**Proof plan.** A direct argument fails because $\partial_t w - \kappa \Delta w \leq 0$ allows equality, which does not produce a contradiction at an interior maximum. The strategy is to perturb $w$ to a strict subsolution $u_\varepsilon := w - \varepsilon t$ satisfying $\partial_t u_\varepsilon - \kappa \Delta u_\varepsilon < 0$ strictly (Step 1), show by contradiction using first- and second-order necessary conditions that $u_\varepsilon$ cannot attain its maximum in the parabolic interior (Claim 1), then send $\varepsilon \to 0$ to recover the result for $w$ (Step 3).
**Step 1 (Strict subsolution).**
Define $u_\varepsilon(t,x) := w(t,x) - \varepsilon t$ for $(t,x) \in \overline{Q}_T$ and $\varepsilon > 0$. Since $\partial_t(\varepsilon t) = \varepsilon$ and $\Delta(\varepsilon t) = 0$:
\begin{align*}
\partial_t u_\varepsilon - \kappa \Delta u_\varepsilon &= (\partial_t w - \varepsilon) - \kappa \Delta w = (\partial_t w - \kappa \Delta w) - \varepsilon = q - \varepsilon.
\end{align*}
Since $q \leq 0$ and $\varepsilon > 0$, we have $\partial_t u_\varepsilon - \kappa \Delta u_\varepsilon = q - \varepsilon \leq -\varepsilon < 0$ throughout $Q_T$.
**Step 2 (Interior maximum is impossible for $u_\varepsilon$).**
[claim:No Interior Maximum For Strict Subsolution]
Let $v \in C^1_t C^2_x(Q_T) \cap C(\overline{Q}_T)$ satisfy $\partial_t v - \kappa \Delta v < 0$ in $Q_T$ with $\kappa > 0$. Then $v$ cannot attain its maximum over $\overline{Q}_T$ at any point $(t_0, x_0)$ with $x_0 \in \Omega$ and $t_0 \in (0, T]$.
[/claim]
[proof]
Suppose for contradiction that $v$ attains $\max_{\overline{Q}_T} v$ at $(t_0, x_0)$ with $x_0 \in \Omega$ and $t_0 \in (0,T]$. We derive the signs of $\partial_t v$ and $\Delta v$ at this point.
**Time [derivative](/page/Derivative).** Since $x_0 \in \Omega$ is fixed, consider the [function](/page/Function) $t \mapsto v(t, x_0)$ on $[0, T]$. This function attains its maximum at $t_0$ (because $v(t_0, x_0) = \max_{\overline{Q}_T} v \geq v(t, x_0)$ for all $t$). If $t_0 \in (0, T)$, then $t_0$ is an interior maximum of a $C^1$ function on $(0,T)$, so $\partial_t v(t_0, x_0) = 0$. If $t_0 = T$, then $t_0$ is a right-endpoint maximum, and for all $h > 0$ sufficiently small:
\begin{align*}
\frac{v(T, x_0) - v(T - h, x_0)}{h} \geq 0,
\end{align*}
so taking $h \to 0^+$ gives $\partial_t v(T, x_0) \geq 0$ (this is the left-hand derivative, which equals $\partial_t v$ since $v \in C^1_t$). In both cases:
\begin{align*}
\partial_t v(t_0, x_0) \geq 0.
\end{align*}
**Laplacian.** Since $t_0$ is fixed and $x_0 \in \Omega$ is an interior maximum of $x \mapsto v(t_0, x)$, the second-order necessary condition for an interior maximum gives $\partial_{x_i x_i} v(t_0, x_0) \leq 0$ for each $i = 1, \dots, n$. Indeed, for each coordinate direction $e_i$, the one-variable function $s \mapsto v(t_0, x_0 + s\,e_i)$ has a local maximum at $s = 0$ (since $x_0$ is a global spatial maximum), so its second derivative satisfies $\partial_{x_i x_i} v(t_0, x_0) \leq 0$. Summing over $i$:
\begin{align*}
\Delta v(t_0, x_0) = \sum_{i=1}^{n} \partial_{x_i x_i} v(t_0, x_0) \leq 0.
\end{align*}
**Contradiction.** Combining the two inequalities with $\kappa > 0$:
\begin{align*}
\partial_t v(t_0, x_0) - \kappa \Delta v(t_0, x_0) \geq 0 - \kappa \cdot 0 = 0.
\end{align*}
More precisely, $\partial_t v(t_0, x_0) \geq 0$ and $-\kappa \Delta v(t_0, x_0) \geq 0$ (since $\kappa > 0$ and $\Delta v \leq 0$), so their sum is $\geq 0$. This contradicts $\partial_t v - \kappa \Delta v < 0$ at $(t_0, x_0)$.
[/proof]
By Claim 1, $u_\varepsilon$ attains its maximum over $\overline{Q}_T$ on the parabolic [boundary](/page/Boundary) $\partial Q_T$:
\begin{align*}
\max_{\overline{Q}_T} u_\varepsilon = \max_{\partial Q_T} u_\varepsilon.
\end{align*}
**Step 3 (Pass to the [limit](/page/Limit) $\varepsilon \to 0$).**
Since $u_\varepsilon = w - \varepsilon t$ and $0 \leq t \leq T$ on $\overline{Q}_T$, we have $w - \varepsilon T \leq u_\varepsilon \leq w$ pointwise. In particular, for any $(t,x) \in \overline{Q}_T$:
\begin{align*}
w(t,x) &= u_\varepsilon(t,x) + \varepsilon t \leq u_\varepsilon(t,x) + \varepsilon T \leq \max_{\overline{Q}_T} u_\varepsilon + \varepsilon T = \max_{\partial Q_T} u_\varepsilon + \varepsilon T.
\end{align*}
Since $u_\varepsilon \leq w$ everywhere, $\max_{\partial Q_T} u_\varepsilon \leq \max_{\partial Q_T} w$. Taking the supremum over $(t,x) \in \overline{Q}_T$ on the left-hand side:
\begin{align*}
\max_{\overline{Q}_T} w \leq \max_{\partial Q_T} w + \varepsilon T.
\end{align*}
Since $\varepsilon > 0$ was arbitrary, sending $\varepsilon \to 0$ gives $\max_{\overline{Q}_T} w \leq \max_{\partial Q_T} w$. The reverse inequality $\max_{\partial Q_T} w \leq \max_{\overline{Q}_T} w$ holds because $\partial Q_T \subseteq \overline{Q}_T$, so equality follows.
