Let $u$ be a solution (classical or weak as convenient) of $$-\Delta u + cu = f$$ $$u|_{\partial\Omega} = g$$ where $f \geq f_0 > 0$ and $g \geq g_0 > 0$ and $f_0, g_0$ are constants. Also $c$ is a positive function away.
Is it possible to obtain $$u \geq k_0>0$$ for some constant $k_0$ depending on $f$ and $g$? I.e. do we get positivity of solutions?
Maximum principles in eg. Evans do not give us this.
Let $u_m = \min u$ be the minimum of $u$. If the minimum is attained at the boundary, then $u_m \ge g_0$. If not, then at the point $x_0$ where $u_m$ is attained, $\Delta u(x_0) \ge 0$ and thus
$$ cu_m \ge -\Delta u(x_0) + cu(x_0) = f(x_0) \ge f_0.$$
In particular, one has $u_m \ge \min\{ g_0, f_0/\max c\}$. Thus the minimum is positive.
Asking that $k_0$ is independent of $c$ is not possible. Here's an antificial example: Let $\Omega = [-1, 1]$,
$$u_n : [-1, 1] \to \mathbb R, \ \ \ \ u_n(x) = \left(1-\frac 1n\right) x^2 + \frac 1n,$$
Then $u_n (\pm 1) = 1$ and
$$\tag{1} -u_n '' + c_n u_n = 1,$$ where $$c_n = \frac{1+ u_n''}{u_n} = \frac{3 - 2/n}{\left(1-\frac 1n\right) x^2 + \frac 1n}$$ (so $c_n$ is chosen so that $(1)$ holds). In this case, $f=g=1$, but the minimum of $u_n$ is $1/n$. Thus there is no bound independent of the function $c$.