Let $X$ Banach space. We prove that if $A \colon D(A) \subset X \to X$ is an m-dissipative operator (i.e. $A$ dissipative and $R(\lambda_0-A)=X$ for some $\lambda_0>0$), then $A$ is closed. Is this correct?
Indeed, since $A$ is dissipative we have $$\|(\lambda I-A) x \| \geq \lambda \|x \|, \quad \quad \forall \lambda>0, x \in D(A).$$ Then $\forall \lambda>0$ we have that $\lambda I-A$ is bounded below. Then the operator $$(\lambda I-A)^{-1} \colon R(\lambda I-A) \to X$$ is continuous. Since $A$ is m-dissipative $\exists \lambda_0:$ $R(\lambda_0 I-A)=X$, it follows $$(\lambda_0 I-A)^{-1} \in \mathcal L(X).$$ Thus $(\lambda_0 I-A)^{-1}$ is closed. Then $\lambda_0 I-A$ is closed. Hence $A$ is closed.
Yes. More generally, you can prove that a dissipative operator $A$ is closed if and only if there is $\lambda_0>0$ such that $R(\lambda_0 I-A)$ is closed (hence $R(\lambda I-A)$ is closed for every $\lambda>0$). Then you can conclude about the m-dissipative case.