A theorem about the symmetric operator $ \ker(A^{*}+iI)=R(A-iI)^{\bot}$

Question

Theorem: If $A$ is a symmetric and unbounded operator on the Hilbert space, then $$ \ker(A^{*}+iI)=R(A-iI)^{\bot}.$$

The proof of this step: When assuming $y\in R(A-iI)^{\bot}$, $\forall y\in D(A^{*})$, $\forall x\in D(A)$, $$(x, (A^{*}+iI)y)=0.$$

Since $D(A)$ is dense in H, $y\in \ker(A^{*}+iI)$.

My problem is why it should explain $D(A)$ is dense in $H$? If it could get $y \in \ker(A^{*}+iI)$ directly when I ignore it ?

Answer 1

The denseness of $D(A)$ is to let $A^{\ast}$ to be well-defined. We let $D(A^{\ast})=\{y\in H:\text{for some}~z\in H,~\text{for all}~x\in D(A),\left<Ax,y\right>=\left<x,z\right>\}$ and we need denseness of $D(A)$ to prove the uniquness of such a $z$, and we define $A^{\ast}(y)=z$.