Norm and scalar product of $H_0^1(\Omega)$

Question

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 $$\begin{align} {(u, v)}_{H^1(\Omega)} &= {(u, v)}_{L^2(\Omega)} + \int_\Omega \nabla u \cdot \nabla v\\ {\|v\|}_{H^1(\Omega)}^2 &= {\|v\|}_{L^2(\Omega)}^2 + {\|\:\!|\nabla v|\:\!\|}_{L^2(\Omega)}^2 \end{align}$$

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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 $H^1(\Omega)$?

Answer 1

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.

Answer 2

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.