Let $\Omega \subset \mathbb R^n$. The $H^1(\Omega)$ and $H^1_0(\Omega)$ spaces are defined as follows: $$\begin{align} H^1(\Omega) &= \{v \in L^2(\Omega) \mid \nabla v \in {(L^2(\Omega))}^n\}\\ H_0^1(\Omega) &= \overline{\mathcal C^\infty_C(\Omega)}^{{\|\cdot\|}_{H^1(\Omega)}} \end{align}$$ where the derivatives are to be understood in the distributional sense and $$\begin{align} {(u, v)}_{H^1(\Omega)} &= {(u, v)}_{L^2(\Omega)} + \int_\Omega \nabla u \cdot \nabla v\\ {\|v\|}_{H^1(\Omega)}^2 &= {\|v\|}_{L^2(\Omega)}^2 + {\|\:\!|\nabla v|\:\!\|}_{L^2(\Omega)}^2 \end{align}$$

Now, in my notes I have the definitions $$\begin{align} {(u, v)}_{H^1_0(\Omega)} &= \int_\Omega \nabla u \cdot \nabla v \tag{1}\\ {\|u\|}_{H^1_0(\Omega)} &= {\|\:\!|\nabla u|\:\!\|}_{L^2(\Omega)} \end{align}$$ which make $H^1_0(\Omega)$ a Hilbert space. However, in the teacher's notes I see in multiple places statements that contradict $(1)$.

For example, they say that if $T \in H^{-1}(\Omega)$ (dual space of $H^1_0(\Omega)$), then by Riesz' representation theorem there exist a $\bar u \in H^1_0(\Omega)$ such that $$\langle T, v \rangle = \int_\Omega (\bar uv + \nabla \bar u \cdot \nabla v),\qquad \forall v \in H^1_0(\Omega)$$ which takes me by surprise since it's using ${(\cdot,\cdot)}_{H^1(\Omega)}$ instead of $(1)$.

In another instance, I see $$a(u, v) = \int_0^1 u'v' + \int_0^1 uv \color{red}{=} {(u, v)}_{H^1_0(0, 1)}$$ which does the same thing.

My question is, does it make sense to identify the scalar product of $H^1_0(\Omega)$ with the scalar product of $H^1(\Omega)$?


Normally, one introduce $H_0^1(\Omega)$ as the closure of $C_0^\infty(\Omega)$ with the norm of $H^1(\Omega)$. Thus, $H_0^1(\Omega)$ inherits the topology of $H^1(\Omega)$, i.e. it is a Hilbert space with the scalar product

$$(u,v)_{H^1(\Omega)}=\int_\Omega uv +\nabla u \cdot \nabla v ~dx ~\text{ for all } u,v \in H_0^1(\Omega).$$

So, it indeed makes sense to have this scalar product on $H_0^1(\Omega)$ and this one is also the canonical one on $H_0^1(\Omega)$.

Now by Poincare's inequality we have for all $u \in H_0^1(\Omega)$

$$\|\nabla u\|_{L^2(\Omega)}^2 \leq\|u\|^2_{H^1(\Omega)}=\|u\|_{L^2(\Omega)}^2+\|\nabla u\|_{L^2(\Omega)}^2 \leq C \|\nabla u\|_{L^2(\Omega)}^2.$$

Hence we define $\|u\|_{H_0^1(\Omega)}:=\|\nabla u\|_{L^2(\Omega)}$ which is an equivalent norm to $\|u\|_{H^1(\Omega)}$ on $H_0^1(\Omega)$. Further it induces the scalar product

$$(u,v)_{H_0^1(\Omega)}=\int_\Omega \nabla u \cdot \nabla v ~dx ~\text{ for all } u,v \in H_0^1(\Omega).$$ I'd try to distinguish these different scalar products and not write $(u,v)_{H_0^1(\Omega)}$ for the upper one even though it wouldn't be wrong of course.


Whenever a so called Poincaré type inequality holds you can estimate $||u||_{L^p}$ in terms of $||\nabla u||_{L^p}$ and then the two norms on $H^{1,p}$ and $H_0^{1,p}$ are in fact equivalent (but, in general, not equal). In case of $p=2$ you can then, obviously, also estimate the scalar products against each other.

Equivalence of norms is sufficient to apply the Riesz representation theorem.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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