Let $(\mathcal D(A),A)$ be a closable linear operator on a $\mathbb R$-Banach space $E$ and $(\mathcal D(\overline A),\overline A)$ denote its closure.
If $\mathcal D(\overline A)$ is dense, are we able to conclude that $\mathcal D(A)$ is dense?
I guess it boils down to show that $\mathcal D(A)$ is dense $\mathcal D(\overline A)$. If $x\in E$ and $\varepsilon>0$, then by density of $\mathcal D(\overline A)$, there is a $x'\in\mathcal D(\overline A)$ with $$\left\|x'-x\right\|_E<\frac\varepsilon2\tag1.$$ Now, by definition, $$(x',\overline Ax')\in\overline{\left\{(x'',Ax''):x''\in\mathcal D(A)\right\}}\tag2$$ and hence there is a $x''\in\mathcal D(A)$ with $$\left\|x''-x'\right\|_E+\left\|Ax''-\overline Ax'\right\|_E<\frac\varepsilon2\tag3.$$