About the self-adjoint extension of an operator.

Question

Let $B$ be a selfadjoint extension of an operator $A$ on a Hilbert space $H$. Let $\varphi \in \ker(A^\ast-z_0)$. Then i want to show that $\varphi + (z- z_0)(B-z)^{-1} \varphi \in \ker(A^\ast-z)$. I am not sure what i have to know to solve this exercise. Maybe you can give me hints without writing down a solution. Do i somehow have to use the Cayley-transform of $B$? Thanks.

Answer 1

If $A \preceq B$, then $B=B^{\star} \preceq A^{\star}$. In other words, $A$ and $B$ are restrictions of $A^{\star}$ to their respective domains; i.e., if $x \in \mathcal{D}(B)$, then $A^{\star}x = Bx$, and if $x \in \mathcal{D}(A)$, then $A^{\star}x = Bx = Ax$.

Suppose $A^{\star}\varphi = z_{0} \varphi$, and let $$ \psi = \varphi + (z-z_{0})(B-zI)^{-1}\varphi. $$ $(B-zI)^{-1}\varphi$ is in the domain of $B$ and, hence, also in the domain of $A^{\star}$, with $$ \begin{align} A^{\star}\psi & = A^{\star}\varphi +(z-z_0)A^{\star}(B-zI)^{-1}\varphi \\ & = z_0\varphi + (z-z_0)B(B-zI)^{-1}\varphi \\ & = z_0\varphi + (z-z_0)(B-zI)(B-zI)^{-1}\varphi + (z-z_0)z(B-zI)^{-1}\varphi \\ & = z_0\varphi + (z-z_0)\varphi + (z-z_0)z(B-zI)^{-1}\varphi \\ & = z\varphi+z(z-z_0)(B-zI)^{-1} \\ & = z\psi. \end{align} $$