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.
I think your sketch has hidden the problem. You do the proof for sets of the form $\Omega \times[0,T')$, $T'<T$. Being a continuous function, $u$ certainly achieves some maximum on $\bar\Omega \times [0,T']$, and its possible that the maximum is achieved at time $T'$ (which is not necessarily in the parabolic boundary of $\Omega \times[0,T')$). In this case you get an inequality on the derivative by $$u(x,\tau)- u(x,\tau-\epsilon)\ge 0 \implies u_t(x,\tau)= \lim_{\epsilon\to 0}\frac{u(x,\tau)- u(x,\tau-\epsilon)}\epsilon\ge 0 $$
I previously said that $u$ would be non-increasing at the maximum, but I don't think this is true for an arbitrary $C^1$ function, and its not needed for the proof.