I have a short question regarding some inconsistency I've seen between the statement and actual usage of the maximum principle which I hope you could clarify for me.
Theorem (The Maximum Principle):
Let ${\Omega _T} = \left\{ {\left( {x,t} \right)|a \le x \le b,0 \le t \le T} \right\}$ and suppose $u(x,t)$ satisfies the heat equation in the interior of ${\Omega _T}$ and is continuous on ${\Omega _T}$, then the maximum of $u(x,t)$ in ${\Omega _T}$ is attained on the parabolic boundary $$P = \left\{ {\left( {x,t} \right)|x = a,0 \le t \le T} \right\} \cup \left\{ {\left( {x,t} \right)|x = b,0 \le t \le T} \right\} \cup \left\{ {\left( {x,t} \right)|a \le x \le b,t = 0} \right\}$$
The requirement that $u(x,t)$ is continuous is indeed used in the proof of this theorem as it used to argue, that by Weierstrass' theorem, the function $u(x,t)$ attains it maximum value on (the compact set) ${\Omega _T}$.
But I've seen numerous exercises where the above maximum principle was used in situations where $u(x,t)$ was clearly not continuous.
For example, in a situation like this $$\left\{ \matrix{ {u_t} = {u_{xx}}{\rm{ , }}a < x < b,t > 0 \hfill \cr u\left( {x,0} \right) = 1{\rm{ , }}a \le x \le b \hfill \cr u\left( {a,t} \right) = u\left( {b,t} \right) = 0{\rm{ , }}t \ge 0 \hfill \cr} \right.$$
the maximum principle was used to conclude that $u\left( {x,t} \right) \le 1$ even though it's clear that $u(x,t)$ is not continuous at $(a,0)$ nor is it continuous at $(b,0)$.
So is there some generalization of the maximum principle that I am not aware of?
Thank you kindly in advance.