Is the domain of a closure operator close?

Question

Let $X$ and $Y$ be two Banach spaces. Let $A:D(A)\subset X \to Y$ linear and continuous. Then $A$ is closable. Let $\overline{A}$ be the closure operator.

My question is: is $D(\overline{A})$ close?

Answer 1

If $A$ is a densely defined linear operator $A : D(A) \to Y$ and $A$ is truly continuous, then $A$ uniquely extends to a linear operator $\bar{A} : X \to Y$ with $\|\bar{A}\| = \|A\| < \infty$ (since by assumption $A$ is continuous) where the extension $\bar{A}$ is, in fact, the same as the closure of $A$. So, $D(\bar{A}) = X$. Otherwise, if you only assume that $A$ is closable, but not necessarily continuous, then the closure $\bar{A}$ still has a dense domain $D(\bar{A})$ that need not equal $X$. So, $D(\bar{A})$ need not be closed since otherwise $D(\bar{A}) = \overline{D(\bar{A})} = X$.