Could I write $H_0^1(\Omega) = \mathcal{D}(\Omega) \cap H^1(\Omega)$?

Question

Let space $\mathcal{D}(\Omega)$ of compactly-supported smooth functions on a domain $\Omega$. I have a few difficulties to understand what is the $H_0^1(\Omega)$. Am I right if I write $$H_0^1(\Omega) = \mathcal{D}(\Omega) \cap H^1(\Omega)?$$

Answer 1

No, you can't write that. The following definition is more intuitive and correct:

$$ H_0^1(\Omega) = \{u \in H^1(\Omega) \mid \forall \epsilon > 0\ \exists \varphi \in \mathcal D(\Omega) : \Vert u - \varphi \Vert_{H^1} \leq \epsilon\}$$

That means essentially that $\mathcal D(\Omega)$ is dense in $H_0^1(\Omega)$ and since you construct $H_0^1(\Omega)$ as the closure of $\mathcal D(\Omega)$ respecting the $H^1$-norm. It actually holds that $\mathcal D(\Omega) \cap H^1(\Omega) \subseteq H_0^1(\Omega)$ as we take the closure of $\mathcal D(\Omega)$. I hope that helps you :)