[proofplan] The mean value formula forces the maximum of a caloric function to propagate backwards in time through heat balls: if $u$ attains its maximum $M$ at an interior point, then $u \equiv M$ on every heat ball centred at that point. An open-and-closed argument in the connected parabolic cylinder below the maximum point extends this to the entire connected component. The boundary maximum principle follows: if the maximum is not attained on the parabolic boundary $\Gamma_T$, then $u$ is identically constant. [/proofplan] [step:Use the mean value formula to propagate the maximum through heat balls] Suppose $u$ attains its maximum $M := \max_{\overline{\Omega}_T} u$ at an interior point $(x_0, t_0) \in \Omega_T$. Let $r > 0$ be small enough that $E(x_0, t_0; r) \subseteq \Omega_T$. By the [Parabolic Mean Value Formula](/theorems/559): \begin{align*} M = u(x_0, t_0) = \frac{1}{4r^n} \iint_{E(x_0, t_0; r)} u(y, s) \frac{|x_0 - y|^2}{(t_0 - s)^2} \, d\mathcal{L}^n(y) \, d\mathcal{L}^1(s). \end{align*} Since $u(y, s) \leq M$ throughout $E(x_0, t_0; r)$ and the weight $|x_0 - y|^2/(t_0 - s)^2$ is strictly positive on the interior of $E(x_0, t_0; r)$, the weighted average is at most $M$. [claim:$u \equiv M$ on $E(x_0, t_0; r)$] If the weighted average of $u$ over $E(x_0, t_0; r)$ equals $M$ and $u \leq M$ everywhere in $E(x_0, t_0; r)$, then $u = M$ throughout $E(x_0, t_0; r)$. [/claim] [proof] Suppose for contradiction that $u(y_1, s_1) < M$ at some point $(y_1, s_1)$ in the interior of $E(x_0, t_0; r)$ where the weight is positive. By continuity of $u$, the strict inequality $u < M$ persists in a neighbourhood $B \subseteq E(x_0, t_0; r)$ of $(y_1, s_1)$. Since the weight is continuous and positive at $(y_1, s_1)$, the integral over $B$ contributes strictly less than $M$ times the weight integral over $B$. The integral over $E(x_0, t_0; r) \setminus B$ contributes at most $M$ times the remaining weight. The total weighted average is therefore strictly less than $M$, contradicting the mean value formula. [/proof] [/step] [step:Extend $u \equiv M$ to the entire connected component below $t_0$ via an open-and-closed argument] Define $S := \{(y, s) \in \Omega_T : s \leq t_0,\, u(y, s) = M\}$. The previous step shows $E(x_0, t_0; r) \subseteq S$ for all sufficiently small $r > 0$. The set $S$ is relatively open in $\Omega_{t_0} := \{(y, s) \in \Omega_T : s \leq t_0\}$: for any $(y_1, s_1) \in S$, there exists $r_1 > 0$ with $E(y_1, s_1; r_1) \subseteq \Omega_T$. Applying the previous step at $(y_1, s_1)$ gives $u \equiv M$ on $E(y_1, s_1; r_1)$. Since the heat ball has nonempty interior in $\mathbb{R}^{n+1}$, $S$ contains a neighbourhood of $(y_1, s_1)$ within $\Omega_{t_0}$. The set $S$ is also relatively closed in $\Omega_{t_0}$: it is a level set of the continuous function $u$, so $S = u^{-1}(\{M\}) \cap \Omega_{t_0}$ is closed in the relative topology. Since $\Omega_{t_0}$ is connected (as $\Omega$ is connected), $S$ is a nonempty clopen subset, hence $S = \Omega_{t_0}$. Therefore $u \equiv M$ on the connected component of $\Omega_{t_0}$ containing $(x_0, t_0)$. [/step] [step:Conclude that the maximum is attained on the parabolic boundary] If $u$ is not identically $M$ on $\overline{\Omega}_T$, then by the previous step the maximum cannot be attained at any interior point of $\Omega_T$. Since $u$ is continuous on the compact set $\overline{\Omega}_T$, it attains its maximum somewhere. The only remaining possibility is the parabolic boundary $\Gamma_T$: \begin{align*} \max_{\overline{\Omega}_T} u = \max_{\Gamma_T} u. \end{align*} [/step]