Let $E$ be a $\mathbb R$-Banach space, $A$ be a multi-valued dissipative linear operator on $E$ and $$A_0:=\left\{(x,y)\in\overline A:y\in\overline{\mathcal D(A)}\right\}.$$
Assume $\overline{\mathcal D(A)}\subseteq\overline{\mathcal R(\lambda-A)}=\mathcal R(\lambda-\overline A)$ for all $\lambda>0$.
How can we conclude that $\mathcal D(A_0)$ is dense in $\overline{\mathcal D(A)}$?
Note that $\overline A$ is dissipative, hence $(\lambda-\overline A)^{-1}$ is single-valued and $$\left\|(\lambda-\overline A)^{-1}y\right\|_E\le\frac1\lambda\left\|y\right\|_E\;\;\;\text{for all }y\in\mathcal R(\lambda-\overline A)\;\;\;\text{for all }\lambda>0.\tag1$$ Now we are able to show that $$\left\|\lambda(\lambda-\overline A)^{-1}x-x\right\|_E\xrightarrow{\lambda\to\infty}0\tag2\;\;\;\text{for all }x\in\mathcal D(\overline A).$$
Now let $x\in\overline{\mathcal D(A)}$. If $\mathcal D(\overline A)$ would be dense in $\overline{\mathcal D(A)}$, we would find a $(x_n)_{n\in\mathbb N}\subseteq\mathcal D(\overline A)$ with $\left\|x_n-x\right\|_E\xrightarrow{n\to\infty}0$ and hence \begin{equation}\begin{split}\left\|\lambda(\lambda-\overline A)^{-1}x-x\right\|_E&\le\lambda\left\|(\lambda-\overline A)^{-1}(x_n-x)\right\|_E+\left\|\lambda(\lambda-\overline A)^{-1}x_n-x_n\right\|_E+\left\|x_n-x\right\|_E\\&\le2\left\|x_n-x\right\|_E+\left\|\lambda(\lambda-\overline A)^{-1}x_n-x_n\right\|_E\xrightarrow{\lambda\to\infty}2\left\|x_n-x\right\|_E\xrightarrow{n\to\infty}0\end{split}\tag3\end{equation} by $(1)$ and $(2)$.
Noting that $(\lambda-\overline A)^{-1}\overline{\mathcal D(A)}=(\lambda-\overline A)^{-1}\mathcal R(\lambda-A_0)\subseteq\mathcal D(A_0)$ for all $\lambda>0$, the desired claim would follow from $(3)$.
Is there anything wrong with the argumentation above? And how can we show that $\mathcal D(\overline A)$ is dense in $\overline{\mathcal D(A)}$?
Clearly, $\mathcal D(\overline A)\subseteq\overline{\mathcal D(A)}$.