Let $\Omega\subseteq\mathbb R^d$ be bounded and open, $V:=H_0^1(\Omega)$ and $H:=L^2(\Omega)$. We know that there is a nondecreasing sequence $(\lambda_n)_{n\in\mathbb N}\subseteq(0,\infty)$ with $\sup_{n\in\mathbb N}\lambda_n=\infty$ and an orthonormal basis $(e_n)_{n\in\mathbb N}\subseteq V$ of $H$, which is also a complete orthogonal set in $V$, satisfying $$\mathfrak a(u,e_n)=\lambda_n\langle u,e_n\rangle_H\;\;\;\text{for all }u\in V\tag1$$ for all $n\in\mathbb N$, where $$\mathfrak a(u,v):=\langle\nabla u,\nabla v\rangle_{L^2(\Omega,\:\mathbb R^d)}.$$ Since the following is true in way more generality, we may notice that $$S(t)f:=\sum_{n\in\mathbb N}e^{-\lambda_nt}\langle f,e_n\rangle_He_n\;\;\;\text{for }f\in H\text{ and }t\ge0$$ is a contractive immediately differentiable $C^0$-semigroup with generator $$Af:=-\sum_{n\in\mathbb N}\lambda_n\langle f,e_n\rangle_He_n$$ for $$f\in\mathcal D(A):=\left\{f\in H:\sum_{n\in\mathbb N}\lambda_n^2\left|\langle f,e_n\rangle_H\right|^2<\infty\right\}.$$
Question 1: If $u\in C^2(\Omega)$, can we show that $Au=\Delta u$? Or do we need to impose further assumptions on $u$? In particular, I wonder whether $Au=\Delta u$ can hold at all, unless $\left.u\right|_{\partial\Omega}=0$, since I would be surprised if the "Dirichlet boundary conditions" implicitly involved in the construction of $A$ (by the choice of $V$) wouldn't require that.
Clearly, if $u$ is twice differentiable on the closure $\overline\Omega$, then $\Delta u$ is bounded (and hence belongs to $H$) and $$\mathfrak a(\varphi,u)=-\langle\varphi,\Delta u\rangle_H\;\;\;\text{for all }\varphi\in V\tag2.$$ And since $(e_n)_{n\in\mathbb N}$ is an orthonormal basis of $H$, we should be able to immediately conclude $$Au=\sum_{n\in\mathbb N}\langle\Delta u,e_n\rangle_He_n=\Delta u\tag3;$$ or am I missing something? Is it sufficient to assume $u\in C(\overline Ω)\cap C^2(Ω)$ or even only $u\in C^2(\Omega)$?
Question 2: Can we further characterize $\mathcal D(A)$?
There is also the notion of "the operator associated to $\mathfrak a$ on $H$". It is defined to be the operator $B$ on $H$ satisfing $\mathcal D(B)=\{v\in V:\mathcal Av\in\mathcal R(\Phi)\}$ and $$\Phi Bv=\mathcal Av\;\;\;\text{for all }v\in\mathcal D(B)\tag4,$$ where $$\Phi:H\to V'\;,\;\;\;f\mapsto\left.\langle\;\cdot\;,f\rangle_H\right|_V$$ is the canonical embedding of $H$ into $V'$ and $$\mathcal A:V\to V'\;,\;\;\;v\mapsto\mathfrak a(\;\cdot\;,v).$$
Does it hold $(\mathcal D(A),A)=(\mathcal D(B),B)$? And is $\mathcal D(B)=\{f\in H:\Delta f\in H\}$, where $\Delta f\in H$ has to be understood in the sense that there is a $g\in H$ satisfying $$\mathfrak a(\varphi,f)=-\langle\varphi, g\rangle_H\;\;\;\text{for all }\varphi\in V\tag5?$$