If $A_0=\{(x,y)∈\overline A:y∈\overline{D(A)}\}$, then $R(λ-A_0)=\overline{D(A)}$ if and only if $R(λ-\overline A)⊇\overline{D(A)}$

Question

Let $E$ be a $\mathbb R$-Banach space, $A$ be a multi-valued dissipative linear operator on $E$, $$A_0:=\left\{(x,y)\in\overline A:y\in\overline{\mathcal D(A)}\right\}$$ and $\lambda>0$.

How can we show that $\mathcal R(\lambda-A_0)=\overline{\mathcal D(A)}$ if and only if $\mathcal R(\lambda-\overline A)\supseteq\overline{\mathcal D(A)}$?

Note that $$\mathcal R(\lambda-A_0)=\left\{\lambda x-y:(x,y)\in A_0\right\}=\left\{\lambda x-y:(x,y)\in\overline A\text{ and }y\in\overline{\mathcal D(A)}\right\}$$ is a subset of both $\overline{\mathcal D(A)}$ and $$\left\{\lambda x-y:(x,y)\in\overline A\right\}=\mathcal R(\lambda-\overline A).$$ So, if $\mathcal R(\lambda-A_0)=\overline{\mathcal D(A)}$, we need to have $\overline{\mathcal D(A)}\subseteq\mathcal R(\lambda-\overline A)$.

How can we show the other direction?

Answer 1

Since we always have that $\mathcal{R}(\lambda - A_0) \subseteq \overline{\mathcal{D}(A)}$, to show the other direction we need to check that if $\overline{\mathcal{D}(A)} \subseteq \mathcal{R}(\lambda - \overline{A})$ then also $\overline{\mathcal{D}(A)} \subseteq \mathcal{R}(\lambda - A_0)$.

Let $z \in \overline{\mathcal{D}(A)}$. Then there is $(x,y) \in \overline{A}$ such that $\lambda x - y = z$. Since $(x,y) \in \overline{A}$, $x \in \mathcal{D}(\overline{A}) \subseteq \overline{\mathcal{D}(A)}$. Hence, $y = \lambda x - z \in \overline{\mathcal{D}(A)}$ also. This then implies that $(x,y) \in A_0$ so that $z \in \mathcal{R}(\lambda - A_0)$ as desired.