Let $Z$ be a $\mathbb R$-Banach space and $C$ be a multi-valued dissipative linear operator on $Z$, i.e. $C$ is a subspace of $Z\times Z$ with $$\forall\lambda>0:\forall(z,z')\in C:\left\|\lambda z-z'\right\|_Z\ge\lambda\left\|z\right\|_Z\tag1.$$ By $(1)$, $\lambda-C$ is injective and hence $(\lambda-C)^{-1}$ is single-valued for all $\lambda>0$.
We can show that $$C_0:=\left\{(z,z')\in\overline C:z'\in\overline{\mathcal D(C)}\right\}$$ is single-valued.
Now let $L$ be a closed subspace of $Z$ and assume $$(\lambda-C)^{-1}L\subseteq\mathcal D(C)\cap L\;\;\;\text{for all }\lambda>0\tag2.$$
Are we able to conclude $(\lambda-C_0)^{-1}(\overline{\mathcal D(C)}\cap L)\subseteq\overline{\mathcal D(C)}\cap L$?
Let $\lambda>0$ and $z\in(\lambda-C_0)^{-1}(\overline{\mathcal D(C)}\cap L)$. Then there is a $z'\in\overline{\mathcal D(C)}\cap L$ with $$z=(\lambda-C_0)^{-1}z'.\tag3$$ Clearly, $$z\in\overline{\mathcal D(C)}\tag4.$$ My problem is to show that $z\in L$. Since $(z',z)\in(\lambda-C_0)^{-1}$, there is a $z''\in\overline{\mathcal D(C)}$ with $(z,z'')\in\overline C$ and $$z'=\lambda z-z''\tag5.$$ So, there is a $((z_n,z_n''))_{n\in\mathbb N}\subseteq C$ with $$\left\|(z_n,z_n'')-(z,z'')\right\|_{Z\times Z}\xrightarrow{n\to\infty}0\tag6.$$ Let $$z_n':=\lambda z_n-z_n''\;\;\;\text{for }n\in\mathbb N.$$ Then $$(z_n',z_n)\in(\lambda-C)^{-1}\;\;\;\text{for all }n\in\mathbb N\tag7.$$ However, I don't know how we might be able to conclude from here.