Bounded data means bounded solution to parabolic PDE Let $u_0 \in L^\infty(\Omega)$ and $f \in L^\infty((0,T)\times\Omega).$ Consider
$$u_t - \Delta u = f$$
$$u|_{\partial\Omega} = 0$$
$$u(0) = u_0$$
or more generally replace $\Delta$ with a suitable elliptic operator $A$. How does one show that $u \in L^\infty((0,T)\times\Omega)$?
(My question stems from this paper: http://www.mat.uniroma2.it/~porretta/papers/Blanchard-Porretta-JDE.pdf. See Theorem A.1 in the appendix (page 425). It is a different nonlinear equation but this should still be true).
Thanks
 A: Let $\Omega\subset\mathbb{R}^n,\,n\geqslant 2,$ be a bounded domain 
with Lipschitz boundary. For a cylinder $\,Q_T=\Omega\times(0,T),\,$ denote by 
$S_T$ its side surface $\partial\Omega\times (0,T)$, with notation $\Gamma_T$ 
standing for its parabolic boundary $\{(x,t)\colon\; x\in \overline{\Omega},t=0\}\cup S_T$. Obviously, solution $u$ to the inhomogeneous heat equation $\,u_t-\Delta u=f\,$ in $\,Q_T\,$ with  innhomogeneous boundary condition $u|_{S_T}=\psi$ can be represented 
in the form $u=v+w\,$  where
$$
\begin{cases}
v_t-\Delta v=0\;\;\text{in}\;\;Q_T\,,\\
v|_{\partial\Omega}=\psi,\quad t\in [0,T],\\
v|_{t=0}=u_0,\quad x\in\overline{\Omega},
\end{cases}
\qquad
\begin{cases}
w_t-\Delta w=f\;\;\text{in}\;\;Q_T\,,\\
w|_{\partial\Omega}=0,\quad t\in [0,T],\\
w|_{t=0}=0,\quad x\in\overline{\Omega}.
\end{cases}
$$
Let $v\in H^1(Q_T)$ be a weak solution satisfying the integral identity 
$$
\int\limits_{Q_T}v_t\varphi_t\,dxdt+\int\limits_{Q_T}\nabla v\cdot\nabla\varphi\,\,dxdt
=0\quad \forall\varphi\in H^1(Q_T)\colon\; \varphi|_{S_T}=0\tag{1}
$$
with initial and boundary conditions 
$$
v|_{t=0}=u_0\in H^{\frac{1}{2}}(\Omega),\quad v|_{S_T}=\psi\in H^{\frac{1}{2}}(S_T).
$$
For solution $v$, denote $m\overset{\rm def}{=}
\underset{\Gamma_T}{\rm ess\,sup\,}{v}\,$  
with the finite essential supremum taken over parabolic boundary $\Gamma_T\,$ 
w.r.t.  the $n$-dimensional Lebesgue mesure, and let $\eta_m$
be a Lipschitz function of the form
$$
\eta_m(\xi)=
\begin{cases}
0,\quad \xi\leqslant m,\\
\xi-m, \quad \xi>m.
\end{cases}
$$
It is clear that $\eta_m(v)\in H^1(Q_T)$ while $\eta_m(v)_{S_T}=0$ for any solution $v$. 
Taking in $(1)$ the test function $\varphi=\eta_m(v)$  yields
$$
\int\limits_{Q_T}\frac{\partial\,}{\partial t}\Phi_m(v)\,dxdt+
\int\limits_{Q_T}\eta'_m(v)|\nabla v|^2dxdt=0\tag{2}
$$
due to the chain rule $\nabla\varphi=\eta'_m(v)\nabla v$, where a nonnegative 
function $\Phi_m$ is defined as
$$
\Phi_m(\xi)=
\begin{cases}
0,\quad \xi\leqslant m,\\
\frac{1}{2}(\xi-m)^2, \quad \xi>m.
