Let $u$ be the weak solution on a smooth bounded domain $\Omega \subset \mathbb{R}^n$ (for $n \leq 3$) of
$$u\_t - \Delta u = f$$
$$u(0) = u\_0$$
$$\partial\_\nu u = 0 \quad\text{on $\partial\Omega$}$$
for $u\_0 \in L^\infty(\Omega)$ (non-negative) and $f \in L^\infty(0,T;L^2(\Omega))$. We know that
$$u(t,x) \leq C(\lVert u\_0 \rVert\_\infty + \lVert f \rVert\_{L^\infty(0,T;L^2)})$$
is satisfied almost every $(x,t)$ by De Giorgi's method (eg. Theorem 4.2.2 in "Elliptic and Parabolic Equations" book by Wu, Yin and Wang).


If instead we had the equation
$$u\_t - \Delta u +\alpha u= f$$
for a constant $\alpha > 0$ (and all else is the same as above), can I expect some kind of decay on the $L^\infty$ estimate involving $\alpha$, eg.
$$u(t,x) \leq \frac{C}{\alpha}(\lVert u\_0 \rVert\_\infty + \lVert f \rVert\_{L^\infty(0,T;L^2)})?$$
It is essential for me that $f$ is only in $L^\infty(0,T;L^2)$. 


Of course I tried using an integrating factor, but this gives a bad contribution on the term involving the $f$ norm. I wonder if it can be true.



$L^\infty$ estimate on heat equation with a lower order term
The estimate as stated is clearly false since $u\to u\_0$ as $t\to0$ hence $\|u\|\_{L^\infty}\ge \|u\_0\|\_{L^\infty}$.


Anyway you can write the kernel explicitly and extract the information you need from it. Just define $v=e^{at}u$ so that $v\_t-\Delta v=e^{at}(u\_t-\Delta u+au)=e^{at}f$. Thus
$$
u(t)=c t^{-n/2}\int e^{-\frac{|x-y|^2}{4t}-at}u\_0(y)dy +
\int\_0^t \int (t-s)^{-n/2} e^{-\frac{|x-y|^2}{4s}-as}f(t-s,y)dy ds.
$$


