \in (0,T)$ was arbitrary, every point with time $t < T$ lies in some truncated cylinder $\overline{U} \times [0,\tau]$ with $t < \tau < T$, and hence $w_\varepsilon(x,t) \leq 0$ for all $(x,t) \in \overline{U} \times [0,T)$. Finally, for each fixed $x \in \overline{U}$, continuity of $w_\varepsilon$ on the closed cylinder gives \begin{align*} w_\varepsilon(x,T) = \lim_{t \uparrow T} w_\varepsilon(x,t) \leq 0. \end{align*} Therefore \begin{align*} w_\varepsilon(x,t) \leq 0 \end{align*} for every $(x,t) \in \overline{U} \times [0,T]$. [/guided] [/step] [step:Let the perturbation vanish and recover the comparison inequality] For every $\varepsilon > 0$ and every $(x,t) \in \overline{U} \times [0,T]$, the previous step gives \begin{align*} w(x,t) - \varepsilon t = w_\varepsilon(x,t) \leq 0. \end{align*} Hence \begin{align*} w(x,t) \leq \varepsilon t. \end{align*} Letting $\varepsilon \downarrow 0$ with $(x,t)$ fixed gives \begin{align*} w(x,t) \leq 0. \end{align*} Since $w = u - v$, this is exactly \begin{align*} u(x,t) \leq v(x,t) \end{align*} for every $(x,t) \in \overline{U} \times [0,T]$. [/step]