\end{cases}
$$
But $\,\Phi_m(v)|_{t=0}=0$, hence by $(2)$ follows
$$
\int\limits_{\Omega}\Phi_m\bigl(v(\cdot,T)\bigr)\,dx+\int\limits_{Q_T}|\nabla
\eta_m(v)|^2dxdt=0$$
since $\eta'_m=(\eta'_m)^2$. Therefore $\nabla \eta_m(v)=0\,$ a.e.  in $Q_T$
which implies that $\eta_m(v)=C(t)\,$ a.e.  in $Q_T\,$. 
But $\eta_m(v)|_{S_T}=0$, i.e., $\,C(t)=0\,$ a.e.  in $(0,T)\,$, 
and hence $\eta_m(v)=0\,$ a.e.  in $Q_T\,$. The latter implies 
the weak essential maximum principle 
$$
\underset{Q_T}{\rm ess\,sup\,}{v}\leqslant m=\underset{\Gamma_T}{\rm ess\,sup\,}{v}
\tag{3}
$$
with the essential supremum taken over the cylinder $Q_T\,$ w.r.t.  the 
$(n+1)$-dimensional Lebesgue mesure. To establish
$$
\underset{\Gamma_T}{\rm ess\,inf\,}{v}\leqslant\underset{Q_T}{\rm ess\,inf\,}{v}
\tag{4}
$$
it suffices to substitute $v$ in $(3)$ by $-v$. Inequalities $(3)$ and $(4)$ yield
the weak essential maximum principle for the modulus
$$
\underset{Q_T}{\rm ess\,sup\,}{|v|}\leqslant\underset{\Gamma_T}{\rm ess\,sup\,}{|v|}
\tag{5}
$$
To estimate solution $w$, notice that it can be constructed using Duhamel's principle,
i.e., in the form
$$
w(x,t)=\int\limits_0^t h(x,t,\tau)\,d\tau
$$
with function $h=h(x,s,\tau)$  defined for every $\tau\in (0,T)$ with finite norm
$\|f(\cdot,\tau)\|_{L^{\infty}(\Omega)}$ as solution of the intial boundary value problem 
$$
\begin{cases}
h_{s}-\Delta h=0\;\;\text{in}\;\;\Omega\times (\tau,T)\,,\\
h|_{\partial\Omega}=0,\quad s\in [\tau,T],\\
h|_{s=0}=f(x,\tau),\quad x\in\overline{\Omega}.
\end{cases}
$$
Otherwise, i.e., for $\tau\in (0,T)$ with the infinite norm
$\|f(\cdot,\tau)\|_{L^{\infty}(\Omega)}\,$, a zero option $h(\cdot,\cdot,\tau)=0$ 
is chosen without loss of generality. Hence for almost every $s\in (\tau,T)$ by $(5)$ follows
$$
\underset{\Omega}{\rm ess\,sup\,}{|h(\cdot,s,\tau)|}
\leqslant\underset{\Omega}{\rm ess\,sup\,}{|f(\cdot,\tau)|}
$$
for almost all $\tau\in (0,s)$, whence follows
$$
\underset{\Omega}{\rm ess\,sup\,}|w(\cdot,t)|\leqslant
\int\limits_0^t \underset{\Omega}{\rm ess\,sup\,}|h(\cdot,t,\tau)|\,d\tau
\leqslant\int\limits_0^t \underset{\Omega}{\rm ess\,sup\,}|f(\cdot,\tau)|\,d\tau
$$
for almost all $t\in (0,T)$. For a weak solution $u\in H^1(Q_T)$ to the
inhomogeneous heat equation $u_t-\Delta u=f$, thus holds the following 
weak essential maximum principle for the modulus
$$
\underset{Q_T}{\rm ess\,sup\,}|u|\leqslant\underset{\Gamma_T}{\rm ess\,sup\,}{|u|}
+\int\limits_0^T \underset{\Omega}{\rm ess\,sup\,}|f(\cdot,t)|\,dt.
$$
A: If you have a classical solution, evaluate at the point of maximum of $u$. The laplacian has the appropriate sign and then you get
$$
\frac{d}{dt}\|u(t)\|_{L^\infty}\leq \|f\|_{L^\infty}.
$$
Then integrate in time.
