The proof combines the [Weak Maximum Principle (Parabolic)](/theorems/607) with a propagation-of-maximum argument. The strategy is to show that the [set](/page/Set) $\{(x, t) \in U_T : u(x, t) = M\}$ (where $M = \max_{\overline{U_T}} u \ge 0$) is both relatively open and relatively closed in $\overline{U} \times [0, t_0]$, hence equals all of $\overline{U} \times [0, t_0]$ by connectedness. **Step 1: Setup.** Let $M := \max_{\overline{U_T}} u \ge 0$ and suppose $u(x_0, t_0) = M$ for some $(x_0, t_0) \in U_T$ (an interior point). Define the set $\Sigma := \{(x, t) \in \overline{U} \times [0, t_0] : u(x, t) = M\}$. This set is closed in $\overline{U} \times [0, t_0]$ by [continuity](/page/Continuity) of $u$. We will show it is also open relative to $U \times (0, t_0]$. **Step 2: Backward propagation in time.** [claim:Maximum At Time $t_0$ Implies Maximum On The Entire Spatial Slice] If $u(x_0, t_0) = M$ for some $x_0 \in U$ and $0 < t_0 \le T$, then $u(x, t_0) = M$ for all $x \in U$. [/claim] [proof] Suppose $u(\cdot, t_0)$ is not identically $M$ on $U$. Then the set $U^+ := \{x \in U : u(x, t_0) < M\}$ is nonempty and open. Choose a ball $B(y, r) \subset U^+$ and construct a barrier [function](/page/Function) of the form $w(x, t) = e^{-\alpha|x - y|^2} - e^{-\alpha r^2}$ for $\alpha > 0$ large. On the annular region $\Omega := B(y, r) \setminus B(y, r/2)$ and the time interval $[t_0 - \varepsilon, t_0]$ (for small $\varepsilon > 0$), one can verify that $w_t + Lw > 0$ for $\alpha$ sufficiently large (the dominant term is $\alpha^2 \sum a_{ij} (x_i - y_i)(x_j - y_j) e^{-\alpha|x-y|^2}$, which is controlled by the ellipticity constant $\theta$). Define $v := u - M + \delta w$ for small $\delta > 0$. On the parabolic [boundary](/page/Boundary) of the cylinder $\Omega \times [t_0 - \varepsilon, t_0]$: on $\partial B(y, r)$, $w = 0$ and $u \le M$, so $v \le 0$; on $\partial B(y, r/2)$, $w > 0$ but $\delta$ can be chosen so small that $v \le 0$; at $t = t_0 - \varepsilon$, $u$ is close to $M$ (by continuity) and $\delta w$ is small. By the [Weak Maximum Principle](/theorems/607), $v \le 0$ in $\Omega \times [t_0 - \varepsilon, t_0]$, i.e., $u \le M - \delta w$. But at the point $x_0$ (which lies in $U$, hence can be connected to $\Omega$ by a chain of balls), this contradicts $u(x_0, t_0) = M$ if $x_0 \in \Omega$ and $w(x_0) > 0$. A chain-of-balls argument extends this to all of $U$. [/proof] **Step 3: Propagation to earlier times.** If $u(\cdot, t_0) \equiv M$ on $U$, then in particular $u(x, t_0) = M$ for all $x$. Consider the time interval $[t_0 - \varepsilon, t_0]$ for small $\varepsilon > 0$. On the lateral boundary $\partial U \times [t_0 - \varepsilon, t_0]$, $u \le M$ by continuity. On the bottom face $U \times \{t_0 - \varepsilon\}$, if $u < M$ somewhere, the argument of Step 2 (applied backward from $t_0$ to $t_0 - \varepsilon$) would show $u < M$ at $t_0$ as well — a contradiction. Therefore $u(\cdot, t_0 - \varepsilon) \equiv M$ on $U$. **Step 4: Conclusion.** Iterating Step 3 backward in time, $u \equiv M$ on $U \times [0, t_0]$. By continuity, $u \equiv M$ on $\overline{U} \times [0, t_0]$.