# 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.

-
This follows from the equivalence of the two norms for bounded domains, i.e. for $\Omega$ bounded, $\|u\|_{H^1}=\|u\|_{L^2}+\|Du\|_{L^2}$ and $\|u\|_{H^1_0} = \|Du\|_{L^2}$ are equivalent. One inequality is trivial, the other is Poincaré – user8 Jan 12 '13 at 16:40
So one can use $\| u\|_{H^1_{0}} = \| Du\|_{L^2}$ instead of $\| u\|_{H^1_{0}} = \| u\|_{L^2} + \| Du\|_{L^2}$ because they are equivalent? – user55556 Jan 12 '13 at 16:54
if $\Omega \subset\subset \mathbb{R}^n$, then you have more generally the equivalence of the norms $\|\cdot\|_{W^{1,p}}$ and $\|\cdot\|_{W^{1,p}_0}$ – user8 Jan 12 '13 at 16:58
Ah, think I follow. Thanks! Does $\| u\|_{H_0^1} = \| Du\|_{L^2}$ originate from an inner product? – user55556 Jan 12 '13 at 17:15
Of course, you can define the inner product on $H^1_0$ as $\langle u,v\rangle = \int Du Dv dx$. This is the whole motivation, since you want to use Riesz to find a unique "weak" solution of $-\Delta u = f$ and zero on the boundary – user8 Jan 12 '13 at 17:20

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.

-