Proof of $\hat{\mathrm{O}}$ta's theorem

Question

I'm trying to prove $\hat{\mathrm{O}}$ta's theorem :

Let $A$ be a closed operator on a Hilbert space $H$ and $\overline{\mathcal{D}(A)}=H$. Suppose that $A\mathcal{D}(A)\subset \mathcal{D}(A)$ and the set $\{\lambda \in \mathbb{C} |\ \mathcal{R}(\lambda I-A) \mathrm{\ is \ closed}\}$ is unbounded. Then $A$ is bounded.

I have a problem with the following part of the proof of this theorem :

Let $\hat H:=\left(\mathcal{D}(A),\|\cdot\|_A\right)$, where $\|x\|_A^2=\|x\|^2+\|Ax\|^2$. How to prove that if $|z|>\|A\|_{\hat H}$ (where $A:\hat H\rightarrow \hat H$ we treat as a closed operator on $\hat H$) then $\mathcal{D}(A)=\mathcal{R}(zI-A)$ ?

Answer 1

As you pointed out in your comment, $\hat{A} : \hat{H}\rightarrow\hat{H}$ is bounded. Therefore $\hat{A}-\lambda I$ is invertible for $|\lambda| > \|\hat{A}\|_{\hat{H}}$, which guarantees that $\mathcal{R}(\hat{A}-\lambda I)=\hat{H}$ for such $\lambda$, or $(A-\lambda I)\mathcal{D}(A)=\mathcal{D}(A)$ for such $\lambda$.