The Dirichlet problem for second order elliptic operators I am studying functional analysis on my own. Here's an excerpt from the book Topics in Functional analysis by S. Kesavan which I am following: 

I have some doubts regarding this:-


*

*Why do we need a classical solution $u$ to be in $C^2(\overline{\Omega})$? Shouldn't it be simply in $C^2(\Omega)$?

*How does $u \in C^2(\overline{\Omega})$ and $u=0$ on $\partial \Omega$ implies $u \in H_{0}^{1}(\Omega)$? The text says this follows from Theorem $2.7.3$ but I don't see how.

*I don't follow the claim made in $3.2.3$. What does it mean when the text says "thus by density"?
 A: *

*Why do we need a classical solution $u$ to be in $C^2(\overline{\Omega})$? Shouldn't it be simply in $C^2(\Omega)$?
Answer: This is simply a technical assumption, in general, we say $u\in C(\overline{\Omega})\cap C^2(\Omega)$ for Dirichlet problem and $u\in C^1(\overline{\Omega})\cap C^2(\Omega)$ for Neumann problem to be a classical solution. And there are many regularity theories for such elliptic operator.  

*How does $u \in C^2(\overline{\Omega})$ and $u=0$ on $\partial \Omega$ implies $u \in H_{0}^{1}(\Omega)$? The text says this follows from Theorem $2.7.3$ but I don't see how.
Answer: This is just by definition of $H_{0}^{1}(\Omega)(=$ the closure of $\mathscr{D}(\Omega)$ under the $H^1(\Omega)$ norm, which equals to $\left(\sum\limits_{n=0}^2\|(\cdot)^{(n)}\|_{L^2(\Omega)}^2\right)^{\frac12})$. Note that $\mathscr{D}(\Omega)= C_c^\infty(\Omega)=C_c^\infty(\overline{\Omega})$ dense in $C^2_0(\Omega)=\{u \in C^2(\overline{\Omega})\mid u|_{\partial \Omega}=0\}$ equipped with the $C^2(\overline{\Omega})$ norm, which equals to $\max\limits_{n\in\{0,1,2\}}\sup\limits_{x\in\overline{\Omega}}|u^{(n)}(x)|$. Hence for any $u\in C^2_0(\Omega)$, there exists a sequence of $\{\phi_n\}_{n=1}^{\infty}\subset\mathscr{D}(\Omega)$ such that $\|\phi_n-u\|_{C^2(\overline{\Omega})}\to 0$ as $n\to\infty$. Therefore, $$\|\phi_n-u\|_{H^1(\Omega)}\leq\sqrt{|\Omega|}\|\phi_n-u\|_{C^2(\overline{\Omega})}\to 0$$ as $n\to\infty$, this implies $u \in H_{0}^{1}(\Omega)$.

*I don't follow the claim made in $3.2.3$. What does it mean when the text says "thus by density"?
Answer: This is a general argument of dense subspace. Note that $\mathscr{D}(\Omega)= C_c^\infty(\Omega)$ dense in $H_{0}^{1}(\Omega)$ by definition, hence for any $v\in H_{0}^{1}(\Omega)$, there exists a sequence of $\{\phi_n\}_{n=1}^{\infty}\subset\mathscr{D}(\Omega)$ such that $\|\phi_n-v\|_{H^1(\Omega)}\to 0$ as $n\to\infty$. Since $$\int_{\Omega}\nabla u\cdot\nabla\phi_n=\int_{\Omega}f\phi_n,$$ and strong convergence implies weak convergence, by sending $n\to\infty$, we have
$$\int_{\Omega}\nabla u\cdot\nabla v=\int_{\Omega}fv.$$
