Consider $u(x,t)$ as the solution of the equation
$$\frac{\partial}{\partial t} u-\Delta u=f(x,t)$$ for $(x,t) \in \Omega \times [0,T]$
Using the weak maximum principle and its prerequisites I need to show that for $f(x,t) <0$ in $\Omega \times [0,T]$, u attains its maximum on $\partial(\Omega \times [0,T])$.
I think the differential operator is given as $Lu=\Delta u=\frac{\partial}{\partial t}u-f(x,t)$. It's clear that $-f(x,t)$ is positive but how could I show that the derivative of $u$ is positive as well. This should give me the desired result?
Suppose that $u \in C^2_1(\Omega_T) \cap C(\overline \Omega_T)$ where $\Omega_T:=\Omega \times (0,T]$. Since $f(x,t)<0$ in $\Omega_T$, $u$ is a subsolution of the heat equation. Hence, the maximum principle tells us $$\max_{\overline \Omega_T} u = \max_{\Gamma_T}u$$ where $\Gamma_T$ is the parabolic boundary of $\Omega_T$. That is, $u$, attains its maximum on $ \Gamma_T \subset \partial (\Omega \times (0,T))$.