I'd like to know when the reverse Poincare inequality is true: Given a bounded domain $\Omega$, and $f \in L^2(\Omega)$, under what conditions on $f$ (neglecting the trivial constant case) and/or $\Omega$ is it true that there is a constant $C(\Omega)$ such that $\|\nabla f\|_{L^2} \leq C \|f\|_{L^2}$?
Thanks in advance.
I don't think such a thing can be true without some extreme restrictions. Just take the simple domain $[0,1]$ and $f_n(x):=\frac{1}{n} \sin(n^2x)$ then you can see that $f_n\to 0$ in $L^2$ but $\|Df_n\|$ will diverge so the inequality can't hold.
Such an inequality won't be true in general. However, if a function $u$ satisfies a PDE of a certain class (the simplest case would be an elliptic equation with smooth coefficients, e.g. the Laplace equation $\Delta u = 0$), it is possible to bound the $L^2$-norm of $\nabla u$ in a domain by the $L^2$-norm of $u$ in a bigger domain. This is, for $\Omega\subset\subset\Omega'$, $$\Vert\nabla u\Vert_{L^2(\Omega)}\leq \Vert u\Vert_{L^2(\Omega')}.$$ This is usually called interior regularity, Cacciopoli's inequality or, as you called it, reverse Poincaré inequality. In L.C. Evans' book $\textit{Partial Differential Equations}$, section 6.3.1, Theorem 1, they actually manage to bound the $H^2$-norm, which is a stronger result. However, the inequality I just wrote is proven in point 7 of the proof.
This inequality can actually be proven for more general PDE's with not-so-smooth coefficients, but having to do a little bit of technical work.
As far as I have come across such an equality, I can say that it holds in a finite dimensional domain. If the domain is divided into quasi-uniform triangulation then such inequality holds and is called "inverse inequality". See Thomee, 2006, Galerkin Finite Element Method for Parabolic Equations.
The reverse Poincare inequality holds, if f is harmonic i.e. if $\Delta f(x) = 0$ for all $x \in \Omega$.
