Weak maximum principle for the heat equation

Fix $\varepsilon>0$ and define $v(x,t) = u(x,t) - \varepsilon t$, so $v_t - \alpha v_{xx} = u_t - \varepsilon - \alpha u_{xx} = -\varepsilon < 0$ everywhere on the open rectangle. Suppose, for contradiction, that $v$ attains its maximum over the closed rectangle at an interior point $(x_0,t_0)$ with $0<x_0<L$ and $0<t_0\le T$ — i.e. off the parabolic boundary.

Because $x_0$ is an interior spatial maximum, the second-derivative test gives $v_{xx}(x_0,t_0)\le 0$. Because $t_0$ is where the maximum over $t\in[0,T]$ is reached (either an interior critical time, giving $v_t(x_0,t_0)=0$, or the endpoint $t_0=T$ approached from the left, giving $v_t(x_0,t_0)\ge 0$), in either case $v_t(x_0,t_0)\ge 0$.

Combining the two inequalities, $v_t(x_0,t_0) - \alpha v_{xx}(x_0,t_0) \ge 0 - \alpha\cdot 0 = 0$, which contradicts $v_t-\alpha v_{xx}=-\varepsilon<0$ at every point. Hence no such interior maximum exists, so $v$'s maximum lies on the parabolic boundary $\Gamma$ (the initial edge together with the two lateral edges), giving $u(x,t) \le \varepsilon t + \max_{\Gamma} u$ everywhere. Letting $\varepsilon \to 0^+$ proves the maximum principle for $u$ itself.