Let $A:D(A)\subseteq H\to H$ be a densely defined, closable linear operator on the Hilbert space, $H$. We denote it's closure by $\overline A$. Suppose that $D(A)$ is also a core for $\overline A$.
Let $B:D(B)\subseteq H\to H$, a closed, potentially unbounded linear operator
We denote by $R(\cdot)$ and $N(\cdot)$ the range and kernel, respectively. Suppose that $R(A)\subseteq N(B)$. Then, $\overline{R(A)}\subseteq N(B)$, since the kernel of a closed linear operator is closed.
My problem: How exactly is it that one deduces
$$\overline{R(A)}\subseteq N(B)\implies\overline{R(\color{red}{\overline A})}\subseteq N(B)$$
from the fact that $D(A)$ is a core for $\overline A$? In addition, it seems me that we actually have the equality $\overline{R(A)}=\overline{R(\overline A)}$. I know that I should use that $D(A)$ is a core for $\overline A$ in order to pass from the range of $A$ to the range of $\overline A$, but I'm not sure on how it actually works.
Best Answer
Suppose $A$ is a close-able operator, then $$\overline{R(A)}\supseteq R\left(\overline A\right).$$ This follows directly from the definition of the closure, if $x\in D(\overline A)$ then $\overline A x$ is defined to be $\lim_n A x_n$ for any sequence $x_n\in D(A)$ converging to $x$ for which $Ax_n$ converges. Thus it is clear than any point in the range of $\overline A$ can be approximated by points $A x_n$ in $R(A)$, giving the above inclusion.
The other inclusion, namely $$R(A)\subseteq R\left(\overline A\right),$$ follows automatically from $\overline A$ being an extension of $A$. This means $D(\overline A)\supseteq D(A)$ and $\overline A\lvert_{D(A)} = A$, hence $$R\left(\overline A\right) = \overline A(D\left(\overline A\right)) \supseteq \overline A(D(A)) = A(D(A)) = R(A).$$
The first inclusion is the one that is relevant to your question. For if $R(A)\subseteq \ker(B)$ and $\ker(B)$ is closed, then $$\ker(B) \supseteq \overline{R(A)}\supseteq R\left(\overline A\right)$$ which is what you want.
The second inclusion is not relevant to your question, if $A'$ is some arbitrary extension of $A$ the very same argument will give you $R(A)\subseteq R(A')$. However $\ker(B)\supseteq R(A)$ does not need to imply $\ker(B)\supseteq R(A')$ if $A'$ is an arbitrary extension of $A$ (if $x\notin D(A)$ and $v\notin \ker(B)$ then define $D(A') = \mathrm{span}\{ x, D(A)\}$ and $A'(\lambda x+ \xi) = \lambda v+A(\xi)$ for $\xi\in D(A)$).