Poincaré's lemma with norm in $H_{0}^{1}$ I wonder why $\| u\|_{H_0^1} = \int_{\Omega} |Du|^2$ for u in $H_0^1(\Omega)$, with $\Omega = (-1,1)$? I might be wrong but isn't $\| u\|_{H^1} = \| u\|_{L^2} + \| Du\|_{L^2}$? How come that $\| u\|_{L^2}$ drops out when the compact support is added? I suspect this has something to do with Poincaré's lemma but haven't found hany derivation. It might be trivial but I can't see it, so some help would be much appreciated. 
 A: I decided to put all the comments into an answer. Suppose you have $\Omega \subset\subset \mathbb{R}^n$, then the $\|\cdot\|_{W^{1,p}_0}$ and $\|\cdot\|_{W^{1,p}_0}$ are equivalent. So for $H^1_0$ and $H^1$ you get $\|u\|_{H^1}=\|u\|_{L^2}+\|Du\|_{L^2}$ and $\|u\|_{H^1_0}=\|Du\|_{L^2}$ are equivalent. For this case you just use Poincaré's inequality for the "non trivial" inequality. The general fact above, follows from a similar inequality. 
For the motivation: If you want to solve $-\Delta u = f$ and $u=0$ on the boundary, take a test function $\phi\in C^\infty_c$, multiply both sides with $\phi$ and integrate the equality:
$$-\int \Delta u \phi=\int\phi f$$
Integration by parts yiels
$$\int Du D\phi=\int\phi f$$
Defining the space $H^1_0$ as the closure of $C^\infty_c$ with respect to $\|u\|_{H^1_0}=\|Du\|_{L^2}$, one can prove that $H^1_0$ is a Hilbert space with inner product $(u,v)=\int Du Dv dx$. Furthermore the map $\phi\mapsto \int f\phi$ can be continuously extended to a mapping $l^*\in (H^1_0)^*$, where the latter denotes the dual space of $H^1_0$. 
Riezs gives you the following theorem: For every $f\in L^2$ it exists exactly one $u\in H^1_0$ s.t. 
$$\int Du Dv = \int f v$$
for all $v\in H^1_0$. Such a $u$ is called a weak solution of class $H^1_0$. Even more, one can show that $u\in H^1_0$ implies $u=0$ on $\partial \Omega$ in a suitable sense, see trace operator.
