Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 1 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.