Norm equivalence between $H^1(D)$ and $H^1_0(D)$

Question

Let $D\subset\mathbb{R}^d$ be a domain and we define $H^1_0(D)=\{v\in H^1(D):v=0 \text{ on }\partial D\}$. The norm for the Sobolev space $H^1(D)$ is defined as $$ \|v\|^2_{H^1(D)} := \|v\|^2_{L^2(D)}+ \|\nabla v\|^2_{L^2(D)}, $$ and many books and literatures often define $\|v\|^2_{H^1_0(D)}:=\|\nabla v\|^2_{L^2(D)}$, but I cannot tell the norm equivalence between the two.

Well, by Friedrich's inequality, I can see that $\|v\|^2_{H^1(D)}\leq C\|v\|^2_{H^1_0(D)}$, for some C>0. But, what about the other part of norm equivalence, i.e. $\|v\|^2_{H^1(D)}\geq c\|v\|^2_{H^1_0(D)}$.

Answer 1

Since $\|v\|^2_{L^2(D)}\geq 0$, $$ \|v\|^2_{H^1(D)} = \|v\|^2_{L^2(D)} + \|\nabla v\|^2_{L^2(D)} \\ \geq \|\nabla v\|^2_{L^2(D)} = \|v\|^2_{H^1_0(D)}. $$