Let $u$ satisfy $u_t = k\nabla^2 u$ in a space-time cylinder $\Omega \times (0, T]$ where $\Omega \subset \mathbb{R}^n$ is a bounded domain. Define the parabolic boundary as: $\Gamma = (\overline{\Omega} \times \{0\}) \cup (\partial\Omega \times [0, T])$

Claim: If $u$ attains its maximum $M$ at some interior point $(x_0, t_0)$ with $t_0 > 0$, then $u \equiv M$ in $\Omega \times [0, t_0]$.

Step 1: Weak Maximum Principle

First, we establish that $\max_{\Omega \times [0,T]} u = \max_\Gamma u$.

Suppose for contradiction that $u$ attains a maximum $M$ at an interior point $(x_1, t_1)$ with $t_1 > 0$ and $M > \max_\Gamma u$. Define: $v(x,t) = u(x,t) + \epsilon(|x|^2 + t)$ for small $\epsilon > 0$.

Then $v$ also attains its maximum at some interior point $(x_\epsilon, t_\epsilon)$, and for small enough $\epsilon$: $v_t(x_\epsilon, t_\epsilon) \geq 0, \quad \nabla v(x_\epsilon, t_\epsilon) = 0, \quad \nabla^2 v(x_\epsilon, t_\epsilon) \leq 0$

At $(x_\epsilon, t_\epsilon)$: $v_t = u_t + \epsilon$ $\nabla^2 v = \nabla^2 u + 2n\epsilon$

Since $u$ satisfies the heat equation: $\epsilon = v_t - u_t = k(\nabla^2 v - \nabla^2 u) - u_t = k \cdot 2n\epsilon - k\nabla^2 u$

But $\nabla^2 v \leq 0$ at a maximum, so $\nabla^2 u \leq -2n\epsilon < 0$. This gives: $u_t = k\nabla^2 u < -2nk\epsilon < 0$

contradicting $v_t \geq 0$ at the maximum. Thus $\max_{\Omega \times [0,T]} u = \max_\Gamma u$.

Step 2: Strong Form

Now suppose $u$ attains its maximum $M$ at an interior point $(x_0, t_0)$ with $t_0 > 0$.

Define $S = \{t \in [0, t_0] : u(x_0, t) = M\}$. Since $u$ is continuous and achieves $M$ at $t_0$, we have $S \neq \emptyset$ and $t_0 \in S$.

Let $t^* = \inf S$. We claim $t^* = 0$.

If $t^* > 0$, then $u(x_0, t^*) = M$ (by continuity) and $u(x_0, t) < M$ for $t \in [t^* - \delta, t^*)$ for some $\delta > 0$.

Consider $u$ in a small ball $B_r(x_0) \times [t^*, t^* + \tau]$. By the weak maximum principle on this cylinder: $\max_{B_r(x_0) \times [t^*, t^*+\tau]} u = M$

But at $(x_0, t^*)$, we have $u_t(x_0, t^*) \geq 0$ (since $u$ is increasing at $x_0$ from $t^*$) and $\nabla^2 u(x_0, t^*) \leq 0$ (since $u(x_0, t^*) = M$).

The heat equation gives: $0 \leq u_t = k\nabla^2 u \leq 0$

Thus $\nabla^2 u(x_0, t^*) = 0$, which means $u(x, t^*) = M$ in a neighborhood of $x_0$. By continuity arguments (repeated application in space), $u \equiv M$ in $\Omega \times \{t^*\}$.

By similar reasoning backward in time, $u \equiv M$ in $\Omega \times [0, t^*]$, contradicting the definition of $t^*$ unless $t^* = 0$.