I have some problems to understand the definition of the closure of an operator.
Let's $H$ be a Hilbert space with the norm $||\cdot||_H$, and let $A$ be a linear and symmetric operator defined on the set $D(A) \subset H$, $A\colon D(A) \rightarrow H$. On $D(A)$ we define the norm $$||x||_A = ||Ax||_H + ||x||_H, \forall\, x \in D(A) \subset H$$ Now, we define the closure of the operator $A$ (and we denote it by $\bar A$) as the extension of $A$ in the set $\overline{D(A)}^{||\cdot||_A}$, i.e.: $$\bar A \colon \, \overline{D(A)}^{||\cdot||_A} \rightarrow H$$ and $$Ax = \bar A x, \forall \, x \in D(A)$$
My problem is: how can $\overline{D(A)}^{||\cdot||_A}$ be well-defined? By definition, such a set contains every $x \in H$ that can be approximated by a sequence $\{x_n\}_{n \in \mathbb{N}} \subset D(A)$ through the norm $||\cdot||_A$, i.e. every $x$ such that exists a sequence $\{x_n\}_{n \in \mathbb{N}} \subset D(A)$ and such that $||x_n - x||_A \rightarrow 0$ as $n \rightarrow \infty$. However, $x_n-x$ is not in $D(A)$ in general, and $||\cdot||_A$ is only defined on $D(A)$.
Any help would be much appreciated, thank you.