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

Solution 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.

Solution 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.