I meant to assign to my class the following homework problem:



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> 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 $(0,T)\times \Omega$. 
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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 $(0,T\color{red}{]} \times \Omega$. 
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**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). 



Maximum principle for heat equation, low regularity case
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.


