Somewhat L2 against H1 estimate; an inequality in H1 somehow I'm a little slow on this one: Let $\Omega = [0,1]^2 \subseteq \mathbb{R}^2$ and $\emptyset \neq D \subsetneq \Omega$. Do constants $c_1,c_2,c_2\in\mathbb{R}_{\geq 0}$ exist such that
$$ c_1 \int_\Omega (u(x))^2 \,\mathrm{d}x \leq c_2 \int_\Omega \Vert \nabla u(x) \Vert^2 \,\mathrm{d}x + c_3 \int_{\Omega\setminus D} (u(x))^2 \,\mathrm{d}x$$
for all $u\in H^{1,2}(\Omega)$ with $|u| \leq 1$ almost everywhere (i. e. $\Vert u \Vert_{L^\infty} \leq 1$)? Further assumptions on $D$ may be made if necessary.
I'm having trouble finding a counterexample (I'd already be happy if I had one in 1D) and I'm not even sure if this statement is false or true. I'm afraid I'm not experienced enough yet to verify it. Any help would be appreciated.
 A: The statement is true and the bound on $u$ is not needed. The following argument are standard in deriving these "Poincare type" inequality. First of all, we assume that $\Omega\setminus D$ is of positive measure (If it is measure zero, the inequality is obviously false by considering constant function). Also $c_1 >0$ or the inequality is trivial. So we ask whether there is $C>0$ such that 
$$\int_\Omega u^2 \leq C \bigg( \int_\Omega |\nabla u|^2 + \int_{\Omega \setminus D} u^2 \bigg)$$
for all $u\in H^{1, 2}(\Omega)$.
Suppose such a $C$ does not exist. Then for all $n\in \mathbb N$, there is $u_n \in H^{1, 2}(\Omega)$ such that 
$$(1) \ \int_\Omega u_n^2 > n \bigg( \int_\Omega |\nabla u_n|^2 + \int_{\Omega \setminus D} u_n^2 \bigg) >  n \int_\Omega |\nabla u_n|^2\ .$$
By multiplying a constant if necessary, we assume $||u_n||_{L^2(\Omega)} = 1$. Thus the above inequality implies 
$$ (2) \ \int_\Omega |\nabla u_n|^2 < \frac{1}{n} <1$$
Thus $||u_n ||_{H^{1, 2}(\Omega)} \leq 2$ and the Sobolev embedding theorem implies then $u_n$ converges weakly in $H^{1, 2}(\Omega)$ and strongly in $L^2(\Omega)$ to $u \in H^{1, 2} (\Omega)$. These two implies that 
$$\int_\Omega |\nabla u|^2 \leq  \liminf_n \int_\Omega |\nabla u_n|^2 = 0$$
(The last equality is by (2)). Thus $u$ is constant and is nonzero as $||u||_{L^2(\Omega)} \neq 0$. But by (1), 
$$\frac{1}{n} \geq  \int_\Omega |\nabla u_n|^2 + \int_{\Omega \setminus D} u_n^2 $$
so when $n\to \infty$ we have 
$$\int_{\Omega\setminus D} u^2 = 0\ ,$$
which is impossible as $u$ is nonzero constant. This contradiction proves that such a positive $C$ exist. 
