Suppose \(\Omega \subset \mathbb{R}^d\) is open and bounded, \(T > 0\), \(\Omega_T = \Omega \times (0,T]\) and

\[u \in C^{(2,1)}(\Omega_T) \cap C^0(\overline{\Omega_T})\]

with

\[\partial_t u - \Delta u \leq 0 \quad \text{in } \Omega_T.\]

Then

\[ \max_{\overline{\Omega_T}} u = \max_{\partial \Omega_T} u. \]
Proof

Step 1: Assume \(\partial_t u - \Delta u < 0\). If the maximum is inside, calculus implies

\[\partial_t u - \Delta u \geq 0.\]

Step 2: Consider \(v_\varepsilon(x,t) = u(x,t) - \varepsilon t\). Then apply Step 1 onto \(v_\varepsilon\) and send \(\varepsilon \to 0\).

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