A relation between the domain of $A$ and the domain of $\bar A$

Question

Let $A$ be an operator: $$ A:D(A)\to R(A) $$ where $D(A)$ and $R(A)$ are respectively the domain and the range of $A$ and they are subspaces of a Hilbert spcae $(H,\|\|)$.

Suppose that $A$ is a closable operator, prove that the domain $D(\bar A)$ can be obtained as the closure of $D(A)$ by the norm $(\|Au\|^2+\|u\|^2)^{\frac 1 2}$ where $\bar A$ is the smallest closed extension of the operator $A$. My problem is that I can't figure out how the domain $D(\bar A)$ can be obtained from an other domain $D(A)$, what can be the relation ? any help or simply a hint will be great thank you for your time.

Answer 1

By definition, the graph of $\overline{A}$ is the closure of the graph of $A$. That means $(x,y)$ is in the graph of $\overline{A}$ iff there is a sequence $x_n \in D(A)$ such that ...

Answer 2

What I am going to say isn't anything new, but sometimes it helps to look more abstractly to see the big picture. Consider a linear map $A : \mathcal{D}(A) \subseteq X \rightarrow Y$ where $X$ and $Y$ are Banach spaces. The graph of $A$ is $\mathscr{G}(A)=\{ (x,Ax) \in X \times Y : x \in \mathcal{D}(A) \}$. The graph is a linear subspace of $X\times Y$.

What's interesting is that any linear subspace $\mathcal{M}$ of $X\times Y$ is a graph $\mathscr{G}(B)$ of some $B:\mathcal{D}(B)\subseteq X\rightarrow Y$ iff the only element of the form $(0,y)$ which is in $\mathcal{M}$ is the one where $y=0$. The domain $\mathcal{D}(B)$ in this case is $\{ x \in X : \exists y \in Y \mbox{ such that } (x,y) \in \mathcal{M} \}$. For $\mathcal{D}(B)$ as defined, there exists a unique $y \in Y$ such that $(x,y) \in \mathcal{M}$ because $(x,y),(x,y')\in\mathcal{M}$ implies $(0,y-y')\in\mathcal{M}$ and, hence, $y-y'=0$. Furthermore this unique correspondence $x\in\mathcal{D}(B)\mapsto y$ is linear because $\mathcal{M}$ is linear (easy exercise.)

When you form the closure $\mathscr{G}(A)^{c}$ in $X\times Y$ of the graph $\mathscr{G}(A)$ of a linear operator $A : \mathcal{D}(A)\subseteq X\rightarrow Y$, you get another graph iff the only element $(0,y) \in\mathscr{G}(A)^{c}$ is where $y=0$. If that is the case, then $\mathscr{G}(A)^{c}=\mathscr{G}(\overline{A})$ for a unique operator $\overline{A} : \mathcal{D}(\overline{A})\subseteq X \rightarrow Y$. The domain of $\overline{A}$ is the set of all first coordinates in the subspace $\mathscr{G}(A)^{c}$, which is the most that you can say in general.