$L^\infty$ bound on solution of $u_t -\Delta u =f$ where $f \in L^\infty$ and $u_0 \in L^\infty$?? Let $u_0 \in L^\infty(\Omega)$ and $f \in L^\infty((0,T)\times\Omega)$ and consider
$$u_t - \Delta u = f$$
$$u(0) = u_0$$
with some boundary conditions.  Consider the weak form
$$\int_{\Omega}u_t\varphi + \int_{\Omega}\nabla u \nabla \varphi = \int_{\Omega}f\varphi $$
How can I show that $u \in {L^\infty((0,T)\times \Omega)}$ or $u \in {L^\infty(0,T;L^\infty(\Omega))}$, without using classical approach (just a weak approach)?
If the classical theory cannot be avoid then I don't mind using it. I would have thought we should test with $(u(t) - \Vert u_0 \Vert_{L^\infty})^+$ but don't know what do with the term involving $f$.

Consider some $v(t) = \alpha t + \beta$. Then $v(t)$ will satisfy
$$
v_t - \Delta v = \alpha,\\
v(0) = \beta.
$$
Take now $\alpha$ such that $\alpha \geq f$ in $(0, T)\times \Omega$ and $\beta$ such that $\beta \geq u(0)$. (it is possible due to the boundedness of $f$ and $u_0$).
Therefore, the Weak maximum principle for $w = v - u$ will give you that $w \geq 0$. Hence, $u \leq v$, but $v$ is bounded (for $T < +\infty$). Thus, $u$ is also bounded.
