Let $U,H$ be complex Hilbert spaces such that $U$ is densely embedded into $H$, $\mathfrak a$ be a bounded coercive symmetric sesquilinear form on $U$, $$\mathcal D(A):=\left\{v\in U\mid\exists a\in H:\forall u\in U:\mathfrak a(u,v)=\langle u,a\rangle_H\right\}$$ and $A:\mathcal D(A)\to H$ with $$Av=a\Leftrightarrow\forall u\in U:\mathfrak a(u,v)=\langle u,a\rangle_H\tag1$$ for all $v\in\mathcal D(A)$ and $a\in H$. It's easy to show that $A$ is a symmetric linear operator on $H$. Moreover, $A$ is bounded below$^1$.
We can show that there is an orthonormal basis $(f_n)_{n\in\mathbb N}\subseteq\mathcal D(A)$ of $H$ with $$\langle u,Af_n\rangle_H=\mathfrak a(u,f_n)=\lambda_n\langle u,f_n\rangle_H\tag2\;\;\;\text{for all }u\in U$$ for some nondecreasing $(\lambda_n)_{n\in\mathbb N}\subseteq(0,\infty)$ with $\lambda_n\xrightarrow{n\to\infty}\infty$ and $$e_n:=\frac1{\sqrt{\lambda_n}}f_n\;\;\;\text{for }n\in\mathbb N$$ is an orthonormal basis of $U$ with respect to the inner product $\mathfrak a$, which is (by the given assumptions) equivalent to $\langle\;\cdot\;,\;\cdot\;\rangle_U$.
Now, if $\alpha\in\mathbb R$, one usually defines $$\mathcal D(A^\alpha):=\left\{x\in H:\sum_{n\in\mathbb N}\lambda_n^{2\alpha}\left|\langle u,f_n\rangle_H\right|^2<\infty\right\}$$ and $$A^\alpha x:=\sum_{n\in\mathbb N}\lambda_n^\alpha\langle x,f_n\rangle_Hf_n\;\;\;\text{for }x\in\mathcal D(A).$$
Question 1: How do we see that for $\alpha=1$ the operator $(\mathcal D(A^\alpha),A^\alpha)$ is actually equal to $(\mathcal D(A),A)$ or at least an extension of it?
We may note that $\mathcal D(A)$ (and I really mean $\mathcal D(A)$, not $\mathcal D(A^1)$) is equal to $$\left\{v\in U:\mathfrak a(\;\cdot\;,v)\text{ is }\left\|\;\cdot\;\right\|_H\text{-continuous}\right\}.\tag2$$
Remark: I'm particularly interested in the case $U=H_0^1(\Omega)$ for some bounded open $\Omega\subseteq\mathbb R^d$, $d\in\mathbb N$, and $H=L^2(\Omega)$ with $$\mathfrak a(u,v)=\langle\nabla u,\nabla v\rangle_{H_0^1(\Omega,\:\mathbb C^d)}\;\;\;\text{for }u,v\in H_0^1(\Omega).$$ Then, in the context of distribution theory, once can rewrite the definition of$^2$ $\mathcal D(A)$ as $$\mathcal D(A)=\left\{u\in H_0^1(\Omega):\Delta u\in L^2(\Omega)\right\}$$ and I really would like to show that $\mathcal D(A)=H_0^1(\Omega)\cap H^2(\Omega)$ (while it's clear that "$\supseteq$" holds) and that this set is actually equal to $\mathcal D(A^1)$.
Question 2: In this particular case, does it follow that the inner product $$\langle u,v\rangle_\alpha:=\langle A^\alpha u,A^\alpha v\rangle_{L^2(\Omega)}\;\;\;\text{for }u,v\in\mathcal D(A)$$ is equivalent to the inner product $\langle\;\cdot\;,\;\cdot\;\rangle_{H^2(\Omega)}$ on $H_0^1(\Omega)\cap H^2(\Omega)$?
$^1$ i.e. $\exists c>0$ with $\left\|Av\right\|_H\ge c\left\|v\right\|_H$ for all $v\in\mathcal D(A)$.
$^2$ though note that $Au=-\Delta u$ for all $u\in H_0^1(\Omega)\cap H^2(\Omega)$.