Consider the solution of the following heat equation in $\mathbb{R}^n\times (0,T)$: $$\left\{\begin{array}{rcl} \partial_tu-\Delta u + c(x) u&=& 0\\ u(x,0)&=& f \end{array} \right.$$ In the case without coefficient (i.e., $c(x)=0$), according to this post, the solution is bounded provided that the initial condition $f$ is bounded. The proof is based on the homogenity of the equation ($c(x)=0$) and properties of fundamental solution.
My question is: do we have the same result in the case of bounded coefficient $c(x)\neq 0$? Can we get a result like $$|u(x,t)|\le C, \quad for \; a.e \; (x,t) \in \mathbb{R}^n\times (0,T),$$ where $C$ is a positive constant depending on $T, \|c\|_\infty$ and $\|f\|_\infty$?
Any hint or reference including some ideas to prove boundedness of solution of such equation would be welcome.
This question depends on a lot of parameters, especially on the spaces in which $u,c$ and $f$ live. Therefore, let us start with a simple example: take $c(x)$ to be a constant, $c(x)=\bar c$, and do the same for $f$, $f(x)=\bar f>0$. Then the problem reduces to the ODE \begin{align*} \frac{d}{dt}u(t)&=\bar c u,\\ u(0)&=\bar f. \end{align*} It is clear that the solution grows when $\bar c>0$. It will be bounded for finite time but certainly not for $T\to \infty$. If you want to study this problem more generally, you can write the equation as \begin{align*} \partial_t u&=\mathcal{L}u,\\ \mathcal{L}u&=\Delta u+c(x)u\\ u(0)&=f. \end{align*} That means: the answer to your question lies in studying the properties of the linear operator $\mathcal{L}$. However, this also forces you to choose function spaces where your solutions will live and hence where your operator will act. If you, for example, choose $L^2(\mathbb{R})$, you can prove that the spectrum of the linear operator depends mainly on $c(-\infty)$ and $c(\infty)$. If both values are negative, your solutions will probably remain bounded, but in these kind of problems the devil is always in the details.