Maximum principle for heat equation, low regularity case I meant to assign to my class the following homework problem:

If $u\in C^2((0,T)\times \Omega) \cap C^0([0,T]\times\bar{\Omega})$ where $\Omega$ is an open, bounded domain, is such that $\partial_t u - \triangle u \leq - \epsilon < 0$ for some constant $\epsilon > 0$, then $u$ cannot have a local maximum on the set $(0,T)\times \Omega$. 

This follows from simple second derivative considerations. I made a typo, however, and asked my students to prove that

... $u$ cannot have a local maximum on the set $(0,T\color{red}{]} \times \Omega$. 

Question: Is the version with the typo still true? Or is there a counterexample?
Remark: If we assume that $\nabla u$ and $\nabla^2 u$ (the spatial gradient and Hessian) both extend continuously to $\{T\} \times \Omega$, then the second derivative test argument also goes through. So any potential counterexample must be non-regular at time $T$. 
Remark 2: the potential lack of regularity also means we cannot directly apply the mean value integral (e.g. that which is given in Evans' textbook). 
 A: This version is still true: if $u$ had a local maximum at $(x,\,T)$, say with $u(x,\,T) = 0$, then $u \leq 0$ in a small parabolic cylinder centered at $(x,\,T)$. After rescaling we can assume that $u \leq 0$ in $\overline{B_1} \times [0,\,1]$, with $u(0,\,1) = 0$.
Replacing $u$ with $u - \frac{\epsilon}{4n}t(|x|^2-1) - \frac{\epsilon}{8n}$ we may assume that $(\partial_t - \Delta)u < 0$, with $u \leq -\epsilon / 8n$ on $(B_1 \times \{0\}) \cup (\partial B_1 \times [0,1])$ and $u(0,1) = \epsilon / 8n$. However, by the maximum principle, on the cylinders $\overline{B_1} \times [0,1-\delta]$, $u$ achieves its maximum on the sides or bottom, so $u \leq 0$ on all such cylinders. Taking $\delta \rightarrow 0$ we get a contradiction.
A: A potential-theoretic argument seems to be applicable. I come from probability, so I use the language of probabilistic potential theory below. However, everything can be translated into the language of harmonic (or ``caloric'') measures.
Let $X_t$ be the standard Brownian motion, generated by the Laplacian $\Delta$, and let $\mathbb{E}^x$ be the expectation corresponding to $X_0 = x$. For an open $D$ let $\tau_D$ be the time of first exit from $D$, and let $t \wedge s$ stand for the minimum of $t$ and $s$.
By standard arguments (Dynkin's formula or Itô's lemma, depending on where about in probability you come from), we have
$$ u(t,x) \leqslant \mathbb{E}^x(u(t - (\tau_U \wedge s), X_{\tau_U \wedge s}) - \varepsilon (\tau_U \wedge s)) $$
whenever $0 < s < t < T$ and $U \Subset \Omega$. By continuity, this estimate extends to $t = T$. If $u$ had a local maximum at $(T,x)$, we would have
$$ u(t,x) \geqslant \mathbb{E}^x u(t - (\tau_U \wedge s), X_{\tau_U \wedge s}) $$
if $s$ is sufficiently small and $U$ is a sufficiently small neighbourhood of $x$, contradicting the previous display. Thus, $u$ has no local minima in $(0, T] \times \Omega$.
