I'm confused about one step in the proof of the weak maximum principle for the heat equation in McOwen.
Theorem (Weak Maximum Principle): Let $u\in C^{2;1}(U)\cap C(\overline{U})$ satisfy $\Delta u \geq u_t$ in $U$. Then $u$ achieves its maximum on the parabolic boundary of $u$:
$$\max_{(x,t)\in~\overline{U}}u(x,t)=\max_{(x,t)\in~\Gamma} u(x,t)$$
Sketch of first part of proof:
Assume $u_t < \Delta u$.
By contradiction, suppose that u has a maximum point at $(x,\tau)$ for some $0<\tau<T$ (in the interior) and $x\in\Omega$.
Then, at $(x,\tau)$,
$$u_t(x,\tau)\geq 0$$
since $\tau$ is a maximum. By the second derivative test from calculus,
$$\Delta u(x,\tau) \leq 0$$
So this means that $u_t \geq \Delta u$ in direct contradiction to our hypothesis that $u_t < \Delta u$. Therefore,
$$\max_{(x,t)\in~\overline{U}}u(x,t)=\max_{(x,t)\in~\Gamma} u(x,t)$$
Question: Why can we conclude that $u_t(x,\tau)\geq 0$ at $(x,\tau)$? Since the point $(x,\tau)$ is a maximum, shouldn't it be that $u_t(x,\tau)= 0$? I may be forgetting something that I learned in calculus.